Stop and go problem

Question

I'm accelerating my ship with:dx += Math.cos(theta)*acceleration;
dy += Math.sin(theta)*acceleration;I want my deceleration to work in such a way that no matter what direction I'm facing, the speed is reduced toward 0. I also want this to be just as strong as my engines are in accelerating. So I figure the logic is to find the angle of the velocity vector, and reduce speed along that angle. So I find the angle with:double velocityAngle = Math.atan2(dy, dx);and then reduce my velocity's magnitude with:dx -= Math.cos(velocityAngle)*acceleration;
dy -= Math.sin(velocityAngle)*acceleration;This seems fine in my head logically, but in testing I've found that if I accelerate in one direction for 3 seconds, it only takes 2 seconds to stop. There are no other forces acting on my speed. Why are the times different?

Answer 1

Times like this, I'd kill for either a chaulkboard or a whiteboard.
I think what bothered me is that theta, as defined in the original problem, appears to be the direction that the ship is pointing in, as opposed to the direction that the ship is currently moving in. In other words, Malohkan's logic makes sense if the spaceship accelerates in one direction and then pivots in place such that theta no longer correspondes to the direction of motion.
Malohkan's solution then is to take the existing velocity vector and recompute what angle that vector points in. Once he has that angle, he can reduce the speed using the sin and cosine functions. The only problem with this approach is that he looses a second somewhere. I still suspect this is a result of the rounding errors that occur from finding the arctangent and then the sin and cosine - a glance at a trignometric table will reveal that these tend to be non-terminating decimals.
What we're trying to point out to Malohkan is that if he has the vector, it is not necessary to calculate the angle - in other words, it's not necessary to go from vector to angle to reverse angle to reverse vector... arg, writing out the formulae is much easier than trying to explain it via words. You will just have to take it on faith for now that mathematically speaking, the angle stuff always cancels itself out and you're left with the vector - reverse vector formula.
An alternate solution is to keep track of the current direction of motion as an angle separately from the current nose orientation - so when the spaceship pivots in place, he still has the angle of motion and doesn't need to recalculate that. That would remove one of the two potential rounding errors (due to the non-terminating decimals) and his answer would then be more accurate (but not totally so).