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我正在尝试使用 Tkinter 在 Python 中创建一个 GUI，该 GUI 使用 Runge-Kutta 方法绘制质量的轨道路径。我的 GUI 工作正常，但我的问题是，无论我向 GUI 输入什么值，它都只会绘制一条直线。我希望有人可以向我展示我在 GUI 中的功能有什么问题，以便它真正正确地绘制轨道。

def calcPath(self):
    M = float(self.entM.get())
    m = float(self.entm.get())
    G = 6.67e-11
    c = 3e8



    velocity = np.array([float(self.entvx.get()),float(self.entvy.get()),float(self.entvz.get())])
    pos = np.array([float(self.entx.get()), float(self.enty.get()), float(self.entz.get())])

    Force = lambda pos: (G*m*M/(pos**2))

    #assigning variables empty lists to append x, y and z values to
    a = [] 
    b = []
    c = []
    t = 0.0
    tf = float(self.enttf.get())
    dt = float(self.entdt.get())

    while t < tf: # creating a while loop to trace the path for a set period of time
        #using the Runge-kutta formula from the problem sheet to assign variables at different steps and half steps 
        k1v=(dt*Force(pos))/m
        k2v=(dt*Force(pos+(k1v/2.0)))/m 
        k3v=(dt*Force(pos+(k2v/2.0)))/m 
        k4v=(dt*Force(pos+k3v))/m
        velocity=velocity+(k1v/6.0)+(k2v/3.0)+(k3v/3.0)+(k4v/6.0) #calaculating the weighted average of the k values
        pos=pos+velocity*dt #velocity is not a function of space or time so it will be identical at all 4 k values

        a.append(pos[0]) # appending the lists for each vaiable to produce a plot
        b.append(pos[1])
        c.append(pos[2])
        t = t+dt # setting the time steps

    #generating the path plot figure and formatting it    
    ax = Axes3D(self.fig) #creating a 3D figure 
    ax.set_title("Path of planetary mass") #produces title on the figure
    ax.plot3D(a,b,c, color='darkviolet', label='Runge-kutta path method') 
    ax.set_xlabel('x')
    ax.set_ylabel('y')
    ax.set_zlabel('z')
    ax.legend(loc='lower left') #selecting the location of legend
    self.canvas.show()
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1 回答 1

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你的物理是错误的。你对运动方程的处理是错误的。

力计算为

Force = lambda pos: -G*m*M/(norm2(pos)**3)*pos

运动方程为

m·pos''(t) = 力(pos(t))

在哪里

pos''(t) = d²/dt² pos(t)

您必须将其转换为一阶系统

[ pos' , vel' ] = [ vel, Force(pos)/m ]

对于这个一阶系统，您现在可以应用 RK4 方法。

    k1p = dt *       vel
    k1v = dt * Force(pos        ) / m
    k2p = dt *      (vel+0.5*k1v) 
    k2v = dt * Force(pos+0.5*k1p) / m 
    k3p = dt *      (vel+0.5*k2v) 
    k3v = dt * Force(pos+0.5*k2p) / m 
    k4p = dt *      (vel+    k3v)
    k4v = dt * Force(pos+    k3p) / m
    pos = pos + (k1p+2*k2p+2*k3p+k4p)/6.0 
    vel = vel + (k1v+2*k2v+2*k3v+k4v)/6.0

注意k向量是如何在位置和速度之间交错的。

或者您可以使用自动执行正确的 RK4 实现

System = lambda pos, vel : vel, Force(pos)/m

然后在时间循环内

    k1p, k1v = System( pos           , vel           )
    k2p, k2v = System( pos+0.5*k1p*dt, vel+0.5*k1v*dt )
    k3p, k3v = System( pos+0.5*k2p*dt, vel+0.5*k2v*dt )
    k3p, k3v = System( pos+    k3p*dt, vel+    k3v*dt )
    pos = pos + (k1p+2*k2p+2*k3p+k4p)*dt/6.0 
    vel = vel + (k1v+2*k2v+2*k3v+k4v)*dt/6.0

如果你只想要数值解，即不必证明你可以实现 RK4，那么使用scipy.integrate.odeor scipy.integrate.odeint。

于 2016-12-01T21:47:09.647 回答