Can you tell be how to set the equations to describe the motion of this machine in movie "Contact":


This is gyroscope with gimbal, am I right?

  • 1
    $\begingroup$ It is just the gimbal. Normally the purpose of a gimbal is to allow a gyroscope to point in a fixed direction when the object holding the gimbal twists and turns. In the movie, the purpose of the gimbal is to look interesting, complicated, and scientific. No equations needed for that. $\endgroup$
    – mmesser314
    Apr 10, 2014 at 13:04

2 Answers 2


As pointed out in the comments, the 'machine' in the movie Contact is just the gimbal part without the gyroscope. The point of this gimbal is to allow the thing being held at the center to be completely free to rotate in all directions. So, if they are working well they have no effect on the system except in the translational degrees of freedom, and they therefore don't contribute to the equations of motion in the rotational degrees of freedom while keeping the object rigidly fixed in the translational degrees of freedom.

They often have a gyroscope placed in the center which acts as an absolute reference due to conservation of angular momentum. The idea is that if you mount one to a plane, for instance, you can know which way the plane is pointing regardless of your ability to see any external references.

If you want to experience what it is like to be inside one of these gimbals, there is a carnival ride which I often see (pictured below) based on putting you inside a three axis gimbal and setting you spinning.

Gyro Ride Pic


It is best to read a book on dynamics to get all the details you need, but here is an overview. I assume all the rotation axes intersect at the origin, and the payload center of mass is located at a distance $\vec{c}$ from the last gimbal (in local coordinates). The mass moment of inertia of the payload is $I_{body}=\begin{bmatrix}I_x&0&0\\0&I_y&0\\0&0&I_z\end{bmatrix}\\$ also on local coordinates (fixed to the last gimbal).

  1. Nomenclature The outside fixed frame is 0. The first rotating frame about global $x$-axis is 1, the second about the local $y$-axis is 2 and the inside frame rotating about the local $z$-axis is 3. The payload is attached to frame 3.
  2. Kinematics of Frames The 3×3 orientation matrix $E_i$, rotational velocity $\vec{\omega}_i$ and rotational acceleration $\vec{\alpha}_i$ vector is described in terms of the rotation angles $\theta_i$ of the three gimbal frames $i=1\ldots3$. Also their 1st and 2nd time derivatives are $\dot\theta_i$ and $\ddot\theta_i$.
    • Orientations $$ \begin{align} \vec{E}_1 & = {\rm Rot}(\hat{i},\theta_1)= \begin{bmatrix}1&0&0\\0&\cos\theta_1&-\sin\theta_1\\0&\sin\theta_1&\cos\theta_1\end{bmatrix} \\ \vec{E}_2 & = E_1 {\rm Rot}(\hat{j},\theta_2) \\ \vec{E}_3 & = E_2 {\rm Rot}(\hat{k},\theta_3) \\ \end{align} $$
    • Rot. Velocity $$ \begin{align} \vec{\omega}_1 = \hat{i} \dot\theta_1 = \begin{pmatrix} \dot\theta_1 \\ 0 \\ 0 \end{pmatrix} \\ \vec{\omega}_2 = \vec{\omega}_1 + E_1 \hat{j} \dot\theta_2 =\begin{pmatrix} \dot\theta_1 \\ \dot\theta_2\cos\theta_1 \\ \dot\theta_2\sin\theta_1 \end{pmatrix} \\ \vec{\omega}_3 = \vec{\omega}_2 + E_2 \hat{k} \dot\theta_3 =\begin{pmatrix} \dot\theta_1 + \theta_3 \sin\theta_2 \\ \dot\theta_2\cos\theta_1 - \dot\theta_3 \sin\theta_1 \cos\theta_2 \\ \dot\theta_2\sin\theta_1 + \dot\theta_3 \cos\theta_1 \cos\theta_2 \end{pmatrix} \\ \end{align} $$
    • Rot. Acceleration $$ \begin{align} \vec{\alpha}_1 = \hat{i} \ddot\theta_1 \\ \vec{\alpha}_2 = \vec{\alpha}_1 + E_1 \hat{j} \ddot\theta_2 + \vec{\omega}_2 \times \vec{\omega}_1 \\ \vec{\alpha}_3 = \vec{\alpha}_2 + E_2 \hat{k} \ddot\theta_3 + \vec{\omega}_3 \times \vec{\omega}_2\\ \end{align} $$
  3. Kinematics of Payload
    • Position of Center of Mass $$\vec{r}_C = E_3 \vec{c}$$
    • Linear Velocity of Center of Mass $$\vec{v}_C = \vec{\omega}_3 \times \vec{r}_C$$
    • Linear Acceleration of Center of Mass $$\vec{a}_C = \vec{\alpha}_3 \times \vec{r}_C + \vec{\omega}_3 \times \vec{v}_C $$
    • Mass moment of inertia tensor in world coordinates $$I_C = E_3 I_{body} E_3^\top$$ (with $^\top$ the matrix transpose operator)
  4. Forces and moments on frame 3
    • Joint Forces $\vec{F} = (F_x,F_y,F_z)$
    • Joint Moments $\vec{\tau} = (0,0,0)$
  5. Forces on payload
    • Gravity Forces $m \vec{g} = (0,-m g,0)$
  6. Sum of forces and moments at payload center of mass

    • Forces $$ \sum\vec{F}_C = \vec{F} + m \vec{g} $$
    • Moments $$ \sum\vec{\tau}_C = \vec{\tau} + (-\vec{r}_C) \times \vec{F} $$
  7. Equations of Motion of payload $$\begin{aligned} \sum \vec{F}_C & = m \vec{a}_C \\ \sum \vec{\tau}_C &= I_C \vec\alpha + \vec{\omega}_3\times I_C \vec{\omega}_3 \end{aligned} $$

  8. Solution There are 6 equations of motion (3 linear, 3 angular) and 6 unknowns (3 reaction forces $F_x$, $F_y$ and $F_z$ and 3 joint accelerations $\ddot{\theta}_i$.

As you can see above, there are no simple problems in robotics and 3D dynamics. Good luck.

  • $\begingroup$ What I am missing above is the torques at the joints which accelerate the gimbal. Their result goes into joint moments $\vec{\tau} = \hat{i} \tau_1 + E_1 \hat{j} \tau_2 + E_2 \hat{k} \tau_3$ $\endgroup$ Apr 12, 2014 at 15:33
  • $\begingroup$ Thanks for detail. I needed this. Is there quantum mechanical analogy of gimbal? $\endgroup$
    – vejn
    Apr 14, 2014 at 10:01

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