{"id":11245,"date":"2025-07-01T22:10:38","date_gmt":"2025-07-02T03:10:38","guid":{"rendered":"http:\/\/blog.espol.edu.ec\/analisisnumerico\/?p=11245"},"modified":"2026-04-05T19:55:53","modified_gmt":"2026-04-06T00:55:53","slug":"s1eva2025paoi_t2-cables-camara-aerea","status":"publish","type":"post","link":"https:\/\/blog.espol.edu.ec\/algoritmos101\/mn-s1eva30\/s1eva2025paoi_t2-cables-camara-aerea\/","title":{"rendered":"s1Eva2025PAOI_T2 Cables de c\u00e1mara a\u00e9rea"},"content":{"rendered":"\n

Ejercicio<\/strong>:\u00a01Eva2025PAOI_T2 Cables de c\u00e1mara a\u00e9rea<\/a><\/p>\n\n\n\n

literal a. matriz aumentada y pivoteo parcial por filas<\/h2>\n\n\n\n

Las ecuaciones que conforman el sistema de ecuaciones son:<\/p>\n\n\n\\frac{55}{56.75} T_{AB} - \\frac{60}{71.36} T_{AD}=0 <\/span>\n\n\n\\frac{14}{56.75} T_{AB} + \\frac{14}{34.93} T_{AC} + \\frac{14}{71.36} T_{AD}-490=0 <\/span>\n\n\n-\\frac{32}{34.93} T_{AC} + \\frac{36}{71.36} T_{AD}=0 <\/span>\n\n\n\n

Se reordenan las ecuaciones para la forma A.X=B, con las constantes del lado derecho.<\/p>\n\n\n\\frac{55}{56.75} T_{AB} - \\frac{60}{71.36} T_{AD}=0 <\/span>\n\n\n\\frac{14}{56.75} T_{AB} + \\frac{14}{34.93} T_{AC} + \\frac{14}{71.36} T_{AD} =490 <\/span>\n\n\n-\\frac{32}{34.93} T_{AC} + \\frac{36}{71.36} T_{AD}=0 <\/span>\n\n\n\n

la matriz A y vector B ser\u00e1n:<\/p>\n\n\n A = \\begin{pmatrix} \\frac{55}{56.75} &0& - \\frac{60}{71.36} \\\\ \\frac{14}{56.75} &\\frac{14}{34.93} &\\frac{14}{71.36} \\\\ 0 &-\\frac{32}{34.93}& \\frac{36}{71.36}\\end{pmatrix} <\/span>\n\n\n B =[0,490,0]<\/span>\n\n\n\n

Matriz Aumentada<\/p>\n\n\n \\left( \\begin{array}{rrr|r} \\frac{55}{56.75} &0& - \\frac{60}{71.36} & 0\\\\ \\frac{14}{56.75} & \\frac{14}{34.93} & \\frac{14}{71.36} &490\\\\ 0 &-\\frac{32}{34.93} & \\frac{36}{71.36} & 0\\end{array} \\right) <\/span>\n\n\n\n

Pivoteo parcial por filas<\/p>\n\n\n\n

columna = 0, no requiere cambios, la mayor magnitud se encuentra en la diagonal.<\/p>\n\n\n\n

columna = 1<\/p>\n\n\n \\left( \\begin{array}{rrr|r} \\frac{55}{56.75} &0& - \\frac{60}{71.36} & 0\\\\ 0 &-\\frac{32}{34.93} & \\frac{36}{71.36} & 0 \\\\ \\frac{14}{56.75} & \\frac{14}{34.93} & \\frac{14}{71.36} &490\\end{array} \\right) <\/span>\n\n\n\n

literal b. Expresiones para m\u00e9todo Jacobi<\/h2>\n\n\n\n

fila = 0<\/strong>
\\frac{55}{56.75} x + 0 y - \\frac{60}{71.36} z =0<\/span><\/p>\n\n\n \\frac{55}{56.75} x = \\frac{60}{71.36} z<\/span>\n\n\n x = \\frac{60}{71.36} z \\left(\\frac{1}{\\frac{55}{56.75}}\\right)<\/span>\n\n\n x = \\frac{60(56.75)}{71.36(55)}z =0.86756 z<\/span>\n\n\n\n

fila = 1<\/strong><\/p>\n\n\n 0 x -\\frac{32}{34.93} y + \\frac{36}{71.36} z = 0<\/span>\n\n\n -\\frac{32}{34.93} y = - \\frac{36}{71.36} z<\/span>\n\n\n y = \\frac{36}{71.36} z \\left(\\frac{1}{\\frac{32}{34.93}}\\right)<\/span>\n\n\n y = \\frac{36(34.93)}{71.36(32)} z = 0.55067 z\/<\/span>\n\n\n\n

fila = 2<\/strong><\/p>\n\n\n \\frac{14}{56.75} x + \\frac{14}{34.93} y + \\frac{14}{71.36} z = 490 <\/span>\n\n\n \\frac{14}{71.36} z = 490 -\\frac{14}{56.75} x -\\frac{14}{34.93} y <\/span>\n\n\n z = \\left( 490 -\\frac{14}{56.75} x -\\frac{14}{34.93} y \\right) \\frac{1}{ \\frac{14}{71.36}}<\/span>\n\n\n z = \\left( 490 -\\frac{14}{56.75} x -\\frac{14}{34.93} y \\right) \\frac{71.36}{14}<\/span>\n\n\n\n

expresiones para el m\u00e9todo:<\/p>\n\n\n x = 0.86756 z<\/span>\n\n\n y = 0.55067 z <\/span>\n\n\n z = \\left( 490 -\\frac{14}{56.75} x -\\frac{14}{34.93} y \\right) (5.09714)<\/span>\n\n\n\n

literal c. Iteraciones Jacobi<\/h2>\n\n\n\n

Si la c\u00e1mara tiene un peso de 490, cuelga de 3 cables, en el mejor de los casos cada cable tendr\u00eda una tensi\u00f3n de 1\/3 del peso. Por lo que el vector inicial X0=[490\/3,490\/3,490\/3]<\/p>\n\n\n x = 0.86756 z<\/span>\n\n\n y = 0.55067 z <\/span>\n\n\n z = \\left( 490 -\\frac{14}{56.75} x -\\frac{14}{34.93} y \\right) (5.09714)<\/span>\n\n\n\n

itera = 0<\/strong><\/p>\n\n\n\n

X0=[490\/3,490\/3,490\/3]<\/p>\n\n\n x = 0.86756 (490\/3) = 141.7014<\/span>\n\n\n y = 0.55067 (490\/3) = 89.9427<\/span>\n\n\n z = \\left( 490 -\\frac{14}{56.75} (490\/3) -\\frac{14}{34.93} (490\/3) \\right) (5.09714)<\/span>\n\n\n z =1958.53<\/span>\n\n\n\n

X1 = [141.7014, 89.9427, 1958.5366]<\/p>\n\n\n\n

 X1:  [141.7014, 89.9427, 1958.5366]\n-X0: -[490\/3,    490\/3,   490\/3    ]\n____________________________________\ndif:  [21.63185  73.38956 1795.2033 ]\nerrado = max(abs(diferencia) = 1795.20<\/code><\/pre>\n\n\n\n
itera = 1<\/strong><\/p>\n\n\n x = 0.86756 (1958.5366)=1699.1483 <\/span>\n\n\n y = 0.55067 (1958.5366)= 1078.51941 <\/span>\n\n\n z = \\left( 490 -\\frac{14}{56.75} (141.7014) -\\frac{14}{34.93} (89.9427) \\right) (5.09714) <\/span>\n\n\n z=2135.66818 <\/span>\n\n\n\n
X2= [1699.1483, 1078.51941, 2135.66818]<\/p>\n\n\n\n
 X2:  [1699.1483, 1078.51941, 2135.66818]\n-X1: -[ 141.7014,   89.9427,  1958.5366 ]\n_________________________________________\ndif:  [1557.44681  988.57564   177.13155]\nerrado = max(abs(diferencia) = 1557.44681<\/code><\/pre>\n\n\n\n
itera = 2<\/strong><\/p>\n\n\n x = 0.86756 (2135.66818)=1852.82057<\/span>\n\n\n y = 0.55067 (2135.66818)=1176.06153<\/span>\n\n\n z = \\left( 490 -\\frac{14}{56.75} (1699.1483) -\\frac{14}{34.93} (1078.51941) \\right) (5.09714)<\/span>\n\n\n z =-1842.33913<\/span>\n\n\n\n
X3= [1852.82057, 1176.06153, -1842.33913]<\/p>\n\n\n\n
 X3:  [1852.82057, 1176.06153, -1842.33913]\n-X2: -[1699.1483,  1078.51941,  2135.66818]\n____________________________________________\ndif:  [ 153.67227   97.54212 3978.00731]\nerrado = max(abs(diferencia) = 3978.00731<\/code><\/pre>\n\n\n\n
literal d. convergencia<\/h2>\n\n\n\n
Se observa que el error aumenta en cada iteraci\u00f3n, por lo que el m\u00e9todo NO converge.<\/p>\n\n\n\n
En la tercera iteraci\u00f3n, la Tensi\u00f3n AD es negativa, lo que no tiene sentido en el contexto del ejercicio.<\/p>\n\n\n\n
literal e. N\u00famero de condici\u00f3n<\/h2>\n\n\n\n
el n\u00famero de condici\u00f3n calculado es:<\/p>\n\n\n\n
numero de condici\u00f3n: 3.254493400285106<\/code><\/pre>\n\n\n\n
literal f. resultados con algoritmo<\/h2>\n\n\n\n
los resultados de las iteraciones con el algoritmo son:<\/p>\n\n\n\n
Matriz aumentada\n[[ 9.6916299e-01  0.0000000e+00 -8.4080717e-01  0.000e+00]\n [ 2.4669603e-01  4.0080160e-01  1.9618834e-01  4.900e+02]\n [ 0.0000000e+00 -9.1611795e-01  5.0448430e-01  0.000e+00]]\nPivoteo parcial:\n  1 intercambiar filas:  1 y 2\nAB\nIteraciones Jacobi\nitera,[X]\n     ,errado,|diferencia|\n0 [163.33333333 163.33333333 163.33333333]\n  nan\n1 [ 141.70149   89.94377 1958.53663]\n  1795.203299403506 [  21.63185   73.38956 1795.2033 ]\n2 [1699.1483  1078.51941 2135.66818]\n  1557.446808619277 [1557.44681  988.57564  177.13155]\n3 [ 1852.82057  1176.06153 -1842.33913]\n  3978.007311178223 [ 153.67227   97.54212 3978.00731]\n4 [-1598.33998 -1014.53222 -2234.84654]\n  3451.1605418268064 [3451.16054 2190.59375  392.50741]\n5 [-1938.86376 -1230.67669  6580.05603]\n  8814.902564542654 [ 340.52378  216.14447 8814.90256]\n6 [5708.59426 3623.47991 7449.81676]\n  7647.458018820763 [7647.45802 4854.1566   869.76074]\n7 [  6463.164     4102.43641 -12083.20597]\n  19533.02272824792 [  754.56974   478.95649 19533.02273]\n8 [-10482.90774  -6653.93333 -14010.51669]\n  16946.071746250553 [16946.07175 10756.36974  1927.31073]\n9 [-12154.96569  -7715.25738  29272.88595]\n  43283.40263645846 [ 1672.05794  1061.32405 43283.40264]\n10 [25395.98875 16119.88011 33543.6313 ]\n  37550.95443771429 [37550.95444 23835.13748  4270.74536]\n11 [ 29101.11715  18471.67771 -62368.45408]\n  95912.08538760681 [ 3705.1284   2351.79761 95912.08539]\n12 [-54108.38416 -34344.82012 -71832.03755]\n  83209.50131084416 [83209.50131 52816.49783  9463.58346]\n13 [-62318.61187 -39556.18982 140700.42439]\n  212532.4619384469 [  8210.22771   5211.3697  212532.46194]\n14 [122066.07854  77480.36788 161670.86502]\n  184384.6904047115 [184384.6904  117036.5577   20970.44063]\n15 [ 140259.19675   89028.28937 -309281.7495 ]\n  470952.61452269484 [ 18193.11821  11547.92149 470952.61452]\n16 [-268320.51494 -170314.0828  -355750.33954]\n  408579.7116922584 [408579.71169 259342.37218  46468.59004]\nNo converge,iteramax superado\nMetodo de Jacobi\nnumero de condici\u00f3n: 3.254493400285106\nX:  nan\nerrado: 408579.7116922584\niteraciones: 16<\/code><\/pre>\n\n\n\n
las gr\u00e1ficas de las iteraciones son:<\/p>\n\n\n\n
\"c\u00e1mara<\/figure>\n\n\n\n
la gr\u00e1fica de errores por iteraci\u00f3n<\/p>\n\n\n\n
\"c\u00e1mara<\/figure>\n\n\n\n
 <\/p>\n","protected":false},"excerpt":{"rendered":"
Ejercicio:\u00a01Eva2025PAOI_T2 Cables de c\u00e1mara a\u00e9rea literal a. matriz aumentada y pivoteo parcial por filas Las ecuaciones que conforman el sistema de ecuaciones son: Se reordenan las ecuaciones para la forma A.X=B, con las constantes del lado derecho. la matriz A y vector B ser\u00e1n: Matriz Aumentada Pivoteo parcial por filas columna = 0, no requiere cambios, […]<\/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":[46],"tags":[58,54],"class_list":["post-11245","post","type-post","status-publish","format-standard","hentry","category-mn-s1eva30","tag-ejemplos-python","tag-mnumericos"],"_links":{"self":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/11245","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=11245"}],"version-history":[{"count":3,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/11245\/revisions"}],"predecessor-version":[{"id":23820,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/posts\/11245\/revisions\/23820"}],"wp:attachment":[{"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/media?parent=11245"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/categories?post=11245"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.espol.edu.ec\/algoritmos101\/wp-json\/wp\/v2\/tags?post=11245"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}