{"id":21167,"date":"2026-01-22T08:00:00","date_gmt":"2026-01-22T13:00:00","guid":{"rendered":"https:\/\/blog.espol.edu.ec\/algoritmos101\/?p=21167"},"modified":"2026-04-05T20:23:18","modified_gmt":"2026-04-06T01:23:18","slug":"s2eva2025paoii_t2-edo-trayectoria-avion-papel","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s2eva30\/s2eva2025paoii_t2-edo-trayectoria-avion-papel\/","title":{"rendered":"s2Eva2025PAOII_T2 EDO trayectoria avi\u00f3n de papel"},"content":{"rendered":"\n

Ejercicio<\/strong>: 2Eva2025PAOII_T2 EDO trayectoria avi\u00f3n de papel<\/a><\/p>\n\n\n\n

\"avi\u00f3n<\/figure>\n\n\n\n \\frac{dv}{dt} = -0.1388 v^2 - g \\sin(\\omega)<\/span>\n\n\n\n \\frac{d\\omega}{dt} = \\frac{0.7654 v^2 - g\\cos(\\omega)}{v}<\/span>\n\n\n\n

Los valores iniciales para el ejercicio se consideran como: g=9.8 m\/s2<\/sup>, t0<\/sub> = 0, v0<\/sub>=4, \u03c90<\/sub> =0, x0<\/sub>=0, y0<\/sub>=2<\/p>\n\n\n\n

literal a<\/h2>\n\n\n\n

Se indica que el planteamiento es para v<\/strong>(t), \u03c9<\/strong>(t). En el literal b se indica el tama\u00f1o de paso h=0.1<\/p>\n\n\n\n f(t,v,\\omega) = -0.1388 v^2 - 9.8 \\sin(\\omega)<\/span>\n\n\n\n g(t,v,\\omega) = \\frac{0.7654 v^2 - 9.8 \\cos(\\omega)}{v}<\/span>\n\n\n\n

Usando Runge-Kutta de 2do orden para sistemas de ecuaciones:<\/p>\n\n\n\n K1v = h f(t,v,\\omega) = 0.1 (-0.1388 v^2 - 9.8 \\sin(\\omega)) <\/span>\n\n\n\n K1\\omega = h g(t,v,\\omega) = 0.1 \\frac{0.7654 v^2 - 9.8 \\cos(\\omega)}{v} <\/span>\n\n\n\n K2v = h f(t+h,v+K1v,\\omega+K1\\omega) = 0.1 (-0.1388 (v+K1v)^2 - 9.8 \\sin(\\omega+K1\\omega) <\/span>\n\n\n\n K2\\omega = h g(t+h,v+K1v,\\omega+K1\\omega) = 0.1 \\frac{0.7654 (v+K1v)^2 - 9.8 \\cos(\\omega+K1\\omega)}{v+K1v} <\/span>\n\n\n\nv[i+1] = v[i] + \\frac{K1v+K2v}{2} <\/span>\n\n\n\n\\omega[i+1] = \\omega[i] + \\frac{K1\\omega+K2\\omega}{2} <\/span>\n\n\n\nt[i+1] =t[i] + h <\/span>\n\n\n\n

literal b<\/h2>\n\n\n\n

iteraci\u00f3n = 0, t0<\/sub> = 0, v0<\/sub>=4, \u03c90<\/sub> =0<\/strong><\/p>\n\n\n\n

Usando Runge-Kutta de 2do orden para sistemas de ecuaciones:<\/p>\n\n\n\n K1v = 0.1 (-0.1388 (4)^2 - 9.8\\sin(0)) =-0.2220 <\/span>\n\n\n\n K1\\omega = 0.1 \\frac{0.7654 (4)^2 - 9.8 \\cos(0)}{4} =0.26116<\/span>\n\n\n\n K2v = 0.1 (-0.1388 (4-0.2220)^2 - 9.8 \\sin(0+0.26116)) =-0.258004 <\/span>\n\n\n\n K2\\omega = 0.1 \\frac{0.7654 (4-0.2220)^2 - 9.8 \\cos(0+0.26116)}{4-0.2220} =0.03024<\/span>\n\n\n\nv[1] = 4 + \\frac{-0.2220-0.258004}{2} =3.7599<\/span>\n\n\n\n\\omega[1] = 0 + \\frac{0.26116+0.03024}{2} =0.04570<\/span>\n\n\n\nt[1] =0 + 0.1 = 0.1 <\/span>\n\n\n\n

iteraci\u00f3n = 1, t0<\/sub> = 0.1, v0<\/sub>=3.7599, \u03c90<\/sub> =0.04570<\/strong><\/p>\n\n\n\n K1v = 0.1 (-0.1388 (3.7599)^2 - 9.8\\sin(0.04570)) =-0.2409 <\/span>\n\n\n\n K1\\omega = 0.1 \\frac{0.7654 (3.7599)^2 - 9.8 \\cos(0.04570)}{3.7599} =0.02741<\/span>\n\n\n\n K2v = 0.1 (-0.1388 (3.7599-0.2409)^2 - 9.8 \\sin(0.04570+0.02741) =-0.2434<\/span>\n\n\n\n K2\\omega = 0.1 \\frac{0.7654 (3.7599-0.2409)^2 - 9.8 \\cos(0.04570+0.02741)}{3.7599-0.2409}=-0.008406 <\/span>\n\n\n\nv[2] = 3.7599 + \\frac{-0.2409-0.2434}{2} =3.5177<\/span>\n\n\n\n\\omega[2] = 0.04570 + \\frac{0.02741-0.008406}{2}=0.05520 <\/span>\n\n\n\nt[2] =0.1 + 0.1 =0.2 <\/span>\n\n\n\n

iteraci\u00f3n = 2, t0<\/sub> = 0.2, v0<\/sub>=3.5177, \u03c90<\/sub> =0.05520<\/strong><\/p>\n\n\n\n K1v = 0.1 (-0.1388 (3.5177)^2 - 9.8 \\sin(0.05520 )) =-0.2258<\/span>\n\n\n\n K1\\omega = 0.1 \\frac{0.7654 (3.5177)^2 - 9.8 \\cos(0.05520 )}{3.5177} =-0.1957<\/span>\n\n\n\n K2v = 0.1 (-0.1388 (3.5177-0.2258)^2 - 9.8 \\sin(0.05520 -0.1957) =-0.008918<\/span>\n\n\n\n K2\\omega = 0.1 \\frac{0.7654 (3.5177-0.2258)^2 - 9.8 \\cos(0.05520 -0.1957)}{3.5177-0.2258} =-0.04542<\/span>\n\n\n\nv[3] = 3.5177 + \\frac{-0.2258-0.008918}{2} <\/span>\n\n\n\n\\omega[3] = 0.05520 + \\frac{-0.1957-0.04542}{2} <\/span>\n\n\n\nt[3] =0.2 + 0.1 = 0.3 <\/span>\n\n\n\n

literal c<\/h2>\n\n\n\n

usando el algoritmo:<\/p>\n\n\n\n

i  [ xi,  yi,  zi ]\n   [ K1y,  K1z,  K2y,  K2z ]\n0 [0. 4. 0.]\n  [0. 0. 0. 0.]\n1 [0.1      3.759908 0.045703]\n  [-0.222133  0.061164 -0.25805   0.030241]\n2 [0.2      3.517638 0.055201]\n  [-0.24104   0.027415 -0.2435   -0.008417]\n3 [0.3      3.306824 0.028019]\n  [-0.225859 -0.008928 -0.195769 -0.045437]\n4 [ 0.4       3.156694 -0.030508]\n  [-0.179271 -0.043133 -0.12099  -0.073922]\n5 [ 0.5       3.086496 -0.108154]\n  [-0.10845  -0.06869  -0.031945 -0.0866  ]\n6 [ 0.6       3.099629 -0.188073]\n  [-0.026475 -0.079413  0.05274  -0.080425]\n7 [ 0.7       3.182333 -0.254509]\n  [ 0.04984  -0.073343  0.115569 -0.059528]\n8 [ 0.8       3.309327 -0.297833]\n  [ 0.106135 -0.054451  0.147852 -0.032198]\n...\n35 [ 3.5       3.526395 -0.162191]\n   [-0.019245 -0.001804 -0.015611 -0.004688]\n36 [ 3.6       3.514804 -0.167548]\n   [-0.014394 -0.004343 -0.008789 -0.006371]\n37 [ 3.7       3.50996  -0.173924]\n   [-0.008082 -0.00589  -0.001606 -0.006863]\n38 [ 3.8       3.511635 -0.180218]\n   [-0.001452 -0.006337  0.004803 -0.00625 ]\n39 [ 3.9       3.518653 -0.185496]\n   [ 0.004456 -0.005769  0.00958  -0.004787]\n40 [ 4.        3.529203 -0.189115]\n   [ 0.008857 -0.004416  0.012243 -0.002822]<\/strong><\/code><\/pre>\n\n\n\n
\"trayectoria<\/figure>\n\n\n\n
literal d<\/h2>\n\n\n\n
Se observa que la velocidad y el \u00e1ngulo oscilan en la trayectoria del avi\u00f3n durante el descenso, lo que se muestra en la gr\u00e1fica de trayectoria del tema 1.<\/p>\n\n\n\n
El algoritmo permite realizar el c\u00e1lculo con otros valores de velocidad inicial y \u00e1ngulo w.<\/p>\n\n\n\n
El desarrollo presentado debe limitarse hasta cuando se toca el suelo en y=0, esto se logra determinar al incorporar al ejercicio las cuatro ecuaciones del modelo presentado o usando integraci\u00f3n para x, y a partir de velocidad y \u00e1ngulo desde el punto de partida.<\/p>\n\n\n\n
Algoritmo en Python<\/h2>\n\n\n
\n# 2Eva2025PAOII_T2 EDO trayectoria avi\u00f3n de papel v,w\n# Sistemas EDO con Runge Kutta de 2do Orden\nimport numpy as np\n \ndef rungekutta2_fg(f,g,x0,y0,z0,h,muestras,\n                   vertabla=False, precision=6):\n    ''' solucion a EDO d2y\/dx2 con Runge-Kutta 2do Orden,\n    f(x,y,z) = z #= y'\n    g(x,y,z) = expresion d2y\/dx2 con z=y'\n    tambien es solucion a sistemas edo f() y g()\n    x0,y0,z0 son valores iniciales, h es tamano de paso,\n    muestras es la cantidad de puntos a calcular.\n    '''\n    tamano = muestras + 1\n    tabla = np.zeros(shape=(tamano,3+4),dtype=float)\n    # incluye el punto [x0,y0,z0,K1y,K1z,K2y,K2z]\n    tabla[0] = [x0,y0,z0,0,0,0,0]\n     \n    xi = x0 # valores iniciales\n    yi = y0\n    zi = z0\n    for i in range(1,tamano,1):\n        K1y = h * f(xi,yi,zi)\n        K1z = h * g(xi,yi,zi)\n         \n        K2y = h * f(xi+h, yi + K1y, zi + K1z)\n        K2z = h * g(xi+h, yi + K1y, zi + K1z)\n \n        yi = yi + (K1y+K2y)\/2\n        zi = zi + (K1z+K2z)\/2\n        xi = xi + h\n         \n        tabla[i] = [xi,yi,zi,K1y,K1z,K2y,K2z]\n         \n    if vertabla==True:\n        np.set_printoptions(precision)\n        print('EDO f,g con Runge-Kutta 2 Orden')\n        print('i ','[ xi,  yi,  zi',']')\n        print('   [ K1y,  K1z,  K2y,  K2z ]')\n        for i in range(0,tamano,1):  \n            txt = ' '\n            if i>=10:\n                txt = '  '\n            print(str(i),tabla[i,0:3])\n            print(txt,tabla[i,3:])\n     \n    return(tabla)\n \n# PROGRAMA ------------------\n \n# INGRESO\n# Ecuaciones\nf = lambda t,v,w: -0.1388*(v**2) - 9.8*np.sin(w)\ng = lambda t,v,w: (0.7654*(v**2) - 9.8*np.cos(w))\/v\n \n# Condiciones iniciales\nt0 = 0. #tiempo\nv0 = 4.0 # velocidad\nw0 = 0. # \u00e1ngulo\n \n# par\u00e1metros del algoritmo\nh = 0.1\nmuestras = 40\n \n# PROCEDIMIENTO\ntabla = rungekutta2_fg(f,g,t0,v0,w0,h,muestras,vertabla=True)\nti = tabla[:,0]\nxi = tabla[:,1]\nyi = tabla[:,2]\n\n# SALIDA\nprint('Sistemas EDO: Trayectoria avion de papel')\nnp.set_printoptions(precision=6)\nprint(' [ ti, vi, wi,K1v,K1w,K2v,K2w]')\nprint(tabla[:,0:7])\n\n# GRAFICA tiempos vs poblaci\u00f3n\nimport matplotlib.pyplot as plt\n \nfig_t, (graf1,graf2) = plt.subplots(2)\nfig_t.suptitle('EDO Trayectoria Avion de papel')\ngraf1.plot(ti,xi, color='blue',label='dv\/dt')\n \n#graf1.set_xlabel('t tiempo')\ngraf1.set_ylabel('v')\ngraf1.legend()\ngraf1.grid()\n \ngraf2.plot(ti,yi, color='orange',label='dw\/dt')\ngraf2.set_xlabel('t tiempo')\ngraf2.set_ylabel('dw\/dt')\ngraf2.legend()\ngraf2.grid()\n \n# gr\u00e1fica xi vs yi\nfig_xy, graf3 = plt.subplots()\ngraf3.plot(xi,yi)\n \ngraf3.set_title('EDO Trayectoria Avion de papel')\ngraf3.set_xlabel('v')\ngraf3.set_ylabel('w')\ngraf3.grid()\nplt.show()\n<\/pre><\/div>","protected":false},"excerpt":{"rendered":"
Ejercicio: 2Eva2025PAOII_T2 EDO trayectoria avi\u00f3n de papel Los valores iniciales para el ejercicio se consideran como: g=9.8 m\/s2, t0 = 0, v0=4, \u03c90 =0, x0=0, y0=2 literal a Se indica que el planteamiento es para v(t), \u03c9(t). En el literal b se indica el tama\u00f1o de paso h=0.1 Usando Runge-Kutta de 2do orden para sistemas […]<\/p>\n","protected":false},"author":8043,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"wp-custom-template-entrada-mn-ejemplo","format":"standard","meta":{"footnotes":""},"categories":[49],"tags":[58,54],"class_list":["post-21167","post","type-post","status-publish","format-standard","hentry","category-mn-s2eva30","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/21167","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/users\/8043"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/comments?post=21167"}],"version-history":[{"count":9,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/21167\/revisions"}],"predecessor-version":[{"id":23880,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/21167\/revisions\/23880"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=21167"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=21167"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=21167"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}