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.

Riemann Flow

STEP #5

Example Type I Ellipsoid (see also) $\frac{b}{a}=\frac{a_2}{a_1}$ 1.25 $\frac{c}{a}=\frac{a_3}{a_1}$ 0.4703 ${\mathrm{\Omega }}_2$ 0.3639 ${\mathrm{\Omega }}_3$ 0.6633 ${\tan }^{-1}\left[\frac{{\mathrm{\Omega }}_3}{{\mathrm{\Omega }}_2}\right]$ 61.25° ${\zeta }_2$ -2.2794 ${\zeta }_3$ -1.9637 ${\tan }^{-1}\left[\frac{{\zeta }_3}{{\zeta }_2}\right]$ 40.74° ${\beta }_{+}$ 1.13449 (1.13332) ${\gamma }_{+}$ 1.8052

As we have summarized in an accompanying discussion of Riemann Type 1 ellipsoids — see also our separate discussion — [EFE] provides an expression for the velocity vector of each fluid element, given its instantaneous body-coordinate position (x, y, z) = (x₁, x₂, x₃) — see his Eq. (154), Chapter 7, §51 (p. 156). As viewed from the rotating body coordinate frame, the three component expressions are,

$\dot{x}=u_1=\hat{\mathit{ı}}\cdot \mathit{u}$	$=$	$(\frac{a}{b}{)}^2\gamma {\mathrm{\Omega }}_{3}y-(\frac{a}{c}{)}^2\beta {\mathrm{\Omega }}_{2}z$	$=$	$-\left[\frac{a^2}{a^2+b^2}\right]{\zeta }_{3}y+\left[\frac{a^2}{a^2+c^2}\right]{\zeta }_{2}z,$
$\dot{y}=u_2=\hat{\mathit{ȷ}}\cdot \mathit{u}$	$=$	$-\gamma {\mathrm{\Omega }}_{3}x$	$=$	$+\left[\frac{b^2}{a^2+b^2}\right]{\zeta }_{3}x,$
$\dot{z}=u_3=\hat{\mathit{k}}\cdot \mathit{u}$	$=$	$+\beta {\mathrm{\Omega }}_{2}x$	$=$	$-\left[\frac{c^2}{a^2+c^2}\right]{\zeta }_{2}x,$

where,

$\beta$

$=$

$-\left[\frac{c^2}{a^2+c^2}\right]\frac{{\zeta }_2}{{\mathrm{\Omega }}_2}$

and,

$\gamma$

$=$

$-\left[\frac{b^2}{a^2+b^2}\right]\frac{{\zeta }_3}{{\mathrm{\Omega }}_3}.$

In order to transform Riemann's velocity vector from the body frame (unprimed) to the "tipped orbit" frame (primed coordinates), we use the following mappings of the three unit vectors:

$\hat{\mathit{ı}}$ $\to$ ${\hat{\mathit{ı}}}^{\prime },$ $\hat{\mathit{ȷ}}$ $\to$ ${\hat{\mathit{ȷ}}}^{\prime }\cos \theta -{\hat{\mathit{k}}}^{\prime }\sin \theta ,$ $\hat{\mathit{k}}$ $\to$ ${\hat{\mathit{ȷ}}}^{\prime }\sin \theta +{\hat{\mathit{k}}}^{\prime }\cos \theta .$

$x$ $\to$ $x^{\prime },$ $y$ $\to$ $y^{\prime }\cos \theta -z^{\prime }\sin \theta ,$ $z-z_0$ $\to$ $y^{\prime }\sin \theta +z^{\prime }\cos \theta .$

In the tipped frame, we find,

${{\mathit{u}}^{\prime }}_{\mathrm{E}\mathrm{F}\mathrm{E}}$	$=$	${\hat{\mathit{ı}}}^{\prime }\left[\right(\frac{a}{b}{)}^2\gamma {\mathrm{\Omega }}_3(y^{\prime }\cos \theta -z^{\prime }\sin \theta )-(\frac{a}{c}{)}^2\beta {\mathrm{\Omega }}_2(z_0+y^{\prime }\sin \theta +z^{\prime }\cos \theta )]+[{\hat{\mathit{k}}}^{\prime }\sin \theta -{\hat{\mathit{ȷ}}}^{\prime }\cos \theta ]\left[\gamma {\mathrm{\Omega }}_3{x}^{\prime }\right]+[{\hat{\mathit{ȷ}}}^{\prime }\sin \theta +{\hat{\mathit{k}}}^{\prime }\cos \theta ]\left[\beta {\mathrm{\Omega }}_2{x}^{\prime }\right]$
	$=$	${\hat{\mathit{ı}}}^{\prime }\left[\right(\frac{a}{b}{)}^2\gamma {\mathrm{\Omega }}_3(y^{\prime }\cos \theta -z^{\prime }\sin \theta )-(\frac{a}{c}{)}^2\beta {\mathrm{\Omega }}_2(z_0+y^{\prime }\sin \theta +z^{\prime }\cos \theta )]+{\hat{\mathit{ȷ}}}^{\prime }[\beta {\mathrm{\Omega }}_2{x}^{\prime }\cdot \sin \theta -\gamma {\mathrm{\Omega }}_3{x}^{\prime }\cdot \cos \theta ]+{\hat{\mathit{k}}}^{\prime }[\beta {\mathrm{\Omega }}_2{x}^{\prime }\cdot \cos \theta +\gamma {\mathrm{\Omega }}_3{x}^{\prime }\cdot \sin \theta ].$

In order for the ${\mathit{k}}^{\prime }$ component to be zero in the tipped plane, we must choose the tipping angle such that,

$\tan \theta$

$=$

$-\frac{\beta {\mathrm{\Omega }}_2}{\gamma {\mathrm{\Omega }}_3}=-0.344793\Rightarrow \theta =-19.0238^{\circ }.$

And if we examine the flow only in the tipped x'-y' plane, then we should set $z^{\prime }=-z_0/\cos \theta$. These two constraints lead to the velocity expression,

${{\mathit{u}}^{\prime }}_{\mathrm{E}\mathrm{F}\mathrm{E}}$

$=$

${\hat{\mathit{ı}}}^{\prime }\left[\right(\frac{a}{b}{)}^2\gamma {\mathrm{\Omega }}_3(y^{\prime }\cos \theta +z_0\tan \theta )-(\frac{a}{c}{)}^2\beta {\mathrm{\Omega }}_2(y^{\prime }\sin \theta )]+{\hat{\mathit{ȷ}}}^{\prime }[\beta {\mathrm{\Omega }}_2\sin \theta -\gamma {\mathrm{\Omega }}_3\cos \theta ]x^{\prime }.$

Graphically Compare Velocities

Using Excel, choose a particular elliptical orbit trajectory in the tipped plane, and plot the ratio of the Cartesian components of the fluid velocity as specified: (a) by EFE; and (b) by the tangent to the elliptical trajectory.

Adopting the axis-ratios, $b/a=1.25$ and $c/a=0.4703$, we can refer to the table labeled "Example Type I Ellipsoid" immediately above (under STEP #5) to obtain and/or check values of ${\mathrm{\Omega }}_2,{\mathrm{\Omega }}_3,{\zeta }_2,{\zeta }_3,{\beta }^{+},{\gamma }^{+}$. From these tabulated values, we determine that the tip angle, $\tan \theta =-0.344793\Rightarrow \theta =-0.33203$.

We then know that z₀ lies within the range,

$z_0^2$	$\le$	$c^2+b^2{\tan }^2\theta =0.40694$
$\Rightarrow -0.63792\le z_0\le +0.63792.$

As our hand-determined example, let's choose, $z_0=-0.4310$, which corresponds to the solid green ellipse that is displayed in Figure 3c of an accompanying discussion of this problem. NOTE:   Just above the referenced "Figure 3," we have stated that the limits are, z₀ = ± 0.650165; not sure why that is!

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)