Page:Mémoires de l’Académie des sciences, Tome 9.djvu/449

Polaire.                      $\varepsilon$ de la grande Ourse.

${\begin{alignedat}{2}{\text{Pass. supér. }}t&=\ \ 1^{h}\ 0'34''{,}\ 0\qquad &{\text{infér. }}\tau &=\ \ 0^{h}46'19''{,}39\\{\text{infér. }}t'&=12.58.37\ {,}22&{\text{supér. }}\tau '&=12.\ 58.\ 24{,}\ 45\\&{\text{–––––––––––––}}&&{\text{–––––––––––––}}\\t'-t&=11.58.\ \ 3\ {,}22&\tau '-\tau &=10.\ \ \ 0.\ \ \ 5{,}06\\&&-(t'-t)&=11.\ 58.\ \ \ 3{,}22\\&&&{\text{–––––––––––––}}\\&&&\qquad \quad \ \ 2'\ 1''{,}84=121''{,}84\end{alignedat}}$

${\begin{alignedat}{2}\Delta &=\ \ 1^{\circ }37'\ \ 8''&{\text{de la polaire}}.\qquad \\\Delta '&=33.\ \ \ 1.57&\varepsilon {\text{ de la grande Ourse.}}\\&{\text{–––––––––––}}&\log .121''{,}84&=2.08583\\\Delta +\Delta '&=34.39.\ \ 5&\sin .\Delta &=8.45104\\{\frac {1}{2}}(\Delta +\Delta ')&=17.19.32&\sin .\Delta '&=9.73650\\&{\text{–––––––––––}}\\\Delta '-\Delta &=31.24.49&c.2\cos .\mathrm {H} &=9.88058\\{\frac {1}{2}}(\Delta '-\Delta )&=15.42.24&c.\sin .(\Delta '+\Delta )&=0.24522\\&&&{\text{–––––––––––}}\\&&\operatorname {l.} 1^{\mathrm {er} }{\text{ terme}}&=0.39917=2''{,}507{\text{ en temps.}}\\&&&\quad \ \ 1.17609\\&&&{\text{–––––––––––}}\\&&&\quad \ \ 1.57526=37''{,}607{\text{ en arc.}}\end{alignedat}}$

Partant

$x=37''{,}61+\beta (1{,}1436)-(\alpha +y)(1{,}5319)\,;$

par Arcturus

$x=37''{,}5+\beta (1{,}1436)-(\alpha +y)(2{,}0796).$

Mais si l’on a toujours eu le soin de maintenir horizontalement l’axe de rotation supposé parfaitement calibré, on a alors $\beta =0\,;$ ainsi

${\begin{aligned}x&=37''{,}61-(\alpha +y)(1{,}5319)\\x&=37''{,}50-(\alpha +y)(2{,}0796)\,;\end{aligned}}$

soustrayant il vient

$\alpha +y={\frac {-0''{,}11}{0{,}5477}}=-0''{,}2008\,;$

d’ailleurs

$\alpha =0''{,}30622\cos .\mathrm {H} =0''{,}20157\,;$