1
$\begingroup$

I found an equation to calculate angle of attack based on the velocity vector in the body frame of the aircraft.

$\arctan(v_x/v_z) = \alpha$

1) Is this correct?

When I input in a $v_z$ approaching zero, the value approaches $\infty$ which means that angle of attack moving only in the x direction is 90 degrees since $\arctan(x)_{x\rightarrow \infty}= \pi/2$

2) Is there supposed to be an assumption that you never calculate Aoa of zero with this?

3) Am I misinterpreting what the velocities and angles represent?

Overall I am just a bit confused, because I kind of expected something that wouldn't mess up at level flight.

$\endgroup$
3
$\begingroup$

It is exactly the other way round:

$$ \alpha = \mathrm{arctan} \left( \frac{v_z}{v_x} \right) $$

You can simply derive it from the trigonometry in a triangle: sketch In this image the aircraft is moving to the right (along the arrow labelled v) and the $x$-axis is pointing along the aircraft body.

The resulting function looks like this: angle of attack

  • For $ v_z = 0 $, you get $ \alpha = \arctan(0) = 0 $.

  • For $ v_x \to 0 $, you get $ \alpha \to \frac{\pi}{2} = 90^\circ $.

$\endgroup$
11
  • $\begingroup$ It looks like you are using a Vx and Vy defined relative to the horizon, not the aircraft body frame. It looks like what you are describing is the climb or glide angle, not the angle-of-attack. You can fix the first diagram by relabelling your V as Vx, and draw your Vz perpendicular to Vx and pointing generally downward (and a little forward) , and relabel your Vx as V. This portrays the AoA of an aircraft in the illustrated pitch attitude in level flight ( i.e. horizontal trajectory). Of course the illustrated AoA would be rather extreme. $\endgroup$ – quiet flyer Aug 1 '19 at 18:02
  • $\begingroup$ Anyway for a diagram redrawn this way, your comments about the correct formula are correct- it needs to be arctan (Vz/ Vx). The comment about an angle-of-attack of 90 degrees seems a little odd, though technically true. $\endgroup$ – quiet flyer Aug 1 '19 at 18:03
  • $\begingroup$ Reference for direction of x, y, and z in aircraft body frame: en.m.wikipedia.org/wiki/… $\endgroup$ – quiet flyer Aug 1 '19 at 18:04
  • $\begingroup$ @quietflyer You are absolutely right! I fixed the sketch. Thanks :) $\endgroup$ – Bianfable Aug 1 '19 at 18:12
  • $\begingroup$ Also if Vz is zero, using Vz as defined in question (aircraft body frame), this means angle-of-attack is zero. Says nothing about whether or not aircraft is in level flight. So, this answer could use some improvement, even though it did give the correct formula for angle-of-attack. $\endgroup$ – quiet flyer Aug 1 '19 at 18:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.