{"id":1323,"date":"2010-11-06T20:59:04","date_gmt":"2010-11-06T19:59:04","guid":{"rendered":"https:\/\/medianbox.fr\/wordpress\/?p=1323"},"modified":"2014-08-13T19:06:28","modified_gmt":"2014-08-13T17:06:28","slug":"cyclisme-velo-et-equations-du-mouvement","status":"publish","type":"post","link":"https:\/\/medianbox.fr\/wordpress\/?p=1323","title":{"rendered":"MeDiaN@Sports – \u00ab L’Equation de mouvement \u00e0 v\u00e9lo \u00bb"},"content":{"rendered":"

Pr\u00e9sentation g\u00e9n\u00e9rale<\/h3>\n

Dans cet article<\/a> pr\u00e9c\u00e9dent, nous avons \u00e9tablis la relation donnant l’\u00e9nergie cin\u00e9tique d’un cycliste se d\u00e9pla\u00e7ant en ligne droite.<\/p>\n

Rappel :<\/em> T = \\frac{1}{2} . K . \\dot{\\theta _{p}^{2}}=\\frac{1}{2} . K \\left (\\frac{\\pi . N}{30} \\right)^{2} avec inertie \u00e9quivalente ramen\u00e9e au p\u00e9dalier K=a^{2}.t^{2} \\left ( M+2. \\left (M_{1}+M_{2} \\right) + m_{P} + 2. m_{p}+M_{Cycliste} \\right)+0,13^{2} \\left (m_{P} + 2. m_{p} \\right) <\/p>\n

En se basant sur le r\u00e9sultat de ce calcul, nous allons maintenant cherch\u00e9 \u00e0 \u00e9tablir gr\u00e2ce aux multiplicateurs de Lagrange l’\u00e9quation du mouvement r\u00e9gissant le d\u00e9placement de ce m\u00eame v\u00e9lo le long d’une pente.<\/p>\n

Mod\u00e9lisation du probl\u00e8me<\/h3>\n

Sch\u00e9matisation<\/h4>\n

\"\"<\/a><\/p>\n

Description<\/h4>\n

Un v\u00e9lo de centre de gravit\u00e9 G (en\u00a0 [\\xi , \\nu, 0]) avance le long d’une pente formant un angle \\alpha avec l’horizontale dans les m\u00eames conditions que celles \u00e9nonc\u00e9es dans l’article pr\u00e9c\u00e9dent.<\/p>\n

R\u00e9solution<\/h3>\n

Equation de liaison<\/h4>\n

\\dot{\\theta _{1}}=\\dot{\\theta _{2}}=\\frac{-\\dot{\\xi }}{a}=\\dot{\\theta_{p}}.t=\\frac{\\pi .N.t}{30}<\/p>\n

Calcul des multiplicateurs de Lagrange : Qi<\/h4>\n

Comme tous les param\u00e8tres d\u00e9pendent de \\theta_{p}, on exprimera tous les multiplicateurs en fonctions de cette m\u00eame variable pour simplifier les d\u00e9rivations \u00e0 venir.<\/p>\n

    \n
  • Pesanteur :\u00a0 V =-m_{i} . \\vec{O_{o}G_{i}}.\\vec{g}+C^{te}\n
      \n
    • Pour le cadre : \\vec{O_{o}G^{\\prime}} = \\vec{OG}+\\vec{GG^{\\prime}}=\\xi.\\vec{xo}+\\mu.\\vec{yo}+b.\\vec{x}+c.\\vec{y} et comme (\\vec{x}\/\/\\vec{xo}),\\vec{O_{o}G^{\\prime}}=\\left (\\xi+b \\right )\\vec{xo}+\\left (\\mu+c\\right )\\vec{yo}.
      \nDans Rg, on a
      \n\\vec{xo}=cos(\\alpha).\\vec{X0}+sin(\\alpha).\\vec{Yo} et \\vec{yo}=-sin(\\alpha).\\vec{X0}+cos(\\alpha).\\vec{Yo}<\/em>.
      \nDonc
      \n\\vec{O_{o}G^{\\prime}}.\\vec{Yo} = \\left (\\xi+b \\right ).sin{(\\alpha)}+\\left (\\mu+c\\right ).cos{(\\alpha)}
      \nFinalement
      \nV_{cadre} =M.g . \\left [ \\left (\\xi+b \\right ).sin{(\\alpha)}+\\left (\\mu+c\\right ).cos{(\\alpha)} \\right ]+C^{te}
      \net en fonction de \\theta_{p} :
      \nV_{cadre} =M.g . \\left [ \\left (a.t.\\theta_{p}+b+k \\right ).sin{(\\alpha)}+\\left (\\mu+c\\right ).cos{(\\alpha)} \\right ]+C^{te}<\/p>\n

      Coefficient de Lagrange : \\left (Q_{cadre}\\right )_\\theta= -\\frac{\\partial V_{cadre} }{\\partial \\theta }=-M.g.a.t.sin{\\alpha }<\/p>\n<\/li>\n

    • Pour la roue avant : \\vec{O_{o}O_{1}} = \\vec{OG}+\\vec{GO_{1}}=\\xi.\\vec{xo}+\\mu.\\vec{yo}+d.\\vec{x}+e.\\vec{y}
      \nDe la m\u00eame fa\u00e7on : \\left (Q_{Rav}\\right )_\\theta= -\\frac{\\partial V_{Rav} }{\\partial \\theta }=-M_{1}.g.a.t.sin{\\alpha }<\/li>\n
    • Pour la roue arri\u00e8re : \\left (Q_{Rar}\\right )_\\theta= -\\frac{\\partial V_{Rar} }{\\partial \\theta }=-M_{2}.g.a.t.sin{\\alpha }<\/li>\n
    • Pour le p\u00e9dalier : \\left (Q_{Pedalier}\\right )_\\theta= -\\frac{\\partial V_{Pedalier} }{\\partial \\theta }=-\\left (m_{P}+2m_{p}\\right ) .g.a.t.sin{\\alpha }<\/li>\n
    • Pour le cycliste : \\left (Q_{Cycliste}\\right )_\\theta= -\\frac{\\partial V_{Cycliste} }{\\partial \\theta }=-M_{Cycliste}.g.a.t.sin{\\alpha }<\/li>\n<\/ul>\n

      Finalement, \\left (Q_{Pesanteur}\\right )_\\theta= -\\frac{\\partial V_{total} }{\\partial \\theta }=-\\left (M+M_{1}+M_{2}+m_{P}+2m_{p}+M_{Cycliste}\\right ) .g.a.t.sin{\\alpha }<\/p>\n<\/li>\n

    • Actionneurs :\n
        \n
          On suppose que le cycliste applique un couple C_{m} sur le p\u00e9dalier et que les pertes dues aux frottements au niveau de la transmission sont directement proportionnelles \u00e0 la vitesse de rotation du p\u00e9dalier \u00e0 un facteur \\lambda_{1} pr\u00e8s : C_{r}=\\lambda_{1} \\dot{\\theta_{p}} .<\/ul>\n<\/ul>\n
            \n
              D’o\u00f9 une puissance virtuelle donn\u00e9e par :<\/ul>\n<\/ul>\n
                \n
                  P_{(cycliste\/pedalier)}^{*}=\\left \\{\\tau _{(cycliste\/pedalier)}\\right \\} \\otimes \\left \\{\\nu _{(pedalier\/cycliste)}^{*}\\right \\}<\/ul>\n<\/ul>\n
                    \n
                      Avec<\/ul>\n<\/ul>\n
                        \n
                          \\left \\{\\tau _{(cycliste\/pedalier)}\\right \\} = \\vec{M_{O_{p}}(Cycliste\/pedalier)}= (C_{m}-C_{r}).\\vec{Z_{0}}<\/ul>\n<\/ul>\n
                            \n
                              Et<\/ul>\n<\/ul>\n
                                \n
                                  \\left \\{\\nu _{(pedalier\/cycliste)}^{*}\\right \\} = \\vec{\\Omega (pedalier\/cycliste)}^{*} = \\dot{\\theta_{p}}^{*}.\\vec{Z_{0}}<\/ul>\n<\/ul>\n
                                    \n
                                      Donc<\/ul>\n<\/ul>\n
                                        P_{(cycliste\/pedalier)}^{*}= (C_{m}-C_{r}).\\dot{\\theta_{p}}^{*}= (C_{m}-\\lambda_{1} \\dot{\\theta_{p}}).\\dot{\\theta_{p}}^{*} <\/ul>\n

                                        Finalement : \\left (Q_{cycliste\/pedalier}\\right )_\\theta= -\\frac{\\partial P_{(cycliste\/pedalier)}^{*} }{\\partial \\theta^{*} }=C_{m}-\\lambda_{1} \\dot{\\theta_{p}}<\/li>\n

                                      • Liaisons : Les roulements constituant les moyeux des roues n’\u00e9tant pas parfaits, il conduisent \u00e0 des pertes par frottement. Comme ci-dessus, elles sont directement proportionnelles \u00e0 la vitesse de rotation des roues \u00e0 un facteur \\lambda_{2} pr\u00e8s. On a donc :
                                        \n\\left (Q_{moyeu_{roues}}\\right )_\\theta= -\\frac{\\partial P_{(moyeu_{roues})}^{*} }{\\partial \\theta^{*} }=-2. \\lambda_{2} \\dot{\\theta_{1}}=-2. \\lambda_{2} \\dot{\\theta_{p}}.t<\/li>\n<\/ul>\n

                                        Bilan <\/strong>
                                        \n\\left (Q_{Systeme}\\right )_\\theta = \\left (Q_{Pesanteur}\\right )_\\theta + \\left (Q_{cycliste\/pedalier}\\right )_\\theta + \\left (Q_{moyeu_{roues}}\\right )_\\theta
                                        \n\\left (Q_{Systeme}\\right )_\\theta = -\\left (M_{tot} \\right ) .g.a.t.sin(\\alpha) + C_{m}-\\lambda_{1} \\dot{\\theta_{p}} - 2. \\lambda_{2} \\dot{\\theta_{p}}t<\/p>\n

                                        D\u00e9termination de l’\u00e9quation g\u00e9n\u00e9rale du mouvement<\/h4>\n

                                        \\left (Q_{Systeme}\\right )_\\theta = \\frac{\\mathrm{d} }{\\mathrm{d} t}\\left ( \\frac{\\partial T}{\\partial \\dot{\\theta_{p}}} \\right )-\\frac{\\partial T}{\\partial \\theta_{p}} avec \\frac{\\partial T}{\\partial \\dot{\\theta_{p}}} = K \\dot{\\theta_{p}} et \\frac{\\partial T}{\\partial \\theta_{p}} = 0
                                        \nD’o\u00f9 \\left (Q_{Systeme}\\right )_\\theta = K\\ddot{\\theta_{p} } - 0
                                        \nFinalement -\\left (M_{tot} \\right ) .g.a.sin(\\alpha) + C_{m}-\\lambda_{1} \\dot{\\theta_{p}} - 2. \\lambda_{2} \\dot{\\theta_{p}}t = K\\ddot{\\theta_{p} }<\/p>\n

                                        Conclusion
                                        \n<\/strong>C_{m} = K\\ddot{\\theta_{p} } + \\left ( \\lambda_{1}+2. \\lambda_{2}t \\right ) \\dot{\\theta_{p}} + \\left (M_{tot} \\right ) .g.a.t.sin(\\alpha)
                                        \nAvec K=a^{2}.t^{2} \\left ( M+2. \\left (M_{1}+M_{2} \\right) + m_{P} + 2. m_{p}+M_{Cycliste} \\right)+0,13^{2} \\left (m_{P} + 2. m_{p} \\right)
                                        \nM_{tot} = \\left (M+M_{1}+M_{2}+m_{P}+2m_{p}+M_{Cycliste} \\right )
                                        \n\\alpha = Arctan (\\frac{Denivele}{100})
                                        \nt=\\frac{Ndents_{plateau} }{Ndents_{pignon}}<\/p>\n

                                        Le couple moteur que doit fournir le cycliste \u00e0 vitesse constante en cote (\\ddot{\\theta_{p} }=0) doit compenser l’effet de la pesanteur (qui sera d’autant plus important que la masse du cycliste+v\u00e9lo sera \u00e9lev\u00e9e et la pente\u00a0 accentu\u00e9e) ainsi que les diff\u00e9rentes pertes dans la transmission et les roulements.<\/p>\n

                                        On constate \u00e9galement que l’inertie intervient quand \u00e0 elle d\u00e8s que le cycliste cherche \u00e0 acc\u00e9l\u00e9rer sa vitesse. Un v\u00e9lo \u00e0 faible inertie favorisera donc particuli\u00e8rement les changements de rythme.<\/p>\n

                                        La vitesse maximale que l’on pourrait calculer \u00e0 partir de cette \u00e9quation serait toutefois gigantesque car l’\u00e9quation ne prend absolument pas en compte l’action qu’exerce l’environnement sur le cycliste (r\u00e9sistance au roulement, facteurs a\u00e9rodynamique,…).<\/p>\n","protected":false},"excerpt":{"rendered":"

                                        Approximation de l’\u00e9quation du mouvement d’un v\u00e9lo montant une pente en ligne droite.<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"ngg_post_thumbnail":0,"footnotes":""},"categories":[60],"tags":[],"class_list":["post-1323","post","type-post","status-publish","format-standard","hentry","category-mediansports"],"yoast_head":"\nMeDiaN@Sports - \u00ab L'Equation de mouvement \u00e0 v\u00e9lo \u00bb - MeDiaN@Tour<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/medianbox.fr\/wordpress\/?p=1323\" \/>\n<meta property=\"og:locale\" content=\"fr_FR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"MeDiaN@Sports - 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