Editing ThreeDimensionalConfigurations/ChallengesPt5

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=Challenges Constructing Ellipsoidal-Like Configurations (Pt. 5)=

This chapter extends the accompanying chapters titled, [[ThreeDimensionalConfigurations/Challenges|''Construction Challenges (Pt. 1)'']], [[ThreeDimensionalConfigurations/ChallengesPt2|''(Pt. 2)'']], [[ThreeDimensionalConfigurations/ChallengesPt3|''(Pt. 3)'']], and [[ThreeDimensionalConfigurations/ChallengesPt4|''(Pt. 4)'']].  

==Tilted Plane Intersects Ellipsoid==

In a [[ThreeDimensionalConfigurations/RiemannTypeI#TippedPlane|an early subsection of the accompanying discussion]], we have pointed out that the intersection of each Lagrangian fluid element's tipped orbital plane with the surface of the (purple) ellipsoidal surface is given by the (unprimed) body-frame coordinates that simultaneously satisfy the expressions,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{x}{a}\biggr)^2 + \biggl( \frac{y}{b}\biggr)^2 + \biggl( \frac{z}{c}\biggr)^2 </math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~z</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~y \tan\theta + z_0 \, ,</math>
  </td>
</tr>
</table>
where z<sub>0</sub> is the location where the tipped plane intersects the z-axis of the body frame.  Combining these two expressions, we see that an intersection between the tipped plane and the ellipsoidal surface will occur at (x, y)-coordinate pairs that satisfy what we will henceforth refer to as the,
<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>
<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>

In the equatorial plane of the ''tipped'' coordinate system &#8212; that is, after mapping <math>~x \rightarrow x'</math> and <math>~y \rightarrow (y' \cos\theta - z'\sin\theta)</math>, then setting <math>~z' = 0</math> &#8212; this intersection expression becomes,
<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)^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>

The light-blue curve in the right-hand panel of the following animation is a plot of this <math>~x'(y')</math>function for various values of <math>~z_0</math> (as indicated by the light-blue numerical value in the upper-right corner of the figure's left-hand panel.

[[File:RiemannType1Tipped01.gif|center|600px|Animation of Intersection Curves]]

As it turns out &#8212; see our [[ThreeDimensionalConfigurations/ChallengesPt4#Tipped_Plane|accompanying discussion]] &#8212; this expression can be rewritten as,
<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>
demonstrating that, as viewed from the x'-y' plane, the (light-blue) intersection curve is always an off-center ellipse.  See also our [[ThreeDimensionalConfigurations/ChallengesPt2#COLLADA-Based_Representation|COLLADA-based representation]] of these curves.

==Trajectory of Lagrangian Fluid Elements==

This subsection borrows heavily from [[ThreeDimensionalConfigurations/ChallengesPt4#Step4|an accompanying discussion]].

===Old Way of Thinking===
It seems reasonable to assume that this ''off-center ellipse'' expression will properly describe the orbital path of various Lagrangian fluid elements that make up the uniform-density ellipsoid.  Assuming that, when viewed from the rotating-and-tipped coordinate frame, each fluid element's motion along this trajectory is oscillatory, it is reasonable to assume that the time-dependent x'-y' coordinate position of each fluid element is 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{surf}\cos(\dot\varphi t)</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  <td align="right">
<math>~y' - y'_\mathrm{center}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~y'_\mathrm{surf}\sin(\dot\varphi t) \, .</math>
  </td>
</tr>
</table>

In this case, as viewed from the rotating-and-tipped coordinate frame, the corresponding velocity components are,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\dot{x}'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- x_\mathrm{surf}~ \dot\varphi \cdot \sin(\dot\varphi t) = (y'_\mathrm{center} - y') \biggl[ \frac{x_\mathrm{surf}}{y'_\mathrm{surf}} \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{surf}~\dot\varphi \cdot \cos(\dot\varphi t) = x' \biggl[ \frac{y'_\mathrm{surf}}{x_\mathrm{surf}}\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'_\mathrm{center} - y') \biggl( \frac{x_\mathrm{surf}}{y'_\mathrm{surf}} \biggr) \dot\varphi \biggr] 
+ 
\boldsymbol{\hat\jmath'} \biggl[ x' \biggl( \frac{y'_\mathrm{surf}}{x_\mathrm{surf}}\biggr) \dot\varphi \biggr] \, .
</math>
  </td>
</tr>
</table>

===New Thoughts===

In our ''Old Way of Thinking'', the hypothesized velocity flow-field was symmetric (in both directions) about the center of the elliptical trajectory.  This hypothesized Lagrangian motion isn't (and cannot be) correct because an examination of EFE's derived Riemann (Eulerian) flow-field is not symmetric about the x'-axis.  Instead, the Eulerian flow-field displays a noticeable m = 1 contribution.  Here we present an alternate hypothesis with two new features:  (1) The flow is described by circulation about an center that is shifted along the y'-axis away from the center of the ellipse; (2) The trajectory of Lagrangian fluid elements is described by motion in a cylindrical-coordinate system such that motion in the angular coordinate is uniform.

We will still insist that the trajectory of Lagrangian fluid elements is that of an ellipse 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>
Now we will introduce a <math>~\varpi - \varphi </math> cylindrical coordinate system that is related to the x'-y' coordinate system such that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\varpi \cos\varphi</math>
  </td>
<td align="centner">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  <td align="right">
<math>~y' - y_\varpi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\varpi \sin\varphi \, ,</math>
  </td>
</tr>
</table>
with <math>~|y_\varpi| < |v'_\mathrm{center}|</math>.  Mapping the other direction gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\varpi^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(x')^2 + (y' - y_\varpi)^2</math>
  </td>
<td align="centner">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  <td align="right">
<math>~\tan\varphi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(y' - y_\varpi)}{x'} \, .</math>
  </td>
</tr>
</table>

Using the (constraint) ellipse expression to eliminate y' from these last two expressions, we find,
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ (y' - y_\varpi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(y'_\mathrm{center} - y_\varpi) + y'_\mathrm{surf} \biggl[ 1 - \frac{(x')^2}{x_\mathrm{surf}^2} \biggr]^{1 / 2} \, .</math>
  </td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\varpi^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(x')^2 
+ 
\biggl\{ (y'_\mathrm{center} - y_\varpi) + y'_\mathrm{surf} \biggl[ 1 - \frac{(x')^2}{x_\mathrm{surf}^2} \biggr]^{1 / 2}  \biggr\}^2 \, ,
</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x' \tan\varphi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(y'_\mathrm{center} - y_\varpi) + y'_\mathrm{surf} \biggl[ 1 - \frac{(x')^2}{x_\mathrm{surf}^2} \biggr]^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ x' \tan\varphi + (y_\varpi - y'_\mathrm{center} )</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{y'_\mathrm{surf}}{x_\mathrm{surf}} \biggl[ x_\mathrm{surf}^2 - (x')^2 \biggr]^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ (x' )^2\tan^2\varphi + 2x' \tan\varphi(y_\varpi - y'_\mathrm{center} )+ (y_\varpi - y'_\mathrm{center} )^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
 \biggl( \frac{y'_\mathrm{surf}}{x_\mathrm{surf}} \biggr)^2 \biggl[ x_\mathrm{surf}^2 - (x')^2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ (x' )^2 \biggl[ \tan^2\varphi + \biggl( \frac{y'_\mathrm{surf}}{x_\mathrm{surf}} \biggr)^2 \biggr] 
+ 
x'  \biggl[ 2\tan\varphi(y_\varpi - y'_\mathrm{center} ) \biggr] 
+ 
\biggl[ (y_\varpi - y'_\mathrm{center} )^2 - \biggl( \frac{y'_\mathrm{surf}}{x_\mathrm{surf}} \biggr)^2 x_\mathrm{surf}^2\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
0 
</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>~x'</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} 
= 
\biggl[ \underbrace{ x_\mathrm{surf}^2 \tan^2\varphi + (y'_\mathrm{surf})^2 }_{A} \biggr]
\biggl[ \underbrace{x_\mathrm{surf}^2(y_\varpi - y'_\mathrm{center} )^2 
- 
(y'_\mathrm{surf})^2 x_\mathrm{surf}^2 }_{C} \biggr]
\biggl[\underbrace{ (y_\varpi - y'_\mathrm{center} )x_\mathrm{surf}^2 \tan\varphi }_{B/2} \biggr]^{-2}
</math>
</div>

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


After <math>~x'</math> has been evaluated for a given value of <math>~\varphi</math>, the accompanying value of <math>~y'</math> can be obtained, in principle, from either of the expressions:
<table border="0" cellpadding="5" align="center">

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

<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" colspan="12">
'''1<sup>st</sup> EXAMPLE:'''<br />  <br /><math>~x_\mathrm{surf} = 0.9200; ~~~y_\mathrm{surf} = 0.8968; ~~~y'_\mathrm{center} = 0.3501; ~~~y_\varpi = -0.1</math>
  </td>
</tr>
<tr>
  <td align="center" rowspan="2"><math>~\varphi</math><br /> <br /><math>~[ \Delta\varphi = 9^\circ]</math></td>
  <td align="center" rowspan="2"><math>~A</math></td>
  <td align="center" rowspan="2"><math>~B</math></td>
  <td align="center" rowspan="2"><math>~C</math></td>
  <td align="center" rowspan="2"><math>~\frac{4AC}{B^2}</math></td>
  <td align="center" colspan="3">"plus"</td>
  <td align="center" colspan="3">"minus"</td>
  <td align="center" rowspan="2">Expression used to obtain y'</td>
</tr>
<tr>
  <td align="center" colspan="1"><math>~x'</math></td>
  <td align="center" colspan="1"><math>~y'</math></td>
  <td align="center" colspan="1"><math>~\mathrm{ATAN2}[x', (y' - y_\varpi)]</math></td>
  <td align="center" colspan="1"><math>~x'</math></td>
  <td align="center" colspan="1"><math>~y'</math></td>
  <td align="center" colspan="1"><math>~\mathrm{ATAN2}[x', (y' - y_\varpi)]</math></td>
</tr>
<tr>
  <td align="center">0.15708</td>
  <td align="center">0.82552</td>
  <td align="center">-0.12068</td>
  <td align="center">-0.50929</td>
  <td align="center">-1.1548 &times; 10<sup>+2</sup></td>
  <td align="center">-0.71575</td>
  <td align="center">+0.91355</td>
  <td align="center">-0.95593 = <math>~\varphi - \pi</math></td>
  <td align="center">+0.86194</td>
  <td align="center">+0.66367 </td>
  <td align="center">0.65616</td>
  <td align="left">
<math>~y' = y_\mathrm{center} +
y'_\mathrm{surf} \biggl[ 1 - \frac{(x')^2}{x_\mathrm{surf}^2} \biggr]^{1 / 2} \,.
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="5">&nbsp;</td>
  <td align="center" bgcolor="lightblue"><font color="white">-0.71575</font></td>
  <td align="center" bgcolor="lightblue"><font color="white">-0.21336</font></td>
  <td align="center">-2.9845 = &#x03D5; - &pi;</td>
  <td align="center" bgcolor="purple"><font color="white">+0.86194</font></td>
  <td align="center" bgcolor="purple"><font color="white">+0.03652</font></td>
  <td align="center">0.15708 = &#x03D5;</td>
  <td align="left">
<math>~y' = y_\mathrm{center} -
y'_\mathrm{surf} \biggl[ 1 - \frac{(x')^2}{x_\mathrm{surf}^2} \biggr]^{1 / 2} \,.
</math>
  </td>-0.41267</tr>
</table>




We will assume that <math>~\varphi = \dot{\varphi} t</math>, with <math>~\dot\varphi</math> constant, and then determine how <math>~\varpi</math> depends on <math>~\varphi</math> and therefore, also, how it varies with time.

First, we note that transforming from the primed-Cartesian system to the cylindrical-coordinate system is accomplished via the relations,

===Simpler Example===
====Positions of Lagrangian Fluid Elements====

Let's determine the points of intersection of the following two expression:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~y</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~y_\varpi + x \tan\varphi \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; 
  <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{surf}} \biggr)^2 \, .</math>
  </td>
</tr>
</table>

Eliminating <math>~x</math>, then solving for <math>~y(\varphi)</math>, 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>~\frac{y-y_\varpi}{\tan\varphi} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{y-y_\varpi}{x_\mathrm{surf}\cdot \tan\varphi} \biggr)^2 +  \biggl(\frac{y}{y_\mathrm{surf}} \biggr)^2 - 1 </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{y^2 - 2y y_\varpi + y_\varpi^2}{x^2_\mathrm{surf} \cdot \tan^2\varphi} \biggr] +  \biggl(\frac{y}{y_\mathrm{surf}} \biggr)^2 - 1 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
y^2 - 2y y_\varpi + y_\varpi^2
+ 
x^2_\mathrm{surf} \cdot \tan^2\varphi \biggl(\frac{y}{y_\mathrm{surf}} \biggr)^2 - x^2_\mathrm{surf} \cdot \tan^2\varphi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
y^2 \biggl[ \underbrace{1 + \biggl( \frac{x_\mathrm{surf}}{y_\mathrm{surf}}\biggr)^2 \tan^2\varphi}_{A} \biggr] + y \biggl[\underbrace{ - 2 y_\varpi }_{B}\biggr]  
+ \biggl[ \underbrace{y_\varpi^2 - x^2_\mathrm{surf} \cdot \tan^2\varphi}_{C} \biggr]
</math>
  </td>
</tr>
</table>
Given that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{4AC}{B^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{y^2_\varpi y^2_\mathrm{surf} }\biggl[ y^2_\mathrm{surf} + x_\mathrm{surf}^2 \cdot \tan^2\varphi \biggr]\biggl[ y_\varpi^2 - x^2_\mathrm{surf} \cdot \tan^2\varphi \biggr] \, ,
</math>
  </td>
</tr>
</table>
the pair of roots are given by the expression,
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{B}{2A} 
\biggl\{\pm \biggl[1 + \frac{1}{y^2_\varpi y^2_\mathrm{surf} }\biggl( y^2_\mathrm{surf} + x_\mathrm{surf}^2 \cdot \tan^2\varphi \biggr)\biggl( x^2_\mathrm{surf} \cdot \tan^2\varphi - y_\varpi^2 \biggr) \biggr]^{1 / 2} -1 \biggr\} \, .
</math>
  </td>
</tr>
</table>

As long as <math>~\varphi \ne (2\pi n)</math> &#8212; where <math>~n</math> is an integer &#8212; the "x" coordinate that corresponds to each value of "y" can be obtained from the expression,
<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>~\frac{y - y_\varpi}{\tan\varphi} \, ;</math>
  </td>
</tr>
</table>
for the case of <math>~\varphi = 2\pi n</math>,
<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{surf} \biggl[ 1 - \frac{y^2}{y^2_\mathrm{surf}} \biggr]^{1 / 2} \, .</math>
  </td>
</tr>
</table>

====Associated Velocities (1<sup>st</sup> Try)====

If we set <math>~\varphi = \dot\varphi t</math>, with <math>~\dot\varphi</math> constant, we appreciate that both of the coefficients, <math>~A</math> and <math>~C</math> will be functions of time.  The time-derivatives of the fluid-element positions will therefore depend on the time-derivatives of these two coefficients.  We find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\dot{A} \equiv \frac{dA}{dt}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\biggl( \frac{x_\mathrm{surf}}{y_\mathrm{surf}}\biggr)^2 \tan\varphi \cdot \frac{d \tan\varphi}{dt} 
=
2 \dot\varphi \biggl( \frac{x_\mathrm{surf}}{y_\mathrm{surf}}\biggr)^2 \frac{\tan\varphi}{\cos^2\varphi} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\dot{C} \equiv \frac{dC}{dt}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{dt} \biggl[
y_\varpi^2 - x^2_\mathrm{surf} \cdot \tan^2\varphi
\biggr]
=
- 2 x^2_\mathrm{surf}\tan\varphi \cdot \frac{d \tan\varphi}{dt}
=
- 2 \dot\varphi x^2_\mathrm{surf}\biggl( \frac{\tan\varphi}{\cos^2\varphi} \biggr) 
=
-\dot{A} y^2_\mathrm{surf} \, .
</math>
  </td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\dot{y} \equiv \frac{dy}{dt}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{B }{2A^2} \biggl\{\pm \biggl[1 - \frac{4AC}{B^2}  \biggr]^{1 / 2} -1 \biggr\} \dot{A}
+
\frac{B}{4A} \biggl\{\pm \biggl[1 - \frac{4AC}{B^2}  \biggr]^{-1 / 2}  \biggl[- \frac{4}{B^2}(C\dot{A} + A\dot{C} )  \biggr]\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{B }{4A^2} \biggl\{2~ \mp~ 2\biggl[1 - \frac{4AC}{B^2}  \biggr]^{1 / 2} 
\mp~ \biggl[1 - \frac{4AC}{B^2}  \biggr]^{-1 / 2}  \biggl[ \frac{4A}{B^2}(C - A y^2_\mathrm{surf} )  \biggr]\biggr\} \dot{A}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{B }{2A^2}\biggl[1 - \frac{4AC}{B^2}  \biggr]^{-1 / 2}   \biggl\{ \biggl[1 - \frac{4AC}{B^2}  \biggr]^{1 / 2}  ~ \mp~ \biggl[1 - \frac{4AC}{B^2}  \biggr] 
\mp~ \frac{2AC}{B^2} \biggl(1 - \frac{A y^2_\mathrm{surf}}{C} \biggr) \biggr\} \dot{A}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{B }{2A^2}\biggl[1 - \frac{4AC}{B^2}  \biggr]^{-1 / 2}   \biggl\{ \biggl[1 - \frac{4AC}{B^2}  \biggr]^{1 / 2}     
~\pm~ \frac{2AC}{B^2} \biggl(1 + \frac{A y^2_\mathrm{surf}}{C} \biggr) \biggr\} \dot{A}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{B }{2A^2}  \biggl\{ 1     
~\pm~ \frac{2AC}{B^2} \biggl(1 + \frac{A B^2}{4C} \biggr)\biggl[1 - \frac{4AC}{B^2}  \biggr]^{-1 / 2}  \biggr\} \dot{A} \, .
</math>
  </td>
</tr>
</table>
And, <font color="red">[NOTE:  The last term in this next expression was corrected on 5/21/2021.  Needs to be incorporated into Excel spreadsheet.]</font>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\dot{x} \equiv \frac{dx}{dt}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\dot{y}}{\tan\varphi} + \frac{(y_\varpi - y)}{\tan^2\varphi} \cdot \frac{d\tan\varphi}{dt}
=
\frac{\dot{y}}{\tan\varphi} 
+ \frac{\dot\varphi (y_\varpi - y)}{\sin^2\varphi} 
+ \biggl[ \cancelto{\mathrm{corrected}}{\frac{\dot\varphi (y_\varpi - y)}{\sin\varphi \cos\varphi}}\biggr] \, .
</math>
  </td>
</tr>
</table>

====Associated Velocities (2<sup>nd</sup> Try)====

Let's simplify notation.  Specifically, let's define,

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

<tr>
  <td align="right">
<math>~F^2</math>
  </td>
  <td align="center">
<math>~\equiv </math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{x_\mathrm{surf}}{y_\mathrm{surf}} \biggr)^2 \tan^2\varphi \, .
</math>
  </td>
</tr>
</table>
Then we can write,

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

<tr>
  <td align="right">
<math>~A</math>
  </td>
  <td align="center">
<math>~\equiv </math>
  </td>
  <td align="left">
<math>~
1 + F^2 \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~B</math>
  </td>
  <td align="center">
<math>~\equiv </math>
  </td>
  <td align="left">
<math>~
-2y_\varpi \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~C</math>
  </td>
  <td align="center">
<math>~\equiv </math>
  </td>
  <td align="left">
<math>~
y_\mathrm{surf}^2 \biggl[ \frac{y_\varpi^2}{y_\mathrm{surf}^2} - F^2 \biggr] \, ,
</math>
  </td>
</tr>
</table>
in which case,

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

<tr>
  <td align="right">
<math>~\frac{4AC}{B^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{v_\mathrm{surf}^2}{y_\varpi^2} \biggl[ (1 + F^2)\biggl( \frac{y_\varpi^2}{y_\mathrm{surf}^2}- F^2 \biggr) \biggr] 
\, .
</math>
  </td>
</tr>
</table>

The pair of roots (desired values of "y") are therefore given by the expression,
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~\frac{y_\varpi}{(1 + F^2)} 
\biggl\{\pm \biggl[1 + \frac{ y_\mathrm{surf}^2 }{y_\varpi^2} \biggl(1 + F^2 \biggr) \biggl( F^2 - \frac{y_\varpi^2}{y_\mathrm{surf}^2}  \biggr) \biggr]^{1 / 2} -1 \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{y_\varpi}{(1 + F^2)} 
\biggl\{ 1 \mp \biggl[1 + \frac{ y_\mathrm{surf}^2 }{y_\varpi^2} \biggl(1 + F^2 \biggr) \biggl( F^2 - \frac{y_\varpi^2}{y_\mathrm{surf}^2}  \biggr) \biggr]^{1 / 2}  \biggr\} \, .
</math>
  </td>
</tr>
</table>

Let's examine the time-derivative of y under the assumption that <math>~\varphi = \dot\varphi t</math>.  First, note that,

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

<tr>
  <td align="right">
<math>~\frac{d(F^2)}{dt}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\tan\varphi
\biggl( \frac{x_\mathrm{surf}}{y_\mathrm{surf}} \biggr)^2 \frac{d}{dt}\biggl[ \frac{ \sin(\dot\varphi t ) }{\cos(\dot\varphi t )} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \dot\varphi \tan\varphi \biggl( \frac{x_\mathrm{surf}}{y_\mathrm{surf}} \biggr)^2 \biggl[ 1 + \tan^2 \varphi  \biggr]
=
\frac{2 \dot\varphi \sin\varphi }{\cos^3 \varphi }\biggl( \frac{x_\mathrm{surf}}{y_\mathrm{surf}} \biggr)^2 \, .
</math>
  </td>
</tr>
</table>
Hence,

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

<tr>
  <td align="right">
<math>\frac{\dot{y}}{y_\varpi} \equiv \frac{1}{y_\varpi} \cdot \frac{dy}{dt}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~\frac{ 1 }{(1 + F^2)^2} 
\biggl\{ 1 \mp \biggl[1 + \frac{ y_\mathrm{surf}^2 }{y_\varpi^2} \biggl(1 + F^2 \biggr) \biggl( F^2 - \frac{y_\varpi^2}{y_\mathrm{surf}^2}  \biggr) \biggr]^{1 / 2}  \biggr\} \frac{d(F^2)}{dt}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\mp \frac{ 1 }{(1 + F^2)} 
 \frac{d}{dt}\biggl\{\biggl[1 + \frac{ y_\mathrm{surf}^2 }{y_\varpi^2} \biggl(1 + F^2 \biggr) \biggl( F^2 - \frac{y_\varpi^2}{y_\mathrm{surf}^2}  \biggr) \biggr]^{1 / 2}  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~\frac{ 1 }{(1 + F^2)^2} 
\biggl\{ 1 \mp \biggl[1 + \frac{ y_\mathrm{surf}^2 }{y_\varpi^2} \biggl(1 + F^2 \biggr) \biggl( F^2 - \frac{y_\varpi^2}{y_\mathrm{surf}^2}  \biggr) \biggr]^{1 / 2}  \biggr\} \frac{d(F^2)}{dt}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\mp \frac{ 1 }{2(1 + F^2)} 
\biggl\{ \frac{ y_\mathrm{surf}^2 }{y_\varpi^2}  \biggl[1 + \frac{ y_\mathrm{surf}^2 }{y_\varpi^2} \biggl(1 + F^2 \biggr) \biggl( F^2 - \frac{y_\varpi^2}{y_\mathrm{surf}^2}  \biggr) \biggr]^{-1 / 2}  \biggr\}
\frac{d}{dt}\biggl\{\biggl(1 + F^2 \biggr) \biggl( F^2 - \frac{y_\varpi^2}{y_\mathrm{surf}^2}  \biggr)   \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~\frac{ 1 }{2(1 + F^2)^2} 
\biggl\{ \frac{2y(1+F^2)}{y_\varpi}  \biggr\} \frac{d(F^2)}{dt}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{ 1 }{2(1 + F^2)^2} 
\biggl\{ \biggl[ \frac{y_\varpi (1 + F^2)}{y(1+F^2) - y_\varpi} \biggr]\frac{y^2_\mathrm{surf}}{y^2_\varpi}
\biggl[1 + 2F^2 - \frac{y_\varpi^2}{y_\mathrm{surf}^2}   \biggr]  \biggr\} \frac{d(F^2)}{dt}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 1 }{2(1 + F^2)^2} 
\biggl\{ \biggl[ \frac{y_\varpi (1 + F^2)}{y(1+F^2) - y_\varpi} \biggr]\frac{y^2_\mathrm{surf}}{y^2_\varpi}
\biggl[1 + 2F^2 - \frac{y_\varpi^2}{y_\mathrm{surf}^2}   \biggr] 
-  \frac{2y(1+F^2)}{y_\varpi} 
\biggr\} \frac{d(F^2)}{dt}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \dot{y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 1 }{2 (1 + F^2)} 
\biggl\{ \biggl[ \frac{1 }{y(1+F^2) - y_\varpi} \biggr]
\biggl[ y^2_\mathrm{surf} (1 + 2F^2) - y_\varpi^2   \biggr] 
-  2y 
\biggr\} \frac{d(F^2)}{dt}
</math>
  </td>
</tr>
</table>

=See Also=

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

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


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