I've recently enrolled in a Coursera aerial robotics course and i'm trying to understand how the rate of change of the angular momentum is computed from the inertia frame to the body-fixed frame.

According to the course:

enter image description here

From this picture, we can see that the rate of change of the angular momentum in the inertial frame (denoted with the super-script A) of the center of mass (denoted with the sub-script C) is equal to the net moment generated by a quad-rotor.

By computing this quantity in the body-fixed frame, this is translated to the rate of change of the angular momentum in the body-fixed frame (denoted with super-script B) plus a correction factor.

This correction factor is equal to the cross product of the angular velocity of the body B (denoted with the super-script B) in the inertial frame (denoted with the super-script A) with the angular momentum itself.

The professor talks about this correction factor as a well-known correction factor used in mechanics / kinematics and this i can't find anywhere in the net and doesn't seem to be very obvious to me.

Can you explain me the origin of that correction factor or maybe give me a resource i can document myself on ?

Thanks in advance.


1 Answer 1


The cross-product correction factor shows up any time we take the time derivative of a vector described in a moving frame.

When I teach this topic in my classes, I generally introduce it when talking about taking time derivatives of position vectors, because it's easier to visualize in that situation:

Start by considering the positions of two points $A$ and $B$, denoted by vectors $\vec{r}_{A}$ and $\vec{r}_{B}$. We can then take the relative position of $B$ with respect to $A$ as a vector $\vec{r}_{B/A} = \vec{r}_{B}-\vec{r}_{A}$ such that $\vec{r}_{B} = \vec{r}_{A} + \vec{r}_{B/A}$.

Now let's say that the vector $\vec{r}_{B/A}$ is fundamentally defined in a frame whose orientation is encoded by a rotation matrix $R_{AB}$, such that $\vec{r}_{B/A}=R_{AB}\ \vec{r}_{B/A}'$ (where $'$ indicates a vector defined in the rotated frame).

If we take the time derivative of the position of $B$, we then get \begin{equation} \frac{d}{dt} \vec{r}_{B} = \frac{d}{dt} \vec{r}_{A} + \left(\frac{d}{dt} R_{AB}\right)\ \vec{r}_{B/A}' + R_{AB} \left(\frac{d}{dt} \vec{r}_{B/A}'\right). \end{equation}

The time derivative of a rotation matrix is the product of its rotational velocity with itself, $\frac{d}{dt} R_{AB} = \vec{\Omega}_{AB} \times R_{AB}$, such that the total time derivative can be regrouped as $$ \frac{d}{dt} \vec{r}_{B} = \frac{d}{dt} \vec{r}_{A} + \vec{\Omega}_{AB} \times \left(\vphantom{\frac{d}{dt}}R_{AB}\ \vec{r}_{B/A}'\right) + R_{AB} \left(\frac{d}{dt} \vec{r}_{B/A}'\right). $$ If you think of $A$ and $B$ as being points on a piece of paper, the first term in the expansion says how the paper is moving at point $A$, which transfers to how it is moving at point $B$. The second term describes how rotation of the paper makes point $B$ orbit around $A$, and the third term describes how point $B$ moves on the paper relative to point $A$.

For your angular momentum problem, you can think of the angular momentum as a "free vector", i.e., saying that we only want the $\vec{r}_{B/A}$ component and not the $\vec{r}_{A}$ component. When we take the time derivative, we then get the latter two terms from the full equation, and thus the "time derivative as seen in the frame" plus the "frame-rotation cross product" terms.

In the standard expression from within the body frame, we then multiply all the terms by ${R}^{-1}_{AB}$. By a bit of algebra that I won't go into right now, it works out that $$ {R}^{-1}_{AB}\left[\vec{\Omega}_{AB} \times \left(\vphantom{\frac{d}{dt}}R_{AB}\ \vec{r}_{B/A}'\right)\right] = \vec{\Omega}_{AB}' \times \vec{r}_{B/A}' $$ so that $$ {R}^{-1}_{AB} \left(\frac{d}{dt} \vec{r}_{B}\right) = {R}^{-1}_{AB}\left(\frac{d}{dt} \vec{r}_{A}\right) + \vec{\Omega}_{AB}' \times \vec{r}_{B/A}' +\frac{d}{dt} \vec{r}_{B/A}', $$ in which the last two terms are the standard expression for the derivative of a free vector expressed in a rotating set of coordinates.

  • $\begingroup$ Very clear. Thanks you so much and sorry for the late reply $\endgroup$
    – Mssm
    Feb 27, 2021 at 8:09

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