2a:null 2b:["$","title",null,{"children":"Distance temporelle | linard-y | Observable"}] 31:T42b,Cette équation a formellement 4 solutions. Cependant, une seule d'entre elles est le temps physique le plus court pour un joueur pour atteindre l'origine. Pour savoir laquelle des solutions choisir, il faut que le temps soit la plus petite racine positive et réelle de l'équation (8). Le modèle proposé contient un paramètre de contrôle ajustable, la valeur de ${tex `F`} . Cette valeur devrait d'après Fujimura et _al_. (2015) être de l'ordre de 10 m/s. En principe, puisque nous n'avons pas limité la vitesse d'un défenseur, des vitesses non physiquement grandes peuvent être développées. C'est pourquoi, dans le modèle de Fujimura et _al_. (2015) la dynamique a été affinée en introduisant une traînée, qui limite l'augmentation de la vitesse. Cependant, pour notre cas, le temps qu'il faut à un défenseur pour atteindre un tireur dépasse rarement 1 seconde, de sorte que les vitesses ne pas atteindre des valeurs non physiques. Par conséquent, et par souci de simplicité, nous avons choisi de ne pas affiner davantage le modèle. 32:T1d19,equationSolver4 = (aq, bq, cq, dq, eq) => { //Permet de résoudre une équation algébrique d'ordre 4. /* This script and many more are available free online at The JavaScript Source!! http://www.javascriptsource.com Created by: Brian Kieffer | http://www.freewebs.com/brianjs/ */ var aq2 = aq; // Keeps Orignial AQ value var bq2 = bq;// Keeps Orignial BQ Value var perfect = 0; var perfectbiquadratic = 0; // The Bi-Quadratic 2 Perfect Squares that are negative test if (cq*cq - 4*aq*eq == 0 && cq > 0) { perfectbiquadratic = 1; } // Divide Equation by the X^4 Coefficent to make equation in the form of X^4 + AX^3 + BX^2 + CX + D bq /= aq; cq /= aq; dq /= aq; eq /= aq; aq = 1; var f2 = cq - (3*bq*bq / 8); var g2 = dq + (bq*bq*bq/8) - (bq*cq/2); var h2 = eq - (3*bq*bq*bq*bq/256) + (bq*bq*(cq/16)) - (bq*dq/4); var a = 1; var b = f2/2; var c = (f2*f2 - (4*h2)) / 16; var d = -1*((g2*g2)/64); if (b == 0 && c == 0 && d == 0) { perfect = 1; } // Cubic routine starts here….. var f = (((3*c) / a) - ((b*b) / (a*a))) / 3; var g = (((2*b*b*b) / (a*a*a)) - ((9*b*c) / (a*a)) + ((27*d) / a)) / 27; var h = eval(((g*g)/4) + ((f*f*f)/27)); var z = 1/3; var i; var j; var k; var l; var m; var n; var p; var xoneterm; var xtwoterm; var xthreeterm; var alreadydone; var alreadydone2 = 0; var ipart = 0; var p = 0; var q = 0; var r = 0; var s = 0; if (h <= 0) { var exec = 2; i = Math.sqrt(((g*g) / 4) - h); j = Math.pow(i,z); k = Math.acos(-1 * (g / (2*i))); l = -1*j; m = Math.cos(k / 3); n = Math.sqrt(3) * Math.sin(k / 3); p = (b / (3*a)) * -1; xoneterm = (2*j) * Math.cos(k/3) - (b / (3*a)); xtwoterm = l * (m + n) + p; xthreeterm = l * (m - n) + p; } if (h > 0) { var exec = 1; var R = (-1*(g / 2)) + Math.sqrt(h); if (R < 0) { var S = -1*(Math.pow((-1*R),z)); } else { var S = Math.pow(R,z); } var T = (-1*(g / 2)) - Math.sqrt(h); if (T < 0) { var U = -1*(Math.pow((-1*T),z)); } else { var U = Math.pow(T,z); } xoneterm = (S + U) - (b / (3*a)); xtwoterm = (-1*(S+U)/2) - (b / (3*a)); var ipart = ((S-U) * Math.sqrt(3)) / 2; xthreeterm = xtwoterm; } if (f == 0 && g == 0 && h == 0) { if ((d/a) < 0 ) { xoneterm = (Math.pow((-1*(d/a)),z)); xtwoterm = xoneterm; xthreeterm = xoneterm; } else { xoneterm = -1*(Math.pow((d/a),z)); xtwoterm = xoneterm; xthreeterm = xoneterm; } } // ….and ends here. // Return to solving the Quartic. if (ipart == 0 && xoneterm.toFixed(10) == 0) { var alreadydone2 = 1; var p2 = Math.sqrt(xtwoterm); var q = Math.sqrt(xthreeterm); var r = -g2 / (8*p2*q); var s = bq2/(4*aq2); } if (ipart == 0 && xtwoterm.toFixed(10) == 0 && alreadydone2 == 0 && alreadydone2 != 1) { var alreadydone2 = 2; var p2 = Math.sqrt(xoneterm); var q = Math.sqrt(xthreeterm); var r = -g2 / (8*p2*q); var s = bq2/(4*aq2); } if (ipart == 0 && xthreeterm.toFixed(10) == 0 && alreadydone2 == 0 && alreadydone2 != 1 && alreadydone2 != 2) { var alreadydone2 = 3; var p2 = Math.sqrt(xoneterm); var q = Math.sqrt(xtwoterm); var r = -g2 / (8*p2*q); var s = bq2/(4*aq2); } if (alreadydone2 == 0 && ipart == 0) { if (xthreeterm.toFixed(10) < 0) { var alreadydone2 = 4; var p2 = Math.sqrt(xoneterm); var q = Math.sqrt(xtwoterm); var r = -g2 / (8*p2*q); var s = bq2/(4*aq2); } else { var alreadydone2 = 5; var p2 = Math.sqrt(xoneterm.toFixed(10)); var q = Math.sqrt(xthreeterm.toFixed(10)); var r = -g2 / (8*p2*q); var s = bq2/(4*aq2); } } if (ipart != 0) { var p2 = isquareroot(xtwoterm,ipart,0); var p2ipart = isquareroot(xtwoterm,ipart,1); var q = isquareroot(xthreeterm,-ipart,0); var qipart = isquareroot(xthreeterm,-ipart,1); var mult = calcmult(p2,p2ipart,q,qipart,0); var r = -g2/(8*mult); var s = bq2/(4*aq2); } if (ipart == 0 && xtwoterm.toFixed(10) < 0 && xthreeterm.toFixed(10) < 0) { xtwoterm /= -1 xthreeterm /= -1 var p2 = 0; var q = 0; var p2ipart = Math.sqrt(xtwoterm); var qipart = Math.sqrt(xthreeterm); var mult = calcmult(p2,p2ipart,q,qipart,0); var r = -g2/(8*mult); var s = bq2/(4*aq2); var ipart = 1; } if (xoneterm.toFixed(10) > 0 && xtwoterm.toFixed(10) < 0 && xthreeterm.toFixed(10) == 0 && ipart == 0) { xtwoterm /= -1; var p2 = Math.sqrt(xoneterm); var q = 0; var p2ipart = 0; var qipart = Math.sqrt(xtwoterm); var mult = calcmult(p2,p2ipart,q,qipart,0); var mult2 = calcmult(p2,p2ipart,q,qipart,1); var r = -g2/(8*mult); if (mult2 != 0) { var ripart = g2/(8*mult2); var r = 0; } var s = bq2/(4*aq2); var ipart = 1; } if (xtwoterm.toFixed(10) == 0 && xthreeterm.toFixed(10) == 0 && ipart == 0) { var p2 = Math.sqrt(xoneterm); var q = 0; var r = 0; var s = bq2/(4*aq2); } if (ipart == 0) { var x1 = eval((p2 + q + r - s).toFixed(10)); var x2 = eval((p2 - q - r - s).toFixed(10)); var x3 = eval((-p2 + q - r - s).toFixed(10)); var x4 = eval((-p2 - q + r - s).toFixed(10)); var x1i = 0; var x2i = 0; var x3i = 0; var x4i = 0; } if (perfect == 1) { var x1 = -bq/4; var x2 = -bq/4; var x3 = -bq/4; var x4 = -bq/4; var x1i = 0; var x2i = 0; var x3i = 0; var x4i = 0; } if (ipart == 0 && xtwoterm.toFixed(10) < 0 && xthreeterm.toFixed(10) < 0) { xtwoterm /= -1; xthreeterm /= -1; var p2 = 0; var q = 0; var p2ipart = Math.sqrt(xtwoterm); var qipart = Math.sqrt(xthreeterm); var mult = calcmult(p2,p2ipart,q,qipart,0); var r = -g2/(8*mult); var s = bq2/(4*aq2); var ipart = 1; } if (xoneterm.toFixed(10) > 0 && xtwoterm.toFixed(10) < 0 && xthreeterm.toFixed(10) == 0 && ipart == 0) { xtwoterm /= -1 var p2 = Math.sqrt(xoneterm); var q = 0; var p2ipart = 0; var qipart = Math.sqrt(xtwoterm); var mult = calcmult(p2,p2ipart,q,qipart,0); var mult2 = calcmult(p2,p2ipart,q,qipart,1); var r = -g2/(8*mult); if (mult2 != 0) { var ripart = g2/(8*mult2); var r = 0; } var s = bq2/(4*aq2); var ipart = 1; } if (xtwoterm.toFixed(10) == 0 && xthreeterm.toFixed(10) == 0 && ipart == 0) { var p2 = Math.sqrt(xoneterm); var q = 0; var r = 0; var s = bq2/(4*aq2); } if (ipart != 0) { var x1 = eval((p2 + q + r - s).toFixed(10)); var x2 = eval((p2 - q - r - s).toFixed(10)); var x3 = eval((-p2 + q - r - s).toFixed(10)); var x4 = eval((-p2 - q + r - s).toFixed(10)); var x1i = eval((p2ipart + qipart).toFixed(10)); var x2i = eval((p2ipart - qipart).toFixed(10)); var x3i = eval((-p2ipart + qipart).toFixed(10)); var x4i = eval((-p2ipart - qipart).toFixed(10)); } if (perfectbiquadratic == 1) { var x1 = 0; var x2 = 0; var x3 = 0; var x4 = 0; var x1i = eval(Math.sqrt(cq/2).toFixed(10)); var x2i = eval(Math.sqrt(cq/2).toFixed(10)); var x3i = -eval(Math.sqrt(cq/2).toFixed(10)); var x4i = -eval(Math.sqrt(cq/2).toFixed(10)); } return ({'x1' : x1 + ' ' + x1i, 'x2' : x2 + ' ' + x2i, 'x3' : x3 + ' ' + x3i, 'x4' : x4 + ' ' + x4i}) };33:[] 2e:{"id":"4305516c1544e441","slug":"distance-temporelle","title":"Distance temporelle","thumbnail":"51b45ee56a1b3ed22ac2b5db93414050c66ad6d09005a32f2600dfe0b15ba162","access_level":"public","create_time":"$D2021-12-20T09:27:35.342Z","update_time":"$D2021-12-29T20:02:57.516Z","view_time":null,"first_public_time":"$D2021-12-20T18:36:03.645Z","fork":false,"stars":1,"starred":false,"owner":{"type":"team","id":"9784867a3e59436b","login":"linard-y","name":"","avatar_url":"https://avatars-next.observableusercontent.com/0ce44546efae24745b35d9327acbc99a3f3fdaeac3595375f92848f133b06ded","tier":"starter_2024"},"version":215,"publish_version":215,"role":null,"authored":false,"authors":[{"type":"individual","id":"c6e84137b8a39d94","login":"linard-y","name":"","avatar_url":"https://avatars-next.observableusercontent.com/0ce44546efae24745b35d9327acbc99a3f3fdaeac3595375f92848f133b06ded","tier":null}],"members":[{"type":"individual","id":"c6e84137b8a39d94","login":"linard-y","name":"","avatar_url":"https://avatars-next.observableusercontent.com/0ce44546efae24745b35d9327acbc99a3f3fdaeac3595375f92848f133b06ded","tier":null,"role":"editor"}],"comments":[],"custom_thumbnail":false,"trashed":false,"subscribed":null,"forks":1,"collection":null,"type":"notebook","body":{"title":"Distance temporelle","cells":[{"id":0,"value":"# Distance temporelle","pinned":false,"hidden":false,"mode":"md"},{"id":209,"value":"_(à voir également, le notebook de [juleallegre](https://observablehq.com/@julesallegre/4-th-order-polynomial-equation-resolution) sur le même sujet)_","pinned":false,"hidden":false,"mode":"md"},{"id":3,"value":"Nous voulons calculer combien de temps il faut à un joueur en mouvement sur un terrain de sport, pour atteindre un point ${tex `(x, y)`}.\nPour généraliser, nous déterminerons le temps qu’il faut à un joueur ayant une position initiale donnée, ${tex `(x_0, y_0)`}, et une vitesse initiale donnée, ${tex `(u_0, v_0)`}, pour atteindre le point ${tex `(x, y)=(0, 0)`}. Nous considérerons que le joueur utilisera une force constante (par unité de masse) dans une direction donnée, d’intensité ${tex `F = \\sqrt{F_x^2 + F_y^2}`}. C’est une hypothèse qui permet de trouver une solution analytique simple. En particulier, il permet de considérer les deux directions séparément.\n","pinned":false,"hidden":false,"mode":"md"},{"id":18,"value":"D'après la deuxième loi de Newton (dynamique), on a :\n","pinned":false,"hidden":false,"mode":"md"},{"id":23,"value":"\\sum{\\vec{F}} = m.\\vec{a} \\tag{1}","pinned":false,"hidden":false,"mode":"tex"},{"id":29,"value":"soit","pinned":false,"hidden":false,"mode":"md"},{"id":32,"value":"\\left\\{ \n \\begin{array}{ll}\n F_x = a_x = \\frac{d^2 x}{dt} \\\\\n \\\\\n F_y = a_y = \\frac{d^2 y}{dt} \n \\end{array}\n\\right \\tag{2}.","pinned":false,"hidden":false,"mode":"tex"},{"id":47,"value":"et, par integrations successives, on obtient:","pinned":false,"hidden":false,"mode":"md"},{"id":53,"value":"\\left\\{ \n \\begin{array}{ll}\n x(t) = x_0 + u_0 t + \\frac{1}{2} F_x t^2 \\\\\n \\\\\n y(t) = y_0 + v_0 t + \\frac{1}{2} F_y t^2 \n \\end{array}\n\\right \\tag{3}.","pinned":false,"hidden":false,"mode":"tex"},{"id":60,"value":"Evaluons d'abord cette expression pour ${tex `x(t)=0`} et ${tex `y(t)=0`} et déterminons le temps ${tex `t`} où ce point est atteint. Déterminons d'abord la force par unité de masse :","pinned":false,"hidden":false,"mode":"md"},{"id":66,"value":"\\left\\{ \n \\begin{array}{ll}\n F_x = -2 \\frac{x_0 + u_0 t}{t^2} \\\\\n \\\\\n F_y = -2 \\frac{y_0 + v_0 t}{t^2} \n \\end{array}\n\\right \\tag{4}.","pinned":false,"hidden":false,"mode":"tex"},{"id":71,"value":"soit","pinned":false,"hidden":false,"mode":"md"},{"id":75,"value":"F = \\sqrt{F_x^2 + F_y^2} = \\frac{2}{t^2} \\sqrt{(x_0+u_0t)^2+(y_0+v_0t)^2} \\tag{5}","pinned":false,"hidden":false,"mode":"tex"},{"id":84,"value":"En élevant au carré, on obtient:","pinned":false,"hidden":false,"mode":"md"},{"id":88,"value":"F^2 = \\frac{4}{t^4} [(x_0+u_0t)^2 + (y_0+v_0t)^2] \\tag{6}","pinned":false,"hidden":false,"mode":"tex"},{"id":91,"value":"d'où on déduit:","pinned":false,"hidden":false,"mode":"md"},{"id":97,"value":"F^2t^4-4x_0^2-8x_0u_0t-4(u_0t)^2-4y_0^2-8y_0v_0t-4(v_0t)^2 = 0 \\tag{7}","pinned":false,"hidden":false,"mode":"tex"},{"id":100,"value":"soit la forme polynomiale du 4ième degré :","pinned":false,"hidden":false,"mode":"md"},{"id":105,"value":"\\frac{1}{4}F^2t^4-(u_0+v_0)t^2-2(x_0u_0+y_0v_0)t-(x_0^2+y_0^2) = 0 \\tag{8}","pinned":false,"hidden":false,"mode":"tex"},{"id":124,"value":"$31","pinned":false,"hidden":false,"mode":"md"},{"id":128,"value":"[1]- Fujimura, Akira, et Kokichi Sugihara. “Geometric Analysis and Quantitative Evaluation of Sport \nTeamwork”. Systems and Computers in Japan, vol. 36, no 6, 2005, p. 49-58. Wiley Online Library, \ndoi:10.1002/scj.20254 .","pinned":false,"hidden":false,"mode":"md"},{"id":152,"value":"Nous reproduisons ci-dessous le code proposé par [Brian Kieffer](https://javascriptsource.com/quartic-equation-solver/) pour la résolution d'une équation polynomiale d'ordre 4.","pinned":false,"hidden":false,"mode":"md"},{"id":140,"value":"$32","pinned":false,"hidden":false,"mode":"ojs"},{"id":143,"value":"solutionFilter = (solution) => {\n \n var newSolution;\n \n let x1 = solution.x1.split(\" \");\n let x2 = solution.x2.split(\" \");\n let x3 = solution.x3.split(\" \");\n let x4 = solution.x4.split(\" \");\n \n let n = 0; //to get one solution\n let x = [x1, x2, x3, x4];\n x.forEach((d, j) => {\n if (+d[1] == 0){\n if (+d[0] >= 0 && n == 0){\n n++;\n newSolution = +d[0];\n }\n }\n \n });\n \n return newSolution;\n};","pinned":false,"hidden":false,"mode":"ojs"},{"id":146,"value":"function calcmult(a2,b2,c2,d2,e2) {\n var real = a2*c2 - b2*d2;\n var img = b2*c2 + a2*d2;\n\n if (e2 == 0) {\n return real;\n } else {\n return img;\n }\n};","pinned":false,"hidden":false,"mode":"ojs"},{"id":149,"value":"function isquareroot(a1,b1,n1) {\n var y = Math.sqrt((a1*a1) + (b1*b1));\n var y1 = Math.sqrt((y - a1) / 2);\n var x1 = b1 / (2*y1);\n\n if (n1 == 0) {\n return x1;\n } else {\n return y1;\n }\n};","pinned":false,"hidden":false,"mode":"ojs"},{"id":156,"value":"Reprenons notre modèle avec les valeurs numériques ci-dessous et résolvons l’équation (8).\n\n","pinned":false,"hidden":false,"mode":"md"},{"id":161,"value":"\\left\\{ \n \\begin{array}{ll}\n x_0=4\n \\\\\n y_0=0\n \\\\\n u_0 = 1\n \\\\\n v_0=0\n \\\\\n F=4\n \\end{array}\n\\right.","pinned":false,"hidden":false,"mode":"tex"},{"id":176,"value":"4t^4-t^2-8t-16 = 0 \\tag{9}","pinned":false,"hidden":false,"mode":"tex"},{"id":180,"value":"sol = equationSolver4(4, 0, -1, -8, -16)","pinned":false,"hidden":false,"mode":"ojs"},{"id":184,"value":"La fonction `equationSolver4` retourne la partie réelle et la partie imaginaire des 4 solutions notées x1, x2, x3 et x4.","pinned":false,"hidden":false,"mode":"md"},{"id":189,"value":"La solution à retenir est ici évidente mais nous pouvons aussi utiliser la fonction 'solutionFilter' pour vérifier.","pinned":false,"hidden":false,"mode":"md"},{"id":194,"value":"solutionFilter(sol)","pinned":true,"hidden":false,"mode":"ojs"},{"id":199,"value":"Nous créons ci-dessous une fonction de calcul de la distance temporelle qui pourra être réutilisée dans d'autre notebook.","pinned":false,"hidden":false,"mode":"md"},{"id":203,"value":"t_dist = (xi, yi, xf, yf, u, v, F) => {\n\n let X = xi - xf;\n let Y = yi - yf;\n let a = 1;\n let b = 0;\n let c = -4 * (u**2 + v**2) / (F**2);\n let d = -8 * (u * X + v * Y) / (F**2);\n let e = -4 * (X**2 + Y**2) / (F**2);\n var temp = equationSolver4(a, b, c, d, e);\n temp = solutionFilter(temp); // on ne garde que les valeurs réels positives.\n \n return temp;\n};","pinned":false,"hidden":false,"mode":"ojs"},{"id":206,"value":"t_dist(4, 0, 0, 0, 1, 0, 4)","pinned":true,"hidden":false,"mode":"ojs"}],"files":"$W33","stdlib":"1","titling":"auto"}} 2f:"$2e:body"