How are coefficients of equation of sideslip angle in the paper “A general Solution to the Aircraft Trim Problem” calculated?

I am sure many of you guys must have read the paper, "A General solution to the Aircraft Trim Problem" by Marco, Duke and Bernt. I am working with the turning of the Aircraft and I am not able to understand the calculation of Beta (sideslip angle) given in equation 72.

What I am able to understand is that the author describes the wind frame and body frame by 2 simple plane equations and then takes a dot product of the two planes to calculate beta.

The problem I am facing is with the calculation of the coefficients of the plane. They say that the coefficients can be expressed according to equation 67, which is nothing but axis transformation equations. Can anyone help me with the detailed calculation of those coefficients.

I have attached the above mentioned part as link to image file. Please have a look into it.

Link: A general solution to the Aircraft trim problem

• If you don't have an answer here in few days, you can try on Physics.SE. They are more used to read research paper and to deal with forces and frame of reference changes in Newtonian mechanics. – Manu H May 30 '20 at 14:32

First, some note on this paper: from my cursory reading, it always assumes zero atmsohperic wind relative to the ground. That's why the wind axis represents the true motion of the aircraft. In general, this is not true, therefore $$\psi_W$$ and $$\theta_W$$ are not the track and flight path angles.

Now onto your question. When you consider zero wind, then the angular difference between the y wind axis ($$\hat{j}_{W}$$) and the y body axis ($$\hat{j}_{B}$$) is the sideslip angle, $$\beta$$. This is consistent when you consider flow incidences as 3-2-1 Euler rotation from the body axis, where yaw (sideslip) is the last rotation:

$$C_{BW} = \begin{bmatrix}\cos\alpha\cos\beta & -\cos\alpha\sin\beta & -\sin\alpha \\ \sin\beta & \cos\beta & 0 \\ \sin\alpha\cos\beta & -\sin\alpha\sin\beta & \cos\alpha \end{bmatrix}$$

Mathematically, this can be immediately seen when you rewrite the rotation matrix in its equivalent direction cosine form:

$$C_{BW} = \begin{bmatrix}\hat{i}_W \cdot \hat{i}_B & \hat{j}_W \cdot \hat{i}_B & \hat{k}_W \cdot \hat{i}_B \\ \hat{i}_W \cdot \hat{j}_B & \hat{j}_W \cdot \hat{j}_B & \hat{k}_W \cdot \hat{j}_B \\ \hat{i}_W \cdot \hat{k}_B & \hat{j}_W \cdot \hat{k}_B & \hat{k}_W \cdot \hat{k}_B\end{bmatrix}$$

So: $$\cos{\beta}=\hat{j}_W \cdot \hat{j}_B$$.

The gradient of the Cartesian equation of a plane is its unit normal, when normalized. For the $$x_W z_W$$ and $$x_B z_B$$ planes, the normal is simply their y axis (i.e. $$\hat{j}_W$$ and $$\hat{j}_B$$, respectively.

Lastly, eqn (72) can be worked out when $$\begin{bmatrix}x & y & z\end{bmatrix}^T$$ in the planar equations are in the inertial frame ($$_V$$), and transformed to the respective frames. The coefficients can then be derived and massaged to obtain (72).

• The angular difference between wind axis and body axis is the side-slip angle even if you consider non-zero wind. There is not much point in considering non-zero wind unless the aircraft is taking off or landing anyway—the ground is not relevant, so you don't ever need to calculate in the ground reference fame at all. – Jan Hudec May 30 '20 at 20:59
• @JanHudec Yeah, I've never said the winds make any difference to sideslip. They, however, make all the difference to the track and flight path angle, which receives a lot of attention in the paper. – JZYL May 30 '20 at 21:34
• Hey buddy thanks for the answer, but I already understood the part that you tried to explain , I am facing more issues in deriving the coefficients of eqn 72. Can you help me with the maths regarding that? Or just guide me through the steps? Thanks – jmoriarty Jun 1 '20 at 8:29