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NOTES DU TRADUCTEUR.
Remarquons maintenant, qu’on a aussi
![{\displaystyle k=a_{1}\mathrm {M} ={\frac {a_{1}d_{1}\sin b'}{\sin(b'\!-b)}}={\frac {\mathrm {R} \sin(\Theta '\!-\Theta )\sin b'}{\sin(b'\!-b)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d48b06410f164c2e334b1d91e24628b8de89d43a)
il vient alors, en introduisant cette valeur dans l’équation (1),
![{\displaystyle {\begin{aligned}\rho ''={\frac {\cos b''.\sin(b'\!-b)}{\sin(\Theta '\!-\alpha '')\sin(b''\!-b')}}&\left({\frac {r''\sin(v''\!-v)\rho \sin(\Theta '\!-\alpha )}{r\sin(v'\!-v)\cos b}}\right.\\&+\left.{\frac {\mathrm {R} \sin(\Theta '\!-\Theta )\sin b'}{\sin(b'\!-b)}}.\mathrm {U} \right),\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e457f7cd95a9ea4311fc75b378ae486993515412)
équation que nous pouvons mettre sous la forme
![{\displaystyle {\begin{aligned}\rho ''={\frac {t''}{t'}}.{\frac {\cos b''}{\cos b}}&.{\frac {\sin(b'\!-b)}{\sin(b''\!-b')}}.{\frac {\sin(\Theta '\!-\alpha )}{\sin(\Theta '\!-\alpha '')}}\left({\frac {r''\sin(v''\!-v)}{r\sin(v'\!-v)}}\,\rho .{\frac {t'}{t''}}\right.\\&+\left.{\frac {t'}{t''}}.{\frac {\mathrm {R} \sin(\Theta '\!-\Theta )\sin b'\cos b}{\rho \sin(\Theta '\!-\alpha )\sin(b'\!-b)}}\mathrm {U} \right).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf76464fd2c2e9862cd4eb35f74707540f9d4eef)
En la divisant par
nous avons,
![{\displaystyle {\begin{aligned}\mathrm {M} '={\frac {t''}{t'}}.{\frac {\cos b''}{\cos b}}&.{\frac {\sin(b'\!-b)}{\sin(b''\!-b')}}.{\frac {\sin(\Theta '\!-\alpha )}{\sin(\Theta '\!-\alpha '')}}\left({\frac {r''\sin(v''\!-v)}{r\sin(v'\!-v)}}.{\frac {t'}{t''}}\right.\\&+\left.{\frac {t'}{t''}}.{\frac {\mathrm {R} \sin(\Theta '\!-\Theta )\sin b'\cos b}{\rho \sin(\Theta '\!-\alpha )\sin(b'\!-b)}}\mathrm {U} \right).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b127b76f6f78b339eb25549036067556c8c96199)
Mais nous avions pris pour
la valeur
![{\displaystyle \mathrm {M} ={\frac {t''}{t'}}.{\frac {\cos b''}{\cos b}}.{\frac {\sin(b'\!-b)}{\sin(b''\!-b')}}.{\frac {\sin(\Theta '\!-\alpha )}{\sin(\Theta '\!-\alpha '')}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a2945d70f6e066c38f8e010bd32116b1eceac64a)
on a donc,
![{\displaystyle {\frac {\mathrm {M} '}{\mathrm {M\,} }}={\frac {r''\sin(v''\!-v)}{r\sin(v'\!-v)}}.{\frac {t'}{t''}}+{\frac {\mathrm {R} \sin(\Theta '\!-\Theta ).\sin b'\cos b\,\mathrm {U} }{\sin(\Theta '\!-\alpha )\sin(b'\!-b)}}.{\frac {t'}{t''.\rho }},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5d1d636aad3d9edb8dccf5c3be3a9cb256a5e19)
ou, en remarquant que
et
![{\displaystyle \operatorname {tang} b\sin(\Theta '\!-\alpha )=\operatorname {tang} \beta ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2302301deb05d70ab151f33472bbeef133c2d9fc)
il vient enfin,
![{\displaystyle {\frac {\mathrm {M} '}{\mathrm {M\,} }}={\frac {r''\sin(v''\!-v)}{r\sin(v'\!-v)}}.{\frac {t'}{t''}}+{\frac {\mathrm {R} \sin(\Theta '\!-\Theta ).\operatorname {tang} b'.\mathrm {U} }{\operatorname {tang} b'\sin(\Theta '\!-\alpha )-\operatorname {tang} \beta }}.{\frac {t'}{t''.\rho }},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f91d31f90d192f5e2640d8e5f9aa1722152fa2b)
Pour calculer ce rapport
on se servira des valeurs
et
trouvées d’après les éléments fournis par la méthode donnée précédemment.