如何避免粒子在弹性粒子碰撞模拟器中一起出现毛刺？ [英] How to avoid particles glitching together in an elastic particle collision simulator?

问题描述

这是一个短视频，其中可以看到两组（也许更多）粒子相互干扰。

Here is a short video in which can be seen that groups of two (maybe more) particles are glitched into each other.

可以在这里。

这是我用来计算碰撞的代码：

And this is the code I use to calculate the collisions with:

function collisionPhysics()
{
    for (var i = 0; i < N - 1; ++i)
    {
        for (var j = i + 1; j < N; ++j)
        {
            var Dx = objects[j].x - objects[i].x;    // Difference in X direction between objects[i] and objects[j].
            var Dy = objects[j].y - objects[i].y;    // Difference in Y direction between objects[i] and objects[j].
            var D2 = Dx * Dx + Dy * Dy;              // Distance between objects[i] and objects[j] squared.

            if (D2 <= (objects[i].rad + objects[j].rad) * (objects[i].rad + objects[j].rad))    // Colision check could be inserted here, reusing D2.
            {
                var delta = 2 * (Dx * (objects[i].Vx - objects[j].Vx) + Dy * (objects[i].Vy - objects[j].Vy)) / (D2 * (objects[i].m + objects[j].m));
                objects[i].Vx += -objects[i].m * delta * Dx;
                objects[i].Vy += -objects[i].m * delta * Dy;
                objects[j].Vx +=  objects[j].m * delta * Dx;
                objects[j].Vy +=  objects[j].m * delta * Dy;
            }
        }
    }
}

编辑2013/04/06：
nwellcome提到的问题正在引起奇怪的行为。并且此调整后的函数应该可以解决此问题，不确定是否可以明智地提高性能，但是它可以正常工作：

Edit 2013/04/06: The issue nwellcome mentioned was causing the strange behavior. And this adjusted function should solve this, not sure if it could be improve performance wise but it works:

function collisionPhysics()
{
    for (var i = 0; i < N - 1; ++i)
    {
        for (var j = i + 1; j < N; ++j)
        {
            var Dx = objects[j].x - objects[i].x + timeStep * (objects[j].u - objects[i].u);
            var Dy = objects[j].y - objects[i].y + timeStep * (objects[j].v - objects[i].v);
            var D2 = Dx * Dx + Dy * Dy;              // Distance between objects[i] and objects[j] squared.

            if (D2 <= (objects[i].r + objects[j].r) * (objects[i].r + objects[j].r))    // Colision check could be inserted here, reusing D2.
            {
                objects[i].col = true;
                objects[j].col = true;
                var dx = objects[j].x - objects[i].x;
                var dy = objects[j].y - objects[i].y;
                var du = objects[j].u - objects[i].u;
                var dv = objects[j].v - objects[i].v;
                var dr = objects[j].r + objects[i].r;
                var dt = (-Math.sqrt(2 * dx * du * dy * dv - du * du * (dy * dy - dr * dr) - dv * dv * (dx * dx - dr * dr)) - dx * du - dy * dv) / (du * du + dv * dv);
                Dx = objects[j].x - objects[i].x + dt * (objects[j].u - objects[i].u);
                Dy = objects[j].y - objects[i].y + dt * (objects[j].v - objects[i].v);
                D2 = Dx * Dx + Dy * Dy;
                var delta = 2 * (Dx * (objects[i].u - objects[j].u) + Dy * (objects[i].v - objects[j].v)) / (D2 * (objects[i].m + objects[j].m));
                objects[i].u += -objects[i].m * delta * Dx;
                objects[i].v += -objects[i].m * delta * Dy;
                objects[j].u +=  objects[j].m * delta * Dx;
                objects[j].v +=  objects[j].m * delta * Dy;
                objects[i].x += (timeStep - dt) * objects[i].u;
                objects[i].y += (timeStep - dt) * objects[i].v; 
                objects[j].x += (timeStep - dt) * objects[j].u;
                objects[j].y += (timeStep - dt) * objects[j].v;
            }
        }
    }
}

推荐答案

问题是您的冲突解决方法不能确保粒子在下一帧开始时仍不会相交。

The problem is your method of collision resolution doesn't ensure that the particles won't still intersect at the beginning of the next frame.

找到碰撞时，您需要向后移动到粒子边界之间发生碰撞的时间点，然后从那里解决碰撞。有关类似方法的问题，请参考此游戏开发答案，以了解类似方法。

When you find a collision, you need to work backward to the point in time when collision between boundary of the particles happened and resolve the collision from there. Refer to this game development answer to a similar question for a method of doing that.

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