Angle of attack from velocity vector 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.
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:

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:



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