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The solution is actually simpler than expected. The trick is to use Minkowski subtraction before your hexagon technique. Here are your rectangles A and B, with their velocities vA and vB. Note that v …

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 …

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. …

You guessed properly. The orthographic projection you came up with is the one used for most of the objects. You can tell because parallel lines in the X, Y or Z directions all become parallel lines, i …

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

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) …

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 …

There are a lot of different ways to store this information. I would suggest using something standard, though. For instance a lot of opensource games use the TMX map format as used by the Tiled editor …

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 …

Here: perlin.lerp(v, perlin.lerp(u, perlin.grad(self[AB + 2], x , y , z - 1), This should have AA + 2 instead.

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 …

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 …

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 …