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To avoid the camera jerkiness, you should keep the floating-point coordinates camera, but clamp the OpenGL projection to the nearest pixel.

The derivative of the ball position is: dx/dt = vx dy/dt = vy + gravity * t So the derivative of the parabola arc length s is, using Pythagora’s theorem: ds/dt = sqrt(vx² + (vy + gravity * t)²) …

Why not a PRNG, actually? It looks like a simple and good tool to me. Some pseudocode using srand/rand that you should be able to adapt to any other PRNG: int shuffle(int seed, int x, int y) { i …

I think what you are trying to do is the 2D image space equivalent to double rounding. …

If I was the ball, I think I would try to 1. maximise the surface of contact with the ground and 2. maximise the vertical compression so that the release will generate forces that are normal to the gr …

Adapted from my answer to "Which Side Was Hit?": I suggest computing the Minkowski sum of B and A, which is a new rectangle, and checking where the centre of rectangle A lies relatively to that new r …

Here is a method that works: get a vector normal to the triangle, for instance p1p2 ^ p1p3 for each edge, take the result of its cross product with that vector normalize If your triangles are in a 2D …

Maybe this: elif sides & LEFT: Should actually be this: if sides & LEFT: Otherwise the left/right branches are never visited when there is a top/bottom collision. The first part o …

In this line: double targetY = circle1.getLayoutX() + Math.sin(angle) * minDistance; Certainly you meant getLayoutY. As a side note, it is useless to compute angle = atan2(y,x) if the only thing y …

You are not using the right variable to perform the collision time calculation. distanceXin is the distance between the two objects’ centres, whereas what you actually want is the distance between the …

Your question is not precise enough. An array of points is only « clockwise » or « anti-clockwise » relative to a reference point. Otherwise, any array of three points can always be either CW or CCW. …

If you are going to render Y = f(X) graphs only, then I suggest trying the following method. The curve data is passed as texture data, making it persistent, and allowing for partial updates through g …

If you have three sprites (or if they have very specific relative positions, which almost never happens) there is a special point called the circumcentre which is at an equal distance of each sprite. …

Here is a simple proposal: you could try modulate the fog intensity for each cell. Here is an example of applying a simple convolution filter: float fog = 0.0; for (int dj = -3; dj <= 3; dj++) for (i …

One possible way is to split the integer into two values between 0 and 29 (this is done using integer division and modulus) then offset them by -15: for (int n = 0; n < 30 * 30; ++n) { if (Array[ …