Latest revision as of 18:13, 23 July 2021

Concentric Ellipsoidal (T6) Coordinates (Part 2)

Orthogonal Coordinates

Speculation5

Spherical Coordinates

$r \cos θ$	$=$	$z,$
$r \sin θ$	$=$	$(x^{2} + y^{2})^{1 / 2},$
$\tan φ$	$=$	$\frac{y}{x} .$

Use λ₁ Instead of r

Here, as above, we define,

$λ_{1}^{2}$

$\equiv$

$x^{2} + q^{2} y^{2} + p^{2} z^{2}$

Using this expression to eliminate "x" (in favor of λ₁) in each of the three spherical-coordinate definitions, we obtain,

$r^{2} \equiv x^{2} + y^{2} + z^{2}$	$=$	$λ_{1}^{2} - y^{2} (q^{2} - 1) - z^{2} (p^{2} - 1);$
$\tan^{2} θ \equiv \frac{x^{2} + y^{2}}{z^{2}}$	$=$	$\frac{1}{z^{2}} [λ_{1}^{2} - y^{2} (q^{2} - 1) - p^{2} z^{2}];$
$\frac{1}{\tan^{2} φ} \equiv \frac{x^{2}}{y^{2}}$	$=$	$\frac{λ_{1}^{2} - p^{2} z^{2}}{y^{2}} - q^{2} .$

After a bit of additional algebraic manipulation, we find that,

$\frac{z^{2}}{λ_{1}^{2}}$	$=$	$\frac{(1 + \tan^{2} φ)}{𝒟^{2}},$
$\frac{y^{2}}{λ_{1}^{2}}$	$=$	$[\frac{𝒟^{2} \tan^{2} φ - p^{2} \tan^{2} φ (1 + \tan^{2} φ)}{(1 + q^{2} \tan^{2} φ) 𝒟^{2}}],$
$\frac{x^{2}}{λ_{1}^{2}}$	$=$	$1 - q^{2} (\frac{y^{2}}{λ_{1}^{2}}) - p^{2} (\frac{z^{2}}{λ_{1}^{2}}),$

where,

$𝒟^{2}$

$\equiv$

$[(1 + q^{2} \tan^{2} φ) (p^{2} + \tan^{2} θ) - p^{2} (q^{2} - 1) \tan^{2} φ] .$

As a check, let's set $q^{2} = p^{2} = 1$ , which should reduce to the normal spherical coordinate system.

$λ_{1}^{2}$

$\to$

$r^{2},$

and,

$𝒟^{2}$

$\to$

$[(1 + \tan^{2} φ) (1 + \tan^{2} θ)] .$

$\Rightarrow \frac{z^{2}}{λ_{1}^{2}}$	$\to$	$\frac{1}{1 + \tan^{2} θ} = \cos^{2} θ = \frac{z^{2}}{r^{2}};$
$\frac{y^{2}}{λ_{1}^{2}}$	$\to$	$[\frac{(1 + \tan^{2} φ) (1 + \tan^{2} θ) \tan^{2} φ - \tan^{2} φ (1 + \tan^{2} φ)}{(1 + \tan^{2} φ) (1 + \tan^{2} φ) (1 + \tan^{2} θ)}]$
	$=$	$\frac{\tan^{2} φ}{(1 + \tan^{2} φ)} [\frac{\tan^{2} θ}{(1 + \tan^{2} θ)}] = \sin^{2} θ \sin^{2} φ = \frac{y^{2}}{r^{2}};$
$\frac{x^{2}}{λ_{1}^{2}}$	$\to$	$1 - (\frac{y^{2}}{λ_{1}^{2}}) - (\frac{z^{2}}{λ_{1}^{2}}),$
	$\to$	$1 - \sin^{2} θ \sin^{2} φ - \cos^{2} θ = - \sin^{2} θ \sin^{2} φ + \sin^{2} θ = \sin^{2} θ \cos^{2} φ = \frac{x^{2}}{r^{2}} .$

Relationship To T3 Coordinates

If we set, $q = 1$ , but continue to assume that $p > 1$ , we expect to see a representation that resembles our previously discussed, T3 Coordinates. Note, for example, that the new "radial" coordinate is,

$λ_{1}^{2}$

$\to$

$(ϖ^{2} + p^{2} z^{2}),$

and,

$𝒟^{2}$

$\to$

$[(1 + \tan^{2} φ) (p^{2} + \tan^{2} θ)] .$

$\Rightarrow \frac{p^{2} z^{2}}{λ_{1}^{2}}$	$\to$	$\frac{p^{2}}{(p^{2} + \tan^{2} θ)} = \frac{1}{(1 + p^{- 2} \tan^{2} θ)},$
$\frac{ϖ^{2}}{λ_{1}^{2}} = \frac{x^{2}}{λ_{1}^{2}} + \frac{y^{2}}{λ_{1}^{2}}$	$\to$	$1 - p^{2} (\frac{z^{2}}{λ_{1}^{2}}) = [1 - \frac{1}{(1 + p^{- 2} \tan^{2} θ)}] .$

We also see that,

$\frac{ϖ^{2}}{p^{2} z^{2}}$

$\to$

$(1 + p^{- 2} \tan^{2} θ) [1 - \frac{1}{(1 + p^{- 2} \tan^{2} θ)}] = p^{- 2} \tan^{2} θ .$

Again Consider Full 3D Ellipsoid

Let's try to replace everywhere, $[ϖ / (p z)]^{2} = p^{- 2} \tan^{2} θ$ with $λ_{2}$ . This gives,

$\frac{𝒟^{2}}{p^{2}}$

$\equiv$

$[(1 + q^{2} \tan^{2} φ) λ_{2} - (q^{2} - 1) \tan^{2} φ] .$

which means that,

$\frac{p^{2} z^{2}}{λ_{1}^{2}}$	$=$	$\frac{(1 + \tan^{2} φ)}{[(1 + q^{2} \tan^{2} φ) λ_{2} - (q^{2} - 1) \tan^{2} φ]}$
	$=$	$\frac{(1 + \tan^{2} φ) / \tan^{2} φ}{[q^{2} λ_{2} (1 + q^{2} \tan^{2} φ) / (q^{2} \tan^{2} φ) - (q^{2} - 1)]} = \frac{1 / \sin^{2} φ}{[q^{2} λ_{2} Q^{2} - (q^{2} - 1)]},$
$\frac{q^{2} y^{2}}{λ_{1}^{2}}$	$=$	$\frac{q^{2} \tan^{2} φ}{(1 + q^{2} \tan^{2} φ)} - \frac{q^{2} \tan^{2} φ (1 + \tan^{2} φ)}{(1 + q^{2} \tan^{2} φ) [(1 + q^{2} \tan^{2} φ) λ_{2} - (q^{2} - 1) \tan^{2} φ]}$
	$=$	$\frac{q^{2} \tan^{2} φ}{(1 + q^{2} \tan^{2} φ)} {1 - \frac{(1 + \tan^{2} φ)}{[(1 + q^{2} \tan^{2} φ) λ_{2} - (q^{2} - 1) \tan^{2} φ]}}$
	$=$	$\frac{q^{2} \tan^{2} φ}{(1 + q^{2} \tan^{2} φ)} {1 - \frac{(1 + \tan^{2} φ) / \tan^{2} φ}{[q^{2} λ_{2} (1 + q^{2} \tan^{2} φ) / (q^{2} \tan^{2} φ) - (q^{2} - 1)]}}$
	$=$	$\frac{1}{Q^{2}} {1 - \frac{1 / \sin^{2} φ}{[q^{2} λ_{2} Q^{2} - (q^{2} - 1)]}} = \frac{1}{Q^{2} []} [[] - \frac{1}{\sin^{2} φ}],$
$\frac{x^{2}}{λ_{1}^{2}}$	$=$	$1 - \frac{q^{2} \tan^{2} φ}{(1 + q^{2} \tan^{2} φ)} {1 - \frac{(1 + \tan^{2} φ)}{[(1 + q^{2} \tan^{2} φ) λ_{2} - (q^{2} - 1) \tan^{2} φ]}}$
		$- \frac{(1 + \tan^{2} φ)}{[(1 + q^{2} \tan^{2} φ) λ_{2} - (q^{2} - 1) \tan^{2} φ]}$
	$=$	$1 - \frac{q^{2} \tan^{2} φ}{(1 + q^{2} \tan^{2} φ)} - {\frac{(1 + \tan^{2} φ)}{[(1 + q^{2} \tan^{2} φ) λ_{2} - (q^{2} - 1) \tan^{2} φ]}}$
	$=$	$1 - \frac{1}{Q^{2}} - {\frac{1 / \sin^{2} φ}{[q^{2} λ_{2} Q^{2} - (q^{2} - 1)]}} = \frac{1}{Q^{2} []} {Q^{2} [] - [] - \frac{Q^{2}}{\sin^{2} φ}},$
$\frac{x^{2} + q^{2} y^{2}}{λ_{1}^{2}}$	$=$	$1 - [1 + \frac{q^{2} \tan^{2} φ}{(1 + q^{2} \tan^{2} φ)}] {\frac{(1 + \tan^{2} φ)}{[(1 + q^{2} \tan^{2} φ) λ_{2} - (q^{2} - 1) \tan^{2} φ]}} .$

Now, notice that,

$\frac{q^{2} y^{2} Q^{2}}{λ_{1}^{2}}$

$=$

$1 - \frac{1}{[] \sin^{2} φ},$

and,

$\frac{x^{2}}{λ_{1}^{2}} + \frac{1}{Q^{2}}$

$=$

$1 - \frac{1}{[] \sin^{2} φ} .$

Hence,

$\frac{x^{2}}{λ_{1}^{2}} + \frac{1}{Q^{2}}$	$=$	$\frac{q^{2} y^{2} Q^{2}}{λ_{1}^{2}}$
$\Rightarrow 0$	$=$	$Q^{4} (\frac{q^{2} y^{2}}{λ_{1}^{2}}) - Q^{2} (\frac{x^{2}}{λ_{1}^{2}}) - 1$
	$=$	$Q^{4} - Q^{2} (\frac{x^{2}}{q^{2} y^{2}}) - (\frac{λ_{1}^{2}}{q^{2} y^{2}}),$

where,

$Q^{2}$

$\equiv$

$\frac{1 + q^{2} \tan^{2} φ}{q^{2} \tan^{2} φ} .$

Solving the quadratic equation, we have,

$Q^{2}$	$=$	$\frac{1}{2} {(\frac{x^{2}}{q^{2} y^{2}}) \pm [(\frac{x^{2}}{q^{2} y^{2}})^{2} + 4 (\frac{λ_{1}^{2}}{q^{2} y^{2}})]^{1 / 2}}$
	$=$	$(\frac{x^{2}}{2 q^{2} y^{2}}) {1 \pm [1 + 4 (\frac{λ_{1}^{2} q^{2} y^{2}}{x^{4}})]^{1 / 2}} .$

Tentative Summary

$λ_{1}$	$\equiv$	$(x^{2} + q^{2} y^{2} + p^{2} z^{2})^{1 / 2},$
$λ_{2}$	$\equiv$	$\frac{(x^{2} + y^{2})^{1 / 2}}{p z},$
$λ_{3} = Q^{2}$	$\equiv$	$(\frac{x^{2}}{2 q^{2} y^{2}}) {1 \pm [1 + 4 (\frac{λ_{1}^{2} q^{2} y^{2}}{x^{4}})]^{1 / 2}} .$

Partial Derivatives & Scale Factors

First Coordinate

$\frac{\partial λ_{1}}{\partial x}$	$=$	$\frac{x}{λ_{1}},$
$\frac{\partial λ_{1}}{\partial y}$	$=$	$\frac{q^{2} y}{λ_{1}},$
$\frac{\partial λ_{1}}{\partial z}$	$=$	$\frac{p^{2} z}{λ_{1}} .$

$h_{1}^{- 2}$	$=$	$(\frac{\partial λ_{1}}{\partial x})^{2} + (\frac{\partial λ_{1}}{\partial y})^{2} + (\frac{\partial λ_{1}}{\partial z})^{2}$	$=$	$(\frac{x}{λ_{1}})^{2} + (\frac{q^{2} y}{λ_{1}})^{2} + (\frac{p^{2} z}{λ_{1}})^{2} .$
$\Rightarrow h_{1}$	$=$	$λ_{1} ℓ_{3 D},$

where,

ℓ_{3 D} \equiv (x^{2} + q^{4} y^{2} + p^{4} z^{2})^{- 1 / 2} .

As a result, the associated unit vector is,

${\hat{e}}_{1}$	$=$	$\hat{ı} h_{1} (\frac{\partial λ_{1}}{\partial x}) + \hat{ȷ} h_{1} (\frac{\partial λ_{1}}{\partial y}) + \hat{k} h_{1} (\frac{\partial λ_{1}}{\partial z})$
	$=$	$\hat{ı} x ℓ_{3 D} + \hat{ȷ} q^{2} y ℓ_{3 D} + \hat{k} p^{2} z ℓ_{3 D} .$

Notice that,

${\hat{e}}_{1} \cdot {\hat{e}}_{1}$

$=$

$(x^{2} + q^{4} y^{2} + p^{4} z^{2}) ℓ_{3 D}^{2} = 1 .$

Second Coordinate (1^st Try)

$\frac{\partial λ_{2}}{\partial x}$	$=$	$\frac{1}{p z} [\frac{x}{(x^{2} + y^{2})^{1 / 2}}],$
$\frac{\partial λ_{2}}{\partial y}$	$=$	$\frac{1}{p z} [\frac{y}{(x^{2} + y^{2})^{1 / 2}}],$
$\frac{\partial λ_{2}}{\partial z}$	$=$	$- \frac{(x^{2} + y^{2})^{1 / 2}}{p z^{2}} .$

$h_{2}^{- 2}$	$=$	$(\frac{\partial λ_{2}}{\partial x})^{2} + (\frac{\partial λ_{2}}{\partial y})^{2} + (\frac{\partial λ_{2}}{\partial z})^{2}$
	$=$	${\frac{1}{p z} [\frac{x}{(x^{2} + y^{2})^{1 / 2}}]}^{2} + {\frac{1}{p z} [\frac{y}{(x^{2} + y^{2})^{1 / 2}}]}^{2} + {\frac{(x^{2} + y^{2})^{1 / 2}}{p z^{2}}}^{2}$
	$=$	${[\frac{x^{2}}{(x^{2} + y^{2}) p^{2} z^{2}}]} + {[\frac{y^{2}}{(x^{2} + y^{2}) p^{2} z^{2}}]} + {\frac{(x^{2} + y^{2})}{p^{2} z^{4}}}$
	$=$	$\frac{1}{p^{2} z^{2}} + \frac{(x^{2} + y^{2})}{p^{2} z^{4}} = \frac{(x^{2} + y^{2} + z^{2})}{p^{2} z^{4}}$
$\Rightarrow h_{2}$	$=$	$\frac{p z^{2}}{r}$

As a result, the associated unit vector is,

${\hat{e}}_{2}$	$=$	$\hat{ı} h_{2} (\frac{\partial λ_{2}}{\partial x}) + \hat{ȷ} h_{2} (\frac{\partial λ_{2}}{\partial y}) + \hat{k} h_{2} (\frac{\partial λ_{2}}{\partial z})$
	$=$	$\hat{ı} [\frac{x z}{r (x^{2} + y^{2})^{1 / 2}}] + \hat{ȷ} [\frac{y z}{r (x^{2} + y^{2})^{1 / 2}}] - \hat{k} [\frac{(x^{2} + y^{2})^{1 / 2}}{r}] .$

Notice that,

${\hat{e}}_{2} \cdot {\hat{e}}_{2}$

$=$

$[\frac{x^{2} z^{2}}{r^{2} (x^{2} + y^{2})}] + [\frac{y^{2} z^{2}}{r^{2} (x^{2} + y^{2})}] + [\frac{(x^{2} + y^{2})}{r^{2}}] = 1 .$

Let's check to see if this "second" unit vector is orthogonal to the "first."

${\hat{e}}_{1} \cdot {\hat{e}}_{2}$	$=$	$x ℓ_{3 D} [\frac{x z}{r (x^{2} + y^{2})^{1 / 2}}] + q^{2} y ℓ_{3 D} [\frac{y z}{r (x^{2} + y^{2})^{1 / 2}}] - p^{2} z ℓ_{3 D} [\frac{(x^{2} + y^{2})^{1 / 2}}{r}]$
	$=$	$ℓ_{3 D} {[\frac{x^{2} z}{r (x^{2} + y^{2})^{1 / 2}}] + [\frac{q^{2} y^{2} z}{r (x^{2} + y^{2})^{1 / 2}}] - [\frac{p^{2} z (x^{2} + y^{2})^{1 / 2}}{r}]}$
	$=$	$\frac{z ℓ_{3 D}}{r (x^{2} + y^{2})^{1 / 2}} {[x^{2}] + [q^{2} y^{2}] - [p^{2} (x^{2} + y^{2})]}$
	$\neq$	$0 .$

Second Coordinate (2^nd Try)

Let's try,

$λ_{2}$	$=$	$[\frac{(x^{2} + q^{2} y^{2} + 𝔣 \cdot p^{2} z^{2})^{1 / 2}}{p z}],$
$\Rightarrow \frac{\partial λ_{2}}{\partial x}$	$=$	$\frac{x}{p z (x^{2} + q^{2} y^{2} + 𝔣 \cdot p^{2} z^{2})^{1 / 2}} = \frac{x}{p^{2} z^{2} λ_{2}},$
$\frac{\partial λ_{2}}{\partial y}$	$=$	$\frac{q^{2} y}{p z (x^{2} + q^{2} y^{2} + 𝔣 \cdot p^{2} z^{2})^{1 / 2}} = \frac{q^{2} y}{p^{2} z^{2} λ_{2}},$
$\frac{\partial λ_{2}}{\partial z}$	$=$	$\frac{𝔣 \cdot p^{2} z}{p z (x^{2} + q^{2} y^{2} + 𝔣 \cdot p^{2} z^{2})^{1 / 2}} - \frac{(x^{2} + q^{2} y^{2} + 𝔣 \cdot p^{2} z^{2})^{1 / 2}}{p z^{2}} = \frac{1}{p^{2} z^{2} λ_{2}} (𝔣 \cdot p^{2} z) - \frac{λ_{2}}{z} = \frac{1}{p^{2} z^{2} λ_{2}} (𝔣 \cdot p^{2} z - p^{2} z λ_{2}^{2}) .$

Hence,

$h_{2}^{- 2}$	$=$	$(\frac{\partial λ_{2}}{\partial x})^{2} + (\frac{\partial λ_{2}}{\partial y})^{2} + (\frac{\partial λ_{2}}{\partial z})^{2}$
	$=$	$[\frac{x}{p^{2} z^{2} λ_{2}}]^{2} + [\frac{q^{2} y}{p^{2} z^{2} λ_{2}}]^{2} + [\frac{𝔣}{z λ_{2}} - \frac{λ_{2}}{z}]^{2}$
	$=$	$[\frac{x^{2} + q^{4} y^{2}}{p^{4} z^{4} λ_{2}^{2}}] + [\frac{1}{z λ_{2}} (𝔣 - λ_{2}^{2})]^{2}$
	$=$	$\frac{1}{p^{4} z^{4} λ_{2}^{2}} [x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣 - λ_{2}^{2})^{2}]$
$\Rightarrow h_{2}$	$=$	$\frac{p^{2} z^{2} λ_{2}}{[x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣 - λ_{2}^{2})^{2}]^{1 / 2}} .$

So, the associated unit vector is,

${\hat{e}}_{2}$	$=$	$\hat{ı} h_{2} (\frac{\partial λ_{2}}{\partial x}) + \hat{ȷ} h_{2} (\frac{\partial λ_{2}}{\partial y}) + \hat{k} h_{2} (\frac{\partial λ_{2}}{\partial z})$
	$=$	$\hat{ı} {\frac{x}{[x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣 - λ_{2}^{2})^{2}]^{1 / 2}}} + \hat{ȷ} {\frac{q^{2} y}{[x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣 - λ_{2}^{2})^{2}]^{1 / 2}}} + \hat{k} {\frac{p^{2} z (𝔣 - λ_{2}^{2})}{[x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣 - λ_{2}^{2})^{2}]^{1 / 2}}} .$

Checking orthogonality …

${\hat{e}}_{1} \cdot {\hat{e}}_{2}$	$=$	$x ℓ_{3 D} {\frac{x}{[x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣 - λ_{2}^{2})^{2}]^{1 / 2}}} + q^{2} y ℓ_{3 D} {\frac{q^{2} y}{[x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣 - λ_{2}^{2})^{2}]^{1 / 2}}} + p^{2} z ℓ_{3 D} {\frac{p^{2} z (𝔣 - λ_{2}^{2})}{[x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣 - λ_{2}^{2})^{2}]^{1 / 2}}}$
	$=$	$\frac{ℓ_{3 D}}{[x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣 - λ_{2}^{2})^{2}]^{1 / 2}} {x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣 - λ_{2}^{2})} .$
	$=$	$\frac{ℓ_{3 D}}{[x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣 - λ_{2}^{2})^{2}]^{1 / 2}} {x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣 - λ_{2}^{2})} .$

If $𝔣 = 0$ , we have …

$p^{2} z (𝔣 - λ_{2}^{2})$

$\to$

$[- p^{2} z λ_{2}^{2}]_{𝔣 = 0} = - p^{2} z [\frac{(x^{2} + q^{2} y^{2} + {𝔣 \cdot}^{0} p^{2} z^{2})^{1 / 2}}{p z}]^{2} = - \frac{(x^{2} + q^{2} y^{2})}{z},$

which, in turn, means …

$[x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣^{0} - λ_{2}^{2})^{2}]^{1 / 2}$	$=$	$[x^{2} + q^{4} y^{2} + p^{4} z^{2} λ_{2}^{4}]^{1 / 2}$
	$=$	${x^{2} + q^{4} y^{2} + p^{4} z^{2} [\frac{(x^{2} + q^{2} y^{2} + {𝔣 \cdot}^{0} p^{2} z^{2})^{1 / 2}}{p z}]^{4}}^{1 / 2}$
	$=$	${x^{2} + q^{4} y^{2} + [\frac{(x^{2} + q^{2} y^{2})^{2}}{z^{2}}]}^{1 / 2}$
	$=$	$(x^{2} + q^{4} y^{2})^{1 / 2} [1 + \frac{(x^{2} + q^{2} y^{2})}{z^{2}}]^{1 / 2}$
	$=$	$\frac{(x^{2} + q^{4} y^{2})^{1 / 2}}{z} [z^{2} + (x^{2} + q^{2} y^{2})]^{1 / 2},$

and,

${\hat{e}}_{1} \cdot {\hat{e}}_{2}$

$=$

$\frac{ℓ_{3 D}}{[x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣 - λ_{2}^{2})^{2}]^{1 / 2}} {x^{2} + q^{4} y^{2} + p^{4} z^{2} (𝔣 - λ_{2}^{2})} .$

Speculation6

Determine λ₂

This is very similar to the above, Speculation2. Try,

$λ_{2}$

$=$

$\frac{x y^{1 / q^{2}}}{z^{2 / p^{2}}},$

in which case,

$\frac{\partial λ_{2}}{\partial x}$	$=$	$\frac{λ_{2}}{x},$
$\frac{\partial λ_{2}}{\partial y}$	$=$	$\frac{x}{z^{2 / p^{2}}} (\frac{1}{q^{2}}) y^{1 / q^{2} - 1} = \frac{λ_{2}}{q^{2} y},$
$\frac{\partial λ_{2}}{\partial z}$	$=$	$- \frac{2 λ_{2}}{p^{2} z} .$

The associated scale factor is, then,

$h_{2}$	$=$	$[(\frac{\partial λ_{2}}{\partial x})^{2} + (\frac{\partial λ_{2}}{\partial y})^{2} + (\frac{\partial λ_{2}}{\partial z})^{2}]^{- 1 / 2}$
	$=$	$[(\frac{λ_{2}}{x})^{2} + (\frac{λ_{2}}{q^{2} y})^{2} + (- \frac{2 λ_{2}}{p^{2} z})^{2}]^{- 1 / 2}$
	$=$	$\frac{1}{λ_{2}} [\frac{1}{x^{2}} + \frac{1}{q^{4} y^{2}} + \frac{4}{p^{4} z^{2}}]^{- 1 / 2}$
	$=$	$\frac{1}{λ_{2}} [\frac{(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2}}{x^{2} q^{4} y^{2} p^{4} z^{2}}]^{- 1 / 2}$
	$=$	$\frac{1}{λ_{2}} [\frac{x q^{2} y p^{2} z}{𝒟}] .$

where,

$𝒟$

$\equiv$

$(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2})^{1 / 2} .$

The associated unit vector is, then,

${\hat{e}}_{2}$	$=$	$\hat{ı} h_{2} (\frac{\partial λ_{2}}{\partial x}) + \hat{ȷ} h_{2} (\frac{\partial λ_{2}}{\partial y}) + \hat{k} h_{2} (\frac{\partial λ_{2}}{\partial z})$
	$=$	$\frac{x q^{2} y p^{2} z}{𝒟} {\hat{ı} (\frac{1}{x}) + \hat{ȷ} (\frac{1}{q^{2} y}) + \hat{k} (- \frac{2}{p^{2} z})} .$

Recalling that the unit vector associated with the "first" coordinate is,

${\hat{e}}_{1}$

$\equiv$

$\hat{ı} (x ℓ_{3 D}) + \hat{ȷ} (q^{2} y ℓ_{3 D}) + \hat{k} (p^{2} z ℓ_{3 D}),$

where,

$ℓ_{3 D}$

$=$

$(x^{2} + q^{4} y^{2} + p^{4} z^{2})^{- 1 / 2},$

let's check to see whether the "second" unit vector is orthogonal to the "first."

${\hat{e}}_{1} \cdot {\hat{e}}_{2}$

$=$

$\frac{(x q^{2} y p^{2} z) ℓ_{3 D}}{𝒟} [1 + 1 - 2] = 0 .$

Hooray!

Direction Cosines for Third Unit Vector

Now, what is the unit vector, ${\hat{e}}_{3}$ , that is simultaneously orthogonal to both these "first" and the "second" unit vectors?

${\hat{e}}_{3} \equiv {\hat{e}}_{1} \times {\hat{e}}_{2}$	$=$	$\hat{ı} [(e_{1 y}) (e_{2 z}) - (e_{2 y}) (e_{1 z}))] + \hat{ȷ} [(e_{1 z}) (e_{2 x}) - (e_{2 z}) (e_{1 x}))] + \hat{k} [(e_{1 x}) (e_{2 y}) - (e_{2 x}) (e_{1 y}))]$
	$=$	$\frac{(x q^{2} y p^{2} z) ℓ_{3 D}}{𝒟} {\hat{ı} [(- \frac{2 q^{2} y}{p^{2} z}) - (\frac{p^{2} z}{q^{2} y})] + \hat{ȷ} [(\frac{p^{2} z}{x}) - (- \frac{2 x}{p^{2} z})] + \hat{k} [(\frac{x}{q^{2} y}) - (\frac{q^{2} y}{x})]}$
	$=$	$\frac{(x q^{2} y p^{2} z) ℓ_{3 D}}{𝒟} {- \hat{ı} [\frac{2 q^{4} y^{2} + p^{4} z^{2}}{q^{2} y p^{2} z}] + \hat{ȷ} [\frac{p^{4} z^{2} + 2 x^{2}}{x p^{2} z}] + \hat{k} [\frac{x^{2} - q^{4} y^{2}}{x q^{2} y}]}$
	$=$	$\frac{ℓ_{3 D}}{𝒟} {- \hat{ı} [x (2 q^{4} y^{2} + p^{4} z^{2})] + \hat{ȷ} [q^{2} y (p^{4} z^{2} + 2 x^{2})] + \hat{k} [p^{2} z (x^{2} - q^{4} y^{2})]} .$

Is this a valid unit vector? First, note that …

$(\frac{ℓ_{3 D}}{𝒟})^{- 2}$	$=$	$(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2}) (x^{2} + q^{4} y^{2} + p^{4} z^{2})$
	$=$	$(x^{2} q^{4} y^{2} p^{4} z^{2} + x^{4} p^{4} z^{2} + 4 x^{4} q^{4} y^{2}) + (q^{8} y^{4} p^{4} z^{2} + x^{2} q^{4} y^{2} p^{4} z^{2} + 4 x^{2} q^{8} y^{4}) + (q^{4} y^{2} p^{8} z^{4} + x^{2} p^{8} z^{4} + 4 x^{2} q^{4} y^{2} p^{4} z^{2})$
	$=$	$6 x^{2} q^{4} y^{2} p^{4} z^{2} + x^{4} (p^{4} z^{2} + 4 q^{4} y^{2}) + q^{8} y^{4} (p^{4} z^{2} + 4 x^{2}) + p^{8} z^{4} (x^{2} + q^{4} y^{2}) .$

Then we have,

$(\frac{ℓ_{3 D}}{𝒟})^{- 2} {\hat{e}}_{3} \cdot {\hat{e}}_{3}$	$=$	$[x (2 q^{4} y^{2} + p^{4} z^{2})]^{2} + [q^{2} y (p^{4} z^{2} + 2 x^{2})]^{2} + [p^{2} z (x^{2} - q^{4} y^{2})]^{2}$
	$=$	$x^{2} (4 q^{8} y^{4} + 4 q^{4} y^{2} p^{4} z^{2} + p^{8} z^{4}) + q^{4} y^{2} (p^{8} z^{4} + 4 x^{2} p^{4} z^{2} + 4 x^{4}) + p^{4} z^{2} (x^{4} - 2 x^{2} q^{4} y^{2} + q^{8} y^{4})$
	$=$	$4 x^{2} q^{8} y^{4} + 4 x^{2} q^{4} y^{2} p^{4} z^{2} + x^{2} p^{8} z^{4} + q^{4} y^{2} p^{8} z^{4} + 4 x^{2} q^{4} y^{2} p^{4} z^{2} + 4 x^{4} q^{4} y^{2} + x^{4} p^{4} z^{2} - 2 x^{2} q^{4} y^{2} p^{4} z^{2} + q^{8} y^{4} p^{4} z^{2}$
	$=$	$6 x^{2} q^{4} y^{2} p^{4} z^{2} + p^{8} z^{4} (x^{2} + q^{4} y^{2}) + x^{4} (4 q^{4} y^{2} + p^{4} z^{2}) + q^{8} y^{4} (4 x^{2} + p^{4} z^{2})$
	$=$	$(\frac{ℓ_{3 D}}{𝒟})^{- 2},$

which means that, ${\hat{e}}_{3} \cdot {\hat{e}}_{3} = 1$ . Hooray! Again (11/11/2020)!

Direction Cosine Components for T6 Coordinates

n

λ_{n}

h_{n}

\frac{\partial λ_{n}}{\partial x}

\frac{\partial λ_{n}}{\partial y}

\frac{\partial λ_{n}}{\partial z}

γ_{n 1}

γ_{n 2}

γ_{n 3}

1

(x^{2} + q^{2} y^{2} + p^{2} z^{2})^{1 / 2}

λ_{1} ℓ_{3 D}

\frac{x}{λ_{1}}

\frac{q^{2} y}{λ_{1}}

\frac{p^{2} z}{λ_{1}}

(x) ℓ_{3 D}

(q^{2} y) ℓ_{3 D}

(p^{2} z) ℓ_{3 D}

2

\frac{x y^{1 / q^{2}}}{z^{2 / p^{2}}}

\frac{1}{λ_{2}} [\frac{x q^{2} y p^{2} z}{𝒟}]

\frac{λ_{2}}{x}

\frac{λ_{2}}{q^{2} y}

- \frac{2 λ_{2}}{p^{2} z}

\frac{x q^{2} y p^{2} z}{𝒟} (\frac{1}{x})

\frac{x q^{2} y p^{2} z}{𝒟} (\frac{1}{q^{2} y})

\frac{x q^{2} y p^{2} z}{𝒟} (- \frac{2}{p^{2} z})

3

---

- \frac{ℓ_{3 D}}{𝒟} [x (2 q^{4} y^{2} + p^{4} z^{2})]

\frac{ℓ_{3 D}}{𝒟} [q^{2} y (p^{4} z^{2} + 2 x^{2})]

\frac{ℓ_{3 D}}{𝒟} [p^{2} z (x^{2} - q^{4} y^{2})]

$ℓ_{3 D}$	$\equiv$	$(x^{2} + q^{4} y^{2} + p^{4} z^{2})^{- 1 / 2},$
$𝒟$	$\equiv$	$(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2})^{1 / 2} .$

Let's double-check whether this "third" unit vector is orthogonal to both the "first" and the "second" unit vectors.

${\hat{e}}_{1} \cdot {\hat{e}}_{3}$	$=$	$\frac{ℓ_{3 D}^{2}}{𝒟} {- x [x (2 q^{4} y^{2} + p^{4} z^{2})] + q^{2} y [q^{2} y (p^{4} z^{2} + 2 x^{2})] + p^{2} z [p^{2} z (x^{2} - q^{4} y^{2})]}$
	$=$	$\frac{ℓ_{3 D}^{2}}{𝒟} {- (2 x^{2} q^{4} y^{2} + x^{2} p^{4} z^{2}) + (q^{4} y^{2} p^{4} z^{2} + 2 x^{2} q^{4} y^{2}) + (x^{2} p^{4} z^{2} - q^{4} y^{2} p^{4} z^{2})}$
	$=$	$0,$

and,

${\hat{e}}_{2} \cdot {\hat{e}}_{3}$	$=$	$\frac{ℓ_{3 D}}{𝒟} \cdot \frac{x q^{2} y p^{2} z}{𝒟} {- [(2 q^{4} y^{2} + p^{4} z^{2})] + [(p^{4} z^{2} + 2 x^{2})] - [2 (x^{2} - q^{4} y^{2})]}$
	$=$	$0 .$

Q. E. D.

Search for Third Coordinate Expression

Let's try …

$λ_{3}$	$=$	$𝒟^{n} ℓ_{3 D}^{m}$
	$=$	$(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2})^{n / 2} (x^{2} + q^{4} y^{2} + p^{4} z^{2})^{- m / 2}$
$\Rightarrow \frac{\partial λ_{3}}{\partial x_{i}}$	$=$	$ℓ_{3 D}^{m} [\frac{n}{2} (q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2})^{n / 2 - 1}] \frac{\partial}{\partial x_{i}} [(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2})]$
		$+ 𝒟^{n} [- \frac{m}{2} (x^{2} + q^{4} y^{2} + p^{4} z^{2})^{- m / 2 - 1}] \frac{\partial}{\partial x_{i}} [(x^{2} + q^{4} y^{2} + p^{4} z^{2})]$
	$=$	$𝒟^{n} ℓ_{3 D}^{m} [\frac{n}{2} (q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2})^{- 1}] \frac{\partial}{\partial x_{i}} [(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2})]$
		$+ 𝒟^{n} ℓ_{3 D}^{m} [- \frac{m}{2} (x^{2} + q^{4} y^{2} + p^{4} z^{2})^{- 1}] \frac{\partial}{\partial x_{i}} [(x^{2} + q^{4} y^{2} + p^{4} z^{2})] .$

Hence,

$\frac{1}{𝒟^{n} ℓ_{3 D}^{m}} \cdot \frac{\partial λ_{3}}{\partial x}$	$=$	$\frac{n}{2 𝒟^{2}} \frac{\partial}{\partial x} [q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2}] - \frac{m ℓ_{3 D}^{2}}{2} \frac{\partial}{\partial x} [x^{2} + q^{4} y^{2} + p^{4} z^{2}]$
	$=$	$x {\frac{n}{𝒟^{2}} [p^{4} z^{2} + 4 q^{4} y^{2}] - m ℓ_{3 D}^{2}}$
	$=$	$x {\frac{n (p^{4} z^{2} + 4 q^{4} y^{2})}{(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2})} - \frac{m}{(x^{2} + q^{4} y^{2} + p^{4} z^{2})}}$

This is overly cluttered! Let's try, instead …

$A \equiv ℓ_{3 D}^{- 2}$

$=$

$(x^{2} + q^{4} y^{2} + p^{4} z^{2}),$

and,

$B \equiv 𝒟^{2}$

$=$

$(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2})$

$\Rightarrow \frac{\partial A}{\partial x}$	$=$	$2 x,$
$\frac{\partial A}{\partial y}$	$=$	$2 q^{4} y,$
$\frac{\partial A}{\partial z}$	$=$	$2 p^{4} z;$
$\frac{\partial B}{\partial x}$	$=$	$2 x (4 q^{4} y^{2} + p^{4} z^{2}),$
$\frac{\partial B}{\partial y}$	$=$	$2 q^{4} y (p^{4} z^{2} + 4 x^{2}),$
$\frac{\partial B}{\partial z}$	$=$	$2 p^{4} z (q^{4} y^{2} + x^{2}) .$

Now, let's assume that,

$λ_{3}$	$\equiv$	$(\frac{A}{B})^{1 / 2},$
$\Rightarrow \frac{\partial λ_{3}}{\partial x_{i}}$	$=$	$\frac{1}{2 (A B)^{1 / 2}} \cdot \frac{\partial A}{\partial x_{i}} - \frac{A^{1 / 2}}{2 B^{3 / 2}} \cdot \frac{\partial B}{\partial x_{i}}$
	$=$	$\frac{λ_{3}}{2 A B} [B \cdot \frac{\partial A}{\partial x_{i}} - A \cdot \frac{\partial B}{\partial x_{i}}] .$
$\Rightarrow [\frac{2 A B}{λ_{3}}] \frac{\partial λ_{3}}{\partial x_{i}}$	$=$	$(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2}) \cdot \frac{\partial A}{\partial x_{i}} - (x^{2} + q^{4} y^{2} + p^{4} z^{2}) \cdot \frac{\partial B}{\partial x_{i}} .$

Looking ahead …

$h_{3}^{- 2}$	$=$	${\frac{λ_{3}}{2 A B} [B \cdot \frac{\partial A}{\partial x} - A \cdot \frac{\partial B}{\partial x}]}^{2} + {\frac{λ_{3}}{2 A B} [B \cdot \frac{\partial A}{\partial y} - A \cdot \frac{\partial B}{\partial y}]}^{2} + {\frac{λ_{3}}{2 A B} [B \cdot \frac{\partial A}{\partial z} - A \cdot \frac{\partial B}{\partial z}]}^{2}$
$\Rightarrow [\frac{2 A B}{λ_{3}}]^{2} h_{3}^{- 2}$	$=$	$[B \cdot \frac{\partial A}{\partial x} - A \cdot \frac{\partial B}{\partial x}]^{2} + [B \cdot \frac{\partial A}{\partial y} - A \cdot \frac{\partial B}{\partial y}]^{2} + [B \cdot \frac{\partial A}{\partial z} - A \cdot \frac{\partial B}{\partial z}]^{2}$
$\Rightarrow [\frac{λ_{3}}{2 A B}] h_{3}$	$=$	${[B \cdot \frac{\partial A}{\partial x} - A \cdot \frac{\partial B}{\partial x}]^{2} + [B \cdot \frac{\partial A}{\partial y} - A \cdot \frac{\partial B}{\partial y}]^{2} + [B \cdot \frac{\partial A}{\partial z} - A \cdot \frac{\partial B}{\partial z}]^{2}}^{- 1 / 2}$

Then, for example,

$γ_{31} \equiv h_{3} (\frac{\partial λ_{3}}{\partial x})$

$=$

$[B \cdot \frac{\partial A}{\partial x} - A \cdot \frac{\partial B}{\partial x}] {[B \cdot \frac{\partial A}{\partial x} - A \cdot \frac{\partial B}{\partial x}]^{2} + [B \cdot \frac{\partial A}{\partial y} - A \cdot \frac{\partial B}{\partial y}]^{2} + [B \cdot \frac{\partial A}{\partial z} - A \cdot \frac{\partial B}{\partial z}]^{2}}^{- 1 / 2}$

As a result, we have,

$\Rightarrow [\frac{2 A B}{λ_{3}}] \frac{\partial λ_{3}}{\partial x}$	$=$	$2 x [(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2}) - (x^{2} + q^{4} y^{2} + p^{4} z^{2}) (4 q^{4} y^{2} + p^{4} z^{2})]$
	$=$	$2 x [(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2}) - (4 x^{2} q^{4} y^{2} + x^{2} p^{4} z^{2} + 4 q^{8} y^{4} + q^{4} y^{2} p^{4} z^{2} + 4 q^{4} y^{2} p^{4} z^{2} + p^{8} z^{4})]$
	$=$	$2 x [- (4 q^{8} y^{4} + 4 q^{4} y^{2} p^{4} z^{2} + p^{8} z^{4})]$
	$=$	$- 2 x (2 q^{4} y^{2} + p^{4} z^{2})^{2}$
	$=$	$- 8 x (q^{4} y^{2} + \frac{p^{4} z^{2}}{2})^{2}$
$\Rightarrow [A B] \frac{\partial \ln λ_{3}}{\partial \ln x}$	$=$	$- [2 x (q^{4} y^{2} + \frac{p^{4} z^{2}}{2})]^{2};$

and,

$\Rightarrow [\frac{2 A B}{λ_{3}}] \frac{\partial λ_{3}}{\partial y}$	$=$	$2 q^{4} y [(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2}) - (x^{2} + q^{4} y^{2} + p^{4} z^{2}) (p^{4} z^{2} + 4 x^{2})]$
	$=$	$2 q^{4} y [(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2}) - (x^{2} p^{4} z^{2} + 4 x^{4} + q^{4} y^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2} + p^{8} z^{4} + 4 x^{2} p^{4} z^{2})]$
	$=$	$- 2 q^{4} y (4 x^{4} + p^{8} z^{4} + 4 x^{2} p^{4} z^{2})$
	$=$	$- 2 q^{4} y (2 x^{2} + p^{4} z^{2})^{2}$
$\Rightarrow [A B] \frac{\partial \ln λ_{3}}{\partial \ln y}$	$=$	$- [2 q^{2} y (x^{2} + \frac{p^{4} z^{2}}{2})]^{2};$

and,

$\Rightarrow [\frac{2 A B}{λ_{3}}] \frac{\partial λ_{3}}{\partial z}$	$=$	$2 p^{4} z [(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2}) - (x^{2} + q^{4} y^{2} + p^{4} z^{2}) (q^{4} y^{2} + x^{2})]$
	$=$	$2 p^{4} z [(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2}) - (x^{2} q^{4} y^{2} + x^{4} + q^{8} y^{4} + x^{2} q^{4} y^{2} + q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2})]$
	$=$	$2 p^{4} z [(2 x^{2} q^{4} y^{2}) - (x^{4} + q^{8} y^{4})]$
	$=$	$- 2 p^{4} z [x^{4} + q^{8} y^{4} - 2 x^{2} q^{4} y^{2}]$
	$=$	$- 2 p^{4} z (x^{2} - q^{4} y^{2})^{2}$
$\Rightarrow [A B] \frac{\partial \ln λ_{3}}{\partial \ln z}$	$=$	$- 4 [(\frac{p^{4} z^{2}}{4}) (x^{2} - q^{4} y^{2})^{2}]$
	$=$	$- [2 (\frac{p^{2} z}{2}) (x^{2} - q^{4} y^{2})]^{2} .$

Wow! Really close! (13 November 2020)

Just for fun, let's see what we get for $h_{3}$ . It is given by the expression,

$h_{3}^{- 2}$	$=$	$(\frac{\partial λ_{3}}{\partial x})^{2} + (\frac{\partial λ_{3}}{\partial y})^{2} + (\frac{\partial λ_{3}}{\partial z})^{2}$
	$=$	${\frac{λ_{3}}{A B x} [2 x (q^{4} y^{2} + \frac{p^{4} z^{2}}{2})]^{2}}^{2} + {\frac{λ_{3}}{A B y} [2 q^{2} y (x^{2} + \frac{p^{4} z^{2}}{2})]^{2}}^{2} + {\frac{λ_{3}}{A B z} [2 (\frac{p^{2} z}{2}) (x^{2} - q^{4} y^{2})]^{2}}^{2}$
$\Rightarrow [\frac{A B}{λ_{3}}]^{2} h_{3}^{- 2}$	$=$	${\frac{1}{x^{2}} [2 x (q^{4} y^{2} + \frac{p^{4} z^{2}}{2})]^{4}} + {\frac{1}{y^{2}} [2 q^{2} y (x^{2} + \frac{p^{4} z^{2}}{2})]^{4}} + {\frac{1}{z^{2}} [2 (\frac{p^{2} z}{2}) (x^{2} - q^{4} y^{2})]^{4}}$

Fiddle Around

Let …

$ℒ_{x}$	$\equiv$	$- [B \cdot \frac{\partial A}{\partial x} - A \cdot \frac{\partial B}{\partial x}]$	$=$	$8 x (q^{4} y^{2} + \frac{p^{4} z^{2}}{2})^{2}$	$=$	$\frac{2}{x} [2 x (q^{4} y^{2} + \frac{p^{4} z^{2}}{2})]^{2}$	$=$	$8 x 𝔉_{x} (y, z)$
$ℒ_{y}$	$\equiv$	$- [B \cdot \frac{\partial A}{\partial y} - A \cdot \frac{\partial B}{\partial y}]$	$=$	$8 q^{4} y (x^{2} + \frac{p^{4} z^{2}}{2})^{2}$	$=$	$\frac{2}{y} [2 q^{2} y (x^{2} + \frac{p^{4} z^{2}}{2})]^{2}$	$=$	$8 y 𝔉_{y} (x, z)$
$ℒ_{z}$	$\equiv$	$- [B \cdot \frac{\partial A}{\partial z} - A \cdot \frac{\partial B}{\partial z}]$	$=$	$2 p^{4} z (x^{2} - q^{4} y^{2})^{2}$	$=$	$\frac{2}{z} [p^{2} z (x^{2} - q^{4} y^{2})]^{2}$	$=$	$8 z 𝔉_{z} (x, y)$

With this shorthand in place, we can write,

${\hat{e}}_{3}$	$=$	$\frac{ℓ_{3 D}}{𝒟} {- \hat{ı} [x (2 q^{4} y^{2} + p^{4} z^{2})] + \hat{ȷ} [q^{2} y (p^{4} z^{2} + 2 x^{2})] + \hat{k} [p^{2} z (x^{2} - q^{4} y^{2})]}$
	$=$	$\frac{1}{(A B)^{1 / 2}} {- \hat{ı} [\frac{x ℒ_{x}}{2}]^{1 / 2} + \hat{ȷ} [\frac{y ℒ_{y}}{2}]^{1 / 2} + \hat{k} [\frac{z ℒ_{z}}{2}]^{1 / 2}} .$

We therefore also recognize that,

$h_{3} (\frac{\partial λ_{3}}{\partial x})$	$=$	$- \frac{1}{(A B)^{1 / 2}} [\frac{x ℒ_{x}}{2}]^{1 / 2}$	$=$	$- \frac{1}{(A B)^{1 / 2}} [4 x^{2} 𝔉_{x}]^{1 / 2},$
$h_{3} (\frac{\partial λ_{3}}{\partial y})$	$=$	$\frac{1}{(A B)^{1 / 2}} [\frac{y ℒ_{y}}{2}]^{1 / 2}$	$=$	$\frac{1}{(A B)^{1 / 2}} [4 y^{2} 𝔉_{y}]^{1 / 2},$
$h_{3} (\frac{\partial λ_{3}}{\partial z})$	$=$	$\frac{1}{(A B)^{1 / 2}} [\frac{z ℒ_{z}}{2}]^{1 / 2}$	$=$	$\frac{1}{(A B)^{1 / 2}} [4 z^{2} 𝔉_{z}]^{1 / 2} .$

Now, if — and it is a BIG "if" — $h_{3} = h_{0} (A B)^{- 1 / 2}$ , then we have,

$h_{0} (\frac{\partial λ_{3}}{\partial x})$	$=$	$- [4 x^{2} 𝔉_{x}]^{1 / 2}$	$=$	$- 2 x [𝔉_{x}]^{1 / 2},$
$h_{0} (\frac{\partial λ_{3}}{\partial y})$	$=$	$[4 y^{2} 𝔉_{y}]^{1 / 2}$	$=$	$2 y [𝔉_{y}]^{1 / 2},$
$h_{0} (\frac{\partial λ_{3}}{\partial z})$	$=$	$[4 z^{2} 𝔉_{z}]^{1 / 2}$	$=$	$2 z [𝔉_{z}]^{1 / 2},$

$\Rightarrow h_{0} λ_{3}$

$=$

$- x^{2} [𝔉_{x}]^{1 / 2} + y^{2} [𝔉_{y}]^{1 / 2} + z^{2} [𝔉_{z}]^{1 / 2} .$

But if this is the correct expression for $λ_{3}$ and its three partial derivatives, then it must be true that,

$h_{3}^{- 2}$	$=$	$(\frac{\partial λ_{3}}{\partial x})^{2} + (\frac{\partial λ_{3}}{\partial y})^{2} + (\frac{\partial λ_{3}}{\partial z})^{2}$
$\Rightarrow (\frac{h_{3}}{h_{0}})^{- 2}$	$=$	$4 x^{2} [𝔉_{x}] + 4 y^{2} [𝔉_{y}] + 4 z^{2} [𝔉_{z}]$
	$=$	$4 x^{2} [(q^{4} y^{2} + \frac{p^{4} z^{2}}{2})^{2}] + 4 y^{2} [q^{4} (x^{2} + \frac{p^{4} z^{2}}{2})^{2}] + 4 z^{2} [\frac{p^{4}}{4} (x^{2} - q^{4} y^{2})^{2}]$
	$=$	$x^{2} (2 q^{4} y^{2} + p^{4} z^{2})^{2} + q^{4} y^{2} (2 x^{2} + p^{4} z^{2})^{2} + p^{4} z^{2} (x^{2} - q^{4} y^{2})^{2}$

Well … the right-hand side of this expression is identical to the right-hand side of the above expression, where we showed that it equals $(ℓ_{3 D} / 𝒟)^{- 2}$ . That is to say, we are now showing that,

$(\frac{h_{3}}{h_{0}})^{- 2}$	$=$	$(\frac{ℓ_{3 D}}{𝒟})^{- 2} = [A B]$
$\Rightarrow \frac{h_{3}}{h_{0}}$	$=$	$(A B)^{- 1 / 2} .$

And this is precisely what, just a few lines above, we hypothesized the functional expression for $h_{3}$ ought to be. EUREKA!

Summary

In summary, then …

$λ_{3}$	$\equiv$	$- x^{2} [𝔉_{x}]^{1 / 2} + y^{2} [𝔉_{y}]^{1 / 2} + z^{2} [𝔉_{z}]^{1 / 2}$
	$=$	$- x^{2} [(q^{4} y^{2} + \frac{p^{4} z^{2}}{2})^{2}]^{1 / 2} + y^{2} [q^{4} (x^{2} + \frac{p^{4} z^{2}}{2})^{2}]^{1 / 2} + z^{2} [\frac{p^{4}}{4} (x^{2} - q^{4} y^{2})^{2}]^{1 / 2}$
	$=$	$- x^{2} (q^{4} y^{2} + \frac{p^{4} z^{2}}{2}) + q^{2} y^{2} (x^{2} + \frac{p^{4} z^{2}}{2}) + \frac{p^{2} z^{2}}{2} (x^{2} - q^{4} y^{2}),$

and,

$h_{3}$	$=$	$(A B)^{- 1 / 2}$
	$=$	$[(x^{2} + q^{4} y^{2} + p^{4} z^{2}) (q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2})]^{- 1 / 2} .$

No! Once again this does not work. The direction cosines — and, hence, the components of the ${\hat{e}}_{3}$ unit vector — are not correct!

Speculation7

Direction Cosine Components for T6 Coordinates

n

λ_{n}

h_{n}

\frac{\partial λ_{n}}{\partial x}

\frac{\partial λ_{n}}{\partial y}

\frac{\partial λ_{n}}{\partial z}

γ_{n 1}

γ_{n 2}

γ_{n 3}

1

(x^{2} + q^{2} y^{2} + p^{2} z^{2})^{1 / 2}

λ_{1} ℓ_{3 D}

\frac{x}{λ_{1}}

\frac{q^{2} y}{λ_{1}}

\frac{p^{2} z}{λ_{1}}

(x) ℓ_{3 D}

(q^{2} y) ℓ_{3 D}

(p^{2} z) ℓ_{3 D}

2

\frac{x y^{1 / q^{2}}}{z^{2 / p^{2}}}

\frac{1}{λ_{2}} [\frac{x q^{2} y p^{2} z}{𝒟}]

\frac{λ_{2}}{x}

\frac{λ_{2}}{q^{2} y}

- \frac{2 λ_{2}}{p^{2} z}

\frac{x q^{2} y p^{2} z}{𝒟} (\frac{1}{x})

\frac{x q^{2} y p^{2} z}{𝒟} (\frac{1}{q^{2} y})

\frac{x q^{2} y p^{2} z}{𝒟} (- \frac{2}{p^{2} z})

3

---

- \frac{ℓ_{3 D}}{𝒟} [x (2 q^{4} y^{2} + p^{4} z^{2})]

\frac{ℓ_{3 D}}{𝒟} [q^{2} y (p^{4} z^{2} + 2 x^{2})]

\frac{ℓ_{3 D}}{𝒟} [p^{2} z (x^{2} - q^{4} y^{2})]

$ℓ_{3 D}$	$\equiv$	$(x^{2} + q^{4} y^{2} + p^{4} z^{2})^{- 1 / 2},$
$𝒟$	$\equiv$	$(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2})^{1 / 2} .$

On my white-board I have shown that, if

$λ_{3} \equiv ℓ_{3 D} 𝒟,$

then everything will work out as long as,

$ℒ$

$=$

$(\frac{𝒟}{ℓ_{3 D}})^{2} \frac{1}{ℓ_{3 D}^{4}},$

where,

$ℒ$	$\equiv$	$x^{2} (2 q^{4} y^{2} + p^{4} z^{2})^{4} + q^{8} y^{2} (2 x^{2} + p^{4} z^{2})^{4} + p^{8} z^{2} (x^{2} - q^{4} y^{2})^{4}$
	$=$	$x^{2} (4 q^{8} y^{4} + 4 q^{4} y^{2} p^{4} z^{2} + p^{8} z^{4})^{2} + q^{8} y^{2} (4 x^{4} + 4 x^{2} p^{4} z^{2} + p^{8} z^{4})^{2} + p^{8} z^{2} (x^{4} - 2 x^{2} q^{4} y^{2} + q^{8} y^{4})^{2}$
	$=$	$x^{2} [16 q^{16} y^{8} + 16 q^{12} y^{6} p^{4} z^{2} + 4 q^{8} y^{4} p^{8} z^{4} + 16 q^{12} y^{6} p^{4} z^{2} + 16 q^{8} y^{4} p^{8} z^{4} + 4 q^{4} y^{2} p^{12} z^{6} + 4 q^{8} y^{4} p^{8} z^{4} + 4 q^{4} y^{2} p^{12} z^{6} + p^{16} z^{8}]$
		$+ q^{8} y^{2} [16 x^{8} + 16 x^{6} p^{4} z^{2} + 4 x^{4} p^{8} z^{4} + 16 x^{6} p^{4} z^{2} + 16 x^{4} p^{8} z^{4} + 4 x^{2} p^{12} z^{6} + 4 x^{4} p^{8} z^{4} + 4 x^{2} p^{12} z^{6} + p^{16} z^{8}]$
		$+ p^{8} z^{2} [x^{8} - 2 x^{6} q^{4} y^{2} + x^{4} q^{8} y^{4} - 2 x^{6} q^{4} y^{2} + 4 x^{4} q^{8} y^{4} - 2 x^{2} q^{12} y^{6} + x^{4} q^{8} y^{4} - 2 x^{2} q^{12} y^{6} + q^{16} y^{8}]$
	$=$	$x^{2} [16 q^{16} y^{8} + 32 q^{12} y^{6} p^{4} z^{2} + 24 q^{8} y^{4} p^{8} z^{4} + 8 q^{4} y^{2} p^{12} z^{6} + p^{16} z^{8}]$
		$+ q^{8} y^{2} [16 x^{8} + 32 x^{6} p^{4} z^{2} + 24 x^{4} p^{8} z^{4} + 8 x^{2} p^{12} z^{6} + p^{16} z^{8}]$
		$+ p^{8} z^{2} [x^{8} - 4 x^{6} q^{4} y^{2} + 6 x^{4} q^{8} y^{4} - 4 x^{2} q^{12} y^{6} + q^{16} y^{8}]$

Let's check this out.

$R H S \equiv (\frac{𝒟}{ℓ_{3 D}})^{2} \frac{1}{ℓ_{3 D}^{4}}$	$=$	$(q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2}) (x^{2} + q^{4} y^{2} + p^{4} z^{2})^{3}$
	$=$	$(x^{2} + q^{4} y^{2} + p^{4} z^{2}) (q^{4} y^{2} p^{4} z^{2} + x^{2} p^{4} z^{2} + 4 x^{2} q^{4} y^{2}) [x^{4} + 2 x^{2} q^{4} y^{2} + 2 x^{2} p^{4} z^{2} + q^{8} y^{4} + 2 q^{4} y^{2} p^{4} z^{2} + p^{8} z^{4}]$
	$=$	$[(6 x^{2} q^{4} y^{2} p^{4} z^{2} + x^{4} p^{4} z^{2} + 4 x^{4} q^{4} y^{2}) + (q^{8} y^{4} p^{4} z^{2} + 4 x^{2} q^{8} y^{4}) + (q^{4} y^{2} p^{8} z^{4} + x^{2} p^{8} z^{4})]$
		$\times [x^{4} + 2 x^{2} q^{4} y^{2} + 2 x^{2} p^{4} z^{2} + q^{8} y^{4} + 2 q^{4} y^{2} p^{4} z^{2} + p^{8} z^{4}]$
	$=$	$[6 x^{2} q^{4} y^{2} p^{4} z^{2} + x^{4} (p^{4} z^{2} + 4 q^{4} y^{2}) + q^{8} y^{4} (p^{4} z^{2} + 4 x^{2}) + p^{8} z^{4} (q^{4} y^{2} + x^{2})]$
		$\times [x^{4} + 2 x^{2} q^{4} y^{2} + 2 x^{2} p^{4} z^{2} + q^{8} y^{4} + 2 q^{4} y^{2} p^{4} z^{2} + p^{8} z^{4}]$

Appendix/Ramblings/T6CoordinatesPt2: Difference between revisions

Latest revision as of 18:13, 23 July 2021

Contents

Concentric Ellipsoidal (T6) Coordinates (Part 2)

Orthogonal Coordinates

Speculation5

Spherical Coordinates

Use λ₁ Instead of r

Relationship To T3 Coordinates

Again Consider Full 3D Ellipsoid

Partial Derivatives & Scale Factors

First Coordinate

Second Coordinate (1^st Try)

Second Coordinate (2^nd Try)

Speculation6

Determine λ₂

Direction Cosines for Third Unit Vector

Search for Third Coordinate Expression

Fiddle Around

Summary

Speculation7

Navigation menu

@@ Line 1: / Line 1: @@
 __FORCETOC__
 <!-- __NOTOC__ will force TOC off -->
-=Concentric Ellipsoidal (T6) Coordinates=
+=Concentric Ellipsoidal (T6) Coordinates (Part 2)=
+==Orthogonal Coordinates==
-==Background==
+===Speculation5===
-Building on our [[Appendix/Ramblings/DirectionCosines|general introduction to ''Direction Cosines'']] in the context of orthogonal curvilinear coordinate systems, and on our previous development of [[User:Tohline/Appendix/Ramblings/T3Integrals|T3]] (concentric oblate-spheroidal) and [[Appendix/Ramblings/EllipticCylinderCoordinates#T5_Coordinates|T5]] (concentric elliptic) coordinate systems, here we explore the creation of a concentric ellipsoidal (T6) coordinate system.  This is motivated by our [[ThreeDimensionalConfigurations/Challenges#Trial_.232|desire to construct a fully analytically prescribable model of a nonuniform-density ellipsoidal configuration that is an analog to Riemann S-Type ellipsoids]].
-==Orthogonal Coordinates==
+====Spherical Coordinates====
-===Primary (''radial-like'') Coordinate===
-We start by defining a "radial" coordinate whose values identify various concentric ellipsoidal shells,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\lambda_1</math>
+<math>~r\cos\theta</math>
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} \, .</math>
+<math>~z \, ,</math>
    </td>
 </tr>
-</table>
-When <math>~\lambda_1 = a</math>, we obtain the standard definition of an ellipsoidal surface, it being understood that, <math>~q^2 = a^2/b^2</math> and <math>~p^2 = a^2/c^2</math>.  (We will assume that <math>~a > b > c</math>, that is, <math>~p^2 > q^2 > 1</math>.)
-A vector, <math>~\bold{\hat{n}}</math>, that is normal to the <math>~\lambda_1</math> = constant surface is given by the gradient of the function,
+<tr>
-<table border="0" cellpadding="5" align="center">
+  <td align="right">
+<math>~r\sin\theta</math>
+  </td>
+  <td align="center">
+<math>~=</math>
+  </td>
+  <td align="left">
+<math>~(x^2 + y^2)^{1 / 2} \, ,</math>
+  </td>
+</tr>
 <tr>
    <td align="right">
-<math>~F(x, y, z)</math>
+<math>~\tan\varphi</math>
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} - \lambda_1 \, .</math>
+<math>~\frac{y}{x} \, .</math>
    </td>
 </tr>
 </table>
-In Cartesian coordinates, this means,
+====Use &lambda;<sub>1</sub> Instead of r ====
+Here, as above, we define,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\bold{\hat{n}}(x, y, z)</math>
+<math>~\lambda_1^2</math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
-<math>~
+<math>~x^2 + q^2 y^2 + p^2 z^2 </math>
-\hat\imath \biggl( \frac{\partial F}{\partial x} \biggr)
-+ \hat\jmath \biggl( \frac{\partial F}{\partial y} \biggr)
-+ \hat{k} \biggl( \frac{\partial F}{\partial z} \biggr)
-</math>
    </td>
 </tr>
+</table>
+Using this expression to eliminate "x" (in favor of &lambda;<sub>1</sub>) in each of the three spherical-coordinate definitions, we obtain,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-&nbsp;
+<math>~r^2 \equiv x^2 + y^2 + z^2</math>
    </td>
    <td align="center">
@@ Line 70: / Line 77: @@
    </td>
    <td align="left">
-<math>~
+<math>~\lambda_1^2 - y^2(q^2-1) - z^2(p^2-1) \, ;</math>
-\hat\imath \biggl[ x(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr]
-+ \hat\jmath \biggl[ q^2y(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr]
-+ \hat{k}\biggl[ p^2 z(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr]
-</math>
    </td>
 </tr>
@@ Line 80: / Line 83: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~\tan^2\theta \equiv \frac{x^2 + y^2}{z^2}</math>
    </td>
    <td align="center">
@@ Line 86: / Line 89: @@
    </td>
    <td align="left">
-<math>~
+<math>~\frac{1}{z^2}\biggl[
-\hat\imath \biggl( \frac{x}{\lambda_1} \biggr)
+\lambda_1^2 -y^2(q^2-1) -p^2z^2
-+ \hat\jmath \biggl( \frac{q^2y}{\lambda_1} \biggr)
+\biggr] \, ;</math>
-+ \hat{k}\biggl(\frac{p^2 z}{\lambda_1} \biggr) \, ,
-</math>
    </td>
 </tr>
-</table>
-where it is understood that this expression is only to be evaluated at points, <math>~(x, y, z)</math>, that lie on the selected <math>~\lambda_1</math> surface &#8212; that is, at points for which the function, <math>~F(x,y,z) = 0</math>.  The length of this normal vector is given by the expression,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~[ \bold{\hat{n}} \cdot \bold{\hat{n}} ]^{1 / 2}</math>
+<math>~\frac{1}{\tan^2\varphi} \equiv \frac{x^2}{y^2}</math>
    </td>
    <td align="center">
@@ Line 106: / Line 104: @@
    <td align="left">
 <math>~
-\biggl[ \biggl( \frac{\partial F}{\partial x} \biggr)^2 + \biggl( \frac{\partial F}{\partial y} \biggr)^2 + \biggl( \frac{\partial F}{\partial z} \biggr)^2 \biggr]^{1 / 2}
+\frac{\lambda_1^2 - p^2z^2}{y^2} - q^2 \, .
 </math>
    </td>
 </tr>
+</table>
+After a bit of additional algebraic manipulation, we find that,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-&nbsp;
+<math>\frac{z^2}{\lambda_1^2}</math>
    </td>
    <td align="center">
@@ Line 120: / Line 122: @@
    <td align="left">
 <math>~
-\biggl[ \biggl( \frac{x}{\lambda_1} \biggr)^2
+\frac{ (1+\tan^2\varphi)}{\mathcal{D}^2} \, ,
-+ \biggl( \frac{q^2y}{\lambda_1} \biggr)^2
-+ \biggl(\frac{p^2 z}{\lambda_1} \biggr)^2 \biggr]^{1 / 2}
 </math>
    </td>
@@ Line 129: / Line 129: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~\frac{y^2}{\lambda_1^2}</math>
    </td>
    <td align="center">
@@ Line 136: / Line 136: @@
    <td align="left">
 <math>~
-\frac{1}{\lambda_1 \ell_{3D}}
+\biggl[\frac{ \mathcal{D}^2 \tan^2\varphi -  p^2 \tan^2\varphi (1+\tan^2\varphi)}{(1+q^2\tan^2\varphi) \mathcal{D}^2}  \biggr]
+\, ,
 </math>
    </td>
 </tr>
-</table>
-where,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\ell_{3D}</math>
+<math>~\frac{x^2}{\lambda_1^2}</math>
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~\biggl[ x^2 + q^4y^2 + p^4 z^2 \biggr]^{- 1 / 2} \, .</math>
+<math>~
+- q^2\biggl(\frac{y^2}{\lambda_1^2} \biggr) - p^2\biggl(\frac{z^2}{\lambda_1^2}\biggr)
+\, ,
+</math>
    </td>
 </tr>
 </table>
-It is therefore clear that the ''properly normalized'' normal unit vector that should be associated with any <math>~\lambda_1</math> = constant ellipsoidal surface is,
+where,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_1 </math>
+<math>~\mathcal{D}^2</math>
    </td>
    <td align="center">
@@ Line 169: / Line 170: @@
    <td align="left">
 <math>~
-\frac{ \bold\hat{n} }{ [ \bold{\hat{n}} \cdot \bold{\hat{n}} ]^{1 / 2} }
+\biggl[
-=
+(1 + q^2\tan^2\varphi)(p^2 + \tan^2\theta) - p^2(q^2-1)\tan^2\varphi
-\hat\imath (x \ell_{3D}) + \hat\jmath (q^2y \ell_{3D}) + \hat{k} (p^2 z \ell_{3D}) \, .
+\biggr] \, .
 </math>
    </td>
 </tr>
 </table>
-From our [[Appendix/Ramblings/DirectionCosines#Scale_Factors|accompanying discussion of direction cosines]], it is clear, as well, that the scale factor associated with the <math>~\lambda_1</math> coordinate is,
+<table border="1" cellpadding="8" align="center" width="80%"><tr><td align="left">
+As a check, let's set <math>~q^2 = p^2 = 1</math>, which should reduce to the normal spherical coordinate system.
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~h_1^2</math>
+<math>~\lambda_1^2</math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\rightarrow</math>
    </td>
    <td align="left">
-<math>~\lambda_1^2 \ell_{3D}^2 \, .</math>
+<math>~
+r^2 \, ,
+</math>
    </td>
-</tr>
+<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
-</table>
+   <td align="right">
-We can also fill in the top line of our direction-cosines table, namely,
+<math>~\mathcal{D}^2</math>
-<table border="1" cellpadding="8" align="center" width="60%">
-<tr>
-   <td align="center" colspan="4">
-'''Direction Cosines for T6 Coordinates'''
-<br />
-<math>~\gamma_{ni} = h_n \biggl( \frac{\partial \lambda_n}{\partial x_i}\biggr)</math>
    </td>
-</tr>
-<tr>
-  <td align="center" width="10%"><math>~n</math></td>
-  <td align="center" colspan="3"><math>~i = x, y, z</math>
-</tr>
-<tr>
-  <td align="center"><math>~1</math></td>
-  <td align="center">&nbsp;<br />
-<math>~x\ell_{3D}</math><br />&nbsp;
-  <td align="center"><math>~q^2 y \ell_{3D}</math>
-  <td align="center"><math>~p^2 z \ell_{3D}</math>
-</tr>
-<tr>
-  <td align="center"><math>~2</math></td>
    <td align="center">
-&nbsp;<br />
+<math>~\rightarrow</math>
----
-<br />&nbsp;
-  <td align="center">
-&nbsp;<br />
----
-<br />&nbsp;
-  <td align="center">
-&nbsp;<br />
----
-<br />&nbsp;
    </td>
-</tr>
+   <td align="left">
-<tr>
+<math>~
-   <td align="center"><math>~3</math></td>
+\biggl[
-  <td align="center">
+(1 + \tan^2\varphi)(1 + \tan^2\theta)
-&nbsp;<br />
+\biggr] \, .
----
+</math>
-<br />&nbsp;
-  </td>
-  <td align="center">
-&nbsp;<br />
----
-<br />&nbsp;
-  </td>
-  <td align="center">
-&nbsp;<br />
----
-<br />&nbsp;
    </td>
 </tr>
 </table>
-===Other Coordinate Pair in the Tangent Plane===
-Let's focus on a particular point on the <math>\lambda_1</math> = constant surface, <math>(x_0, y_0, z_0)</math>, that necessarily satisfies the function, <math>F(x_0, y_0, z_0) = 0</math>.  We have already derived the expression for the unit vector that is normal to the ellipsoidal surface at this point, namely,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>\hat{e}_1 </math>
+<math>~\Rightarrow ~~~ \frac{z^2}{\lambda_1^2}</math>
    </td>
    <td align="center">
-<math>\equiv</math>
+<math>~\rightarrow</math>
    </td>
    <td align="left">
-<math>
+<math>~
-\hat\imath (x_0 \ell_{3D}) + \hat\jmath (q^2y_0 \ell_{3D}) + \hat\jmath (p^2 z_0 \ell_{3D}) \, ,
+\frac{1}{1+\tan^2\theta} = \cos^2\theta = \frac{z^2}{r^2} \, ;
 </math>
    </td>
 </tr>
-</table>
-where, for this specific point on the surface,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>\ell_{3D}</math>
+<math>~\frac{y^2}{\lambda_1^2}</math>
    </td>
    <td align="center">
-<math>=</math>
+<math>~\rightarrow</math>
    </td>
    <td align="left">
-<math>\biggl[ x_0^2 + q^4y_0^2 + p^4 z_0^2 \biggr]^{- 1 / 2} \, .</math>
+<math>~
+\biggl[\frac{(1 + \tan^2\varphi)(1 + \tan^2\theta) \tan^2\varphi -  \tan^2\varphi (1+\tan^2\varphi)}{(1+\tan^2\varphi) (1 + \tan^2\varphi)(1 + \tan^2\theta)}  \biggr]
+</math>
    </td>
 </tr>
-</table>
+<tr>
-<table border="1" align="center" width="80%" cellpadding="10"><tr><td align="left">
+   <td align="right">
-<div align="center">
+&nbsp;
-'''Tangent Plane'''<br />
-[See, for example, [http://math.furman.edu/~dcs/courses/math21/ Dan Sloughter's] ([https://www.furman.edu Furman University]) 2001 Calculus III  class lecture notes &#8212; specifically [http://math.furman.edu/~dcs/courses/math21/lectures/l-15.pdf Lecture 15]]
-</div>
-----
-The two-dimensional plane that is tangent to the <math>\lambda_1</math> = constant surface ''at this point'' is given by the expression,
-<table border="0" cellpadding="5" align="center">
-<tr>
-   <td align="right">
-<math>0</math>
    </td>
    <td align="center">
-<math>=</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>
+<math>~\frac{\tan^2\varphi}{(1 + \tan^2\varphi)}
-(x - x_0) \biggl[ \frac{\partial \lambda_1}{\partial x} \biggr]_0
+\biggl[\frac{\tan^2\theta }{ (1 + \tan^2\theta)}  \biggr]
-+ (y - y_0) \biggl[\frac{\partial \lambda_1}{\partial y} \biggr]_0
+= \sin^2\theta \sin^2\varphi
-+ (z - z_0) \biggl[\frac{\partial \lambda_1}{\partial z} \biggr]_0
+= \frac{y^2}{r^2} \,;
 </math>
    </td>
@@ Line 313: / Line 258: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~\frac{x^2}{\lambda_1^2}</math>
    </td>
    <td align="center">
-<math>=</math>
+<math>~\rightarrow</math>
    </td>
    <td align="left">
-<math>
+<math>~
-(x - x_0) \biggl[ \frac{\partial F}{\partial x} \biggr]_0
+- \biggl(\frac{y^2}{\lambda_1^2} \biggr) - \biggl(\frac{z^2}{\lambda_1^2}\biggr)
-+ (y - y_0) \biggl[\frac{\partial F}{\partial y} \biggr]_0
+\, ,
-+ (z - z_0) \biggl[ \frac{\partial F}{\partial z} \biggr]_0
 </math>
    </td>
@@ Line 332: / Line 276: @@
    </td>
    <td align="center">
-<math>=</math>
+<math>~\rightarrow</math>
    </td>
    <td align="left">
-<math>
+<math>~
-(x - x_0) \biggl( \frac{x}{\lambda_1}\biggr)_0 + (y - y_0)\biggl( \frac{q^2 y  }{ \lambda_1 } \biggr)_0 + (z - z_0)\biggl( \frac{ p^2z }{ \lambda_1 } \biggr)_0
+- \sin^2\theta \sin^2\varphi - \cos^2\theta
+=
+- \sin^2\theta \sin^2\varphi + \sin^2\theta
+=
+\sin^2\theta \cos^2\varphi = \frac{x^2}{r^2}
+\, .
 </math>
    </td>
 </tr>
+</table>
+</td></tr></table>
+====Relationship To T3 Coordinates====
+If we set, <math>~q = 1</math>, but continue to assume that <math>~p > 1</math>, we expect to see a representation that resembles our previously discussed, [[User:Tohline/Appendix/Ramblings/T3Integrals#Integrals_of_Motion_in_T3_Coordinates|T3 Coordinates]].  Note, for example, that the new "radial" coordinate is,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>\Rightarrow~~~
+<math>~\lambda_1^2</math>
-x \biggl( \frac{x}{\lambda_1}\biggr)_0 + y \biggl( \frac{q^2 y  }{ \lambda_1 } \biggr)_0 + z \biggl( \frac{ p^2z }{ \lambda_1 } \biggr)_0
-</math>
    </td>
    <td align="center">
-<math>=</math>
+<math>~\rightarrow</math>
    </td>
    <td align="left">
-<math>
+<math>~
-x_0 \biggl( \frac{x}{\lambda_1}\biggr)_0 + y_0 \biggl( \frac{q^2 y  }{ \lambda_1 } \biggr)_0 + z_0 \biggl( \frac{ p^2z }{ \lambda_1 } \biggr)_0
+(\varpi^2 + p^2z^2) \, ,
 </math>
    </td>
-</tr>
+<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
-<tr>
    <td align="right">
-<math>\Rightarrow~~~
+<math>~\mathcal{D}^2</math>
-x x_0  + q^2 y y_0  + p^2 z z_0
-</math>
    </td>
    <td align="center">
-<math>=</math>
+<math>~\rightarrow</math>
    </td>
    <td align="left">
-<math>
+<math>~
-x_0^2  +  q^2 y_0^2 + p^2 z_0^2
+\biggl[
+(1 + \tan^2\varphi)(p^2 + \tan^2\theta)
+\biggr]  \, .
 </math>
    </td>
 </tr>
+</table>
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>\Rightarrow~~~
+<math>~\Rightarrow ~~~ \frac{p^2z^2}{\lambda_1^2}</math>
-x x_0  + q^2 y y_0  + p^2 z z_0
-</math>
    </td>
    <td align="center">
-<math>=</math>
+<math>~\rightarrow</math>
    </td>
    <td align="left">
-<math>
+<math>~\frac{ p^2}{(p^2 + \tan^2\theta)} = \frac{ 1}{(1 + p^{-2} \tan^2\theta)}\, ,</math>
-(\lambda_1^2)_0 \, .
-</math>
    </td>
 </tr>
-</table>
-</td></tr></table>
-Fix the value of <math>\lambda_1</math>.  This means that the relevant ellipsoidal surface is defined by the expression,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>\lambda_1^2</math>
+<math>~\frac{\varpi^2}{\lambda_1^2} = \frac{x^2}{\lambda_1^2} + \frac{y^2}{\lambda_1^2}</math>
    </td>
    <td align="center">
-<math>=</math>
+<math>~\rightarrow</math>
    </td>
    <td align="left">
-<math>x^2 + q^2y^2 + p^2z^2 \, .</math>
+<math>~1 -  p^2 \biggl( \frac{z^2}{\lambda_1^2}\biggr)
+=
+\biggl[1 -  \frac{ 1}{(1 + p^{-2}\tan^2\theta)} \biggr] \, .</math>
    </td>
 </tr>
 </table>
-If <math>z = 0</math>, the semi-major axis of the relevant x-y ellipse is <math>\lambda_1</math>, and the square of the semi-minor axis is <math>\lambda_1^2/q^2</math>.  At any other value, <math>z = z_0 < c</math>, the square of the semi-major axis of the relevant x-y ellipse is, <math>~(\lambda_1^2 - p^2z_0^2)</math> and the square of the corresponding semi-minor axis is, <math>(\lambda_1^2 - p^2z_0^2)/q^2</math>.  Now, for any chosen <math>x_0^2 \le (\lambda_1^2 - p^2z_0^2)</math>, the y-coordinate of the point on the <math>~\lambda_1</math> surface is given by the expression,
+We also see that,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>y_0^2</math>
+<math>~\frac{\varpi^2}{p^2z^2}</math>
    </td>
    <td align="center">
-<math>=</math>
+<math>~\rightarrow</math>
    </td>
    <td align="left">
-<math>\frac{1}{q^2}\biggl[ \lambda_1^2 - p^2 z_0 -x_0^2 \biggr] \, .</math>
+<math>~
+(1 + p^{-2}\tan^2\theta)\biggl[1 -  \frac{ 1}{(1 + p^{-2}\tan^2\theta)} \biggr]
+=
+p^{-2}\tan^2\theta \, .
+</math>
    </td>
 </tr>
 </table>
-The slope of the line that lies in the <math>z = z_0</math> plane and that is tangent to the ellipsoidal surface at <math>(x_0, y_0)</math> is,
+====Again Consider Full 3D Ellipsoid====
+Let's try to replace everywhere, <math>~[\varpi/(pz)]^2 = p^{-2}\tan^2\theta</math> with <math>~\lambda_2</math>.  This gives,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>m \equiv \frac{dy}{dx}\biggr|_{z_0}</math>
+<math>~\frac{\mathcal{D}^2}{p^2}</math>
    </td>
    <td align="center">
-<math>=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
-<math>- \frac{x_0}{q^2y_0}</math>
+<math>~
+\biggl[
+(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi
+\biggr] \, .
+</math>
    </td>
 </tr>
 </table>
-===Speculation1===
+which means that,
-Building on our experience developing [[User:Tohline/Appendix/Ramblings/T3Integrals#Integrals_of_Motion_in_T3_Coordinates|T3 Coordinates]] and, more recently, [[User:Tohline/Appendix/Ramblings/EllipticCylinderCoordinates#T5_Coordinates|T5 Coordinates]], let's define the two "angles,"
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\Zeta</math>
+<math>\frac{p^2 z^2}{\lambda_1^2}</math>
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~\sinh^{-1}\biggl(\frac{qy}{x} \biggr)</math>
+<math>~
+\frac{ (1+\tan^2\varphi)}{[(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi]}
+</math>
    </td>
-<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
+</tr>
+<tr>
    <td align="right">
-<math>~\Upsilon</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~\sinh^{-1}\biggl(\frac{pz}{x} \biggr) \, ,</math>
+<math>~
+\frac{ (1+\tan^2\varphi)/\tan^2\varphi}{[q^2 \lambda_2 (1 + q^2\tan^2\varphi)/(q^2\tan^2\varphi) - (q^2-1)]}
+= \frac{1/\sin^2\varphi}{[q^2\lambda_2 Q^2 - (q^2-1) ]}
+\, ,
+</math>
    </td>
 </tr>
-</table>
-in which case we can write,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\lambda_1^2</math>
+<math>~\frac{q^2y^2}{\lambda_1^2}</math>
    </td>
    <td align="center">
@@ Line 475: / Line 437: @@
    </td>
    <td align="left">
-<math>~x^2(\cosh^2\Zeta + \sinh^2\Upsilon)\, .</math>
+<math>~
+\frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) }
+-
+\frac{ q^2 \tan^2\varphi (1+\tan^2\varphi)}{(1+q^2\tan^2\varphi) [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] }
+</math>
    </td>
 </tr>
-</table>
-We speculate that the other two orthogonal coordinates may be defined by the expressions,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\lambda_2</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~x \biggl[ \sinh\Zeta \biggr]^{1/(1-q^2)}
+<math>~
-=
+\frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) } \biggl\{1
-x \biggl[ \frac{qy}{x}\biggr]^{1/(1-q^2)}
+-
-=
+\frac{ (1+\tan^2\varphi)}{ [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] }
-x \biggl[ \frac{x}{qy}\biggr]^{1/(q^2-1)}
+\biggr\}
-=
+</math>
-\biggl[ \frac{x^{q^2}}{qy}\biggr]^{1/(q^2-1)}
-\, ,</math>
    </td>
 </tr>
@@ Line 503: / Line 464: @@
 <tr>
    <td align="right">
-<math>~\lambda_3</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~x \biggl[ \sinh\Upsilon \biggr]^{1/(1-p^2)}
+<math>~
-=
+\frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) } \biggl\{1
-x \biggl[ \frac{pz}{x}\biggr]^{1/(1-p^2)}
+-
-=
+\frac{ (1+\tan^2\varphi)/\tan^2\varphi}{ [q^2\lambda_2(1 + q^2\tan^2\varphi)/(q^2\tan^2\varphi) - (q^2-1) ] }
-x \biggl[ \frac{x}{pz}\biggr]^{1/(p^2-1)}
+\biggr\}
-=
+</math>
-\biggl[ \frac{x^{p^2}}{pz}\biggr]^{1/(p^2-1)}
-\, .</math>
    </td>
 </tr>
-</table>
-Some relevant partial derivatives are,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial x}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 532: / Line 487: @@
    </td>
    <td align="left">
-<math>~\biggl[ \frac{1}{qy}\biggr]^{1/(q^2-1)} \biggl[ \frac{q^2}{q^2-1} \biggr]x^{1/(q^2-1)}
+<math>~
-=
+\frac{ 1}{Q^2 } \biggl\{1
-\biggl[ \frac{q^2}{q^2-1} \biggr]\biggl[ \frac{x}{qy}\biggr]^{1/(q^2-1)}
+-
-=
+\frac{ 1/\sin^2\varphi}{ [q^2\lambda_2 Q^2 - (q^2-1) ] }
-\biggl[ \frac{q^2}{q^2-1} \biggr]\frac{\lambda_2}{x}
+\biggr\}
-\, ;
+= \frac{1}{Q^2[~~]} \biggl[ [~~] - \frac{1}{\sin^2\varphi} \biggr]
+\, ,
 </math>
    </td>
@@ Line 544: / Line 500: @@
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial y}</math>
+<math>~\frac{x^2}{\lambda_1^2}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\biggl[ \frac{x^{q^2}}{q}\biggr]^{1/(q^2-1)} \biggl[ \frac{1}{1-q^2} \biggr] y^{q^2/(1-q^2)}
-=
-- \biggl[ \frac{1}{q^2-1} \biggr] \frac{\lambda_2}{y}
-\, ;</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial \lambda_3}{\partial x}</math>
    </td>
    <td align="center">
@@ Line 566: / Line 507: @@
    <td align="left">
 <math>~
-\biggl[ \frac{p^2}{p^2-1} \biggr]\frac{\lambda_3}{x}
-\, ;
+- \frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) } \biggl\{1
+-
+\frac{ (1+\tan^2\varphi)}{ [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] }
+\biggr\}
 </math>
    </td>
@@ Line 574: / Line 518: @@
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_3}{\partial z}</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~=</math>
+&nbsp;
    </td>
    <td align="left">
 <math>~
-- \biggl[ \frac{1}{p^2-1} \biggr] \frac{\lambda_3}{z}
+- \frac{ (1+\tan^2\varphi)}{[(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi]}
-\, .</math>
+</math>
    </td>
 </tr>
-</table>
-And the associated scale factors are,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~h_2^2</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 598: / Line 539: @@
    <td align="left">
 <math>~
-\biggl\{ \biggl[ \biggl( \frac{q^2}{q^2-1} \biggr)\frac{\lambda_2}{x} \biggr]^2 + \biggl[ - \biggl( \frac{1}{q^2-1} \biggr) \frac{\lambda_2}{y} \biggr]^2 \biggr\}^{-1}
+- \frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) }
+- \biggl\{\frac{ (1+\tan^2\varphi)}{ [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] }
+\biggr\}
 </math>
    </td>
@@ Line 612: / Line 556: @@
    <td align="left">
 <math>~
-\biggl\{ \biggl( \frac{q^2}{q^2-1} \biggr)^2 \frac{\lambda_2^2}{x^2}  + \biggl( \frac{1}{q^2-1} \biggr)^2 \frac{\lambda_2^2}{y^2} \biggr\}^{-1}
+- \frac{ 1 }{ Q^2 }
+- \biggl\{\frac{ 1/\sin^2\varphi}{ [q^2\lambda_2 Q^2 - (q^2-1) ] }
+\biggr\}
+=
+\frac{1}{Q^2 [~~] } \biggl\{
+Q^2[~~] - [~~] - \frac{Q^2}{\sin^2\varphi}
+\biggr\}
+\, ,
 </math>
    </td>
@@ Line 619: / Line 571: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~\frac{x^2 + q^2y^2}{\lambda_1^2}</math>
    </td>
    <td align="center">
@@ Line 625: / Line 577: @@
    </td>
    <td align="left">
-<math>~
+<math>~1
-\biggl\{x^2 + q^4 y^2 \biggr\}^{-1}
+- \biggl[1 + \frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) }\biggr] \biggl\{
-\biggl[ \frac{(q^2 - 1)^2x^2 y^2}{\lambda_2^2} \biggr] \, ;
+\frac{ (1+\tan^2\varphi)}{ [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] }
+\biggr\} \, .
 </math>
    </td>
 </tr>
+</table>
+Now, notice that,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~h_3^2</math>
+<math>~\frac{q^2 y^2 Q^2}{\lambda_1^2}</math>
    </td>
    <td align="center">
@@ Line 640: / Line 597: @@
    </td>
    <td align="left">
-<math>~
+<math>~1 - \frac{1}{[~~]\sin^2\varphi} \, ,</math>
-\biggl\{x^2 + p^4 z^2 \biggr\}^{-1}
-\biggl[ \frac{(p^2 - 1)^2x^2 z^2}{\lambda_3^2} \biggr] \, .
-</math>
    </td>
 </tr>
 </table>
-We can now fill in the rest of our direction-cosines table, namely,
+and,
+<table border="0" cellpadding="5" align="center">
-<table border="1" cellpadding="8" align="center" width="60%">
 <tr>
-   <td align="center" colspan="4">
+   <td align="right">
-'''Direction Cosines for T6 Coordinates'''
+<math>~\frac{x^2}{\lambda_1^2} + \frac{1}{Q^2}</math>
-<br />
-<math>~\gamma_{ni} = h_n \biggl( \frac{\partial \lambda_n}{\partial x_i}\biggr)</math>
    </td>
-</tr>
-<tr>
-  <td align="center" width="10%"><math>~n</math></td>
-  <td align="center" colspan="3"><math>~i = x, y, z</math>
-</tr>
-<tr>
-  <td align="center"><math>~1</math></td>
-  <td align="center">&nbsp;<br />
-<math>~x\ell_{3D}</math><br />&nbsp;
-  <td align="center"><math>~q^2 y \ell_{3D}</math>
-  <td align="center"><math>~p^2 z \ell_{3D}</math>
-</tr>
-<tr>
-  <td align="center"><math>~2</math></td>
    <td align="center">
-<math>~q^2 y \ell_q </math>
+<math>~=</math>
-   <td align="center">
+   </td>
-<math>~-x\ell_q</math>
+   <td align="left">
-   <td align="center">
+<math>~1 - \frac{1}{[~~]\sin^2\varphi} \, .</math>
-<math>~0</math>
    </td>
 </tr>
+</table>
+Hence,
+<table border="0" cellpadding="5" align="center">
 <tr>
-   <td align="center"><math>~3</math></td>
+   <td align="right">
-  <td align="center">
+<math>~\frac{x^2}{\lambda_1^2} + \frac{1}{Q^2}</math>
-<math>~p^2 z \ell_p</math>
    </td>
    <td align="center">
-<math>~0</math>
+<math>~=</math>
    </td>
-   <td align="center">
+   <td align="left">
-<math>~-x\ell_p</math>
+<math>~\frac{q^2 y^2 Q^2}{\lambda_1^2}</math>
    </td>
 </tr>
-</table>
-Hence,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_2</math>
+<math>~\Rightarrow~~~ 0</math>
    </td>
    <td align="center">
@@ Line 702: / Line 639: @@
    </td>
    <td align="left">
-<math>~
+<math>~Q^4 \biggl( \frac{q^2 y^2}{\lambda_1^2} \biggr) - Q^2 \biggl( \frac{x^2}{\lambda_1^2} \biggr) - 1</math>
-\hat\imath \gamma_{21}
-+ \hat\jmath \gamma_{22}
-+\hat{k} \gamma_{23}
-=
-\hat\imath (q^2y\ell_q)
-- \hat\jmath (x\ell_q) \, ;
-</math>
    </td>
 </tr>
@@ Line 715: / Line 645: @@
 <tr>
    <td align="right">
-<math>~\hat{e}_3</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 721: / Line 651: @@
    </td>
    <td align="left">
-<math>~
+<math>~Q^4  - Q^2 \biggl( \frac{x^2}{q^2 y^2} \biggr) - \biggl( \frac{\lambda_1^2}{q^2 y^2} \biggr) \, ,</math>
-\hat\imath \gamma_{31}
-+ \hat\jmath \gamma_{32}
-+\hat{k} \gamma_{33}
-=
-\hat\imath (p^2z\ell_p)
--\hat{k} (x\ell_p) \, .
-</math>
    </td>
 </tr>
 </table>
+where,
-Check:
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_2 \cdot \hat{e}_2</math>
+<math>~Q^2</math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
-<math>~
+<math>~\frac{1 + q^2\tan^2\varphi}{q^2\tan^2\varphi} \, .</math>
-(q^2y\ell_q)^2
-+ (x\ell_q)^2
-=
-\, ;
-</math>
    </td>
 </tr>
+</table>
+Solving the quadratic equation, we have,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_3 \cdot \hat{e}_3</math>
+<math>~Q^2</math>
    </td>
    <td align="center">
@@ Line 761: / Line 681: @@
    </td>
    <td align="left">
-<math>~
+<math>~\frac{1}{2} \biggl\{ \biggl( \frac{x^2}{q^2 y^2} \biggr) \pm \biggl[ \biggl( \frac{x^2}{q^2 y^2} \biggr)^2 + 4\biggl( \frac{\lambda_1^2}{q^2 y^2} \biggr) \biggr]^{1 / 2}  \biggr\}</math>
-(p^2z\ell_p)^2
-+ (x\ell_p)^2
-=
-\, ;
-</math>
    </td>
 </tr>
@@ Line 772: / Line 687: @@
 <tr>
    <td align="right">
-<math>~\hat{e}_2 \cdot \hat{e}_3</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 778: / Line 693: @@
    </td>
    <td align="left">
-<math>~
+<math>~\biggl( \frac{x^2}{2q^2 y^2} \biggr) \biggl\{ 1  \pm \biggl[ 1 + 4\biggl( \frac{\lambda_1^2 q^2 y^2}{x^4} \biggr) \biggr]^{1 / 2}  \biggr\} \, .</math>
-(q^2y\ell_q)(p^2z\ell_p) \ne 0 \, .
-</math>
    </td>
 </tr>
 </table>
-===Speculation2===
+<table border="1" align="center" width="80%" cellpadding="10"><tr><td align="left">
+<div align="center">'''Tentative Summary'''</div>
-Try,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\lambda_2</math>
+<math>~\lambda_1</math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
-<math>~
+<math>~(x^2 + q^2y^2 + p^2 z^2)^{1 / 2} \, ,</math>
-\frac{x}{y^{1/q^2} z^{1/p^2}} \, ,
-</math>
    </td>
 </tr>
-</table>
-in which case,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial x}</math>
+<math>~\lambda_2</math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
 <math>~
-\frac{\lambda_2}{x} \, ,
+\frac{(x^2 + y^2)^{1 / 2}}{pz}
-</math>
+\, ,</math>
    </td>
 </tr>
@@ Line 823: / Line 730: @@
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial y}</math>
+<math>~\lambda_3 = Q^2</math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
-<math>~
+<math>~\biggl( \frac{x^2}{2q^2 y^2} \biggr) \biggl\{ 1  \pm \biggl[ 1 + 4\biggl( \frac{\lambda_1^2 q^2 y^2}{x^4} \biggr) \biggr]^{1 / 2}  \biggr\} \, .</math>
-\frac{x}{z^{1/p^2}} \biggl(-\frac{1}{q^2}\biggr) y^{-1/q^2 - 1}
-=
--\frac{\lambda_2}{q^2 y}
-\, ,
-</math>
    </td>
 </tr>
+</table>
+</td></tr></table>
+====Partial Derivatives &amp; Scale Factors====
+=====First Coordinate=====
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial z}</math>
+<math>~\frac{\partial \lambda_1}{\partial x}</math>
    </td>
    <td align="center">
@@ Line 846: / Line 756: @@
    </td>
    <td align="left">
-<math>~
+<math>~\frac{x}{\lambda_1} \, ,</math>
--\frac{\lambda_2}{p^2 z}
-\, .
-</math>
    </td>
 </tr>
-</table>
-The associated scale factor is, then,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~h_2^2</math>
+<math>~\frac{\partial \lambda_1}{\partial y}</math>
    </td>
    <td align="center">
@@ Line 864: / Line 768: @@
    </td>
    <td align="left">
-<math>~\biggl[
+<math>~\frac{q^2 y}{\lambda_1} \, ,</math>
-\biggl( \frac{\partial \lambda_2}{\partial x} \biggr)^2
+   </td>
-+
-\biggl( \frac{\partial \lambda_2}{\partial y} \biggr)^2
-+
-\biggl( \frac{\partial \lambda_2}{\partial z} \biggr)^2
-\biggr]^{-1}
-</math>
-   </td>
 </tr>
 <tr>
    <td align="right">
-&nbsp;
+<math>~\frac{\partial \lambda_1}{\partial z}</math>
    </td>
    <td align="center">
@@ Line 883: / Line 780: @@
    </td>
    <td align="left">
-<math>~\biggl[
+<math>~\frac{p^2 z}{\lambda_1} \, .</math>
-\biggl( \frac{ \lambda_2}{x} \biggr)^2
-+
-\biggl( -\frac{\lambda_2}{q^2y} \biggr)^2
-+
-\biggl( - \frac{\lambda_2}{p^2z} \biggr)^2
-\biggr]^{-1}
-</math>
    </td>
 </tr>
 </table>
-===Speculation3===
-Try,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\lambda_2</math>
+<math>~h_1^{-2}</math>
    </td>
    <td align="center">
@@ Line 909: / Line 796: @@
    <td align="left">
 <math>~
-\frac{(x+p^2 z)^{1 / 2}}{y^{1/q^2} } \, ,
+\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^2
++ \biggl( \frac{\partial \lambda_1}{\partial y} \biggr)^2
++ \biggl( \frac{\partial \lambda_1}{\partial z} \biggr)^2
 </math>
-  </td>
-</tr>
-</table>
-in which case,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial x}</math>
    </td>
    <td align="center">
@@ Line 926: / Line 806: @@
    <td align="left">
 <math>~
-\frac{1}{2y^{1/q^2}}\biggl(x + p^2z\biggr)^{- 1 / 2}
+\biggl( \frac{x}{\lambda_1} \biggr)^2
-=
++ \biggl( \frac{q^2 y}{\lambda_1} \biggr)^2
-\frac{\lambda_2}{2(x + p^2z) }
++ \biggl( \frac{p^2 z}{\lambda_1} \biggr)^2
-\, ,
+\, .</math>
-</math>
    </td>
 </tr>
@@ Line 936: / Line 815: @@
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial y}</math>
+<math>~\Rightarrow ~~~ h_1</math>
    </td>
    <td align="center">
 <math>~=</math>
    </td>
-   <td align="left">
+   <td align="left" colspan="3">
 <math>~
--\frac{\lambda_2}{q^2y}
+\lambda_1 \ell_{3D} \, ,
-\, ,
 </math>
    </td>
 </tr>
+</table>
-<tr>
+where,
+<div align="center"><math>\ell_{3D} \equiv (x^2 + q^4y^2 + p^4z^2)^{-1 / 2} \, .</math></div>
+As a result, the associated unit vector is,
+<table border="0" cellpadding="5" align="center">
+<tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial z}</math>
+<math>~\hat{e}_1</math>
    </td>
    <td align="center">
@@ Line 958: / Line 841: @@
    <td align="left">
 <math>~
+\hat{\imath} h_1 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)
-\, .
++ \hat{\jmath} h_1 \biggl( \frac{\partial \lambda_1}{\partial y} \biggr)
++ \hat{k} h_1 \biggl( \frac{\partial \lambda_1}{\partial z} \biggr)
 </math>
    </td>
 </tr>
-</table>
-===Speculation4===
-====Development====
-Here we stick with the [[#Primary_.28radial-like.29_Coordinate|primary (radial-like) coordinate as defined above]]; for example,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\lambda_1</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} \, ,</math>
+<math>~
+\hat{\imath} x \ell_{3D}
++ \hat{\jmath} q^2 y\ell_{3D}
++ \hat{k} p^2 z \ell_{3D}
+\, .
+</math>
    </td>
 </tr>
+</table>
+Notice that,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~h_1</math>
+<math>~\hat{e}_1 \cdot \hat{e}_1</math>
    </td>
    <td align="center">
@@ Line 992: / Line 877: @@
    </td>
    <td align="left">
-<math>~\lambda_1 \ell_{3D} \, ,</math>
+<math>~
+(x^2 + q^4 y^2 + p^4 z^2) \ell_{3D}^2 = 1 \, .
+</math>
    </td>
 </tr>
+</table>
+=====Second Coordinate (1<sup>st</sup> Try)=====
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\ell_{3D}</math>
+<math>~\frac{\partial \lambda_2}{\partial x}</math>
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~[ x^2 + q^4y^2 + p^4 z^2 ]^{- 1 / 2} \, ,</math>
+<math>~
+\frac{1}{pz} \biggl[ \frac{x}{(x^2 + y^2)^{1 / 2}} \biggr] \, ,
+</math>
    </td>
 </tr>
@@ Line 1,010: / Line 904: @@
 <tr>
    <td align="right">
-<math>~\hat{e}_1 </math>
+<math>~\frac{\partial \lambda_2}{\partial y}</math>
    </td>
    <td align="center">
@@ Line 1,017: / Line 911: @@
    <td align="left">
 <math>~
-\hat\imath (x \ell_{3D}) + \hat\jmath (q^2y \ell_{3D}) + \hat{k} (p^2 z \ell_{3D}) \, .
+\frac{1}{pz} \biggl[ \frac{y}{(x^2 + y^2)^{1 / 2}} \biggr]
-</math>
+\, ,</math>
    </td>
 </tr>
-</table>
-<table border="1" width="80%" align="center" cellpadding="10"><tr><td align="left">
-Note that, <math>~\hat{e}_1 \cdot \hat{e}_1 = 1</math>, which means that this is, indeed, a properly normalized ''unit'' vector.
-</td></tr></table>
-Then, drawing from our [https://www.phys.lsu.edu/astro/H_Book.current/Appendices/Mathematics/operators.tohline1.pdf earliest discussions of "T1 Coordinates"], we'll try defining the ''second'' coordinate as,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\lambda_3</math>
+<math>~\frac{\partial \lambda_2}{\partial z}</math>
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
 <math>~
-\tan^{-1} u \, ,
+- \frac{(x^2 + y^2)^{1 / 2}}{pz^2}
-</math>
+\, .</math>
-&nbsp; &nbsp; &nbsp; where,
    </td>
 </tr>
+</table>
-<tr>
-  <td align="right">
-<math>~u</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>\frac{y^{1/q^2}}{x} \, .</math>
-  </td>
-</tr>
-</table>
-The relevant partial derivatives are,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_3}{\partial x}</math>
+<math>~h_2^{-2}</math>
    </td>
    <td align="center">
@@ Line 1,069: / Line 942: @@
    <td align="left">
 <math>~
-\frac{1}{1 + u^2} \biggl[ - \frac{y^{1/q^2}}{x^2} \biggr]
+\biggl( \frac{\partial \lambda_2}{\partial x} \biggr)^2
-=
++ \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)^2
-- \biggl[ \frac{u}{1 + u^2}\biggr]\frac{1}{x}
++ \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)^2
-=
-- \frac{\sin\lambda_3 \cos\lambda_3}{x}
-\, ,
 </math>
    </td>
@@ Line 1,081: / Line 951: @@
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_3}{\partial y}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 1,088: / Line 958: @@
    <td align="left">
 <math>~
-\frac{1}{1 + u^2} \biggl[ \frac{y^{(1/q^2-1)}}{q^2x} \biggr]
+\biggl\{ \frac{1}{pz} \biggl[ \frac{x}{(x^2 + y^2)^{1 / 2}} \biggr] \biggr\}^2
-=
++ \biggl\{ \frac{1}{pz} \biggl[ \frac{y}{(x^2 + y^2)^{1 / 2}} \biggr] \biggr\}^2
-\biggl[ \frac{u}{1 + u^2}\biggr]\frac{1}{q^2y}
++ \biggl\{ \frac{(x^2 + y^2)^{1 / 2}}{pz^2} \biggr\}^2
-=
-\frac{\sin\lambda_3 \cos\lambda_3}{q^2y} \, ,
 </math>
    </td>
 </tr>
-</table>
-which means that,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~h_3^2</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 1,109: / Line 974: @@
    <td align="left">
 <math>~
-\biggl[ \biggl( \frac{\partial \lambda_3}{\partial x} \biggr)^2 + \biggl( \frac{\partial \lambda_3}{\partial y} \biggr)^2 \biggr]^{-1}
+\biggl\{ \biggl[ \frac{x^2}{(x^2 + y^2)p^2 z^2} \biggr] \biggr\}
++ \biggl\{ \biggl[ \frac{y^2}{(x^2 + y^2)p^2 z^2} \biggr] \biggr\}
++ \biggl\{ \frac{(x^2 + y^2)}{p^2 z^4} \biggr\}
 </math>
    </td>
@@ Line 1,123: / Line 990: @@
    <td align="left">
 <math>~
-\biggl[ \frac{u}{1 + u^2}\biggr]^{-2} \biggl[ \frac{1}{x^2} + \frac{1}{q^4y^2} \biggr]^{-1}
+\frac{1}{p^2 z^2}
++ \frac{(x^2 + y^2)}{p^2 z^4}
+=
+\frac{(x^2 + y^2 + z^2)}{p^2 z^4}
 </math>
    </td>
@@ Line 1,130: / Line 1,000: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~\Rightarrow~~~h_2</math>
    </td>
    <td align="center">
@@ Line 1,137: / Line 1,007: @@
    <td align="left">
 <math>~
-\biggl[ \frac{1 + u^2}{u}\biggr]^{2} \biggl[ \frac{x^2 + q^4y^2}{x^2q^4y^2} \biggr]^{-1}
+\frac{p z^2}{r }
 </math>
    </td>
 </tr>
+</table>
+As a result, the associated unit vector is,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\Rightarrow~~~h_3</math>
+<math>~\hat{e}_2</math>
    </td>
    <td align="center">
@@ Line 1,151: / Line 1,024: @@
    <td align="left">
 <math>~
-\biggl[ \frac{1 + u^2}{u}\biggr]xq^2 y \ell_q = \frac{xq^2 y \ell_q}{\sin\lambda_3 \cos\lambda_3} \, ,
+\hat{\imath} h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)
++ \hat{\jmath} h_2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)
++ \hat{k} h_2 \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)
 </math>
-&nbsp; &nbsp; &nbsp; where,
    </td>
 </tr>
@@ Line 1,159: / Line 1,033: @@
 <tr>
    <td align="right">
-<math>~\ell_q</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~[x^2 + q^4 y^2]^{-1 / 2} \, .</math>
+<math>~
+\hat{\imath}  \biggl[ \frac{xz}{r(x^2 + y^2)^{1 / 2}} \biggr]
++ \hat{\jmath}  \biggl[ \frac{yz}{r(x^2 + y^2)^{1 / 2}} \biggr]
+- \hat{k} \biggl[ \frac{(x^2 + y^2)^{1 / 2}}{r} \biggr] \, .
+</math>
    </td>
-</tr>
 </table>
-The third row of direction cosines can now be filled in to give,
+Notice that,
+<table border="0" cellpadding="5" align="center">
-<table border="1" cellpadding="8" align="center" width="60%">
 <tr>
-   <td align="center" colspan="4">
+   <td align="right">
-'''Direction Cosines for T6 Coordinates'''
+<math>~\hat{e}_2 \cdot \hat{e}_2</math>
-<br />
-<math>~\gamma_{ni} = h_n \biggl( \frac{\partial \lambda_n}{\partial x_i}\biggr)</math>
    </td>
-</tr>
-<tr>
-  <td align="center" width="10%"><math>~n</math></td>
-  <td align="center" colspan="3"><math>~i = x, y, z</math>
-</tr>
-<tr>
-  <td align="center"><math>~1</math></td>
-  <td align="center">&nbsp;<br />
-<math>~x\ell_{3D}</math><br />&nbsp;
-  <td align="center"><math>~q^2 y \ell_{3D}</math>
-  <td align="center"><math>~p^2 z \ell_{3D}</math>
-</tr>
-<tr>
-  <td align="center"><math>~2</math></td>
    <td align="center">
-&nbsp;<br />
+<math>~=</math>
----
+  </td>
-<br />&nbsp;
+   <td align="left">
-   <td align="center">
+<math>~
-&nbsp;<br />
+\biggl[ \frac{x^2 z^2}{r^2(x^2 + y^2)} \biggr]
----
++ \biggl[ \frac{y^2 z^2}{r^2(x^2 + y^2)} \biggr]
-<br />&nbsp;
++ \biggl[ \frac{(x^2 + y^2)}{r^2} \biggr]
-  <td align="center">
+= 1 \, .
-&nbsp;<br />
+</math>
----
-<br />&nbsp;
    </td>
 </tr>
-<tr>
+</table>
-  <td align="center"><math>~3</math></td>
-  <td align="center">
+Let's check to see if this "second" unit vector is orthogonal to the "first."
-<math>~-q^2 y \ell_q</math>
-  </td>
-  <td align="center">
-<math>~x \ell_q</math>
-  </td>
-  <td align="center">
-<math>~0</math>
-  </td>
-</tr>
-</table>
-which means that the associated unit vector is,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_3 </math>
+<math>~\hat{e}_1 \cdot \hat{e}_2</math>
    </td>
    <td align="center">
@@ Line 1,232: / Line 1,080: @@
    <td align="left">
 <math>~
--\hat\imath (q^2 y \ell_{q}) + \hat\jmath (x \ell_{q})  \, .
+x\ell_{3D} \biggl[ \frac{xz}{r(x^2 + y^2)^{1 / 2}} \biggr]
++ q^2 y\ell_{3D} \biggl[ \frac{yz}{r(x^2 + y^2)^{1 / 2}} \biggr]
+- p^2 z \ell_{3D} \biggl[ \frac{(x^2 + y^2)^{1 / 2}}{r} \biggr]
 </math>
    </td>
 </tr>
-</table>
-<table border="1" width="80%" align="center" cellpadding="10"><tr><td align="left">
-Note that, <math>~\hat{e}_3 \cdot \hat{e}_3 = 1</math>, which means that this also is a properly normalized ''unit'' vector.  Note, as well, that the dot product between our "first" and "third" unit vectors should be zero if they are indeed orthogonal to each other.  Let's see &hellip;
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_3 \cdot \hat{e}_1</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 1,251: / Line 1,096: @@
    <td align="left">
 <math>~
-(- q^2y \ell_q)x\ell_{3D} + (x\ell_q) q^2y\ell_{3D} = 0 \, .
+\ell_{3D} \biggl\{ \biggl[ \frac{x^2z}{r(x^2 + y^2)^{1 / 2}} \biggr]
++ \biggl[ \frac{q^2 y^2 z}{r(x^2 + y^2)^{1 / 2}} \biggr]
+-  \biggl[ \frac{p^2 z(x^2 + y^2)^{1 / 2}}{r} \biggr]
+\biggr\}
 </math>
    </td>
 </tr>
-</table>
-Q.E.D.
-</td></tr></table>
-Now, even though we have not yet determined the proper expression for the "second" orthogonal coordinate, <math>~\lambda_2</math>, we should be able to obtain an expression for its associated unit vector from the cross product of the "third" and "first" unit vectors.  Specifically we find,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_2 \equiv \hat{e}_3 \times \hat{e}_1</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 1,271: / Line 1,113: @@
    <td align="left">
 <math>~
-\hat\imath \biggl[ (e_3)_2 (e_1)_3 - (e_3)_3(e_1)_2 \biggr]
+\frac{z\ell_{3D}}{r (x^2 + y^2)^{1 / 2}} \biggl\{ \biggl[ x^2\biggr]
-+
++ \biggl[ q^2 y^2 \biggr]
-\hat\jmath \biggl[ (e_3)_3 (e_1)_1 - (e_3)_1(e_1)_3 \biggr]
+-  \biggl[ p^2 (x^2 + y^2) \biggr]
-+
+\biggr\}
-\hat{k} \biggl[ (e_3)_1 (e_1)_2 - (e_3)_2(e_1)_1 \biggr]
 </math>
    </td>
@@ Line 1,285: / Line 1,126: @@
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\ne</math>
    </td>
    <td align="left">
 <math>~
-\hat\imath \biggl[ (x \ell_q) (p^2 z \ell_{3D}) - 0 \biggr]
+\ .
-+
-\hat\jmath \biggl[ 0 - (-q^2y \ell_q)(p^2z \ell_{3D}) \biggr]
-+
-\hat{k} \biggl[ (-q^2y \ell_q) (q^2 y \ell_{3D}) - (x\ell_q)(x\ell_{3D}) \biggr]
 </math>
    </td>
 </tr>
+</table>
+=====Second Coordinate (2<sup>nd</sup> Try)=====
+Let's try,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-&nbsp;
+<math>~\lambda_2</math>
    </td>
    <td align="center">
@@ Line 1,306: / Line 1,150: @@
    </td>
    <td align="left">
-<math>~\ell_q \ell_{3D}\biggl[
+<math>~
-\hat\imath ( xp^2 z ) + \hat\jmath ( q^2y p^2z ) - \hat{k} ( x^2 + q^4 y^2 )
+\biggl[\frac{(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2}}{pz}  \biggr]
-\biggr]
+\, ,
 </math>
    </td>
@@ Line 1,315: / Line 1,159: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~\Rightarrow~~~\frac{\partial \lambda_2}{\partial x}</math>
    </td>
    <td align="center">
@@ Line 1,321: / Line 1,165: @@
    </td>
    <td align="left">
-<math>~\ell_q \ell_{3D}\biggl[
+<math>~
-\hat\imath ( xp^2 z ) + \hat\jmath ( q^2y p^2z ) - \hat{k} \biggl( \frac{1}{\ell_q^2}  \biggr)
+\frac{x}{pz(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2} } = \frac{x}{p^2 z^2 \lambda_2}
-\biggr] \, .
+\, ,
 </math>
    </td>
 </tr>
-</table>
-<table border="1" width="80%" align="center" cellpadding="10"><tr><td align="left">
+<tr>
-Note that,
+  <td align="right">
-<table border="0" cellpadding="5" align="center">
+<math>~\frac{\partial \lambda_2}{\partial y}</math>
+  </td>
+  <td align="center">
+<math>~=</math>
+  </td>
+  <td align="left">
+<math>~
+\frac{q^2 y}{pz(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2} } = \frac{q^2y}{p^2 z^2 \lambda_2}
+\, ,
+</math>
+  </td>
+</tr>
 <tr>
    <td align="right">
-<math>~\hat{e}_3 \cdot \hat{e}_2</math>
+<math>~\frac{\partial \lambda_2}{\partial z}</math>
    </td>
    <td align="center">
@@ Line 1,341: / Line 1,195: @@
    </td>
    <td align="left">
-<math>~\ell_q^2 \ell_{3D} \biggl[
+<math>~
-(- q^2y )x p^2 z
+\frac{\mathfrak{f}\cdot p^2z}{pz(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2} }
-+ (x) q^2y p^2 z
+-
-\biggr] = 0 \, ;
+\frac{(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2}}{pz^2}
+= \frac{1}{p^2z^2 \lambda_2 } \biggl( \mathfrak{f}\cdot p^2z \biggr) - \frac{\lambda_2 }{z}
+= \frac{1}{p^2z^2 \lambda_2 } \biggl( \mathfrak{f}\cdot p^2z  - p^2z \lambda_2^2 \biggr)
+\, .
 </math>
    </td>
 </tr>
 </table>
-and,
+Hence,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_1 \cdot \hat{e}_2</math>
+<math>~h_2^{-2}</math>
    </td>
    <td align="center">
@@ Line 1,361: / Line 1,220: @@
    <td align="left">
 <math>~
-(x\ell_{3D})xp^2z \ell_q \ell_{3D}
+\biggl( \frac{\partial \lambda_2}{\partial x} \biggr)^2
-+ (q^2y \ell_{3D}) q^2yp^2 z \ell_q \ell_{3D}
++ \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)^2
-- (x^2 + q^4 y^2)\ell_q \ell_{3D} (p^2 z \ell_{3D} )
++ \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)^2
 </math>
    </td>
@@ Line 1,376: / Line 1,235: @@
    </td>
    <td align="left">
-<math>~ \ell_q \ell_{3D}^2 \biggl[
+<math>~
-x^2p^2z
+\biggl[ \frac{x}{p^2 z^2 \lambda_2} \biggr]^2
-+ (q^4y^2 ) p^2 z
++ \biggl[ \frac{q^2y}{p^2 z^2 \lambda_2} \biggr]^2
-- (x^2 + q^4 y^2) (p^2 z )
++ \biggl[ \frac{ \mathfrak{f} }{z \lambda_2 }  - \frac{\lambda_2 }{z}\biggr]^2
-\biggr] = 0 \, .
 </math>
    </td>
 </tr>
-</table>
-We conclude, therefore, that <math>~\hat{e}_2</math> is perpendicular to both of the other unit vectors. <font color="red">'''Hooray!'''</font>
-</td></tr></table>
-Filling in the second row of the direction cosines table gives,
-<table border="1" cellpadding="8" align="center" width="60%">
 <tr>
-   <td align="center" colspan="4">
+   <td align="right">
-'''Direction Cosines for T6 Coordinates'''
+&nbsp;
-<br />
+  </td>
-<math>~\gamma_{ni} = h_n \biggl( \frac{\partial \lambda_n}{\partial x_i}\biggr)</math>
+  <td align="center">
+<math>~=</math>
+  </td>
+  <td align="left">
+<math>~
+\biggl[ \frac{x^2 + q^4 y^2}{p^4 z^4 \lambda_2^2} \biggr]
++ \biggl[ \frac{1}{z\lambda_2}\biggl( \mathfrak{f} - \lambda_2^2 \biggr) \biggr]^2
+</math>
    </td>
-</tr>
-<tr>
-  <td align="center" width="10%"><math>~n</math></td>
-  <td align="center" colspan="3"><math>~i = x, y, z</math>
 </tr>
 <tr>
-   <td align="center"><math>~1</math></td>
+   <td align="right">
-  <td align="center">&nbsp;<br />
+&nbsp;
-<math>~x\ell_{3D}</math><br />&nbsp;
+   </td>
-  <td align="center"><math>~q^2 y \ell_{3D}</math>
-   <td align="center"><math>~p^2 z \ell_{3D}</math>
-</tr>
-<tr>
-  <td align="center"><math>~2</math></td>
    <td align="center">
-<math>~x p^2 z\ell_q \ell_{3D}</math>
+<math>~=</math>
-   <td align="center">
+  </td>
-<math>~q^2y p^2 z\ell_q \ell_{3D}</math>
+   <td align="left">
-  <td align="center">
+<math>~ \frac{1}{p^4 z^4 \lambda_2^2}
-<math>~-(x^2 + q^4y^2)\ell_q \ell_{3D} = - \ell_{3D}/\ell_q</math>
+\biggl[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 \biggr]
+</math>
    </td>
 </tr>
 <tr>
-   <td align="center"><math>~3</math></td>
+   <td align="right">
-  <td align="center">
+<math>~\Rightarrow ~~~ h_2</math>
-<math>~-q^2 y \ell_q</math>
    </td>
    <td align="center">
-<math>~x \ell_q</math>
+<math>~=</math>
    </td>
-   <td align="center">
+   <td align="left">
-<math>~0</math>
+<math>~
+\frac{p^2 z^2 \lambda_2}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \, .
+</math>
    </td>
 </tr>
 </table>
+So, the associated unit vector is,
-====Analysis====
+<table border="0" cellpadding="5" align="center">
-Let's break down each direction cosine into its components.
-<table border="1" cellpadding="8" align="center">
-<tr>
-  <td align="center" colspan="9">'''Direction Cosine Components for T6 Coordinates'''</td>
-</tr>
-<tr>
-  <td align="center"><math>~n</math></td>
-  <td align="center"><math>~\lambda_n</math></td>
-  <td align="center"><math>~h_n</math></td>
-  <td align="center"><math>~\frac{\partial \lambda_n}{\partial x}</math></td>
-  <td align="center"><math>~\frac{\partial \lambda_n}{\partial y}</math></td>
-  <td align="center"><math>~\frac{\partial \lambda_n}{\partial z}</math></td>
-  <td align="center"><math>~\gamma_{n1}</math></td>
-  <td align="center"><math>~\gamma_{n2}</math></td>
-  <td align="center"><math>~\gamma_{n3}</math></td>
-</tr>
-<tr>
-  <td align="center"><math>~1</math></td>
-  <td align="center"><math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math></td>
-  <td align="center"><math>~\lambda_1 \ell_{3D}</math></td>
-  <td align="center"><math>~\frac{x}{\lambda_1}</math></td>
-  <td align="center"><math>~\frac{q^2 y}{\lambda_1}</math></td>
-  <td align="center"><math>~\frac{p^2 z}{\lambda_1}</math></td>
-  <td align="center"><math>~(x) \ell_{3D}</math></td>
-  <td align="center"><math>~(q^2 y)\ell_{3D}</math></td>
-  <td align="center"><math>~(p^2z) \ell_{3D}</math></td>
-</tr>
-<tr>
-  <td align="center"><math>~2</math></td>
-  <td align="center">---</td>
-  <td align="center">---</td>
-  <td align="center">---</td>
-  <td align="center">---</td>
-  <td align="center">---</td>
-  <td align="center"><math>~\ell_q \ell_{3D} (xp^2z)</math></td>
-  <td align="center"><math>~\ell_q \ell_{3D} (q^2 y p^2z) </math></td>
-  <td align="center"><math>~- (x^2 + q^4y^2)\ell_q \ell_{3D}</math></td>
-</tr>
-<tr>
-  <td align="center"><math>~3</math></td>
-  <td align="center"><math>~\tan^{-1}\biggl( \frac{y^{1/q^2}}{x} \biggr)</math></td>
-  <td align="center"><math>~\frac{xq^2 y \ell_q}{\sin\lambda_3 \cos\lambda_3}</math></td>
-  <td align="center"><math>~-\frac{\sin\lambda_3 \cos\lambda_3}{x}</math></td>
-  <td align="center"><math>~+\frac{\sin\lambda_3 \cos\lambda_3}{q^2y}</math></td>
-  <td align="center"><math>~0</math></td>
-  <td align="center"><math>~-q^2 y \ell_q</math></td>
-  <td align="center"><math>~x\ell_q</math></td>
-  <td align="center"><math>~0</math></td>
-</tr>
-</table>
-Try,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\lambda_2</math>
+<math>~\hat{e}_2</math>
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
 <math>~
-\tan^{-1} w \, ,
+\hat{\imath} h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)
++ \hat{\jmath} h_2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)
++ \hat{k} h_2 \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)
 </math>
-&nbsp; &nbsp; &nbsp; where,
    </td>
 </tr>
@@ Line 1,513: / Line 1,307: @@
 <tr>
    <td align="right">
-<math>~w</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>\frac{(x^2 + q^2y^2)^{1 / 2}}{z^{1/p^2}}
+<math>~
-~~~\Rightarrow~~~\frac{1}{z^{1 / p^2} } = \frac{w}{(x^2 + q^2 y^2)^{1 / 2}}
+\hat{\imath}  \biggl\{ \frac{x}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\}
-\, .</math>
++ \hat{\jmath}  \biggl\{ \frac{q^2 y}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\}
++ \hat{k} \biggl\{ \frac{p^2z(\mathfrak{f}-\lambda_2^2)}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\} \, .
+</math>
    </td>
-</tr>
 </table>
-The relevant partial derivatives are,
+Checking orthogonality &hellip;
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial x}</math>
+<math>~\hat{e}_1 \cdot \hat{e}_2</math>
    </td>
    <td align="center">
@@ Line 1,538: / Line 1,333: @@
    <td align="left">
 <math>~
-\frac{1}{1 + w^2} \biggl[ \frac{x}{(x^2 + q^2y^2)^{1 / 2}~z^{1/p^2}} \biggr]
+x\ell_{3D} \biggl\{ \frac{x}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\}
-=
++ q^2 y\ell_{3D} \biggl\{ \frac{q^2 y}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\}
-\frac{w}{1 + w^2} \biggl[ \frac{x}{(x^2 + q^2y^2)} \biggr]
++ p^2 z \ell_{3D} \biggl\{ \frac{p^2z(\mathfrak{f}-\lambda_2^2)}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\}
-\, ,
 </math>
    </td>
@@ Line 1,548: / Line 1,342: @@
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial y}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 1,554: / Line 1,348: @@
    </td>
    <td align="left">
-<math>~
+<math>~\frac{\ell_{3D}}{ [ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} }
-\frac{1}{1 + w^2} \biggl[ \frac{q^2y}{(x^2 + q^2y^2)^{1 / 2}~z^{1/p^2}} \biggr]
+\biggl\{ x^2 + q^4y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)\biggr\} \, .
-=
-\frac{w}{1 + w^2} \biggl[ \frac{q^2y}{(x^2 + q^2y^2)} \biggr]
-\, ,
 </math>
    </td>
@@ Line 1,565: / Line 1,356: @@
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial z}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 1,571: / Line 1,362: @@
    </td>
    <td align="left">
-<math>~
+<math>~\frac{\ell_{3D}}{ [ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} }
-\frac{w}{1 + w^2} \biggl[- \frac{1}{p^2 z} \biggr]
+\biggl\{ x^2 + q^4y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)\biggr\} \, .
-\, ,
 </math>
    </td>
@@ Line 1,579: / Line 1,369: @@
 </table>
-which means that,
+If <math>~\mathfrak{f} = 0</math>, we have &hellip;
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~h_2^{-2}</math>
+<math>~p^2 z (\mathfrak{f} - \lambda_2^2) </math>
    </td>
    <td align="center">
-<math>~=</math>
+&nbsp; &nbsp; <math>~~~\rightarrow ~~~ </math>&nbsp; &nbsp;
    </td>
    <td align="left">
 <math>~
-\biggl( \frac{\partial \lambda_2}{\partial x} \biggr)^2
+\biggl[- p^2 z \lambda_2^2\biggr]_{\mathfrak{f} = 0}
-+
+=
-\biggl( \frac{\partial \lambda_2}{\partial y} \biggr)^2
+- p^2 z\biggl[\frac{(x^2 + q^2y^2 + \cancelto{0}{\mathfrak{f} \cdot }p^2 z^2 )^{1 / 2}}{pz}  \biggr]^2
-+
+=
-\biggl( \frac{\partial \lambda_2}{\partial z} \biggr)^2
+- \frac{(x^2 + q^2y^2  )}{z} \, ,</math>
-</math>
    </td>
 </tr>
+</table>
+which, in turn, means &hellip;
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-&nbsp;
+<math>~[ x^2 + q^4 y^2 + p^4 z^2 (\cancelto{0}{\mathfrak{f}} - \lambda_2^2)^2 ]^{1 / 2}</math>
    </td>
    <td align="center">
@@ Line 1,609: / Line 1,401: @@
    <td align="left">
 <math>~
-\biggl[ \biggl( \frac{w}{1 + w^2}\biggr)^2 \frac{x^2}{(x^2 + q^2y^2)^2}  \biggr]
+[ x^2 + q^4 y^2 + p^4 z^2 \lambda_2^4 ]^{1 / 2}
-+
-\biggl[ \biggl( \frac{w}{1 + w^2}\biggr)^2 \frac{q^4 y^2}{(x^2 + q^2y^2)^2}  \biggr]
-+
-\biggl[ \biggl( \frac{w}{1 + w^2}\biggr)^2 \frac{1}{p^4 z^2} \biggr]
 </math>
    </td>
@@ Line 1,626: / Line 1,414: @@
    </td>
    <td align="left">
-<math>~\biggl( \frac{w}{1 + w^2}\biggr)^2
+<math>~
-\biggl[ \frac{(x^2 + q^4y^2)(p^4 z^2) + (x^2 + q^2y^2)^2}{(x^2 + q^2y^2)^2~p^4 z^2}  \biggr]
+\biggl\{ x^2 + q^4 y^2 + p^4 z^2 \biggl[\frac{(x^2 + q^2y^2 + \cancelto{0}{\mathfrak{f}\cdot} p^2 z^2)^{1 / 2}}{pz}  \biggr]^4 \biggr\}^{1 / 2}
 </math>
    </td>
@@ Line 1,634: / Line 1,422: @@
 <tr>
    <td align="right">
-<math>~\Rightarrow~~~ h_2</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 1,641: / Line 1,429: @@
    <td align="left">
 <math>~
-\biggl( \frac{1 + w^2}{w}\biggr) \biggl\{ \frac{(x^2 + q^2y^2)~p^2 z}{ \mathcal{D}}  \biggr\}
+\biggl\{ x^2 + q^4 y^2 + \biggl[\frac{(x^2 + q^2y^2 )^{2}}{z^2}  \biggr] \biggr\}^{1 / 2}
-\, ,
 </math>
    </td>
 </tr>
-</table>
-where,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\mathcal{D}</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~[(x^2 + q^4y^2)(p^4 z^2) + (x^2 + q^2y^2)^2]^{1 / 2} \, .</math>
+<math>~
+(x^2 + q^4 y^2)^{1 / 2} \biggl[ 1 + \frac{(x^2 + q^2y^2 )}{z^2} \biggr]^{1 / 2}
+</math>
    </td>
 </tr>
-</table>
-Hence, the trio of associated direction cosines are,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\gamma_{21} = h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 1,674: / Line 1,457: @@
    <td align="left">
 <math>~
-\biggl( \frac{1 + w^2}{w}\biggr) \biggl\{ \frac{(x^2 + q^2y^2)~p^2 z}{ \mathcal{D}}  \biggr\}\frac{w}{1 + w^2} \biggl[ \frac{x}{(x^2 + q^2y^2)} \biggr]
+\frac{(x^2 + q^4 y^2)^{1 / 2}}{z} \biggl[ z^2 + (x^2 + q^2y^2 ) \biggr]^{1 / 2} \, ,
-=
-\biggl\{ \frac{x~p^2 z}{ \mathcal{D}}  \biggr\} \, ,
 </math>
    </td>
 </tr>
+</table>
+and,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\gamma_{22} = h_2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)</math>
+<math>~\hat{e}_1 \cdot \hat{e}_2</math>
    </td>
    <td align="center">
@@ Line 1,689: / Line 1,473: @@
    </td>
    <td align="left">
-<math>~
+<math>~\frac{\ell_{3D}}{ [ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} }
-\biggl( \frac{1 + w^2}{w}\biggr) \biggl\{ \frac{(x^2 + q^2y^2)~p^2 z}{ \mathcal{D}}  \biggr\} \frac{w}{1 + w^2} \biggl[ \frac{q^2y}{(x^2 + q^2y^2)} \biggr]
+\biggl\{ x^2 + q^4y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)\biggr\} \, .
-=
-\biggl\{ \frac{q^2 y~p^2 z}{ \mathcal{D}}  \biggr\} \, ,
 </math>
    </td>
 </tr>
+</table>
+===Speculation6===
+====Determine &lambda;<sub>2</sub>====
+This is very similar to the [[#Speculation2|above, Speculation2]].
+Try,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\gamma_{23} = h_2 \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)</math>
+<math>~\lambda_2</math>
    </td>
    <td align="center">
@@ Line 1,706: / Line 1,496: @@
    <td align="left">
 <math>~
-\biggl( \frac{1 + w^2}{w}\biggr) \biggl\{ \frac{(x^2 + q^2y^2)~p^2 z}{ \mathcal{D}}  \biggr\}\frac{w}{1 + w^2} \biggl[- \frac{1}{p^2 z} \biggr]
+\frac{x y^{1/q^2}}{ z^{2/p^2}} \, ,
-=
-\biggl\{- \frac{(x^2 + q^2y^2)}{ \mathcal{D}}  \biggr\} \, .
 </math>
    </td>
 </tr>
 </table>
+in which case,
-<font color="red">'''VERY close!'''</font>
-Let's examine the function, <math>~\mathcal{D}^2</math>.
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{1}{\ell_{3D}^2 \ell_d^2}</math>
+<math>~\frac{\partial \lambda_2}{\partial x}</math>
    </td>
    <td align="center">
@@ Line 1,728: / Line 1,513: @@
    <td align="left">
 <math>~
-(x^2 + q^4 y^2)(x^2 + q^4 y + p^4 z)
+\frac{\lambda_2}{x} \, ,
-=
-(x^2 + q^4 y^2)p^4 z + (x^2 + q^4 y^2)^2 \, .
 </math>
    </td>
 </tr>
-</table>
-===Eureka (NOT!)===
-Try,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\lambda_2</math>
+<math>~\frac{\partial \lambda_2}{\partial y}</math>
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
 <math>~
-\tan^{-1} w \, ,
+\frac{x}{z^{2/p^2}} \biggl(\frac{1}{q^2}\biggr) y^{1/q^2 - 1}
+=
+\frac{\lambda_2}{q^2 y}
+\, ,
 </math>
-&nbsp; &nbsp; &nbsp; where,
    </td>
 </tr>
@@ Line 1,759: / Line 1,537: @@
 <tr>
    <td align="right">
-<math>~w</math>
+<math>~\frac{\partial \lambda_2}{\partial z}</math>
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>\frac{(x^2 + q^2y^2)^{1 / 2}}{p^2 z}
+<math>~
-~~~\Rightarrow~~~\frac{1}{p^2 z } = \frac{w}{(x^2 + q^2 y^2)^{1 / 2}}
+-\frac{2\lambda_2}{p^2 z}
-\, .</math>
+\, .
+</math>
    </td>
 </tr>
 </table>
+The associated scale factor is, then,
-The relevant partial derivatives are,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial x}</math>
+<math>~h_2</math>
    </td>
    <td align="center">
@@ Line 1,783: / Line 1,561: @@
    </td>
    <td align="left">
-<math>~
+<math>~\biggl[
-\frac{1}{1 + w^2} \biggl[ \frac{x}{(x^2 + q^2y^2)^{1 / 2}~p^2 z} \biggr]
+\biggl( \frac{\partial \lambda_2}{\partial x} \biggr)^2
-=
++
-\frac{w}{1 + w^2} \biggl[ \frac{x}{(x^2 + q^2y^2)} \biggr]
+\biggl( \frac{\partial \lambda_2}{\partial y} \biggr)^2
-\, ,
++
+\biggl( \frac{\partial \lambda_2}{\partial z} \biggr)^2
+\biggr]^{-1 / 2}
 </math>
    </td>
@@ Line 1,794: / Line 1,574: @@
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial y}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 1,800: / Line 1,580: @@
    </td>
    <td align="left">
-<math>~
+<math>~\biggl[
-\frac{1}{1 + w^2} \biggl[ \frac{q^2y}{(x^2 + q^2y^2)^{1 / 2}~p^2z} \biggr]
+\biggl( \frac{ \lambda_2}{x} \biggr)^2
-=
++
-\frac{w}{1 + w^2} \biggl[ \frac{q^2y}{(x^2 + q^2y^2)} \biggr]
+\biggl( \frac{\lambda_2}{q^2y} \biggr)^2
-\, ,
++
+\biggl( - \frac{2\lambda_2}{p^2z} \biggr)^2
+\biggr]^{-1 / 2}
 </math>
    </td>
@@ Line 1,811: / Line 1,593: @@
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial z}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 1,817: / Line 1,599: @@
    </td>
    <td align="left">
-<math>~
+<math>~\frac{1}{\lambda_2}\biggl[
-\frac{1}{1 + w^2} \biggl[- \frac{(x^2 + q^2y^2)^{1 / 2}}{~p^2z^2} \biggr]
+\frac{ 1}{x^2}
-=
++
-\frac{w}{1 + w^2} \biggl[- \frac{1}{z} \biggr]
+\frac{1}{q^4y^2}
-\, ,
++
+\frac{4}{p^4z^2}
+\biggr]^{-1 / 2}
 </math>
    </td>
 </tr>
-</table>
-which means that,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~h_2^{-2}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 1,838: / Line 1,618: @@
    </td>
    <td align="left">
-<math>~
+<math>~\frac{1}{\lambda_2}\biggl[
-\biggl( \frac{\partial \lambda_2}{\partial x} \biggr)^2
+\frac{ (q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2}{x^2 q^4 y^2 p^4 z^2}
-+
+\biggr]^{-1 / 2}
-\biggl( \frac{\partial \lambda_2}{\partial y} \biggr)^2
-+
-\biggl( \frac{\partial \lambda_2}{\partial z} \biggr)^2
 </math>
    </td>
@@ Line 1,856: / Line 1,633: @@
    </td>
    <td align="left">
-<math>~\biggl[ \frac{w}{1 + w^2} \biggr]^2 \biggl\{
+<math>~
-\biggl[  \frac{x}{(x^2 + q^2y^2)} \biggr]^2
+\frac{1}{\lambda_2}\biggl[
-+
+\frac{x q^2 y p^2 z}{ \mathcal{D}}
-\biggl[ \frac{q^2y}{(x^2 + q^2y^2)}  \biggr]^2
+\biggr] \, .
-+
-\biggl[ - \frac{1}{z} \biggr]^2
-\biggr\}
 </math>
    </td>
 </tr>
+</table>
+where,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-&nbsp;
+<math>~\mathcal{D}</math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
-<math>~\biggl[ \frac{w}{1 + w^2} \biggr]^2 \biggl\{
+<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>
-\frac{x^2 + q^4y^2}{(x^2 + q^2y^2)^2}
-+
-\frac{1}{z^2}
-\biggr\}
-</math>
    </td>
 </tr>
 </table>
-===Speculation5===
+The associated unit vector is, then,
-====Spherical Coordinates====
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~r\cos\theta</math>
+<math>~\hat{e}_2</math>
    </td>
    <td align="center">
@@ Line 1,899: / Line 1,668: @@
    </td>
    <td align="left">
-<math>~z \, ,</math>
+<math>~
+\hat{\imath} h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)
++ \hat{\jmath} h_2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)
++ \hat{k} h_2 \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)
+</math>
    </td>
 </tr>
@@ Line 1,905: / Line 1,678: @@
 <tr>
    <td align="right">
-<math>~r\sin\theta</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 1,911: / Line 1,684: @@
    </td>
    <td align="left">
-<math>~(x^2 + y^2)^{1 / 2} \, ,</math>
+<math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl\{
+\hat{\imath} \biggl( \frac{1}{x} \biggr)
++ \hat{\jmath}  \biggl( \frac{1}{q^2 y} \biggr)
++ \hat{k} \biggl( -\frac{2}{p^2 z} \biggr)
+\biggr\} \ .
+</math>
    </td>
 </tr>
+</table>
+Recalling that the unit vector associated with the "first" coordinate is,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\tan\varphi</math>
+<math>~\hat{e}_1 </math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
-<math>~\frac{y}{x} \, .</math>
+<math>~
+\hat\imath (x \ell_{3D}) + \hat\jmath (q^2y \ell_{3D}) + \hat{k} (p^2 z \ell_{3D}) \, ,
+</math>
    </td>
 </tr>
 </table>
+where,
-====Use &lambda;<sub>1</sub> Instead of r ====
-Here, as above, we define,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\lambda_1^2</math>
+<math>~\ell_{3D}</math>
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~x^2 + q^2 y^2 + p^2 z^2 </math>
+<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} \, ,</math>
    </td>
 </tr>
 </table>
+let's check to see whether the "second" unit vector is orthogonal to the "first."
-Using this expression to eliminate "x" (in favor of &lambda;<sub>1</sub>) in each of the three spherical-coordinate definitions, we obtain,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~r^2 \equiv x^2 + y^2 + z^2</math>
+<math>~\hat{e}_1 \cdot \hat{e}_2</math>
    </td>
    <td align="center">
@@ Line 1,958: / Line 1,737: @@
    </td>
    <td align="left">
-<math>~\lambda_1^2 - y^2(q^2-1) - z^2(p^2-1) \, ;</math>
+<math>~
+\frac{(x q^2 y p^2 z) \ell_{3D}}{\mathcal{D}} \biggl[
++ 1 - 2
+\biggr] = 0 \, .
+</math>
    </td>
 </tr>
+</table>
+<font color="red">'''Hooray!'''</font>
+====Direction Cosines for <i>Third</i> Unit Vector====
+Now, what is the unit vector, <math>~\hat{e}_3</math>, that is simultaneously orthogonal to both these "first" and the "second" unit vectors?
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\tan^2\theta \equiv \frac{x^2 + y^2}{z^2}</math>
+<math>~\hat{e}_3 \equiv \hat{e}_1 \times \hat{e}_2</math>
    </td>
    <td align="center">
@@ Line 1,970: / Line 1,761: @@
    </td>
    <td align="left">
-<math>~\frac{1}{z^2}\biggl[
+<math>~
-\lambda_1^2 -y^2(q^2-1) -p^2z^2
+\hat\imath \biggl[ ( e_{1y} )( e_{2z}) - ( e_{2y} )( e_{1z}) ) \biggl]
-\biggr] \, ;</math>
++ \hat\jmath \biggl[ ( e_{1z} )( e_{2x}) - ( e_{2z} )( e_{1x}) ) \biggl]
++ \hat{k} \biggl[ ( e_{1x} )( e_{2y}) - ( e_{2x} )( e_{1y}) ) \biggl]
+</math>
    </td>
 </tr>
@@ Line 1,978: / Line 1,771: @@
 <tr>
    <td align="right">
-<math>~\frac{1}{\tan^2\varphi} \equiv \frac{x^2}{y^2}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 1,984: / Line 1,777: @@
    </td>
    <td align="left">
-<math>~
+<math>~\frac{(x q^2 y p^2 z) \ell_{3D}}{\mathcal{D}}
-\frac{\lambda_1^2 - p^2z^2}{y^2} - q^2 \, .
+\biggl\{
+\hat\imath \biggl[ \biggl( -\frac{2 q^2y}{p^2 z} \biggr) - \biggl( \frac{p^2z}{q^2y} \biggr)  \biggl]
++ \hat\jmath \biggl[ \biggl( \frac{p^2z}{x} \biggr) - \biggl(-\frac{2x}{p^2z} \biggr)  \biggl]
++ \hat{k} \biggl[ \biggl( \frac{x}{q^2y} \biggr) - \biggl( \frac{q^2y}{x} \biggr)  \biggl]
+\biggr\}
 </math>
    </td>
 </tr>
-</table>
-After a bit of additional algebraic manipulation, we find that,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>\frac{z^2}{\lambda_1^2}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 2,002: / Line 1,795: @@
    </td>
    <td align="left">
-<math>~
+<math>~\frac{(x q^2 y p^2 z) \ell_{3D}}{\mathcal{D}}
-\frac{ (1+\tan^2\varphi)}{\mathcal{D}^2} \, ,
+\biggl\{
+-\hat\imath \biggl[ \frac{2 q^4y^2 + p^4z^2}{q^2 y p^2 z} \biggl]
++ \hat\jmath \biggl[ \frac{p^4z^2 + 2x^2}{xp^2 z}  \biggl]
++ \hat{k} \biggl[ \frac{x^2 - q^4y^2}{x q^2y} \biggl]
+\biggr\}
 </math>
    </td>
@@ Line 2,010: / Line 1,807: @@
 <tr>
    <td align="right">
-<math>~\frac{y^2}{\lambda_1^2}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 2,016: / Line 1,813: @@
    </td>
    <td align="left">
-<math>~
+<math>~\frac{\ell_{3D}}{\mathcal{D}}
-\biggl[\frac{ \mathcal{D}^2 \tan^2\varphi -  p^2 \tan^2\varphi (1+\tan^2\varphi)}{(1+q^2\tan^2\varphi) \mathcal{D}^2}  \biggr]
+\biggl\{
-\, ,
+-\hat\imath \biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]
++ \hat\jmath \biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]
++ \hat{k} \biggl[ p^2z( x^2 - q^4y^2 ) \biggl]
+\biggr\} \, .
 </math>
    </td>
 </tr>
+</table>
+Is this a valid unit vector?  First, note that &hellip;
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{x^2}{\lambda_1^2}</math>
+<math>~\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2}</math>
    </td>
    <td align="center">
@@ Line 2,032: / Line 1,836: @@
    <td align="left">
 <math>~
-- q^2\biggl(\frac{y^2}{\lambda_1^2} \biggr) - p^2\biggl(\frac{z^2}{\lambda_1^2}\biggr)
+(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
-\, ,
+(x^2 + q^4y^2 + p^4 z^2 )
 </math>
    </td>
 </tr>
-</table>
-where,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\mathcal{D}^2</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
 <math>~
-\biggl[
+(x^2 q^4 y^2 p^4 z^2 + x^4 p^4 z^2 + 4x^4q^4y^2)
-(1 + q^2\tan^2\varphi)(p^2 + \tan^2\theta) - p^2(q^2-1)\tan^2\varphi
++ (q^8 y^4 p^4 z^2 + x^2 q^4y^2 p^4 z^2 + 4x^2q^8y^4)
-\biggr] \, .
++(q^4 y^2 p^8 z^4 + x^2 p^8 z^4 + 4x^2q^4y^2 p^4 z^2)
 </math>
    </td>
 </tr>
-</table>
-<table border="1" cellpadding="8" align="center" width="80%"><tr><td align="left">
-As a check, let's set <math>~q^2 = p^2 = 1</math>, which should reduce to the normal spherical coordinate system.
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\lambda_1^2</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\rightarrow</math>
+<math>~=</math>
    </td>
    <td align="left">
 <math>~
-r^2 \, ,
+x^2 q^4 y^2 p^4 z^2 + x^4(p^4 z^2 + 4 q^4y^2)
-</math>
++ q^8 y^4(p^4 z^2  + 4x^2)
-  </td>
++p^8z^4(x^2 + q^4 y^2 )\, .
-<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
-  <td align="right">
-<math>~\mathcal{D}^2</math>
-  </td>
-  <td align="center">
-<math>~\rightarrow</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl[
-(1 + \tan^2\varphi)(1 + \tan^2\theta)
-\biggr] \, .
 </math>
    </td>
 </tr>
 </table>
+<span id="Eureka">Then we have,</span>
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\Rightarrow ~~~ \frac{z^2}{\lambda_1^2}</math>
+<math>~\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2}\hat{e}_3 \cdot \hat{e}_3</math>
    </td>
    <td align="center">
-<math>~\rightarrow</math>
+<math>~=</math>
    </td>
    <td align="left">
 <math>~
-\frac{1}{1+\tan^2\theta} = \cos^2\theta = \frac{z^2}{r^2} \, ;
+\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]^2
++
+\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]^2
++
+\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]^2
 </math>
    </td>
@@ Line 2,109: / Line 1,898: @@
 <tr>
    <td align="right">
-<math>~\frac{y^2}{\lambda_1^2}</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\rightarrow</math>
+<math>~=</math>
    </td>
    <td align="left">
 <math>~
-\biggl[\frac{(1 + \tan^2\varphi)(1 + \tan^2\theta) \tan^2\varphi -  \tan^2\varphi (1+\tan^2\varphi)}{(1+\tan^2\varphi) (1 + \tan^2\varphi)(1 + \tan^2\theta)}  \biggr]
+x^2(4 q^8y^4 + 4q^4y^2p^4z^2 + p^8z^4 )
++
+q^4 y^2(p^8z^4 + 4x^2p^4z^2 + 4x^4 )
++
+p^4z^2( x^4 - 2x^2q^4 y^2 + q^8y^4 )
 </math>
    </td>
@@ Line 2,129: / Line 1,922: @@
    </td>
    <td align="left">
-<math>~\frac{\tan^2\varphi}{(1 + \tan^2\varphi)}
+<math>~
-\biggl[\frac{\tan^2\theta }{ (1 + \tan^2\theta)}  \biggr]
+x^2 q^8y^4 + 4x^2 q^4y^2p^4z^2 + x^2 p^8z^4
-= \sin^2\theta \sin^2\varphi
++
-= \frac{y^2}{r^2} \,;
+q^4 y^2p^8z^4 + 4x^2q^4 y^2p^4z^2 + 4x^4q^4 y^2
++
+x^4p^4z^2 - 2x^2q^4 y^2p^4z^2 + q^8y^4p^4z^2
 </math>
    </td>
@@ Line 2,139: / Line 1,934: @@
 <tr>
    <td align="right">
-<math>~\frac{x^2}{\lambda_1^2}</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\rightarrow</math>
+<math>~=</math>
    </td>
    <td align="left">
 <math>~
-- \biggl(\frac{y^2}{\lambda_1^2} \biggr) - \biggl(\frac{z^2}{\lambda_1^2}\biggr)
+x^2 q^4y^2p^4z^2
-\, ,
++  p^8z^4 (x^2  +q^4 y^2)
++  x^4(4q^4 y^2 + p^4z^2)
++ q^8 y^4(4 x^2  + p^4z^2 )
 </math>
    </td>
@@ Line 2,157: / Line 1,954: @@
    </td>
    <td align="center">
-<math>~\rightarrow</math>
+<math>~=</math>
    </td>
    <td align="left">
 <math>~
-- \sin^2\theta \sin^2\varphi - \cos^2\theta
+\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2} \, ,
-=
-- \sin^2\theta \sin^2\varphi + \sin^2\theta
-=
-\sin^2\theta \cos^2\varphi = \frac{x^2}{r^2}
-\, .
 </math>
    </td>
 </tr>
 </table>
-</td></tr></table>
+which means that, <math>~\hat{e}_3\cdot \hat{e}_3 = 1</math>.  &nbsp; &nbsp;<font color="red">'''Hooray! Again (11/11/2020)!'''</font>
-====Relationship To T3 Coordinates====
+<table border="1" cellpadding="8" align="center">
+<tr>
-If we set, <math>~q = 1</math>, but continue to assume that <math>~p > 1</math>, we expect to see a representation that resembles our previously discussed, [[User:Tohline/Appendix/Ramblings/T3Integrals#Integrals_of_Motion_in_T3_Coordinates|T3 Coordinates]].  Note, for example, that the new "radial" coordinate is,
+  <td align="center" colspan="9">'''Direction Cosine Components for T6 Coordinates'''</td>
-<table border="0" cellpadding="5" align="center">
+</tr>
+<tr>
+  <td align="center"><math>~n</math></td>
+  <td align="center"><math>~\lambda_n</math></td>
+  <td align="center"><math>~h_n</math></td>
+  <td align="center"><math>~\frac{\partial \lambda_n}{\partial x}</math></td>
+  <td align="center"><math>~\frac{\partial \lambda_n}{\partial y}</math></td>
+  <td align="center"><math>~\frac{\partial \lambda_n}{\partial z}</math></td>
+  <td align="center"><math>~\gamma_{n1}</math></td>
+  <td align="center"><math>~\gamma_{n2}</math></td>
+  <td align="center"><math>~\gamma_{n3}</math></td>
+</tr>
 <tr>
-   <td align="right">
+   <td align="center"><math>~1</math></td>
-<math>~\lambda_1^2</math>
+  <td align="center"><math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math></td>
+  <td align="center"><math>~\lambda_1 \ell_{3D}</math></td>
+  <td align="center"><math>~\frac{x}{\lambda_1}</math></td>
+  <td align="center"><math>~\frac{q^2 y}{\lambda_1}</math></td>
+  <td align="center"><math>~\frac{p^2 z}{\lambda_1}</math></td>
+  <td align="center"><math>~(x) \ell_{3D}</math></td>
+  <td align="center"><math>~(q^2 y)\ell_{3D}</math></td>
+  <td align="center"><math>~(p^2z) \ell_{3D}</math></td>
+</tr>
+<tr>
+  <td align="center"><math>~2</math></td>
+  <td align="center"><math>~\frac{x y^{1/q^2}}{ z^{2/p^2}}</math></td>
+  <td align="center"><math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y p^2 z}{ \mathcal{D}}\biggr] </math></td>
+  <td align="center"><math>~\frac{\lambda_2}{x}</math></td>
+  <td align="center"><math>~\frac{\lambda_2}{q^2 y}</math></td>
+  <td align="center"><math>~-\frac{2\lambda_2}{p^2 z}</math></td>
+  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{x}\biggr)</math></td>
+  <td align="center"><math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{q^2y}\biggr)</math></td>
+  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(-\frac{2}{p^2z}\biggr)</math></td>
+</tr>
+<tr>
+  <td align="center"><math>~3</math></td>
+  <td align="center">---</td>
+  <td align="center">---</td>
+  <td align="center">---</td>
+  <td align="center">---</td>
+  <td align="center">---</td>
+  <td align="center"><math>~-\frac{\ell_{3D}}{\mathcal{D}}\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]</math></td>
+  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]</math></td>
+  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]</math></td>
+</tr>
+<tr>
+  <td align="left" colspan="9">
+<table border="0" cellpadding="8" align="center">
+<tr>
+  <td align="right">
+<math>~\ell_{3D}</math>
    </td>
    <td align="center">
-<math>~\rightarrow</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
-<math>~
+<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} \, ,</math>
-(\varpi^2 + p^2z^2) \, ,
-</math>
    </td>
-<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
+</tr>
+<tr>
    <td align="right">
-<math>~\mathcal{D}^2</math>
+<math>~\mathcal{D}</math>
    </td>
    <td align="center">
-<math>~\rightarrow</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
-<math>~
+<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>
-\biggl[
+  </td>
-(1 + \tan^2\varphi)(p^2 + \tan^2\theta)
+</tr>
-\biggr]  \, .
-</math>
+</table>
    </td>
 </tr>
 </table>
+Let's double-check whether this "third" unit vector is orthogonal to both the "first" and the "second" unit vectors.
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\Rightarrow ~~~ \frac{p^2z^2}{\lambda_1^2}</math>
+<math>~\hat{e}_1 \cdot \hat{e}_3</math>
    </td>
    <td align="center">
-<math>~\rightarrow</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~\frac{ p^2}{(p^2 + \tan^2\theta)} = \frac{ 1}{(1 + p^{-2} \tan^2\theta)}\, ,</math>
+<math>~\frac{\ell_{3D}^2}{\mathcal{D}}
+\biggl\{
+-x \biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]
++ q^2 y  \biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]
++ p^2 z  \biggl[ p^2z( x^2 - q^4y^2 ) \biggl]
+\biggr\}
+</math>
    </td>
 </tr>
@@ Line 2,222: / Line 2,075: @@
 <tr>
    <td align="right">
-<math>~\frac{\varpi^2}{\lambda_1^2} = \frac{x^2}{\lambda_1^2} + \frac{y^2}{\lambda_1^2}</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\rightarrow</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~1 -  p^2 \biggl( \frac{z^2}{\lambda_1^2}\biggr)
+<math>~\frac{\ell_{3D}^2}{\mathcal{D}}
-=
+\biggl\{
-\biggl[1 -  \frac{ 1}{(1 + p^{-2}\tan^2\theta)} \biggr] \, .</math>
+- (2 x^2q^4y^2 + x^2p^4z^2 )
++ (q^4 y^2 p^4z^2 + 2x^2 q^4 y^2)
++ ( x^2p^4z^2  - q^4y^2 p^4z^2  )
+\biggr\}
+</math>
    </td>
 </tr>
-</table>
-We also see that,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{\varpi^2}{p^2z^2}</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\rightarrow</math>
+<math>~=</math>
    </td>
    <td align="left">
 <math>~
-(1 + p^{-2}\tan^2\theta)\biggl[1 -  \frac{ 1}{(1 + p^{-2}\tan^2\theta)} \biggr]
+\, ,
-=
-p^{-2}\tan^2\theta \, .
 </math>
    </td>
@@ Line 2,255: / Line 2,106: @@
 </table>
+and,
-====Again Consider Full 3D Ellipsoid====
-Let's try to replace everywhere, <math>~[\varpi/(pz)]^2 = p^{-2}\tan^2\theta</math> with <math>~\lambda_2</math>.  This gives,
 <table border="0" cellpadding="5" align="center">
@@ Line 2,263: / Line 2,112: @@
 <tr>
    <td align="right">
-<math>~\frac{\mathcal{D}^2}{p^2}</math>
+<math>~\hat{e}_2 \cdot \hat{e}_3</math>
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~
+<math>~\frac{\ell_{3D}}{\mathcal{D}} \cdot \frac{x q^2 y p^2 z}{\mathcal{D}}
-\biggl[
+\biggl\{
-(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi
+- \biggl[ (2 q^4y^2 + p^4z^2 ) \biggl]
-\biggr] \, .
++  \biggl[ (p^4z^2 + 2x^2 )  \biggl]
+- \biggl[ 2( x^2 - q^4y^2 ) \biggl]
+\biggr\}
 </math>
    </td>
 </tr>
-</table>
-which means that,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>\frac{p^2 z^2}{\lambda_1^2}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 2,290: / Line 2,137: @@
    <td align="left">
 <math>~
-\frac{ (1+\tan^2\varphi)}{[(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi]}
+\, .
 </math>
    </td>
 </tr>
+</table>
+<font color="red">'''Q. E. D.'''</font>
+<!-- TEST -->
+====Search for <i>Third</i> Coordinate Expression====
+Let's try &hellip;
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-&nbsp;
+<math>~\lambda_3</math>
    </td>
    <td align="center">
@@ Line 2,303: / Line 2,158: @@
    </td>
    <td align="left">
-<math>~
+<math>~\mathcal{D}^n \ell_{3D}^m </math>
-\frac{ (1+\tan^2\varphi)/\tan^2\varphi}{[q^2 \lambda_2 (1 + q^2\tan^2\varphi)/(q^2\tan^2\varphi) - (q^2-1)]}
-= \frac{1/\sin^2\varphi}{[q^2\lambda_2 Q^2 - (q^2-1) ]}
-\, ,
-</math>
    </td>
 </tr>
@@ Line 2,313: / Line 2,164: @@
 <tr>
    <td align="right">
-<math>~\frac{q^2y^2}{\lambda_1^2}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 2,320: / Line 2,171: @@
    <td align="left">
 <math>~
-\frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) }
+(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{n / 2} (x^2 + q^4y^2 + p^4 z^2 )^{- m / 2}
--
-\frac{ q^2 \tan^2\varphi (1+\tan^2\varphi)}{(1+q^2\tan^2\varphi) [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] }
 </math>
    </td>
@@ Line 2,329: / Line 2,178: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~\Rightarrow ~~~ \frac{\partial \lambda_3}{\partial x_i}</math>
    </td>
    <td align="center">
@@ Line 2,336: / Line 2,185: @@
    <td align="left">
 <math>~
-\frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) } \biggl\{1
+\ell_{3D}^m \biggl[ \frac{n}{2} \biggl(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 \biggr)^{n / 2 - 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)\biggr]
--
-\frac{ (1+\tan^2\varphi)}{ [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] }
-\biggr\}
 </math>
    </td>
@@ Line 2,349: / Line 2,195: @@
    </td>
    <td align="center">
-<math>~=</math>
+&nbsp;
    </td>
    <td align="left">
 <math>~
-\frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) } \biggl\{1
++ \mathcal{D}^n \biggl[ - \frac{m}{2} (x^2 + q^4y^2 + p^4 z^2 )^{- m / 2 - 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(x^2 + q^4y^2 + p^4 z^2 )\biggr]
--
-\frac{ (1+\tan^2\varphi)/\tan^2\varphi}{ [q^2\lambda_2(1 + q^2\tan^2\varphi)/(q^2\tan^2\varphi) - (q^2-1) ] }
-\biggr\}
 </math>
    </td>
@@ Line 2,370: / Line 2,213: @@
    <td align="left">
 <math>~
-\frac{ 1}{Q^2 } \biggl\{1
+\mathcal{D}^n \ell_{3D}^m \biggl[ \frac{n}{2} \biggl(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 \biggr)^{- 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)\biggr]
--
-\frac{ 1/\sin^2\varphi}{ [q^2\lambda_2 Q^2 - (q^2-1) ] }
-\biggr\}
-= \frac{1}{Q^2[~~]} \biggl[ [~~] - \frac{1}{\sin^2\varphi} \biggr]
-\, ,
 </math>
    </td>
@@ Line 2,382: / Line 2,220: @@
 <tr>
    <td align="right">
-<math>~\frac{x^2}{\lambda_1^2}</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~=</math>
+&nbsp;
    </td>
    <td align="left">
 <math>~
++ \mathcal{D}^n \ell_{3D}^m \biggl[ - \frac{m}{2} (x^2 + q^4y^2 + p^4 z^2 )^{- 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(x^2 + q^4y^2 + p^4 z^2 )\biggr] \, .
-- \frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) } \biggl\{1
--
-\frac{ (1+\tan^2\varphi)}{ [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] }
-\biggr\}
 </math>
    </td>
 </tr>
+</table>
+Hence,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-&nbsp;
+<math>~\frac{1}{\mathcal{D}^n \ell_{3D}^m} \cdot \frac{\partial \lambda_3}{\partial x}</math>
    </td>
    <td align="center">
-&nbsp;
+<math>~=</math>
    </td>
    <td align="left">
 <math>~
-- \frac{ (1+\tan^2\varphi)}{[(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi]}
+\frac{n}{2\mathcal{D}^2}\frac{\partial}{\partial x} \biggl[q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 \biggr]
+-
+\frac{m \ell_{3D}^2}{2} \frac{\partial}{\partial x} \biggl[ x^2 + q^4y^2 + p^4 z^2 \biggr]
 </math>
    </td>
@@ Line 2,420: / Line 2,261: @@
    </td>
    <td align="left">
-<math>~
+<math>~x \biggl\{
+\frac{n}{\mathcal{D}^2}\biggl[p^4 z^2 + 4q^4y^2 \biggr]
-- \frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) }
+-
-- \biggl\{\frac{ (1+\tan^2\varphi)}{ [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] }
+m \ell_{3D}^2
 \biggr\}
 </math>
    </td>
@@ Line 2,437: / Line 2,278: @@
    </td>
    <td align="left">
-<math>~
+<math>~x \biggl\{
+\frac{n (p^4 z^2 + 4q^4y^2)}{ ( q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 ) }
-- \frac{ 1 }{ Q^2 }
+-
-- \biggl\{\frac{ 1/\sin^2\varphi}{ [q^2\lambda_2 Q^2 - (q^2-1) ] }
+\frac{m}{ ( x^2 + q^4y^2 + p^4 z^2 ) }
-\biggr\}
-=
-\frac{1}{Q^2 [~~] } \biggl\{
-Q^2[~~] - [~~] - \frac{Q^2}{\sin^2\varphi}
 \biggr\}
-\, ,
 </math>
    </td>
 </tr>
+</table>
+This is overly cluttered!  Let's try, instead &hellip;
+<table border="1" cellpadding="8" align="center" width="80%"><tr><td align="left">
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{x^2 + q^2y^2}{\lambda_1^2}</math>
+<math>~A \equiv \ell_{3D}^{-2}</math>
    </td>
    <td align="center">
@@ Line 2,459: / Line 2,301: @@
    </td>
    <td align="left">
-<math>~1
+<math>~(x^2 + q^4y^2 + p^4 z^2 ) \, ,</math>
-- \biggl[1 + \frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) }\biggr] \biggl\{
-\frac{ (1+\tan^2\varphi)}{ [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] }
-\biggr\} \, .
-</math>
    </td>
-</tr>
+<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
-</table>
+   <td align="right">
+<math>~B \equiv \mathcal{D}^2</math>
-Now, notice that,
-<table border="0" cellpadding="5" align="center">
-<tr>
-   <td align="right">
-<math>~\frac{q^2 y^2 Q^2}{\lambda_1^2}</math>
    </td>
    <td align="center">
@@ Line 2,479: / Line 2,311: @@
    </td>
    <td align="left">
-<math>~1 - \frac{1}{[~~]\sin^2\varphi} \, ,</math>
+<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)</math>
    </td>
 </tr>
 </table>
-and,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{x^2}{\lambda_1^2} + \frac{1}{Q^2}</math>
+<math>~\Rightarrow ~~~ \frac{\partial A}{\partial x}</math>
    </td>
    <td align="center">
@@ Line 2,494: / Line 2,326: @@
    </td>
    <td align="left">
-<math>~1 - \frac{1}{[~~]\sin^2\varphi} \, .</math>
+<math>~2x \, ,</math>
    </td>
 </tr>
-</table>
-Hence,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{x^2}{\lambda_1^2} + \frac{1}{Q^2}</math>
+<math>~\frac{\partial A}{\partial y}</math>
    </td>
    <td align="center">
@@ Line 2,509: / Line 2,338: @@
    </td>
    <td align="left">
-<math>~\frac{q^2 y^2 Q^2}{\lambda_1^2}</math>
+<math>~2q^4 y \, ,</math>
    </td>
 </tr>
@@ Line 2,515: / Line 2,344: @@
 <tr>
    <td align="right">
-<math>~\Rightarrow~~~ 0</math>
+<math>~\frac{\partial A}{\partial z}</math>
    </td>
    <td align="center">
@@ Line 2,521: / Line 2,350: @@
    </td>
    <td align="left">
-<math>~Q^4 \biggl( \frac{q^2 y^2}{\lambda_1^2} \biggr) - Q^2 \biggl( \frac{x^2}{\lambda_1^2} \biggr) - 1</math>
+<math>~ 2p^4 z\, ;</math>
    </td>
 </tr>
@@ Line 2,527: / Line 2,356: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~ \frac{\partial B}{\partial x}</math>
    </td>
    <td align="center">
@@ Line 2,533: / Line 2,362: @@
    </td>
    <td align="left">
-<math>~Q^4  - Q^2 \biggl( \frac{x^2}{q^2 y^2} \biggr) - \biggl( \frac{\lambda_1^2}{q^2 y^2} \biggr) \, ,</math>
+<math>~2x( 4q^4y^2 + p^4 z^2 ) \, ,</math>
    </td>
 </tr>
-</table>
-where,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~Q^2</math>
+<math>~\frac{\partial B}{\partial y}</math>
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~\frac{1 + q^2\tan^2\varphi}{q^2\tan^2\varphi} \, .</math>
+<math>~2q^4 y (p^4 z^2 + 4x^2) \, ,</math>
    </td>
 </tr>
-</table>
-Solving the quadratic equation, we have,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~Q^2</math>
+<math>~\frac{\partial B}{\partial z}</math>
    </td>
    <td align="center">
@@ Line 2,563: / Line 2,386: @@
    </td>
    <td align="left">
-<math>~\frac{1}{2} \biggl\{ \biggl( \frac{x^2}{q^2 y^2} \biggr) \pm \biggl[ \biggl( \frac{x^2}{q^2 y^2} \biggr)^2 + 4\biggl( \frac{\lambda_1^2}{q^2 y^2} \biggr) \biggr]^{1 / 2}  \biggr\}</math>
+<math>~ 2p^4 z(q^4 y^2 + x^2)\, .</math>
    </td>
 </tr>
+</table>
+</td></tr></table>
+Now, let's assume that,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-&nbsp;
+<math>~\lambda_3</math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
-<math>~\biggl( \frac{x^2}{2q^2 y^2} \biggr) \biggl\{ 1  \pm \biggl[ 1 + 4\biggl( \frac{\lambda_1^2 q^2 y^2}{x^4} \biggr) \biggr]^{1 / 2}  \biggr\} \, .</math>
+<math>~\biggl( \frac{A}{B} \biggr)^{1 / 2} \, ,</math>
    </td>
 </tr>
-</table>
-<table border="1" align="center" width="80%" cellpadding="10"><tr><td align="left">
-<div align="center">'''Tentative Summary'''</div>
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\lambda_1</math>
+<math>~\Rightarrow~~~ \frac{ \partial \lambda_3}{\partial x_i}</math>
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~(x^2 + q^2y^2 + p^2 z^2)^{1 / 2} \, ,</math>
+<math>~
+\frac{1}{2 (AB)^{1 / 2}} \cdot \frac{\partial A}{\partial x_i}
+-
+\frac{A^{1 / 2}}{2 B^{3 / 2}} \cdot \frac{\partial B}{\partial x_i}
+</math>
    </td>
 </tr>
@@ Line 2,598: / Line 2,427: @@
 <tr>
    <td align="right">
-<math>~\lambda_2</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~
+<math>~\frac{\lambda_3}{2AB}
-\frac{(x^2 + y^2)^{1 / 2}}{pz}
+\biggl[
-\, ,</math>
+B \cdot \frac{\partial A}{\partial x_i}
+-
+A \cdot \frac{\partial B}{\partial x_i}
+\biggr]
+\, .
+</math>
    </td>
 </tr>
@@ Line 2,612: / Line 2,446: @@
 <tr>
    <td align="right">
-<math>~\lambda_3 = Q^2</math>
+<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x_i}</math>
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~\biggl( \frac{x^2}{2q^2 y^2} \biggr) \biggl\{ 1  \pm \biggl[ 1 + 4\biggl( \frac{\lambda_1^2 q^2 y^2}{x^4} \biggr) \biggr]^{1 / 2}  \biggr\} \, .</math>
+<math>~
+(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) \cdot \frac{\partial A}{\partial x_i}
+-
+(x^2 + q^4y^2 + p^4 z^2 ) \cdot \frac{\partial B}{\partial x_i} \, .
+</math>
    </td>
 </tr>
 </table>
-</td></tr></table>
+<table border="1" cellpadding="8" align="center" width="80%"><tr><td align="left">
+Looking ahead &hellip;
-====Partial Derivatives &amp; Scale Factors====
-=====First Coordinate=====
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_1}{\partial x}</math>
+<math>~h_3^{-2}</math>
    </td>
    <td align="center">
@@ Line 2,638: / Line 2,473: @@
    </td>
    <td align="left">
-<math>~\frac{x}{\lambda_1} \, ,</math>
+<math>~
+\biggl\{ \frac{\lambda_3}{2AB} \biggl[
+B \cdot \frac{\partial A}{\partial x}
+-
+A \cdot \frac{\partial B}{\partial x}
+\biggr] \biggr\}^2
++
+\biggl\{ \frac{\lambda_3}{2AB} \biggl[
+B \cdot \frac{\partial A}{\partial y}
+-
+A \cdot \frac{\partial B}{\partial y}
+\biggr] \biggr\}^2
++
+\biggl\{ \frac{\lambda_3}{2AB} \biggl[
+B \cdot \frac{\partial A}{\partial z}
+-
+A \cdot \frac{\partial B}{\partial z}
+\biggr] \biggr\}^2
+</math>
    </td>
 </tr>
@@ Line 2,644: / Line 2,497: @@
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_1}{\partial y}</math>
+<math>~\Rightarrow ~~~ \biggl[\frac{2AB}{\lambda_3}  \biggr]^2 h_3^{-2}</math>
    </td>
    <td align="center">
@@ Line 2,650: / Line 2,503: @@
    </td>
    <td align="left">
-<math>~\frac{q^2 y}{\lambda_1} \, ,</math>
+<math>~
-   </td>
+\biggl[
-</tr>
+B \cdot \frac{\partial A}{\partial x}
+-
-<tr>
+A \cdot \frac{\partial B}{\partial x}
-   <td align="right">
+\biggr]^2
-<math>~\frac{\partial \lambda_1}{\partial z}</math>
++
+\biggl[
+B \cdot \frac{\partial A}{\partial y}
+-
+A \cdot \frac{\partial B}{\partial y}
+\biggr]^2
++
+\biggl[
+B \cdot \frac{\partial A}{\partial z}
+-
+A \cdot \frac{\partial B}{\partial z}
+\biggr]^2
+</math>
+   </td>
+</tr>
+<tr>
+   <td align="right">
+<math>~\Rightarrow ~~~ \biggl[\frac{\lambda_3}{2AB}  \biggr] h_3</math>
    </td>
    <td align="center">
@@ Line 2,662: / Line 2,533: @@
    </td>
    <td align="left">
-<math>~\frac{p^2 z}{\lambda_1} \, .</math>
+<math>~
+\biggl\{\biggl[
+B \cdot \frac{\partial A}{\partial x}
+-
+A \cdot \frac{\partial B}{\partial x}
+\biggr]^2
++
+\biggl[
+B \cdot \frac{\partial A}{\partial y}
+-
+A \cdot \frac{\partial B}{\partial y}
+\biggr]^2
++
+\biggl[
+B \cdot \frac{\partial A}{\partial z}
+-
+A \cdot \frac{\partial B}{\partial z}
+\biggr]^2
+\biggr\}^{-1 / 2}
+</math>
    </td>
 </tr>
 </table>
+Then, for example,
 <table border="0" cellpadding="5" align="center">
@@ Line 2,671: / Line 2,562: @@
 <tr>
    <td align="right">
-<math>~h_1^{-2}</math>
+<math>~\gamma_{31} \equiv h_3 \biggl(\frac{\partial \lambda_3}{\partial x} \biggr)</math>
    </td>
    <td align="center">
@@ Line 2,677: / Line 2,568: @@
    </td>
    <td align="left">
-<math>~
+<math>~\biggl[
-\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^2
+B \cdot \frac{\partial A}{\partial x}
-+ \biggl( \frac{\partial \lambda_1}{\partial y} \biggr)^2
+-
-+ \biggl( \frac{\partial \lambda_1}{\partial z} \biggr)^2
+A \cdot \frac{\partial B}{\partial x}
+\biggr]
+\biggl\{\biggl[
+B \cdot \frac{\partial A}{\partial x}
+-
+A \cdot \frac{\partial B}{\partial x}
+\biggr]^2
++
+\biggl[
+B \cdot \frac{\partial A}{\partial y}
+-
+A \cdot \frac{\partial B}{\partial y}
+\biggr]^2
++
+\biggl[
+B \cdot \frac{\partial A}{\partial z}
+-
+A \cdot \frac{\partial B}{\partial z}
+\biggr]^2
+\biggr\}^{-1 / 2}
 </math>
+  </td>
+</tr>
+</table>
+</td></tr></table>
+As a result, we have,
+<table border="0" cellpadding="5" align="center">
+<tr>
+  <td align="right">
+<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}</math>
    </td>
    <td align="center">
@@ Line 2,687: / Line 2,609: @@
    </td>
    <td align="left">
-<math>~
+<math>~2x \biggl[
-\biggl( \frac{x}{\lambda_1} \biggr)^2
+(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
-+ \biggl( \frac{q^2 y}{\lambda_1} \biggr)^2
+-
-+ \biggl( \frac{p^2 z}{\lambda_1} \biggr)^2
+(x^2 + q^4y^2 + p^4 z^2 ) ( 4q^4y^2 + p^4 z^2 ) \biggr]
-\, .</math>
+</math>
    </td>
 </tr>
@@ Line 2,697: / Line 2,619: @@
 <tr>
    <td align="right">
-<math>~\Rightarrow ~~~ h_1</math>
+&nbsp;
    </td>
    <td align="center">
 <math>~=</math>
    </td>
-   <td align="left" colspan="3">
+   <td align="left">
-<math>~
+<math>~2x \biggl[
-\lambda_1 \ell_{3D} \, ,
+(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
+-
+(4x^2q^4y^2 + x^2 p^4 z^2 + 4q^8y^4 + q^4y^2 p^4z^2 + 4q^4y^2 p^4 z^2 + p^8z^4) \biggr]
 </math>
    </td>
 </tr>
-</table>
-where,
-<div align="center"><math>\ell_{3D} \equiv (x^2 + q^4y^2 + p^4z^2)^{-1 / 2} \, .</math></div>
-As a result, the associated unit vector is,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_1</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 2,722: / Line 2,641: @@
    </td>
    <td align="left">
-<math>~
+<math>~2x \biggl[
-\hat{\imath} h_1 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)
+-
-+ \hat{\jmath} h_1 \biggl( \frac{\partial \lambda_1}{\partial y} \biggr)
+(4q^8y^4  + 4q^4y^2 p^4 z^2 + p^8z^4) \biggr]
-+ \hat{k} h_1 \biggl( \frac{\partial \lambda_1}{\partial z} \biggr)
 </math>
    </td>
@@ Line 2,738: / Line 2,656: @@
    </td>
    <td align="left">
-<math>~
+<math>~-2x (2q^4y^2  + p^4z^2)^2
-\hat{\imath} x \ell_{3D}
-+ \hat{\jmath} q^2 y\ell_{3D}
-+ \hat{k} p^2 z \ell_{3D}
-\, .
 </math>
    </td>
 </tr>
-</table>
-Notice that,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_1 \cdot \hat{e}_1</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 2,759: / Line 2,669: @@
    </td>
    <td align="left">
-<math>~
+<math>~-8x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2
-(x^2 + q^4 y^2 + p^4 z^2) \ell_{3D}^2 = 1 \, .
 </math>
    </td>
 </tr>
-</table>
-=====Second Coordinate (1<sup>st</sup> Try)=====
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial x}</math>
+<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln\lambda_3}{\partial \ln{x}}</math>
    </td>
    <td align="center">
@@ Line 2,778: / Line 2,682: @@
    </td>
    <td align="left">
-<math>~
+<math>~-\biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^2 \, ;
-\frac{1}{pz} \biggl[ \frac{x}{(x^2 + y^2)^{1 / 2}} \biggr] \, ,
 </math>
    </td>
 </tr>
+</table>
+and,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial y}</math>
+<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}</math>
    </td>
    <td align="center">
@@ Line 2,793: / Line 2,699: @@
    <td align="left">
 <math>~
-\frac{1}{pz} \biggl[ \frac{y}{(x^2 + y^2)^{1 / 2}} \biggr]
+q^4y\biggl[
-\, ,</math>
+(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
+-
+(x^2 + q^4y^2 + p^4 z^2 ) (p^4 z^2 + 4x^2)
+\biggr]
+</math>
    </td>
 </tr>
@@ Line 2,800: / Line 2,710: @@
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial z}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 2,807: / Line 2,717: @@
    <td align="left">
 <math>~
-- \frac{(x^2 + y^2)^{1 / 2}}{pz^2}
+q^4y\biggl[
-\, .</math>
+(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
+-
+( x^2p^4z^2 + 4x^4 + q^4y^2p^4z^2 + 4x^2q^4y^2 + p^8z^4 + 4x^2p^4z^2)
+\biggr]
+</math>
    </td>
 </tr>
-</table>
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~h_2^{-2}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 2,824: / Line 2,735: @@
    <td align="left">
 <math>~
-\biggl( \frac{\partial \lambda_2}{\partial x} \biggr)^2
+-2q^4y( 4x^4 + p^8z^4 + 4x^2p^4z^2)
-+ \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)^2
-+ \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)^2
 </math>
    </td>
@@ Line 2,840: / Line 2,749: @@
    <td align="left">
 <math>~
-\biggl\{ \frac{1}{pz} \biggl[ \frac{x}{(x^2 + y^2)^{1 / 2}} \biggr] \biggr\}^2
+-2q^4y( 2x^2 + p^4z^2 )^2
-+ \biggl\{ \frac{1}{pz} \biggl[ \frac{y}{(x^2 + y^2)^{1 / 2}} \biggr] \biggr\}^2
-+ \biggl\{ \frac{(x^2 + y^2)^{1 / 2}}{pz^2} \biggr\}^2
 </math>
    </td>
@@ Line 2,849: / Line 2,756: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln\lambda_3}{\partial \ln{y} }</math>
    </td>
    <td align="center">
@@ Line 2,856: / Line 2,763: @@
    <td align="left">
 <math>~
-\biggl\{ \biggl[ \frac{x^2}{(x^2 + y^2)p^2 z^2} \biggr] \biggr\}
+- \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^2 \, ;
-+ \biggl\{ \biggl[ \frac{y^2}{(x^2 + y^2)p^2 z^2} \biggr] \biggr\}
-+ \biggl\{ \frac{(x^2 + y^2)}{p^2 z^4} \biggr\}
 </math>
    </td>
 </tr>
+</table>
+and,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-&nbsp;
+<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math>
    </td>
    <td align="center">
@@ Line 2,871: / Line 2,781: @@
    </td>
    <td align="left">
-<math>~
+<math>~2p^4 z \biggl[
-\frac{1}{p^2 z^2}
+(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
-+ \frac{(x^2 + y^2)}{p^2 z^4}
+-
-=
+(x^2 + q^4y^2 + p^4 z^2 ) (q^4 y^2 + x^2)
-\frac{(x^2 + y^2 + z^2)}{p^2 z^4}
+\biggr]
 </math>
    </td>
@@ Line 2,882: / Line 2,792: @@
 <tr>
    <td align="right">
-<math>~\Rightarrow~~~h_2</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 2,888: / Line 2,798: @@
    </td>
    <td align="left">
-<math>~
+<math>~2p^4 z \biggl[
-\frac{p z^2}{r }
+(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
+-
+(x^2q^4y^2 + x^4 + q^8y^4 + x^2q^4y^2 + q^4y^2p^4z^2 + x^2p^4z^2)
+\biggr]
 </math>
    </td>
 </tr>
-</table>
-As a result, the associated unit vector is,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_2</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 2,905: / Line 2,815: @@
    </td>
    <td align="left">
-<math>~
+<math>~2p^4 z \biggl[
-\hat{\imath} h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)
+( 2x^2q^4y^2)
-+ \hat{\jmath} h_2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)
+-
-+ \hat{k} h_2 \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)
+( x^4 + q^8y^4 )
+\biggr]
 </math>
    </td>
@@ Line 2,921: / Line 2,832: @@
    </td>
    <td align="left">
-<math>~
+<math>~-2p^4 z \biggl[
-\hat{\imath}  \biggl[ \frac{xz}{r(x^2 + y^2)^{1 / 2}} \biggr]
+x^4 + q^8y^4
-+ \hat{\jmath}  \biggl[ \frac{yz}{r(x^2 + y^2)^{1 / 2}} \biggr]
+- 2x^2q^4y^2
-- \hat{k} \biggl[ \frac{(x^2 + y^2)^{1 / 2}}{r} \biggr] \, .
+\biggr]
 </math>
    </td>
-</table>
+</tr>
-Notice that,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_2 \cdot \hat{e}_2</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 2,940: / Line 2,848: @@
    </td>
    <td align="left">
-<math>~
+<math>~-2p^4 z (x^2 - q^4y^2 )^2
-\biggl[ \frac{x^2 z^2}{r^2(x^2 + y^2)} \biggr]
-+ \biggl[ \frac{y^2 z^2}{r^2(x^2 + y^2)} \biggr]
-+ \biggl[ \frac{(x^2 + y^2)}{r^2} \biggr]
-= 1 \, .
 </math>
    </td>
 </tr>
-</table>
-Let's check to see if this "second" unit vector is orthogonal to the "first."
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_1 \cdot \hat{e}_2</math>
+<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln \lambda_3}{\partial \ln{z}}</math>
    </td>
    <td align="center">
@@ Line 2,961: / Line 2,861: @@
    </td>
    <td align="left">
-<math>~
+<math>~-4 \biggl[ \biggl( \frac{p^4 z^2}{4} \biggr) (x^2 - q^4y^2 )^2 \biggr]
-x\ell_{3D} \biggl[ \frac{xz}{r(x^2 + y^2)^{1 / 2}} \biggr]
-+ q^2 y\ell_{3D} \biggl[ \frac{yz}{r(x^2 + y^2)^{1 / 2}} \biggr]
-- p^2 z \ell_{3D} \biggl[ \frac{(x^2 + y^2)^{1 / 2}}{r} \biggr]
 </math>
    </td>
@@ Line 2,977: / Line 2,874: @@
    </td>
    <td align="left">
-<math>~
+<math>~-\biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^2
-\ell_{3D} \biggl\{ \biggl[ \frac{x^2z}{r(x^2 + y^2)^{1 / 2}} \biggr]
+\, .
-+ \biggl[ \frac{q^2 y^2 z}{r(x^2 + y^2)^{1 / 2}} \biggr]
--  \biggl[ \frac{p^2 z(x^2 + y^2)^{1 / 2}}{r} \biggr]
-\biggr\}
 </math>
    </td>
 </tr>
+</table>
+<font color="red">'''Wow! &nbsp; Really close!''' (13 November 2020)</font>
+Just for fun, let's see what we get for <math>~h_3</math>.  It is given by the expression,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-&nbsp;
+<math>~h_3^{-2}</math>
    </td>
    <td align="center">
@@ Line 2,995: / Line 2,895: @@
    <td align="left">
 <math>~
-\frac{z\ell_{3D}}{r (x^2 + y^2)^{1 / 2}} \biggl\{ \biggl[ x^2\biggr]
+\biggl( \frac{\partial \lambda_3}{\partial x} \biggr)^2
-+ \biggl[ q^2 y^2 \biggr]
++\biggl( \frac{\partial \lambda_3}{\partial y} \biggr)^2
--  \biggl[ p^2 (x^2 + y^2) \biggr]
++\biggl( \frac{\partial \lambda_3}{\partial z} \biggr)^2
-\biggr\}
 </math>
    </td>
@@ Line 3,008: / Line 2,907: @@
    </td>
    <td align="center">
-<math>~\ne</math>
+<math>~=</math>
    </td>
    <td align="left">
 <math>~
-\ .
+\biggl\{ \frac{\lambda_3}{ABx} \biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^2 \biggr\}^2
++\biggl\{ \frac{\lambda_3}{ABy} \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^2 \biggr\}^2
++\biggl\{ \frac{\lambda_3}{ABz} \biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^2 \biggr\}^2
 </math>
    </td>
 </tr>
-</table>
+<tr>
-=====Second Coordinate (2<sup>nd</sup> Try)=====
+   <td align="right">
+<math>~\Rightarrow~~~ \biggl[ \frac{AB}{\lambda_3}\biggr]^2 h_3^{-2}</math>
-Let's try,
-<table border="0" cellpadding="5" align="center">
-<tr>
-   <td align="right">
-<math>~\lambda_2</math>
    </td>
    <td align="center">
@@ Line 3,033: / Line 2,927: @@
    <td align="left">
 <math>~
-\biggl[\frac{(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2}}{pz}  \biggr]
+\biggl\{ \frac{1}{x^2} \biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^4 \biggr\}
-\, ,
++\biggl\{ \frac{1}{y^2} \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^4 \biggr\}
++\biggl\{ \frac{1}{z^2} \biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^4 \biggr\}
 </math>
    </td>
 </tr>
+</table>
+====Fiddle Around====
+Let &hellip;
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\Rightarrow~~~\frac{\partial \lambda_2}{\partial x}</math>
+<math>~\mathcal{L}_x</math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
 <math>~
-\frac{x}{pz(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2} } = \frac{x}{p^2 z^2 \lambda_2}
+- \biggl[
-\, ,
+B \cdot \frac{\partial A}{\partial x}
+-
+A \cdot \frac{\partial B}{\partial x}
+\biggr]
 </math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial y}</math>
    </td>
    <td align="center">
@@ Line 3,062: / Line 2,960: @@
    </td>
    <td align="left">
-<math>~
+<math>~8x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2
-\frac{q^2 y}{pz(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2} } = \frac{q^2y}{p^2 z^2 \lambda_2}
-\, ,
 </math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial z}</math>
    </td>
    <td align="center">
@@ Line 3,077: / Line 2,967: @@
    </td>
    <td align="left">
-<math>~
+<math>~\frac{2}{x} \biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^2
-\frac{\mathfrak{f}\cdot p^2z}{pz(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2} }
--
-\frac{(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2}}{pz^2}
-= \frac{1}{p^2z^2 \lambda_2 } \biggl( \mathfrak{f}\cdot p^2z \biggr) - \frac{\lambda_2 }{z}
-= \frac{1}{p^2z^2 \lambda_2 } \biggl( \mathfrak{f}\cdot p^2z  - p^2z \lambda_2^2 \biggr)
-\, .
 </math>
-  </td>
-</tr>
-</table>
-Hence,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~h_2^{-2}</math>
    </td>
    <td align="center">
@@ Line 3,101: / Line 2,974: @@
    </td>
    <td align="left">
-<math>~
+<math>~8x~\mathfrak{F}_x(y,z)
-\biggl( \frac{\partial \lambda_2}{\partial x} \biggr)^2
-+ \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)^2
-+ \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)^2
 </math>
    </td>
@@ Line 3,111: / Line 2,981: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~\mathcal{L}_y</math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
 <math>~
-\biggl[ \frac{x}{p^2 z^2 \lambda_2} \biggr]^2
+- \biggl[
-+ \biggl[ \frac{q^2y}{p^2 z^2 \lambda_2} \biggr]^2
+B \cdot \frac{\partial A}{\partial y}
-+ \biggl[ \frac{ \mathfrak{f} }{z \lambda_2 }  - \frac{\lambda_2 }{z}\biggr]^2
+-
+A \cdot \frac{\partial B}{\partial y}
+\biggr]
 </math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
    </td>
    <td align="center">
@@ Line 3,133: / Line 2,999: @@
    </td>
    <td align="left">
 <math>~
-\biggl[ \frac{x^2 + q^4 y^2}{p^4 z^4 \lambda_2^2} \biggr]
+q^4y\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2
-+ \biggl[ \frac{1}{z\lambda_2}\biggl( \mathfrak{f} - \lambda_2^2 \biggr) \biggr]^2
 </math>
    </td>
-</tr>
+  <td align="center">
+<math>~=</math>
-<tr>
+  </td>
-   <td align="right">
+   <td align="left">
-&nbsp;
+<math>~
+\frac{2}{y} \biggl[ 2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr)\biggr]^2
+</math>
    </td>
    <td align="center">
@@ Line 3,148: / Line 3,015: @@
    </td>
    <td align="left">
-<math>~ \frac{1}{p^4 z^4 \lambda_2^2}
+<math>~8y~\mathfrak{F}_y(x,z)
-\biggl[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 \biggr]
 </math>
    </td>
@@ Line 3,156: / Line 3,022: @@
 <tr>
    <td align="right">
-<math>~\Rightarrow ~~~ h_2</math>
+<math>~\mathcal{L}_z</math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
 <math>~
-\frac{p^2 z^2 \lambda_2}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \, .
+-\biggl[
+B \cdot \frac{\partial A}{\partial z}
+-
+A \cdot \frac{\partial B}{\partial z}
+\biggr]
 </math>
-  </td>
-</tr>
-</table>
-So, the associated unit vector is,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\hat{e}_2</math>
    </td>
    <td align="center">
@@ Line 3,180: / Line 3,041: @@
    <td align="left">
 <math>~
-\hat{\imath} h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)
+p^4 z \biggl(x^2 - q^4y^2 \biggr)^2
-+ \hat{\jmath} h_2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)
-+ \hat{k} h_2 \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)
 </math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
    </td>
    <td align="center">
@@ Line 3,196: / Line 3,049: @@
    <td align="left">
 <math>~
-\hat{\imath}  \biggl\{ \frac{x}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\}
+\frac{2}{z} \biggl[p^2 z \biggl(x^2 - q^4y^2 \biggr)\biggr]^2
-+ \hat{\jmath}  \biggl\{ \frac{q^2 y}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\}
-+ \hat{k} \biggl\{ \frac{p^2z(\mathfrak{f}-\lambda_2^2)}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\} \, .
 </math>
-  </td>
-</table>
-Checking orthogonality &hellip;
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\hat{e}_1 \cdot \hat{e}_2</math>
    </td>
    <td align="center">
@@ Line 3,214: / Line 3,056: @@
    </td>
    <td align="left">
-<math>~
+<math>~8z~\mathfrak{F}_z(x,y)
-x\ell_{3D} \biggl\{ \frac{x}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\}
-+ q^2 y\ell_{3D} \biggl\{ \frac{q^2 y}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\}
-+ p^2 z \ell_{3D} \biggl\{ \frac{p^2z(\mathfrak{f}-\lambda_2^2)}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\}
 </math>
    </td>
 </tr>
+</table>
+With this shorthand in place, we can write,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-&nbsp;
+<math>~\hat{e}_3</math>
    </td>
    <td align="center">
@@ Line 3,230: / Line 3,073: @@
    </td>
    <td align="left">
-<math>~\frac{\ell_{3D}}{ [ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} }
+<math>~\frac{\ell_{3D}}{\mathcal{D}}
-\biggl\{ x^2 + q^4y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)\biggr\} \, .
+\biggl\{
+-\hat\imath \biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]
++ \hat\jmath \biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]
++ \hat{k} \biggl[ p^2z( x^2 - q^4y^2 ) \biggl]
+\biggr\}
 </math>
    </td>
@@ Line 3,244: / Line 3,091: @@
    </td>
    <td align="left">
-<math>~\frac{\ell_{3D}}{ [ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} }
+<math>~
-\biggl\{ x^2 + q^4y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)\biggr\} \, .
+\frac{1}{(AB)^{1 / 2}}
+\biggl\{
+-\hat\imath \biggl[ \frac{x \mathcal{L}_x}{2} \biggl]^{1 / 2}
++ \hat\jmath \biggl[ \frac{y \mathcal{L}_y}{2} \biggl]^{1 / 2}
++ \hat{k} \biggl[ \frac{z \mathcal{L}_z}{2} \biggl]^{1 / 2}
+\biggr\}
+\, .
 </math>
    </td>
@@ Line 3,251: / Line 3,104: @@
 </table>
-If <math>~\mathfrak{f} = 0</math>, we have &hellip;
+We therefore also recognize that,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~p^2 z (\mathfrak{f} - \lambda_2^2) </math>
+<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial x}\biggr)</math>
    </td>
    <td align="center">
-&nbsp; &nbsp; <math>~~~\rightarrow ~~~ </math>&nbsp; &nbsp;
+<math>~=</math>
    </td>
    <td align="left">
 <math>~
-\biggl[- p^2 z \lambda_2^2\biggr]_{\mathfrak{f} = 0}
+-\frac{1}{(AB)^{1 / 2}} \biggl[ \frac{x \mathcal{L}_x}{2} \biggl]^{1 / 2}
-=
+</math>
-- p^2 z\biggl[\frac{(x^2 + q^2y^2 + \cancelto{0}{\mathfrak{f} \cdot }p^2 z^2 )^{1 / 2}}{pz}  \biggr]^2
-=
-- \frac{(x^2 + q^2y^2  )}{z} \, ,</math>
-  </td>
-</tr>
-</table>
-which, in turn, means &hellip;
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~[ x^2 + q^4 y^2 + p^4 z^2 (\cancelto{0}{\mathfrak{f}} - \lambda_2^2)^2 ]^{1 / 2}</math>
    </td>
    <td align="center">
@@ Line 3,283: / Line 3,124: @@
    <td align="left">
 <math>~
-[ x^2 + q^4 y^2 + p^4 z^2 \lambda_2^4 ]^{1 / 2}
+-\frac{1}{(AB)^{1 / 2}} \biggl[ 4x^2 ~\mathfrak{F}_x \biggl]^{1 / 2} \, ,
 </math>
    </td>
@@ Line 3,290: / Line 3,131: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial y}\biggr)</math>
    </td>
    <td align="center">
@@ Line 3,297: / Line 3,138: @@
    <td align="left">
 <math>~
-\biggl\{ x^2 + q^4 y^2 + p^4 z^2 \biggl[\frac{(x^2 + q^2y^2 + \cancelto{0}{\mathfrak{f}\cdot} p^2 z^2)^{1 / 2}}{pz}  \biggr]^4 \biggr\}^{1 / 2}
+\frac{1}{(AB)^{1 / 2}} \biggl[ \frac{y \mathcal{L}_y}{2} \biggl]^{1 / 2}
 </math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
    </td>
    <td align="center">
@@ Line 3,311: / Line 3,146: @@
    <td align="left">
 <math>~
-\biggl\{ x^2 + q^4 y^2 + \biggl[\frac{(x^2 + q^2y^2 )^{2}}{z^2}  \biggr] \biggr\}^{1 / 2}
+\frac{1}{(AB)^{1 / 2}} \biggl[ 4y^2~\mathfrak{F}_y \biggl]^{1 / 2} \, ,
 </math>
    </td>
@@ Line 3,318: / Line 3,153: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial z}\biggr)</math>
    </td>
    <td align="center">
@@ Line 3,325: / Line 3,160: @@
    <td align="left">
 <math>~
-(x^2 + q^4 y^2)^{1 / 2} \biggl[ 1 + \frac{(x^2 + q^2y^2 )}{z^2} \biggr]^{1 / 2}
+\frac{1}{(AB)^{1 / 2}} \biggl[ \frac{z \mathcal{L}_z}{2} \biggl]^{1 / 2}
 </math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
    </td>
    <td align="center">
@@ Line 3,339: / Line 3,168: @@
    <td align="left">
 <math>~
-\frac{(x^2 + q^4 y^2)^{1 / 2}}{z} \biggl[ z^2 + (x^2 + q^2y^2 ) \biggr]^{1 / 2} \, ,
+\frac{1}{(AB)^{1 / 2}} \biggl[ 4z^2~\mathfrak{F}_z \biggl]^{1 / 2} \, .
 </math>
    </td>
 </tr>
 </table>
-and,
+Now, if &#8212; and it is a BIG "if" &#8212; <math>~h_3 = h_0(AB)^{-1 / 2}</math>, then we have,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_1 \cdot \hat{e}_2</math>
+<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial x}\biggr)</math>
    </td>
    <td align="center">
@@ Line 3,355: / Line 3,185: @@
    </td>
    <td align="left">
-<math>~\frac{\ell_{3D}}{ [ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} }
+<math>~
-\biggl\{ x^2 + q^4y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)\biggr\} \, .
+-\biggl[ 4x^2 ~\mathfrak{F}_x \biggl]^{1 / 2}
 </math>
-  </td>
-</tr>
-</table>
-===Speculation6===
-====Determine &lambda;<sub>2</sub>====
-This is very similar to the [[#Speculation2|above, Speculation2]].
-Try,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\lambda_2</math>
    </td>
    <td align="center">
@@ Line 3,378: / Line 3,194: @@
    <td align="left">
 <math>~
-\frac{x y^{1/q^2}}{ z^{2/p^2}} \, ,
+-2x \biggl[ \mathfrak{F}_x \biggl]^{1 / 2} \, ,
 </math>
    </td>
 </tr>
-</table>
-in which case,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial x}</math>
+<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial y}\biggr)</math>
    </td>
    <td align="center">
@@ Line 3,395: / Line 3,208: @@
    <td align="left">
 <math>~
-\frac{\lambda_2}{x} \, ,
+\biggl[ 4y^2~\mathfrak{F}_y \biggl]^{1 / 2}
 </math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial y}</math>
    </td>
    <td align="center">
@@ Line 3,409: / Line 3,216: @@
    <td align="left">
 <math>~
-\frac{x}{z^{2/p^2}} \biggl(\frac{1}{q^2}\biggr) y^{1/q^2 - 1}
+y \biggl[ \mathfrak{F}_y \biggl]^{1 / 2} \, ,
-=
-\frac{\lambda_2}{q^2 y}
-\, ,
 </math>
    </td>
@@ Line 3,419: / Line 3,223: @@
 <tr>
    <td align="right">
-<math>~\frac{\partial \lambda_2}{\partial z}</math>
+<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial z}\biggr)</math>
    </td>
    <td align="center">
@@ Line 3,426: / Line 3,230: @@
    <td align="left">
 <math>~
--\frac{2\lambda_2}{p^2 z}
+\biggl[ 4z^2~\mathfrak{F}_z \biggl]^{1 / 2}
-\, .
 </math>
-  </td>
-</tr>
-</table>
-The associated scale factor is, then,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~h_2</math>
    </td>
    <td align="center">
@@ Line 3,443: / Line 3,237: @@
    </td>
    <td align="left">
-<math>~\biggl[
+<math>~
-\biggl( \frac{\partial \lambda_2}{\partial x} \biggr)^2
+z\biggl[ \mathfrak{F}_z \biggl]^{1 / 2} \, ,
-+
-\biggl( \frac{\partial \lambda_2}{\partial y} \biggr)^2
-+
-\biggl( \frac{\partial \lambda_2}{\partial z} \biggr)^2
-\biggr]^{-1 / 2}
 </math>
    </td>
 </tr>
+</table>
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-&nbsp;
+<math>~\Rightarrow~~~ h_0 \lambda_3</math>
    </td>
    <td align="center">
@@ Line 3,462: / Line 3,254: @@
    </td>
    <td align="left">
-<math>~\biggl[
+<math>~
-\biggl( \frac{ \lambda_2}{x} \biggr)^2
+-x^2 \biggl[ \mathfrak{F}_x \biggl]^{1 / 2}
-+
++ y^2 \biggl[ \mathfrak{F}_y \biggl]^{1 / 2}
-\biggl( \frac{\lambda_2}{q^2y} \biggr)^2
++ z^2 \biggl[ \mathfrak{F}_z \biggl]^{1 / 2} \, .
-+
-\biggl( - \frac{2\lambda_2}{p^2z} \biggr)^2
-\biggr]^{-1 / 2}
 </math>
    </td>
 </tr>
+</table>
+But if this is the correct expression for <math>~\lambda_3</math> and its three partial derivatives, then it must be true that,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-&nbsp;
+<math>~h_3^{-2}</math>
    </td>
    <td align="center">
@@ Line 3,481: / Line 3,273: @@
    </td>
    <td align="left">
-<math>~\frac{1}{\lambda_2}\biggl[
+<math>~
-\frac{ 1}{x^2}
+\biggl(\frac{\partial \lambda_3}{\partial x}\biggr)^2
 +
-\frac{1}{q^4y^2}
+\biggl(\frac{\partial \lambda_3}{\partial y}\biggr)^2
 +
-\frac{4}{p^4z^2}
+\biggl(\frac{\partial \lambda_3}{\partial z}\biggr)^2
-\biggr]^{-1 / 2}
 </math>
    </td>
@@ Line 3,494: / Line 3,285: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~\Rightarrow ~~~ \biggl( \frac{h_3}{h_0}\biggr)^{-2}</math>
    </td>
    <td align="center">
@@ Line 3,500: / Line 3,291: @@
    </td>
    <td align="left">
-<math>~\frac{1}{\lambda_2}\biggl[
+<math>~
-\frac{ (q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2}{x^2 q^4 y^2 p^4 z^2}
+x^2 \biggl[ \mathfrak{F}_x \biggl]
-\biggr]^{-1 / 2}
++
+y^2 \biggl[ \mathfrak{F}_y \biggl]
++
+z^2\biggl[ \mathfrak{F}_z \biggl]
 </math>
    </td>
@@ Line 3,516: / Line 3,310: @@
    <td align="left">
 <math>~
-\frac{1}{\lambda_2}\biggl[
+x^2 \biggl[ \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2 \biggl]
-\frac{x q^2 y p^2 z}{ \mathcal{D}}
++
-\biggr] \, .
+y^2 \biggl[ q^4\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2\biggl]
++
+z^2\biggl[ \frac{p^4}{4}\biggl(x^2 - q^4y^2 \biggr)^2 \biggl]
 </math>
    </td>
 </tr>
-</table>
-where,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\mathcal{D}</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>
+<math>~
+x^2 (2q^4y^2  + p^4z^2 )^2
++
+q^4 y^2( 2x^2 + p^4z^2 )^2
++
+p^4 z^2 (x^2 - q^4y^2 )^2
+</math>
    </td>
 </tr>
 </table>
-The associated unit vector is, then,
+Well &hellip; the right-hand side of this expression is identical to the right-hand side of the [[#Eureka|above expression]], where we showed that it equals <math>~(\ell_{3D}/\mathcal{D})^{-2}</math>.  That is to say, we are now showing that,
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_2</math>
+<math>~\biggl( \frac{h_3}{h_0}\biggr)^{-2}</math>
    </td>
    <td align="center">
@@ Line 3,550: / Line 3,349: @@
    </td>
    <td align="left">
-<math>~
+<math>~\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2} = [AB]</math>
-\hat{\imath} h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)
-+ \hat{\jmath} h_2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)
-+ \hat{k} h_2 \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)
-</math>
    </td>
 </tr>
@@ Line 3,560: / Line 3,355: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~\Rightarrow ~~~ \frac{h_3}{h_0}</math>
    </td>
    <td align="center">
@@ Line 3,566: / Line 3,361: @@
    </td>
    <td align="left">
-<math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl\{
+<math>~(AB)^{-1 / 2} \, .</math>
-\hat{\imath} \biggl( \frac{1}{x} \biggr)
-+ \hat{\jmath}  \biggl( \frac{1}{q^2 y} \biggr)
-+ \hat{k} \biggl( -\frac{2}{p^2 z} \biggr)
-\biggr\} \ .
-</math>
    </td>
 </tr>
 </table>
+And this is ''precisely'' what, just a few lines above, we hypothesized the functional expression for <math>~h_3</math> ought to be.  <font color="red">'''EUREKA!'''</font>
+====Summary====
-Recalling that the unit vector associated with the "first" coordinate is,
+In summary, then &hellip;
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_1 </math>
+<math>~\lambda_3</math>
    </td>
    <td align="center">
@@ Line 3,588: / Line 3,381: @@
    <td align="left">
 <math>~
-\hat\imath (x \ell_{3D}) + \hat\jmath (q^2y \ell_{3D}) + \hat{k} (p^2 z \ell_{3D}) \, ,
+-x^2 \biggl[ \mathfrak{F}_x \biggl]^{1 / 2}
++ y^2 \biggl[ \mathfrak{F}_y \biggl]^{1 / 2}
++ z^2 \biggl[ \mathfrak{F}_z \biggl]^{1 / 2}
 </math>
    </td>
 </tr>
-</table>
-where,
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\ell_{3D}</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 3,604: / Line 3,396: @@
    </td>
    <td align="left">
-<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} \, ,</math>
+<math>~
+-x^2 \biggl[\biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2 \biggl]^{1 / 2}
++ y^2 \biggl[ q^4\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2\biggl]^{1 / 2}
++ z^2 \biggl[ \frac{p^4}{4} \biggl(x^2 - q^4y^2 \biggr)^2 \biggl]^{1 / 2}
+</math>
    </td>
 </tr>
-</table>
-let's check to see whether the "second" unit vector is orthogonal to the "first."
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_1 \cdot \hat{e}_2</math>
+&nbsp;
    </td>
    <td align="center">
@@ Line 3,620: / Line 3,413: @@
    <td align="left">
 <math>~
-\frac{(x q^2 y p^2 z) \ell_{3D}}{\mathcal{D}} \biggl[
+-x^2 \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)
-+ 1 - 2
++ q^2y^2 \biggl( x^2 + \frac{p^4z^2}{2} \biggr)
-\biggr] = 0 \, .
++ \frac{p^2 z^2}{2}  \biggl(x^2 - q^4y^2 \biggr) \, ,
 </math>
    </td>
 </tr>
 </table>
+and,
-<font color="red">'''Hooray!'''</font>
-====Direction Cosines for <i>Third</i> Unit Vector====
-Now, what is the unit vector, <math>~\hat{e}_3</math>, that is simultaneously orthogonal to both these "first" and the "second" unit vectors?
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_3 \equiv \hat{e}_1 \times \hat{e}_2</math>
+<math>~h_3</math>
    </td>
    <td align="center">
@@ Line 3,642: / Line 3,433: @@
    </td>
    <td align="left">
-<math>~
+<math>~(AB)^{-1 / 2} </math>
-\hat\imath \biggl[ ( e_{1y} )( e_{2z}) - ( e_{2y} )( e_{1z}) ) \biggl]
-+ \hat\jmath \biggl[ ( e_{1z} )( e_{2x}) - ( e_{2z} )( e_{1x}) ) \biggl]
-+ \hat{k} \biggl[ ( e_{1x} )( e_{2y}) - ( e_{2x} )( e_{1y}) ) \biggl]
-</math>
    </td>
 </tr>
@@ Line 3,658: / Line 3,445: @@
    </td>
    <td align="left">
-<math>~\frac{(x q^2 y p^2 z) \ell_{3D}}{\mathcal{D}}
+<math>~\biggl[
-\biggl\{
+(x^2 + q^4y^2 + p^4 z^2 )(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
-\hat\imath \biggl[ \biggl( -\frac{2 q^2y}{p^2 z} \biggr) - \biggl( \frac{p^2z}{q^2y} \biggr)  \biggl]
+\biggr]^{-1 / 2} \, .</math>
-+ \hat\jmath \biggl[ \biggl( \frac{p^2z}{x} \biggr) - \biggl(-\frac{2x}{p^2z} \biggr)  \biggl]
-+ \hat{k} \biggl[ \biggl( \frac{x}{q^2y} \biggr) - \biggl( \frac{q^2y}{x} \biggr)  \biggl]
-\biggr\}
-</math>
    </td>
 </tr>
+</table>
+No!  Once again this does not work.  The direction cosines &#8212; and, hence, the components of the <math>~\hat{e}_3</math> unit vector &#8212; are not correct!
+===Speculation7===
+<table border="1" cellpadding="8" align="center">
+<tr>
+  <td align="center" colspan="9">'''Direction Cosine Components for T6 Coordinates'''</td>
+</tr>
 <tr>
-   <td align="right">
+   <td align="center"><math>~n</math></td>
-&nbsp;
+   <td align="center"><math>~\lambda_n</math></td>
-   </td>
+   <td align="center"><math>~h_n</math></td>
-   <td align="center">
+   <td align="center"><math>~\frac{\partial \lambda_n}{\partial x}</math></td>
-<math>~=</math>
+  <td align="center"><math>~\frac{\partial \lambda_n}{\partial y}</math></td>
-  </td>
+  <td align="center"><math>~\frac{\partial \lambda_n}{\partial z}</math></td>
-   <td align="left">
+  <td align="center"><math>~\gamma_{n1}</math></td>
-<math>~\frac{(x q^2 y p^2 z) \ell_{3D}}{\mathcal{D}}
+  <td align="center"><math>~\gamma_{n2}</math></td>
-\biggl\{
+  <td align="center"><math>~\gamma_{n3}</math></td>
--\hat\imath \biggl[ \frac{2 q^4y^2 + p^4z^2}{q^2 y p^2 z} \biggl]
-+ \hat\jmath \biggl[ \frac{p^4z^2 + 2x^2}{xp^2 z}  \biggl]
-+ \hat{k} \biggl[ \frac{x^2 - q^4y^2}{x q^2y} \biggl]
-\biggr\}
-</math>
-  </td>
 </tr>
 <tr>
-   <td align="right">
+   <td align="center"><math>~1</math></td>
-&nbsp;
+  <td align="center"><math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math></td>
-   </td>
+  <td align="center"><math>~\lambda_1 \ell_{3D}</math></td>
-   <td align="center">
+  <td align="center"><math>~\frac{x}{\lambda_1}</math></td>
-<math>~=</math>
+  <td align="center"><math>~\frac{q^2 y}{\lambda_1}</math></td>
-   </td>
+  <td align="center"><math>~\frac{p^2 z}{\lambda_1}</math></td>
-   <td align="left">
+  <td align="center"><math>~(x) \ell_{3D}</math></td>
-<math>~\frac{\ell_{3D}}{\mathcal{D}}
+  <td align="center"><math>~(q^2 y)\ell_{3D}</math></td>
-\biggl\{
+  <td align="center"><math>~(p^2z) \ell_{3D}</math></td>
--\hat\imath \biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]
+</tr>
-+ \hat\jmath \biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]
-+ \hat{k} \biggl[ p^2z( x^2 - q^4y^2 ) \biggl]
+<tr>
-\biggr\} \, .
+   <td align="center"><math>~2</math></td>
-</math>
+   <td align="center"><math>~\frac{x y^{1/q^2}}{ z^{2/p^2}}</math></td>
-  </td>
+  <td align="center"><math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y p^2 z}{ \mathcal{D}}\biggr] </math></td>
+  <td align="center"><math>~\frac{\lambda_2}{x}</math></td>
+   <td align="center"><math>~\frac{\lambda_2}{q^2 y}</math></td>
+   <td align="center"><math>~-\frac{2\lambda_2}{p^2 z}</math></td>
+  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{x}\biggr)</math></td>
+  <td align="center"><math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{q^2y}\biggr)</math></td>
+  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(-\frac{2}{p^2z}\biggr)</math></td>
+</tr>
+<tr>
+  <td align="center"><math>~3</math></td>
+  <td align="center">---</td>
+  <td align="center">---</td>
+  <td align="center">---</td>
+  <td align="center">---</td>
+  <td align="center">---</td>
+  <td align="center"><math>~-\frac{\ell_{3D}}{\mathcal{D}}\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]</math></td>
+  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]</math></td>
+  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]</math></td>
 </tr>
-</table>
-Is this a valid unit vector?  First, note that &hellip;
+<tr>
-<table border="0" cellpadding="5" align="center">
+  <td align="left" colspan="9">
+<table border="0" cellpadding="8" align="center">
 <tr>
    <td align="right">
-<math>~\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2}</math>
+<math>~\ell_{3D}</math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
-<math>~
+<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} \, ,</math>
-(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
-(x^2 + q^4y^2 + p^4 z^2 )
-</math>
    </td>
 </tr>
@@ Line 3,725: / Line 3,526: @@
 <tr>
    <td align="right">
-&nbsp;
+<math>~\mathcal{D}</math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
-<math>~
+<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>
-(x^2 q^4 y^2 p^4 z^2 + x^4 p^4 z^2 + 4x^4q^4y^2)
+  </td>
-+ (q^8 y^4 p^4 z^2 + x^2 q^4y^2 p^4 z^2 + 4x^2q^8y^4)
+</tr>
-+(q^4 y^2 p^8 z^4 + x^2 p^8 z^4 + 4x^2q^4y^2 p^4 z^2)
-</math>
+</table>
    </td>
 </tr>
+</table>
+On my white-board I have shown that, if
+<div align="center">
+<math>~\lambda_3 \equiv \ell_{3D} \mathcal{D} \, ,</math>
+</div>
+then everything will work out as long as,
+<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-&nbsp;
+<math>~\mathcal{L}</math>
    </td>
    <td align="center">
@@ Line 3,748: / Line 3,558: @@
    <td align="left">
 <math>~
-x^2 q^4 y^2 p^4 z^2 + x^4(p^4 z^2 + 4 q^4y^2)
+\biggl(\frac{\mathcal{D}}{\ell_{3D}} \biggr)^2 \frac{1}{\ell_{3D}^4} \, ,
-+ q^8 y^4(p^4 z^2  + 4x^2)
-+p^8z^4(x^2 + q^4 y^2 )\, .
 </math>
    </td>
 </tr>
 </table>
+where,
-<span id="Eureka">Then we have,</span>
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2}\hat{e}_3 \cdot \hat{e}_3</math>
+<math>~\mathcal{L}</math>
    </td>
    <td align="center">
-<math>~=</math>
+<math>~\equiv</math>
    </td>
    <td align="left">
 <math>~
-\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]^2
+x^2 (2q^4 y^2 + p^4z^2 )^4
 +
-\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]^2
+q^8 y^2 (2x^2 + p^4 z^2)^4
 +
-\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]^2
+p^8z^2( x^2 - q^4y^2)^4
 </math>
    </td>
@@ Line 3,786: / Line 3,593: @@
    <td align="left">
 <math>~
-x^2(4 q^8y^4 + 4q^4y^2p^4z^2 + p^8z^4 )
+x^2 (4q^8 y^4 + 4q^4y^2p^4z^2 + p^8z^4 )^2
 +
-q^4 y^2(p^8z^4 + 4x^2p^4z^2 + 4x^4 )
+q^8 y^2 (4x^4 + 4x^2p^4z^2 + p^8 z^4)^2
 +
-p^4z^2( x^4 - 2x^2q^4 y^2 + q^8y^4 )
+p^8z^2( x^4 - 2x^2q^4y^2 + q^8y^4)^2
 </math>
    </td>
@@ Line 3,804: / Line 3,611: @@
    <td align="left">
 <math>~
-x^2 q^8y^4 + 4x^2 q^4y^2p^4z^2 + x^2 p^8z^4
+x^2
+[16q^{16}y^8 + 16q^{12}y^6p^4z^2 + 4q^8y^4p^8z^4 + 16q^{12}y^6p^4z^2 + 16q^8y^4p^8z^4 + 4q^4y^2p^{12}z^6 + 4q^8y^4p^8z^4 + 4q^4y^2 p^{12}z^6 + p^{16}z^8]
+</math>
+  </td>
+</tr>
+<tr>
+  <td align="right">
+&nbsp;
+  </td>
+  <td align="center">
+&nbsp;
+  </td>
+  <td align="left">
+<math>~
 +
-q^4 y^2p^8z^4 + 4x^2q^4 y^2p^4z^2 + 4x^4q^4 y^2
+q^8 y^2
-+
+[16x^8 + 16x^6p^4z^2 + 4x^4p^8z^4 + 16x^6p^4z^2 + 16x^4p^8z^4 + 4x^2p^{12}z^6 + 4x^4p^8z^4 + 4x^2p^{12}z^6 + p^{16}z^8]
-x^4p^4z^2 - 2x^2q^4 y^2p^4z^2 + q^8y^4p^4z^2
 </math>
    </td>
@@ Line 3,818: / Line 3,638: @@
    </td>
    <td align="center">
-<math>~=</math>
+&nbsp;
    </td>
    <td align="left">
 <math>~
-x^2 q^4y^2p^4z^2
++
-+  p^8z^4 (x^2  +q^4 y^2)
+p^8z^2
-+  x^4(4q^4 y^2 + p^4z^2)
+[x^8 - 2x^6q^4y^2 + x^4q^8y^4 - 2x^6q^4y^2 + 4x^4q^8y^4 - 2x^2q^{12}y^6 + x^4q^8y^4 - 2x^2q^{12}y^6 + q^{16}y^8]
-+ q^8 y^4(4 x^2  + p^4z^2 )
 </math>
    </td>
@@ Line 3,839: / Line 3,658: @@
    <td align="left">
 <math>~
-\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2} \, ,
+x^2
+[16q^{16}y^8 + 32q^{12}y^6p^4z^2 + 24q^8y^4p^8z^4 + 8q^4y^2p^{12}z^6 + p^{16}z^8]
 </math>
    </td>
 </tr>
-</table>
-which means that, <math>~\hat{e}_3\cdot \hat{e}_3 = 1</math>.  &nbsp; &nbsp;<font color="red">'''Hooray! Again (11/11/2020)!'''</font>
-<table border="1" cellpadding="8" align="center">
 <tr>
-   <td align="center" colspan="9">'''Direction Cosine Components for T6 Coordinates'''</td>
+   <td align="right">
-</tr>
+&nbsp;
-<tr>
+   </td>
-   <td align="center"><math>~n</math></td>
+   <td align="center">
-   <td align="center"><math>~\lambda_n</math></td>
+&nbsp;
-   <td align="center"><math>~h_n</math></td>
+   </td>
-   <td align="center"><math>~\frac{\partial \lambda_n}{\partial x}</math></td>
+   <td align="left">
-  <td align="center"><math>~\frac{\partial \lambda_n}{\partial y}</math></td>
+<math>~
-  <td align="center"><math>~\frac{\partial \lambda_n}{\partial z}</math></td>
++
-  <td align="center"><math>~\gamma_{n1}</math></td>
+q^8 y^2
-  <td align="center"><math>~\gamma_{n2}</math></td>
+[16x^8 + 32x^6p^4z^2 + 24x^4p^8z^4  + 8x^2p^{12}z^6 + p^{16}z^8]
-   <td align="center"><math>~\gamma_{n3}</math></td>
+</math>
+   </td>
 </tr>
 <tr>
-   <td align="center"><math>~1</math></td>
+   <td align="right">
-   <td align="center"><math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math></td>
+&nbsp;
-   <td align="center"><math>~\lambda_1 \ell_{3D}</math></td>
+   </td>
-   <td align="center"><math>~\frac{x}{\lambda_1}</math></td>
+   <td align="center">
-   <td align="center"><math>~\frac{q^2 y}{\lambda_1}</math></td>
+&nbsp;
-  <td align="center"><math>~\frac{p^2 z}{\lambda_1}</math></td>
+   </td>
-  <td align="center"><math>~(x) \ell_{3D}</math></td>
+   <td align="left">
-  <td align="center"><math>~(q^2 y)\ell_{3D}</math></td>
+<math>~
-   <td align="center"><math>~(p^2z) \ell_{3D}</math></td>
++
+p^8z^2
+[x^8 - 4x^6q^4y^2 + 6x^4q^8y^4 - 4x^2q^{12}y^6 + q^{16}y^8]
+</math>
+   </td>
 </tr>
+</table>
+Let's check this out.
+<table border="0" cellpadding="5" align="center">
 <tr>
-   <td align="center"><math>~2</math></td>
+   <td align="right">
-  <td align="center"><math>~\frac{x y^{1/q^2}}{ z^{2/p^2}}</math></td>
+<math>~\mathrm{RHS}~\equiv \biggl(\frac{\mathcal{D}}{\ell_{3D}} \biggr)^2 \frac{1}{\ell_{3D}^4}</math>
-  <td align="center"><math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y p^2 z}{ \mathcal{D}}\biggr] </math></td>
+   </td>
-  <td align="center"><math>~\frac{\lambda_2}{x}</math></td>
+   <td align="center">
-  <td align="center"><math>~\frac{\lambda_2}{q^2 y}</math></td>
+<math>~=</math>
-  <td align="center"><math>~-\frac{2\lambda_2}{p^2 z}</math></td>
+   </td>
-  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{x}\biggr)</math></td>
+   <td align="left">
-  <td align="center"><math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{q^2y}\biggr)</math></td>
+<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)(x^2 + q^4y^2 + p^4 z^2 )^3</math>
-  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(-\frac{2}{p^2z}\biggr)</math></td>
+  </td>
-</tr>
-<tr>
-   <td align="center"><math>~3</math></td>
-   <td align="center">---</td>
-  <td align="center">---</td>
-  <td align="center">---</td>
-   <td align="center">---</td>
-   <td align="center">---</td>
-  <td align="center"><math>~-\frac{\ell_{3D}}{\mathcal{D}}\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]</math></td>
-  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]</math></td>
-  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]</math></td>
 </tr>
-<tr>
-  <td align="left" colspan="9">
-<table border="0" cellpadding="8" align="center">
 <tr>
    <td align="right">
-<math>~\ell_{3D}</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} \, ,</math>
+<math>~(x^2 + q^4y^2 + p^4 z^2 )(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)[x^4 + 2x^2q^4y^2 + 2x^2p^4z^2 +q^8y^4 + 2q^4y^2p^4z^2 + p^8z^4]</math>
    </td>
 </tr>
@@ Line 3,916: / Line 3,726: @@
 <tr>
    <td align="right">
-<math>~\mathcal{D}</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~\equiv</math>
+<math>~=</math>
    </td>
    <td align="left">
-<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>
+<math>~
-  </td>
+[(6x^2q^4 y^2 p^4 z^2 + x^4 p^4 z^2 + 4x^4q^4y^2)
-</tr>
++
+(q^8 y^4 p^4 z^2 + 4x^2q^8 y^4)
-</table>
++
+(q^4 y^2 p^8 z^4 + x^2 p^8 z^4 )]
+</math>
    </td>
 </tr>
-</table>
-Let's double-check whether this "third" unit vector is orthogonal to both the "first" and the "second" unit vectors.
-<table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~\hat{e}_1 \cdot \hat{e}_3</math>
+&nbsp;
    </td>
    <td align="center">
-<math>~=</math>
+&nbsp;
    </td>
    <td align="left">
-<math>~\frac{\ell_{3D}^2}{\mathcal{D}}
+<math>~\times [x^4 + 2x^2q^4y^2 + 2x^2p^4z^2 +q^8y^4 + 2q^4y^2p^4z^2 + p^8z^4]</math>
-\biggl\{
--x \biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]
-+ q^2 y  \biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]
-+ p^2 z  \biggl[ p^2z( x^2 - q^4y^2 ) \biggl]
-\biggr\}
-</math>
    </td>
 </tr>
@@ Line 3,962: / Line 3,762: @@
    </td>
    <td align="left">
-<math>~\frac{\ell_{3D}^2}{\mathcal{D}}
+<math>~
-\biggl\{
+[6x^2q^4 y^2 p^4 z^2 + x^4(p^4 z^2 + 4q^4y^2)
-- (2 x^2q^4y^2 + x^2p^4z^2 )
++
-+ (q^4 y^2 p^4z^2 + 2x^2 q^4 y^2)
+q^8 y^4(p^4 z^2 + 4x^2)
-+ ( x^2p^4z^2  - q^4y^2 p^4z^2  )
++
-\biggr\}
+p^8 z^4 (q^4 y^2 + x^2 )]
 </math>
    </td>
@@ Line 3,977: / Line 3,777: @@
    </td>
    <td align="center">
-<math>~=</math>
+&nbsp;
    </td>
    <td align="left">
-<math>~
+<math>~\times [x^4 + 2x^2q^4y^2 + 2x^2p^4z^2 +q^8y^4 + 2q^4y^2p^4z^2 + p^8z^4]</math>
-\, ,
-</math>
    </td>
 </tr>
 </table>
-and,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\hat{e}_2 \cdot \hat{e}_3</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\frac{\ell_{3D}}{\mathcal{D}} \cdot \frac{x q^2 y p^2 z}{\mathcal{D}}
-\biggl\{
-- \biggl[ (2 q^4y^2 + p^4z^2 ) \biggl]
-+  \biggl[ (p^4z^2 + 2x^2 )  \biggl]
-- \biggl[ 2( x^2 - q^4y^2 ) \biggl]
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\, .
-</math>
-  </td>
-</tr>
-</table>
-<font color="red">'''Q. E. D.'''</font>
-====Search for <i>Third</i> Coordinate Expression====
-Let's try &hellip;
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\lambda_3</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\mathcal{D}^n \ell_{3D}^m </math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{n / 2} (x^2 + q^4y^2 + p^4 z^2 )^{- m / 2}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~ \frac{\partial \lambda_3}{\partial x_i}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\ell_{3D}^m \biggl[ \frac{n}{2} \biggl(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 \biggr)^{n / 2 - 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
-+ \mathcal{D}^n \biggl[ - \frac{m}{2} (x^2 + q^4y^2 + p^4 z^2 )^{- m / 2 - 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(x^2 + q^4y^2 + p^4 z^2 )\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\mathcal{D}^n \ell_{3D}^m \biggl[ \frac{n}{2} \biggl(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 \biggr)^{- 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
-+ \mathcal{D}^n \ell_{3D}^m \biggl[ - \frac{m}{2} (x^2 + q^4y^2 + p^4 z^2 )^{- 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(x^2 + q^4y^2 + p^4 z^2 )\biggr] \, .
-</math>
-  </td>
-</tr>
-</table>
-Hence,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\frac{1}{\mathcal{D}^n \ell_{3D}^m} \cdot \frac{\partial \lambda_3}{\partial x}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{n}{2\mathcal{D}^2}\frac{\partial}{\partial x} \biggl[q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 \biggr]
--
-\frac{m \ell_{3D}^2}{2} \frac{\partial}{\partial x} \biggl[ x^2 + q^4y^2 + p^4 z^2 \biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~x \biggl\{
-\frac{n}{\mathcal{D}^2}\biggl[p^4 z^2 + 4q^4y^2 \biggr]
--
-m \ell_{3D}^2
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~x \biggl\{
-\frac{n (p^4 z^2 + 4q^4y^2)}{ ( q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 ) }
--
-\frac{m}{ ( x^2 + q^4y^2 + p^4 z^2 ) }
-\biggr\}
-</math>
-  </td>
-</tr>
-</table>
-This is overly cluttered!  Let's try, instead &hellip;
-<table border="1" cellpadding="8" align="center" width="80%"><tr><td align="left">
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~A \equiv \ell_{3D}^{-2}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~(x^2 + q^4y^2 + p^4 z^2 ) \, ,</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
-  <td align="right">
-<math>~B \equiv \mathcal{D}^2</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)</math>
-  </td>
-</tr>
-</table>
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~ \frac{\partial A}{\partial x}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2x \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial A}{\partial y}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2q^4 y \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial A}{\partial z}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~ 2p^4 z\, ;</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~ \frac{\partial B}{\partial x}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2x( 4q^4y^2 + p^4 z^2 ) \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial B}{\partial y}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2q^4 y (p^4 z^2 + 4x^2) \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial B}{\partial z}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~ 2p^4 z(q^4 y^2 + x^2)\, .</math>
-  </td>
-</tr>
-</table>
-</td></tr></table>
-Now, let's assume that,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\lambda_3</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~\biggl( \frac{A}{B} \biggr)^{1 / 2} \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow~~~ \frac{ \partial \lambda_3}{\partial x_i}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{1}{2 (AB)^{1 / 2}} \cdot \frac{\partial A}{\partial x_i}
--
-\frac{A^{1 / 2}}{2 B^{3 / 2}} \cdot \frac{\partial B}{\partial x_i}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\frac{\lambda_3}{2AB}
-\biggl[
-B \cdot \frac{\partial A}{\partial x_i}
--
-A \cdot \frac{\partial B}{\partial x_i}
-\biggr]
-\, .
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x_i}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) \cdot \frac{\partial A}{\partial x_i}
--
-(x^2 + q^4y^2 + p^4 z^2 ) \cdot \frac{\partial B}{\partial x_i} \, .
-</math>
-  </td>
-</tr>
-</table>
-<table border="1" cellpadding="8" align="center" width="80%"><tr><td align="left">
-Looking ahead &hellip;
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~h_3^{-2}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl\{ \frac{\lambda_3}{2AB} \biggl[
-B \cdot \frac{\partial A}{\partial x}
--
-A \cdot \frac{\partial B}{\partial x}
-\biggr] \biggr\}^2
-+
-\biggl\{ \frac{\lambda_3}{2AB} \biggl[
-B \cdot \frac{\partial A}{\partial y}
--
-A \cdot \frac{\partial B}{\partial y}
-\biggr] \biggr\}^2
-+
-\biggl\{ \frac{\lambda_3}{2AB} \biggl[
-B \cdot \frac{\partial A}{\partial z}
--
-A \cdot \frac{\partial B}{\partial z}
-\biggr] \biggr\}^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~ \biggl[\frac{2AB}{\lambda_3}  \biggr]^2 h_3^{-2}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl[
-B \cdot \frac{\partial A}{\partial x}
--
-A \cdot \frac{\partial B}{\partial x}
-\biggr]^2
-+
-\biggl[
-B \cdot \frac{\partial A}{\partial y}
--
-A \cdot \frac{\partial B}{\partial y}
-\biggr]^2
-+
-\biggl[
-B \cdot \frac{\partial A}{\partial z}
--
-A \cdot \frac{\partial B}{\partial z}
-\biggr]^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~ \biggl[\frac{\lambda_3}{2AB}  \biggr] h_3</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl\{\biggl[
-B \cdot \frac{\partial A}{\partial x}
--
-A \cdot \frac{\partial B}{\partial x}
-\biggr]^2
-+
-\biggl[
-B \cdot \frac{\partial A}{\partial y}
--
-A \cdot \frac{\partial B}{\partial y}
-\biggr]^2
-+
-\biggl[
-B \cdot \frac{\partial A}{\partial z}
--
-A \cdot \frac{\partial B}{\partial z}
-\biggr]^2
-\biggr\}^{-1 / 2}
-</math>
-  </td>
-</tr>
-</table>
-Then, for example,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\gamma_{31} \equiv h_3 \biggl(\frac{\partial \lambda_3}{\partial x} \biggr)</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\biggl[
-B \cdot \frac{\partial A}{\partial x}
--
-A \cdot \frac{\partial B}{\partial x}
-\biggr]
-\biggl\{\biggl[
-B \cdot \frac{\partial A}{\partial x}
--
-A \cdot \frac{\partial B}{\partial x}
-\biggr]^2
-+
-\biggl[
-B \cdot \frac{\partial A}{\partial y}
--
-A \cdot \frac{\partial B}{\partial y}
-\biggr]^2
-+
-\biggl[
-B \cdot \frac{\partial A}{\partial z}
--
-A \cdot \frac{\partial B}{\partial z}
-\biggr]^2
-\biggr\}^{-1 / 2}
-</math>
-  </td>
-</tr>
-</table>
-</td></tr></table>
-As a result, we have,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2x \biggl[
-(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
--
-(x^2 + q^4y^2 + p^4 z^2 ) ( 4q^4y^2 + p^4 z^2 ) \biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2x \biggl[
-(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
--
-(4x^2q^4y^2 + x^2 p^4 z^2 + 4q^8y^4 + q^4y^2 p^4z^2 + 4q^4y^2 p^4 z^2 + p^8z^4) \biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2x \biggl[
--
-(4q^8y^4  + 4q^4y^2 p^4 z^2 + p^8z^4) \biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~-2x (2q^4y^2  + p^4z^2)^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~-8x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln\lambda_3}{\partial \ln{x}}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~-\biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^2 \, ;
-</math>
-  </td>
-</tr>
-</table>
-and,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-q^4y\biggl[
-(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
--
-(x^2 + q^4y^2 + p^4 z^2 ) (p^4 z^2 + 4x^2)
-\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-q^4y\biggl[
-(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
--
-( x^2p^4z^2 + 4x^4 + q^4y^2p^4z^2 + 4x^2q^4y^2 + p^8z^4 + 4x^2p^4z^2)
-\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
--2q^4y( 4x^4 + p^8z^4 + 4x^2p^4z^2)
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
--2q^4y( 2x^2 + p^4z^2 )^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln\lambda_3}{\partial \ln{y} }</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-- \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^2 \, ;
-</math>
-  </td>
-</tr>
-</table>
-and,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2p^4 z \biggl[
-(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
--
-(x^2 + q^4y^2 + p^4 z^2 ) (q^4 y^2 + x^2)
-\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2p^4 z \biggl[
-(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
--
-(x^2q^4y^2 + x^4 + q^8y^4 + x^2q^4y^2 + q^4y^2p^4z^2 + x^2p^4z^2)
-\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2p^4 z \biggl[
-( 2x^2q^4y^2)
--
-( x^4 + q^8y^4 )
-\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~-2p^4 z \biggl[
-x^4 + q^8y^4
-- 2x^2q^4y^2
-\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~-2p^4 z (x^2 - q^4y^2 )^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln \lambda_3}{\partial \ln{z}}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~-4 \biggl[ \biggl( \frac{p^4 z^2}{4} \biggr) (x^2 - q^4y^2 )^2 \biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~-\biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^2
-\, .
-</math>
-  </td>
-</tr>
-</table>
-<font color="red">'''Wow! &nbsp; Really close!''' (13 November 2020)</font>
-Just for fun, let's see what we get for <math>~h_3</math>.  It is given by the expression,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~h_3^{-2}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl( \frac{\partial \lambda_3}{\partial x} \biggr)^2
-+\biggl( \frac{\partial \lambda_3}{\partial y} \biggr)^2
-+\biggl( \frac{\partial \lambda_3}{\partial z} \biggr)^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl\{ \frac{\lambda_3}{ABx} \biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^2 \biggr\}^2
-+\biggl\{ \frac{\lambda_3}{ABy} \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^2 \biggr\}^2
-+\biggl\{ \frac{\lambda_3}{ABz} \biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^2 \biggr\}^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow~~~ \biggl[ \frac{AB}{\lambda_3}\biggr]^2 h_3^{-2}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl\{ \frac{1}{x^2} \biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^4 \biggr\}
-+\biggl\{ \frac{1}{y^2} \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^4 \biggr\}
-+\biggl\{ \frac{1}{z^2} \biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^4 \biggr\}
-</math>
-  </td>
-</tr>
-</table>
-====Fiddle Around====
-Let &hellip;
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\mathcal{L}_x</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~
-- \biggl[
-B \cdot \frac{\partial A}{\partial x}
--
-A \cdot \frac{\partial B}{\partial x}
-\biggr]
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~8x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\frac{2}{x} \biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^2
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~8x~\mathfrak{F}_x(y,z)
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\mathcal{L}_y</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~
-- \biggl[
-B \cdot \frac{\partial A}{\partial y}
--
-A \cdot \frac{\partial B}{\partial y}
-\biggr]
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-q^4y\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{2}{y} \biggl[ 2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr)\biggr]^2
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~8y~\mathfrak{F}_y(x,z)
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\mathcal{L}_z</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~
--\biggl[
-B \cdot \frac{\partial A}{\partial z}
--
-A \cdot \frac{\partial B}{\partial z}
-\biggr]
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-p^4 z \biggl(x^2 - q^4y^2 \biggr)^2
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{2}{z} \biggl[p^2 z \biggl(x^2 - q^4y^2 \biggr)\biggr]^2
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~8z~\mathfrak{F}_z(x,y)
-</math>
-  </td>
-</tr>
-</table>
-With this shorthand in place, we can write,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\hat{e}_3</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\frac{\ell_{3D}}{\mathcal{D}}
-\biggl\{
--\hat\imath \biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]
-+ \hat\jmath \biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]
-+ \hat{k} \biggl[ p^2z( x^2 - q^4y^2 ) \biggl]
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{1}{(AB)^{1 / 2}}
-\biggl\{
--\hat\imath \biggl[ \frac{x \mathcal{L}_x}{2} \biggl]^{1 / 2}
-+ \hat\jmath \biggl[ \frac{y \mathcal{L}_y}{2} \biggl]^{1 / 2}
-+ \hat{k} \biggl[ \frac{z \mathcal{L}_z}{2} \biggl]^{1 / 2}
-\biggr\}
-\, .
-</math>
-  </td>
-</tr>
-</table>
-We therefore also recognize that,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial x}\biggr)</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
--\frac{1}{(AB)^{1 / 2}} \biggl[ \frac{x \mathcal{L}_x}{2} \biggl]^{1 / 2}
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
--\frac{1}{(AB)^{1 / 2}} \biggl[ 4x^2 ~\mathfrak{F}_x \biggl]^{1 / 2} \, ,
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial y}\biggr)</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{1}{(AB)^{1 / 2}} \biggl[ \frac{y \mathcal{L}_y}{2} \biggl]^{1 / 2}
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{1}{(AB)^{1 / 2}} \biggl[ 4y^2~\mathfrak{F}_y \biggl]^{1 / 2} \, ,
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial z}\biggr)</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{1}{(AB)^{1 / 2}} \biggl[ \frac{z \mathcal{L}_z}{2} \biggl]^{1 / 2}
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{1}{(AB)^{1 / 2}} \biggl[ 4z^2~\mathfrak{F}_z \biggl]^{1 / 2} \, .
-</math>
-  </td>
-</tr>
-</table>
-Now, if &#8212; and it is a BIG "if" &#8212; <math>~h_3 = h_0(AB)^{-1 / 2}</math>, then we have,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial x}\biggr)</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
--\biggl[ 4x^2 ~\mathfrak{F}_x \biggl]^{1 / 2}
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
--2x \biggl[ \mathfrak{F}_x \biggl]^{1 / 2} \, ,
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial y}\biggr)</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl[ 4y^2~\mathfrak{F}_y \biggl]^{1 / 2}
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-y \biggl[ \mathfrak{F}_y \biggl]^{1 / 2} \, ,
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial z}\biggr)</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl[ 4z^2~\mathfrak{F}_z \biggl]^{1 / 2}
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-z\biggl[ \mathfrak{F}_z \biggl]^{1 / 2} \, ,
-</math>
-  </td>
-</tr>
-</table>
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\Rightarrow~~~ h_0 \lambda_3</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
--x^2 \biggl[ \mathfrak{F}_x \biggl]^{1 / 2}
-+ y^2 \biggl[ \mathfrak{F}_y \biggl]^{1 / 2}
-+ z^2 \biggl[ \mathfrak{F}_z \biggl]^{1 / 2} \, .
-</math>
-  </td>
-</tr>
-</table>
-But if this is the correct expression for <math>~\lambda_3</math> and its three partial derivatives, then it must be true that,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~h_3^{-2}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl(\frac{\partial \lambda_3}{\partial x}\biggr)^2
-+
-\biggl(\frac{\partial \lambda_3}{\partial y}\biggr)^2
-+
-\biggl(\frac{\partial \lambda_3}{\partial z}\biggr)^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~ \biggl( \frac{h_3}{h_0}\biggr)^{-2}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-x^2 \biggl[ \mathfrak{F}_x \biggl]
-+
-y^2 \biggl[ \mathfrak{F}_y \biggl]
-+
-z^2\biggl[ \mathfrak{F}_z \biggl]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-x^2 \biggl[ \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2 \biggl]
-+
-y^2 \biggl[ q^4\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2\biggl]
-+
-z^2\biggl[ \frac{p^4}{4}\biggl(x^2 - q^4y^2 \biggr)^2 \biggl]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-x^2 (2q^4y^2  + p^4z^2 )^2
-+
-q^4 y^2( 2x^2 + p^4z^2 )^2
-+
-p^4 z^2 (x^2 - q^4y^2 )^2
-</math>
-  </td>
-</tr>
-</table>
-Well &hellip; the right-hand side of this expression is identical to the right-hand side of the [[#Eureka|above expression]], where we showed that it equals <math>~(\ell_{3D}/\mathcal{D})^{-2}</math>.  That is to say, we are now showing that,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\biggl( \frac{h_3}{h_0}\biggr)^{-2}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2} = [AB]</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~ \frac{h_3}{h_0}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~(AB)^{-1 / 2} \, .</math>
-  </td>
-</tr>
-</table>
-And this is ''precisely'' what, just a few lines above, we hypothesized the functional expression for <math>~h_3</math> ought to be.  <font color="red">'''EUREKA!'''</font>
-====Summary====
-In summary, then &hellip;
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\lambda_3</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~
--x^2 \biggl[ \mathfrak{F}_x \biggl]^{1 / 2}
-+ y^2 \biggl[ \mathfrak{F}_y \biggl]^{1 / 2}
-+ z^2 \biggl[ \mathfrak{F}_z \biggl]^{1 / 2}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
--x^2 \biggl[\biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2 \biggl]^{1 / 2}
-+ y^2 \biggl[ q^4\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2\biggl]^{1 / 2}
-+ z^2 \biggl[ \frac{p^4}{4} \biggl(x^2 - q^4y^2 \biggr)^2 \biggl]^{1 / 2}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
--x^2 \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)
-+ q^2y^2 \biggl( x^2 + \frac{p^4z^2}{2} \biggr)
-+ \frac{p^2 z^2}{2}  \biggl(x^2 - q^4y^2 \biggr) \, ,
-</math>
-  </td>
-</tr>
-</table>
-and,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~h_3</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~(AB)^{-1 / 2} </math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\biggl[
-(x^2 + q^4y^2 + p^4 z^2 )(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
-\biggr]^{-1 / 2} \, .</math>
-  </td>
-</tr>
-</table>
-No!  Once again this does not work.  The direction cosines &#8212; and, hence, the components of the <math>~\hat{e}_3</math> unit vector &#8212; are not correct!
-===Speculation7===
-<table border="1" cellpadding="8" align="center">
-<tr>
-  <td align="center" colspan="9">'''Direction Cosine Components for T6 Coordinates'''</td>
-</tr>
-<tr>
-  <td align="center"><math>~n</math></td>
-  <td align="center"><math>~\lambda_n</math></td>
-  <td align="center"><math>~h_n</math></td>
-  <td align="center"><math>~\frac{\partial \lambda_n}{\partial x}</math></td>
-  <td align="center"><math>~\frac{\partial \lambda_n}{\partial y}</math></td>
-  <td align="center"><math>~\frac{\partial \lambda_n}{\partial z}</math></td>
-  <td align="center"><math>~\gamma_{n1}</math></td>
-  <td align="center"><math>~\gamma_{n2}</math></td>
-  <td align="center"><math>~\gamma_{n3}</math></td>
-</tr>
-<tr>
-  <td align="center"><math>~1</math></td>
-  <td align="center"><math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math></td>
-  <td align="center"><math>~\lambda_1 \ell_{3D}</math></td>
-  <td align="center"><math>~\frac{x}{\lambda_1}</math></td>
-  <td align="center"><math>~\frac{q^2 y}{\lambda_1}</math></td>
-  <td align="center"><math>~\frac{p^2 z}{\lambda_1}</math></td>
-  <td align="center"><math>~(x) \ell_{3D}</math></td>
-  <td align="center"><math>~(q^2 y)\ell_{3D}</math></td>
-  <td align="center"><math>~(p^2z) \ell_{3D}</math></td>
-</tr>
-<tr>
-  <td align="center"><math>~2</math></td>
-  <td align="center"><math>~\frac{x y^{1/q^2}}{ z^{2/p^2}}</math></td>
-  <td align="center"><math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y p^2 z}{ \mathcal{D}}\biggr] </math></td>
-  <td align="center"><math>~\frac{\lambda_2}{x}</math></td>
-  <td align="center"><math>~\frac{\lambda_2}{q^2 y}</math></td>
-  <td align="center"><math>~-\frac{2\lambda_2}{p^2 z}</math></td>
-  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{x}\biggr)</math></td>
-  <td align="center"><math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{q^2y}\biggr)</math></td>
-  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(-\frac{2}{p^2z}\biggr)</math></td>
-</tr>
-<tr>
-  <td align="center"><math>~3</math></td>
-  <td align="center">---</td>
-  <td align="center">---</td>
-  <td align="center">---</td>
-  <td align="center">---</td>
-  <td align="center">---</td>
-  <td align="center"><math>~-\frac{\ell_{3D}}{\mathcal{D}}\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]</math></td>
-  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]</math></td>
-  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]</math></td>
-</tr>
-<tr>
-  <td align="left" colspan="9">
-<table border="0" cellpadding="8" align="center">
-<tr>
-  <td align="right">
-<math>~\ell_{3D}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\mathcal{D}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>
-  </td>
-</tr>
-</table>
-  </td>
-</tr>
-</table>
-On my white-board I have shown that, if
-<div align="center">
-<math>~\lambda_3 \equiv \ell_{3D} \mathcal{D} \, ,</math>
-</div>
-then everything will work out as long as,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\mathcal{L}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl(\frac{\mathcal{D}}{\ell_{3D}} \biggr)^2 \frac{1}{\ell_{3D}^4} \, ,
-</math>
-  </td>
-</tr>
-</table>
-where,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\mathcal{L}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~
-x^2 (2q^4 y^2 + p^4z^2 )^4
-+
-q^8 y^2 (2x^2 + p^4 z^2)^4
-+
-p^8z^2( x^2 - q^4y^2)^4
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-x^2 (4q^8 y^4 + 4q^4y^2p^4z^2 + p^8z^4 )^2
-+
-q^8 y^2 (4x^4 + 4x^2p^4z^2 + p^8 z^4)^2
-+
-p^8z^2( x^4 - 2x^2q^4y^2 + q^8y^4)^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-x^2
-[16q^{16}y^8 + 16q^{12}y^6p^4z^2 + 4q^8y^4p^8z^4 + 16q^{12}y^6p^4z^2 + 16q^8y^4p^8z^4 + 4q^4y^2p^{12}z^6 + 4q^8y^4p^8z^4 + 4q^4y^2 p^{12}z^6 + p^{16}z^8]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
-+
-q^8 y^2
-[16x^8 + 16x^6p^4z^2 + 4x^4p^8z^4 + 16x^6p^4z^2 + 16x^4p^8z^4 + 4x^2p^{12}z^6 + 4x^4p^8z^4 + 4x^2p^{12}z^6 + p^{16}z^8]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
-+
-p^8z^2
-[x^8 - 2x^6q^4y^2 + x^4q^8y^4 - 2x^6q^4y^2 + 4x^4q^8y^4 - 2x^2q^{12}y^6 + x^4q^8y^4 - 2x^2q^{12}y^6 + q^{16}y^8]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-x^2
-[16q^{16}y^8 + 32q^{12}y^6p^4z^2 + 24q^8y^4p^8z^4 + 8q^4y^2p^{12}z^6 + p^{16}z^8]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
-+
-q^8 y^2
-[16x^8 + 32x^6p^4z^2 + 24x^4p^8z^4  + 8x^2p^{12}z^6 + p^{16}z^8]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
-+
-p^8z^2
-[x^8 - 4x^6q^4y^2 + 6x^4q^8y^4 - 4x^2q^{12}y^6 + q^{16}y^8]
-</math>
-  </td>
-</tr>
-</table>
-Let's check this out.
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\mathrm{RHS}~\equiv \biggl(\frac{\mathcal{D}}{\ell_{3D}} \biggr)^2 \frac{1}{\ell_{3D}^4}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)(x^2 + q^4y^2 + p^4 z^2 )^3</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~(x^2 + q^4y^2 + p^4 z^2 )(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)[x^4 + 2x^2q^4y^2 + 2x^2p^4z^2 +q^8y^4 + 2q^4y^2p^4z^2 + p^8z^4]</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-[(6x^2q^4 y^2 p^4 z^2 + x^4 p^4 z^2 + 4x^4q^4y^2)
-+
-(q^8 y^4 p^4 z^2 + 4x^2q^8 y^4)
-+
-(q^4 y^2 p^8 z^4 + x^2 p^8 z^4 )]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~\times [x^4 + 2x^2q^4y^2 + 2x^2p^4z^2 +q^8y^4 + 2q^4y^2p^4z^2 + p^8z^4]</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-[6x^2q^4 y^2 p^4 z^2 + x^4(p^4 z^2 + 4q^4y^2)
-+
-q^8 y^4(p^4 z^2 + 4x^2)
-+
-p^8 z^4 (q^4 y^2 + x^2 )]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~\times [x^4 + 2x^2q^4y^2 + 2x^2p^4z^2 +q^8y^4 + 2q^4y^2p^4z^2 + p^8z^4]</math>
-  </td>
-</tr>
-</table>
-==Best Thus Far==
-===Part A===
-<table border="1" cellpadding="8" align="center">
-<tr>
-  <td align="center" colspan="9">'''Direction Cosine Components for T6 Coordinates'''</td>
-</tr>
-<tr>
-  <td align="center"><math>~n</math></td>
-  <td align="center"><math>~\lambda_n</math></td>
-  <td align="center"><math>~h_n</math></td>
-  <td align="center"><math>~\frac{\partial \lambda_n}{\partial x}</math></td>
-  <td align="center"><math>~\frac{\partial \lambda_n}{\partial y}</math></td>
-  <td align="center"><math>~\frac{\partial \lambda_n}{\partial z}</math></td>
-  <td align="center"><math>~\gamma_{n1}</math></td>
-  <td align="center"><math>~\gamma_{n2}</math></td>
-  <td align="center"><math>~\gamma_{n3}</math></td>
-</tr>
-<tr>
-  <td align="center"><math>~1</math></td>
-  <td align="center"><math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math></td>
-  <td align="center"><math>~\lambda_1 \ell_{3D}</math></td>
-  <td align="center"><math>~\frac{x}{\lambda_1}</math></td>
-  <td align="center"><math>~\frac{q^2 y}{\lambda_1}</math></td>
-  <td align="center"><math>~\frac{p^2 z}{\lambda_1}</math></td>
-  <td align="center"><math>~(x) \ell_{3D}</math></td>
-  <td align="center"><math>~(q^2 y)\ell_{3D}</math></td>
-  <td align="center"><math>~(p^2z) \ell_{3D}</math></td>
-</tr>
-<tr>
-  <td align="center"><math>~2</math></td>
-  <td align="center"><math>~\frac{x y^{1/q^2}}{ z^{2/p^2}}</math></td>
-  <td align="center"><math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y p^2 z}{ \mathcal{D}}\biggr] </math></td>
-  <td align="center"><math>~\frac{\lambda_2}{x}</math></td>
-  <td align="center"><math>~\frac{\lambda_2}{q^2 y}</math></td>
-  <td align="center"><math>~-\frac{2\lambda_2}{p^2 z}</math></td>
-  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{x}\biggr)</math></td>
-  <td align="center"><math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{q^2y}\biggr)</math></td>
-  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(-\frac{2}{p^2z}\biggr)</math></td>
-</tr>
-<tr>
-  <td align="center"><math>~3</math></td>
-  <td align="center">---</td>
-  <td align="center">---</td>
-  <td align="center">---</td>
-  <td align="center">---</td>
-  <td align="center">---</td>
-  <td align="center"><math>~-\frac{\ell_{3D}}{\mathcal{D}}\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]</math></td>
-  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]</math></td>
-  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]</math></td>
-</tr>
-<tr>
-  <td align="left" colspan="9">
-<table border="0" cellpadding="8" align="center">
-<tr>
-  <td align="right">
-<math>~\ell_{3D}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\mathcal{D}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>
-  </td>
-</tr>
-</table>
-  </td>
-</tr>
-</table>
-<span id="ABderivatives">&nbsp;</span>
-<table border="1" cellpadding="8" align="center" width="80%"><tr><td align="left">
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~A \equiv \ell_{3D}^{-2}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~(x^2 + q^4y^2 + p^4 z^2 ) \, ,</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
-  <td align="right">
-<math>~B \equiv \mathcal{D}^2</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)</math>
-  </td>
-</tr>
-</table>
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~ \frac{\partial A}{\partial x}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2x \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial A}{\partial y}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2q^4 y \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial A}{\partial z}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~ 2p^4 z\, ;</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~ \frac{\partial B}{\partial x}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2x( 4q^4y^2 + p^4 z^2 ) \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial B}{\partial y}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2q^4 y (p^4 z^2 + 4x^2) \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial B}{\partial z}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~ 2p^4 z(q^4 y^2 + x^2)\, .</math>
-  </td>
-</tr>
-</table>
-</td></tr></table>
-Try &hellip;
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\lambda_3</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl(\frac{B}{A}\biggr)^{m/2} = (D \ell_{3D})^m
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~\biggl[ \frac{AB}{m\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x_i}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{1}{2}\biggl\{
-A \cdot \frac{\partial B}{\partial x_i} - B \cdot \frac{\partial A}{\partial x_i}
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~\biggl[ \frac{AB}{m} \biggr]  \frac{\partial \ln \lambda_3}{\partial \ln x_i}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{x_i}{2}\biggl\{
-A \cdot \frac{\partial B}{\partial x_i} - B \cdot \frac{\partial A}{\partial x_i}
-\biggr\} \, .
-</math>
-  </td>
-</tr>
-</table>
-In this case we find,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\frac{x}{2}\biggl\{~~\biggr\}_x</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-x^2(2q^4 y^2 + p^4z^2)^2 \, ,
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{y}{2}\biggl\{~~\biggr\}_y</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-q^4y^2(2x^2 + p^4 z^2)^2 \, ,
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{z}{2}\biggl\{~~\biggr\}_z</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-p^4 z^2(x^2 - q^4y^2)^2 \, .
-</math>
-  </td>
-</tr>
-</table>
-The scale factor is, then,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~h_3^{-2}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\sum_{i=1}^3 \biggl( \frac{\partial\lambda_3}{\partial x_i}\biggr)^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\sum_{i=1}^3 \biggl\{
-\biggl[ \frac{m\lambda_3}{2AB} \biggr] \biggl[
-A \cdot \frac{\partial B}{\partial x_i} - B \cdot \frac{\partial A}{\partial x_i}
-\biggr]
-\biggr\}^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl[ \frac{m\lambda_3}{AB} \biggr]^2
-\biggl\{
-\biggl[
-x(2q^4 y^2 + p^4z^2)^2
-\biggr]^2
-+
-\biggl[
-q^4y(2x^2 + p^4 z^2)^2
-\biggr]^2
-+
-\biggl[
-p^4 z(x^2 - q^4y^2)^2
-\biggr]^2
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow~~~h_3</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl[ \frac{AB}{m\lambda_3} \biggr]
-\biggl\{
-\biggl[
-x(2q^4 y^2 + p^4z^2)^2
-\biggr]^2
-+
-\biggl[
-q^4y(2x^2 + p^4 z^2)^2
-\biggr]^2
-+
-\biggl[
-p^4 z(x^2 - q^4y^2)^2
-\biggr]^2
-\biggr\}^{-1 / 2} \, .
-</math>
-  </td>
-</tr>
-</table>
-===Part B (25 February 2021)===
-Now, from [[#Eureka|above]], we know that,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\biggl( \frac{\mathcal{D}}{\ell_{3D}} \biggr)^{2} = AB</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]^2
-+
-\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]^2
-+
-\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]^2 \, .
-</math>
-  </td>
-</tr>
-</table>
-<table border="1" align="center" cellpadding="8">
-<tr>
-  <td align="center" colspan="4">
-'''Example:'''  <br /><math>~(q^2, p^2) = (2, 5)</math> &nbsp; &nbsp; and &nbsp; &nbsp; <math>~(x, y, z) = (0.7, \sqrt{0.23}, 0.1)~~\Rightarrow~~ \lambda_1 = 1.0</math>
-  </td>
-</tr>
-<tr>
-  <td align="center"><math>~\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]^2</math></td>
-  <td align="center"><math>~\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]^2</math></td>
-  <td align="center"><math>~\biggl[ p^2z( x^2 - q^4y^2 )  \biggl]^2</math></td>
-  <td align="center"><math>~\biggl( \frac{\mathcal{D}}{\ell_{3D}} \biggr)^{2}</math></td>
-</tr>
-<tr>
-  <td align="center">2.14037</td>
-  <td align="center">1.39187</td>
-  <td align="center">0.04623</td>
-  <td align="center">3.57847</td>
-</tr>
-</table>
-As an aside, note that,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~AB</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl[ \frac{AB}{m} \biggr]  \frac{\partial \ln \lambda_3}{\partial \ln x}
-+
-\biggl[ \frac{AB}{m} \biggr]  \frac{\partial \ln \lambda_3}{\partial \ln y}
-+
-\biggl[ \frac{AB}{m} \biggr]  \frac{\partial \ln \lambda_3}{\partial \ln z}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~ m</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{\partial \ln \lambda_3}{\partial \ln x}
-+
-\frac{\partial \ln \lambda_3}{\partial \ln y}
-+
-\frac{\partial \ln \lambda_3}{\partial \ln z} \, .
-</math>
-  </td>
-</tr>
-</table>
-We realize that this ratio of lengths may also be written in the form,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\biggl( \frac{\mathcal{D}}{\ell_{3D}} \biggr)^{2} </math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-x^2 q^4y^2 p^4 z^2
-+
-x^4(4q^4y^2 + p^4z^2)
-+
-q^8y^4(4x^2 + p^4z^2)
-+
-p^8z^4(x^2 + q^4y^2) \, .
-</math>
-  </td>
-</tr>
-</table>
-<table border="1" align="center" cellpadding="8">
-<tr>
-  <td align="center" colspan="5">
-'''Same Example, but Different Expression:'''  <br /><math>~(q^2, p^2) = (2, 5)</math> &nbsp; &nbsp; and &nbsp; &nbsp; <math>~(x, y, z) = (0.7, \sqrt{0.23}, 0.1)~~\Rightarrow~~ \lambda_1 = 1.0</math>
-  </td>
-</tr>
-<tr>
-  <td align="center"><math>~6x^2 q^4y^2 p^4 z^2</math></td>
-  <td align="center"><math>~x^4(4q^4y^2 + p^4z^2)</math></td>
-  <td align="center"><math>~q^8y^4(4x^2 + p^4z^2)</math></td>
-  <td align="center"><math>~p^8z^4(x^2 + q^4y^2)</math></td>
-  <td align="center"><math>~\biggl( \frac{\mathcal{D}}{\ell_{3D}} \biggr)^{2}</math></td>
-</tr>
-<tr>
-  <td align="center">0.67620</td>
-  <td align="center">0.94359</td>
-  <td align="center">1.87054</td>
-  <td align="center">0.08813</td>
-  <td align="center">3.57847</td>
-</tr>
-</table>
-Let's try &hellip;
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\frac{\partial \lambda_5}{\partial x}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~-x(2 q^4y^2 + p^4z^2 ) \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial \lambda_5}{\partial y}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~q^2 y(p^4z^2 + 2x^2 ) \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial \lambda_5}{\partial z}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~p^2z( x^2 - q^4y^2 ) \, .</math>
-  </td>
-</tr>
-</table>
-This means that the relevant scale factor is,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~h_5^{-2}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl[ -x(2 q^4y^2 + p^4z^2 ) \biggr]^2
-+
-\biggl[ q^2 y(p^4z^2 + 2x^2 ) \biggr]^2
-+
-\biggl[ p^2z( x^2 - q^4y^2 ) \biggr]^2
-=
-\biggl( \frac{\mathcal{D}}{\ell_{3D}} \biggr)^{2}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow~~~h_5</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr) \, ,
-</math>
-  </td>
-</tr>
-</table>
-and the three associated direction cosines are,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\gamma_{51} = h_5 \biggl( \frac{\partial \lambda_5}{\partial x} \biggr)</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~-x(2 q^4y^2 + p^4z^2 )\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr) \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\gamma_{52} = h_5 \biggl( \frac{\partial \lambda_5}{\partial y} \biggr)</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~q^2 y(p^4z^2 + 2x^2 )\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr) \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\gamma_{53} = h_5 \biggl( \frac{\partial \lambda_5}{\partial z} \biggr)</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~p^2z( x^2 - q^4y^2 )\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr) \, .</math>
-  </td>
-</tr>
-</table>
-<span id="PartBCoordinatesT10">These direction cosines</span> exactly match what is required in order to ensure that the coordinate, <math>~\lambda_5</math>, is everywhere orthogonal to both <math>~\lambda_1</math> and <math>~\lambda_4</math>.  <font color="red">'''GREAT!'''</font>  The resulting summary table is, therefore:
-<table border="1" cellpadding="8" align="center">
-<tr>
-  <td align="center" colspan="9">'''Direction Cosine Components for T10 Coordinates'''</td>
-</tr>
-<tr>
-  <td align="center"><math>~n</math></td>
-  <td align="center"><math>~\lambda_n</math></td>
-  <td align="center"><math>~h_n</math></td>
-  <td align="center"><math>~\frac{\partial \lambda_n}{\partial x}</math></td>
-  <td align="center"><math>~\frac{\partial \lambda_n}{\partial y}</math></td>
-  <td align="center"><math>~\frac{\partial \lambda_n}{\partial z}</math></td>
-  <td align="center"><math>~\gamma_{n1}</math></td>
-  <td align="center"><math>~\gamma_{n2}</math></td>
-  <td align="center"><math>~\gamma_{n3}</math></td>
-</tr>
-<tr>
-  <td align="center"><math>~1</math></td>
-  <td align="center"><math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math></td>
-  <td align="center"><math>~\lambda_1 \ell_{3D}</math></td>
-  <td align="center"><math>~\frac{x}{\lambda_1}</math></td>
-  <td align="center"><math>~\frac{q^2 y}{\lambda_1}</math></td>
-  <td align="center"><math>~\frac{p^2 z}{\lambda_1}</math></td>
-  <td align="center"><math>~(x) \ell_{3D}</math></td>
-  <td align="center"><math>~(q^2 y)\ell_{3D}</math></td>
-  <td align="center"><math>~(p^2z) \ell_{3D}</math></td>
-</tr>
-<tr>
-  <td align="center"><math>~4</math></td>
-  <td align="center"><math>~\frac{x y^{1/q^2}}{ z^{2/p^2}}</math></td>
-  <td align="center"><math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y p^2 z}{ \mathcal{D}}\biggr] </math></td>
-  <td align="center"><math>~\frac{\lambda_2}{x}</math></td>
-  <td align="center"><math>~\frac{\lambda_2}{q^2 y}</math></td>
-  <td align="center"><math>~-\frac{2\lambda_2}{p^2 z}</math></td>
-  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{x}\biggr)</math></td>
-  <td align="center"><math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{q^2y}\biggr)</math></td>
-  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(-\frac{2}{p^2z}\biggr)</math></td>
-</tr>
-<tr>
-  <td align="center"><math>~5</math></td>
-  <td align="center">---</td>
-  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}</math></td>
-  <td align="center"><math>~-x(2 q^4y^2 + p^4z^2)</math></td>
-  <td align="center"><math>~q^2 y(p^4z^2 + 2x^2 )</math></td>
-  <td align="center"><math>~p^2z( x^2 - q^4y^2 )</math></td>
-  <td align="center"><math>~-\frac{\ell_{3D}}{\mathcal{D}}\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]</math></td>
-  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]</math></td>
-  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]</math></td>
-</tr>
-<tr>
-  <td align="left" colspan="9">
-<table border="0" cellpadding="8" align="center">
-<tr>
-  <td align="right">
-<math>~\ell_{3D}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} \, ,</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\mathcal{D}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>
-  </td>
-</tr>
-</table>
-  </td>
-</tr>
-</table>
-Try &hellip;
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\lambda_5</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-x^{-2q^4} \cdot y^{2q^2}
-+
-y^{q^2p^4} \cdot z^{-q^4p^2}
-+
-x^{-p^4} \cdot z^{p^2}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{ y^{2q^2} }{ x^{2q^4} }
-+
-\frac{ y^{q^2p^4} }{ z^{q^4p^2} }
-+
-\frac{ z^{p^4} }{ x^{p^2} }
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{1}{x^{2q^4 + p^2} z^{q^4p^2} }\biggl\{
-[x^{p^2}] y^{2q^2} [z^{q^4p^2}]
-+
-[x^{2q^4 + p^2}]y^{q^2p^4}
-+
-[x^{2q^4}]z^{p^4+q^4p^2}
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{1}{x^{2q^4 + p^2} z^{q^4p^2} }\biggl\{
-\mathfrak{F_5}
-\biggr\} \, .
-</math>
-  </td>
-</tr>
-</table>
-This gives,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\frac{\partial \lambda_5}{\partial x}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
--\frac{2q^4}{x}\biggl( \frac{ y^{2q^2} }{ x^{2q^4} } \biggr)
-- \frac{p^4}{x}\biggl( \frac{ z^{p^2} }{ x^{p^2} } \biggr)
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-- \frac{1}{x^{2q^4 + p^2 + 1}}\biggl[
-q^4y^{2q^2} x^{p^2}
-+ p^4 x^{2q^4}z^{p^2}
-\biggr] \, .
-</math>
-  </td>
-</tr>
-</table>
-Or, given that,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~x^{2q^4 + p^2 + 1} </math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{x}{ z^{q^4p^2} }\biggl\{
-\frac{\mathfrak{F_5}}{\lambda_5}
-\biggr\}
-\, ,
-</math>
-  </td>
-</tr>
-</table>
-we can also write,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\frac{\partial \lambda_5}{\partial x}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-- \frac{ z^{q^4p^2} }{x}\biggl\{
-\frac{\lambda_5}{\mathfrak{F_5}}
-\biggr\}
-\biggl[ 2q^4y^{2q^2} x^{p^2} + p^4 x^{2q^4}z^{p^2} \biggr]
-</math>
-  </td>
-</tr>
-</table>
-Similarly,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\frac{\partial \lambda_5}{\partial y}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{2q^2}{y} \biggl( \frac{ y^{2q^2} }{ x^{2q^4} } \biggr)
-+
-\frac{q^2p^4}{y} \biggl( \frac{ y^{q^2p^4} }{ z^{q^4p^2} } \biggr)
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\frac{1}{ x^{2q^4} z^{q^4p^2} }
-\biggl[ \frac{2q^2}{y} \biggl( y^{2q^2} z^{q^4p^2} \biggr)
-+
-\frac{q^2p^4}{y} \biggl( y^{q^2p^4} x^{2q^4} \biggr)\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\frac{x^{p^2}}{ y } \biggl\{ \frac{\lambda_5}{\mathfrak{F_5}} \biggr\}
-\biggl[ 2q^2 \biggl( y^{2q^2} z^{q^4p^2} \biggr)
-+
-q^2p^4 \biggl( y^{q^2p^4} x^{2q^4} \biggr)\biggr] \, ;
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\frac{\partial \lambda_5}{\partial z}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-- \frac{q^4p^2}{z} \biggl( \frac{ y^{q^2p^4} }{ z^{q^4p^2} } \biggr)
-+
-\frac{p^4}{z} \biggl( \frac{ z^{p^4} }{ x^{p^2} } \biggr)
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{1}{ x^{p^2} z^{q^4p^2 }} \biggl[
-\frac{p^4}{z} \biggl( z^{p^4+q^4p^2}  \biggr)
-- \frac{q^4p^2}{z} \biggl( x^{p^2} y^{q^2p^4} \biggr) \biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~ \frac{x^{2q^4} }{z} \biggl\{ \frac{\lambda_5}{\mathfrak{F_5}} \biggr\} \biggl[
-p^4 \biggl( z^{p^4 + q^4p^2 }  \biggr)
--
-q^4p^2 \biggl( x^{p^2} y^{q^2p^4}  \biggr)\biggr]
-</math>
-  </td>
-</tr>
-</table>
-===Understanding the Volume Element===
-Let's see if the expression for the volume element makes sense; that is, does
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~(h_1 h_4 h_5) d\lambda_1 d\lambda_4 d\lambda_5</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~dx dy dz \, ?</math>
-  </td>
-</tr>
-</table>
-First, let's make sure that we understand how to relate the components of the Cartesian line element with the components of our T10 coordinates.
-====Line Element====
-MF53 claim that the following relation gives the various expressions for the scale factors; we will go ahead and incorporate the expectation that, since our coordinate system is orthogonal, the off-diagonal elements are zero.
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~ds^2 = dx^2 + dy^2 + dz^2</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\sum_{i=1,4,5} h_i^2 d\lambda_i^2 \, .
-</math>
-  </td>
-</tr>
-</table>
-Let's see.  The first term on the RHS is,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~h_1^2 d\lambda_1^2</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-h_1^2 \biggl[
-\biggl( \frac{\partial \lambda_1}{\partial x}\biggr)dx
-+
-\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)dy
-+
-\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)dz
-\biggr]^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-h_1^2 \biggl[
-\biggl( \frac{\partial \lambda_1}{\partial x}\biggr)^2 dx^2
-+
-\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 dy^2
-+
-\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 dz^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
-+
-\cancel{2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)} dx~dy
-+
-\cancel{2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)} dx~dz
-+
-\cancel{2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)} dy~dz
-\biggr] \, ;
-</math>
-  </td>
-</tr>
-</table>
-the other two terms assume easily deduced, similar forms.  When put together and after regrouping terms, we can write,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~
-\sum_{i=1,4,5} h_i^2 d\lambda_i^2
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl[
-h_1^2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)^2
-+
-h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2
-+
-h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2
-\biggr] dx^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
-+ \biggl[
-h_1^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-+
-h_4^2\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2
-+
-h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2
-\biggr] dy^2
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
-+ \biggl[
-h_1^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
-+
-h_4^2\biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2
-+
-h_5^2 \biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2
-\biggr] dz^2 \, .
-</math>
-  </td>
-</tr>
-</table>
-Given that this summation should also equal the square of the Cartesian line element, <math>~(dx^2 + dy^2 + dz^2)</math>, we conclude that the three terms enclosed inside each of the pair of brackets must sum to unity.  Specifically, from the coefficient of <math>~dx^2</math>, we can write,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~h_1^2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)^2
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
--
-h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2
--
-h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \, .
-</math>
-  </td>
-</tr>
-</table>
-Using this relation to replace <math>~h_1^2</math> in each of the other two bracketed expressions, we find for the coefficients of <math>~dy^2</math> and <math>~dz^2</math>, respectively,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\biggl[
-- h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2
-\biggr]
-\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl[ 1
--
-h_4^2\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2
--
-h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2
-\biggr]\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \, ;
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\biggl[
-- h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2
-\biggr]
-\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl[ 1
--
-h_4^2\biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2
--
-h_5^2 \biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2
-\biggr]\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \, .
-</math>
-  </td>
-</tr>
-</table>
-We can use the first of these two expressions to solve for <math>~h_4^2</math> in terms of <math>~h_5^2</math>, namely,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-- h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-- h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
--
-h_4^2\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
--
-h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2
-\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~
-h_4^2
-\biggl[
-\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-\biggr]
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
--
-\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-+ h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
--
-h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-</math>
-  </td>
-</tr>
-</table>
-Analogously, the second of these two expressions gives,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~
-h_4^2 \biggl[ \biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
-\biggr]
- </math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
--
-\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
-+ h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
--
-h_5^2 \biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2
-\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-</math>
-  </td>
-</tr>
-</table>
-Eliminating <math>~h_4</math> between the two gives the desired overall expression for <math>~h_5</math>, namely,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~0
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\biggl[
-\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
--
-\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
-+ h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
--
-h_5^2 \biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2
-\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-\biggr]
-\biggl[
-\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
--
-\biggl[
-\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
--
-\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-+ h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
--
-h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-\biggr]
-\biggl[ \biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
-\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-h_5^2 \biggl\{
-\biggl[ \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
--
-\biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \biggr]
-\biggl[ \biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
-\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
--
-\biggl[ \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
--
-\biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2
-\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \biggr]
-\biggl[
-\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-\biggr]
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
--\biggl[ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
--
-\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 \biggr]
-\biggl[
-\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
-+
-\biggl[ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
--
-\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2  \biggr]
-\biggl[ \biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
-\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{h_5^2}{h_1^4 h_4^2 h_5^2}\biggl\{
-\biggl[ \gamma_{51}^2 \gamma_{12}^2
--
-\gamma_{52}^2 \gamma_{11}^2 \biggr]
-\biggl[ \gamma_{43}^2 \gamma_{11}^2
-- \gamma_{41}^2 \gamma_{13}^2
-\biggr]
--
-\biggl[ \gamma_{51}^2 \gamma_{13}^2
--
-\gamma_{53}^2 \gamma_{11}^2
-\biggr]
-\biggl[
-\gamma_{42}^2 \gamma_{11}^2
-- \gamma_{41}^2 \gamma_{12}^2
-\biggr]
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
-+ \frac{1}{h_4^2 h_1^2}\biggl\{
-\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-\biggl[ \gamma_{43}^2 \gamma_{11}^2
-- \gamma_{41}^2 \gamma_{13}^2
-\biggr]
--
-\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-\biggl[ \gamma_{43}^2\gamma_{11}^2
--~ \gamma_{41}^2 \gamma_{13}^2
-\biggr]
-- \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-\biggl[
-\gamma_{42}^2 \gamma_{11}^2
-- \gamma_{41}^2 \gamma_{12}^2
-\biggr]
-+ \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
-\biggl[
-\gamma_{42}^2 \gamma_{11}^2
-- \gamma_{41}^2\gamma_{12}^2
-\biggr]
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{h_5^2}{h_1^4 h_4^2 h_5^2}\biggl\{
-\biggl[ \gamma_{51}^2 \gamma_{12}^2
--
-\gamma_{52}^2 \gamma_{11}^2 \biggr]
-\biggl[ \gamma_{43} \gamma_{11}
-+ \gamma_{41} \gamma_{13}
-\biggr]
-\biggl[ \gamma_{43} \gamma_{11}
-- \gamma_{41} \gamma_{13}
-\biggr]
--
-\biggl[ \gamma_{51}^2 \gamma_{13}^2
--
-\gamma_{53}^2 \gamma_{11}^2
-\biggr]
-\biggl[
-\gamma_{42} \gamma_{11}
-+ \gamma_{41} \gamma_{12}
-\biggr]
-\biggl[
-\gamma_{42} \gamma_{11}
-- \gamma_{41} \gamma_{12}
-\biggr]
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
-+ \frac{1}{h_4^2 h_1^2}\biggl\{
-\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-\biggl[ \gamma_{43} \gamma_{11}
-+ \gamma_{41} \gamma_{13}
-\biggr]
-\biggl[ \gamma_{43} \gamma_{11}
-- \gamma_{41} \gamma_{13}
-\biggr]
--
-\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-\biggl[ \gamma_{43}\gamma_{11}
-+ \gamma_{41} \gamma_{13}
-\biggr]
-\biggl[ \gamma_{43}\gamma_{11}
--~ \gamma_{41} \gamma_{13}
-\biggr]
-- \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-\biggl[
-\gamma_{42} \gamma_{11}
-+ \gamma_{41} \gamma_{12}
-\biggr]
-\biggl[
-\gamma_{42} \gamma_{11}
-- \gamma_{41} \gamma_{12}
-\biggr]
-+ \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
-\biggl[
-\gamma_{42} \gamma_{11}
-+ \gamma_{41}\gamma_{12}
-\biggr]
-\biggl[
-\gamma_{42} \gamma_{11}
-- \gamma_{41}\gamma_{12}
-\biggr]
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{1}{h_1^4 h_4^2 }\biggl\{
--
-\biggl[ \gamma_{51} \gamma_{12}
-+
-\gamma_{52} \gamma_{11} \biggr]
-\gamma_{43}
-\biggl[ \gamma_{43} \gamma_{11}
-+ \gamma_{41} \gamma_{13}
-\biggr]
-\gamma_{52}
-+
-\biggl[ \gamma_{51} \gamma_{13}
-+
-\gamma_{53} \gamma_{11}
-\biggr]
-\gamma_{42}
-\biggl[
-\gamma_{42} \gamma_{11}
-+ \gamma_{41} \gamma_{12}
-\biggr] \gamma_{53}
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-&nbsp;
-  </td>
-  <td align="left">
-<math>~
-+ \frac{1}{ h_1^4 h_4^2}\biggl\{
-\gamma_{12}^2
-\biggl[ \gamma_{43}\gamma_{11}
-+ \gamma_{41} \gamma_{13}
-\biggr]\gamma_{52}
--\gamma_{11}^{2}
-\biggl[ \gamma_{43} \gamma_{11}
-+ \gamma_{41} \gamma_{13}
-\biggr]\gamma_{52}
-- \gamma_{11}^{2}
-\biggl[
-\gamma_{42} \gamma_{11}
-+ \gamma_{41} \gamma_{12}
-\biggr]\gamma_{53}
-+ \gamma_{13}^2
-\biggl[
-\gamma_{42} \gamma_{11}
-+ \gamma_{41}\gamma_{12}
-\biggr]\gamma_{53}
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{1}{h_1^4 h_4^2 }\biggl\{
-\biggl[
-(- \gamma_{51} \gamma_{12}
--
-\gamma_{52} \gamma_{11} )
-\gamma_{43}
-+
-\gamma_{12}^2
--\gamma_{11}^{2}
-\biggr]
-(\gamma_{43} \gamma_{11}
-+ \gamma_{41} \gamma_{13} )
-\gamma_{52}
-+
-\biggl[
-(\gamma_{51} \gamma_{13}
-+
-\gamma_{53} \gamma_{11} )
-\gamma_{42}
-- \gamma_{11}^{2}
-+ \gamma_{13}^2
-\biggr]
-(\gamma_{42} \gamma_{11}
-+ \gamma_{41}\gamma_{12} )
-\gamma_{53}
-\biggr\}
-</math>
-  </td>
-</tr>
-</table>
-&hellip; Not sure this is headed anywhere useful!
-====Volume Element====
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~(h_1 h_4 h_5) d\lambda_1 d\lambda_4 d\lambda_5</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-(h_1 h_4 h_5)
-\biggl[
-\biggl( \frac{\partial \lambda_1}{\partial x}\biggr) dx
-+ \biggl( \frac{\partial \lambda_1}{\partial y}\biggr) dy
-+ \biggl( \frac{\partial \lambda_1}{\partial z}\biggr) dz
-\biggr]
-\biggl[
-\biggl( \frac{\partial \lambda_4}{\partial x}\biggr) dx
-+ \biggl( \frac{\partial \lambda_4}{\partial y}\biggr) dy
-+ \biggl( \frac{\partial \lambda_4}{\partial z}\biggr) dz
-\biggr]
-\biggl[
-\biggl( \frac{\partial \lambda_5}{\partial x}\biggr) dx
-+ \biggl( \frac{\partial \lambda_5}{\partial y}\biggr) dy
-+ \biggl( \frac{\partial \lambda_5}{\partial z}\biggr) dz
-\biggr]
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\biggl[
-\biggl( \gamma_{11}\biggr) dx
-+ \biggl( \gamma_{12}\biggr) dy
-+ \biggl( \gamma_{13}\biggr) dz
-\biggr]
-\biggl[
-\biggl( \gamma_{41}\biggr) dx
-+ \biggl( \gamma_{42}\biggr) dy
-+ \biggl( \gamma_{43}\biggr) dz
-\biggr]
-\biggl[
-\biggl( \gamma_{51} \biggr) dx
-+ \biggl( \gamma_{52} \biggr) dy
-+ \biggl( \gamma_{53} \biggr) dz
-\biggr]
-</math>
-  </td>
-</tr>
-</table>
-=COLLADA=
-Here we try to use the 3D-visualization capabilities of COLLADA to test whether or not the three coordinates associated with the T6 Coordinate system are indeed orthogonal to one another.  We begin by making a copy of the '''Inertial17.dae''' text file, which we obtain from [[ThreeDimensionalConfigurations/MeetsCOLLADAandOculusRiftS#The_COLLADA_Code_.26_Initial_3D_Scene|an accompanying discussion]].  When viewed with the Mac's '''Preview''' application, this group of COLLADA-based instructions displays a purple ellipsoid with axis ratios, (b/a, c/a) = (0.41, 0.385).  This means that we are dealing with an ellipsoid for which,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>q \equiv \frac{a}{b}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2.44</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
-  <td align="right">
-<math>~p \equiv \frac{a}{c}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2.60 \, .</math>
-  </td>
-</tr>
-</table>
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\lambda_1</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} \, ;</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\lambda_2</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{x y^{1/q^2}}{ z^{2/p^2}} \, ;
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\ell_{3D}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~\biggl[ x^2 + q^4y^2 + p^4 z^2 \biggr]^{- 1 / 2} \, ;</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\mathcal{D}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>
-  </td>
-</tr>
-</table>
-== First Trial==
-<table border="1" align="center" width="80%" cellpadding="8">
-<tr>
-  <td align="center" colspan="6">'''First Trial'''<br />(specified variable values have bgcolor="pink")</td>
-</tr>
-<tr>
-  <td align="center">x</td>
-  <td align="center">y</td>
-  <td align="center">z</td>
-  <td align="center"><math>~\lambda_1</math></td>
-  <td align="center"><math>~\ell_{3D}</math></td>
-  <td align="center"><math>~\mathcal{D}</math></td>
-</tr>
-<tr>
-  <td align="center" bgcolor="pink">0.5</td>
-  <td align="center">0.35493</td>
-  <td align="center" bgcolor="pink">0.00000</td>
-  <td align="center" bgcolor="pink">1</td>
-  <td align="center">0.46052</td>
-  <td align="center">2.11310</td>
-</tr>
-</table>
-===Unit Vectors===
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\hat{e}_1 </math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\hat\imath (x_0 \ell_{3D}) + \hat\jmath (q^2y_0 \ell_{3D}) + \hat{k} (p^2 z_0 \ell_{3D})
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\hat\imath (0.23026) + \hat\jmath (0.97313)  \, ;
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\hat{e}_2</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~ \frac{x_0 q^2 y_0 p^2 z_0}{\mathcal{D}} \biggl\{
-\hat{\imath} \biggl( \frac{1}{x_0} \biggr)
-+ \hat{\jmath}  \biggl( \frac{1}{q^2 y_0} \biggr)
-+ \hat{k} \biggl( -\frac{2}{p^2 z_0} \biggr)
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~-\hat{k} ~\biggl(  \frac{2x_0 q^2 y_0  }{\mathcal{D}} \biggr)
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~-\hat{k} ~\biggl(  1 \biggr)
-\ ;
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\hat{e}_3 </math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\frac{\ell_{3D}}{\mathcal{D}}
-\biggl\{
--\hat\imath \biggl[ x_0(2 q^4y_0^2 + p^4z_0^2 ) \biggl]
-+ \hat\jmath \biggl[ q^2 y_0(p^4z_0^2 + 2x_0^2 )  \biggl]
-+ \hat{k} \biggl[ p^2z_0( x_0^2 - q^4y_0^2 ) \biggl]
-\biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\frac{2q^2 x_0 y_0\ell_{3D}}{\mathcal{D}}
-\biggl\{
--\hat\imath ( q^2y_0 )
-+ \hat\jmath  (x_0)
-\biggr\}
-=
-\biggl(1\biggr)\ell_{3D}\biggl\{ -\hat\imath ( q^2y_0 ) + \hat\jmath  (x_0) \biggr\}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
--\hat\imath (0.97313 ) + \hat\jmath  (0.23026) \, .
-</math>
-  </td>
-</tr>
-</table>
-===Tangent Plane===
-From our [[Appendix/Ramblings/ConcentricEllipsoidalCoordinates#Other_Coordinate_Pair_in_the_Tangent_Plane|above derivation]], the plane that is tangent to the ellipsoid's surface at <math>~(x_0, y_0, z_0)</math> is given by the expression,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~
-x x_0  + q^2 y y_0  + p^2 z z_0
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-(\lambda_1^2)_0 \, .
-</math>
-  </td>
-</tr>
-</table>
-For this ''First Trial,'' we have (for all values of <math>~z</math>, given that <math>~z_0 = 0</math>) &hellip;
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~
-(0.5)x   + (2.11310)y
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~ y
-</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{(1 - 0.5x)}{2.11310} \, .
-</math>
-  </td>
-</tr>
-</table>
-So let's plot a segment of the tangent plane whose four corners are given by the coordinates,
-<table border="1" cellpadding="5" align="center">
-<tr>
-  <td align="center">Corner</td>
-  <td align="center">x</td>
-  <td align="center">y</td>
-  <td align="center">z</td>
-</tr>
-<tr>
-  <td align="center">A</td>
-  <td align="center" bgcolor="pink">x_0 - 0.25 = +0.25</td>
-  <td align="center">0.41408</td>
-  <td align="center" bgcolor="pink">-0.25</td>
-</tr>
-<tr>
-  <td align="center">B</td>
-  <td align="center" bgcolor="pink">x_0 + 0.25 = +0.75</td>
-  <td align="center">0.29577</td>
-  <td align="center" bgcolor="pink">-0.25</td>
-</tr>
-<tr>
-  <td align="center">C</td>
-  <td align="center" bgcolor="pink">x_0 - 0.25 = +0.25</td>
-  <td align="center">0.41408</td>
-  <td align="center" bgcolor="pink">+0.25</td>
-</tr>
-<tr>
-  <td align="center">D</td>
-  <td align="center" bgcolor="pink">x_0 + 0.25 = +0.75</td>
-  <td align="center">0.29577</td>
-  <td align="center" bgcolor="pink">+0.25</td>
-</tr>
-</table>
-Now, in order to give some thickness to this tangent-plane, let's adjust the four corner locations by a distance of <math>~\pm 0.1</math> in the <math>~\hat{e}_1</math> direction.
-===Eight Corners of Tangent Plane===
-Corner 1:  Shift surface-point location <math>~(x_0, y_0, z_0)</math> by <math>~(+\Delta e_1)</math> in the <math>~\hat{e}_1</math> direction,  by <math>~(+\Delta e_2)</math> in the <math>~\hat{e}_2</math> direction, and by  by <math>~(+\Delta e_3)</math> in the <math>~\hat{e}_3</math> direction.  This gives &hellip;
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~x_1</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~x_0 + (\Delta e_1)0.23026 - (\Delta e_2)0.97313</math>
-  </td>
-</tr>
-</table>
-==Second Trial==
-<table border="1" align="center" width="80%" cellpadding="8">
-<tr>
-  <td align="center" colspan="6">'''Second Trial''' &hellip; <math>~(q = 2.44, p = 2.60)</math><br />[specified variable values have bgcolor="pink"]</td>
-</tr>
-<tr>
-  <td align="center">x_0</td>
-  <td align="center">y_0</td>
-  <td align="center">z_0</td>
-  <td align="center"><math>~\lambda_1</math></td>
-  <td align="center"><math>~\ell_{3D}</math></td>
-  <td align="center"><math>~\mathcal{D}</math></td>
-</tr>
-<tr>
-  <td align="center" bgcolor="pink">0.5</td>
-  <td align="center">0.35493</td>
-  <td align="center" bgcolor="pink">0.00000</td>
-  <td align="center" bgcolor="pink">1</td>
-  <td align="center">0.46052</td>
-  <td align="center">2.11310</td>
-</tr>
-</table>
-===Generic Unit Vector Expressions===
-Let's adopt the notation,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\hat{e}_i</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\hat\imath ~[e_{ix}] + \hat\jmath ~[e_{iy}] + \hat{k} ~[e_{iz}]</math>
-  </td>
-  <td align="center">&nbsp; &nbsp; &nbsp; for, &nbsp; &nbsp; &nbsp;</td>
-  <td align="center"><math>~i = 1,3 \, .</math></td>
-</tr>
-</table>
-Then, for the T6 Coordinate system, we have,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~e_{1x}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~x_0 \ell_{3D} \, ;</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp;</td>
-  <td align="right">
-<math>~e_{1y}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~q^2y_0 \ell_{3D} \, ;</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp;</td>
-  <td align="right">
-<math>~e_{1z}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~p^2 z_0 \ell_{3D} \, ;</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~e_{2x}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~ \frac{q^2 y_0 p^2 z_0}{\mathcal{D}}\, ;</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp;</td>
-  <td align="right">
-<math>~e_{2y}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~\frac{x_0 p^2 z_0}{\mathcal{D}} \, ;</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp;</td>
-  <td align="right">
-<math>~e_{2z}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~- \frac{2x_0 q^2 y_0}{\mathcal{D}}\, ;</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~e_{3x}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~-x_0(2 q^4y_0^2 + p^4z_0^2 )\frac{\ell_{3D}}{\mathcal{D}}  \, ;</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp;</td>
-  <td align="right">
-<math>~e_{3y}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~q^2 y_0(p^4z_0^2 + 2x_0^2 )\frac{\ell_{3D}}{\mathcal{D}}   \, ;</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp;</td>
-  <td align="right">
-<math>~e_{3z}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~p^2z_0( x_0^2 - q^4y_0^2 )\frac{\ell_{3D}}{\mathcal{D}} \, .</math>
-  </td>
-</tr>
-</table>
-<table border="1" align="center" cellpadding="8">
-<tr>
-  <td align="center" colspan="4">'''Second Trial'''</td>
-</tr>
-<tr>
-  <td align="center"> &nbsp;</td>
-  <td align="center"> x</td>
-  <td align="center"> y</td>
-  <td align="center"> z</td>
-</tr>
-<tr>
-  <td align="center"> <math>~e_1</math></td>
-  <td align="center">0.23026</td>
-  <td align="center">0.97313</td>
-  <td align="center"> 0.0</td>
-</tr>
-<tr>
-  <td align="center"> <math>~e_2</math></td>
-  <td align="center">0.0</td>
-  <td align="center">0.0</td>
-  <td align="center">-1.0</td>
-</tr>
-<tr>
-  <td align="center"> <math>~e_3</math></td>
-  <td align="center">- 0.97313</td>
-  <td align="center">0.23026</td>
-  <td align="center"> 0.0</td>
-</tr>
-</table>
-What are the coordinates of the eight corners of a thin tangent-plane?  Let's say that we want the plane to extend &hellip;
-<ul>
-  <li>From <math>~(-\Delta_1)</math> to <math>~(+\Delta_1)</math> in the <math>~\hat{e}_1</math> direction &hellip; here we set <math>~\Delta_1 = 0.05</math>;</li>
-  <li>From <math>~(-\Delta_2)</math> to <math>~(+\Delta_2)</math> in the <math>~\hat{e}_2</math> direction &hellip; here we set <math>~\Delta_2 = 0.25</math>;</li>
-  <li>From <math>~(-\Delta_3)</math> to <math>~(+\Delta_3)</math> in the <math>~\hat{e}_3</math> direction &hellip; here we set <math>~\Delta_3 = 0.25</math>.</li>
-</ul>
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\Delta_x</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\Delta_1 e_{1x} + \Delta_2 e_{2x} + \Delta_3 e_{3x} = -0.23177 \, ;</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Delta_y</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\Delta_1 e_{1y} + \Delta_2 e_{2y} + \Delta_3 e_{3y} = +0.10622 \, ;</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Delta_z</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\Delta_1 e_{1z} + \Delta_2 e_{2z} + \Delta_3 e_{3z} = -0.25000 \, .</math>
-  </td>
-</tr>
-</table>
-<table border="0" align="center" cellpadding="8">
-<tr>
-  <td align="left">[[File:TangentPlaneSchematic.png|Tangent Plane Schematic]]</td>
-  <td align="left">
-<table border="1" cellpadding="5" align="center">
-  <tr>
-    <td align="center">vertex</td>
-    <td align="center">x</td>
-    <td align="center">y</td>
-    <td align="center">z</td>
-    <td align="center" rowspan="9" bgcolor="lightgray">&nbsp;</td>
-    <td align="center">x</td>
-    <td align="center">y</td>
-    <td align="center">z</td>
-  </tr>
-  <tr>
-    <td align="center">0<sup>&dagger;</sup></td>
-    <td align="center"><math>~x_0 - |\Delta_x|</math></td>
-    <td align="center"><math>~y_0 - |\Delta_y|</math></td>
-    <td align="center"><math>~z_0 - |\Delta_z|</math></td>
-    <td align="center">0.26823</td>
-    <td align="center">0.24871</td>
-    <td align="center">-0.25</td>
-  </tr>
-  <tr>
-    <td align="center">1</td>
-    <td align="center"><math>~x_0 - |\Delta_x|</math></td>
-    <td align="center"><math>~y_0 + |\Delta_y|</math></td>
-    <td align="center"><math>~z_0 - |\Delta_z|</math></td>
-    <td align="center">0.26823</td>
-    <td align="center">0.46115</td>
-    <td align="center">-0.25</td>
-  </tr>
-  <tr>
-    <td align="center">2</td>
-    <td align="center"><math>~x_0 - |\Delta_x|</math></td>
-    <td align="center"><math>~y_0 - |\Delta_y|</math></td>
-    <td align="center"><math>~z_0 + |\Delta_z|</math></td>
-    <td align="center">0.26823</td>
-    <td align="center"> 0.24871 </td>
-    <td align="center">0.25</td>
-  </tr>
-  <tr>
-    <td align="center">3</td>
-    <td align="center"><math>~x_0 - |\Delta_x|</math></td>
-    <td align="center"><math>~y_0 + |\Delta_y|</math></td>
-    <td align="center"><math>~z_0 + |\Delta_z|</math></td>
-    <td align="center">0.26823</td>
-    <td align="center"> 0.46115 </td>
-    <td align="center">0.25</td>
-  </tr>
-  <tr>
-    <td align="center">4</td>
-    <td align="center"><math>~x_0 + |\Delta_x|</math></td>
-    <td align="center"><math>~y_0 - |\Delta_y|</math></td>
-    <td align="center"><math>~z_0 - |\Delta_z|</math></td>
-    <td align="center">0.73177</td>
-    <td align="center"> 0.24871 </td>
-    <td align="center">-0.25</td>
-  </tr>
-  <tr>
-    <td align="center">5</td>
-    <td align="center"><math>~x_0 + |\Delta_x|</math></td>
-    <td align="center"><math>~y_0 + |\Delta_y|</math></td>
-    <td align="center"><math>~z_0 - |\Delta_z|</math></td>
-    <td align="center"> 0.73177 </td>
-    <td align="center"> 0.46115 </td>
-    <td align="center">-0.25</td>
-  </tr>
-  <tr>
-    <td align="center">6</td>
-    <td align="center"><math>~x_0 + |\Delta_x|</math></td>
-    <td align="center"><math>~y_0 - |\Delta_y|</math></td>
-    <td align="center"><math>~z_0 + |\Delta_z|</math></td>
-    <td align="center"> 0.73177 </td>
-    <td align="center"> 0.24871 </td>
-    <td align="center">0.25</td>
-  </tr>
-  <tr>
-    <td align="center">7</td>
-    <td align="center"><math>~x_0 + |\Delta_x|</math></td>
-    <td align="center"><math>~y_0 + |\Delta_y|</math></td>
-    <td align="center"><math>~z_0 + |\Delta_z|</math></td>
-    <td align="center"> 0.73177 </td>
-    <td align="center"> 0.46115 </td>
-    <td align="center">0.25</td>
-  </tr>
-</table>
-  </td>
-</tr>
-<tr>
-  <td align="left" colspan="2">
-<sup>&dagger;</sup>In the figure on the left, vertex 0 is hidden from view behind the (yellow) solid rectangle.
-  </td>
-</tr>
-</table>
-==Third Trial==
-===GoodPlane01===
-<table border="1" align="center" width="80%" cellpadding="8">
-<tr>
-  <td align="center" colspan="6">'''Third Trial''' &hellip; <math>~(q = 2.44, p = 2.60)</math><br />[specified variable values have bgcolor="pink"]</td>
-</tr>
-<tr>
-  <td align="center">x_0</td>
-  <td align="center">y_0</td>
-  <td align="center">z_0</td>
-  <td align="center"><math>~\lambda_1</math></td>
-  <td align="center"><math>~\ell_{3D}</math></td>
-  <td align="center"><math>~\mathcal{D}</math></td>
-</tr>
-<tr>
-  <td align="center" bgcolor="pink">0.8</td>
-  <td align="center">0.24600</td>
-  <td align="center" bgcolor="pink">0.00000</td>
-  <td align="center" bgcolor="pink">1</td>
-  <td align="center">0.59959</td>
-  <td align="center">2.34146</td>
-</tr>
-</table>
-Again, for the T6 Coordinate system, we have,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~e_{1x}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~x_0 \ell_{3D} \, ;</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp;</td>
-  <td align="right">
-<math>~e_{1y}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~q^2y_0 \ell_{3D} \, ;</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp;</td>
-  <td align="right">
-<math>~e_{1z}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~p^2 z_0 \ell_{3D} \, ;</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~e_{2x}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~ \frac{q^2 y_0 p^2 z_0}{\mathcal{D}}\, ;</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp;</td>
-  <td align="right">
-<math>~e_{2y}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~\frac{x_0 p^2 z_0}{\mathcal{D}} \, ;</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp;</td>
-  <td align="right">
-<math>~e_{2z}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~- \frac{2x_0 q^2 y_0}{\mathcal{D}}\, ;</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~e_{3x}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~-x_0(2 q^4y_0^2 + p^4z_0^2 )\frac{\ell_{3D}}{\mathcal{D}}  \, ;</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp;</td>
-  <td align="right">
-<math>~e_{3y}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~q^2 y_0(p^4z_0^2 + 2x_0^2 )\frac{\ell_{3D}}{\mathcal{D}}   \, ;</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp;</td>
-  <td align="right">
-<math>~e_{3z}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~p^2z_0( x_0^2 - q^4y_0^2 )\frac{\ell_{3D}}{\mathcal{D}} \, .</math>
-  </td>
-</tr>
-</table>
-<table border="1" align="center" cellpadding="8">
-<tr>
-  <td align="center" colspan="5">'''Third Trial'''</td>
-</tr>
-<tr>
-  <td align="center"> &nbsp;</td>
-  <td align="center"> x</td>
-  <td align="center"> y</td>
-  <td align="center"> z</td>
-  <td align="center"> <math>~\Delta_\mathrm{TP}</math></td>
-</tr>
-<tr>
-  <td align="center"> <math>~e_1</math></td>
-  <td align="center">0.47967</td>
-  <td align="center">0.87745</td>
-  <td align="center"> 0.0</td>
-  <td align="center"> 0.02</td>
-</tr>
-<tr>
-  <td align="center"> <math>~e_2</math></td>
-  <td align="center">0.0</td>
-  <td align="center">0.0</td>
-  <td align="center">-1.0</td>
-  <td align="center"> 0.25</td>
-</tr>
-<tr>
-  <td align="center"> <math>~e_3</math></td>
-  <td align="center">- 0.87753</td>
-  <td align="center"> 0.47952 </td>
-  <td align="center">0.0</td>
-  <td align="center"> 0.25</td>
-</tr>
-</table>
-In constructing the Tangent-Plane (TP) for a 3D COLLADA display, we first move from the point that is on the surface of the ellipsoid, <math>~\vec{x}_0 = (x_0, y_0, z_0) = (0.8, 0.246, 0.0)</math>, to
-<table border="1" cellpadding="8" align="center">
-<tr>
-  <td align="center" rowspan="2">vertex <br />"m"</td>
-  <td align="center" rowspan="2"><math>~\vec{P}_m</math></td>
-  <td align="center" colspan="3">Components</td>
-</tr>
-<tr>
-  <td align="center"><math>~x_m = \hat\imath \cdot \vec{P}_m</math></td>
-  <td align="center"><math>~y_m = \hat\jmath \cdot \vec{P}_m</math></td>
-  <td align="center"><math>~z_m = \hat{k} \cdot \vec{P}_m</math></td>
-</tr>
-<tr>
-  <td align="center">0</td>
-  <td align="center"><math>~\vec{x}_0 - \Delta_1\hat{e}_1 - \Delta_2\hat{e}_2 - \Delta_3\hat{e}_3</math></td>
-  <td align="center">
-<math>~
-x_0 - \Delta_1 e_{1x} - \Delta_2 e_{2x} - \Delta_3 e_{3x}
-</math>
-  </td>
-  <td align="center">
-<math>~
-y_0 - \Delta_1 e_{1y} - \Delta_2 e_{2y} - \Delta_3 e_{3y}
-</math>
-  </td>
-  <td align="center">
-<math>~
-z_0 - \Delta_1 e_{1z} - \Delta_2 e_{2z} - \Delta_3 e_{3z}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="center">&nbsp;</td>
-  <td align="center">&nbsp;</td>
-  <td align="center">
-.8 - 0.02 (0.47952)  - 0.25 (-0.87752) = 1.00979
-  </td>
-  <td align="center">
-.24590 - 0.02 (0.87753)  - 0.25 (0.47952) = 0.10847
-  </td>
-  <td align="center">
-- 0.25 (-1.0) = + 0.25
-  </td>
-</tr>
-<tr>
-  <td align="center">1</td>
-  <td align="center"><math>~\vec{x}_0 - \Delta_1\hat{e}_1 + \Delta_2\hat{e}_2 - \Delta_3\hat{e}_3</math></td>
-  <td align="center">
-<math>~
-x_0 - \Delta_1 e_{1x} + \Delta_2 e_{2x} - \Delta_3 e_{3x}
-</math>
-  </td>
-  <td align="center">
-<math>~
-y_0 - \Delta_1 e_{1y} + \Delta_2 e_{2y} - \Delta_3 e_{3y}
-</math>
-  </td>
-  <td align="center">
-<math>~
-z_0 - \Delta_1 e_{1z} + \Delta_2 e_{2z} - \Delta_3 e_{3z}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="center">&nbsp;</td>
-  <td align="center">&nbsp;</td>
-  <td align="center">
-.8 - 0.02 (0.47952)  - 0.25 (-0.87752) =  1.00979
-  </td>
-  <td align="center">
-.24590 - 0.02 (0.87753)  - 0.25 (0.47952) = 0.10847
-  </td>
-  <td align="center">
-+ 0.25 (-1.0) = - 0.25
-  </td>
-</tr>
-<tr>
-  <td align="center">2</td>
-  <td align="center"><math>~\vec{x}_0 - \Delta_1\hat{e}_1 - \Delta_2\hat{e}_2 + \Delta_3\hat{e}_3</math></td>
-  <td align="center">
-<math>~
-x_0 - \Delta_1 e_{1x} - \Delta_2 e_{2x} + \Delta_3 e_{3x}
-</math>
-  </td>
-  <td align="center">
-<math>~
-y_0 - \Delta_1 e_{1y} - \Delta_2 e_{2y} + \Delta_3 e_{3y}
-</math>
-  </td>
-  <td align="center">
-<math>~
-z_0 - \Delta_1 e_{1z} - \Delta_2 e_{2z} + \Delta_3 e_{3z}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="center">&nbsp;</td>
-  <td align="center">&nbsp;</td>
-  <td align="center">
-.8 - 0.02 (0.47952)  + 0.25 (-0.87752) =   0.56307
-  </td>
-  <td align="center">
-.24590 - 0.02 (0.87753)  + 0.25 (0.47952) = 0.34823
-  </td>
-  <td align="center">
-- 0.25 (-1.0) = + 0.25
-  </td>
-</tr>
-<tr>
-  <td align="center">3</td>
-  <td align="center"><math>~\vec{x}_0 - \Delta_1\hat{e}_1 + \Delta_2\hat{e}_2 + \Delta_3\hat{e}_3</math></td>
-  <td align="center">
-<math>~
-x_0 - \Delta_1 e_{1x} + \Delta_2 e_{2x} + \Delta_3 e_{3x}
-</math>
-  </td>
-  <td align="center">
-<math>~
-y_0 - \Delta_1 e_{1y} + \Delta_2 e_{2y} + \Delta_3 e_{3y}
-</math>
-  </td>
-  <td align="center">
-<math>~
-z_0 - \Delta_1 e_{1z} + \Delta_2 e_{2z} + \Delta_3 e_{3z}
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="center">&nbsp;</td>
-  <td align="center">&nbsp;</td>
-  <td align="center">
-.8 - 0.02 (0.47952)  + 0.25 (-0.87752) = 0.57103
-  </td>
-  <td align="center">
-.24590 - 0.02 (0.87753)  + 0.25 (0.47952) = 0.34823
-  </td>
-  <td align="center">
-+ 0.25 (-1.0) = - 0.25
-  </td>
-</tr>
-<tr>
-  <td align="center">4</td>
-  <td align="center">&nbsp;</td>
-  <td align="center">
-.8 + 0.02 (0.47952)  - 0.25 (-0.87752) =   1.0290
-  </td>
-  <td align="center">
-.24590 + 0.02 (0.87753)  - 0.25 (0.47952) = 0.1436
-  </td>
-  <td align="center">
-- 0.25 (-1.0) = + 0.25
-  </td>
-</tr>
-<tr>
-  <td align="center">5</td>
-  <td align="center">&nbsp;</td>
-  <td align="center">
-.8 + 0.02 (0.47952)  - 0.25 (-0.87752) =  1.0290
-  </td>
-  <td align="center">
-.24590 + 0.02 (0.87753)  - 0.25 (0.47952) = 0.1436
-  </td>
-  <td align="center">
-+ 0.25 (-1.0) = - 0.25
-  </td>
-</tr>
-<tr>
-  <td align="center">6</td>
-  <td align="center">&nbsp;</td>
-  <td align="center">
-.8 + 0.02 (0.47952)  + 0.25 (-0.87752) =   0.59021
-  </td>
-  <td align="center">
-.24590 + 0.02 (0.87753)  + 0.25 (0.47952) = 0.38333
-  </td>
-  <td align="center">
-- 0.25 (-1.0) = + 0.25
-  </td>
-</tr>
-<tr>
-  <td align="center">7</td>
-  <td align="center">&nbsp;</td>
-  <td align="center">
-.8 + 0.02 (0.47952)  + 0.25 (-0.87752) = 0.59021
-  </td>
-  <td align="center">
-.24590 + 0.02 (0.87753)  + 0.25 (0.47952) = 0.38333
-  </td>
-  <td align="center">
-+ 0.25 (-1.0) = - 0.25
-  </td>
-</tr>
-</table>
-<table border="1" align="center" cellpadding="8">
-<tr>
-  <td align="left">[[File:TangentPlaneSchematic.png|Tangent Plane Schematic]]</td>
-  <td align="left">[[File:ExcelVertices080.png|Vertex Locations via Excel]]</td>
-</tr>
-<tr>
-<td align="center" colspan="2"><math>~x_0 = 0.8, z_0 = 0.0, y_0 = 0.246, \lambda_1 = 1.0</math></td>
-</tr>
-<tr>
-  <td align="center" colspan="2" bgcolor="lightgray">[[File:GoodPlane01.png|400px|Tangent Plane Schematic]]</td>
-</tr>
-<tr>
-<td align="center" colspan="2"><math>~\Delta_1 = 0.02, \Delta_2 = 0.25, \Delta_3 = 0.25</math></td>
-</tr>
-</table>
-===GoodPlane02===
-<table border="1" align="center" cellpadding="8">
-<tr>
-<td align="center"><math>~x_0 = 0.075, z_0 = 0.0, y_0 = 0.4089, \lambda_1 = 1.0</math></td>
-</tr>
-<tr>
-  <td align="left" bgcolor="lightgray">[[File:GoodPlane02.png|400px|Tangent Plane Schematic]]</td>
-</tr>
-<tr>
-<td align="center" colspan="2"><math>~\Delta_1 = 0.02, \Delta_2 = 0.25, \Delta_3 = 0.25</math></td>
-</tr>
-</table>
-===GoodPlane03===
-<table border="1" align="center" cellpadding="8">
-<tr>
-<td align="center" colspan="2"><math>~x_0 = 0.25, z_0 = 0.20, y_0 = 0.33501, \lambda_1 = 1.0</math></td>
-</tr>
-<tr>
-  <td align="left" bgcolor="lightgray">[[File:GoodPlane03.png|400px|Tangent Plane Schematic]]</td>
-  <td align="left" bgcolor="lightgray">[[File:GoodPlane03B.png|400px|Tangent Plane Schematic]]</td>
-</tr>
-<tr>
-<td align="center" colspan="1"><math>~\Delta_1 = 0.02, \Delta_2 = 0.25, \Delta_3 = 0.25</math></td>
-<td align="center" colspan="1"><math>~\Delta_1 = 0.02, \Delta_2 = 0.10, \Delta_3 = 0.25</math></td>
-</tr>
-<tr>
-  <td align="left" colspan="2">
-CAPTION: &nbsp; The image on the right differs from the image on the left in only one way &#8212; <math>~\Delta_2</math> = 0.1 instead of 0.25.  It illustrates more clearly that the <math>~\hat{e}_3</math> (longest) coordinate axis is not parallel to the z-axis when <math>~z_0 \ne 0.</math>
-  </td>
-</tr>
-</table>
-===GoodPlane04===
-<table border="1" align="center" cellpadding="8">
-<tr>
-<td align="center" colspan="1"><math>~x_0 = 0.25, z_0 = 1/3, y_0 = 0.1777, \lambda_1 = 1.0</math></td>
-</tr>
-<tr>
-  <td align="left" bgcolor="lightgray">[[File:GoodPlane04A.png|400px|Tangent Plane Schematic]]</td>
-</tr>
-<tr>
-<td align="center" colspan="1"><math>~\Delta_1 = 0.02, \Delta_2 = 0.10, \Delta_3 = 0.25</math></td>
-</tr>
-</table>
-=Further Exploration=
-Let's set:  <math>~x_0 = 0.25, y_0 = 0.33501, z_0 = 0.2 ~~~\Rightarrow ~~~\lambda_1 = 1.00000, \lambda_2 = 0.33521 \, .</math>
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~q \equiv \frac{a}{b}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2.43972</math>
-  </td>
-<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
-  <td align="right">
-<math>~p \equiv \frac{a}{c}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~2.5974 \, .</math>
-  </td>
-</tr>
-</table>
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\lambda_1</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} \, ;</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\lambda_2</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~
-\frac{x y^{1/q^2}}{ z^{2/p^2}} \, ;
-</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\ell_{3D}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~\biggl[ x^2 + q^4y^2 + p^4 z^2 \biggr]^{- 1 / 2} \, ;</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\mathcal{D}</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
-  <td align="left">
-<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>
-  </td>
-</tr>
-</table>
-Next, let's examine the curve that results from varying <math>~z</math> while <math>~\lambda_1</math> and <math>~\lambda_2</math> are held fixed.  From the expression for <math>~\lambda_2</math> we appreciate that,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~x</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\frac{\lambda_2 z^{2/p^2}}{y^{1/q^2}} \, ;</math>
-  </td>
-</tr>
-</table>
-and from the expression for <math>~\lambda_1</math> we have,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~x^2</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\lambda_1^2 - q^2y^2 - p^2z^2 \, .</math>
-  </td>
-</tr>
-</table>
-Hence, the relationship between <math>~y</math> and <math>~z</math> is,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\biggl[ \frac{\lambda_2 z^{2/p^2}}{y^{1/q^2}} \biggr]^2</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\lambda_1^2 - q^2y^2 - p^2z^2</math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow ~~~ \lambda_2^2 z^{4/p^2} </math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~y^{2/q^2} \biggl[ \lambda_1^2 - q^2y^2 - p^2z^2\biggr] \, .</math>
-  </td>
-</tr>
-</table>
-<table border="1" align="center" cellpadding="8" width="80%">
-<tr><td align="left">
-Alternatively,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~y</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\biggl[ \frac{\lambda_2 z^{2/p^2}}{x}\biggr]^{q^2} \, .</math>
-  </td>
-</tr>
-</table>
-Hence, the relationship between <math>~x</math> and <math>~z</math> is,
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~x^2</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\lambda_1^2 - p^2z^2 - q^2\biggl[ \frac{\lambda_2 z^{2/p^2}}{x}\biggr]^{2q^2} </math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow~~~ x^{2(q^2+1)}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~x^{2q^2} \biggl[\lambda_1^2 - p^2z^2\biggr] - q^2\biggl[ \lambda_2 z^{2/p^2}\biggr]^{2q^2} </math>
-  </td>
-</tr>
-<tr>
-  <td align="right">
-<math>~\Rightarrow~~~ x^{2}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>~\biggl\{ x^{2q^2} \biggl[\lambda_1^2 - p^2z^2\biggr] - q^2\biggl[ \lambda_2 z^{2/p^2}\biggr]^{2q^2} \biggr\}^{1/(q^2+1)}</math>
-  </td>
-</tr>
-</table>
-</td></tr>
-</table>
-Here are some example values &hellip;
-<table border="1" align="center" cellpadding="8">
-<tr>
-<td align="center" colspan="7"><math>~\lambda_1 = 1</math>&nbsp; &nbsp; and, &nbsp; &nbsp; <math>~\lambda_2 = 0.33521</math></td>
-</tr>
-<tr>
-  <td align="center" rowspan="2"><math>~z</math></td>
-  <td align="center" colspan="2">1<sup>st</sup> Solution</td>
-  <td align="center" rowspan="20" bgcolor="lightgray">&nbsp;</td>
-  <td align="center" colspan="2">2<sup>nd</sup> Solution</td>
-  <td align="center" rowspan="20" bgcolor="white">[[File:Lambda3ImageA.png|450px|lambda_3 coordinate]]</td>
-</tr>
-<tr>
-  <td align="center"><math>~y_1</math></td>
-  <td align="center"><math>~x_1</math></td>
-  <td align="center"><math>~y_2</math></td>
-  <td align="center"><math>~x_2</math></td>
-</tr>
-<tr>
-  <td align="center">0.01</td>
-  <td align="center">0.407825695</td>
-  <td align="center">0.0995168</td>
-  <td align="center">-</td>
-  <td align="center">-</td>
-</tr>
-<tr>
-  <td align="center">0.03</td>
-  <td align="center">0.40481851</td>
-  <td align="center">0.138</td>
-  <td align="center">-</td>
-  <td align="center">-</td>
-</tr>
-<tr>
-  <td align="center">0.04</td>
-  <td align="center">0.40309223</td>
-  <td align="center">0.1503934</td>
-  <td align="center">-</td>
-  <td align="center">-</td>
-</tr>
-<tr>
-  <td align="center">0.08</td>
-  <td align="center">0.393779065</td>
-  <td align="center">0.1854283</td>
-  <td align="center">-</td>
-  <td align="center">-</td>
-</tr>
-<tr>
-  <td align="center">0.12</td>
-  <td align="center">0.37990705</td>
-  <td align="center">0.2103761</td>
-  <td align="center">-</td>
-  <td align="center">-</td>
-</tr>
-<tr>
-  <td align="center">0.16</td>
-  <td align="center">0.36067787</td>
-  <td align="center">0.23111</td>
-  <td align="center">1.04123&times;10<sup>-4</sup></td>
-  <td align="center">0.9095546</td>
-</tr>
-<tr>
-  <td align="center">0.2</td>
-  <td align="center">0.33500747</td>
-  <td align="center">0.2500033</td>
-  <td align="center">2.23778&times;10<sup>-4</sup></td>
-  <td align="center">0.85448</td>
-</tr>
-<tr>
-  <td align="center">0.22</td>
-  <td align="center">0.31923525</td>
-  <td align="center">0.2592611</td>
-  <td align="center">3.36653 &times;10<sup>-4</sup></td>
-  <td align="center">0.82065</td>
-</tr>
-<tr>
-  <td align="center">0.24</td>
-  <td align="center">0.30106924</td>
-  <td align="center">0.2686685</td>
-  <td align="center">5.2327 &times;10<sup>-4</sup></td>
-  <td align="center">0.78192</td>
-</tr>
-<tr>
-  <td align="center">0.26</td>
-  <td align="center">0.2799962</td>
-  <td align="center">0.2784963</td>
-  <td align="center">8.53243 &times;10<sup>-4</sup></td>
-  <td align="center">0.73752</td>
-</tr>
-<tr>
-  <td align="center">0.28</td>
-  <td align="center">0.25521147</td>
-  <td align="center">0.2891526</td>
-  <td align="center">1.491545 &times;10<sup>-3</sup></td>
-  <td align="center">0.68634</td>
-</tr>
-<tr>
-  <td align="center">0.3</td>
-  <td align="center">0.22530908</td>
-  <td align="center">0.3013752</td>
-  <td align="center">2.89262 &times;10<sup>-3</sup></td>
-  <td align="center">0.62671</td>
-</tr>
-<tr>
-  <td align="center">0.32</td>
-  <td align="center">0.1873233</td>
-  <td align="center">0.3168808</td>
-  <td align="center">6.6223 &times;10<sup>-3</sup></td>
-  <td align="center">0.55579</td>
-</tr>
-<tr>
-  <td align="center">0.34</td>
-  <td align="center">0.13149897</td>
-  <td align="center">0.3423994</td>
-  <td align="center">2.09221 &times;10<sup>-2</sup></td>
-  <td align="center">0.46637</td>
-</tr>
-<tr>
-  <td align="center">0.343</td>
-  <td align="center">0.1191543</td>
-  <td align="center">0.3490285</td>
-  <td align="center">0.026458</td>
-  <td align="center">0.4496</td>
-</tr>
-<tr>
-  <td align="center">0.344</td>
-  <td align="center">0.1145</td>
-  <td align="center">0.3517</td>
-  <td align="center">0.02880</td>
-  <td align="center">0.4435</td>
-</tr>
-<tr>
-  <td align="center">0.345</td>
-  <td align="center">0.1093972</td>
-  <td align="center">0.354688</td>
-  <td align="center">0.03155965</td>
-  <td align="center">0.4371186</td>
-</tr>
-<tr>
-  <td align="center">0.3485</td>
-  <td align="center">0.0847372</td>
-  <td align="center">0.3713588</td>
-  <td align="center">0.0480478</td>
-  <td align="center">0.4085204</td>
-</tr>
-</table>
-=See Also=
-<ul>
-  <li>
-[[User:Tohline/Appendix/Ramblings/DirectionCosines|Direction Cosines]]
-  </li>
-</ul>
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Appendix/Ramblings/T6CoordinatesPt2: Difference between revisions

Latest revision as of 18:13, 23 July 2021

Concentric Ellipsoidal (T6) Coordinates (Part 2)

Orthogonal Coordinates

Speculation5

Spherical Coordinates

Use λ1 Instead of r

Relationship To T3 Coordinates

Again Consider Full 3D Ellipsoid

Partial Derivatives & Scale Factors

First Coordinate

Second Coordinate (1st Try)

Second Coordinate (2nd Try)

Speculation6

Determine λ2

Direction Cosines for Third Unit Vector

Search for Third Coordinate Expression

Fiddle Around

Summary

Speculation7

Navigation menu

Search

Use λ₁ Instead of r

Second Coordinate (1^st Try)

Second Coordinate (2^nd Try)

Determine λ₂