碰撞响应中如何固定圆和矩形重叠？

Question

由于在数字世界中几乎不会发生真正的碰撞，我们总会出现“碰撞”的圆圈与矩形重叠的情况。

在与矩形完美碰撞而不重叠的情况下，如何将圆形放回原处？

假设矩形停止（零速度）并对轴。

Answer 1

多年来一直盯着这个问题，从来没有想出一个完美的解决方案，我终于做到了！

这是一个非常简单的算法，不需要循环和近似。

这是它在更高层次上的工作方式：

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float L = bounds.left;
float T = bounds.top;
float R = bounds.right;
float B = bounds.bottom;
float dx = end.x - start.x;
float dy = end.y - start.y;

float ltime = Float.MAX_VALUE;
float rtime = Float.MAX_VALUE;
float ttime = Float.MAX_VALUE;
float btime = Float.MAX_VALUE;

if (start.x - radius < L && end.x + radius > L) {
   ltime = ((L - radius) - start.x) / dx;
}
if (start.x + radius > R && end.x - radius < R) {
   rtime = (start.x - (R + radius)) / -dx;
}
if (start.y - radius < T && end.y + radius > T) {
   ttime = ((T - radius) - start.y) / dy;
}
if (start.y + radius > B && end.y - radius < B) {
   btime = (start.y - (B + radius)) / -dy;
}

现在我们试着看看它是否是一个严格的边交叉点（而不是拐角）。如果碰撞点在侧面，则返回交叉点：

if (ltime >= 0.0f && ltime <= 1.0f) {
   float ly = dy * ltime + start.y;
   if (ly >= T && ly <= B) {
      return new Intersection( dx * ltime + start.x, ly, ltime, -1, 0 );
   }
}
else if (rtime >= 0.0f && rtime <= 1.0f) {
   float ry = dy * rtime + start.y;
   if (ry >= T && ry <= B) {
      return new Intersection( dx * rtime + start.x, ry, rtime, 1, 0 );
   }
}

if (ttime >= 0.0f && ttime <= 1.0f) {
   float tx = dx * ttime + start.x;
   if (tx >= L && tx <= R) {
      return new Intersection( tx, dy * ttime + start.y, ttime, 0, -1 );
   }
}
else if (btime >= 0.0f && btime <= 1.0f) {
   float bx = dx * btime + start.x;
   if (bx >= L && bx <= R) {
      return new Intersection( bx, dy * btime + start.y, btime, 0, 1 );
   }
}

我们已经走了这么远所以我们知道不是没有交叉点，就是和一个拐角相撞了。我们需要确定拐角：

float cornerX = Float.MAX_VALUE;
float cornerY = Float.MAX_VALUE;

if (ltime != Float.MAX_VALUE) {
   cornerX = L;
} else if (rtime != Float.MAX_VALUE) {
   cornerX = R;
}

if (ttime != Float.MAX_VALUE) {
   cornerY = T;
} else if (btime != Float.MAX_VALUE) {
   cornerY = B;
}

// Account for the times where we don't pass over a side but we do hit it's corner
if (cornerX != Float.MAX_VALUE && cornerY == Float.MAX_VALUE) {
   cornerY = (dy > 0.0f ? B : T);
}

if (cornerY != Float.MAX_VALUE && cornerX == Float.MAX_VALUE) {
   cornerX = (dx > 0.0f ? R : L);
}

现在我们有足够的信息来解三角形。这使用了距离公式，找到两个向量之间的角度，和正弦定律（两次）：

double inverseRadius = 1.0 / radius;
double lineLength = Math.sqrt( dx * dx + dy * dy );
double cornerdx = cornerX - start.x;
double cornerdy = cornerY - start.y;
double cornerdist = Math.sqrt( cornerdx * cornerdx + cornerdy * cornerdy );
double innerAngle = Math.acos( (cornerdx * dx + cornerdy * dy) / (lineLength * cornerdist) );
double innerAngleSin = Math.sin( innerAngle );
double angle1Sin = innerAngleSin * cornerdist * inverseRadius;

// The angle is too large, there cannot be an intersection
if (Math.abs( angle1Sin ) > 1.0f) {
   return null;
}

double angle1 = Math.PI - Math.asin( angle1Sin );
double angle2 = Math.PI - innerAngle - angle1;
double intersectionDistance = radius * Math.sin( angle2 ) / innerAngleSin;

// Solve for time
float time = (float)(intersectionDistance / lineLength);

// If time is outside the boundaries, return null. This algorithm can 
// return a negative time which indicates the previous intersection. 
if (time > 1.0f || time < 0.0f) {
   return null;
}

// Solve the intersection and normal
float ix = time * dx + start.x;
float iy = time * dy + start.y;
float nx = (float)((ix - cornerX) * inverseRadius);
float ny = (float)((iy - cornerY) * inverseRadius);

return new Intersection( ix, iy, time, nx, ny );

Intersection inter = handleIntersection( bounds, start, end, radius );

if (inter != null) 
{
   // Project Future Position
   float remainingTime = 1.0f - inter.time;
   float dx = end.x - start.x;
   float dy = end.y - start.y;
   float dot = dx * inter.nx + dy * inter.ny;
   float ndx = dx - 2 * dot * inter.nx;
   float ndy = dy - 2 * dot * inter.ny;
   float newx = inter.x + ndx * remainingTime;
   float newy = inter.y + ndy * remainingTime;
   // new circle position = {newx, newy}
 }

编辑：您可以修改这为一个旋转矩形通过使用该矩形的角度取消旋转的矩形与圆的开始和结束点。您将执行交点检查，然后旋转得到的点和法线。

编辑：我修改了pastebin上的代码，如果圆的路径的边界卷不与矩形相交，就提前退出。