# How to derive the bank angle of an aircraft from its roll angle and pitch angle?

From Young (2017) (https://onlinelibrary.wiley.com/doi/book/10.1002/9781118534786) it is stated that we can define the bank angle ($$\Phi$$) of an aircraft as the angle between its Y body axis and the horizontal plane. He then states the following equivalence:

$$\sin(\Phi)=\sin(\phi)\cos(\theta)$$

where $$\theta$$ is the pitch angle of the aircraft and $$\phi$$ is its roll angle. I want to derive this equivalence and this is my attempt so far which yields an alternative expression:

I define a global axis system $$E=(O,X_e,Y_e,Z_e)$$ where $$O=\begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}$$ is the origin of the system, and $$X_e=\begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix}$$, $$Y_e=\begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix}$$ and $$Z_e=\begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix}$$ are orthonormal unit vectors which define North, East and 'down' respectively.

I also define an aircraft body axis system, $$B=(O,X_b,Y_b,Z_b)$$, whose starting orientation and position is coincident with $$E$$.

I first rotate $$B$$ about the $$Y_b$$ axis an angle $$\theta$$, the pitch rotation, and then I rotate this rotated body axis system about its new $$Xb$$ axis an angle $$\phi$$, the roll rotation. The relevant rotation matrices are:

$$R_Y(\theta)$$=$$\begin{bmatrix} \cos(\theta) & 0 & \sin(\theta) \\ 0 & 1 & 0 \\ -\sin(\theta) & 0 & \cos(\theta) \end{bmatrix}$$

$$R_X(\phi)$$=$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos(\phi) & -\sin(\phi) \\ 0 & \sin(\phi) & \cos(\phi) \end{bmatrix}$$

which when applied in the correct order to specify the rotations outlined in the text above yields the composite matrix:

$$R_Y(\theta)R_X(\phi)=R=$$$$\begin{bmatrix} \cos(\theta) & \sin(\theta)\sin(\phi) & \sin(\theta)\cos(\phi) \\ 0 & \cos(\phi) & -\sin(\phi) \\ -\sin(\theta) & \cos(\theta)\sin(\phi) & \cos(\theta)\cos(\phi) \end{bmatrix}$$

I then have $$B2=RB=R$$ which is the body axis system after the rotations. $$Y_{B2}= \begin{bmatrix} \sin(\theta)\sin(\phi) \\ \cos(\phi) \\ \cos(\theta)\sin(\phi) \end{bmatrix}$$, the y axis of B2, and its projection onto the horizontal plane is $$Y_{pB2}= \begin{bmatrix} \sin(\theta)\sin(\phi) \\ \cos(\phi) \\ 0 \end{bmatrix}$$. I can then say that the cosine of the angle between them, the bank angle $$\Phi$$, is the normalised dot product of the two vectors:

\begin{align} \cos(\Phi)=\frac{Y_{B2}\cdot Y_{pB2}}{|Y_{B2}||Y_{pB2}|} \end{align}

Doing the computation I am left with:

\begin{align} \cos(\Phi)=\sin(\theta)\sin(\phi) + \cos(\phi) \end{align}

Whereas I want:

\begin{align} \sin(\Phi)=\sin(\phi)\cos(\theta) \end{align}

Any advice on where I have gone wrong would be much appreciated.

• \begin{align*} \text{you can use this }\\ &\tan(\Phi)=\frac{Y_z}{\sqrt{Y_x^2+Y_y^2}}\\ &\text{where} \\ &Y=\vec{Y}_{B2} \end{align*}
– Eli
Jan 5 at 15:05
• Thank you for your response Eli, I see how one can get the bank angle from this. However, I do not see how it is possible to move from your expression to Young's statement of $sin(\Phi)=sin(\phi)cos(\theta)$. Do you know how to do this? I also wonder if you have any comments with my own method. I'd rather like to know why it isn't working. Jan 5 at 17:35

$$\vec{y'}=\left[ \begin {array}{c} \sin \left( \theta \right) \sin \left( \phi \right) \\\cos \left( \phi \right) \\ \cos \left( \theta \right) \sin \left( \phi \right) \end {array} \right]$$

$$\Rightarrow$$

$$\tan(\Phi)=\frac{\sin(\Phi)}{\cos(\Phi)}=\frac{\vec{y'}_x}{\vec{y'}_y}=\frac{\sin(\theta)\,\sin(\phi)}{\cos(\phi)}$$

thus:

$$\sin(\Phi)=\sin(\theta)\,\sin(\phi)$$

but this angle is not the angle between y‘ and the plane xy

• Thanks for the further comments Eli. Is it possible that the discrepancy comes from a difference in Euler angle conventions? Jan 7 at 16:16
• I don’t think so because we have the same y’ vector components
– Eli
Jan 7 at 16:35