# The relation between Euler-angle rate and body-axis rates?

A paper about quadcopter dynamics and control shows the relation between Euler-angle rate and body-axis rates as:  Q1: why are there $$R_{v2}^{b}(\dot{\phi})$$, $$R_{v1}^{v2}(\dot{\theta})$$, and $$R_{v}^{v1}(\dot{\psi})$$?
I understand these are rotational matrices, but how could we use the $$\dot{\phi}$$, $$\dot{\theta}$$ and $$\dot{\psi}$$ to replace $$\phi$$, $$\theta$$ and $$\psi$$?

Q2: is the eqn(7) the approximate representation of the relation between Euler-angle rate and body-axis rates? if YES, what are accurate representation?

Let me use the following graph to try to clarify your confusion, which is excerpted from Etkins, Dynamics of Flight: The body axis is rotated from the [flat] Earth axis via a series of Euler angle rotations, $$\psi$$ (heading), $$\theta$$ (pitch), $$\phi$$ (roll), in this sequence. In the nomenclature of the graph, we've rotated from the Earth axis: $$(x_E, y_E, z_E)=(x_1, y_1, z_1)$$ to the body axis: $$(x, y, z)$$.

Now angular velocity is always expressed in the body axis, so the roll rate $$p$$ is about $$x$$, the pitch rate $$q$$ is about $$y$$ and yaw rate $$r$$ is about $$z$$. Our goal is to relate an instantaneous change in Euler angles ($$\dot{\phi}$$, $$\dot{\theta}$$, $$\dot{\psi}$$) to the angular velocity ($$p$$, $$q$$, $$r$$). Since the rotation is instantaneous, it does not add to the existing Euler angles.

We start off backwards, with the last rotation, $$\dot{\phi}$$. This rotation is about $$x=x_3$$, which is the same unit vector as $$p$$. Therefore, no frame transformation is required. i.e. trivially $$\dot{\phi} = p$$. That's why your rotation matrix, $$R_{v2}^b$$, is identity.

Next, let's look at $$\dot{\theta}$$. This rotation is about the $$y_3=y_2$$ axis, which lags behind the body axis by a roll rotation ($$\phi$$). So to bring it to the body axis, we need to pre-multiply by $$R_{v1}^{v2}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos{\phi} & \sin{\phi} \\ 0 & -\sin{\phi} & \cos{\phi} \end{bmatrix}$$.

Last, $$\dot{\psi}$$ is about $$z_2=z_1$$, which lags behind by two rotations: $$R_{v1}^{v2}R_v^{v1}=\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos{\phi} & \sin{\phi} \\ 0 & -\sin{\phi} & \cos{\phi} \end{bmatrix}\begin{bmatrix} \cos{\theta} & 0 & -\sin{\theta} \\ 0 & 1 & 0 \\ \sin{\theta} & 0 & \cos{\theta} \end{bmatrix}$$.

When you add everything together, we have the kinematic relation:

$$\begin{bmatrix}p \\ q \\ r\end{bmatrix} = \begin{bmatrix}1 & 0 & -\sin{\theta} \\ 0 & \cos{\phi} & \sin{\phi}\cos{\theta} \\ 0 & -\sin{\phi} & \cos{\phi}\cos{\theta}\end{bmatrix} \begin{bmatrix}\dot{\phi} \\ \dot{\theta} \\ \dot{\psi}\end{bmatrix}$$

To answer your second question, no, this is not an approximation because the angular velocity is, by definition, an instantaneous change. Magic of calculus. The paper is incorrect to say that the rates are small; the magnitudes of the rates have no bearing on the derivation of the kinematic relation.

• Your answer completely answered my doubts and thank you Feb 2, 2021 at 14:41
• +1 for a most comprehensive answer. Sep 1, 2021 at 0:16
• this result contradict stengel.mycpanel.princeton.edu/Quaternions.pdf where the phi theta and psi dot are inversed with p, q, r, did I missed something ?
– yota
Sep 15, 2021 at 12:41
• @yota What do you mean inversed?
– JZYL
Sep 15, 2021 at 18:19
• I think the final formula here, with this matrix, should be [p q r] = THIS_MATRIX [phidot, thetadot, psidot], as we can find in the litterature (like in the link posted in my previous comment)
– yota
Sep 16, 2021 at 13:50