Editing Apps/MaclaurinSpheroids/GoogleBooks (section)

===Generate Curve AKG===

In &sect;644 (p. 122 of Volume II) of his Treatise, Maclaurin states that "the gravity at the pole A towards the spheroid ADBE will be measured by [the ratio],"
<div align="center">
<math>\frac{\mathrm{AKGC}}{AC} \, ,</math>
</div>
where the numerator, "AKGC," is the area under the curve, "AKG," and the denominator is the length of the semi-major axis, <math>~a_1</math>.  So a key element of Maclaurin's derivation is the proper construction of the curve, "AKG."  The key sentence from his &sect;644 appears to be the following:

[[File:Vol2Par644KeyPhrase.png|400px|thumb|center|Extracted directly from &sect;644 of Maclaurin's Book 1, as digitized by Google]]

We interpret this phrase to mean that, as the angle, <math>~\alpha</math>, varies from <math>~\tfrac{\pi}{2}</math> to <math>~0</math>, the curve "AKG" is generated by plotting, as the ordinate, the length of the line-segment "AQ" and, as the abscissa, the <math>~z</math>-coordinate location of the vertical line, "RN" &#8212; we'll label this, <math>~z_\mathrm{RN}</math>.  In our terminology, as [[Apps/MaclaurinSpheroids/GoogleBooks#Intersection_With_Circle|derived above]], <math>~z_\mathrm{RN}</math> is the <math>~z</math>-coordinate of the intersection of the red line-segment with the green quarter-circle, that is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~z_\mathrm{RN}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\cos\alpha - 1 \, .</math>
  </td>
</tr>
</table>
</div>
And the magnitude of "AQ" is obtained from the <math>~z</math>-coordinate location of the vertical line, "MQ" &#8212; that is, the <math>~z</math>-coordinate of the intersection of the red line-segment with the blue ellipse &#8212; as [[Apps/MaclaurinSpheroids/GoogleBooks#Intersection_With_Ellipse|derived above]].  Specifically, we interpret Maclaurin's phrase to mean that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\mathrm{AQ}}{a_1} \equiv 1 + \frac{z}{a_1}\biggr|_+</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - ~\frac{[\tan^2\alpha - (1-e^2) ]}{[\tan^2\alpha+ (1-e^2) ]} \, .</math>
  </td>
</tr>
</table>
</div>

Using the first of these two expressions, we can rewrite the tangent function directly in terms of the ordinate, <math>~z_\mathrm{RN}</math>; specifically,
<div align="center">
<math>\tan^2\alpha = \frac{1- \cos^2\alpha}{\cos^2\alpha} = \frac{1- (1+z_\mathrm{RN})^2}{(1+z_\mathrm{RN})^2}  \, .</math>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\mathrm{AQ}}{a_1} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - ~\frac{(1+z_\mathrm{RN})^2[\tan^2\alpha - (1-e^2) ]}{(1+z_\mathrm{RN})^2[\tan^2\alpha+ (1-e^2) ]}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - ~\frac{[1-(1+z_\mathrm{RN})^2 - (1-e^2)(1+z_\mathrm{RN})^2 ]}{[1-(1+z_\mathrm{RN})^2+ (1-e^2) (1+z_\mathrm{RN})^2]}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - ~\frac{[1+(1+z_\mathrm{RN})^2 (-1 - (1-e^2) ]}{[1+(1+z_\mathrm{RN})^2(-1+1-e^2)]}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - ~\frac{[1-(2-e^2)(1+z_\mathrm{RN})^2 ]}{[1-e^2(1+z_\mathrm{RN})^2]}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{[1-e^2(1+z_\mathrm{RN})^2]-[1-(2-e^2)(1+z_\mathrm{RN})^2 ]}{[1-e^2(1+z_\mathrm{RN})^2]}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{[2(1-e^2)(1+z_\mathrm{RN})^2 ]}{[1-e^2(1+z_\mathrm{RN})^2]} \, .</math>
  </td>
</tr>
</table>
</div>

In our above figure, this function, <math>~\mathrm{AQ}(z_\mathrm{RN})</math> normalized to <math>~a_1</math>, is delineated by the orange triangles over the coordinate range, <math>~-1 \le z_\mathrm{RN} \le 0</math>.  If our interpretation of Maclaurin's discussion is correct, this is precisely the curve in Maclauin's Figure 291, No. 2 that is labeled, "AKG."