In classical mechanics, a double pendulum is a pendulum attached to the end of another pendulum. Its equations of motion are often written using the Lagrangian formulation of mechanics and solved numerically, which is the approach taken here. The dynamics of the double pendulum are chaotic and complex, as illustrated below.

The code:

import sys import numpy as np from scipy.integrate import odeint import matplotlib.pyplot as plt from matplotlib.patches import Circle # Pendulum rod lengths (m), bob masses (kg). L1, L2 = 1, 1 m1, m2 = 1, 1 # The gravitational acceleration (m.s-2). g = 9.81 def deriv(y, t, L1, L2, m1, m2): """Return the first derivatives of y = theta1, z1, theta2, z2.""" theta1, z1, theta2, z2 = y c, s = np.cos(theta1-theta2), np.sin(theta1-theta2) theta1dot = z1 z1dot = (m2*g*np.sin(theta2)*c - m2*s*(L1*z1**2*c + L2*z2**2) - (m1+m2)*g*np.sin(theta1)) / L1 / (m1 + m2*s**2) theta2dot = z2 z2dot = ((m1+m2)*(L1*z1**2*s - g*np.sin(theta2) + g*np.sin(theta1)*c) + m2*L2*z2**2*s*c) / L2 / (m1 + m2*s**2) return theta1dot, z1dot, theta2dot, z2dot def calc_E(y): """Return the total energy of the system.""" th1, th1d, th2, th2d = y.T V = -(m1+m2)*L1*g*np.cos(th1) - m2*L2*g*np.cos(th2) T = 0.5*m1*(L1*th1d)**2 + 0.5*m2*((L1*th1d)**2 + (L2*th2d)**2 + 2*L1*L2*th1d*th2d*np.cos(th1-th2)) return T + V # Maximum time, time point spacings and the time grid (all in s). tmax, dt = 30, 0.01 t = np.arange(0, tmax+dt, dt) # Initial conditions: theta1, dtheta1/dt, theta2, dtheta2/dt. y0 = np.array([3*np.pi/7, 0, 3*np.pi/4, 0]) # Do the numerical integration of the equations of motion y = odeint(deriv, y0, t, args=(L1, L2, m1, m2)) # Check that the calculation conserves total energy to within some tolerance. EDRIFT = 0.05 # Total energy from the initial conditions E = calc_E(y0) if np.max(np.sum(np.abs(calc_E(y) - E))) > EDRIFT: sys.exit('Maximum energy drift of {} exceeded.'.format(EDRIFT)) # Unpack z and theta as a function of time theta1, theta2 = y[:,0], y[:,2] # Convert to Cartesian coordinates of the two bob positions. x1 = L1 * np.sin(theta1) y1 = -L1 * np.cos(theta1) x2 = x1 + L2 * np.sin(theta2) y2 = y1 - L2 * np.cos(theta2) # Plotted bob circle radius r = 0.05 # Plot a trail of the m2 bob's position for the last trail_secs seconds. trail_secs = 1 # This corresponds to max_trail time points. max_trail = int(trail_secs / dt) def make_plot(i): # Plot and save an image of the double pendulum configuration for time # point i. # The pendulum rods. ax.plot([0, x1[i], x2[i]], [0, y1[i], y2[i]], lw=2, c='k') # Circles representing the anchor point of rod 1, and bobs 1 and 2. c0 = Circle((0, 0), r/2, fc='k', zorder=10) c1 = Circle((x1[i], y1[i]), r, fc='b', ec='b', zorder=10) c2 = Circle((x2[i], y2[i]), r, fc='r', ec='r', zorder=10) ax.add_patch(c0) ax.add_patch(c1) ax.add_patch(c2) # The trail will be divided into ns segments and plotted as a fading line. ns = 20 s = max_trail // ns for j in range(ns): imin = i - (ns-j)*s if imin < 0: continue imax = imin + s + 1 # The fading looks better if we square the fractional length along the # trail. alpha = (j/ns)**2 ax.plot(x2[imin:imax], y2[imin:imax], c='r', solid_capstyle='butt', lw=2, alpha=alpha) # Centre the image on the fixed anchor point, and ensure the axes are equal ax.set_xlim(-L1-L2-r, L1+L2+r) ax.set_ylim(-L1-L2-r, L1+L2+r) ax.set_aspect('equal', adjustable='box') plt.axis('off') plt.savefig('frames/_img{:04d}.png'.format(i//di), dpi=72) plt.cla() # Make an image every di time points, corresponding to a frame rate of fps # frames per second. # Frame rate, s-1 fps = 10 di = int(1/fps/dt) fig = plt.figure(figsize=(8.3333, 6.25), dpi=72) ax = fig.add_subplot(111) for i in range(0, t.size, di): print(i // di, '/', t.size // di) make_plot(i)

The images are saved to a subdirectory, `frames/`

and can be converted into an animated gif, for example with ImageMagick's `convert`

utility.

The derivation of the double pendulum equations of motion using the Lagrangian formulation has become a standard exercise in introductory classical mechanics, but an outline is given below. There are many, many similar derivations on the internet. We use the following coordinate system:

The two degrees of freedom are taken to be $\theta_1$ and $\theta_2$, the angle of each pendulum rod from the vertical. The components of the bob positions and velocities are

\begin{align*}
x_1 & = l_1\sin\theta_1 & \dot{x}_1 &= l_1\dot{\theta}_1\cos\theta_1\\
y_1 & = -l_1\cos\theta_1 & \dot{y}_1 &= l_1\dot{\theta}_1\sin\theta_1\\
x_2 & = l_1\sin\theta_1 + l_2\sin\theta_2 & \dot{x}_2 &= l_1\dot{\theta}_1\cos\theta_1 + l_2\dot{\theta}_2\cos\theta_2\\
y_2 & = -l_1\cos\theta_1 - l_2\cos\theta_2 & \dot{y}_2 &= l_1\dot{\theta}_1\sin\theta_1 + l_2\dot{\theta}_2\sin\theta_2
\end{align*}

The potential and kinetic energies are then

\begin{align*}
V &= m_1gy_1 + m_2gy_2 = -(m_1 + m_2)l_1g\cos\theta_1 - m_2l_2g\cos\theta_2\\
T &= \tfrac{1}{2}m_1v_1^2 + \tfrac{1}{2}m_2v_2^2 = \tfrac{1}{2}m_1(\dot{x}_1^2 + \dot{y}_1^2) + \tfrac{1}{2}m_2(\dot{x}_2^2 + \dot{y}_2^2)\\
&= \tfrac{1}{2}m_1l_1^2\dot{\theta}_1^2 + \tfrac{1}{2}m_2[l_1^2\dot{\theta}_1^2 + l_2^2\dot{\theta}_2^2 + 2l_1l_2\dot{\theta}_1\dot{\theta}_2\cos(\theta_1 - \theta_2)]
\end{align*}

The Lagrangian, $\mathcal{L} = T - V$ is therefore:

\begin{align*}
\mathcal{L} = \tfrac{1}{2}(m_1+m_2)l_1^2\dot{\theta}_1^2 + \tfrac{1}{2}m_2l_2^2\dot{\theta}_2^2 + m_1l_1l_2\dot{\theta}_1\dot{\theta}_2\cos(\theta_1 - \theta_2) + (m_1+m_2)l_1g\cos\theta_1 + m_2gl_2\cos\theta_2.
\end{align*}

The Euler-Lagrange equations are:

\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial\mathcal{L}}{\partial \dot{q}_i}\right) - \frac{\partial \mathcal{L}}{\partial q_i} = 0 \quad \mathrm{for}\;q_i = \theta_1, \theta_2.
\end{align*}

For these coordinates, after some calculus and algebra, we get:

\begin{align*}
(m_1 + m_2)l_1 \ddot{\theta}_1 + m_2l_2\ddot{\theta}_2\cos(\theta_1 - \theta_2) + m_2l_2\dot{\theta}_2^2\sin(\theta_1 - \theta_2) + (m_1+m_2)g\sin\theta_1 &= 0,\\
m_2l_2\ddot{\theta}_2 + m_2l_1\ddot{\theta}_1\cos(\theta_1 - \theta_2) - m_2l_1\dot{\theta}_1^2\sin(\theta_1 - \theta_2) + m_2g\sin\theta_2 = 0.
\end{align*}

`scipy`

's ordinary differential equation solver, `integrate.odeint`

needs to work with systems of first-order differential equations, so let $z_1 \equiv \dot{\theta_1} \Rightarrow \ddot{\theta}_1 = \dot{z}_1$ and $z_2 \equiv \dot{\theta_2} \Rightarrow \ddot{\theta}_2 = \dot{z}_2$. After some rearranging, the following expressions for $\dot{z}_1$ and $\dot{z}_2$ are obtained:

\begin{align*}
\dot{z}_1 = \frac{m_2g\sin\theta_2\cos(\theta_1-\theta_2) - m_2\sin(\theta_1 - \theta_2)[l_1z_1^2\cos(\theta_1 - \theta_2) + l_2z_2^2] - (m_1+m_2)g\sin\theta_1}{l_1[m_1 + m_2\sin^2(\theta_1-\theta_2)]},\\
\dot{z}_2 = \frac{(m_1+m_2)[l_1z_1^2\sin(\theta_1-\theta_2) - g\sin\theta_2 + g\sin\theta_1\cos(\theta_1-\theta_2)]+m_2l_2z_2^2\sin(\theta_1-\theta_2)\cos(\theta_1-\theta_2)}{l_2[m_1 + m_2\sin^2(\theta_1-\theta_2)]}
\end{align*}

It is these equations which appear in the function `deriv`

in the code above.

## Comments

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## Matias Koskimies 1 year, 8 months ago

I would love to see how a chaos pendulum with the middle point (knee) fixed would look like.

Link | Reply## christian 1 year, 8 months ago

I'm not sure I follow you – if you fix both the first and second joints, then you have one pendulum. If you fix only the second ("knee"), you have two single pendulums? Or are you interested in the relative motion between the two bobs (theta2 - theta1)?

Link | Reply## Dave F 1 year, 6 months ago

The double pendulum animation at the top of the page starts off slowly then builds up its speed until it is moving very quickly before the animation resets. (On my browser, at least.) I assume it's written in Javascript.

Link | ReplyMy guess is that in discretizing the changes to the angles, a tiny change to the system's total energy in introduced with each discrete step, and this accumulates with time.

## christian 1 year, 6 months ago

Many thanks for noticing this: the code is in Python but there was a bug – a missing factor of cos(theta1-theta2) – which caused the energy to drift. I've corrected it now and added a check for energy conservation.

Link | Reply## Joel 1 year, 6 months ago

Nice.

Link | Reply## Andrey 4 months, 3 weeks ago

Can we slide the joint connecting the second pendulum (or "knee", as one commenter puts it) up and down the first pendulum? E.g., one corner case is that you align the joint connecting the second pendulum to the first with the joint of the first pendulum (the two pendulums are, in effect, independent pendulums).

Link | Reply## christian 4 months, 2 weeks ago

Hi Andrey,

Link | Replyyou can alter the values of L1 and L2 to effectively move the joint connecting the two pendulums: but this keeps the "knee" at the bob of the first pendulum. I think you're describing the situation where the knee is along the rod L1 between the top pivot and the first bob.

No doubt this could be investigated, but it requires a new analysis of the dynamics and an additional variable (the position of the knee).

## John 1 month, 3 weeks ago

Pretty cool simulation.

Link | ReplyHow would you account for friction in the joints?

I'm building a 3d printed double pendulum and wonder if I could use this simulation to model different bearing and their friction, rod lengths, and different rod masses.

Could you use your energy conservation fix to insert a small lose each time?

## christian 1 month, 2 weeks ago

I think you could account for friction by adding a dissipative term to the Euler-Lagrange equations... this would add some complexity to the solution, however.

Link | ReplyMy energy conservation fix was a fix to the implementation of the friction-free equations: I had a bug that I should have detected by checking for energy conservation. I don't think the bug itself was a good way of introducing a dissipation effect: in my case the energy increased...(!)

Don't forget that this is a chaotic system, so modelling in this way may not tell you very much about the precise long-term behaviour of an actual, physical double-pendulum.

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