Editing ThreeDimensionalConfigurations/ChallengesPt6 (section)

=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, [[ThreeDimensionalConfigurations/ChallengesPt4|''Construction Challenges (Pt. 4)'']].

==Intersection Expression==

<font color="red"><b>STEP #1</b></font>

First, we present the mathematical expression that describes the intersection between the surface of an  ellipsoid and a plane having the following properties:
<ul>
<li>The plane cuts through the ellipsoid's z-axis at a distance, <math>~z_0</math>, from the center of the ellipsoid;</li>
<li>The line of intersection is parallel to the x-axis of the ellipsoid; and,</li>
<li>The line that is perpendicular to the plane and passes through the z-axis at <math>~z_0</math> is tipped at an angle, <math>~\theta</math>, to the z-axis.</li>
</ul>
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, <math>~\theta</math>, 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.  
<table border="1" width="50%" cellpadding="8" align="center">
<tr>
  <td align="center" colspan="3"><b>Figure 1</b></td>
</tr>

<tr>
<td align="left">
<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>~x' \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~y</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
y' \cos\theta - z'\sin\theta 
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~(z - z_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
z' \cos\theta + y'\sin\theta 
\, .</math>
  </td>
</tr>
</table>
</td>

<td align="center">[[File:PrimedCoordinates3.png|250px|Primed Coordinates]]</td>
<td align="left">
<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>~x \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~y'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
y \cos\theta + (z - z_0) \sin\theta 
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~z'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(z-z_0) \cos\theta - y \sin\theta 
\, .</math>
  </td>
</tr>
</table>
</td>
</tr>
</table>


As has been shown in [[ThreeDimensionalConfigurations/ChallengesPt2#Intersection_of_Tipped_Plane_With_Ellipsoid_Surface|our accompanying discussion]], we obtain the following,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="3"><font color="maroon">'''Intersection Expression'''</font></td>
</tr>

<tr>
  <td align="right">
<math>~1 - \frac{x^2}{a^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~y^2 \biggl[\frac{c^2 + b^2\tan^2\theta}{b^2c^2} \biggr] + y \biggl[ \frac{2z_0 \tan\theta}{c^2} \biggr] + \frac{z_0^2}{c^2} \, , </math>
  </td>
</tr>
</table>
as long as z<sub>0</sub> lies within the range,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~z_0^2</math>
  </td>
  <td align="center">
<math>~\le</math>
  </td>
  <td align="left">
<math>~c^2 + b^2\tan^2\theta \, .</math>
  </td>
</tr>
</table>
Rewriting this "intersection expression" in terms of the ''tipped'' (primed) coordinate frame gives us,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~1 - \frac{(x')^2}{a^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(y' \cos\theta - z' \sin\theta)^2 \biggl[\frac{c^2 + b^2\tan^2\theta}{b^2c^2} \biggr] + (y' \cos\theta - z' \sin\theta) \biggl[ \frac{2z_0 \tan\theta}{c^2} \biggr] + \frac{z_0^2}{c^2} \, . </math>
  </td>
</tr>
</table>

<span id="Step2"><font color="red"><b>STEP #2</b></font></span>

As viewed from the ''tipped'' coordinated frame, the curve that is identified by this intersection should be an 
<table border="0" cellpadding="5" align="center">
<tr>
<td align="center" colspan="3"><font color="maroon">'''Off-Center Ellipse'''</font></td>
</tr>
<tr>
  <td align="right">
<math>~1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{x'}{x_\mathrm{max}} \biggr]^2 + \biggl[\frac{y' - y_c}{y_\mathrm{max}} \biggr]^2 </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{x'}{x_\mathrm{max}} \biggr]^2 + \biggl[\frac{(y')^2  - 2y' y_c + y_c^2}{y^2_\mathrm{max}} \biggr] \, ,</math>
  </td>
</tr>
</table>

<span id="Result3">that lies in the</span> x'-y' plane &#8212; that is, <math>~z' = 0</math>.  Let's see if the intersection expression can be molded into this form.
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~1 - \frac{z_0^2}{c^2} - \frac{(x')^2}{a^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(y')^2  \biggl[\frac{c^2 + b^2\tan^2\theta}{b^2c^2} \biggr]\cos^2\theta  + 2y' \biggl[ \frac{z_0 \sin\theta}{c^2} \biggr]  </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{c^2 + b^2\tan^2\theta}{b^2c^2} \biggr]\cos^2\theta \biggl\{ (y')^2 - 2y' \biggl[ \frac{-z_0 \sin\theta}{c^2 \cos^2\theta} \biggr]\biggl[\frac{b^2c^2}{c^2 + b^2\tan^2\theta} \biggr]  \biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\kappa^2 \biggl[ (y')^2 - 2y' \underbrace{\biggl( \frac{-z_0 \sin\theta}{c^2 \kappa^2} \biggr)}_{y_c}  \biggr] \, ,</math>
  </td>
</tr>
</table>

<table border="1" align="center" cellpadding="10" width="60%" bordercolor="orange">
<tr><td align="center" bgcolor="lightblue">'''RESULT 3'''<br />(same as [[ThreeDimensionalConfigurations/ChallengesPt2#Result1|Result 1]], but different from [[#Result2|Result 2, below]])
</td></tr>
<tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{y_c}{z_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\sin\theta}{c^2\kappa^2}
</math>
  </td>
</tr>
</table>

</td></tr>
</table>

where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\kappa^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{c^2 \cos^2\theta + b^2 \sin^2\theta}{b^2c^2} \, .
</math>
  </td>
</tr>
</table>
Dividing through by <math>~\kappa^2</math>, then adding <math>~y_c^2</math> to both sides gives,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(y')^2 - 2y' y_c  + y_c^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\underbrace{\biggl[ \frac{1}{\kappa^2} - \frac{z_0^2}{c^2 \kappa^2} + y_c^2 \biggr]}_{y^2_\mathrm{max}} - \frac{(x')^2}{a^2\kappa^2} \, .</math>
  </td>
</tr>
</table>
Finally, we have,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{1}{y^2_\mathrm{max}} \biggl[ (y')^2 - 2y' y_c  + y_c^2 \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - (x')^2 \underbrace{\biggl[ \frac{1}{a^2\kappa^2 y_\mathrm{max}^2} \biggr]}_{ 1/x^2_\mathrm{max} } \, .</math>
  </td>
</tr>
</table>
So &hellip; the intersection expression can be molded into the form of an off-center ellipse if we make the following associations:

<table border="1" cellpadding="8" align="center" width="60%"><tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{y_c}{z_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{\sin\theta}{c^2 \kappa^2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<font color="red">MISTAKE! ([[#FixedMistake|see below]]) &hellip;</font>&nbsp; &nbsp; &nbsp; 
<math>~y_\mathrm{max}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\kappa^2}\biggl[ 1 - \frac{z_0^2}{c^2 } - \frac{z_0 \sin\theta}{c^2} \biggr] \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<font color="red">MISTAKE! ([[#FixedMistake|see below]]) &hellip;</font>&nbsp; &nbsp; &nbsp; 
<math>~x_\mathrm{max}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a^2 \biggl[ 1 - \frac{z_0^2}{c^2 } - \frac{z_0 \sin\theta}{c^2} \biggr] \, .</math>
  </td>
</tr>
</table>
Note as well that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a^2\kappa^2 = \frac{a^2}{b^2 c^2} \biggl[ c^2 \cos^2\theta + b^2 \sin^2\theta \biggr] \, .</math>
  </td>
</tr>
</table>

</td></tr></table>

==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 &#8212; both the amplitude and its vector orientation &#8212; can be derived.

<font color="red"><b>STEP #3</b></font>

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 <math>~(x', y')</math> coordinate location, the slope of the [[#Step2|above-defined off-center ellipse]] is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{dy'}{dx'}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}} \biggr)^2 \frac{x'}{(y_c - y')} \, .
</math>
  </td>
</tr>
</table>

From this expression we deduce that the x'- and y'- components of the velocity vector are, respectively,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\boldsymbol{\hat\imath'} \cdot \boldsymbol{u'} }{ [\boldsymbol{u'}\cdot \boldsymbol{u'}]^{1 / 2} }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{u'_0} \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr) (y_c - y') \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~\frac{\boldsymbol{\hat\jmath'} \cdot \boldsymbol{u'} }{ [\boldsymbol{u'}\cdot \boldsymbol{u'}]^{1 / 2} }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{u'_0} \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) x' \, ,
</math>
  </td>
</tr>
</table>
where the position-dependent &#8212; and, hence also, the time-dependent &#8212; length scale,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~u'_0</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl\{
\biggl[ \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr) (y_c - y') \biggr]^2
+
\biggl[ \frac{1}{|u'|} \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) x' \biggr]^2
\biggr\}^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{x_\mathrm{max} y_\mathrm{max}}
\biggl[
x_\mathrm{max}^4 ( y_c - y')^2 
+
y_\mathrm{max}^4 (x')^2 
\biggr]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>

<font color="red"><b>STEP #4</b></font>

Skipped this step on purpose.

==Riemann Flow==

<font color="red"><b>STEP #5</b></font>

<table border="0" cellpadding="5" align="right">
<tr><td align="left" rowspan="4">
<table border="1" cellpadding="8" align="center">
<tr><th align="center" colspan="2">Example Type I<br />Ellipsoid<br />([[ThreeDimensionalConfigurations/RiemannTypeI#Example_b1.25c0.470|see also]])</th></tr>
<tr>
  <td align="center"><math>~\frac{b}{a} = \frac{a_2}{a_1}</math></td>
  <td align="center">1.25</td>
</tr>
<tr>
  <td align="center"><math>~\frac{c}{a} = \frac{a_3}{a_1}</math></td>
  <td align="center">0.4703</td>
</tr>
<tr>
  <td align="center"><math>~\Omega_2</math></td>
  <td align="center">0.3639</td>
</tr>
<tr>
  <td align="center"><math>~\Omega_3</math></td>
  <td align="center">0.6633</td>
</tr>
<tr>
  <td align="center"><math>~\tan^{-1} \biggl[ \frac{\Omega_3}{\Omega_2} \biggr]</math></td>
  <td align="center">61.25&deg;</td>
</tr>
<tr>
  <td align="center"><math>~\zeta_2</math></td>
  <td align="center">-2.2794</td>
</tr>
<tr>
  <td align="center"><math>~\zeta_3</math></td>
  <td align="center">-1.9637</td>
</tr>
<tr>
  <td align="center"><math>~\tan^{-1} \biggl[ \frac{\zeta_3}{\zeta_2} \biggr]</math></td>
  <td align="center">40.74&deg;</td>
</tr>
<tr>
  <td align="center"><math>~\beta_+</math></td>
  <td align="center">1.13449 (1.13332)</td>
</tr>
<tr>
  <td align="center"><math>~\gamma_+</math></td>
  <td align="center">1.8052</td>
</tr>
</table>
</td>
</tr>
</table>

As we have summarized in an [[ThreeDimensionalConfigurations/RiemannTypeI#EFEvelocities|accompanying discussion]] of Riemann Type 1 ellipsoids &#8212; see also [[ThreeDimensionalConfigurations/ChallengesPt3#Riemann-Derived_Expressions|our separate discussion]] &#8212; [[Appendix/References#EFE|[<font color="red">EFE</font>] ]]  provides an expression for the velocity vector of each fluid element, given its  instantaneous ''body''-coordinate position (x, y, z) = (x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>) &#8212; see his Eq. (154), Chapter 7, &sect;51 (p. 156).  As viewed from the rotating ''body'' coordinate frame, the three component expressions are,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\dot{x} = u_1 = \boldsymbol{\hat\imath} \cdot \boldsymbol{u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 y - \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 z</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[ \frac{a^2}{a^2 + b^2} \biggr] \zeta_3 y + \biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 z \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\dot{y} = u_2 = \boldsymbol{\hat\jmath} \cdot \boldsymbol{u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \gamma \Omega_3 x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~+\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3 x \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\dot{z} = u_3 = \boldsymbol{\hat{k}} \cdot \boldsymbol{u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~+ \beta \Omega_2 x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 x \, ,</math>
  </td>
</tr>
</table>
<span  id="betagamma">where,</span>

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\beta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \frac{\zeta_2}{\Omega_2}
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~\gamma</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl[ \frac{b^2}{a^2 + b^2} \biggr] \frac{\zeta_3}{\Omega_3} \, .
</math>
  </td>
</tr>
</table>


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:
<table border="1" align="center" width="60%" cellpadding="8"><tr>
<td align="left">

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\boldsymbol{\hat\imath}</math>
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~\boldsymbol{\hat\imath'} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\boldsymbol{\hat\jmath}</math>
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~\boldsymbol{\hat\jmath'}\cos\theta - \boldsymbol{\hat{k}'}\sin\theta \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\boldsymbol{\hat{k}}</math>
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~\boldsymbol{\hat\jmath'}\sin\theta + \boldsymbol{\hat{k}'}\cos\theta \, .</math>
  </td>
</tr>
</table>
</td>
<td align="left">

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x</math>
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~x' \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~y</math>
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~y' \cos\theta - z' \sin\theta \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~z - z_0</math>
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~y' \sin\theta + z'\cos\theta \, .</math>
  </td>
</tr>
</table>

</td></tr></table>

In the ''tipped'' frame, we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\boldsymbol{u'}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath'} \biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  y'\cos\theta - z'\sin\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 ( z_0 + y'\sin\theta + z'\cos\theta  ) \biggr]
+
[\boldsymbol{\hat{k}'}\sin\theta -\boldsymbol{\hat\jmath'}\cos\theta  ] \biggl[ \gamma \Omega_3 x' \biggr]
+
[ \boldsymbol{\hat\jmath'}\sin\theta + \boldsymbol{\hat{k}'}\cos\theta] \biggl[ \beta \Omega_2 x' \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath'} \biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  y'\cos\theta - z'\sin\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 ( z_0 + y'\sin\theta + z'\cos\theta  ) \biggr]
+
\boldsymbol{\hat\jmath'}  \biggl[\beta \Omega_2 x' \cdot \sin\theta 
- \gamma \Omega_3 x' \cdot \cos\theta \biggr]
+
\boldsymbol{\hat{k}'}\biggl[ \beta \Omega_2 x' \cdot \cos\theta
+
\gamma \Omega_3 x' \cdot \sin\theta \biggr] \, .
</math>
  </td>
</tr>
</table>
<span id="ThetaDef">In order</span> for the <math>~\boldsymbol{k}'</math> component to be zero in the tipped plane, we must choose the tipping angle such that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tan\theta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\beta\Omega_2}{\gamma \Omega_3} = -0.344793 ~~~\Rightarrow~~~ \theta = -19.0238^\circ \, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" width="80%" cellpadding="5"><tr><td align="left">
<div align="center"><font color="red"><b>NO!!!</b> This is wrong!</font></div>

And if we examine the flow only in the tipped x'-y' plane, then we should set <math>~z' = -z_0/\cos\theta</math>.  These two constraints lead to the velocity expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\boldsymbol{u'}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  y'\cos\theta + z_0\tan\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 (  y'\sin\theta ) \biggr]
+
\boldsymbol{\hat\jmath'}  \biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr]x' \, .
</math>
  </td>
</tr>
</table>

</td></tr></table>

And if we examine the flow only in the tipped x'-y' plane, then we should set <math>~z' = 0</math>.  These two constraints lead to the velocity expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\boldsymbol{u'}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  y'\cos\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 (z_0 +  y'\sin\theta ) \biggr]
+
\boldsymbol{\hat\jmath'}  \biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr]x' \, .
</math>
  </td>
</tr>
</table>

==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, <math>b/a = 1.25</math> and <math>c/a = 0.4703</math>, we can refer to the table labeled "Example Type I Ellipsoid" immediately above (under <font color="red">STEP #5</font>) to obtain and/or check values of <math>\Omega_2, \Omega_3, \zeta_2, \zeta_3, \beta^+, \gamma^+</math>.  From these tabulated values, we determine that the ''tip'' angle, <math>\tan\theta = -0.344793 ~\Rightarrow~ \theta = -0.33203</math>.

We then know that z<sub>0</sub> lies within the range,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~z_0^2</math>
  </td>
  <td align="center">
<math>~\le</math>
  </td>
  <td align="left">
<math>~c^2 + b^2\tan^2\theta = 0.40694</math>
  </td>
</tr>

<tr>
  <td align="center" colspan="3">
<math>\Rightarrow ~~~~- 0.63792 \le z_0 \le + 0.63792 \, .</math>
  </td>
</tr>
</table>
As our hand-determined example, let's choose, <math>z_0 = -0.4310</math>, which corresponds to the solid green ellipse that is displayed in [[ThreeDimensionalConfigurations/RiemannTypeI#Figure3|Figure 3c of an accompanying discussion]] of this problem.  <font color="red">NOTE: &nbsp; Just above the referenced "Figure 3," we have stated that the limits are, z<sub>0</sub> = &pm; 0.650165; not sure why that is!</font>

Next, we find that,  
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\kappa^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{c^2 \cos^2\theta + b^2 \sin^2\theta}{b^2c^2} = 1.05238 \, .
</math>
  </td>
</tr>
</table>

<span id="FixedMistake">So we can evaluate a number of terms that define the relevant elliptical orbit.</span>
<table border="1" cellpadding="8" align="center" width="60%">
<tr><td align="left">
<font color="red">CORRECTION:</font>

<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right"><math>y_\mathrm{max}^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{1}{\kappa^2} - \frac{z_0^2}{c^2 \kappa^2} + y_c^2</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{1}{\kappa^2}\biggl[1 - \frac{z_0^2}{c^2} + \frac{z_0^2 \sin^2\theta}{c^4 \kappa^2}\biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>\frac{1}{c^4\kappa^4}\biggl[c^4 \kappa^2 - z_0^2 c^2\kappa^2 + z_0^2 \sin^2\theta \biggr] = 0.51646</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ y_\mathrm{max}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>0.71865 \, .</math>
  </td>
</tr>
</table>
</td></tr>

<tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{y_c}{z_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{\sin\theta}{c^2 \kappa^2} = +1.40038~ \Rightarrow ~ y_c = -0.60356\, ,</math>
  </td>
</tr>
</table>
Note as well that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a^2\kappa^2 = \frac{a^2}{b^2 c^2} \biggl[ c^2 \cos^2\theta + b^2 \sin^2\theta \biggr] = 1.05238 
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{x_\mathrm{max}}{y_\mathrm{max}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>1.02585
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~x_\mathrm{max}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>0.73723
\, .</math>
  </td>
</tr>
</table>

</td></tr></table>

Select points along the elliptical orbit by first specifying various values of <math>y'</math> over the range,
<div align="center">
<math> 
(y_c - y_\mathrm{max}) \le y'\le (y_c + y_\mathrm{max})
~\Rightarrow~ -1.32221 \le y'\le 0.11509
</math>
</div>
Then, for each specified value of <math>y'</math>, calculate the corresponding value(s) of <math>x'</math> via the relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl[\frac{x'}{x_\mathrm{max}} \biggr]^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>1 - \biggl[\frac{y' - y_c}{y_\mathrm{max}} \biggr]^2 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ x'</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> \pm x_\mathrm{max} \biggl\{1 - \biggl[\frac{y' - y_c}{y_\mathrm{max}} \biggr]^2 \biggr\}^{1 / 2} \, . </math>
  </td>
</tr>
</table>