Editing ThreeDimensionalConfigurations/ChallengesPt6 (section)

=With Help From Excel Spreadsheet=

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

===Velocity Components===

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

===Limits on z'===

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

====Initial Determination====

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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