Rotation Interpolation

Question

NB: I'll present this question in degrees purely for simplicity, radians, degrees, different zero-bearing, the problem is essentially the same.

Does anyone have any ideas on the code behind rotational interpolation? Given a linear interpolation function: Lerp(from, to, amount), where amount is 0...1 which returns a value between from and to, by amount. How could I apply this same function to a rotational interpolation between 0 and 360 degrees? Given that degrees should not be returned outside 0 and 360.

Given this unit circle for degrees:

where from = 45 and to = 315, the algorithm should take the shortest path to the angle, i.e. it should go through zero, to 360 and then to 315 - and not all the way round 90, 180, 270 to 315.

Is there a nice way to achieve this? Or is it going to just be a horrid mess of if() blocks? Am I missing some well understood standard way of doing this? Any help would be appreciated.

Answer 1

I know this is 2 years old, but I've recently been looking around for the same problem and I don't see an elegant solution without ifs posted in here, so here it goes:

    shortest_angle=((((end - start) % 360) + 540) % 360) - 180;     return shortest_angle * amount; 

that's it

ps: of course, % is meaning modulo and shortest_angle is the variable that holds the whole interpolation angle

Answer 2

Sorry, that was a bit convoluted, here's a more concise version:

    public static float LerpDegrees(float start, float end, float amount)     {         float difference = Math.Abs(end - start);         if (difference > 180)         {             // We need to add on to one of the values.             if (end > start)             {                 // We'll add it on to start...                 start += 360;             }             else             {                 // Add it on to end.                 end += 360;             }         }          // Interpolate it.         float value = (start + ((end - start) * amount));          // Wrap it..         float rangeZero = 360;          if (value >= 0 && value <= 360)             return value;          return (value % rangeZero);     } 

Anyone got a more optimised version?

Answer 3

I think a better approach is to interpolate sin and cos since they don't suffer form being multiply defined. Let w = "amount" so that w = 0 is angle A and w = 1 is angle B. Then

CS = (1-w)*cos(A) + w*cos(B);
SN = (1-w)*sin(A) + w*sin(B);
C = atan2(SN,CS);

One has to convert to radians and degrees as needed. One also has to adjust the branch. For atan2 C comes back in the range -pi to pi. If you want 0 to 2pi then just add pi to C.