Editing Appendix/Ramblings/ConcentricEllipsoidalDaringAttack (section)

====Better Organized====

From our above [[#Table1DaringAttack|Daring Attack Table]], we appreciate that the three direction cosines associated with the (as yet unknown) second curvilinear coordinate are,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\gamma_{21}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\ell_q \ell_{3D} (xp^2z) \, ,</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\gamma_{22}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\ell_q \ell_{3D} (q^2 y p^2z) \, ,</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\gamma_{23}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\ell_q \ell_{3D} (x^2 + q^4 y^2) \, .</math>
  </td>
</tr>
</table>

It is easy to see that the desired ''orthogonality'' relationship,
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<tr>
  <td align="right">
<math>~\sum_{i=1}^3 (\gamma_{2i})^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 \, ,</math>
  </td>
</tr>
</table>
is satisfied because,
<table border="0" cellpadding="5" align="center">

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

Now, as we attempt to determine the functional form of the second curvilinear coordinate, <math>~\lambda_2(x, y, z)</math>, a seemingly useful intermediate step is to determine the functional form of each of the three partial derivatives of this key coordinate function, namely, <math>~\partial \lambda_2/\partial x_i</math>, for i = 1, 3.  Here, we will accomplish this intermediate step by ''guessing'' the functional form of the second scale factor, <math>~h_2(x, y, z)</math>, then applying the relation,

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

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial x_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\gamma_{2i}}{h_2} \, .</math>
  </td>
</tr>
</table>
Notice that, without violating the above-state ''orthogonality'' relationship, we can adopt virtually any functional form for <math>~h_2(x, y, z)</math> and deduce that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A(x, y, z) (xp^2z) \, ,</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A(x, y, z) (q^2 y p^2z) \, ,</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-A(x, y, z) (x^2 + q^4 y^2) \, ,</math>
  </td>
</tr>
</table>
as long as,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~A(x, y, z)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{ \ell_q \ell_{3D} }{h_2} \, .
</math>
  </td>
</tr>
</table>

This key, leading coefficient function is unity &#8212; and, hence, is independent of position &#8212; if, as in our [[#First|''First'' speculation]] above, we ''guess'' that <math>~h_2^2 = (\ell_q \ell_{3D})^2</math>.  If, as in our [[#Second|''Second'' speculation]] above, we ''guess'' that <math>~h_2^2 = (\ell_q \ell_{3D})^{-2}</math>, we find that, <math>~A = (\ell_q \ell_{3D})^2</math>.  Our above [[#Third|''Third'' speculation]] is replicated if we ''guess'' that <math>~h_2^2 = (\lambda_2 \ell_q \ell_{3D})^2</math>; we immediately see that, in this ''Third'' case,

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

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial x} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{xp^2 z}{\lambda_2} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial y} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{q^2y p^2z}{\lambda_2} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial z} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{(x^2 + q^4y^2)}{\lambda_2} \, .</math>
  </td>
</tr>
</table>