Pitch=Flight Path Angle + Angle of Attack if bank angle=0?

Question

Question: Is the statement in the title correct?

Recently, I am taking a course about flight mechanics (click for the slides). It states that "If bank angle is null, then pitch angle = flight path angle + angle of attack" in the page 6 of slides. However, I disagree on it. I will first show an example which I plotted in Geogebra (click for the 3D Geogebra model) (screenshot is also posted below). In the example, pitch angle = 45 degree, flight path angle = 24.09 degree and angle of attack = 19.76 degree, so

$$ \text{(pitch angle)}45 \neq \text{(flight path angle)}24.09+\text{(angle of attack)}19.76 $$

Besides the above example, here is my reasoning:

• Since there is no bank angle (means roll = 0, click for reference), so the symmetry plane of aircraft is vertical plane, which is perpendicular to horizontal plane
• Angle of attack is defined as the angle between "projection of velocity vector in symmetry plane of aircraft" and "x_b"
• flight path angle is defined as angle between velocity vector and horizontal plane
• pitch angle is defined as angle between "x_b" and horizontal plane

Based on the above points, it is impossible for pitch angle = flight path angle + angle of attack. I believe if both roll = 0 and slideslip (here is yaw previously, should be slideslip, thanks to @U_flow) = 0, then it is correct that pitch = flight path angle + angle of attack.

Answer 1

You are correct, at least in theory. In order for the rule to be correct, we need to add one more criterion: that the sideslip angle is zero (or negligible).

Here is an even more extreme example. Suppose that an airplane is flying with a pitch of 89.999 degrees, a bank angle of 0 degrees, and a heading of 0 degrees (due north). It's traveling up and due east, with a flight path angle of 45 degrees. Then the angle of attack is very nearly 0, and adding the flight path angle gives approximately 45 degrees, which is quite far away from its pitch of 89.999 degrees.

(Needless to say, it would be very difficult to make an airplane actually do that.)

In practice, I think that the sideslip angle is practically always small enough that the rule has an insignificant amount of error.

Answer 2

I think there's simply a typo in the last sentence of the slides: a "and $\beta$ is null" is missing where it said "If bank angle $\phi$ is null, then:".

Or it has been implicitly supposed that, being $\beta$ normally very small, for all practical purposes it has a negligible impact on that equation. But I agree that this should have been stated anyway.