Editing Appendix/Ramblings/ConcentricEllipsoidalT8Coordinates

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=Concentric Ellipsoidal (T8) Coordinates=

==Background==
Building on our [[User:Tohline/Appendix/Ramblings/DirectionCosines|general introduction to ''Direction Cosines'']] in the context of orthogonal curvilinear coordinate systems, and on our previous development of [[User:Tohline/Appendix/Ramblings/T3Integrals|T3]] (concentric oblate-spheroidal) and [[User:Tohline/Appendix/Ramblings/EllipticCylinderCoordinates#T5_Coordinates|T5]] (concentric elliptic) coordinate systems, here we explore the creation of a concentric ellipsoidal (T8) coordinate system.  This is motivated by our [[User:Tohline/ThreeDimensionalConfigurations/Challenges#Trial_.232|desire to construct a fully analytically prescribable model of a nonuniform-density ellipsoidal configuration that is an analog to Riemann S-Type ellipsoids]].

Note that, in a [[User:Tohline/Appendix/Ramblings/ConcentricEllipsodalCoordinates#Background|separate but closely related discussion]], we made attempts to define this coordinate system, numbering the trials up through "T7."  In this "T7" effort, we were able to define a set of three, mutually orthogonal unit vectors that should work to define a fully three-dimensional, concentric  ellipsoidal coordinate system.  But we were unable to figure out what coordinate function, <math>~\lambda_3(x, y, z)</math>, was associated with the third unit vector.  In addition, we found the <math>~\lambda_2</math> coordinate to be rather strange in that it was not oriented in a manner that resembled the classic spherical coordinate system.  Here we begin by redefining the <math>~\lambda_2</math> coordinate such that its associated <math>~\hat{e}_3</math> unit vector lies parallel to the x-y plane.

==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.  

==Search for the Third Coordinate==

===Cross Product of First Two Unit Vectors===

The cross-product of these two unit vectors should give the third, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_3 = \hat{e}_1 \times \hat{e}_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath~\biggl[ {e}_{1y}{e}_{2z} - {e}_{1z}{e}_{2y} \biggr]
+
\hat\jmath~\biggl[ {e}_{1z}{e}_{2x} - {e}_{1x}{e}_{2z} \biggr]
+
\hat{k}~\biggl[ {e}_{1x}{e}_{2y} - {e}_{1y}{e}_{2x} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath~\biggl[  \frac{x p^2z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} \biggr]
+
\hat\jmath~\biggl[ \frac{q^2 y p^2z \ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}}  \biggr]
-
\hat{k}~\biggl[ \frac{x^2\ell_{3D} }{(x^2 + q^4y^2)^{1 / 2}}  ~+~ \frac{q^4 y^2\ell_{3D} }{(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>~ \frac{\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} \biggl\{~
\hat\imath~( x p^2z )
+
\hat\jmath~(q^2 y p^2z  )
-
\hat{k}~(x^2  + q^4y^2)
~\biggr\} \, .
</math>
  </td>
</tr>
</table>
Inserting these component expressions into the last row of the T8 Direction Cosine table gives &hellip;

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


===Associated h<sub>3</sub> Scale Factor===
<table border="1" align="right" cellpadding="10"><tr><td align="center">
[[File:EUREKA 21Jan2021 sm.png|350px|Whiteboard EUREKA moment]]</td></tr></table>
After working through various scenarios on my whiteboard today (21 January 2021), I propose that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_3}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{xp^2z}{(x^2 + q^4y^2)} \, ;</math>
  </td>
<td align="center>&nbsp; &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_3}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{q^2 y p^2z}{(x^2 + q^4y^2)} \, ;</math>
  </td>
<td align="center>&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_3}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-1 \, .</math>
  </td>
</tr>
</table>

This means that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_3^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sum_{i=1}^3 \biggl( \frac{\partial \lambda_3}{\partial x_i}\biggr)^2</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{p^4z^2(x^2 + q^4y^2)}{(x^2 + q^4y^2)^2} 
+
1
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(x^2 + q^4y^2 +p^4z^2)}{(x^2 + q^4y^2)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\ell_{3D}^2 (x^2 + q^4y^2)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ h_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\ell_{3D} (x^2 + q^4y^2)^{1 / 2} \, . 
</math>
  </td>
</tr>
</table>
This seems to work well because, when combined with the three separate expressions for <math>~\partial \lambda_3/\partial x_i</math>,  this single expression for <math>~h_3</math> generates all three components of the third unit vector, that is, all three direction cosines, <math>~\gamma_{3i}</math>.  All of the elements of this new "EUREKA moment" result have been entered into 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"><math>~\ell_{3D}(x^2 + q^4 y^2)^{1 / 2}</math></td>
  <td align="center"><math>~\frac{xp^2z}{(x^2 + q^4y^2)} </math></td>
  <td align="center"><math>~\frac{q^2 y p^2z}{(x^2 + q^4y^2)}</math></td>
  <td align="center"><math>~-1</math></td>
  <td align="center"><math>~\frac{x p^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math></td>
  <td align="center"><math>~\frac{q^2 y p^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math></td>
  <td align="center"><math>~-\frac{(x^2 + q^4 y^2)\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math></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>


===What is the Third Coordinate Function, &lambda;<sub>3</sub>===

The remaining [https://en.wikipedia.org/wiki/The_$64,000_Question $64,000 question] is, "What is the actual expression for <math>~\lambda_3(x, y, z)</math>&nbsp; ? &nbsp;"  

Notice that the (partial) derivatives of <math>~\lambda_3</math> with respect to <math>~x</math> and <math>~y</math> may be rewritten, respectively, in the form
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{q^2 y}{p^2z} \biggr) \frac{\partial \lambda_3}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{q^2 y}{x(1 + \eta^2)}
=
\frac{\eta}{(1 + \eta^2)} \, ,
</math>
&nbsp; &nbsp; &nbsp; and,
  </td>
</tr>

<tr>
  <td align="right">
<math>~\biggl( \frac{x}{p^2z} \biggr) \frac{\partial \lambda_3}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{q^2y}{x(1 + \eta^2)}
=
\frac{\eta}{(1 + \eta^2)} \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\eta</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{q^2y}{x} ~~~~\Rightarrow~~ \frac{\partial \ln\eta}{\partial \ln x} = -1 \, , ~~~~\frac{\partial \ln\eta}{\partial \ln y} = +1 \, . 
</math>
  </td>
</tr>
</table>

Then, after searching through the [[User:Tohline/Appendix/References#CRC|CRC Mathematical Handbook's]] pages of familiar derivative expressions, we appreciate that

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

<tr>
  <td align="right">
<math>~\frac{d}{dx_i} \biggl[\frac{1}{\cosh\gamma}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[ \frac{\tanh\gamma}{\cosh\gamma} \biggr] \frac{d\gamma}{dx_i} \, .</math>
  </td>
</tr>
</table>

Hence, it will be useful to adopt the mapping,

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

<tr>
  <td align="right">
<math>~\eta</math>
  </td>
  <td align="center">
<math>~~~\rightarrow~~~</math>
  </td>
  <td align="left">
<math>~\sinh \gamma \, ,</math>
  </td>
</tr>
</table>
because the right-hand side of both partial-derivative expressions becomes,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\eta}{(1+\eta^2)}</math>
  </td>
  <td align="center">
<math>~~~\rightarrow~~~</math>
  </td>
  <td align="left">
<math>~\frac{\sinh\gamma}{\cosh^2\gamma} = \frac{\tanh\gamma}{\cosh\gamma} \, .</math>
  </td>
</tr>
</table>

====Guess A====
In particular, this suggests that we set,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{A}{\cosh\gamma} \, ,</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\gamma</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\sinh^{-1}\eta = \pm \cosh^{-1}[\eta^2 + 1]^{1 / 2} \, .</math>
  </td>
</tr>
</table>
In other words,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A[\eta^2 + 1]^{-1 / 2} \, .</math>
  </td>
</tr>
</table>
Let's check the first and second partial derivatives.
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_3}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{A}{2} \biggl[ \frac{2\eta}{(\eta^2 + 1)^{3 / 2}} \biggr] \frac{\partial \eta}{\partial x}
</math>
  </td>
</tr>
</table>

====Guess B====

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

<tr>
  <td align="right">
<math>~\lambda_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2}\ln(1+\eta^2) \, .</math>
  </td>
</tr>
</table>
Then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d\lambda_3}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\eta}{(1+\eta^2)} \, ,
</math>
  </td>
</tr>
</table>
in which case we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial\lambda_3}{\partial x_i} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d\lambda_3}{d\eta} \cdot \frac{\partial\eta}{\partial x_i}</math>
  </td>
</tr>
</table>
which means,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial\lambda_3}{\partial x} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\eta}{(1+\eta^2)}  \biggl[ - \frac{q^2 y}{x^2} \biggr]
=
- \frac{x}{(x^2 + q^4y^2)} \biggl[ \frac{q^4 y^2}{x^2} \biggr] \, .
</math>
  </td>
</tr>
</table>

====Guess C====

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

<tr>
  <td align="right">
<math>~\lambda_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2} \ln(1+\eta^{-2}) \, .</math>
  </td>
</tr>
</table>
Then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d\lambda_3}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-
\frac{\eta^{-3}}{(1+\eta^{-2})} 
=
-\frac{ \eta^{-1}}{(1+\eta^{2})} 
</math>
  </td>
</tr>
</table>
in which case we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial\lambda_3}{\partial x_i} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d\lambda_3}{d\eta} \cdot \frac{\partial\eta}{\partial x_i}</math>
  </td>
</tr>
</table>
which means,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial\lambda_3}{\partial x} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\eta^{-1}}{(1+\eta^2)}  \biggl[ - \frac{q^2 y}{x^2} \biggr]
=
\frac{x}{(x^2 + q^4y^2)}  \, ;
</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial\lambda_3}{\partial y} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\eta^{-1}}{(1+\eta^2)}  \biggl[ \frac{q^2 }{x} \biggr]
=
-\frac{x^2}{y(x^2+ q^4y^2)}  
</math>
  </td>
</tr>
</table>

==Inverting Coordinate Relations==

===In a Plane Perpendicular to the Z-Axis===
====General Case====
At a fixed value of <math>~z = z_0</math>, let's invert the <math>~\lambda_1(x, y)</math> and <math>~\lambda_2(x, y)</math> relations to obtain expressions for <math>~x(\lambda_1, \lambda_2)</math> and <math>~y(\lambda_1, \lambda_2)</math>.  Perhaps this will help us determine what the third coordinate expression should be.  

We start with,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_1</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(x^2 + q^2y^2 + p^2 z_0^2)^{1 / 2} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\lambda_2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{x}{y^{1/q^2}} \, .</math>
  </td>
</tr>
</table>
This means that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\ln y</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q^2(\ln x - \ln\lambda_2 ) \, ,</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~2\ln y</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\ln\biggl\{ \frac{1}{q^2} \biggl[ (\lambda_1^2 - p^2z_0^2) - x^2 \biggr] \biggr\} \, .</math>
  </td>
</tr>
</table>
Together, this gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\ln\biggl\{ \frac{1}{q^2} \biggl[ (\lambda_1^2 - p^2z_0^2) - x^2 \biggr] \biggr\} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2q^2(\ln x - \ln\lambda_2 ) </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\ln\biggl[\frac{x}{\lambda_2} \biggr]^{2q^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~(\lambda_1^2 - p^2z_0^2) - x^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q^2\biggl[\frac{x}{\lambda_2} \biggr]^{2q^2} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ x^2 +q^2\biggl[\frac{x}{\lambda_2} \biggr]^{2q^2} +  (p^2z_0^2 - \lambda_1^2) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>

====What if Axisymmetric (q<sup>2</sup> = 1)====

In an axisymmetric configuration, <math>~q^2 = 1</math> and <math>~(\lambda_1^2 - p^2z_0^2) = \varpi^2</math>, so this general expression for <math>~x</math> 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>~
x^2 + \biggl[\frac{x}{\lambda_2} \biggr]^{2} -\varpi^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2\biggl[\frac{1 + \lambda_2^2}{\lambda_2^2}\biggr]  -\varpi^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\varpi \biggl[\frac{\lambda_2^2}{1 + \lambda_2^2}\biggr]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
Given that, for axisymmetric systems,
<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>
</tr>
</table>
we conclude that when <math>~q^2 = 1</math>,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[\frac{\lambda_2^2}{1 + \lambda_2^2}\biggr]^{1 / 2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\cos\varphi </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\lambda_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\cot^2\varphi </math>
  </td>
</tr>
</table>

====What if q<sup>2</sup> = 2====
For example, if we choose <math>~q^2 = 2</math>, we have a quadratic expression for <math>~x^2</math>, namely,

<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>~x^2 + 2\biggl[\frac{x}{\lambda_2} \biggr]^{4} +  (p^2z_0^2 - \lambda_1^2) </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^{4} + \frac{1}{2} \lambda_2^4~ x^2 +  \frac{1}{2}\lambda_2^4(p^2z_0^2 - \lambda_1^2) </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 2x^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{\lambda_2^4}{2} \pm \biggl[ \frac{\lambda_2^8}{4} - 2\lambda_2^4(p^2z_0^2 - \lambda_1^2) \biggr]^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\lambda_2^4}{2} \biggl\{ \biggl[ 1 - \frac{8(p^2z_0^2 - \lambda_1^2)}{\lambda_2^4} \biggr]^{1 / 2} - 1 \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\lambda_2^4}{4} \biggl\{ \biggl[ 1 - \frac{8(p^2z_0^2 - \lambda_1^2)}{\lambda_2^4} \biggr]^{1 / 2} - 1 \biggr\} \, .
</math>
  </td>
</tr>
</table>
Given that, for <math>~q^2 = 2</math>, one of the two defining expression means, <math>~\lambda_2 = x/\sqrt{y}</math>, we also have,

<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{\lambda_2^2}{4} \biggl\{ \biggl[ 1 - \frac{8(p^2z_0^2 - \lambda_1^2)}{\lambda_2^4} \biggr]^{1 / 2} - 1 \biggr\} \, .
</math>
  </td>
</tr>
</table>

===New 2<sup>nd</sup> Coordinate===

Apparently it will be cleaner to define a new "2<sup>nd</sup> coordinate," <math>~\kappa_2</math>, such that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\kappa_2^{2q^2}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
q^2\biggl[\frac{1}{\lambda_2} \biggr]^{2q^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\kappa_2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
q^{1/q^2}\biggl[\frac{1}{\lambda_2} \biggr] 
= 
q^{1/q^2}\biggl[\frac{y^{1/q^2}}{x} \biggr]
= 
\frac{(qy)^{1/q^2}}{x}  \, .
</math>
  </td>
</tr>
</table>
(With this new definition, <math>~\kappa_2 \sim \tan\varphi</math>; it is exactly this when <math>~q^2 = 1</math>.)  Then we can rewrite the last expression from the [[#General_Case|above general case]] as,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x^2 +(\kappa_2 x)^{2q^2} -  (\lambda_1^2 - p^2z_0^2 ) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>
When <math>~q=1</math> (the axisymmetric case), this gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x^2(1+\kappa_2^2)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ (\lambda_1^2 - p^2z_0^2 ) </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{x^2}{(\lambda_1^2 - p^2z_0^2 ) }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{(1+\kappa_2^2)} \, ,</math>
  </td>
</tr>
</table>
which means that <math>~\kappa_2 = \tan\varphi</math>.  And, for the case of <math>~q^2 = 2</math>, after making the substitution, 
<div align="center"><math>~\lambda_2 \rightarrow (q)^{1/q^2}\kappa_2^{-1} = \frac{2^{1 / 4}}{\kappa_2} \, ,</math></div>
we find,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\lambda_2^4}{4} \biggl\{ \biggl[ 1 - \frac{8(p^2z_0^2 - \lambda_1^2)}{\lambda_2^4} \biggr]^{1 / 2} - 1 \biggr\} \, .
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~ x^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\kappa_2^4} \biggl\{ \biggl[ 1 + 4\kappa_2^4(\lambda_1^2 - p^2z_0^2 ) \biggr]^{1 / 2} - 1 \biggr\} \, ,
</math>
  </td>
</tr>
</table>
and,

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

<tr>
  <td align="right">
<math>~y = \biggl(\frac{\kappa_2^2}{2^{1 / 2}} \biggr) x^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{\kappa_2^2}{2^{1 / 2}} \biggr)
\frac{1}{2\kappa_2^4} \biggl\{ \biggl[ 1 + 4\kappa_2^4(\lambda_1^2 - p^2z_0^2 ) \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{1}{2^{3 / 2}\kappa_2^2} \biggl\{ \biggl[ 1 + 4\kappa_2^4(\lambda_1^2 - p^2z_0^2 ) \biggr]^{1 / 2} - 1 \biggr\} \, .
</math>
  </td>
</tr>
</table>

=Angle Between Unit Vectors=
We begin by restating that the coordinate, scale factor, and unit vector associated with the normal to our ellipsoidal surface are,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_1</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(x^2 + q^2y^2 + p^2z^2)^{1 / 2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~h_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\lambda_1 \ell_{3D} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\hat{e}_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\hat\imath (x\ell_{3D} )
+
\hat\jmath (q^2 y\ell_{3D} )
+
\hat{k} (p^2z\ell_{3D} ) \, .
</math>
  </td>
</tr>
</table>

In the [[#TableKappa8|Table below titled, "Direction Cosines Components for &kappa;8 Coordinates"]], there are two fully-formed unit vectors that are each orthogonal to the (first) unit vector that is normal to the ellipsoid's surface.  Here we will refer to the coordinates of these two fully-formed unit vectors as,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{xy^{1/q^2}}{z^{2/p^2}}</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~\kappa_2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{(qy)^{1/q^2}}{x} \, .</math>
  </td>
</tr>
</table>
The associated scale factors and unit vectors are given by the following expressions:

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

<tr>
  <td align="right">
<math>~h_{\lambda_2}^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl(\frac{\partial\lambda_2}{\partial x} \biggr)^2
+
\biggl(\frac{\partial\lambda_2}{\partial y} \biggr)^2
+
\biggl(\frac{\partial\lambda_2}{\partial z} \biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl(\frac{\lambda_2}{x} \biggr)^2
+
\biggl(\frac{\lambda_2}{q^2y} \biggr)^2
+
\biggl(- \frac{2\lambda_2}{p^2z} \biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \lambda_2^2 \biggl[
\frac{q^4y^2p^4z^2 + x^2p^4z^2 + 4x^2 q^4y^2}{x^2 q^4y^2 p^4z^2} 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ h_{\lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{x q^2y p^2z}{\lambda_2 \mathcal{D}}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \hat{e}_{\lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\hat\imath \biggl[ h_{\lambda_2} \biggl(\frac{\partial \lambda_2}{\partial x}\biggr)\biggr]
+
\hat\jmath \biggl[ h_{\lambda_2} \biggl(\frac{\partial \lambda_2}{\partial y}\biggr)\biggr]
+
\hat{k} \biggl[ h_{\lambda_2} \biggl(\frac{\partial \lambda_2}{\partial z}\biggr)\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 p^2z}{\mathcal{D}} \biggr]
+
\hat\jmath \biggl[ \frac{xp^2z}{\mathcal{D}} \biggr]
-
\hat{k} \biggl[ \frac{2x q^2y }{\mathcal{D}} \biggr] \, .
</math>
  </td>
</tr>
</table>


<table border="1" width="80%" align="center" cellpadding="8"><tr><td align="left">
With regard to orthogonality, note that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_1 \cdot \hat{e}_{\lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[
\hat\imath (x\ell_{3D} )
+
\hat\jmath (q^2 y\ell_{3D} )
+
\hat{k} (p^2z\ell_{3D} ) 
\biggr] \cdot
\biggl\{
\hat\imath \biggl[ \frac{q^2y p^2z}{\mathcal{D}} \biggr]
+
\hat\jmath \biggl[ \frac{xp^2z}{\mathcal{D}} \biggr]
-
\hat{k} \biggl[ \frac{2x q^2y }{\mathcal{D}} \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(x\ell_{3D} )\biggl[ \frac{q^2y p^2z}{\mathcal{D}} \biggr]
+
(q^2 y\ell_{3D} )\biggl[ \frac{xp^2z}{\mathcal{D}} \biggr]
-
(p^2z\ell_{3D} )\biggl[ \frac{2x q^2y }{\mathcal{D}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
0 \, .</math>
  </td>
</tr>
</table>

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


And,

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

<tr>
  <td align="right">
<math>~h_{\kappa_2}^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl(\frac{\partial\kappa_2}{\partial x} \biggr)^2
+
\biggl(\frac{\partial\kappa_2}{\partial y} \biggr)^2
+
\biggl(\frac{\partial\kappa_2}{\partial z} \biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl(-\frac{\kappa_2}{x} \biggr)^2
+
\biggl(\frac{\kappa_2}{q^2 y} \biggr)^2
=
\frac{\kappa_2^2}{x^2q^4y^2}\biggl[x^2 + q^4y^2\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~h_{\kappa_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{xq^2y}{\kappa_2 (x^2 + q^4y^2)^{1 / 2}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\hat{e}_{\kappa_2}</math>
  </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]
+
\hat{k} \biggl[ 0 \biggr] \, .
</math>
  </td>
</tr>
</table>


<table border="1" width="80%" align="center" cellpadding="8"><tr><td align="left">
Again, note that with regard to orthogonality, 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_1 \cdot \hat{e}_{\kappa_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[
\hat\imath (x\ell_{3D} )
+
\hat\jmath (q^2 y\ell_{3D} )
+
\hat{k} (p^2z\ell_{3D} ) 
\biggr] \cdot
\biggl\{
-\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]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~(x\ell_{3D} )\biggl[ \frac{q^2y}{(x^2 + q^4y^2)^{1 / 2}} \biggr]
+
(q^2 y\ell_{3D} )\biggl[ \frac{x}{(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>~
0 \, .</math>
  </td>
</tr>
</table>

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


From this pair of orthogonality checks, we appreciate that both unit vectors always lie in the plane that is tangent to the surface of our ellipsoid.  Next, let's determine the angle, <math>~\alpha</math>, between these two unit vectors as measured in the relevant tangent-plane.

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

<tr>
  <td align="right">
<math>~\cos\alpha \equiv \hat{e}_{\lambda_2} \cdot \hat{e}_{\kappa_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
\hat\imath \biggl[ \frac{q^2y p^2z}{\mathcal{D}} \biggr]
+
\hat\jmath \biggl[ \frac{xp^2z}{\mathcal{D}} \biggr]
-
\hat{k} \biggl[ \frac{2x q^2y }{\mathcal{D}} \biggr] 
\biggr\} 
\biggl\{
-\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]
+
\hat{k} \biggl[ 0 \biggr] 
\biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{p^2 z}{\mathcal{D} (x^2 + q^4y^2)^{1 / 2}} \biggr] \biggl[ x^2 - q^4y^2 \biggr] \, .
</math>
  </td>
</tr>
</table>


----


Let's again visit the unit vector that we know lies in the tangent-plane and is always orthogonal to <math>~\lambda_2</math>, namely,

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

<tr>
  <td align="right">
<math>~\hat{e}_{\lambda_3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
-~\hat\imath \biggl[ x(2q^4 y^2 + p^4z^2 )\biggr]\frac{\ell_{3D}}{\mathcal{D}}
+
\hat\jmath \biggl[q^2y(p^4z^2 + 2x^2) \biggr]\frac{\ell_{3D}}{\mathcal{D}}
+
\hat{k} \biggl[p^2z (x^2 - q^4y^2) \biggr]\frac{\ell_{3D}}{\mathcal{D}} 
\, .
</math>
  </td>
</tr>
</table>

We acknowledge that,

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

<tr>
  <td align="right">
<math>~\hat{e}_{\lambda_2} \cdot \hat{e}_{\lambda_3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
\hat\imath \biggl[ \frac{q^2y p^2z}{\mathcal{D}} \biggr]
+
\hat\jmath \biggl[ \frac{xp^2z}{\mathcal{D}} \biggr]
-
\hat{k} \biggl[ \frac{2x q^2y }{\mathcal{D}} \biggr] 
\biggr\} 
\biggl\{
-~\hat\imath \biggl[ x(2q^4 y^2 + p^4z^2 )\biggr]\frac{\ell_{3D}}{\mathcal{D}}
+
\hat\jmath \biggl[q^2y(p^4z^2 + 2x^2) \biggr]\frac{\ell_{3D}}{\mathcal{D}}
+
\hat{k} \biggl[p^2z (x^2 - q^4y^2) \biggr]\frac{\ell_{3D}}{\mathcal{D}} 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\biggl[ xq^2yp^2z(2q^4 y^2 + p^4z^2 )\biggr]\frac{\ell_{3D}}{\mathcal{D}^2}
+
\biggl[x q^2y p^2z(p^4z^2 + 2x^2) \biggr]\frac{\ell_{3D}}{\mathcal{D}^2}
-
\biggl[2xq^2y p^2z (x^2 - q^4y^2) \biggr]\frac{\ell_{3D}}{\mathcal{D}^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[
-(2q^4 y^2 + p^4z^2 )
+
(p^4z^2 + 2x^2) 
-
2 (x^2 - q^4y^2)  
\biggr] \frac{(x q^2y p^2z) \ell_{3D}}{\mathcal{D}^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
0 \, .
</math>
  </td>
</tr>
</table>

=Kappa (&kappa;8) Coordinates=
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\kappa_1</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(x^2 + q^2y^2 + p^2 z^2)^{1 / 2} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\kappa_3</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\tan^{-1}\biggl[ \frac{(qy)^{1/q^2}}{x} \biggr] \, .</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\frac{\partial\kappa_3}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1 + \frac{(qy)^{2/q^2}}{x^2} \biggr]^{-1} \biggl[- \frac{(qy)^{1/q^2}}{x^2}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\sin^2\kappa_3}{(qy)^{1/q^2}} = -\frac{\sin^2\kappa_3}{x\tan\kappa_3}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\sin\kappa_3 \cos\kappa_3}{x}\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial\kappa_3}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1 + \frac{(qy)^{2/q^2}}{x^2} \biggr]^{-1} \biggl[\frac{q^{1/q^2} y^{1/q^2}}{q^2 x y}  \biggr]
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
 \frac{1}{q^2y}\biggl[\frac{\tan\kappa_3}{1 + \tan^2\kappa_3 }\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
+ \frac{\sin\kappa_3 \cos\kappa_3}{q^2y} \, .
</math>
  </td>
</tr>
</table>
Therefore,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_3^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[-\frac{\sin\kappa_3 \cos\kappa_3}{x} \biggr]^2
+
\biggl[\frac{\sin\kappa_3 \cos\kappa_3}{q^2y} \biggr]^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(x^2 + q^4y^2)
\biggl[\frac{\sin\kappa_3 \cos\kappa_3}{xq^2y} \biggr]^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ h_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(x^2 + q^4y^2)^{-1 / 2}
\biggl[\frac{xq^2y}{\sin\kappa_3 \cos\kappa_3} \biggr] \, .
</math>
  </td>
</tr>
</table>


<span id="TableKappa8">&nbsp;</span>
<table border="1" cellpadding="8" align="center">
<tr>
  <td align="center" colspan="9">'''Direction Cosine Components for &kappa;8 Coordinates'''</td>
</tr>
<tr>
  <td align="center"><math>~n</math></td>
  <td align="center"><math>~\kappa_n</math></td>
  <td align="center"><math>~h_n</math></td>
  <td align="center"><math>~\frac{\partial \kappa_n}{\partial x}</math></td>
  <td align="center"><math>~\frac{\partial \kappa_n}{\partial y}</math></td>
  <td align="center"><math>~\frac{\partial \kappa_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>~\kappa_1 \ell_{3D}</math></td>
  <td align="center"><math>~\frac{x}{\kappa_1}</math></td>
  <td align="center"><math>~\frac{q^2 y}{\kappa_1}</math></td>
  <td align="center"><math>~\frac{p^2 z}{\kappa_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">---</td>
  <td align="center"><math>~\ell_{3D}(x^2 + q^4 y^2)^{1 / 2}</math></td>
  <td align="center"><math>~\frac{xp^2z}{(x^2 + q^4y^2)} </math></td>
  <td align="center"><math>~\frac{q^2 y p^2z}{(x^2 + q^4y^2)}</math></td>
  <td align="center"><math>~-1</math></td>
  <td align="center"><math>~\frac{x p^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math></td>
  <td align="center"><math>~\frac{q^2 y p^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math></td>
  <td align="center"><math>~-\frac{(x^2 + q^4 y^2)\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math></td>
</tr>

<tr>
  <td align="center"><math>~3</math></td>
  <td align="center"><math>~\tan^{-1}\biggl[ \frac{(qy)^{1/q^2}}{x} \biggr]</math></td>
  <td align="center"><math>~\frac{1}{\sin\kappa_3 \cos\kappa_3}\biggl[\frac{x q^2 y }{(x^2 + q^4y^2)^{1 / 2}}\biggr] </math></td>
  <td align="center"><math>~-\frac{\sin\kappa_3 \cos\kappa_3}{x}</math></td>
  <td align="center"><math>~\frac{\sin\kappa_3 \cos\kappa_3}{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="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>

<tr>
  <td align="center"><math>~4</math></td>
  <td align="center"><math>~\frac{x y^{1/q^2}}{ z^{2/p^2}}</math></td>
  <td align="center"><math>~\frac{1}{\kappa_4}\biggl[\frac{x q^2 y p^2 z}{ \mathcal{D}}\biggr] </math></td>
  <td align="center"><math>~\frac{\kappa_4}{x}</math></td>
  <td align="center"><math>~\frac{\kappa_4}{q^2 y}</math></td>
  <td align="center"><math>~-\frac{2\kappa_4}{p^2 z}</math></td>
  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{x}\biggr)</math></td>
  <td align="center"><math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{q^2y}\biggr)</math></td>
  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(-\frac{2}{p^2z}\biggr)</math></td>
</tr>

<tr>
  <td align="center"><math>~5</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"><math>~-\frac{\ell_{3D}}{\mathcal{D}}\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]</math></td>
  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]</math></td>
  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]</math></td>
</tr>

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

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

</table>
Also note &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~AB \equiv \biggl( \frac{\mathcal{D}}{\ell_{3D}} \biggr)^{2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]^2
+
\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]^2
+
\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]^2 \, ,
</math>
  </td>
</tr>
</table>
and the partial derivatives of <math>~A</math> and <math>~B</math> are detailed in [[User:Tohline/Appendix/Ramblings/ConcentricEllipsodalCoordinates#ABderivatives|an accompanying discussion]].
  </td>
</tr>

</table>

The direction-cosines of the second unit vector &#8212; as has already been inserted into the "&kappa;8 coordinates" table &#8212; should be obtainable from the first and third unit vectors via the cross product,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_2 = \hat{e}_3 \times \hat{e}_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath \biggl[ e_{3y}e_{1z} - e_{3z} e_{1y} \biggr]
+
\hat\jmath \biggl[ e_{3z}e_{1x} - e_{3x} e_{1z} \biggr]
+
\hat{k} \biggl[ e_{3x}e_{1y} - e_{3y} e_{1x} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath \biggl[ \frac{x (p^2z) \ell_{3D} }{(x^2 + q^4y^2)^{1 / 2}} \biggr]
+
\hat\jmath \biggl[  \frac{q^2 y  (p^2z) \ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}}\biggr]
-
\hat{k} \biggl[ \frac{(x^2 + q^4y^2) \ell_{3D} }{(x^2 + q^4y^2)^{1 / 2}} \biggr] \, .
</math>
  </td>
</tr>
</table>

The other boxes in the n = 2 row have been drawn from [[User:Tohline/Appendix/Ramblings/ConcentricEllipsodalT8Coordinates#Associated_h3_Scale_Factor|our accompanying EUREKA! moment]] and the n = 3 row of the table that details "Direction Cosine Components for T8 Coordinates."

==Attempt 1==
Let's try &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\kappa_2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\tan^{-1}\biggl[ \frac{x(qy)^{1/q^2}}{(pz)^{1/p^2}} \biggr] \, ,</math>
  </td>
</tr>
</table>
which leads to,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial\kappa_3}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1 + \frac{x^2(qy)^{2/q^2}}{(pz)^{2/p^2}} \biggr]^{-1} \biggl[ \frac{(qy)^{1/q^2}}{(pz)^{1/p^2}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{x}\biggl[1+\tan^2\kappa_2\biggr]^{-1} \tan\kappa_2 = \frac{\sin\kappa_2 \cos\kappa_2}{x} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial\kappa_3}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[1 + \frac{x^2(qy)^{2/q^2}}{(pz)^{2/p^2}} \biggr]^{-1} \biggl[ \frac{xq^{1/q^2}(y)^{1/q^2}}{q^2 y(pz)^{1/p^2}} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{q^2y}\biggl[1+\tan^2\kappa_2\biggr]^{-1} \tan\kappa_2 = \frac{\sin\kappa_2 \cos\kappa_2}{q^2y} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial\kappa_3}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{\sin\kappa_2 \cos\kappa_2}{p^2z} \, .
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~h_2^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sin^2\kappa_2 \cos^2\kappa_2 \biggl[ \frac{1}{x^2} + \frac{1}{q^4y^2} + \frac{1}{p^4z^2} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sin^2\kappa_2 \cos^2\kappa_2 \biggl[ \frac{x^2 + q^4 y^2 + p^4 z^2}{x^2 q^4y^2 p^4 z^2}  \biggr]
=
\biggl[ \frac{\sin^2\kappa_2 \cos^2\kappa_2 }{x^2 q^4y^2 p^4 z^2 \ell_{3D}^2}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ h_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{x q^2y p^2 z \ell_{3D}}{\sin\kappa_2 \cos\kappa_2 }  \biggr] \, .
</math>
  </td>
</tr>
</table>
The three direction-cosines are, then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\gamma_{21} = h_2 \biggl(\frac{\partial \kappa_2}{\partial x}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{x q^2y p^2 z \ell_{3D}}{\sin\kappa_2 \cos\kappa_2 }  \biggr] \frac{\sin\kappa_2 \cos\kappa_2}{x}
=
q^2y p^2z \ell_{3D} \, .
</math>
  </td>
</tr>
</table>

=See Also=

<ul>
  <li>
[[User:Tohline/Appendix/Ramblings/DirectionCosines|Direction Cosines]]
  </li>
  <li>[[User:Tohline/Appendix/Ramblings/ConcentricEllipsodalCoordinates#Background|Trials up through T7 Coordinates]]</li>
</ul>


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