Editing ThreeDimensionalConfigurations/ChallengesPt4 (section)

==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>