Calculate the side slip angle correctly

[ufechner7]

The following line is not giving the correct result:
https://github.com/ufechner7/KiteModels.jl/blob/87e7e7690afbb7f4d619f78dc771f28008754538/src/KPS4.jl#L355

It gives the following result when flying power cycles:

Jelle wrote: I believe side slip is usually between -10 and +10deg

[ufechner7]

Perplexity says:

[1-Bart-1]

The problem is probably that the apparent wind is in global frame here. Rotating it to the body frame should solve your problem.

R_b_w = [e_x ; e_y ; e_z]
va_b = R_b_w' * va

Where R_b_w is the rotation matrix from body to world frame, and R_b_w' = R_w_b

[ufechner7]

We did not define any world frame. The model uses the ENU reference frame for velocities and positions. So I assume you suggest to convert the the apparent wind speed vector from the ENU to the Kite reference frame:
va_k = ENU2K(va).

[1-Bart-1]

Yes

[ufechner7]

Using:

    va_xy2 = va_2 - (va_2 ⋅ z) * z # va at the top particle in the xy plane
    R_k_w = [x, y, z]
    va_k = R_k_w' * va_xy2

    s.side_slip = atan(va_k[2], -va_k[1])

[1-Bart-1]

Why do you calculate va_xy2? And you should not use commas in [x, y, z], use [x y z] instead. This should work:

    R_k_w = [x y z]
    va_k = R_k_w' * va_2

    s.side_slip = atan(va_k[2], -va_k[1])

A different and simpler solution without the rotation matrix that also should work:

s.side_slip = atan(va_2 ⋅ y, va_2 ⋅ x)

[ufechner7]

Fixed. Thanks a lot!