Editing Appendix/Ramblings/ToroidalCoordinates (section)

====Reality Check One====
Let's see if these derived results make sense.  As a first example, let's assign values of various Figure 2 parameters as follows:

<div id="Example1A">
<table align="center" border="1" cellpadding="5">
<tr><th align="center" colspan="4">Example 1A</th></tr>
<tr>
  <td align="center" width="25%"><math>~\varpi_t</math></td>
  <td align="center" width="25%"><math>~r_t</math></td>
  <td align="center" width="25%"><math>~Z_0</math></td>
  <td align="center"><math>~\alpha</math></td>
</tr>
<tr>
  <td align="center"><math>~\tfrac{3}{4}</math></td>
  <td align="center"><math>~\tfrac{1}{4}</math></td>
  <td align="center"><math>~1</math></td>
  <td align="center"><math>~\tfrac{\pi}{6}</math></td>
</tr>
</table>
</div>

(Notice that the first pair of these parameter values aligns with the properties of the pink torus that was sketched in Figure 4 of [http://adsabs.harvard.edu/abs/2012MNRAS.424.2635T Trova, Hur&eacute; &amp; Hersant (2012)] &#8212; as [[#THH12Figure4|reprinted immediately below]] &#8212; and that the chosen value of <math>~Z_0</math> aligns with the z-coordinate of their "Point B.")  

<div align="center" id="THH12Figure4">
<table border="1" cellpadding="8">

<tr><td align="center">
Figure 4 extracted without modification from p. 2640 of [http://adsabs.harvard.edu/abs/2012MNRAS.424.2635T Trova, Hur&eacute; &amp; Hersant (2012)]<p></p>
"''The Potential of Discs from a 'Mean Green Function' ''"<p></p>
Monthly Notices of the Royal Astronomical Society, vol. 424, pp. 2635-2645 &copy; RAS
</td>
</tr>

<tr>
<td align="center">
[[File:Figure4THH2012.png|350px|Figure 4 from Trova, Hur&eacute; &amp; Hersant (2012)]]
</td>
</tr>
<tr>
  <td align="center">
<math>~\varpi_t = \tfrac{3}{4}\, ; ~ r_t = \tfrac{1}{4}</math><p></p>
Point A:  <math>~(\varpi, Z) = (\tfrac{3}{4}, 0)</math><p></p>
Point B:  <math>~(\varpi, Z) = (1, 1)</math><p></p>
Point C:  <math>~(\varpi, Z) = (10, 10)</math>
  </td>
</tr>
</table>
</div>



Taken together, this choice for the values of <math>~\alpha</math> and <math>~Z_0</math> implies: (1) That the hypotenuse of the blue right-triangle in [[#THH12Figure4|our Figure 2]] and, hence, the distance between the centers of the two circles, is 
<div align="center">
<math>~h = \frac{Z_0}{\cos\alpha} = \frac{2\sqrt{3}}{3} \, ;</math>
</div>

and, (2) that the side of the triangle that is opposite the angle, <math>~\alpha</math>, is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\varpi_t - R_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~h \sin\alpha = \frac{\sqrt{3}}{3} \, ,</math>
  </td>
</tr>
</table>
</div>
which, taken together with the choice of <math>~\varpi_t</math>, gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~R_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3}{4} - \frac{\sqrt{3}}{3} = \frac{9-4\sqrt{3}}{12} \approx 0.17265\, .</math>
  </td>
</tr>
</table>
</div>

With this set of parameters held fixed, it is clear that, in order for the <math>~\xi_1</math> = constant circle to make first/final contact with the pink torus, it will need to have a radius,
<div align="center">
<math>~r_\pm = h \mp r_t = \frac{2\sqrt{3}}{3} \mp \frac{1}{4} \, .</math>
</div>

Let's see if this expectation matches the result obtained via the expressions derived above.   Specifically, we find,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Lambda</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{Z_0}{\varpi_t - R_0} = \sqrt{3} \, ;</math>
  </td>
</tr>
</table>
</div>
hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~r_\pm</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(\varpi_t - R_0)[1 + \Lambda^2 ]^{1/2} \mp r_t </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{\sqrt{3}}{3}\biggl[1 + (\sqrt{3})^2 \biggr]^{1/2} \mp \frac{1}{4} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{2\sqrt{3}}{3} \mp \frac{1}{4} \, .</math>
  </td>
</tr>
</table>
</div>
This precisely matches our expectation.