我试图解决一个系统的颂歌基本上类似于这一个,但多了一个Spring和阻尼器==> http://scipy-cookbook.readthedocs.io/items/CoupledSpringMassSystem.html
不过,我有一个小问题,因为我想实现的参数之一是时间相关的。我的第一个尝试是以下一个:
import scipy as sci
import numpy as np
import matplotlib.pyplot as plt
def bump(t):
if t <= (0.25 / 6.9):
return (0.075 * (1 - np.cos(np.pi * 8 * 6.9 * t)))
else:
return 0
def membre_droite(w, t, p):
x1,y1,x2,y2,x3,y3 = w
m1,m2,m3,k1,k2,k3,l1,l2,l3,c2,c3 = p
f = [y1,
(-k1 * (x1 - l1 - bump(t)) + k2 * (x2 - x1 - l2) + c2 * (y2 - y1)) / m1,
y2,
(-c2 * (y2 - y1) - k2 * (x2 - x1 - l2) + k3 * (x3 - x2 - l3) + c3 * (y3 - y2)) / m2,
y3,
(-c3 * (y3 - y2) - k3 * (x3 - x2 - l3)) / m3]
return f
# Initial values
x11 = 0.08
y11 = 0
x22 = 0.35
y22 = 0
x33 = 0.6
y33 = 0
# Parameters
m1 = 90
m2 = 4000
m3 = 105
k1 = 250000
k2 = 25000
k3 = 30000
l1 = 0.08
l2 = x22-x11
l3 = x33-x22
c2 = 2500
c3 = 850
# Initial paramaters regrouped + time array
time=np.linspace(0.0, 5, 1001)
w0 = [x11,y11,x22,y22,x33,y33]
p0 = [m1,m2,m3,k1,k2,k3,l1,l2,l3,c2,c3]
x1,y1,x2,y2,x3,y3 = sci.integrate.odeint(membre_droite, w0, time, args=(p0,)).T
plt.plot(time,x1,'b')
plt.plot(time,x2,'g')
plt.plot(time,x3,'r')
plt.plot(time,y2,'yellow')
plt.plot(time,y3,'black')
plt.xlabel('t')
plt.grid(True)
plt.legend((r'$x_1$', r'$x_2$', r'$x_3$', r'$y_2$', r'$y_3$'))
plt.show()
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我得到的错误是:
if t <= (0.25 / 6.9):
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
型
我已经寻找了类似的情况下,我遇到了这个主题==> Solving a system of odes (with changing constant!) using scipy.integrate.odeint?
然后,我尝试将代码调整为这种格式:
import scipy as sci
import numpy as np
import matplotlib.pyplot as plt
def bump(t):
if t <= (0.25 / 6.9):
return (0.075 * (1 - np.cos(np.pi * 8 * 6.9 * t)))
else:
return 0
def membre_droite(w, t, bump):
x1,y1,x2,y2,x3,y3 = w
f = [y1,
(-250000 * (x1 - x11 - bump(t)) + 25000 * (x2 - x1 - x22 + x11) + 2500 * (y2-y1)) / 90,
y2,
(-2500 * (y2 - y1) - 25000 * (x2 - x1 - x22 + x11) + 30000 * (x3 - x2 - x33 + x22) + 850 * (y3 - y2)) / 4000,
y3,
(-850 * (y3 - y2) - 30000 * (x3 - x2 - x33 + x22)) / 105]
return f
# Initial values
x11 = 0.08
y11 = 0
x22 = 0.35
y22 = 0
x33 = 0.6
y33 = 0
# Initial paramaters regrouped + time array
time = np.linspace(0.0, 5, 1001)
w0 = [x11,y11,x22,y22,x33,y33]
x1,y1,x2,y2,x3,y3 = sci.integrate.odeint(membre_droite, w0, time, args=(bump,)).T
plt.plot(time,x1,'b')
plt.plot(time,x2,'g')
plt.plot(time,x3,'r')
plt.plot(time,y2,'yellow')
plt.plot(time,y3,'black')
plt.xlabel('t')
plt.grid(True)
plt.legend((r'$x_1$', r'$x_2$', r'$x_3$', r'$y_2$', r'$y_3$'))
plt.show()
型
阅读前面的链接,它应该已经工作,但我得到另一个错误:
(-250000 * (x1 - x11 - bump(t)) + 25000 * (x2 - x1 - x22 + x11) + 2500 * (y2 - y1)) / 90,
TypeError: 'list' object is not callable
型
2条答案
按热度按时间xxhby3vn1#
变更为:
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和/或
型
完整编码:
型
x1c 0d1x的数据
hgncfbus2#
通过在每个时间步长设置其值来引入时间相关变量u,例如
u(t_0)= 1,u(t_1)= 1,u(t_2)= 0,u(t_3)= 0,…
然后使用
numpy.interp()
在这些值之间插值。下面是一个最小的例子:
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的数据