Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/161

$r_{3}^{2}$	$=q_{3}^{2}+.7130624$	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \\\ \\\ \\\ \\\ \ \end{matrix}}\right\}\,}$ ${\text{III}}_{3}$ corrected.
	$\alpha _{3}=-[9.3810712](q_{3}+1.5798163)(1+R_{3})$
$R_{3}$	$=[9.5619251]r_{3}^{-3}$
	$\beta _{3}=+[9.6537283](q_{3}-.4630521)(1+R_{3})$
	$\gamma _{3}=+[8.8361231](q_{3}+.5599304)(1+R_{3})$

$\Delta q_{3}$						+ .0003302		+.0000424
$q_{3}$		$+$		2.4036347		2.4039649		2.4040073
$\log r_{3}$		$+$		.4061399		.4061929		.4061998
$\log R_{3}$		$+$		8.3435055		8.3433463		8.3433257
$\log(1+R_{3})$		$+$		.094742		.0094708		.0094704
$\alpha _{3}$		$-$		.9790500		.9791236		.9791329
$\beta _{3}$		$+$		.8935824		.8937277		.8937461
$\gamma _{3}$		$+$		.2076882		.2077097		.2077124
$\log s_{3}$		$+$						.1277120

With these corrected equations the last values of $q_{1},q_{2},q_{3}$ give the residuals

$\alpha =.0001135$$\beta =-.0003934$$\gamma =-.0000326$

These give the corrections

$\Delta q_{1}=.0002887$$\Delta q_{2}=-.0000955$$\Delta q_{3}=.0003302$

The next residuals are

$\alpha =.0000035$$\beta =.0000140$$\gamma =-.0000003$

which give the corrections

$\Delta q_{1}=.-.0000217$$\Delta q_{2}=.0000187$$\Delta q_{3}=.0000424$

The next residuals are

$\alpha =-.0000001$$\beta =.0000003$$\gamma =.0000000$

which must be regarded as entirely insensible.

IX, X.

It remains to determine the ellipse which passes through the points to which the numbers relate in the last columns under the corrected equations ${\text{III}}_{1},{\text{III}}_{2},{\text{III}}_{3},$ and also the time of perihelion passage. The computations are as follows: