loosing quadrant data using atan2 when finding angle between 3 points (in 3D space) in MATLAB

I want to find the angle between O-F1 and O-Tp. Tp is a point that can vary depending on different inputs. So in the attached diagram I have shown 2 points Tp1 and Tp2. As indicated in the attached figure, the angle output by my code gives me an angle in the range of 0 to 180. However I want the range to be in the form of 0 to 360 or -180 to 180.

Some additional info: the 3 points (O, F1, Tp) lie on a plane formed by three other points(F1, F2 and F3) forming an equilateral triangle. F1 is the chosen reference point for measurement of the angle in clockwise direction. O is the centroid of the triangle.

I have tried several alternate equations and thought of using spherical/cylindrical coordinates. Haven't tried looking up quaternions yet. But I don't think this is supposed to be difficult. I feel I am messing something up.

  %calculating unit vectors
    v1 = OF1/norm(OF1);
    v2 = OTp/norm(OTp);

    %Angle Calculation
    A = rad2deg(atan2(norm(cross(v1,v2)), dot(v1,v2)))

Picture - 1 = [diagram showing how the output to my code is giving the same angle for different points instead of showing one of them as -173 degrees or -187 degrees]

Answer 1

[Bruno Luong]

You diagram looks like 2D and not 3D.

In 3D you can turn around and look in opposite direction. The atan2 formula assume you look from the cross vector direction and thus the angle is always > 0.

If you have a preference direction (for example vector normal and come out of your drawing plane) you need to take the dot product of the cross-vector and this normal vector and reverse the sign.

d = [0;0,1] % your preference direction
x = cross(v1,v2);
nx = norm(x);
A = atan2(nx, dot(v1,v2));
x = x / nx;
if dot(x,d) < 0
  x = -x;
  A = -A;
end
% v2 is obtained by rotating v1 about the (unit) x vector of angle A radian  

[Allan Reji]

This is perfect. Thank you very much.

I did notice that norm(x) essentially removed any chance of mentioning the direction. And I understood your solution.

However, I still I dont understand how the cross vector direction makes it always >0. From what I previously understand using atan funcitons is supposed to help distinguish quadrants isnt it?

Isnt the direction of the cross vector determined by right hand thumb rule? So always in the set direction according to the order in which the vectors are crossed? Or am I missing some fundamental understanding?

[Bruno Luong]

Take an example of x, y, z vectors x = [1; 0; 0], y = [0; 1; 0].

The angle (as computed by atan2) betwen x and y vector is pi/2 (anticlock-wise pi 90 degree) if you look from above (meaning from the direction z = cross(x,y))

If you now compute the ange between y and x the formula gives you positiove pi/2 because the cross(y,x)=-z, so if you look from bellow it's pi/2 and not -pi/2. (take a transparent paper, draw x and y vectors, and turn it in the verso side and observe)

The ATAN2(NORM(...)) formula always returns positive angle in [0,pi] since the first argument is postitive, and meaning you look at the vectors from the direction of the cross vector, and angle from vector 1 to 2 never negative. You just go to opposite side.

In 3D you must chose appropriately from where you want to look for to get negative angle.

[Allan Reji]

Thank you very much. I understand now,