Editing Apps/Ostriker64 (section)

====Single Offset Circle====
Now an [[Appendix/Ramblings/ToroidalCoordinates#Off-center_Circle|off-center circle]] whose major and minor radii are, respectively, <math>~(\varpi_0,d)</math>, will be described by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~d^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(\varpi - \varpi_0)^2 + z^2 \, .
</math>
  </td>
</tr>
</table>

<span id="Dsquared">where both <math>~d</math> and <math>~\varpi_0</math> are held constant while mapping out the variation of <math>~z</math> with <math>~\varpi</math>.  If we acknowledge that, in general, <math>~\varpi_0 \ne R_\mathrm{JPO}</math>, then we know how <math>~r</math> varies with <math>~\phi</math> via the relation,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~d^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ R_\mathrm{JPO} + r\cos\phi - \varpi_0\biggr]^2 + r^2\sin^2\phi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(R_\mathrm{JPO}-\varpi_0)^2 
+ 2\biggl[ (R_\mathrm{JPO}-\varpi_0) r\cos\phi \biggr]
+r^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
r^2 + 2r\biggl[ (R_\mathrm{JPO}-\varpi_0) \cos\phi \biggr] + \biggl[(R_\mathrm{JPO}-\varpi_0)^2 - d^2\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ r </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2}\biggl\{
- 2\biggl[ (R_\mathrm{JPO}-\varpi_0) \cos\phi \biggr] \pm 
\sqrt{ 4\biggl[ (R_\mathrm{JPO}-\varpi_0) \cos\phi \biggr]^2 - 4\biggl[(R_\mathrm{JPO}-\varpi_0)^2 - d^2\biggr] }
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2}\biggl\{
2\biggl[ (\varpi_0 - R_\mathrm{JPO}) \cos\phi \biggr] \pm 
\sqrt{ 4\biggl[ (\varpi_0 - R_\mathrm{JPO}) \cos\phi \biggr]^2 - 4\biggl[(\varpi_0 - R_\mathrm{JPO})^2 - d^2\biggr] }
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{r}{ (\varpi_0 - R_\mathrm{JPO}) }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\cos\phi  \pm 
\sqrt{ \cos^2\phi  - 1 + d^2 (\varpi_0 - R_\mathrm{JPO})^{-2} }
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\cos\phi  \pm \sqrt{ d^2 (\varpi_0 - R_\mathrm{JPO})^{-2}-\sin^2\phi }
</math>
  </td>
</tr>
</table>

In order to align this expression with the terminology (and variable labels) that we use in the context of a toroidal coordinate system, we associate the radius of the ''anchor ring'' as <math>~R_\mathrm{JPO}\leftrightarrow a</math>, and we associate the major radius of each circular torus as <math>~\varpi_0 \leftrightarrow R_0</math>.  We therefore have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{r}{ (R_0-a) }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\cos\phi  \pm \sqrt{ d^2 (R_0-a)^{-2}-\sin^2\phi } 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{r}{a}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{R_0}{a}-1 \biggr)
\biggl[ \cos\phi  \pm \sqrt{ \biggl(\frac{d}{a}\biggr)^2 \biggl(\frac{R_0}{a}-1 \biggr)^{-2}-\sin^2\phi } \biggr]
</math>
  </td>
</tr>
</table>

and, the coordinates of points along the surface of the torus <math>~(\varpi,z)</math> are provided by the expressions,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\varpi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
a + (R_0 - a)\cos\phi \biggl[
\cos\phi  \pm \sqrt{ d^2 (R_0 - a)^{-2}-\sin^2\phi }
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~z</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(R_0 - a)\sin\phi \biggl[
\cos\phi  \pm \sqrt{ d^2 (R_0 - a)^{-2}-\sin^2\phi }
\biggr]
</math>
  </td>
</tr>
</table>
We have tested this pair of expressions using Excel and have successfully demonstrated that they do, indeed, trace out a circle of radius, <math>~d</math>, whose center is offset from the symmetry axis by a distance, <math>~R_0</math>.