我想在我的双摆中有摆锤

Question

在这段代码中，我想要像 this. But I dont want the other pendulums that come into picture later. Just the initial one. Currently this is my output 这样的动画。这是动画完成后的图像。在动画中，我想要一个绘制红线的球（blob）和另一个绘制绿线的球。

import numpy as np
from numpy import cos, sin, arange, pi
import matplotlib.pyplot as plt
import matplotlib.animation as animation


h = 0.0002   #the change in runge kutta
figsize = 6
dpi = 1000
N = 200000 # iterations
L1=1    #length 1
L2=1.5  #lenth 2
m1=50  #mass of bob 1
m2=1    #mass of bob2
g = 9.81#gravity
theta_01 = (np.pi/180)*90
theta_02 = (np.pi/180)*60
w_1 = 0
w_2 = 0

# dw/dt function oft theta 1

def funcdwdt1(theta1,theta2,w1,w2):
    cos12 = cos(theta1 - theta2)#for wrirting the main equation in less complex manner
    sin12 = sin(theta1 - theta2)
    sin1 = sin(theta1)
    sin2 = sin(theta2)
    denom = cos12**2*m2 - m1 - m2
    ans = ( L1*m2*cos12*sin12*w1**2 + L2*m2*sin12*w2**2
            - m2*g*cos12*sin2      + (m1 + m2)*g*sin1)/(L1*denom)
    return ans

# dw/dt function oft thetas 2
    
def funcdwdt2(theta2,theta1,w1,w2):
    cos12 = cos(theta1 - theta2)
    sin12 = sin(theta1 - theta2)
    sin1 = sin(theta1)
    sin2 = sin(theta2)
    denom = cos12**2*m2 - m1 - m2
    ans2 = -( L2*m2*cos12*sin12*w2**2 + L1*(m1 + m2)*sin12*w1**2
            + (m1 + m2)*g*sin1*cos12  - (m1 + m2)*g*sin2 )/(L2*denom)
    return  ans2

# d0/dt function for theta 1

def funcd0dt1(w0):
    return w0

# d0/dt function for theta 2
    
def funcd0dt2(w0):
    return w0


X1= []
X2= []
Y1= []
Y2= []

def func(w1,w2, theta1,theta2): 
    for i in range(N):
        k1a = h * funcd0dt1(w1)  # gives theta1
        k1b = h * funcdwdt1(theta1,theta2,w1,w2)  # gives omega1
        k1c = h * funcd0dt2(w2)  # gives theta2
        k1d = h * funcdwdt2(theta2,theta1,w1,w2)   # gives omega2

        k2a = h * funcd0dt1(w1 + (0.5 * k1b))
        k2b = h * funcdwdt1(theta1 + (0.5 * k1a),theta2,w1,w2)
        k2c = h * funcd0dt2(w2 + (0.5 * k1d))
        k2d = h * funcdwdt2(theta2 + (0.5 * k1c),theta1,w1,w2)

        k3a = h * funcd0dt1(w1 + (0.5 * k2b))
        k3b = h * funcdwdt1(theta1 + (0.5 * k2a),theta2,w1,w2)
        k3c = h * funcd0dt2(w2 + (0.5 * k2d))
        k3d = h * funcdwdt2(theta2 + (0.5 * k2c),theta1,w1,w2)

        k4a = h * funcd0dt1(w1 + k3b)
        k4b = h * funcdwdt1(theta1 + k3a,theta2,w1,w2)
        k4c = h * funcd0dt2(w2 + k3d)
        k4d = h * funcdwdt2(theta2 + k3c,theta1,w1,w2)

        #addidng the vakue aftyer the iterartions
        theta1 += 1 / 6 * (k1a + 2 * k2a + 2 * k3a + k4a)  
        w1 +=1 / 6 * (k1b + 2 * k2b + 2 * k3b + k4b)             
        theta2 += + 1 / 6 * (k1c + 2 * k2c + 2 * k3c + k4c)     
        w2 += 1 / 6 * (k1d + 2 * k2d + 2 * k3d + k4d)
        x1 = L1 * sin(theta1)
        y1 = -L1 * cos(theta1)
        x2 = x1 + L2 * sin(theta2)
        y2 = y1 - L2 * cos(theta2)
        X1.append(x1)
        X2.append(x2)
        Y1.append(y1)
        Y2.append(y2)    
    return x1,y1,x2,y2


print(func(w_1, w_2, theta_01, theta_02))

fig, ax = plt.subplots()
l1, = ax.plot([], [])
l2, = ax.plot([],[])
ax.set(xlim=(-3, 3), ylim=(-2,2))

def animate(i):
    l1.set_data(X1[:i], Y2[:i])
    l2.set_data(X2[:i], Y2[:i])
    return l1,l2,

ani = animation.FuncAnimation(fig, animate, interval = 5, frames=len(X1))
# plt.show()
ani.save('save.mp4', writer='ffmpeg')

Answer 1

再添加一行

l3, = ax.plot([],[], '-ob', lw=2, ms=8)

并在动画函数中将其值设置为

    l3.set_data([0,X1[i],X2[i]], [0,Y1[i],Y2[i]])

根据需要调整线宽和标记大小。这应该在摆的位置和原点绘制实心圆，并用直线连接它们。

您应该在 l1 数据中使用 Y1。摆的总长度为 2.5，垂直限制太小。使用

就足够了

h = 0.005   #the change in runge kutta
N = 5000 # iterations

以逼真的速度制作动画。或者为每一帧组合几个 RK4 步骤。对于最小误差，您可以使用 h=1e-3，较小的步长只会导致浮点误差的累积，从而主导方法误差。

我想在我的双摆中有摆锤

I want to have the pendulum blob in my double pendulum

python

simulation

animation

matplotlib

matplotlib-animation