我想写我自己的分离轴定理的实现,但我有一些麻烦得到它的工作尽可能准确,我想要的。我不能肯定地说,但它看起来像是说当形状周围的假想盒子碰到第一个形状时会发生碰撞。但第二个形状完美运作。多边形碰撞检测实现
这里的顶点数据进行平方(确切坐标):
vertsx = [ 200, 220, 220, 200 ]
vertsy = [ 220, 220, 200, 200 ]
下面是测试形状的顶点数据1(相对于小鼠):
vertsx = [ -10, 0, 10, 10, -10 ]
vertsy = [ -10, -50, -10, 10, 10 ]
并且最后,这里的顶点数据对于测试形状2(相对于小鼠):
vertsx = [ -10, 0, 10, 10, -10 ]
vertsy = [ -10, -20, -10, 10, 10 ]
只是为了说明翻译的coo rdinates是被测试的那些,并且这些形状已经按照所示的坐标进行了测试。
这里的实际功能。
function collisionConvexPolygon (vertsax, vertsay, vertsbx, vertsby) {
var alen = vertsax.length;
var blen = vertsbx.length;
// Loop for axes in Shape A
for (var i = 0, j = alen - 1; i < alen; j = i++) {
// Get the axis
var vx = vertsax[ j ] - vertsax[ i ];
var vy = -(vertsay[ j ] - vertsay[ i ]);
var len = Math.sqrt(vx * vx + vy * vy);
vx /= len;
vy /= len;
// Project shape A
var max0 = vertsax[ 0 ] * vx + vertsay[ 0 ] * vy, min0 = max0;
for (k = 1; k < alen; k++) {
var proja = vertsax[ k ] * vx + vertsay[ k ] * vy;
if (proja > max0) {
max0 = proja;
}
else if (proja < min0) {
min0 = proja;
}
}
// Project shape B
var max1 = vertsbx[ 0 ] * vx + vertsby[ 0 ] * vy, min1 = max1;
for (var k = 1; k < blen; k++) {
var projb = vertsbx[ k ] * vx + vertsby[ k ] * vy;
if (projb > max1) {
max1 = projb;
}
else if (projb < min1) {
min1 = projb;
}
}
// Test for gaps
if (!axisOverlap(min0, max0, min1, max1)) {
return false;
}
}
// Loop for axes in Shape B (same as above)
for (var i = 0, j = blen - 1; i < blen; j = i++) {
var vx = vertsbx[ j ] - vertsbx[ i ];
var vy = -(vertsby[ j ] - vertsby[ i ]);
var len = Math.sqrt(vx * vx + vy * vy);
vx /= len;
vy /= len;
var max0 = vertsax[ 0 ] * vx + vertsay[ 0 ] * vy, min0 = max0;
for (k = 1; k < alen; k++) {
var proja = vertsax[ k ] * vx + vertsay[ k ] * vy;
if (proja > max0) {
max0 = proja;
}
else if (proja < min0) {
min0 = proja;
}
}
var max1 = vertsbx[ 0 ] * vx + vertsby[ 0 ] * vy, min1 = max1;
for (var k = 1; k < blen; k++) {
var projb = vertsbx[ k ] * vx + vertsby[ k ] * vy;
if (projb > max1) {
max1 = projb;
}
else if (projb < min1) {
min1 = projb;
}
}
if (!axisOverlap(min0, max0, min1, max1)) {
return false;
}
}
return true;
}
我会尝试其他形状,如果你需要我。
这是我的axisOverlap
函数。
function axisOverlap (a0, a1, b0, b1) {
return !(a0 > b1 || b0 > a1);
}
您可以发布您'axisOverlap()'函数吗? – techfoobar
你能否在不颠倒vy的方向的情况下尝试一下? –
@Asad是的,不幸的是它没有改变结果。 – SpaceFace