Editing Appendix/Ramblings/ConcentricEllipsoidalT8Coordinates (section)

==Realigning the Second Coordinate==

The first coordinate remains the same as before, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^2 + q^2 y^2 + p^2 z^2 \, .</math>
  </td>
</tr>
</table>
This may 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}{a}\biggr)^2 + \biggl( \frac{y}{b}\biggr)^2 + \biggl(\frac{z}{c}\biggr)^2 \, ,</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a = \lambda_1 \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp;&nbsp; &nbsp;</td>
  <td align="center">
<math>~b = \frac{\lambda_1}{q} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp;&nbsp; &nbsp;</td>
  <td align="left">
<math>~c = \frac{\lambda_1}{p} \, .</math>
  </td>
</tr>
</table>

By specifying the value of <math>~z = z_0 < c</math>, as well as the value of <math>~\lambda_1</math>, we are picking a plane that lies parallel to, but a distance <math>~z_0</math> above, the equatorial plane.  The elliptical curve that defines the intersection of the <math>~\lambda_1</math>-constant surface with this plane is defined by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_1^2 - p^2z_0^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^2 + q^2 y^2  </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{x}{a_{2D}}\biggr)^2 + \biggl( \frac{y}{b_{2D}}\biggr)^2 \, ,</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a_{2D} = \biggl(\lambda_1^2 - p^2z_0^2 \biggr)^{1 / 2} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp;&nbsp; &nbsp;</td>
  <td align="left">
<math>~b_{2D} = \frac{1}{q} \biggl(\lambda_1^2 - p^2z_0^2 \biggr)^{1 / 2} \, .</math>
  </td>
</tr>
</table>

At each point along this elliptic curve, the line that is tangent to the curve has a slope that can be determined by simply differentiating the equation that describes the curve, that is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2x dx}{a_{2D}^2} + \frac{2y dy}{b_{2D}^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{dy}{dx}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{2x}{a_{2D}^2} \cdot \frac{b_{2D}^2}{2y} = - \frac{x}{q^2y} \, .</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\Delta y</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl( \frac{x}{q^2y} \biggr)\Delta x \, .</math>
  </td>
</tr>
</table>

The unit vector that lies tangent to any point on this elliptical curve will be described by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath~ \biggl\{ \frac{\Delta x}{[ (\Delta x)^2 + (\Delta y)^2 ]^{1 / 2}} \biggr\}
+
\hat\jmath~ \biggl\{ \frac{\Delta y}{[ (\Delta x)^2 + (\Delta y)^2 ]^{1 / 2}} \biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath~ \biggl\{ \frac{1}{[ 1 + x^2/(q^4y^2) ]^{1 / 2}} \biggr\}
-
\hat\jmath~ \biggl\{ \frac{x/(q^2y)}{[ 1 + x^2/(q^4y^2) ]^{1 / 2}} \biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath~ \biggl\{ \frac{q^2y}{[ x^2 + q^4y^2 ]^{1 / 2}} \biggr\}
-
\hat\jmath~ \biggl\{ \frac{x}{[ x^2 + q^4y^2  ]^{1 / 2}} \biggr\} \, .</math>
  </td>
</tr>
</table>
As we have discovered, the coordinate that gives rise to this unit vector is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{x}{y^{1/q^2}} \, .</math>
  </td>
</tr>
</table>

Other properties that result from this definition of <math>~\lambda_2</math> are presented in the following table.

<table border="1" cellpadding="8" align="center">
<tr>
  <td align="center" colspan="9">'''Direction Cosine Components for T8 Coordinates'''</td>
</tr>
<tr>
  <td align="center"><math>~n</math></td>
  <td align="center"><math>~\lambda_n</math></td>
  <td align="center"><math>~h_n</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial x}</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial y}</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial z}</math></td>
  <td align="center"><math>~\gamma_{n1}</math></td>
  <td align="center"><math>~\gamma_{n2}</math></td>
  <td align="center"><math>~\gamma_{n3}</math></td>
</tr>

<tr>
  <td align="center"><math>~1</math></td>
  <td align="center"><math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math></td>
  <td align="center"><math>~\lambda_1 \ell_{3D}</math></td>
  <td align="center"><math>~\frac{x}{\lambda_1}</math></td>
  <td align="center"><math>~\frac{q^2 y}{\lambda_1}</math></td>
  <td align="center"><math>~\frac{p^2 z}{\lambda_1}</math></td>
  <td align="center"><math>~(x) \ell_{3D}</math></td>
  <td align="center"><math>~(q^2 y)\ell_{3D}</math></td>
  <td align="center"><math>~(p^2z) \ell_{3D}</math></td>
</tr>

<tr>
  <td align="center"><math>~2</math></td>
  <td align="center"><math>~\frac{x}{  y^{1/q^2}}</math></td>
  <td align="center"><math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y }{(x^2 + q^4y^2)^{1 / 2}}\biggr] </math></td>
  <td align="center"><math>~\frac{\lambda_2}{x}</math></td>
  <td align="center"><math>~-\frac{\lambda_2}{q^2 y}</math></td>
  <td align="center"><math>~0</math></td>
  <td align="center"><math>~\frac{q^2 y }{(x^2 + q^4y^2)^{1 / 2}} </math></td>
  <td align="center"><math>~- \frac{x }{(x^2 + q^4y^2)^{1 / 2}} </math></td>
  <td align="center"><math>~0</math></td>
</tr>

<tr>
  <td align="center"><math>~3</math></td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
</tr>

<tr>
  <td align="left" colspan="9">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>~\ell_{3D}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} </math>
  </td>
</tr>

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

The associated unit vector is, then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath~\biggl[ \frac{q^2 y }{(x^2 + q^4y^2)^{1 / 2}}  \biggr] - \hat\jmath~\biggl[ \frac{x }{(x^2 + q^4y^2)^{1 / 2}}  \biggr] \, .
</math>
  </td>
</tr>
</table>

It is easy to see that <math>~\hat{e}_2 \cdot \hat{e}_2 = 1</math>.  We also see that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_1 \cdot \hat{e}_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x \ell_{3D}\biggl[ \frac{q^2 y }{(x^2 + q^4y^2)^{1 / 2}}  \biggr]
-
q^2y \ell_{3D} \biggl[ \frac{x }{(x^2 + q^4y^2)^{1 / 2}}  \biggr] = 0 \, ,
</math>
  </td>
</tr>
</table>
so it is clear that these first two unit vectors are orthogonal to one another.