Editing ThreeDimensionalConfigurations/ChallengesPt6

__FORCETOC__<!--  will force the creation of a Table of Contents -->
<!-- __NOTOC__ will force TOC off -->
=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>

=Orthogonality Check=

Let's see if the velocity vectors associated with the Riemann steady-state flow are orthogonal to the vector that is normal to the surface of the ellipsoid, at all points.

==EFE Rotating Frame==
The relevant "surface" is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>P(x, y, z)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{x}{a}\biggr)^2 + \biggl( \frac{y}{b}\biggr)^2 + \biggl( \frac{z}{c}\biggr)^2 \, ,</math>
  </td>
</tr>
</table>
where the constant, <math>0 \le P(x,y,z) \le 1</math>. And, [[#Riemann_Flow|from above]], the relevant EFE flow field is,

<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>
where,

<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>

We want to determine wether the, 
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="3"><font color="maroon">Orthogonality Condition</font></td>
</tr>
<tr>
  <td align="right">
<math>\mathbf{u} \cdot \nabla P</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0 \, .</math>
  </td>
</tr>
</table>
is satisfied everywhere inside, and on the surface, of the ellipsoid.  First, what is the expression for <math>\nabla P</math>?
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\nabla P</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\boldsymbol{\hat\imath} \biggl( \frac{2x}{a^2} \biggr) 
+ \boldsymbol{\hat\jmath} \biggl( \frac{2y}{b^2}\biggr) 
+ \mathbf{\hat{k}} \biggl( \frac{2z}{c^2}\biggr) \, .</math>
  </td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\mathbf{u} \cdot \nabla P</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{2x}{a^2} \biggr) \biggl\{ - \biggl[ \frac{a^2}{a^2 + b^2} \biggr] \zeta_3 y + \biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 z \biggr\}
+ \biggl( \frac{2y}{b^2}\biggr) \biggl\{ +\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3 x \biggr\}
+ \biggl( \frac{2z}{c^2}\biggr) \biggl\{ - \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 x \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{2x}{a^2} \biggr) \biggl[ \frac{a^2}{a^2 + c^2} \biggr] \zeta_2 z
- \biggl( \frac{2x}{a^2} \biggr) \biggl[ \frac{a^2}{a^2 + b^2} \biggr] \zeta_3 y
+ \biggl( \frac{2y}{b^2}\biggr) \biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3 x
- \biggl( \frac{2z}{c^2}\biggr) \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 x
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{2\zeta_2 ~ x z }{a^2 + c^2} \biggr] 
- \biggl[ \frac{2\zeta_3 ~ xy}{a^2 + b^2} \biggr]
+ \biggl[ \frac{2\zeta_3~ xy}{a^2 + b^2} \biggr] 
- \biggl[ \frac{2\zeta_2 ~xz}{a^2 + c^2} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
0 \, .
</math>
  </td>
</tr>
</table>

==Tipped Frame==

===For Arbitrary Tip Angles===

Given 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>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>
the constraint becomes,
<table border="0" cellpadding="5" align="center">

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

<span id="gradP">Hence,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\nabla P'(x', y', z')</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl(\frac{2x'}{a^2}\biggr)
+ \boldsymbol{\hat\jmath'} \biggl\{ \biggl[\frac{2( y'\cos\theta - z'\sin\theta )\cos\theta}{b^2}\biggr] 
+ \biggl[\frac{2( z_0 + z'\cos\theta + y'\sin\theta) \sin\theta}{c^2}\biggr] \biggr\}

+ \mathbf{\hat{k}'}\biggl\{- \biggl[\frac{2( y'\cos\theta - z'\sin\theta) \sin\theta}{b^2}\biggr] 
+ \biggl[\frac{2 ( z_0 + z'\cos\theta + y'\sin\theta) \cos\theta}{c^2}\biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>

And, 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^2}{a^2+b^2}\biggr] \zeta_3 (y'\cos\theta - z'\sin\theta) 
+ \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 (z_0 + y'\sin\theta + z'\cos\theta)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\biggl[\boldsymbol{\hat\jmath'} \cos\theta  - \mathbf{\hat{k}'} \sin\theta \biggr] \biggl\{ 
\biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 x'
\biggr\}
+
\biggl[\boldsymbol{\hat\jmath'} \sin\theta  + \mathbf{\hat{k}'} \cos\theta \biggr] \biggl\{ 
- 
\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 x'
\biggr\} \, .
</math>
  </td>
</tr>
</table>

As a result, we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\mathbf{u'}_\mathrm{EFE} \cdot \nabla P'</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{2x'}{a^2}\biggr)\biggl\{ 
- \biggl[\frac{a^2}{a^2+b^2}\biggr] \zeta_3 (y'\cos\theta - z'\sin\theta) 
+ \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 (z_0 + y'\sin\theta + z'\cos\theta)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+
\biggl\{
\biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 \cos\theta ~ x' - \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2\sin\theta ~x' 
\biggr\}
\biggl\{ \biggl[\frac{2( y'\cos\theta - z'\sin\theta )\cos\theta}{b^2}\biggr] 
+ \biggl[\frac{2( z_0 + z'\cos\theta + y'\sin\theta) \sin\theta}{c^2}\biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
- \biggl\{\biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 \sin\theta~ x'
+ \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 \cos\theta~ x'
\biggr\}
\biggl\{- \biggl[\frac{2( y'\cos\theta - z'\sin\theta) \sin\theta}{b^2}\biggr] 
+ \biggl[\frac{2 ( z_0 + z'\cos\theta + y'\sin\theta) \cos\theta}{c^2}\biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl[\frac{1}{2x'}  \biggr] \mathbf{u'}_\mathrm{EFE} \cdot \nabla P'</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \biggl[\frac{1}{a^2+b^2}\biggr] \zeta_3 (y'\cos\theta - z'\sin\theta) 
+ \biggl[ \frac{1}{a^2 + c^2}\biggr] \zeta_2 (z_0 + y'\sin\theta + z'\cos\theta)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+
\biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3    
\biggl\{ \biggl[\frac{( y'\cos\theta - z'\sin\theta )\cos^2\theta}{b^2}\biggr] 
+ \biggl[\frac{( z_0 + z'\cos\theta + y'\sin\theta) \sin\theta \cos\theta}{c^2}\biggr] \biggr\}
-
\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 
\biggl\{ \biggl[\frac{( y'\cos\theta - z'\sin\theta )\sin\theta\cos\theta}{b^2}\biggr] 
+ \biggl[\frac{( z_0 + z'\cos\theta + y'\sin\theta) \sin^2\theta}{c^2}\biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 
\biggl\{\biggl[\frac{( y'\cos\theta - z'\sin\theta) \sin^2\theta}{b^2}\biggr] 
- \biggl[\frac{ ( z_0 + z'\cos\theta + y'\sin\theta) \sin\theta \cos\theta}{c^2}\biggr] \biggr\}

+ \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2
\biggl\{\biggl[\frac{( y'\cos\theta - z'\sin\theta) \sin\theta \cos\theta}{b^2}\biggr] 
- \biggl[\frac{ ( z_0 + z'\cos\theta + y'\sin\theta) \cos^2\theta}{c^2}\biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \biggl[\frac{1}{a^2+b^2}\biggr] \zeta_3 (y'\cos\theta - z'\sin\theta) 
+ \biggl[ \frac{1}{a^2 + c^2}\biggr] \zeta_2 (z_0 + y'\sin\theta + z'\cos\theta)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+
\biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3    
\biggl[\frac{( y'\cos\theta - z'\sin\theta )}{b^2}\biggr] 
-
\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 
\biggl[\frac{( z_0 + z'\cos\theta + y'\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{( y'\cos\theta - z'\sin\theta )}{a^2 + b^2}\biggr] \zeta_3    
- \biggl[\frac{1}{a^2+b^2}\biggr] \zeta_3 (y'\cos\theta - z'\sin\theta) 
+ \biggl[ \frac{1}{a^2 + c^2}\biggr] \zeta_2 (z_0 + y'\sin\theta + z'\cos\theta)
-
\biggl[ \frac{( z_0 + z'\cos\theta + y'\sin\theta)}{a^2 + c^2} \biggr] \zeta_2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
0 \, .
</math>
  </td>
</tr>
</table>

Q. E. D.

===For Specific Tip Angle===

So &hellip; we have demonstrated that the velocity vectors are everywhere orthogonal to the normal to the ellipsoid for <b>all</b> values of the "tip" angle, <math>\theta</math>.  So why have we been unable to demonstrate the same result in the case where,
<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} = - \biggl[ \frac{c^2 (a^2 + b^2)}{b^2(a^2 + c^2)}\biggr]\frac{\zeta_2}{\zeta_3} \, .
</math>
  </td>
</tr>
</table>
Remember that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>1 + \tan^2\theta</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\sin^2\theta + \cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \cos^2\theta = \frac{1}{1 + \tan^2\theta}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>
\sin^2\theta = \frac{\tan^2\theta}{1 + \tan^2\theta} \, .
</math>
  </td>
</tr>
</table>

Rearranging terms in the expression for the "tipped plane" Riemann flow velocity, we have,
<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\{ 
- \cos\theta \biggl[\frac{a^2}{a^2+b^2}\biggr] \zeta_3 (y' - z'\tan\theta) 
+ \cos\theta \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 (y'\tan\theta + z')
+ \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 z_0
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\boldsymbol{\hat\jmath'} \biggl\{   
\biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 
-
\tan\theta 
\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 
\biggr\} x' \cos\theta
-
\mathbf{\hat{k}'} \biggl\{
\tan\theta  
\biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 
+
\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 
\biggr\} x' \cos\theta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl\{ 
- \cos\theta \biggl[\frac{a^2}{a^2+b^2}\biggr] \zeta_3 (y' - z'\tan\theta) 
+ \cos\theta \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 (y'\tan\theta + z')
+ \biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 z_0
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\boldsymbol{\hat\jmath'} \biggl\{   
1 
-
\tan\theta 
\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 
\biggl[ \frac{a^2 + b^2}{b^2}\biggr] \frac{1}{\zeta_3}
\biggr\} x' \cos\theta \biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3
-
\mathbf{\hat{k}'} \biggl\{
\tan\theta  
+
\biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 \biggl[ \frac{a^2 + b^2}{b^2}\biggr] \frac{1}{\zeta_3} 
\biggr\} x' \cos\theta \biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3 
</math>
  </td>
</tr>
</table>


<span id="SpecificTipAngle">Then, for this specific tip angle</span>, the Riemann flow velocity is,

<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^2}{a^2 + c^2}\biggr] \zeta_2 z_0
\biggr\}
+
\boldsymbol{\hat\imath'} \biggl\{ 
\zeta_3 (z'\tan\theta - y' ) \cos\theta 
+ \cos\theta \biggl[ \frac{a^2+b^2}{a^2 + c^2}\biggr] \zeta_2 (y'\tan\theta + z')
\biggr\}\biggl[\frac{a^2}{a^2+b^2}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\boldsymbol{\hat\jmath'} \biggl\{   
1 + \tan^2\theta 
\biggr\} x' \cos\theta \biggl[ \frac{b^2}{a^2 + b^2}\biggr] \zeta_3
-
\mathbf{\hat{k}'} \biggl\{
0\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^2}{a^2 + c^2}\biggr] \zeta_2 z_0
\biggr\}
+
\boldsymbol{\hat\imath'} \biggl\{ 
\zeta_3 (z'\tan\theta - y' )\cos\theta 
- \cos\theta \biggl[ \frac{b^2 \zeta_3 \tan\theta}{c^2 \zeta_2}\biggr] \zeta_2 (y'\tan\theta + z')
\biggr\}\biggl[\frac{a^2}{a^2+b^2}\biggr] 
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{b^2}{a^2 + b^2}\biggr] \frac{\zeta_3~x' }{\cos\theta}
</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^2}{a^2 + c^2}\biggr] \zeta_2 z_0
\biggr\}
+
\boldsymbol{\hat\imath'} \biggl\{ 
c^2 (z'\tan\theta - y' )\cos\theta 
- b^2 \sin\theta (y'\tan\theta + z')
\biggr\}\frac{\zeta_3 }{c^2}\biggl[\frac{a^2}{a^2+b^2}\biggr] 
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{b^2}{a^2 + b^2}\biggr] \frac{\zeta_3~x' }{\cos\theta}
</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^2}{a^2 + c^2}\biggr] \zeta_2 z_0
\biggr\}
+
\boldsymbol{\hat\imath'} \biggl\{ 
z' (c^2 - b^2 )\tan\theta  
- y' [c^2  + b^2 \tan^2\theta ]
\biggr\}\frac{\zeta_3 \cos\theta}{c^2}\biggl[\frac{a^2}{a^2+b^2}\biggr] 
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{b^2}{a^2 + b^2}\biggr] \frac{\zeta_3~x' }{\cos\theta}
</math>
  </td>
</tr>
</table>

As we have [[#gradP|already stated]],
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\nabla P'(x', y', z')</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl(\frac{2x'}{a^2}\biggr)
+ \boldsymbol{\hat\jmath'} \biggl\{ \biggl[\frac{2( y'\cos\theta - z'\sin\theta )\cos\theta}{b^2}\biggr] 
+ \biggl[\frac{2( z_0 + z'\cos\theta + y'\sin\theta) \sin\theta}{c^2}\biggr] \biggr\}

+ \mathbf{\hat{k}'}\biggl\{- \biggl[\frac{2( y'\cos\theta - z'\sin\theta) \sin\theta}{b^2}\biggr] 
+ \biggl[\frac{2 ( z_0 + z'\cos\theta + y'\sin\theta) \cos\theta}{c^2}\biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>

<span id="Orthogonal">Hence,</span>

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

<tr>
  <td align="right">
<math>\biggl( \frac{1}{2 x'}\biggr)\boldsymbol{u'}_\mathrm{EFE} \cdot \nabla P'</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{1}{a^2}\biggr)\biggl\{\biggl[ \frac{a^2}{a^2 + c^2}\biggr] \zeta_2 z_0
\biggr\}
+
\biggl(\frac{1}{a^2}\biggr)\biggl\{ 
z' (c^2 - b^2 )\tan\theta  
- y' [c^2  + b^2 \tan^2\theta ]
\biggr\}\frac{\zeta_3 \cos\theta}{c^2}\biggl[\frac{a^2}{a^2+b^2}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+
\biggl[ \frac{b^2}{a^2 + b^2}\biggr] \frac{\zeta_3 }{\cos\theta}
\biggl\{ \biggl[\frac{( y'\cos\theta - z'\sin\theta )\cos\theta}{b^2}\biggr] 
+ \biggl[\frac{( z_0 + z'\cos\theta + y'\sin\theta) \sin\theta}{c^2}\biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \biggl[ \frac{\zeta_3 }{c^2(a^2 + b^2)}\biggr] b^2 z_0 \tan\theta
+
\biggl\{ 
z' (c^2 - b^2 )\tan\theta  
- y' [c^2  + b^2 \tan^2\theta ]
\biggr\}\cos\theta \biggl[\frac{\zeta_3}{c^2(a^2+b^2)}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
+
\biggl[ \frac{\zeta_3 }{c^2 (a^2 + b^2)}\biggr] \frac{1}{\cos\theta}
\biggl\{ \biggl[c^2( y'\cos\theta - z'\sin\theta )\cos\theta\biggr] 
+ \biggl[b^2 ( z_0 + z'\cos\theta + y'\sin\theta) \sin\theta\biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl[ \frac{c^2(a^2 + b^2)}{\zeta_3 }\biggr] \biggl( \frac{1}{2 x'}\biggr)\boldsymbol{u'}_\mathrm{EFE} \cdot \nabla P'</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-  b^2 z_0 \tan\theta
+
\biggl\{ 
z' (c^2 - b^2 )\tan\theta  
- y' [c^2  + b^2 \tan^2\theta ]
\biggr\}\cos\theta  
+
\biggl[c^2( y'\cos\theta - z'\sin\theta )\biggr] 
+ \biggl[b^2 ( z_0 + z'\cos\theta + y'\sin\theta) \tan\theta\biggr] 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>0 \, .</math>
  </td>
</tr>
</table>

=With Help From Excel Spreadsheet=

[[File:DataFileButton02.png|left|60px|file = Dropbox/3Dviewers/RiemannModels/RiemannCalculations.xlsx --- worksheet = TypeI_1d]]Here, as an example, we consider the properties of a Riemann Type I ellipsoid whose semi-axes are <math>(a, b, c) = (1.0000, 1.2500, 0.4703)</math>.  Many properties of this particular model have been stored in the Dropbox location identified by scrolling your cursor over the yellow "Data Files" icon shown here, on the left.

==Body Frame of Riemann Type I Ellipsoid==
We recognize that, in the ''body frame'' of a Riemann ellipsoid, the surface of the configuration is defined by the following expression:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>1</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{x}{a}\biggr)^2 + \biggl( \frac{y}{b}\biggr)^2 + \biggl( \frac{z}{c}\biggr)^2 </math>
  </td>
</tr>
</table>

===Blue (x = 0) Ellipse===
By setting <math>x = z = 0</math>, we find the point where the y-axis intersects the surface of the ellipsoid, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>1</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{y}{b}\biggr)^2 ~~\Rightarrow ~~ y = y_\mathrm{max} = b \, .</math>
  </td>
</tr>
</table>

Similarly, by setting <math>x = y = 0</math>, we find the point where the z-axis intersects the surface of the ellipsoid, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>1</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{z}{c}\biggr)^2 ~~\Rightarrow ~~ z = z_\mathrm{max} = c \, .</math>
  </td>
</tr>
</table>

If we only set, <math>x = 0</math>, this expression generates an ellipse in the y-z plane whose semi-axes are <math>(y_\mathrm{max}, z_\mathrm{max}) = (1.25, 0.4703)</math>.  The <math>(y, z)</math> coordinates of individual points along the ellipse can be determined by choosing values of <math>y</math> in the range,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>- y_\mathrm{max} \le y \le + y_\mathrm{max} \, ,</math>
  </td>
</tr>
</table>
then determining the corresponding pair of values of <math>z</math> via the expression,
<table border="0" cellpadding="5" align="center">

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

This ellipse is identified in Figure 2 by the dotted-blue curve.

<table border="1" align="center" cellpadding="5">
<tr>
  <td align="center" colspan="2">Figure 2:  &nbsp; Y-Z plane(s) of Riemann Type I Ellipsoid[[File:DataFileButton02.png|right|60px|file = Dropbox/3Dviewers/RiemannModels/RiemannCalculations.xlsx --- worksheet = Feb22]]</td>
</tr>
<tr>
  <td align="center">[[File:YZplane3.png|400px|x = +0.70]]</td>
  <td align="center">[[File:YZplaneXm085.png|400px|x = -0.85]]
</tr>

<tr>
  <td align="center">Blue ellipse (x/a = 0.0); Green ellipse (x/a = + 0.70)</td>
  <td align="center">Blue ellipse (x/a = 0.0); Green ellipse (x/a = - 0.85)</td>
</tr>
</table>

===Green (x/a = 0.7) Ellipse===

Next, let's examine the surface-intersection-ellipse that results from a y-z plane that slices through the ellipsoid at <math>x/a = 0.7</math>.  By setting <math>z = 0</math>, we find the point where the y-axis intersects the surface of the ellipsoid, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>1 - (0.7)^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{y}{b}\biggr)^2 ~~\Rightarrow ~~ y = y_\mathrm{max} = b [1 - (0.7)^2]^{1 / 2} = 0.89268\, .</math>
  </td>
</tr>
</table>

Similarly, by setting <math>y = 0</math>, we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>1 - (0.7)^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{z}{c}\biggr)^2 ~~\Rightarrow ~~ z = z_\mathrm{max} = c [1 - (0.7)^2]^{1 / 2} = 0.33586\, .</math>
  </td>
</tr>
</table>
The <math>(y, z)</math> coordinates of individual points along this ellipse can be determined, as before, by choosing values of <math>y</math> in the range,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>- y_\mathrm{max} \le y \le + y_\mathrm{max} \, ,</math>
  </td>
</tr>
</table>
then determining the corresponding pair of values of <math>z</math> via the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>z_\pm</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\pm ~z_\mathrm{max} \biggl[1 - (0.7)^2 - \frac{y^2}{y_\mathrm{max}^2}  \biggr]^{1 / 2}
\, .</math>
  </td>
</tr>
</table>

This ellipse is identified in Figure 2 by the dotted-green curve.  All four of the red arrows (velocity vectors, as  explained below) that are displayed in Figure 2 are anchored on this dotted-green curve; the <math>(x, y, z)_\mathrm{base}</math> coordinates of these anchor positions are listed in the yellow-colored elements of the following Table titled, "Red Arrows."  (There is nothing special about these four chosen anchor positions other than they lie on the dotted-green ellipse.)

<table border="1" align="center" cellpadding="5">
<tr>
  <td align="center" colspan="12">'''Red Arrows'''<br />(Velocity Components in an y-z Plane)</td>
</tr>
<tr>
  <td align="center" rowspan="2">Number</td>
  <td align="center" colspan="4">Base of each Arrow</td>
  <td align="center" colspan="1" rowspan="6" bgcolor="lightgray">&nbsp;</td>
  <td align="center" colspan="2" rowspan="1">Velocity</td>
  <td align="center" colspan="1" rowspan="6" bgcolor="lightgray">&nbsp;</td>
  <td align="center" colspan="3">Arrow Tips</td>
</tr>
<tr>
  <td align="center"><math>x</math></td>
  <td align="center"><math>y</math></td>
  <td align="center"><math>z</math></td>
  <td align="center"><math>\biggl(\frac{x}{a}\biggr)^2 + \biggl(\frac{y}{b}\biggr)^2 + \biggl(\frac{z}{c}\biggr)^2</math></td>
  <td align="center" colspan="1" rowspan="1"><math>\dot{y} = u_2</math></td>
  <td align="center" colspan="1" rowspan="1"><math>\dot{z} = u_3</math></td>
  <td align="center"><math>x</math></td>
  <td align="center"><math>y</math></td>
  <td align="center"><math>z</math></td>
</tr>
<tr>
  <td align="center">1</td>
  <td align="right" bgcolor="yellow">0.7</td>
  <td align="right" bgcolor="yellow">-0.89268</td>
  <td align="right" bgcolor="yellow">0.00000</td>
  <td align="right">1.00000</td>
  <td align="right" rowspan="4">-1.19738 x</td>
  <td align="right" rowspan="4">+0.41285 x</td>
  <td align="right">0.7</td>
  <td align="right">-1.10222</td>
  <td align="right">+0.07225</td>
</tr>
<tr>
  <td align="center">2</td>
  <td align="right" bgcolor="yellow">0.7</td>
  <td align="right" bgcolor="yellow">-0.16396</td>
  <td align="right" bgcolor="yellow">+0.33015</td>
  <td align="right">1.00001</td>
  <td align="right">0.7</td>
  <td align="right">-0.37350</td>
  <td align="right">+0.40240</td>
</tr>
<tr>
  <td align="center">3</td>
  <td align="right" bgcolor="yellow">0.7</td>
  <td align="right" bgcolor="yellow">+0.81981</td>
  <td align="right" bgcolor="yellow">+0.13291</td>
  <td align="right">1.00000</td>
  <td align="right">0.7</td>
  <td align="right">+0.61027</td>
  <td align="right">+0.20516</td>
</tr>
<tr>
  <td align="center">4</td>
  <td align="right" bgcolor="yellow">0.7</td>
  <td align="right" bgcolor="yellow">+0.38258</td>
  <td align="right" bgcolor="yellow">-0.30345</td>
  <td align="right">0.99999</td>
  <td align="right">0.7</td>
  <td align="right">+0.17304</td>
  <td align="right">-0.23121</td>
</tr>
</table>
As a check, we have also included in the "Red Arrows" table a column that tallies,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\biggl[ \biggl(\frac{x}{a}\biggr)^2 + 
\biggl(\frac{y}{b}\biggr)^2 + \biggl(\frac{z}{c}\biggr)^2
\biggr]_\mathrm{base} \, ,</math>
  </td>
</tr>
</table>
which in every case totals 1.000, as it should.

===Velocity Components===

In steady-state, the velocity that is associated with each coordinate location can be ascertained from our [[#Riemann_Flow|above <font color="red">STEP #5</font> discussion of the Riemann Flow]].  Here we are especially interested in the velocity components that are in an x = constant, y-z plane, that is,

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

<tr>
  <td align="right">
<math>\dot{y} = u_2 </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
+\biggl[ \frac{b^2}{a^2 + b^2} \biggr] \zeta_3 x 
=
+\biggl[ \frac{(1.25)^2}{1 + (1.25)^2} \biggr] (-1.9637) x
=
-1.19738 x
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\dot{z} = u_3 </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \biggl[ \frac{c^2}{a^2 + c^2} \biggr] \zeta_2 x 
=
-\biggl[ \frac{(0.4703)^2}{1 + (0.4703)^2} \biggr] (-2.2794) x
=
+0.41285 x
\, .</math>
  </td>
</tr>
</table>
Given that all of the points along the black-dotted ellipse in our ''Red Arrows'' figure are positioned in the <math>x = 0</math>, y-z plane, we appreciate that <math>u_2 = u_3 = 0</math> at all points along this black ellipse.  But along the green-dotted ellipse, for which <math>x/a = 0.7</math>, each fluid element exhibits a nonzero component of motion in the relevant y-z plane; specifically, for all points along the green ellipse,

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

<tr>
  <td align="right">
<math>u_2\biggr|_{x=0.7} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-0.83816
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>u_3\biggr|_{x=0.7} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
+0.28899
\, .</math>
  </td>
</tr>
</table>

The pointed tip of each of the four red arrows in our figure is located at a coordinate position, <math>(y_\mathrm{tip}, z_\mathrm{tip})</math>, determined from the following pair of expressions: 

<table border="0" align="center" cellpadding="5">
<tr>
   <td align="center"><math>y_\mathrm{tip}</math></td>
   <td align="center"><math>=</math></td>
   <td align="center"><math>y_\mathrm{base} + \tfrac{1}{4} u_2\biggr|_{x=0.7} = y_\mathrm{base} - \tfrac{1}{4}(0.83817) \, ,</math></td>
</tr>
<tr>
   <td align="center"><math>z_\mathrm{tip}</math></td>
   <td align="center"><math>=</math></td>
   <td align="center"><math>z_\mathrm{base} + \tfrac{1}{4} u_2\biggr|_{x=0.7} = z_\mathrm{base} + \tfrac{1}{4}(0.28899) \, .</math></td>
</tr>
</table>
That is, each arrow illustrates how far a fluid element would travel away from its ''base'' location if it moved at the prescribed velocity for a time, <math>\Delta t = \tfrac{1}{4} \times [\pi G \rho]^{-1 / 2}</math>.  The four red arrows serve to illustrate that, at every point along the dotted-green ellipse, the component of the velocity that lies in the y-z plane is precisely the same, in both magnitude and direction.

[[File:PrimedCoordinates3.png|250px|right|Primed Coordinates]]Now, if we were to examine in a similar manner the component of the fluid motion in any other x = constant, y-z plane, we would find that the red velocity vectors arising from every ''base point'' along the relevant ellipse in this new y-z plane would be the same &#8212; in both magnitude and direction &#8212; around the entire ellipse.  Relative to the (dotted green) ellipse that lies in the x = 0.7, y-z plane, the magnitudes would be different &#8212; larger for larger values of <math>x</math> and smaller for smaller values of <math>x</math> &#8212; however, all of the ''red arrows'' in the new y-z plane would point in the same ''direction'' as the red arrows displayed in Figure 2.

All of the "red arrow" flow-components are tipped up, out of the x-y plane by an angle that is given by the ratio of the pair of velocity components, <math>(u_2, u_3)</math>.  Specifically,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>\tan\theta</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>\frac{u_3}{u_2} = \frac{+0.41285 x}{- 1.19738 x} = -0.34479 \, ,</math></td>
</tr>
<tr>
  <td align="right"><math>\Rightarrow ~~~ \theta</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>-0.33203</math> radians <math>= -19.024^\circ \, .</math></td>
</tr>
</table>
Borrowing from Figure 1, above, and appreciating that <math>\theta</math> is ''negative'' in the example being used here, we would find that all of the red arrows, in all of the x = constant, y-z planes would lie parallel to the <math>y'</math> axis.  In our Figure 2, above, the black, dashed line serves to illustrate one such <math>y'</math> axis; it has been drawn using the expression,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>z</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>y \tan\theta + z_0 \, ,</math></td>
</tr>
</table>
where we have set <math>\tan\theta = -0.34479</math> and <math>z_0 = 0.12758</math> for <math>-1.25 \le y \le + 1.25</math>.

==Tipped Frame==
Let's continue to examine the x' = x = constant, y'-z' plane, and set <math>\tan\theta = -0.34479</math> and <math>z_0 = 0.12758</math>.

===Draw Tilted Ellipse===
So, for a fixed value of <math>x'</math> and over this range in <math>z'</math>, the value of <math>y'</math> is obtained from the expression,
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
c^2\biggl[y'\cos\theta - z'\sin\theta \biggr]^2 
+ b^2\biggl[z_0 + z'\cos\theta + y'\sin\theta \biggr]^2
+b^2c^2 \biggl[\biggl(\frac{x'}{a}\biggr)^2 -1 \biggr]
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
A (y')^2 
+ B y' 
+ C \, ,
</math>
  </td>
</tr>
</table>
where,

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

<tr>
  <td align="right">
<math>A</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
c^2 \cos^2\theta + b^2 \sin^2\theta \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>B</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
2b^2\sin\theta(z_0 + z'\cos\theta )- 2c^2 z'\sin\theta \cos\theta  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>C</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
b^2(z_0 + z'\cos\theta )^2 +  c^2(z')^2\sin^2\theta + b^2c^2 \biggl[\biggl(\frac{x'}{a}\biggr)^2 -1 \biggr] \, .
</math>
  </td>
</tr>
</table>
In which case,
<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>
\frac{B}{2A}\biggl\{ - 1 \pm \biggl[1 - \frac{4AC}{B^2}  \biggr]^{1 / 2} \biggr\} \, .
</math>
  </td>
</tr>
</table>

<span id="zlimits">At what value(s)</span> of <math>z'</math> do we find that, <math>B^2 = 4AC</math>?

<table border="1" align="center" width="90%" cellpadding="8">
<tr><td align="left">

<table border="0" cellpadding="3" align="center">

<tr>
  <td align="right">
<math>B^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
4AC
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ b^2\sin\theta(z_0 + z'\cos\theta )- c^2 z'\sin\theta \cos\theta \biggr]^2
~-~ 
\biggl[  c^2 \cos^2\theta + b^2 \sin^2\theta \biggr]
\biggl[ b^2(z_0 + z'\cos\theta )^2 +  c^2(z')^2\sin^2\theta - b^2c^2 \epsilon^2  \biggr]
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
b^4\sin^2\theta(z_0 + z'\cos\theta )^2
- 2 b^2 c^2 (z')\sin^2\theta \cos\theta  (z_0 + z'\cos\theta )
+ c^4 (z')^2 \sin^2\theta \cos^2\theta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~-~ 
b^2c^2 \cos^2\theta (z_0 + z'\cos\theta )^2
~-~ 
b^4 \sin^2\theta (z_0 + z'\cos\theta )^2 
~-~ 
c^4 (z')^2 \sin^2\theta\cos^2\theta 
~-~ 
b^2 c^2(z')^2 \sin^4\theta 
~+~ 
b^2c^2 \epsilon^2 (c^2 \cos^2\theta + b^2 \sin^2\theta )
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- 2 b^2 c^2 z_0 (z')\sin^2\theta \cos\theta
- 2 b^2 c^2 (z')^2 \sin^2\theta \cos^2\theta
+ c^4 (z')^2 \sin^2\theta \cos^2\theta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~-~ 
b^2c^2 \cos^2\theta \biggl[ z_0^2 + 2z_0 (z') \cos\theta + (z')^2 \cos^2\theta \biggr]
~-~ 
c^4 (z')^2 \sin^2\theta\cos^2\theta 
~-~ 
b^2 c^2(z')^2 \sin^4\theta 
~+~ 
b^2c^2 \epsilon^2 (c^2 \cos^2\theta + b^2 \sin^2\theta )
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ c^4 \sin^2\theta \cos^2\theta
- 2 b^2 c^2 \sin^2\theta \cos^2\theta ~-~ b^2c^2 \cos^4\theta 
~-~ c^4 \sin^2\theta\cos^2\theta ~-~ b^2 c^2 \sin^4\theta 
\biggr](z')^2 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
~-~ b^2 c^2 \biggl[ 
2 \sin^2\theta \cos^2\theta + \cos^4\theta + \sin^4\theta \biggr](z')^2 
~-~ 2z_0 b^2c^2 \cos\theta (z') 
~+~  b^2c^2 \epsilon^2 (c^2 \cos^2\theta + b^2 \sin^2\theta )
~-~ z_0^2 b^2c^2 \cos^2\theta \, .
</math>
  </td>
</tr>
</table>
Hence, after dividing through by <math>(-b^2c^2)</math>,

<table border="0" cellpadding="3" align="center">

<tr>
  <td align="right">
<math>0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(z')^2 
~+~ 2z_0 \cos\theta (z') 
~+~ z_0^2 \cos^2\theta ~-~  \epsilon^2 (c^2 \cos^2\theta + b^2 \sin^2\theta )
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~ \frac{z'}{z_0\cos\theta}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\pm \biggl\{ 1 -  \frac{z_0^2 \cos^2\theta ~-~  \epsilon^2 (c^2 \cos^2\theta + b^2 \sin^2\theta )}{z_0^2 \cos^2\theta} \biggr\}^{1 / 2}
- 1
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\pm \biggl\{ \frac{ \epsilon^2 (c^2 \cos^2\theta + b^2 \sin^2\theta )}{z_0^2 \cos^2\theta} \biggr\}^{1 / 2}
- 1
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~ z'</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\pm \epsilon (c^2 \cos^2\theta + b^2 \sin^2\theta )^{1 / 2}
- z_0\cos\theta
= - 0.12061 \pm 0.60307
=
( 0.48246  , -0.72369  )
</math>
  </td>
</tr>
</table>

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

===Limits on z'===

The limits on <math>z'</math> will occur where we find that <math>dz'/dy' = 0</math>. We will figure out where, along the tilted ellipse, this happens by differentiating both sides of this last expression.  First, note that,

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

<tr>
  <td align="right">
<math>dB</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ 2b^2\sin\theta \cos\theta  - 2c^2 \sin\theta \cos\theta \biggr]dz'
=
2\sin\theta \cos\theta (b^2  - c^2 )dz' \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>dC</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
b^2\biggl[ 2z_0 \cos\theta + 2(z') \cos^2\theta \biggr]dz' +  2c^2 z'\sin^2\theta dz'
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl[ 2b^2 z_0 \cos\theta + 2z' (b^2 \cos^2\theta  +  c^2 \sin^2\theta )\biggr] dz' \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>2A dy' </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- dB \pm d\biggl[B^2 - 4AC \biggr]^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- dB ~\pm~ \tfrac{1}{2}[B^2 - 4AC ]^{-1 / 2} \biggl[ 2B \cdot dB \biggr]
~\mp~ \tfrac{1}{2}[B^2 - 4AC ]^{-1 / 2} \biggl[ 4A \cdot dC \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~2A(B^2 - 4AC )^{1 / 2} dy'</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \pm~ B  - (B^2 - 4AC )^{1 / 2} \biggr]dB ~\mp~ 2A dC 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl\{
\biggl[ \pm~ B  - (B^2 - 4AC )^{1 / 2} \biggr]2\sin\theta \cos\theta (b^2  - c^2 )
\mp~ 2A \biggl[ 2b^2 z_0 \cos\theta + 2z' (b^2 \cos^2\theta  +  c^2 \sin^2\theta )\biggr] \biggr\} dz' 
\, .
</math>
  </td>
</tr>
</table>

The derivative, <math>dz'/dy'</math>, will go to zero when the coefficient on the LHS goes to zero, that is, when,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>B^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
4AC \, .
</math>
  </td>
</tr>
</table>
From the [[#zlimits|boxed-in derivation, above]], we know that this occurs when,

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

<tr>
  <td align="right">
<math>z'</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\pm \epsilon (c^2 \cos^2\theta + b^2 \sin^2\theta )^{1 / 2}
- z_0\cos\theta \, .
</math>
  </td>
</tr>
</table>
And the corresponding value(s) of <math>y'</math> comes from the relation,
<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>
\frac{B}{2A}\biggl\{ - 1 \pm \cancelto{0}{\biggl[1 - \frac{4AC}{B^2}  \biggr]^{1 / 2}} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\frac{b^2\sin\theta(z_0 + z'\cos\theta )- c^2 z'\sin\theta \cos\theta}{c^2 \cos^2\theta + b^2 \sin^2\theta} \, .
</math>
  </td>
</tr>
</table>
A numerical evaluation for our sample problem gives:
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>(z'_\mathrm{min}, y')</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>(-0.72369, -0.64380)</math></td>
</tr>
<tr>
  <td align="right"><math>(z'_\mathrm{max}, y')</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>(+ 0.48246, +0.72697)</math></td>
</tr>
</table>

<table border="1" align="center" cellpadding="5">
<tr>
  <td align="center" colspan="2">Figure 3:  &nbsp; Tipped Y'-Z' plane(s) of Riemann Type I Ellipsoid[[File:DataFileButton02.png|right|60px|file = Dropbox/3Dviewers/RiemannModels/RiemannCalculations.xlsx --- worksheet = Feb22tip]]</td>
</tr>
<tr>
  <td align="center">[[File:YZtipped01.png|400px|x = +0.70]]</td>
  <td align="center">[[File:YZtippedXm085.png|400px|x = -0.85]]
</tr>

<tr>
  <td align="center">Blue ellipse (x/a = 0.0); Green ellipse (x/a = + 0.70)</td>
  <td align="center">Blue ellipse (x/a = 0.0); Green ellipse (x/a = - 0.85)</td>
</tr>
</table>

===Lagrangian Trajectories in x'-y' Plane===

====Initial Determination====

For a given choice of <math>z_0</math>, let's map out the Lagrangian trajectory in the x'-y' "equatorial" (i.e., z' = 0) plane.  The y'(x') relation is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>(y')_{z_0} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{
\frac{B}{2A}\biggl[ - 1 \pm \biggl(1 - \frac{4AC}{B^2}  \biggr)^{1 / 2} \biggr] \biggr\}_{z'=0} \, ,
</math>
  </td>
</tr>
</table>
where,

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

<tr>
  <td align="right">
<math>A_{z_0} = A</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
c^2 \cos^2\theta + b^2 \sin^2\theta \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>B_{z_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ 2b^2\sin\theta(z_0 + z'\cos\theta )- 2c^2 z'\sin\theta \cos\theta \biggr]_{z'=0} 
=
2b^2\sin\theta(z_0 )  
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>C_{z_0}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ b^2(z_0 + z'\cos\theta )^2 +  c^2(z')^2\sin^2\theta - b^2c^2 \epsilon^2  \biggr]_{z'=0}
=
b^2(z_0 )^2  - b^2c^2 \epsilon^2
\, .
</math>
  </td>
</tr>
</table>
That is to say,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>2( c^2 \cos^2\theta + b^2 \sin^2\theta )(y')_{z_0} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- B_{z_0} \pm \biggl[B^2_{z_0} - 4AC_{z_0}  \biggr]^{1 / 2} 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- z_0 b^2\sin\theta \pm bc\biggl[ 
\epsilon^2 (c^2  \cos^2\theta   
+ b^2 \sin^2\theta )  
- z_0^2 \cos^2\theta   
\biggr]^{1 / 2} 
\, .
</math>
  </td>
</tr>
</table>

For a given choice of <math>z_0</math>, the limits on x' are given by when the argument of the square root is set to zero, that is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\epsilon^2  </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{z_0^2 \cos^2\theta}{(c^2  \cos^2\theta + b^2 \sin^2\theta )}   
</math>
  </td>
</tr>

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

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

<table border="1" align="center" cellpadding="3">
<tr><td align="center">
[[File:XYSlices01.png|450px|center|Various z_0 Slices]]
</td></tr>
</table>

====Are Orbits Exact Circles====
After plotting <math>(y')_{z_0}</math> as a function of <math>x'</math> (between the just-derived limits) for several different values of <math>z_0</math>, we noticed that each Lagrangian trajectory appears to be a circle.  If this is exactly the case &hellip;

<font color="red">1.) RADIUS OF CIRCLE, MEASURED PERPENDICULAR TO x'-AXIS:</font> &nbsp; Given simply by this last expression, namely,

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

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

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


<font color="red">2.) CENTER OF CIRCLE:</font> &nbsp; The y'-coordinate of the center of the circle is the value of <math>(y')_{z_0}</math> obtained when the argument of the square root goes to zero.  That is,

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

<tr>
  <td align="right">
<math>( c^2 \cos^2\theta + b^2 \sin^2\theta )[(y')_{z_0}]_0 </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- z_0 b^2\sin\theta 
\pm 
bc \cancelto{0}{\biggl[ \epsilon^2 (c^2  \cos^2\theta   + b^2 \sin^2\theta )  - z_0^2 \cos^2\theta   \biggr]^{1 / 2} }
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ [(y')_{z_0}]_0 </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-  \frac{z_0 b^2\sin\theta }{ ( c^2 \cos^2\theta + b^2 \sin^2\theta )} \, .
</math>
  </td>
</tr>
</table>
And, along the <math>x'</math>-axis, the inner and outer edges of the circle are identified by the positions at which <math>x'/a = 0 ~\Rightarrow ~ \epsilon = 1</math>.  That is,

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

<tr>
  <td align="right">
<math>( c^2 \cos^2\theta + b^2 \sin^2\theta )[(y')_{z_0}]_\mathrm{limits} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- z_0 b^2\sin\theta 
\pm 
bc \biggl[ \cancelto{1}{\epsilon^2} (c^2  \cos^2\theta   + b^2 \sin^2\theta )  - z_0^2 \cos^2\theta   \biggr]^{1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ [(y')_{z_0}]_\mathrm{limits} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \biggl\{ z_0 b^2\sin\theta 
\mp 
bc \biggl[ (c^2  \cos^2\theta   + b^2 \sin^2\theta )  - z_0^2 \cos^2\theta   \biggr]^{1 / 2}\biggr\}( c^2 \cos^2\theta + b^2 \sin^2\theta )^{-1} \, .
 </math>
  </td>
</tr>
</table>

<font color="red">3.) RADIUS OF CIRCLE, MEASURED PERPENDICULAR TO x'-AXIS:</font> &nbsp; The radius of the "circle" along the <math>x'</math>-axis is,

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

<tr>
  <td align="right">
<math>[(y')_{z_0}]_\mathrm{radius} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
bc \biggl[ (c^2  \cos^2\theta   + b^2 \sin^2\theta )  - z_0^2 \cos^2\theta   \biggr]^{1 / 2} ( c^2 \cos^2\theta + b^2 \sin^2\theta )^{-1}
 </math>
  </td>
</tr>

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

<table border="1" align="center" cellpadding="5">
<tr>
  <td align="center"><math>\frac{z_0}{z_\mathrm{max}}</math></td>
  <td align="center"><math>(x')_\mathrm{radius}</math></td>
  <td align="center"><math>(y')_0</math></td>
  <td align="center"><math>(y')_\mathrm{radius}</math></td>
  <td align="center"><math>\biggl[ \frac{y'}{x'}  \biggr]_\mathrm{radius}</math></td>
</tr>
<tr>
  <td align="right">0.000</td>
  <td align="right">1.000</td>
  <td align="right">0.000</td>
  <td align="right">0.974797</td>
  <td align="right" bgcolor="yellow">0.974797</td>
</tr>
<tr>
  <td align="right">- 0.700</td>
  <td align="right">0.714143</td>
  <td align="right">- 0.625325</td>
  <td align="right">0.696144</td>
  <td align="right" bgcolor="yellow">0.974797</td>
</tr>
<tr>
  <td align="right">- 0.975</td>
  <td align="right">0.222205</td>
  <td align="right">- 0.870989</td>
  <td align="right">0.216605</td>
  <td align="right" bgcolor="yellow">0.974797</td>
</tr>
</table>

How do the two radii compare?  As the (immediately above) table illustrates, each trajectory's x'-radius is slightly larger than that trajectory's y'-radius.  Hence, the orbits are not circular!  However, as the last column (bgcolor="yellow") tabulates, the degree of flattening is very slight and, surprisingly, the ratio of radii is ''identical'' in every case.

Let's examine the analytic expression for the ratio of radii:

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{ ( c^2 \cos^2\theta + b^2 \sin^2\theta )^{1 / 2}}{bc} \, .
</math>
  </td>
</tr>
</table>

From this last expression, we see that the two radii will be the same &#8212; thereby making the LHS unity &#8212; only if,

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

<tr>
  <td align="right">
<math>
b^2 c^2
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
c^2 \cos^2\theta + b^2 \sin^2\theta 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
c^2 \cos^2\theta + b^2 (1 - \cos^2\theta) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ b^2 (c^2 - 1)
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
(c^2 - b^2) \cos^2\theta  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \cos^2\theta
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{b^2 (c^2 - 1)}{(c^2 - b^2) } = 0.907244
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow \theta
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\pm 0.30948 \, .
</math>
  </td>
</tr>
</table>
This is not the case for our example model; its tilt angle is, instead, <math>\theta = -0.332029</math>.

====Plot Off-Center, Slightly Flattened Ellipse====


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

<tr>
  <td align="right">
<math>
\biggl[ \frac{y' - y'_0}{y'_\mathrm{radius}} \biggr]^2
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 - \biggl( \frac{x'}{x'_\mathrm{radius}} \biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \frac{y' - y'_0}{y'_\mathrm{radius}} 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\pm \biggl[ 1 - \biggl( \frac{x'}{x'_\mathrm{radius}} \biggr)^2 \biggr]^{1 / 2}
</math>
  </td>
</tr>
</table>
This seems to work perfectly!  We used Excel to generate trajectories using this expression and the results matched earlier determinations of these trajectories to machine precision.  

Let's now examine the normal to the surface that is obtained from this compact trajectory expression.  Given that <math>y'_0</math>, <math>y'_\mathrm{radius}</math>, and <math>x'_\mathrm{radius}</math> are all constants, we have,

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

<tr>
  <td align="right">
<math>
G'
</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{y' - y'_0}{y'_\mathrm{radius}} \biggr]^2
+
\biggl( \frac{x'}{x'_\mathrm{radius}} \biggr)^2 - 1
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\frac{1}{2} \nabla G'
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl[ \frac{x'}{(x')^2_\mathrm{radius}} \biggr] 
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{y' - y'_0}{(y')^2_\mathrm{radius}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl[ \frac{x'}{a^2} \biggr] 
\biggl[ \frac{(c^2  \cos^2\theta + b^2 \sin^2\theta )}{(c^2 - z_0^2) \cos^2\theta + b^2\sin^2\theta} \biggr]
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{y' }{b^2 c^2} \biggr]
\biggl[ \frac{(c^2  \cos^2\theta + b^2 \sin^2\theta )^2}{(c^2 - z_0^2) \cos^2\theta + b^2\sin^2\theta} \biggr]
-
\boldsymbol{\hat\jmath'} \biggl[ \frac{y'_0}{b^2 c^2} \biggr]
\biggl[ \frac{(c^2  \cos^2\theta + b^2 \sin^2\theta )^2}{(c^2 - z_0^2) \cos^2\theta + b^2\sin^2\theta} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ \biggl[ \frac{ (c^2 - z_0^2) \cos^2\theta + b^2\sin^2\theta}{ (c^2  \cos^2\theta + b^2 \sin^2\theta ) } \biggr] \frac{1}{2} \nabla G'
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl[ \frac{x'}{a^2} \biggr] 
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{y' }{b^2 c^2} \biggr]
\biggl[ (c^2  \cos^2\theta + b^2 \sin^2\theta ) \biggr]
-
\boldsymbol{\hat\jmath'} \biggl[ \frac{y'_0}{b^2 c^2} \biggr]
\biggl[ (c^2  \cos^2\theta + b^2 \sin^2\theta ) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl[ \frac{x'}{a^2} \biggr] 
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{y' }{b^2 c^2} \biggr]
\biggl[ (c^2  \cos^2\theta + b^2 \sin^2\theta ) \biggr]
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{1}{b^2 c^2} \biggr]
\biggl[ z_0 b^2 \sin\theta \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl[ \frac{x'}{a^2} \biggr] 
+
\boldsymbol{\hat\jmath'} \biggl[ \frac{1}{b^2 c^2} \biggr]\biggl\{
y'(c^2  \cos^2\theta + b^2 \sin^2\theta )
+
z_0 b^2 \sin\theta \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl[ \frac{x'}{a^2} \biggr] 
+
\boldsymbol{\hat\jmath'} \biggl\{
\frac{y' \cos^2\theta }{b^2}
+ \frac{(y'\sin\theta +z_0 )\sin\theta}{c^2}
\biggr\} \, .
</math>
  </td>
</tr>
</table>

As we have [[#gradP|already stated]] &#8212; but setting <math>z' = 0</math> and ignoring the <math>\mathbf{\hat{k}'}</math> component because there is no motion in that direction &#8212; 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{1}{2} \nabla P'(x', y', z')</math>
  </td>
  <td align="center">
<math>\rightarrow</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl(\frac{x'}{a^2}\biggr)
+ \boldsymbol{\hat\jmath'} \biggl\{ \biggl[\frac{( y'\cos\theta - \cancelto{0}{z'\sin\theta} )\cos\theta}{b^2}\biggr] 
+ \biggl[\frac{( z_0 + \cancelto{0}{z'\cos\theta} + y'\sin\theta) \sin\theta}{c^2}\biggr] \biggr\}

+ \mathbf{\hat{k}'} \cancelto{0}{\biggl\{- \biggl[\frac{( y'\cos\theta - z'\sin\theta) \sin\theta}{b^2}\biggr] 
+ \biggl[\frac{ ( z_0 + z'\cos\theta + y'\sin\theta) \cos\theta}{c^2}\biggr] \biggr\}} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\boldsymbol{\hat\imath'} \biggl(\frac{x'}{a^2}\biggr)
+ \boldsymbol{\hat\jmath'} \biggl\{ \biggl[\frac{( y'\cos\theta  )\cos\theta}{b^2}\biggr] 
+ \biggl[\frac{( z_0 + y'\sin\theta) \sin\theta}{c^2}\biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>

Since, properly normalized, <math>\nabla G'</math> is identical to <math>\nabla P'</math>, and since we have [[#Orthogonal|already shown]] that <math>\mathbf{u'}_\mathrm{EFE}</math> is everywhere orthogonal to <math>\nabla P'</math>, it must be true that <math>\nabla G'</math> is everywhere orthogonal to <math>\mathbf{u'}_\mathrm{EFE}</math>.

<b><font color="red">Hooray!!</font></b>

=See Also=

* [[ThreeDimensionalConfigurations/DescriptionOfRiemannTypeI|Description of Riemann Type I Ellipsoids]]
* [[ThreeDimensionalConfigurations/RiemannTypeI#Riemann_Type_1_Ellipsoids|Riemann Type 1 Ellipsoids]] (old introduction)
* [[ThreeDimensionalConfigurations/Challenges#Challenges_Constructing_Ellipsoidal-Like_Configurations|Construction Challenges (Pt. 1)]]
* [[ThreeDimensionalConfigurations/ChallengesPt2|Construction Challenges (Pt. 2)]]
* [[ThreeDimensionalConfigurations/ChallengesPt3|Construction Challenges (Pt. 3)]]
* [[ThreeDimensionalConfigurations/ChallengesPt4|Construction Challenges (Pt. 4)]]
* [[ThreeDimensionalConfigurations/ChallengesPt5|Construction Challenges (Pt. 5)]]
* [[ThreeDimensionalConfigurations/ChallengesPt6|Construction Challenges (Pt. 6)]]

* Related discussions of models viewed from a rotating reference frame:
** [[PGE/RotatingFrame#Rotating_Reference_Frame|PGE]]
** <font color="red"><b>NOTE to Eric Hirschmann &amp; David Neilsen...  </b></font>I have moved the earlier contents of this page to a new Wiki location called [[Apps/RiemannEllipsoids_Compressible|Compressible Riemann Ellipsoids]].


{{ SGFfooter }}