I am attempting to simulate a double spherical pendulum, i.e. a combination of the spherical pendulum and the double pendulum.

I understand that the equations of motion can be derived via the Lagrangian and the Euler-Lagrange equations. However this method rapidly becomes very messy.

Is there an alternate method that could be used to simplify the calculations required?


This is an update to specify the Lagrangian of the system as requested in the comments.

We have two spherical pendulums, denoted $1$ and $2$. The lengths of each pendulum $l_1=l_2=1$ whilst the masses $m_1=m_2=1$.

There are two parameters that describe the location of the pendulum mass $\theta$, the angle from the vertical, and $\phi$ the azimuthal angle about the vertical axis. For a diagram see here.

The locations of the masses $r_1 , r_2$ are therefore given by (for a vertical $z$ axis): $$x_1 = sin(\theta_1)cos(\phi_1)$$ $$y_1 = sin(\theta_1)sin(\phi_1)$$ $$z_1 = -cos(\theta_1)$$


$$x_2 = x_1+sin(\theta_2)cos(\phi_2)$$ $$y_2 = y_1+sin(\theta_2)sin(\phi_2)$$ $$z_2 = z_1-cos(\theta_2)$$

The Lagrangian is given by the difference between the kinetic and potential energies $L = E_k -E_p$

$$E_k = \frac{1}{2} \left( \dot{r}_1^2 +\dot{r}_2^2 \right)$$ $$E_p = g (z_1 +z_2)$$

The Lagrangian then follows simply from taking the time derivatives of $r_1$ and $r_2$. (Note that this leads me to a very long and complicated form the Lagrangian)

  • $\begingroup$ I am afraid that this problem actually IS very messy. Traditional way is to use the small amplitude approximation. $\endgroup$ Nov 20, 2015 at 10:54
  • 2
    $\begingroup$ Hi Tom. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$
    – Qmechanic
    Nov 20, 2015 at 11:10
  • 10
    $\begingroup$ ...this method rapidly becomes very messy. welcome to physics ;) $\endgroup$
    – Kyle Kanos
    Nov 20, 2015 at 12:30
  • $\begingroup$ Thanks for all comments. Qmechanic have read the policy now and so accept your edit :). I am happy to accept these comments as an answer. What is the protocol in this situation? Thanks all $\endgroup$
    – Tom
    Nov 20, 2015 at 13:24
  • $\begingroup$ @Tom If you can't be bothered to go through the derivation of the e.o.m. yourself (it's not that tedious), you can do it easily in a program like Mathematica. $\endgroup$
    – JamalS
    Nov 20, 2015 at 16:02

1 Answer 1


Not that I know of. However if you're fine with considering only small oscillations, then you can replace $\sin \theta$ by $\theta$ and $\cos \theta$ by $1-\frac{\theta^2}{2}$. This might make things simpler although the solution you get will be acceptable for small angles.

  • $\begingroup$ This is explained here $\endgroup$
    – Leander
    Sep 30, 2019 at 9:12

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