Page:Scientific Memoirs, Vol. 2 (1841).djvu/94

$u_{0}-u_{0}'$	= 23°	9′
$u_{1}-u_{1}'$	= 47°	42′
$u_{2}-u_{2}'$	71°	48′
$u_{2}''-u_{2}'''$	69°	21′
$u_{1}''-u_{1}'''$	46°	12′
$u_{0}''-u_{0}'''$	22°	27′

$R_{0}$	= 450	millimetres
$R_{1}$	= 350
$R_{2}$	= 300
$t$	= 6″·67
$a$	= 101·0	millimetres.
$b$	= 17·5
$p$	= 142000	milligrammes.

From these may next be calculated,

$v_{0}$	= 14(23° 9′ + 22° 27′)	= 11° 24′·00
$v_{1}$	= 14(47° 42′ + 46° 12′)	= 23° 28′·50
$v_{2}$	= 14(71° 48′ + 69° 21′)	= 35° 17′·25

If now we take the second and the millimetres as the fundamental units of time and space in our calculation, we may deduce from the ascertained values of $R_{0}$, $R_{1}$, $R_{2}$, $v_{0}$, $v_{1}$, $v_{2}$, the following values of $A$, $A'$, $B$, $B'$, $B''$, viz.

${\begin{aligned}A&={\frac {\mathrm {tang} \;11{\mbox{°}}\;24{\mbox{′}}}{450^{3}}}+{\frac {\mathrm {tang} \;23{\mbox{°}}\;28{\mbox{′·}}5}{350^{3}}}+{\frac {\mathrm {tang} \;35{\mbox{°}}\;17{\mbox{′·}}25}{300^{3}}}={\frac {385{\mbox{·}}54}{10^{10}}};\\A^{\prime }&={\frac {\mathrm {tang} \;11{\mbox{°}}\;24{\mbox{′}}}{450^{5}}}+{\frac {\mathrm {tang} \;23{\mbox{°}}\;28{\mbox{′·}}5}{350^{5}}}+{\frac {\mathrm {tang} \;35{\mbox{°}}\;17{\mbox{′·}}25}{300^{5}}}={\frac {384{\mbox{·}}86}{10^{15}}};\\B&={\frac {1}{450^{6}}}+{\frac {1}{350^{6}}}+{\frac {1}{300^{6}}}={\frac {2{\mbox{·}}0362}{10^{15}}};\\B^{\prime }&={\frac {1}{450^{8}}}+{\frac {1}{350^{8}}}+{\frac {1}{300^{8}}}={\frac {2{\mbox{·}}0277}{10^{20}}};\\B^{\prime \prime }&={\frac {1}{450^{10}}}+{\frac {1}{350^{10}}}+{\frac {1}{300^{10}}}={\frac {2{\mbox{·}}0855}{10^{25}}}.\end{aligned}}$

From these $r$ is calculated:

$r={\frac {1}{2}}\cdot {\frac {385{\mbox{·}}54+2{\mbox{·}}0855-384{\mbox{·}}86+2{\mbox{·}}0277}{2{\mbox{·}}0362+2{\mbox{·}}0855-(2{\mbox{·}}0277)^{2}}}\cdot 10^{5}$

or

$r=87650000.$