Question

我正在使用和扩展一个功能，检测两个2d旋转的矩形是否发生碰撞。该函数来自here。

我的问题：该功能错误地检测到以下矩形之间的碰撞。

Rect1 = centreX = 0，centreY = 0，width = 8，height = 4 angle = 0
Rect1 = centreX = 16，centreY = 0，width = 8，height = 4 angle = 0

我错了还是这些矩形不会碰撞？我该如何编辑该功能，以便检测旋转矩形之间的像素完美碰撞？我有一种感觉，我需要编辑最后两行，但我不确定。

int RotRectsCollision(_RotRect * rr1, _RotRect * rr2)
{
 _Vector2D A, B,   // vertices of the rotated rr2
       C,      // center of rr2
       BL, TR; // vertices of rr2 (bottom-left, top-right)

 float ang = rr1->ang - rr2->ang, // orientation of rotated rr1
       cosa = cos(ang),           // precalculated trigonometic -
       sina = sin(ang);           // - values for repeated use

 float t, x, a;      // temporary variables for various uses
 float dx;           // deltaX for linear equations
 float ext1, ext2;   // min/max vertical values

 // move rr2 to make rr1 cannonic
 C = rr2->C;
 SubVectors2D(&C, &rr1->C);

 // rotate rr2 clockwise by rr2->ang to make rr2 axis-aligned
 RotateVector2DClockwise(&C, rr2->ang);

 // calculate vertices of (moved and axis-aligned := 'ma') rr2
 BL = TR = C;
 SubVectors2D(&BL, &rr2->S);
 AddVectors2D(&TR, &rr2->S);

 // calculate vertices of (rotated := 'r') rr1
 A.x = -rr1->S.y*sina; B.x = A.x; t = rr1->S.x*cosa; A.x += t; B.x -= t;
 A.y =  rr1->S.y*cosa; B.y = A.y; t = rr1->S.x*sina; A.y += t; B.y -= t;

 t = sina*cosa;

 // verify that A is vertical min/max, B is horizontal min/max
 if (t < 0)
 {
  t = A.x; A.x = B.x; B.x = t;
  t = A.y; A.y = B.y; B.y = t;
 }

 // verify that B is horizontal minimum (leftest-vertex)
 if (sina < 0) { B.x = -B.x; B.y = -B.y; }

 // if rr2(ma) isn't in the horizontal range of
 // colliding with rr1(r), collision is impossible
 if (B.x > TR.x || B.x > -BL.x) return 0;

 // if rr1(r) is axis-aligned, vertical min/max are easy to get
 if (t == 0) {ext1 = A.y; ext2 = -ext1; }
 // else, find vertical min/max in the range [BL.x, TR.x]
 else
 {
  x = BL.x-A.x; a = TR.x-A.x;
  ext1 = A.y;
  // if the first vertical min/max isn't in (BL.x, TR.x), then
  // find the vertical min/max on BL.x or on TR.x
  if (a*x > 0)
  {
   dx = A.x;
   if (x < 0) { dx -= B.x; ext1 -= B.y; x = a; }
   else       { dx += B.x; ext1 += B.y; }
   ext1 *= x; ext1 /= dx; ext1 += A.y;
  }

  x = BL.x+A.x; a = TR.x+A.x;
  ext2 = -A.y;
  // if the second vertical min/max isn't in (BL.x, TR.x), then
  // find the local vertical min/max on BL.x or on TR.x
  if (a*x > 0)
  {
   dx = -A.x;
   if (x < 0) { dx -= B.x; ext2 -= B.y; x = a; }
   else       { dx += B.x; ext2 += B.y; }
   ext2 *= x; ext2 /= dx; ext2 -= A.y;
  }
 }

 // check whether rr2(ma) is in the vertical range of colliding with rr1(r)
 // (for the horizontal range of rr2)
 return !((ext1 < BL.y && ext2 < BL.y) ||
      (ext1 > TR.y && ext2 > TR.y));
}

Answer 1

我不认为这些矩形应该碰撞。

第一个的中心位于（0,0），宽度为8 =＆gt;它在X轴上从-4跨越到4。

第二个的中心位于（16,0），宽度为8 =＆gt;它在X轴上从16 - 4 = 12到16 + 4 = 20。

因此，第一个和第二个在X轴上不重叠。由于两者都没有旋转，因此它们不会发生碰撞。

我没有检查过代码但是给出了示例矩形，它们似乎没有碰撞。

错误地检测矩形之间的碰撞

1 个答案: