Fin des corrections de quelques équations de balistique

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Guyot 2021-05-05 08:48:07 +02:00
parent 900f44f0fe
commit 0214be657b
6 changed files with 47 additions and 46 deletions

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@ -195,11 +195,11 @@ et on remplace dans la seconde~:
où on a utilisé la définition de la tangente~:
\[\tan(\beta)=\sin(\beta)/\cos(\beta)\]
En mettant en évidence \(T_1\) et le déplaçant à droite de l'équation, on a~:
\[F-P=T_1\cdot (\sin(\alpha)+\sin(\beta))\]
\[F-P=T_1\cdot (\sin(\alpha)+\cos(\alpha)\cdot\tan(\beta))\]
et donc finalement avec \(P=m\cdot g\)~:
\[T_1=\frac{F-m\cdot g}{\sin(\alpha)+\sin(\beta)}\]
\[T_1=\frac{F-m\cdot g}{\sin(\alpha)+\cos(\alpha)\cdot\tan(\beta)}\]
et pour \(T_2\)~:
\[T_2=\frac{(F-m\cdot g)\cdot \cos(\alpha)}{\cos(\beta)\cdot (\sin(\alpha)+\sin(\beta))}\]
\[T_2=\frac{(F-m\cdot g)\cdot \cos(\alpha)}{\cos(\beta)\cdot (\sin(\alpha)+\cos(\alpha)\cdot\tan(\beta))}\]
\subsection{Plan incliné}
\begin{figure}[t]
@ -359,7 +359,7 @@ Car, \(2\cdot \sin(\alpha)\cos(\alpha)=\sin(2\cdot \alpha)\).\\
Une conséquence de ce résultat est qu'on peut déterminer l'angle sous lequel on doit tirer pour que le tir une porté L~:
\begin{align*}
L&=\frac{v_o^2}{g}\cdot \sin(2\cdot \alpha)\;\Rightarrow\;\sin(2\cdot \alpha)=\frac{g\cdot L}{v_o^2}\\
&\Rightarrow\;\alpha = \arcsin(\frac{g\cdot L}{v_o^2})
&\Rightarrow\;\alpha = \arcsin(\frac{g\cdot L}{v_o^2})/2
\end{align*}
\item Pour déterminer l'angle nécessaire pour atteindre le point C de coordonnée \((x_1;y_1)\), il faut partir de l'équation de la parabole appliquée au point C~:
\[y_1=-\frac{1}{2}\cdot g\cdot\left(\frac{x_1}{v_o}\right)^2\cdot \frac{1}{\cos^2(\alpha)}+x_1\cdot \tan(\alpha)\]