This is a double pendulum simulation. Small angles behave almost like a linear system; large angles are nonlinear and chaotic.
Press Space to restart with a new random start.
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Edit the parameters in main.c:
L1, L2, g are constants.m1, m2, and the initial angles are set in init_solver().
Assumptions: the rods are massless and rigid, and the masses are point masses. Below is a direct Newtonian derivation of the equations of motion.
Kinematics of the Double Pendulum Kinematics describes position, velocity, and acceleration without considering forces.
Place the origin at the fixed pivot. Angles $\theta_1, \theta_2$ are measured from vertical (counter-clockwise positive). $T_1, T_2$ are rod tensions.
Positions:
$$x_1 = L_1\sin\theta_1, \qquad y_1 = -L_1\cos\theta_1$$
$$x_2 = x_1 + L_2\sin\theta_2, \qquad y_2 = y_1 - L_2\cos\theta_2$$
Differentiating positions twice:
$$\ddot{x}_1 = -\dot\theta_1^2 L_1\sin\theta_1 + \ddot\theta_1 L_1\cos\theta_1 \tag{1}$$
$$\ddot{y}_1 = \dot\theta_1^2 L_1\cos\theta_1 + \ddot\theta_1 L_1\sin\theta_1 \tag{2}$$
$$\ddot{x}_2 = \ddot{x}_1 - \dot\theta_2^2 L_2\sin\theta_2 + \ddot\theta_2 L_2\cos\theta_2 \tag{3}$$
$$\ddot{y}_2 = \ddot{y}_1 + \dot\theta_2^2 L_2\cos\theta_2 + \ddot\theta_2 L_2\sin\theta_2 \tag{4}$$
Upper mass $m_1$ — acted on by $T_1$ (upper rod), $T_2$ (lower rod), and gravity:
$$m_1\ddot{x}_1 = -T_1\sin\theta_1 + T_2\sin\theta_2 \tag{5}$$
$$m_1\ddot{y}_1 = T_1\cos\theta_1 - T_2\cos\theta_2 - m_1 g \tag{6}$$
Lower mass $m_2$ — acted on by $T_2$ and gravity only:
$$m_2\ddot{x}_2 = -T_2\sin\theta_2 \tag{7}$$
$$m_2\ddot{y}_2 = T_2\cos\theta_2 - m_2 g \tag{8}$$
From (7) and (8): $T_2\sin\theta_2 = -m_2\ddot{x}_2$ and $T_2\cos\theta_2 = m_2\ddot{y}_2 + m_2 g$ .
Substituting into (5) and (6):
$$m_1\ddot{x}_1 = -T_1\sin\theta_1 - m_2\ddot{x}_2 \tag{9}$$
$$m_1\ddot{y}_1 = T_1\cos\theta_1 - m_2\ddot{y}_2 - (m_1+m_2)g \tag{10}$$
To eliminate $T_1$ , multiply (9) by $\cos\theta_1$ and (10) by $\sin\theta_1$ . Both give the same left-hand side $T_1\sin\theta_1\cos\theta_1$ , so setting them equal:
$$\sin\theta_1\bigl(m_1\ddot{y}_1 + m_2\ddot{y}_2 + (m_1+m_2)g\bigr) = -\cos\theta_1\bigl(m_1\ddot{x}_1 + m_2\ddot{x}_2\bigr) \tag{11}$$
Similarly from (7) and (8), multiply by $\cos\theta_2$ and $\sin\theta_2$ and set equal:
$$\sin\theta_2\bigl(m_2\ddot{y}_2 + m_2 g\bigr) = -\cos\theta_2\bigl(m_2\ddot{x}_2\bigr) \tag{12}$$
Substituting Accelerations Plug equations (1–4) into (11) and (12). Using the identity $\sin A\cos B - \cos A\sin B = \sin(A-B)$ , cross terms combine. Let $\Delta = \theta_1 - \theta_2$ .
From (12):
$$\ddot\theta_2, L_2 + \ddot\theta_1, L_1\cos\Delta + \dot\theta_1^2 L_1\sin\Delta + g\sin\theta_2 = 0 \tag{13}$$
From (11):
$$(m_1+m_2)\ddot\theta_1, L_1 + m_2\ddot\theta_2, L_2\cos\Delta - m_2\dot\theta_2^2 L_2\sin\Delta + (m_1+m_2)g\sin\theta_1 = 0 \tag{14}$$
Rewrite as a matrix equation:
$$\begin{pmatrix}(m_1+m_2)L_1 & m_2 L_2\cos\Delta \ L_1\cos\Delta & L_2\end{pmatrix}\begin{pmatrix}\ddot\theta_1\\ddot\theta_2\end{pmatrix} = \begin{pmatrix}m_2 L_2\dot\theta_2^2\sin\Delta - (m_1+m_2)g\sin\theta_1 \ -L_1\dot\theta_1^2\sin\Delta - g\sin\theta_2\end{pmatrix}$$
Applying Cramer's rule. The determinant simplifies using $\cos(2\Delta) = 2\cos^2\Delta - 1$ , giving a shared denominator:
$$D = 2m_1 + m_2 - m_2\cos(2\Delta)$$
$$\ddot\theta_1 = \frac{-g(2m_1+m_2)\sin\theta_1\ -\ m_2 g\sin(\theta_1-2\theta_2)\ -\ 2\sin\Delta\cdot m_2\bigl(\dot\theta_2^2 L_2 + \dot\theta_1^2 L_1\cos\Delta\bigr)}{L_1\cdot D}$$
$$\ddot\theta_2 = \frac{2\sin\Delta\bigl(\dot\theta_1^2 L_1(m_1+m_2)\ +\ g(m_1+m_2)\cos\theta_1\ +\ \dot\theta_2^2 L_2 m_2\cos\Delta\bigr)}{L_2\cdot D}$$
where $\Delta = \theta_1 - \theta_2$ and $D = 2m_1 + m_2 - m_2\cos(2\Delta)$ .
A big thanks to the resources made available from MIT (https://www.myphysicslab.com/pendulum/double-pendulum-en.html ) and Claude for helping me solve the very horrible 2x2 matrix to get the required equations for the motion of the pendulum.
