How do I find the shortest distance around a circle from one degree value to another, simply by adding or subtracting rotational velocity? I am trying to create a movement system for a 2D spaceship game using only the arrow keys. So essentially I have a 2d vector and I want to use that to set a rotation for my spaceship. If the ship starts with an angle of 0, and then I press the up arrow the ship should rotate from 0 to 90 degrees or pi/2 from 2pi based on a rotation velocity. This is the code I have that does this.


    private void updateRotation()
    {
        if (Direction.X < 0)
        {
            setRotationValue(MathHelper.ToRadians(180));
        }
        if( Direction.X > 0)
        {
            if(MathHelper.ToDegrees(rotation) < 180)
            {
                setRotationValue(MathHelper.ToRadians(0));
            }
            else
            {
                setRotationValue(MathHelper.ToRadians(360));
            }
            
        }
        if(Direction.Y < 0)
        {
            setRotationValue(MathHelper.ToRadians(270));
        }
        if(Direction.Y > 0)
        {
            setRotationValue(MathHelper.ToRadians(90));
        }
        
    }

    private void setRotationValue(float target)
    {
        if (target > rotation)
        {
            rotation += MathHelper.ToRadians(rotationVelocity);
        }
        if (target < rotation)
        {
            rotation -= MathHelper.ToRadians(rotationVelocity);

        }
    }

<
However, my problem arises in a situation where the ship is say facing 45 degrees and I want to rotate to 270 degrees. Well if I press the down key the target rotation is greater then the current rotation so it goes from 45 to 270. This is not the quickest path, as it would be much faster to go from 45 back to 0, then start at 360 and go down to 270. I'm not sure the best way to implement this in code, or if my approach is the right way to do this. I have had a long conversation with chat GPT, which did not give me a solid answer. Any help or direction on how to solve this problem would be much appreciated, especially if this problem is even possible or if I need to look into something like Euler angles or quaternions.
 A: I don't fully understand what you want your arrow keys to do, or what you mean by "rotation velocity" in your context, but let me try to restate your goal mathematically:
You have a starting angle $S$, and a desired ending angle $E$, and you want to find an angle $D$, with $-180 \lt D \le +180$, so that $$ E = S + D \mod 360$$
(If that wouldn't solve your problem, please explain further. Also, I'll keep things in degrees, as that's how you've written your question.)
I'll go through the solution in multiple steps, to make it easier to understand, but you certainly can combine equations together in your code if you wish.
First, let's set $$D_0 = E -S$$
And then we'll convert units to "full rotations" $$ d_0 = D_0/360$$
Now, set $$ n = \text{closest_integer_to}(d_0)$$
I'd normally expect this to correspond to a function named round(), but not all computer languages have a "native" round() function that behaves right, especially when handling negative numbers. You should find one where round(-1.7) equals -2, not -1. And when it comes to rounding, there can be a lot of discussion of what will happen with half-integral values, like $3.5$, but we don't care about that here - rounding them either up or down will work for us.
Next, set $$d = d_0 -n$$
Notice that we've modified the amount we rotate by an integral number of full rotations, which doesn't make a difference with angles.
Finally $$ D = 360 * d$$
converting back to degrees to give us the final desired answer.
One nice thing about this is it doesn't matter what range of values your $S$ and $E$ angles are in. If you had been slowly incrementing the value of $S$ and it was currently at, say, $5735$ degrees, you could still use $270$ as your $E$ value and the computed $D$ value would still be between $-180$ and $180$. And of course you can change the two occurences of "$360$" to "$2\pi$" if your angles are kept as radians.
(I gotta confess that I haven't done any 'testing' of this with actual values; it's just what makes sense mathematically. So you might want to run a few test cases after you code it up to make sure I don't have a bug!)
