Page:Annales de mathématiques pures et appliquées, 1815-1816, Tome 6.djvu/274

264

RECHERCHE

Supposons en outre que les observations soient équidistantes et posons

$t'-t=t-t_{l}=i,\quad$d’où$\quad t'-t_{l}=2i\,;$

en substituant toutes ces valeurs dans les formules (7 et 8), elles deviendront

$\left.{\begin{aligned}f_{l}&=i\left\{m_{l}(h_{l}-2h+h')-n_{l}(g_{l}-2g+g')\right\}\operatorname {Tang} .\gamma _{l},\\f\ &=i\left\{m\ (h_{l}-2h+h')-n\ (g_{l}-2g+g')\right\}\operatorname {Tang} .\gamma ,\\f'&=i\left\{m'(h_{l}-2h+h')-n'(g_{l}-2g+g')\right\}\operatorname {Tang} .\gamma '\,;\\\end{aligned}}\right\}$(15)

$k=\left\{(mn'-m'n)-(m_{l}n'-m'n_{l})+(m_{l}n-mn_{l})\right\}\operatorname {Tang} .\gamma '\operatorname {Tang} .\gamma \operatorname {Tang} .\gamma _{l}\,;$(16)

de sorte qu’en posant

$(m-m_{l})(n'-n)-(m'-m)(n-n_{l})=(mn_{l}-m_{l}n)-(m'n_{l}-m_{l}n')+(m'n-mn')=K\,;$(17)

et substituant dans (9), il viendra

$\left.{\begin{aligned}D_{l}&={\frac {(m'-m)(h_{l}-2h+h')-(n'-n)(g_{l}-2g+g')}{K\operatorname {Sin} .\gamma _{l}}}\\D\ &={\frac {(m'-m_{l})(h_{l}-2h+h')-(n'-n_{l})(g_{l}-2g+g')}{K\operatorname {Sin} .\gamma }}\\D'&={\frac {(m-m_{l})(h_{l}-2h+h')-(n-n_{l})(g_{l}-2g+g')}{K\operatorname {Sin} .\gamma '}}\\\end{aligned}}\right\}$(18)

et par suite (1)

$\left.{\begin{aligned}-Z_{l}&={\frac {1}{K}}\left\{(m'-m)(h_{l}-2h+h')-(n'-n)(g_{l}-2g+g')\right\},\\-Z\ &={\frac {1}{2K}}\left\{(m'-m_{l})(h_{l}-2h+h')-(n'-n_{l})(g_{l}-2g+g')\right\},\\-Z'&={\frac {1}{K}}\left\{(m-m_{l})(h_{l}-2h+h')-(n-n_{l})(g_{l}-2g+g')\right\}\,;\\\end{aligned}}\right\}$(19)

ce qui donne encore

$-(Z'-Z_{l})={\tfrac {1}{K}}\left\{(n_{l}-2n+n_{l})(g_{l}-2g+g')-(m_{l}-2m+m')(h_{l}-2h+h_{l})\right\}\,;$(20)

et l’on a ensuite

$X=g+mZ,\qquad Y=h+nZ.\qquad$(21)