ThreeDimensionalConfigurations/ChallengesPt6

Challenges Constructing Ellipsoidal-Like Configurations (Pt. 6)

This chapter has been created in February 2022, after letting this discussion lie dormant for close to one year. We begin the chapter by grabbing large segments of our earlier derivations found primarily in the "Ramblings" chapter titled, Construction Challenges (Pt. 4).

Intersection Expression

STEP #1

First, we present the mathematical expression that describes the intersection between the surface of an ellipsoid and a plane having the following properties:

• The plane cuts through the ellipsoid's z-axis at a distance, $z_0$, from the center of the ellipsoid;
• The line of intersection is parallel to the x-axis of the ellipsoid; and,
• The line that is perpendicular to the plane and passes through the z-axis at $z_0$ is tipped at an angle, $\theta$, to the z-axis.

As is illustrated in Figure 1, we will use the line referenced in this third property description to serve as the z'-axis of a Cartesian grid that is tipped at the angle, $\theta$, with respect to the body frame; and we will align the x' axis with the x-axis, so it should be clear that the z'-axis lies in the y-z plane of the ellipsoid.

Figure 1
$x$ $=$ $x^{\prime },$ $y$ $=$ $y^{\prime }\cos \theta -z^{\prime }\sin \theta ,$ $(z-z_0)$ $=$ $z^{\prime }\cos \theta +y^{\prime }\sin \theta .$		$x^{\prime }$ $=$ $x,$ $y^{\prime }$ $=$ $y\cos \theta +(z-z_0)\sin \theta ,$ $z^{\prime }$ $=$ $(z-z_0)\cos \theta -y\sin \theta .$

As has been shown in our accompanying discussion, we obtain the following,

Intersection Expression
$1-\frac{x^2}{a^2}$	$=$	$y^2\left[\frac{c^2+b^2{\tan }^2\theta }{b^2{c}^2}\right]+y\left[\frac{2z_0\tan \theta }{c^2}\right]+\frac{z_0^2}{c^2},$

as long as z₀ lies within the range,

$z_0^2$

$\le$

$c^2+b^2{\tan }^2\theta .$

Rewriting this "intersection expression" in terms of the tipped (primed) coordinate frame gives us,

$1-\frac{(x^{\prime }{)}^2}{a^2}$

$=$

$(y^{\prime }\cos \theta -z^{\prime }\sin \theta {)}^2[\frac{c^2+b^2{\tan }^2\theta }{b^2{c}^2}]+(y^{\prime }\cos \theta -z^{\prime }\sin \theta )[\frac{2z_0\tan \theta }{c^2}]+\frac{z_0^2}{c^2}.$

STEP #2

As viewed from the tipped coordinated frame, the curve that is identified by this intersection should be an

Off-Center Ellipse
$1$	$=$	$[\frac{x^{\prime }}{x_{\mathrm{m}\mathrm{a}\mathrm{x}}}{]}^2+[\frac{y^{\prime }-y_c}{y_{\mathrm{m}\mathrm{a}\mathrm{x}}}{]}^2$
	$=$	$[\frac{x^{\prime }}{x_{\mathrm{m}\mathrm{a}\mathrm{x}}}{]}^2+[\frac{(y^{\prime }{)}^2-2y^{\prime }{y}_c+y_c^2}{y_{\mathrm{m}\mathrm{a}\mathrm{x}}^2}],$

that lies in the x'-y' plane — that is, $z^{\prime }=0$. Let's see if the intersection expression can be molded into this form.

$1-\frac{z_0^2}{c^2}-\frac{(x^{\prime }{)}^2}{a^2}$	$=$	$(y^{\prime }{)}^2[\frac{c^2+b^2{\tan }^2\theta }{b^2{c}^2}]{\cos }^2\theta +2y^{\prime }[\frac{z_0\sin \theta }{c^2}]$
	$=$	$\left[\frac{c^2+b^2{\tan }^2\theta }{b^2{c}^2}\right]{\cos }^2\theta \{(y^{\prime }{)}^2-2y^{\prime }\left[\frac{-z_0\sin \theta }{c^2{\cos }^2\theta }\right]\left[\frac{b^2{c}^2}{c^2+b^2{\tan }^2\theta }\right]\}$
	$=$	${\kappa }^2[(y^{\prime }{)}^2-2y^{\prime }\underset{y_c}{\underbrace{\left(\frac{-z_0\sin \theta }{c^2{\kappa }^2}\right)}}],$

RESULT 3 (same as Result 1, but different from Result 2, below)

$\frac{y_c}{z_0}$ $=$ $-\frac{\sin \theta }{c^2{\kappa }^2}$

where,

${\kappa }^2$

$\equiv$

$\frac{c^2{\cos }^2\theta +b^2{\sin }^2\theta }{b^2{c}^2}.$

Dividing through by ${\kappa }^2$, then adding $y_c^2$ to both sides gives,

$(y^{\prime }{)}^2-2y^{\prime }{y}_c+y_c^2$

$=$

$\underset{y_{\mathrm{m}\mathrm{a}\mathrm{x}}^2}{\underbrace{[\frac{1}{{\kappa }^2}-\frac{z_0^2}{c^2{\kappa }^2}+y_c^2]}}-\frac{(x^{\prime }{)}^2}{a^2{\kappa }^2}.$

Finally, we have,

$\frac{1}{y_{\mathrm{m}\mathrm{a}\mathrm{x}}^2}[(y^{\prime }{)}^2-2y^{\prime }{y}_c+y_c^2]$

$=$

$1-(x^{\prime }{)}^2\underset{1/x_{\mathrm{m}\mathrm{a}\mathrm{x}}^2}{\underbrace{\left[\frac{1}{a^2{\kappa }^2{y}_{\mathrm{m}\mathrm{a}\mathrm{x}}^2}\right]}}.$

So … the intersection expression can be molded into the form of an off-center ellipse if we make the following associations:

$\frac{y_c}{z_0}$ $=$ $-\frac{\sin \theta }{c^2{\kappa }^2},$ $y_{\mathrm{m}\mathrm{a}\mathrm{x}}^2$ $=$ $\frac{1}{{\kappa }^2}[1-\frac{z_0^2}{c^2}-\frac{z_0\sin \theta }{c^2}],$ $x_{\mathrm{m}\mathrm{a}\mathrm{x}}^2$ $=$ $a^2[1-\frac{z_0^2}{c^2}-\frac{z_0\sin \theta }{c^2}].$ Note as well that, $(\frac{x_{\mathrm{m}\mathrm{a}\mathrm{x}}}{y_{\mathrm{m}\mathrm{a}\mathrm{x}}}{)}^2$ $=$ $a^2{\kappa }^2=\frac{a^2}{b^2{c}^2}[c^2{\cos }^2\theta +b^2{\sin }^2\theta ].$

Lagrangian Trajectory and Velocities

We presume that the off-center ellipse that is defined by the intersection expression identifies the trajectory of a Lagrangian fluid element. If this is the case, there are a couple of ways that the velocity — both the amplitude and its vector orientation — can be derived.

STEP #3

If the intersection expression identifies a Lagrangian trajectory, then the velocity vector must be tangent to the off-center ellipse at every location. At each $(x^{\prime },y^{\prime })$ coordinate location, the slope of the above-defined off-center ellipse is,

$\frac{d{y}^{\prime }}{d{x}^{\prime }}$

$=$

$(\frac{y_{\mathrm{m}\mathrm{a}\mathrm{x}}}{x_{\mathrm{m}\mathrm{a}\mathrm{x}}}{)}^2\frac{x^{\prime }}{(y_c-y^{\prime })}.$

From this expression we deduce that the x'- and y'- components of the velocity vector are, respectively,

$\frac{{\hat{\mathit{ı}}}^{\prime }\cdot {\mathit{u}}^{\prime }}{[{\mathit{u}}^{\prime }\cdot {\mathit{u}}^{\prime }{]}^{1/2}}$

$=$

$\frac{1}{u{\text{'}}_0}\left(\frac{x_{\mathrm{m}\mathrm{a}\mathrm{x}}}{y_{\mathrm{m}\mathrm{a}\mathrm{x}}}\right)(y_c-y^{\prime }),$

and,

$\frac{{\hat{\mathit{ȷ}}}^{\prime }\cdot {\mathit{u}}^{\prime }}{[{\mathit{u}}^{\prime }\cdot {\mathit{u}}^{\prime }{]}^{1/2}}$

$=$

$\frac{1}{u{\text{'}}_0}\left(\frac{y_{\mathrm{m}\mathrm{a}\mathrm{x}}}{x_{\mathrm{m}\mathrm{a}\mathrm{x}}}\right)x^{\prime },$

where the position-dependent — and, hence also, the time-dependent — length scale,

$u{\text{'}}_0$	$\equiv$	$\left\{\right[\left(\frac{x_{\mathrm{m}\mathrm{a}\mathrm{x}}}{y_{\mathrm{m}\mathrm{a}\mathrm{x}}}\right)(y_c-y^{\prime }){]}^2+\left[\frac{1}{|u^{\prime }|}\right(\frac{y_{\mathrm{m}\mathrm{a}\mathrm{x}}}{x_{\mathrm{m}\mathrm{a}\mathrm{x}}})x^{\prime }{]}^2{\}}^{1/2}$
	$=$	$\frac{1}{x_{\mathrm{m}\mathrm{a}\mathrm{x}}{y}_{\mathrm{m}\mathrm{a}\mathrm{x}}}[x_{\mathrm{m}\mathrm{a}\mathrm{x}}^4(y_c-y^{\prime }{)}^2+y_{\mathrm{m}\mathrm{a}\mathrm{x}}^4(x^{\prime }{)}^2{]}^{1/2}.$

STEP #4 Skipped this step on purpose.

See Also

• Riemann Type 1 Ellipsoids
• Construction Challenges (Pt. 1)
• Construction Challenges (Pt. 2)
• Construction Challenges (Pt. 3)
• Construction Challenges (Pt. 4)
• Construction Challenges (Pt. 5)
• Construction Challenges (Pt. 6)