{"id":7481,"date":"2022-04-15T16:43:03","date_gmt":"2022-04-15T14:43:03","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=7481"},"modified":"2022-04-15T16:43:05","modified_gmt":"2022-04-15T14:43:05","slug":"perimetre-dune-ellipse","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2022\/04\/15\/perimetre-dune-ellipse\/","title":{"rendered":"P\u00e9rim\u00e8tre d&rsquo;une ellipse"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Calculer le p\u00e9rim\u00e8tre d&rsquo;une ellipse n&rsquo;est pas chose ais\u00e9e.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">On pourrait croire qu&rsquo;\u00e0 l&rsquo;instar de son cousin le cercle, l&rsquo;ellipse poss\u00e8de une formule math\u00e9matique pour d\u00e9terminer son p\u00e9rim\u00e8tre&#8230; Que nenni !<\/p>\n\n\n\n<!--more-->\n\n\n\n<p class=\"wp-block-paragraph\">Nous allons consid\u00e9rer ici une ellipse d&rsquo;\u00e9quation cart\u00e9sienne:$$\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1$$. Comme vous pouvez le constater, j&rsquo;ai pris la d\u00e9cision de centrer l&rsquo;ellipse en l&rsquo;origine du rep\u00e8re.<\/p>\n\n\n\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_85 counter-hierarchy ez-toc-counter ez-toc-white ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Au menu sur cette page...<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" 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href=\"https:\/\/www.mathweb.fr\/euclide\/2022\/04\/15\/perimetre-dune-ellipse\/#Perimetre_dune_ellipse_formules_de_Ramanujan\" >P\u00e9rim\u00e8tre d&rsquo;une ellipse: formules de Ramanujan<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.mathweb.fr\/euclide\/2022\/04\/15\/perimetre-dune-ellipse\/#Premiere_formule_de_Ramanujan_pour_le_perimetre_dune_ellipse\" >Premi\u00e8re formule de Ramanujan pour le p\u00e9rim\u00e8tre d&rsquo;une ellipse<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.mathweb.fr\/euclide\/2022\/04\/15\/perimetre-dune-ellipse\/#Deuxieme_formule_de_Ramanujan\" >Deuxi\u00e8me formule de Ramanujan<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.mathweb.fr\/euclide\/2022\/04\/15\/perimetre-dune-ellipse\/#Avec_lequation_parametrique_de_lellipse\" >Avec l&rsquo;\u00e9quation param\u00e9trique de l&rsquo;ellipse<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.mathweb.fr\/euclide\/2022\/04\/15\/perimetre-dune-ellipse\/#Approximer_le_perimetre_dune_ellipse_a_laide_de_Python\" >Approximer le p\u00e9rim\u00e8tre d&rsquo;une ellipse \u00e0 l&rsquo;aide de Python<\/a><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Perimetre_dune_ellipse_premiere_approximation\"><\/span>P\u00e9rim\u00e8tre d&rsquo;une ellipse: premi\u00e8re approximation<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Une formule \u00ab\u00a0classique\u00a0\u00bb que l&rsquo;on peut voir ci et l\u00e0 est la suivante:$$P\\approx2\\pi\\sqrt{\\frac{a^2+b^2}{2}}$$<\/p>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\" style=\"text-transform:lowercase\"><table><thead><tr><th class=\"has-text-align-center\" data-align=\"center\">Valeurs de a<\/th><th class=\"has-text-align-center\" data-align=\"center\">Valeurs de b<\/th><th class=\"has-text-align-center\" data-align=\"center\">Valeur approximative du p\u00e9rim\u00e8tre<\/th><\/tr><\/thead><tbody><tr><td class=\"has-text-align-center\" data-align=\"center\">5<\/td><td class=\"has-text-align-center\" data-align=\"center\">7<\/td><td class=\"has-text-align-center\" data-align=\"center\">38,2191241574<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">4<\/td><td class=\"has-text-align-center\" data-align=\"center\">5<\/td><td class=\"has-text-align-center\" data-align=\"center\">28,4483314254<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">15<\/td><td class=\"has-text-align-center\" data-align=\"center\">51<\/td><td class=\"has-text-align-center\" data-align=\"center\">236,184258737<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Pour le moment, dans la mesure o\u00f9 nous n&rsquo;avons pas d&rsquo;autres m\u00e9thodes, nous ne pouvons pas comparer ces valeurs avec d&rsquo;autres qui pourraient \u00eatre plus ou moins pr\u00e9cises&#8230;<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Perimetre_dune_ellipse_formules_de_Ramanujan\"><\/span>P\u00e9rim\u00e8tre d&rsquo;une ellipse: formules de Ramanujan<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Premiere_formule_de_Ramanujan_pour_le_perimetre_dune_ellipse\"><\/span>Premi\u00e8re formule de Ramanujan pour le p\u00e9rim\u00e8tre d&rsquo;une ellipse<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Rappelons que <a href=\"https:\/\/fr.wikipedia.org\/wiki\/Srinivasa_Ramanujan\" target=\"_blank\" rel=\"noreferrer noopener\">Srinivasa Ramanujan<\/a> \u00e9tait un math\u00e9maticien indien d&rsquo;une intuition remarquable qui a \u00e9tabli (on ne sait pas trop comment) une multitude de formules, notamment sur le nombre \\(\\pi\\).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Une de ses formules est une approximation du p\u00e9rim\u00e8tre d&rsquo;une ellipse:$$P\\approx\\pi\\Big(3(a+b)\u2212\\sqrt{(3a+b)(a+3b)}\\Big).$$<\/p>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table><thead><tr><th class=\"has-text-align-center\" data-align=\"center\">valeurs de a<\/th><th class=\"has-text-align-center\" data-align=\"center\">valeurs de b<\/th><th class=\"has-text-align-center\" data-align=\"center\">valeur approximative du p\u00e9rim\u00e8tre<\/th><\/tr><\/thead><tbody><tr><td class=\"has-text-align-center\" data-align=\"center\">5<\/td><td class=\"has-text-align-center\" data-align=\"center\">7<\/td><td class=\"has-text-align-center\" data-align=\"center\">37,9613673288<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">4<\/td><td class=\"has-text-align-center\" data-align=\"center\">5<\/td><td class=\"has-text-align-center\" data-align=\"center\">28,3616677843<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">15<\/td><td class=\"has-text-align-center\" data-align=\"center\">51<\/td><td class=\"has-text-align-center\" data-align=\"center\">223,065447072<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Deuxieme_formule_de_Ramanujan\"><\/span>Deuxi\u00e8me formule de Ramanujan<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$P \\approx \\pi(a+b)\\left(1+\\frac{3h}{10+\\sqrt{4-3h}}\\right)\\quad,\\quad h=\\frac{(a-b)^2}{(a+b)^2}.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Celle-ci, il fallait la trouver!<\/p>\n\n\n\n<figure class=\"wp-block-table is-style-stripes\"><table><thead><tr><th class=\"has-text-align-center\" data-align=\"center\">valeurs de a<\/th><th class=\"has-text-align-center\" data-align=\"center\">valeurs de b<\/th><th class=\"has-text-align-center\" data-align=\"center\">valeur approximative du p\u00e9rim\u00e8tre<\/th><\/tr><\/thead><tbody><tr><td class=\"has-text-align-center\" data-align=\"center\">5<\/td><td class=\"has-text-align-center\" data-align=\"center\">7<\/td><td class=\"has-text-align-center\" data-align=\"center\">37,9613689347<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">4<\/td><td class=\"has-text-align-center\" data-align=\"center\">5<\/td><td class=\"has-text-align-center\" data-align=\"center\">28,361667889<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">15<\/td><td class=\"has-text-align-center\" data-align=\"center\">51<\/td><td class=\"has-text-align-center\" data-align=\"center\">223,078484953<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Avec_lequation_parametrique_de_lellipse\"><\/span>Avec l&rsquo;\u00e9quation param\u00e9trique de l&rsquo;ellipse<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Nous avons vu l&rsquo;\u00e9quation cart\u00e9sienne de l&rsquo;ellipse, mais on peut aussi dire que tout point M(<em>x<\/em>;<em>y<\/em>) de cette ellipse v\u00e9rifie:$$\\begin{cases}x(t) &amp; = a\\cos(t)\\\\y(t)&amp;=b\\sin(t)\\end{cases}\\quad,\\quad 0\\leq t\\leq 2\\pi.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Pour des raisons de sym\u00e9trie par rapport aux deux axes du rep\u00e8re, on peut alors dire que le p\u00e9rim\u00e8tre de l&rsquo;ellipse est:$$P = 4\\int_0^{\\frac{\\pi}{2}} \\sqrt{x'(t)^2+y'(t)^2} \\text{d}t.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Or, \\(x'(t)=-a\\sin(t)\\) et \\(y'(t)=b\\cos(t)\\), ce qui donne, apr\u00e8s simplifications:$$P=4a\\int_0^{\\frac{\\pi}{2}} \\sqrt{1-e^2\\sin^2(t)} \\text{d}t,\\quad e=\\sqrt{1-\\frac{b^2}{a^2}}.$$\\(e\\) d\u00e9signe ici ce que l&rsquo;on appelle l&rsquo;<em>excentricit\u00e9<\/em> de l&rsquo;ellipse (et non le nombre d&rsquo;Euler).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Le probl\u00e8me est que l&rsquo;on ne peut pas d\u00e9terminer la valeur exacte de ce genre d&rsquo;int\u00e9grale&#8230;<\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Approximer_le_perimetre_dune_ellipse_a_laide_de_Python\"><\/span>Approximer le p\u00e9rim\u00e8tre d&rsquo;une ellipse \u00e0 l&rsquo;aide de Python<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Mon id\u00e9e ici est de consid\u00e9rer un point M de l&rsquo;ellipse tel que \\( \\big(\\vec{i};\\vec{OM}\\big)=\\theta\\), avec \\( 0\\leq \\theta\\leq 2\\pi\\).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">La premi\u00e8re chose que j&rsquo;ai envie de faire est d&rsquo;exprimer la longueur OM, longueur que je vais abusivement nomm\u00e9e le <em>module<\/em>, et que je vais noter \\(r_\\theta\\). Je vais d\u00e9j\u00e0 dire que les coordonn\u00e9es (<em>x<\/em> ; <em>y<\/em>) de M v\u00e9rifient:$$\\begin{cases}x &amp; = r_\\theta\\cos\\theta\\\\y &amp; = r_\\theta\\sin\\theta\\end{cases}$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Je remplace maintenant dans l&rsquo;\u00e9quation cart\u00e9sienne de l&rsquo;ellipse <em>x<\/em> et <em>y<\/em>, ce qui donne:$$b^2r^2_\\theta\\cos^2\\theta+a^2r^2\\theta\\sin^2\\theta=a^2b^2$$soit:$$r_\\theta=\\frac{ab}{\\sqrt{b^2\\cos^2\\theta+a^2\\sin^2\\theta}}.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Je sais donc maintenant calculer le \u00ab\u00a0module\u00a0\u00bb d&rsquo;un point d\u00e9finit par son angle avec l&rsquo;axe des abscisses.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\u00c0 pr\u00e9sent, si je consid\u00e8re deux points \\(M_0(x_0;y_0)\\) et \\(M_1(x_1;y_1)\\), d\u00e9finis par leurs angles \\(\\theta\\) et \\(\\theta+\\delta\\), \\(\\delta\\) \u00e9tant un nombre positif tr\u00e8s proche de 0, je peux calculer leur distance:$$M_0M_1=\\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}.$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">L&rsquo;id\u00e9e ici est donc de partir d&rsquo;un angle \\(\\theta=0\\) et de lui ajouter \\(\\delta\\) afin de calculer la distance \\(M_0M_1\\), puis de changer la valeur de \\(\\theta\\) en \\(\\theta+\\delta\\) et de faire de m\u00eame jusqu&rsquo;\u00e0 \\(\\theta=2\\pi-\\delta\\). En ajoutant toutes les longueurs, on aura une approximation du p\u00e9rim\u00e8tre.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Cela donne en Python:<\/p>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"dracula\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">\"\"\"\nEllipse d'\u00e9quation :\nx\u00b2   y\u00b2\n-- + -- = 1\na\u00b2   b\u00b2\n\"\"\"\nfrom math import pi,sin,cos,sqrt\n\ndef module(a,b,angle):\n    return a*b\/sqrt( b*b*(cos(angle)**2) + a*a*(sin(angle)**2) )\n\ndef perimetre_ellipse(a,b,n=10000):\n    theta = 0\n    delta = 2*pi\/n\n    p = 0\n\n    while theta &lt; 2*pi:\n        x_0 = module(a,b,theta) * cos(theta)\n        y_0 = module(a,b,theta) * sin(theta)\n        x_1 = module(a,b,theta+delta) * cos(theta+delta)\n        y_1 = module(a,b,theta+delta) * sin(theta+delta)\n        \n        p += sqrt( (x_0-x_1)**2 + (y_0-y_1)**2 )\n        \n        theta += delta\n        \n    return p<\/pre>\n\n\n\n<pre class=\"wp-block-code\"><code>>>> perimetre_ellipse(5,7)\n37.96136808634785\n>>> perimetre_ellipse(4,5)\n28.361667351049444\n>>> perimetre_ellipse(15,51)\n223.07846961614933<\/code><\/pre>\n","protected":false},"excerpt":{"rendered":"<p>Calculer le p\u00e9rim\u00e8tre d&rsquo;une ellipse n&rsquo;est pas chose ais\u00e9e. On pourrait croire qu&rsquo;\u00e0 l&rsquo;instar de son cousin le cercle, l&rsquo;ellipse poss\u00e8de une formule math\u00e9matique pour d\u00e9terminer son p\u00e9rim\u00e8tre&#8230; Que nenni !<\/p>\n","protected":false},"author":1,"featured_media":7496,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6,5],"tags":[343,342,344],"class_list":["post-7481","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematiques","category-python","tag-conique","tag-ellipse","tag-perimetre"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v28.0 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>P\u00e9rim\u00e8tre d&#039;une ellipse - Mathweb.fr<\/title>\n<meta name=\"description\" content=\"Comment calculer, ou tout du moins d\u00e9terminer une approximation, du p\u00e9rim\u00e8tre d&#039;une ellipse donn\u00e9e ? 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