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Question

假设您有4个微分方程，它们代表动态系统的运动（摆锤），并使用 scipy.integrate.odeint 求解了这些方程10秒，间隔为0.01秒。

最后，您得到大小为（1000,4）的解矩阵。对于每个差异方程，您将获得1000个数据点。到目前为止一切都还好。例如，如果我绘制其中一种运动，则可以获得漂亮的图形。（下图显示了摆杆的运动（摆动））

但是，我不想制作无聊的图形，而是想制作一个动画来显示购物车的运动，就像史蒂夫·布伦顿（Steve Brunton）所做的那样，如下所示使用Matlab链接。这是link of the cart-pend video！

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为了使数字动画，我实际上试图用Python做史蒂夫·布鲁顿（Steve Brunton）在Matlab中所做的事情。但是结果只是冻结的数字，而不是移动的数字。实际上，如果我从Spyder IDE运行此脚本，我将在IPython控制台中获得1000个图形。（每个图形代表系统瞬时运动的快照，这是很好的。但是我只想要一个图形，上面有1000个固定帧。）

这里是snap of frozen cart-pend

我写了两个python脚本。一个仅用于绘制另一个图是解决差分方程并将结果馈送到另一个图。

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此代码用于绘制动画人物。

from math import sqrt, sin, cos
import matplotlib.pyplot as plt
from matplotlib import animation

def draw_cart(states, m, M, L):

    x = states[0]       # Position of the center of the cart
    theta = states[3]   # Angle of the pendulum rod

    #Dimensions
    W = 1*sqrt(M/5)     # Cart width
    H = .5*sqrt(M/5)    # Cart Height
    wr = .2             # Wheel radius
    mr = .3*sqrt(m)     # Mass Radius

    #Positions

    y = wr/2+ H/2       # Cart Vertical Position

    w1x = x-.9*W/2      # Left Wheel x coordinate
    w1y = 0             # Left wheel y coordinate
    w2x = x+(.9*W/2)    # Right Wheel x coordinate
    w2y = 0             # Right Wheel y coordinate

    # Pendulum Mass x-y coordinates
    px = x+(L*sin(theta))
    py = y-(L*cos(theta))



    #Identfying Figure
    plt.figure()
    plt.axes(xlim=(-5, 5), ylim=(-2, 2.5))

    # Plotting the base line
    line = plt.Line2D((-10, 10), (0, 0), color='k', linewidth=2)
    plt.gca().add_line(line)
    plt.hold(True)


    # Shapes
    rectangle1 = plt.Rectangle((x-(W/2), (y-H/2)), W, H, fill=True, color='b') # Cart

    rectangle2= plt.Rectangle((px-(mr/2), py-(mr/2)), mr, mr, fill=True, color='r') # Pendulum mass

    circle2 = plt.Circle((w1x, w1y), wr/2, fill=True, color='g') #Left whell

    circle3 = plt.Circle((w2x, w2y), wr/2, fill=True, color='g')  #Right whell

    plt.plot((x, px), (y, py), 'k', lw=2) #Pendulum rod

    #Adding shapes to the figure
    plt.gca().add_patch(rectangle1)
    plt.gca().add_patch(rectangle2)
    plt.gca().add_patch(circle2) 
    plt.gca().add_patch(circle3) 

    # Showing the figure
    plt.show()

    plt.hold(False)

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这是用于解决差异方程并将解决方案提供给上述代码的其他代码。

from math import pi, sin, cos
import numpy as np
from scipy.integrate import odeint
import draw_cart_pend_rt
import matplotlib.pyplot as plt

# System Parameters
m = 1
M = 5
L = 2
g = -10
d = 1
u = 0


def cart_pend_dynamics(states, tspan):

    Sy = sin(states[2])
    Cy = cos(states[2])
    D = m*L*L*(M+(m*(1-(Cy**2))))

    state_derivatives = np.zeros_like(states)

    state_derivatives[0] = states[1]
    state_derivatives[1] = ((1/D)*(((-m**2)*(L**2)*g*Cy*Sy)+(m*(L**2)*(m*L*(states[3]**2)*Sy-d*(states[1])))))+(m*L*L*(1/D)*u)
    state_derivatives[2] = states[3]
    state_derivatives[3] = ((1/D)*((m+M)*m*g*L*Sy-m*L*Cy*(m*L*(states[3])**2*Sy-d*states[1])))-(m*L*Cy*(1/D)*u)+(0.01*1)

    return state_derivatives



def solution_of_cartpend(dt):

    # Initial conditions to solve diff eqs
    states = np.array([0.0, 0.0, pi, 0.5])  # Left to right, cart; position-velocity, pend mass; angle-angular velocity

    tspan = np.arange(0, 10, dt)

    state_sol = odeint(cart_pend_dynamics, states, tspan)

    return state_sol


# Time Interval
dt = 0.01

solution = solution_of_cartpend(dt)

x_den, y_den = solution.shape



# Validating the solution
plt.axes(xlim=(0,10), ylim=(-10,10))
t = np.arange(0, 10, dt)
plt.gca().plot(t, (solution[:, 2]), 'b', label='theta1')

# Animating the figures
for i in range(x_den):

    draw_cart_pend_rt.draw_cart(solution[i,:], m, M, L)

如何在Python中为动态系统的运动设置动画？

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0 个答案: