Editing ThreeDimensionalConfigurations/ChallengesPt4 (section)

=Challenges Constructing Ellipsoidal-Like Configurations (Pt. 4)=

This chapter extends the accompanying chapters titled, [[ThreeDimensionalConfigurations/Challenges|''Construction Challenges (Pt. 1)'']], [[ThreeDimensionalConfigurations/ChallengesPt2|''(Pt. 2)'']], and [[ThreeDimensionalConfigurations/ChallengesPt3|''(Pt. 3)'']].  The focus here is on firming up our understanding of the relationships between various "tilted" Cartesian coordinate frames.

==The Plan==

===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">
<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">
<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>

<span id="Step4"><font color="red"><b>STEP #4</b></font></span>

As a function of time, the x'-y' coordinates and associated velocity components of each Lagrangian fluid element are given by the expressions,
<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_\mathrm{max}\cos(\dot\varphi t)</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  <td align="right">
<math>~y' - y_c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~y_\mathrm{max}\sin(\dot\varphi t) \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\dot{x}'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- x_\mathrm{max}~ \dot\varphi \cdot \sin(\dot\varphi t) = (y_c - y') \biggl[ \frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr] \dot\varphi </math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  <td align="right">
<math>~\dot{y}' </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~y_\mathrm{max}~\dot\varphi \cdot \cos(\dot\varphi t) = x' \biggl[ \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr] \dot\varphi \, .</math>
  </td>
</tr>
</table>
This means that the (dimensional) velocity vector is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\boldsymbol{u'}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath'} \dot{x}' + \boldsymbol{\hat\jmath'} \dot{y}' 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath'} \biggl[ (y_c - y') \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr) \dot\varphi \biggr] 
+ 
\boldsymbol{\hat\jmath'} \biggl[ x' \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \dot\varphi \biggr]
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(u_0')^2 \dot\varphi^2 \, .
</math>
  </td>
</tr>
</table>

===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>
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'}  \underbrace{\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr]}_{\dot\varphi y_\mathrm{max}/x_\mathrm{max}}x' \, .
</math>
  </td>
</tr>
</table>

As we have indicated, this <math>~\boldsymbol{\hat\jmath'}</math> component will match our <font color="red">Step #4</font> velocity expression if,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl(\frac{y_\mathrm{max}}{x_\mathrm{max}} \biggr)\dot\varphi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{1}{1 + \tan^2\theta}\biggr]^{1 / 2}
\biggl[\beta\Omega_2 \tan\theta - \gamma\Omega_3\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\biggl[\frac{\gamma^2 \Omega_3^2}{\gamma^2 \Omega_3^2 + \beta^2\Omega_2^2}\biggr]^{1 / 2}
\biggl[ \frac{\beta^2\Omega_2^2}{\gamma \Omega_3} + \gamma\Omega_3\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\biggl[ \beta^2\Omega_2^2 + \gamma^2\Omega_3^2\biggr]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
Rewriting the <math>~\boldsymbol{\hat\imath'}</math> component, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\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 )</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
y' \biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 \cos\theta - \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 \sin\theta  \biggr]
+
\biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 ( z_0\tan\theta )
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
y' \biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3  - \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 \tan\theta  \biggr]\biggl[\frac{1}{1 + \tan^2\theta}\biggr]^{1 / 2}
-
\biggl(\frac{a}{b}\biggr)^2 \beta \Omega_2 z_0
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
y' \biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma^2 \Omega_3^2  + \biggl(\frac{a}{c}\biggr)^2 \beta^2 \Omega_2^2  \biggr]
\biggl[\frac{1}{\gamma^2 \Omega_3^2 + \beta^2\Omega_2^2}\biggr]^{1 / 2}
-
\biggl(\frac{a}{b}\biggr)^2 \beta \Omega_2 z_0 \, .
</math>
  </td>
</tr>
</table>
Hence,
<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\{ y' 
\underbrace{\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma^2 \Omega_3^2  + \biggl(\frac{a}{c}\biggr)^2 \beta^2 \Omega_2^2  \biggr] \biggl[\frac{1}{\gamma^2 \Omega_3^2 
+ \beta^2\Omega_2^2}\biggr]^{1 / 2}}_{-\dot\varphi x_\mathrm{max}/y_\mathrm{max}}
-
\biggl(\frac{a}{b}\biggr)^2 \beta \Omega_2 z_0\biggr\}
+
\boldsymbol{\hat\jmath'} \biggl[ \dot\varphi \biggl(\frac{y_\mathrm{max}}{x_\mathrm{max}} \biggr) \biggr]x' \, .
</math>
  </td>
</tr>
</table>
That is to say, we need to set,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\dot\varphi \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~
\biggl[ c^2 \gamma^2 \Omega_3^2  + b^2 \beta^2 \Omega_2^2  \biggr] \biggl[\frac{1}{\gamma^2 \Omega_3^2 
+ \beta^2\Omega_2^2}\biggr]^{1 / 2} \biggl( \frac{a^2}{b^2 c^2}\biggr) \, .
</math>
  </td>
</tr>
</table>
When this is combined with the constraint set by the  <math>~\boldsymbol{\hat\jmath'}</math> component, we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\dot\varphi^2 = \dot\varphi \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr) \cdot \dot\varphi \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~+
\biggl[ c^2 \gamma^2 \Omega_3^2  + b^2 \beta^2 \Omega_2^2  \biggr] 
\biggl( \frac{a^2}{b^2 c^2}\biggr) 
= (1.29930)^2
\, ,
</math>
  </td>
</tr>
</table>
and, hence,
<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>~
\biggl[ c^2 \gamma^2 \Omega_3^2  + b^2 \beta^2 \Omega_2^2  \biggr] \biggl[\frac{1}{\gamma^2 \Omega_3^2 
+ \beta^2\Omega_2^2}\biggr]^{1 / 2} \biggl( \frac{a^2}{b^2 c^2}\biggr) \biggl[ \beta^2\Omega_2^2 + \gamma^2\Omega_3^2\biggr]^{-1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{c^2 \gamma^2 \Omega_3^2  + b^2 \beta^2 \Omega_2^2  }{\gamma^2 \Omega_3^2 
+ \beta^2\Omega_2^2}\biggr] \biggl( \frac{a^2}{b^2 c^2}\biggr) 
= ( 1.02585 )^2
\, .
</math>
  </td>
</tr>
</table>

<span id="Result2">Finally, then,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~y_c \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr) \dot\varphi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\biggl(\frac{a}{b}\biggr)^2 \beta \Omega_2 z_0
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{y_c}{z_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\biggl(\frac{a}{b}\biggr)^2 \beta \Omega_2 
\biggl\{
\biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr) \dot\varphi 
\biggr\}^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
c^2 \beta \Omega_2 
\biggl[ c^2 \gamma^2 \Omega_3^2  + b^2 \beta^2 \Omega_2^2  \biggr]^{-1} 
\biggl[\gamma^2 \Omega_3^2 + \beta^2\Omega_2^2\biggr]^{1 / 2}  
= 0.19823
\, .
</math>
  </td>
</tr>
</table>
<table border="1" align="center" cellpadding="10" width="60%" bordercolor="orange">
<tr><td align="center" bgcolor="lightblue">'''RESULT 2'''<br />(different from [[ThreeDimensionalConfigurations/ChallengesPt2#Result1|Result 1]] and [[#Result3|Result 3, above]])
</td></tr>
<tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{y_0}{z_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\sin\theta}{b^2\kappa^2}
</math>
  </td>
</tr>
</table>

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

==Reconcile ''Results'' Differences==
<span id="Fig1">Before calling upon any of Riemann's model parameters</span>, from geometric considerations alone we should be able to determine exactly what the expression is for any off-center ellipse that results from slicing &#8212; at a tipped angle &#8212; the chosen 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:ExcelAxes02C.png|400px|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>

===Body Frame===
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>

<table border="1" align="center" cellpadding="8" width="80%"><tr><td align="left">
Given the values of <math>~a, b, c, \theta, z_0</math>, we can immediately map this to the tipped plane to obtain the surface-intersection function, <math>~x'(y')</math>.  Specifically we find,
<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>~a \biggl\{ 1 - \biggl[ (y'\cos\theta - \cancelto{0}{z'} \sin\theta)^2 \biggl( \frac{c^2 + b^2\tan^2\theta}{b^2c^2} \biggr) + (y'\cos\theta - \cancelto{0}{z'} \sin\theta) 
\biggl( \frac{2z_0 \tan\theta}{c^2} \biggr) + \frac{z_0^2}{c^2} \biggr] \biggr\}^{1 / 2}</math>
  </td>
</tr>

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

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

Along the ''tipped'' <math>~y'</math> axis, two points will mark the ends of the x'-y' orbit; they are identified by the roots of this ''Intersection Expression'' when x = 0.  That is, by the roots of,
<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>~y^2 \biggl[\frac{\kappa^2}{\cos^2\theta} \biggr] + y \biggl[ \frac{2z_0 \tan\theta}{c^2} \biggr] + \biggl[\frac{z_0^2}{c^2} -1 \biggr] \, ,</math>
  </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>
The roots are &hellip;
<table border="1" align="center" cellpadding="8" width="80%"><tr><td align="left">
Scratch notes:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~y_\mathrm{edge}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{B}{2A} \biggl\{\pm \biggl[1 - \frac{4AC}{B^2}  \biggr]^{1 / 2} -1 \biggr\}</math>
  </td>
</tr>
</table>
where,
<div align="center">
<math>~
\frac{4AC}{B^2} = \frac{4\kappa^2}{\cos^2\theta} \biggl[\frac{z_0^2}{c^2} -1 \biggr] \biggl[ \frac{2z_0 \tan\theta}{c^2} \biggr]^{-2}
=
\frac{\kappa^2}{\cos^2\theta}  \biggl[ \frac{c^2(z_0^2 -  c^2)}{z_0^2 \tan^2\theta} \biggr]
=
\biggl[ \frac{c^2 \kappa^2 (z_0^2 -  c^2)}{z_0^2 \sin^2\theta} \biggr]
</math>
</div>
</td></tr></table>


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

<tr>
  <td align="right">
<math>~y_\mathrm{edge}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{z_0 \sin\theta \cos\theta}{c^2 \kappa^2} \biggr]\biggl\{\pm \biggl[1 - \frac{c^2 \kappa^2 (z_0^2 -  c^2)}{z_0^2 \sin^2\theta} \biggr]^{1 / 2} -1 \biggr\}</math>
  </td>
</tr>

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

Hereafter we will use <math>~y_p</math> to denote the "plus" root, and <math>~y_m</math> to denote the "minus" root; that is,

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

<tr>
  <td align="right">
<math>~y_p</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\cos\theta}{c^2 \kappa^2} \biggr]\biggl\{\biggl[z_0^2 \sin^2\theta  + c^2 \kappa^2 (c^2 - z_0^2 ) \biggr]^{1 / 2} - z_0 \sin\theta  \biggr\} \, ,</math>
  </td>
</tr>

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

<table border="1" align="center" cellpadding="8" width="80%">
<tr>
  <td align="left">

<ul>
  <li>If tan&theta; is positive &hellip;
    <ul>
<li>Quadrant 1 &#x21D2; both sin&theta; and cos&theta; are positive.</li>
<li>Quadrant 3 &rArr; both sin&theta; and cos&theta; are negative.</li>
    </ul>
  </li>
</ul>
<ul>
  <li>If tan&theta; is negative &hellip;
    <ul>
<li>Quadrant 2 &#x21D2; sin&theta; is positive but cos&theta; is negative.</li>
<li>Quadrant 4 &rArr; sin&theta; is negative while cos&theta; are positive.</li>
    </ul>
  </li>
</ul>

Taking numerical values from our [[#Riemann_Flow|chosen Example Model]] &#8212; and using z<sub>0</sub> = &plusmn; 0.25 as our 1<sup>st</sup> test cases &#8212; we have:
<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>

<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>

<tr>
  <td align="right">
<math>~\mathrm{TERM1}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~z_0^2 \sin^2\theta + c^2\kappa^2(c^2-z_0^2) 
=0.043577
</math>&nbsp; &nbsp; (for z<sub>0</sub> = &plusmn; 0.25).
  </td>
</tr>
</table>
z<sub>0</sub> = +0.25 and &theta; in QUADRANT #4 (sin&theta; is negative while cos&theta; is positive):

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

<tr>
  <td align="right">
<math>~y_p</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\cos\theta}{c^2 \kappa^2} \biggr]\biggl\{\biggl[ \mathrm{TERM1} \biggr]^{1 / 2} - z_0 \sin\theta  \biggr\} 
=
1.1788
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~y_m</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~- \biggl[ \frac{\cos\theta}{c^2 \kappa^2} \biggr]\biggl\{\biggl[ \mathrm{TERM1} \biggr]^{1 / 2} + z_0 \sin\theta  \biggr\} 
=
- 0.5169
\, .</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~z_p</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~-c \biggl[ 1 - \biggl( \frac{y_p}{b} \biggr)^2 \biggr]^{1 / 2} 
=
-0.15645
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~z_m</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~c \biggl[ 1 - \biggl( \frac{y_m}{b} \biggr)^2 \biggr]^{1 / 2} 
=
0.42821
\, ,</math>
  </td>
</tr>
</table>

  </td>
  <td align="center">
<table border="1" align="center" cellpadding="5">
  <tr>
<td align="center" colspan="5"><math>~|z_0|_\mathrm{max} = 0.63792</math></td>
  </tr>
  <tr>
<td align="center"><math>~z_0</math></td>
<td align="center"><math>~y_p</math></td>
<td align="center"><math>~z_p</math></td>
<td align="center"><math>~y_m</math></td>
<td align="center"><math>~z_m</math></td>
  </tr>
  <tr>
<td align="right"><math>~+ 0.6379</math></td>
<td align="right"><math>~+ 0.8511</math></td>
<td align="right"><math>~+ 0.3444</math></td>
<td align="right"><math>~+ 0.8379</math></td>
<td align="right"><math>~+ 0.3490</math></td>
  </tr>
  <tr>
<td align="right"><math>~+ 0.60</math></td>
<td align="right"><math>~+ 1.1073</math></td>
<td align="right"><math>~+ 0.2182</math></td>
<td align="right"><math>~+ 0.4814</math></td>
<td align="right"><math>~+ 0.4340</math></td>
  </tr>
  <tr>
<td align="right"><math>~+ 0.55</math></td>
<td align="right"><math>~+ 1.1950</math></td>
<td align="right"><math>~+ 0.1380</math></td>
<td align="right"><math>~+ 0.2613</math></td>
<td align="right"><math>~+ 0.4599</math></td>
  </tr>
  <tr>
<td align="right"><math>~+ 0.50</math></td>
<td align="right"><math>~+ 1.2342</math></td>
<td align="right"><math>~+ 0.0744</math></td>
<td align="right"><math>~+ 0.0897</math></td>
<td align="right"><math>~+ 0.4691</math></td>
  </tr>
  <tr>
<td align="right"><math>~+ 0.45</math></td>
<td align="right"><math>~+ 1.2489</math></td>
<td align="right"><math>~- 0.0194</math></td>
<td align="right"><math>~- 0.0574</math></td>
<td align="right"><math>~+ 0.4698</math></td>
  </tr>
  <tr>
<td align="right"><math>~+ 0.40</math></td>
<td align="right"><math>~+ 1.2474</math></td>
<td align="right"><math>~- 0.0301</math></td>
<td align="right"><math>~- 0.1883</math></td>
<td align="right"><math>~+ 0.4649</math></td>
  </tr>
  <tr>
<td align="right"><math>~+ 0.35</math></td>
<td align="right"><math>~+ 1.2338</math></td>
<td align="right"><math>~- 0.0754</math></td>
<td align="right"><math>~- 0.3071</math></td>
<td align="right"><math>~+ 0.4559</math></td>
  </tr>
  <tr>
<td align="right"><math>~+ 0.30</math></td>
<td align="right"><math>~+ 1.2105</math></td>
<td align="right"><math>~- 0.1174</math></td>
<td align="right"><math>~- 0.4161</math></td>
<td align="right"><math>~+ 0.4435</math></td>
  </tr>
  <tr>
<td align="right" bgcolor="orange">+ 0.25</td>
<td align="right" bgcolor="orange">+ 1.1788</td>
<td align="right" bgcolor="orange">- 0.1564</td>
<td align="right" bgcolor="orange">- 0.5169</td>
<td align="right" bgcolor="orange">+ 0.4282</td>
  </tr>
  <tr>
<td align="right"><math>~+ 0.20</math></td>
<td align="right"><math>~+ 1.1399</math></td>
<td align="right"><math>~- 0.1930</math></td>
<td align="right"><math>~- 0.6103</math></td>
<td align="right"><math>~+ 0.4104</math></td>
  </tr>
  <tr>
<td align="right"><math>~+ 0.15</math></td>
<td align="right"><math>~+ 1.0943</math></td>
<td align="right"><math>~- 0.2273</math></td>
<td align="right"><math>~- 0.6971</math></td>
<td align="right"><math>~+ 0.3904</math></td>
  </tr>
  <tr>
<td align="right"><math>~+ 0.10</math></td>
<td align="right"><math>~+ 1.0426</math></td>
<td align="right"><math>~- 0.2595</math></td>
<td align="right"><math>~- 0.7778</math></td>
<td align="right"><math>~+ 0.3682</math></td>
  </tr>
  <tr>
<td align="right"><math>~+ 0.05</math></td>
<td align="right"><math>~+ 0.9849</math></td>
<td align="right"><math>~- 0.2896</math></td>
<td align="right"><math>~- 0.8525</math></td>
<td align="right"><math>~+ 0.3439</math></td>
  </tr>
  <tr>
<td align="right"><math>~+ 0.00</math></td>
<td align="right"><math>~+ 0.9216</math></td>
<td align="right"><math>~- 0.3177</math></td>
<td align="right"><math>~- 0.9216</math></td>
<td align="right"><math>~+ 0.3177</math></td>
  </tr>
  <tr>
<td align="right"><math>~- 0.05</math></td>
<td align="right"><math>~+ 0.8525</math></td>
<td align="right"><math>~- 0.3439</math></td>
<td align="right"><math>~- 0.9849</math></td>
<td align="right"><math>~+ 0.2895</math></td>
  </tr>
  <tr>
<td align="right"><math>~- 0.10</math></td>
<td align="right"><math>~+ 0.7778</math></td>
<td align="right"><math>~- 0.3682</math></td>
<td align="right"><math>~- 1.0426</math></td>
<td align="right"><math>~+ 0.2595</math></td>
  </tr>
  <tr>
<td align="center" colspan="5">''etc.''
  </tr>
  <tr>
<td align="right"><math>~- 0.6379</math></td>
<td align="right"><math>~- 0.8379</math></td>
<td align="right"><math>~- 0.3490</math></td>
<td align="right"><math>~- 0.8511</math></td>
<td align="right"><math>~- 0.3444</math></td>
  </tr>
</table>
  </td>
</tr>
</table>


The y-coordinate of the center of the orbit will lie halfway between these two "edges".  That is (numerical example given for the case of z<sub>0</sub> = + 0.25),
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~y_\mathrm{center}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2}\biggl[ y_p + y_m \biggr] = +0.3310 \, .</math>
  </td>
</tr>
</table>
Correspondingly, the vertical (z) location of the orbit center in the ''body'' frame is given by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~z_\mathrm{center}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~y_\mathrm{center}\tan\theta + z_0 = +0.1359 \, .</math>
  </td>
</tr>
</table>

The line that runs parallel to the x-axis &#8212; lies perpendicular to the y-z plane &#8212; and that passes through this "center" location intersects the surface of the ellipsoid at a value of x that is obtained by plugging <math>~y_\mathrm{center}</math> into the ''Intersection Expression''.  That is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{x_\mathrm{surf}}{a}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{
1 - y^2_\mathrm{center} \biggl[\frac{\kappa^2}{\cos^2\theta} \biggr] 
- y_\mathrm{center} \biggl[ \frac{2z_0 \tan\theta}{c^2} \biggr] 
- \frac{z_0^2}{c^2}
\biggr\}^{1 / 2} = 0.92001 \, ;</math>
  </td>
</tr>
</table>
along the relevant orbit, this is also the maximum value of x.

===Tipped Plane===

We use the expressions in the right panel of [[#Fig1|Figure 1, above]] to transform (x, y, z) ''body'' coordinates to (x', y', z') ''tipped-frame'' coordiantes.  Given that <math>~x' = x</math>, we appreciate that the x'-coordinate of the orbit center is zero, and the maximum value <math>~(x'_\mathrm{surf})</math> that is encountered along the relevant orbit is the same as the value of <math>~x_\mathrm{surf}</math> as determined in the ''body'' (unprimed) frame.

The z'-coordinate of the orbit center is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~z'_\mathrm{center}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(z_\mathrm{center} - z_0)\cos\theta - y_\mathrm{center}\sin\theta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(y_\mathrm{center}\tan\theta) \cos\theta - y_\mathrm{center}\sin\theta
</math>
  </td>
</tr>

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

Finally, in the ''tipped'' plane, the maximum (subscript_p) and minimum (subscript_m) values of y' are,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~y'_p</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(z_p - z_0)\sin\theta + y_p\cos\theta = 1.2469 \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~y'_m</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(z_m - z_0)\sin\theta + y_m\cos\theta = -0.5467 \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ y'_\mathrm{center}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2}\biggl[ y_p' + y_m'\biggr] = 0.3501\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ y'_\mathrm{surf}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
y'_p - y'_\mathrm{center} = 0.8968\, .
</math>
  </td>
</tr>
</table>
So, in the ''tipped'', x'-y' plane, the elliptical orbit that corresponds to z<sub>0</sub> = + 0.25 should be described by 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'}{x_\mathrm{surf}} \biggr]^2 + \biggl[ \frac{(y' - y'_\mathrm{center} ) }{y'_\mathrm{surf}} \biggr]^2 \, .
</math>
  </td>
</tr>
</table>


NOTE:
*  <math>~y'_\mathrm{center}/z_0 = 0.350096/0.25 = 1.40038 \, .</math>
* <math>~x_\mathrm{surf}/y'_\mathrm{surf} = 0.92001/0.896819 = 1.0259 \, .</math>

At a minimum these numerical values should be compared to the Riemann model parameters [[ThreeDimensionalConfigurations/ChallengesPt2#Example_Equilibrium_Model|computed in our earlier (Pt. 2) examination]].

==Various Coordinate Frames==


===Tipped Orbit Planes===

====Summary====
In a [[ThreeDimensionalConfigurations/RiemannTypeI#Try_Again|separate discussion]], we have shown that, as viewed from a frame that "tumbles" with the (purple) body of a Type 1 Riemann ellipsoid, each Lagrangian fluid element moves along an elliptical path in a plane that is tipped by an angle <math>~\theta</math> about the x-axis of the body.  As viewed from the (primed) coordinates associated with this tipped plane, by definition, z' = constant and dz'/dt = 0, and the planar orbit is defined by the expression for 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(z')}{y_\mathrm{max}} \biggr]^2 \, .</math>
  </td>
</tr>
</table>

<table border="0" cellpadding="10" align="right" width="30%"><tr><td align="center">
<table border="1" align="center" cellpadding="8">
<tr><td align="center">
''Tipped Orbit Frame'' (yellow, primed) <br />
</td>
</tr>
<tr>
  <td align="center">[[File:TippedAxes03.png|350px|Tipped Orbital Planes]]</td>
</tr>
<tr><td align="center">
Given that b/a = 1.25 and c/a = 0.4703 for our chosen [[ThreeDimensionalConfigurations/ChallengesPt2#Example_Equilibrium_Model|Example Type I Ellipsoid]], we find that, <math>~\theta = - 1.18122 ~\mathrm{rad} = -67.68^\circ</math>.
</td>
</tr>
</table>
</td></tr></table>
Notice that the offset, <math>~y_c</math>, is a function of the tipped plane's vertical coordinate, <math>~z'</math>.  As a function of time, the x'-y' coordinates and associated velocity components of each Lagrangian fluid element are given by the expressions,
<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_\mathrm{max}\cos(\dot\varphi t)</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  <td align="right">
<math>~y' - y_c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~y_\mathrm{max}\sin(\dot\varphi t) \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\dot{x}'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- x_\mathrm{max}~ \dot\varphi \cdot \sin(\dot\varphi t) = (y_c - y') \biggl[ \frac{x_\mathrm{max}}{y_\mathrm{max}} \biggr] \dot\varphi </math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  <td align="right">
<math>~\dot{y}' </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~y_\mathrm{max}~\dot\varphi \cdot \cos(\dot\varphi t) = x' \biggl[ \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr] \dot\varphi \, .</math>
  </td>
</tr>
</table>

We have determined that (numerical value given for our [[ThreeDimensionalConfigurations/ChallengesPt2#Example_Equilibrium_Model|chosen example Type I ellipsoid]]),
<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{b^2 \beta \Omega_2}{c^2 \gamma \Omega_3}  
=
-2.43573\, ,
</math>
  </td>
</tr>
<tr><td align="center" colspan="3"><b><font color="red">This definition of tan(&theta;) is inconsistent with all others!</font></b></td>
</tr>
</table>
where, as has also been specified [[ThreeDimensionalConfigurations/ChallengesPt3#betagamma|in an accompanying discussion]],
<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}
=
+1.13451
</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} 
=
+1.80518\, .
</math>
  </td>
</tr>
</table>

We also have determined that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[\frac{x_\mathrm{max}}{y_\mathrm{max}}  \biggr]^4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{a^4 (c^4 \gamma^2 \Omega_3^2 + b^4 \beta^2 \Omega_2^2)}{b^4 c^4(\gamma^2\Omega_3^2 + \beta^2\Omega_2^2)}
~~~\Rightarrow ~~~\frac{x_\mathrm{max}}{y_\mathrm{max}}  = 1.26218  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~{\dot\varphi}^4 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{a^4}{b^4 c^4} \biggl(\gamma^2\Omega_3^2 + \beta^2\Omega_2^2 \biggr) (c^4 \gamma^2 \Omega_3^2 + b^4 \beta^2 \Omega_2^2)
~~~\Rightarrow ~~~ \dot\varphi = 1.59862\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{y_c}{z_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\sin\theta
~~~\Rightarrow~~~~ \frac{y_c}{z_0} = -0.92507
\, .</math>
  </td>
</tr>
</table>

====Demonstration====

In order to transform a vector from the "tipped orbit" frame (primed coordinates) to the "body" frame (unprimed), we use the following mappings of the three unit vectors:
<table border="1" align="center" width="40%" 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></tr></table>

Given that, by design in our "tipped orbit" frame, there is no vertical motion &#8212; that is, <math>~\dot{z}' = 0</math> &#8212; mapping the (primed coordinate) velocity to the body (unprimed) coordinate is particularly straightforward.  Specifically,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\boldsymbol{u'}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath'} \dot{x}'
+
\boldsymbol{\hat\jmath'} \dot{y}'
+
\boldsymbol{\hat{k}'} \cancelto{0}{\dot{z}'}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~~~\rightarrow~~</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath} \dot{x}'
+
[\boldsymbol{\hat\jmath}\cos\theta + \boldsymbol{\hat{k}}\sin\theta] \dot{y}' 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath} \biggl\{
(y_c - y') \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr) \dot\varphi
\biggr\}
+
[\boldsymbol{\hat\jmath}\cos\theta + \boldsymbol{\hat{k}}\sin\theta] \biggl\{
x' \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \dot\varphi
\biggr\} \, .
</math>
  </td>
</tr>
</table>

Recognizing, [[ThreeDimensionalConfigurations/ChallengesPt2#Tipped_Orbital_Plane|as before]], that the relevant coordinate mapping is,

<table border="1" width="50%" cellpadding="8" align="center">
<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>

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

<tr>
  <td align="right">
<math>~\boldsymbol{u'}</math>
  </td>
  <td align="center">
<math>~~~\rightarrow~~~</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath}  \dot\varphi \biggl( \frac{x_\mathrm{max}}{y_\mathrm{max}}\biggr)\biggl\{y_c - y\cos\theta - (z - z_0)\sin\theta\biggr\}
+
\boldsymbol{\hat\jmath}  \dot\varphi \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) \biggr\{ x\cos\theta \biggr\}
+
\boldsymbol{\hat{k}}  \dot\varphi
\biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr)
\biggr\{ x\sin\theta \biggr\} \, .
</math>
  </td>
</tr>
</table>
Therefore, as viewed from the ''body'' frame, the individual velocity components are,

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

<tr>
  <td align="right">
<math>~\boldsymbol{\hat\jmath} \cdot \boldsymbol{u'}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\dot\varphi \biggl( \frac{y_\mathrm{max}}{x_\mathrm{max}}\biggr) x\cos\theta 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x \cdot 
\biggl\{
\frac{b^4 c^4(\gamma^2\Omega_3^2 + \beta^2\Omega_2^2)}{a^4 (c^4 \gamma^2 \Omega_3^2 + b^4 \beta^2 \Omega_2^2)}
\frac{a^4}{b^4 c^4} \biggl(\gamma^2\Omega_3^2 + \beta^2\Omega_2^2 \biggr) (c^4 \gamma^2 \Omega_3^2 + b^4 \beta^2 \Omega_2^2)
\biggr\}^{1 / 4} \frac{{c^2 \gamma \Omega_3} }{[c^4 \gamma^2 \Omega_3^2 + b^4\beta^2 \Omega_2^2]^{1 / 2}} 
</math>
  </td>
</tr>
</table>

==Test Orthogonality==

Let's see if the Riemann fluid velocity vector is everywhere tangent to our identified off-center elliptical (Lagrangian-particle) orbital trajectories.  We can determine this by first deriving an expression for the vector normal to the trajectories, then see if the dot product of this vector and the velocity vector is everywhere zero.

The unit vector that is normal to a trajectory is obtained from the (appropriately normalized) gradient of the function,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~f(x', y')</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{x'}{x_\mathrm{max}} \biggr]^2 + \biggl[\frac{y' - y_c}{y_\mathrm{max}} \biggr]^2 - 1\, .</math>
  </td>
</tr>
</table>
We find, first,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nabla f</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath'} \biggl[\frac{2x'}{x^2_\mathrm{max}} \biggr] 
+ 
\boldsymbol{\hat\jmath'} \biggl[\frac{2(y' - y_c)}{y^2_\mathrm{max}} \biggr] 
- 
\boldsymbol{\hat{k}'} \biggl[\frac{2(y' - y_c)}{y^2_\mathrm{max}} \biggr] \frac{\partial y_c}{\partial z'} \, ,
</math>
  </td>
</tr>
</table>

where, drawing from [[#Step2|Step #2, above]],
<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">
<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>

<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}\biggl[ 1 - \frac{z_0^2}{c^2 } - \frac{z_0 \sin\theta}{c^2} \biggr] \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<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>


Next, after [[#ThetaDef|setting]] <math>~\tan\theta = - (\beta\Omega_2/\gamma\Omega_3)</math>, and pulling from [[#Riemann_Flow|Step #3, above]], 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[ \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>

===Stick With Rotating Frame===
The dot product of these two vectors is,

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

<tr>
  <td align="right">
<math>~\nabla f \cdot \boldsymbol{u'}_\mathrm{EFE}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\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]\biggl[\frac{2x'}{x^2_\mathrm{max}} \biggr] 
+
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr]x' \biggl[\frac{2(y' - y_c)}{y^2_\mathrm{max}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  y'\cos\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 (  y'\sin\theta ) \biggr]\biggl[\frac{2x'}{x^2_\mathrm{max}} \biggr] 
+
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  z_0\tan\theta ) 
\biggr]\biggl[\frac{2x'}{x^2_\mathrm{max}} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr]x' \biggl[\frac{2y' }{y^2_\mathrm{max}} \biggr]
-
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr]x' \biggl[\frac{2y_c }{y^2_\mathrm{max}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2x'y' \biggl\{
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  \cos\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 ( \sin\theta ) \biggr]\biggl[\frac{1}{x^2_\mathrm{max}} \biggr] 
+
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr] \biggl[\frac{1 }{y^2_\mathrm{max}} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2x' \biggl\{
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 \tan\theta  
\biggr]\biggl[\frac{z_0}{x^2_\mathrm{max}} \biggr] 
-
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr]\biggl[\frac{y_c }{y^2_\mathrm{max}} \biggr] 
\biggr\} \, .
</math>
  </td>
</tr>
</table>
Now, drawing from the various above "boxed" relations and examining the two terms separately, we first find,

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

<tr>
  <td align="right">
TERM1
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2x'y'}{x_\mathrm{max}^2} \biggl( \frac{a^2}{b^2 c^2} \biggr) \biggl\{
\biggl[ c^2 \gamma \Omega_3 (  \cos\theta ) 
- b^2 \beta \Omega_2 ( \sin\theta ) \biggr] 
+
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr] \biggl[\frac{x_\mathrm{max}^2 }{y^2_\mathrm{max}} \biggr]\biggl( \frac{b^2 c^2}{a^2} \biggr)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2x'y'}{x_\mathrm{max}^2} \biggl( \frac{a^2}{b^2 c^2} \biggr) \biggl\{
\biggl[ c^2 (  \cos\theta ) 
+ b^2 \tan\theta ( \sin\theta ) \biggr] 
-
\biggl[ \tan\theta  \sin\theta 
+  \cos\theta \biggr] \biggl[ c^2 \cos^2\theta +  b^2\sin^2\theta \biggr]
\biggr\} \gamma \Omega_3
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2x'y'}{x_\mathrm{max}^2} \biggl( \frac{a^2}{b^2 c^2} \biggr) \biggl\{
\biggl[ c^2 + b^2 \tan^2\theta\biggr] 
-
\biggl[ \tan^2\theta  +  1 \biggr] \biggl[ c^2 \cos^2\theta +  b^2\sin^2\theta \biggr]
\biggr\} \gamma \Omega_3 \cos\theta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2x'y'}{x_\mathrm{max}^2} \biggl( \frac{a^2}{b^2 c^2} \biggr) \biggl\{
c^2 + b^2 \tan^2\theta
-
\biggl[ \tan^2\theta  +  1 \biggr] \biggl[ c^2 \cos^2\theta \biggr]
-
\biggl[ \tan^2\theta  +  1 \biggr] \biggl[  b^2\sin^2\theta \biggr]
\biggr\} \gamma \Omega_3 \cos\theta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2x'y'}{x_\mathrm{max}^2} \biggl( \frac{a^2}{b^2 c^2} \biggr) \biggl\{
c^2 + b^2 \tan^2\theta
- c^2 - b^2\tan^2\theta 
\biggr\} \gamma \Omega_3 \cos\theta
</math>
  </td>
</tr>

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

And, in order for the second term to go to zero as well, we need &hellip;

<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>~
\frac{2x'}{ x^2_\mathrm{max} } \biggl\{
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 \tan\theta  
\biggr]z_0 
-
y_c \biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta \biggr]\biggl[\frac{ x^2_\mathrm{max} }{y^2_\mathrm{max}} \biggr] 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2x'}{ x^2_\mathrm{max} } \biggl\{
\biggl[ c^2  \tan\theta  
\biggr]z_0 
+
y_c \biggl[(\tan\theta) \sin\theta 
+  \cos\theta \biggr]( c^2 \cos^2\theta + b^2 \sin^2\theta ) 
\biggr\} \biggl( \frac{a^2}{b^2 c^2}  \biggr)\gamma \Omega_3
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2x'}{ x^2_\mathrm{max} } \biggl\{
\biggl[ c^2  \tan\theta  
\biggr]z_0 
+
y_c \cos\theta( c^2 + b^2 \tan^2\theta ) 
\biggr\} \biggl( \frac{a^2}{b^2 c^2}  \biggr)\gamma \Omega_3
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{y_c}{z_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-  \frac{c^2  \tan\theta}{\cos\theta( c^2 + b^2 \tan^2\theta )}
=
-  \frac{c^2  \tan^2\theta}{\sin\theta( c^2 + b^2 \tan^2\theta )}
=
-  \frac{c^2  \sin\theta}{( c^2\cos^2\theta + b^2 \sin^2\theta )}
=
-  \frac{\sin\theta}{b^2  \kappa^2} \, .
</math>
  </td>
</tr>
</table>

===Try Shifting to (Quasi-) Inertial Frame===

Let's adjust both remaining components of <math>~\boldsymbol{u'}_\mathrm{EFE}</math> to see what their equivalent rotating-frame expressions are.  The relevant shift is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\boldsymbol{v'}_\mathrm{shift} \equiv \boldsymbol{\Omega'} \times \boldsymbol{x'}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat{k}} \Omega' \times \biggl[ \boldsymbol{\hat\imath'}x' +  \boldsymbol{\hat\jmath'}y' \biggr]
=
- \boldsymbol{\hat\imath'} \Omega' y' + \boldsymbol{\hat\jmath'} \Omega' x' \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Omega'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \Omega_2^2 + \Omega_3^2 \biggr]^{1 / 2} \, .</math>
  </td>
</tr>
</table>
We have, then,

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

<tr>
  <td align="right">
<math>~\boldsymbol{u'}_\mathrm{inertial} \equiv \boldsymbol{u'}_\mathrm{EFE} + \boldsymbol{v'}_\mathrm{shift}</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 )  - \Omega' y'\biggr]
+
\boldsymbol{\hat\jmath'}  \biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta + \Omega'\biggr]x' 
</math>
  </td>
</tr>
</table>
The dot product of this expression with <math>~\nabla f</math> is,

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

<tr>
  <td align="right">
<math>~\nabla f \cdot \boldsymbol{u'}_\mathrm{inertial}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\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 ) - \Omega' y' \biggr]\biggl[\frac{2x'}{x^2_\mathrm{max}} \biggr] 
+
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta + \Omega'\biggr]x' \biggl[\frac{2(y' - y_c)}{y^2_\mathrm{max}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  \cos\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 (  \sin\theta ) - \Omega' \biggr]\biggl[\frac{2x' y'}{x^2_\mathrm{max}} \biggr] 
+
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  z_0\tan\theta ) 
\biggr]\biggl[\frac{2x'}{x^2_\mathrm{max}} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta  + \Omega'\biggr]x' \biggl[\frac{2y' }{y^2_\mathrm{max}} \biggr]
-
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta + \Omega' \biggr]x' \biggl[\frac{2y_c }{y^2_\mathrm{max}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2x'y' \biggl\{
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 (  \cos\theta ) 
- \biggl(\frac{a}{c}\biggr)^2 \beta \Omega_2 ( \sin\theta ) - \Omega'\biggr]\biggl[\frac{1}{x^2_\mathrm{max}} \biggr] 
+
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta + \Omega' \biggr] \biggl[\frac{1 }{y^2_\mathrm{max}} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2x' \biggl\{
\biggl[ \biggl(\frac{a}{b}\biggr)^2 \gamma \Omega_3 \tan\theta  
\biggr]\biggl[\frac{z_0}{x^2_\mathrm{max}} \biggr] 
-
\biggl[\beta \Omega_2  \sin\theta 
- \gamma \Omega_3  \cos\theta + \Omega' \biggr]\biggl[\frac{y_c }{y^2_\mathrm{max}} \biggr] 
\biggr\} \, .
</math>
  </td>
</tr>
</table>
Now, drawing from the various above "boxed" relations and examining the two terms separately, we first find,