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\begin{document}

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\section*{test chuba md}

\hypertarget{analyse_des_forces_dans_une_suspension_de_type_double_whishbone}{}\subsection*{{Analyse des forces dans une suspension de type \emph{double whishbone}}}\label{analyse_des_forces_dans_une_suspension_de_type_double_whishbone}

Plan provenant du prototype (jamais construit) de la \href{http://www.automobile-catalog.com/make/maserati/chubasco/chubasco_concept/1990.html}{Chubasco 1990 de Maserati}

\begin{verbatim}import numpy as np
from math import *

def chuba(n, m):
    "retourne le vecteur position reliant le point n vers le point m"
    points = [[6.13,-14.10], [8.57,13.10], [39.68,8.52], [47.51,-13.10],
        [10.02,-14.03], [42.55,20.36], [45.40,12.73], [55.44,16.06],
        [48.46,-13.28], [0.00,-26.17]]
    return Vecteur(points[m-1]) - Vecteur(points[n-1])

def unit(theta):
    "retourne le vecteur unitaire définit par l'angle theta en deg"
    theta_rad = pi*theta/180.
    return Vecteur([cos(theta_rad), sin(theta_rad)])

def v(x, y):
    "retourne un vecteur dont les composantes sont x et y"
    return Vecteur([x,y])

def module(vect):
    "retourne la norme du vecteur vect"
    return np.linalg.norm(vect)

class Vecteur(np.ndarray):

    def __new__(cls, input_array):
        # Input array is an already formed ndarray instance
        # We first cast to be our class type
        obj = np.asarray(input_array).view(cls)
        # add the new attribute to the created instance
        # Finally, we must return the newly created object:
        return obj
    def __invert__(self):
        "~ vect normalise le vecteur"
        norm = np.linalg.norm(self)
        return self/norm

    def __mul__(self, arg):
        "A*B correspond au produit vecoriel"
        return np.cross(self, arg)
\end{verbatim}




L'équilibre en rotation de la pièce isolée (avec pivot au point 1) permet d'écrire l'équation:

\begin{displaymath}
M_{SA} = - M_{BA}
\end{displaymath}
\begin{displaymath}
\overrightarrow{R_{12}} \times \overrightarrow{BA} = - \overrightarrow{R_{110}} \times \overrightarrow{SA}
\end{displaymath}
où $\overrightarrow{BA}$ est le seul vecteur inconnu, donc calculable! Les conventions d'écriture suivantes sont utilisées: * BA est la force \emph{par} la pièce B \emph{sur} la pièce A et SA, la force par S sur A; * $\overrightarrow{R_{12}}$ est le vecteur position qui va du point 1 vers le point 2.

Les outils informatiques de calcul suivants sont disponibles: * {\colorbox[rgb]{1.00,0.93,1.00}{\tt chuba\char40\char49\char44\char50\char41}} génère le vecteur position qui va de la jonction 1 vers la jonction 2 * {\colorbox[rgb]{1.00,0.93,1.00}{\tt \char126AB}} est le vecteur unitaire $\widehat{AB}$ * {\colorbox[rgb]{1.00,0.93,1.00}{\tt AB\char42CD}} évalue le produit vectoriel $\overrightarrow{AB} \times \overrightarrow{CD}$ * {\colorbox[rgb]{1.00,0.93,1.00}{\tt v\char40x\char44y\char41}} donne le vecteur de composantes x et y

\begin{verbatim}SA=v(0,1000)
SA\end{verbatim}
\begin{verbatim}Vecteur([   0, 1000])\end{verbatim}
Comment isoler le module de la force $\overrightarrow{BA}$ dans l'équation précédente? \$$\overrightarrow{R_{12}} \times \overrightarrow{BA} = - \overrightarrow{R_{110}} \times \overrightarrow{SA}$\$

il suffit de dévelloper le vecteur $\overrightarrow{BA}$ comme étant le produit de son module et sa direction:

\begin{displaymath}
\overrightarrow{BA} = BA \cdot \widehat{BA}
\end{displaymath}
\begin{displaymath}
\overrightarrow{R_{12}} \times (BA \cdot \widehat{BA}) = - \overrightarrow{R_{110}} \times \overrightarrow{SA}
\end{displaymath}
\begin{displaymath}
BA \cdot (\overrightarrow{R_{12}} \times  \widehat{BA}) = - \overrightarrow{R_{110}} \times \overrightarrow{SA}
\end{displaymath}
et alors

\begin{displaymath}
BA  = - \frac{\overrightarrow{R_{110}} \times \overrightarrow{SA}} {\overrightarrow{R_{12}} \times  \widehat{BA}}
\end{displaymath}
en module, et pour le vecteur:

\begin{displaymath}
\overrightarrow{BA}  = - \frac{\overrightarrow{R_{110}} \times \overrightarrow{SA}} {\overrightarrow{R_{12}} \times  \widehat{BA}} \widehat{BA}
\end{displaymath}
où $\widehat{BA}$ apparaît deux fois: il est donc utile de le calculer au préalable. $\widehat{BA}$ est la direction de la force BA et donc parallèle à la pièce B (effort axial pur). Cette direction est définie par les points 3 vers 2. On a alors $\widehat{BA}$ = {\colorbox[rgb]{1.00,0.93,1.00}{\tt \char32r\char51\char50d\char32\char61\char32\char126chuba\char40\char51\char44\char50\char41}} :

\begin{verbatim}r32d=~chuba(3,2) # vecteur unitaire en direction de (3,2)
r32d\end{verbatim}
\begin{verbatim}Vecteur([-0.98933623,  0.14564963])\end{verbatim}
Le vecteur force BA \$$\overrightarrow{BA}  = - \frac{\overrightarrow{R_{110}} \times \overrightarrow{SA}} {\overrightarrow{R_{12}} \times  \widehat{BA}} \widehat{BA}$\$

finalement s'évalue ainsi:

\begin{verbatim}BA = chuba(1,10)*SA/(-chuba(1,2)*r32d)*r32d
print(BA)\end{verbatim}
\begin{verbatim}[-222.43013262   32.7460626 ]\end{verbatim}
La force manquante sur la pièce A s'obtient par équilibre de translation:

\begin{displaymath}
\overrightarrow{SA} + \overrightarrow{BA} + \overrightarrow{CA} = 0
\end{displaymath}
\begin{displaymath}
\overrightarrow{CA} = -(\overrightarrow{SA} + \overrightarrow{BA} )
\end{displaymath}
\begin{verbatim}CA = -(BA+SA)
print(CA)\end{verbatim}
\begin{verbatim}[  222.43013262 -1032.7460626 ]\end{verbatim}
On parcourt l'assemblage en résolvant les équations de pièce en pièce\ldots{}

\begin{verbatim}CA\end{verbatim}
\begin{verbatim}Vecteur([  222.43013262, -1032.7460626 ])\end{verbatim}
\begin{verbatim}print(CA)\end{verbatim}
\begin{verbatim}[  222.43013262 -1032.7460626 ]\end{verbatim}


\end{document}