Editing ThreeDimensionalConfigurations/ChallengesPt6 (section)

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

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

<tr>
  <td align="right">
<math>1 </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl(\frac{x}{a}\biggr)^2 + \biggl(\frac{y}{b}\biggr)^2 + \biggl(\frac{z}{c}\biggr)^2
\, .
</math>
  </td>
</tr>
</table>
Given that,
<table border="0" cellpadding="5" align="center">

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

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

<tr>
  <td align="right">
<math>(z - z_0)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
z' \cos\theta + y'\sin\theta 
\, ,</math>
  </td>
</tr>
</table>
the constraint becomes,
<table border="0" cellpadding="5" align="center">

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

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

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

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

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

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

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

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

<tr>
  <td align="right">
<math>C</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
b^2(z_0 + z'\cos\theta )^2 +  c^2(z')^2\sin^2\theta + b^2c^2 \biggl[\biggl(\frac{x'}{a}\biggr)^2 -1 \biggr] \, .
</math>
  </td>
</tr>
</table>
In which case,
<table border="0" cellpadding="5" align="center">

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

===Limits on z'===

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

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

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

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

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

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

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

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

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

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

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

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

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

<tr>
  <td align="right">
<math>z'</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\pm \epsilon (c^2 \cos^2\theta + b^2 \sin^2\theta )^{1 / 2}
- z_0\cos\theta \, .
</math>
  </td>
</tr>
</table>
And the corresponding value(s) of <math>y'</math> comes from the relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>y' </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{B}{2A}\biggl\{ - 1 \pm \cancelto{0}{\biggl[1 - \frac{4AC}{B^2}  \biggr]^{1 / 2}} \biggr\}
</math>
  </td>
</tr>

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

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

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

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

====Initial Determination====

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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