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[1.7.10] Getting the pitch and yaw from look vector


Thornack

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Hi Everyone,

 

I want to get the pitch and the yaw fromt he look vector. I think I have the yaw done correctly but the pitch is eluding me i think. How would you do it. Currently I am doing it liek this and I am not sure what my mistake is.

 

Pitch Code - I think I buggered up something in this one

public static float getPitchFromVector(Vec3 vec3) {
        double dx = vec3.xCoord;
        double dz = vec3.zCoord;
        double pitch = 0;
        // Set pitch
        if (dx != 0) {
            // Set pitch i think start value based on dx
            if (dx < 0) {
                pitch = 1.5 * Math.PI;
            } else {
                pitch = 0.5 * Math.PI;
            }
            pitch -= Math.sin(vec3.yCoord/Math.sqrt((vec3.xCoord*vec3.xCoord + vec3.zCoord*vec3.zCoord)));
        } else if (dz < 0) {
            pitch = Math.PI;
        }
        return (float) (-pitch * 180 / Math.PI - 90);
    }

 

Yaw Code - I think this one is correct

public static float getYawFromVector(Vec3 vec3) {
        double dx = vec3.xCoord;
        double dz = vec3.zCoord;
        double yaw = 0;
        // Set yaw
        if (dx != 0) {
            // Set yaw start value based on dx
            if (dx < 0) {
                yaw = 1.5 * Math.PI;
            } else {
                yaw = 0.5 * Math.PI;
            }
            yaw -= Math.atan(dz / dx);
        } else if (dz < 0) {
            yaw = Math.PI;
        }
        return (float) (-yaw * 180 / Math.PI - 90);
    }

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If my vector math is correct, it should be this:

double pitch = -Math.asin(vec3.yCoord) * (180 / Math.PI);

 

Edit:

A faster way to calculate rotational head yaw.

double rotationYawHead = -Math.atan2(vec3.xCoord, vec3.zCoord) * 180.0F / (float)Math.PI;

 

Since you are getting this from the look vector, the yaw is for the head, not the body, which differs slightly.

 

Another Edit: After looking at it again, the above concerning the yaw appears to be only mostly correct. It gets the correct base value, but it seems random how many times 360 must be added or subtracted. That shouldn't make a difference, I think, since if you rotate 360 degrees you are back where you started from. I could be wrong though.

With all due respect, sir: I do, what I do, the way I do it. ~ MacGyver

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