在过去的几天里，我一直在搜索，并研究Vectors，但仍然无法完全理解数学。
我有两个AABB's。发生碰撞时，我希望我的方法返回一个Vector，然后可以将其添加到Vector位置以使我的对象回到范围内。
这是我当前的代码：
（位置向量是AABB的中心）
public Vector2f collide(Sprite other) {
if(!collideBoolean(other)) return null; //no collision
float xAxis = Math.abs(this.position.x - other.getX()); //distance between centers on x axis
float yAxis = Math.abs(this.position.y - other.getY()); //distance between centers on y axis
//these combined values show the minimum distance apart the objects need to be to not collide
int cw = this.getHalfWidth() + other.getHalfWidth(); //combined width
int ch = this.getHalfHeight() + other.getHalfHeight(); //combined height
float ox = Math.abs(xAxis - cw); //overlap on x
float oy = Math.abs(yAxis - ch); //overlap on y
//direction
Vector2f dir = new Vector2f(this.position);
dir.sub(other.getPosition()); //subtract other.position from this.position
dir.normalise();
return new Vector2f(dir.x * ox, dir.y * oy);
}
（自我解释，但这也是collideBoolean（Sprite other）的代码）
public boolean collideBoolean(Sprite other) {
//code using halfwidths and centers
if(Math.abs(this.position.x - other.getX()) > (this.getHalfWidth() + other.getHalfWidth())) return false;
if(Math.abs(this.position.y - other.getY()) > (this.getHalfHeight() + other.getHalfHeight())) return false;
return true;
}
我当前的代码或多或少都起作用。但是，与“其他”冲突的（this）对象被推向“其他”的最靠近角并且朝向另一侧。
我想我真的很亲近。当然，对于另一双眼睛来说，这将是令人目眩的显而易见的东西，但我无法完全解决。任何帮助将不胜感激！
谢谢，
编辑：
将其添加到collide（Sprite other）方法的末尾是可行的，除了不是我想要的那么整洁。同样，当仅沿一个轴移动到“另一”身体中时，它的效果很好，但是如果以一定角度推入身体，您将进入形状，并且将您随机甩出。
这可能只是因为我每步移动了太多像素，但是我应该扫描碰撞
（此代码查看投影向量以查看哪个分量更大，然后0表示最大分量。这意味着我仅沿最短路径投影出形状）
....
//direction
....
Vector2f projection = new Vector2f(dir.x * (ox+1), dir.y * (oy+1));
if(Math.abs(projection.x) > Math.abs(projection.y)) projection.x = 0;
else if(Math.abs(projection.y) > Math.abs(projection.x)) projection.y = 0;
return projection;
}
编辑两个
随着Ishtar答案的实施，情况看起来很好。但是我发现，如果我正在将一个小物体与一个宽物体碰撞，它可以准确地将碰撞固定在中心附近，但是当您靠近拐角处时，就会陷入该形状中。
像这样：
_
_______l_l_________
| |
|______OK___________|
_--________________
| -- |
|_____SINKS IN______|
编辑三
当前的碰撞代码：
public class Collision {
/** fix collision based on mass */
public static void collide(Sprite s1, Sprite s2) {
float xAxis = Math.abs(s1.getX() - s2.getX()); //distance between centers
float yAxis = Math.abs(s1.getY() - s2.getY()); //distance between centers
int cw = s1.getHalfWidth() + s2.getHalfWidth(); //combined width
int ch = s1.getHalfHeight() + s2.getHalfHeight(); //combined height
//early exit
if(xAxis > cw) return;
if(yAxis > ch) return;
float ox = Math.abs(xAxis - cw); //overlap on x
float oy = Math.abs(yAxis - ch); //overlap on y
if(s1.getMass() <= s2.getMass())
fixCollision(s1, s2, ox+1, oy+1); //the +1's make you get out of the shape instead of
else //if(s1.getMass() > s2.getMass()) //correcting you onto the edge where you'll be in constant collision
fixCollision(s2, s1, ox+1, oy+1);
}
/**
* Fixes the collision
* @param s1 : this gets pushed out (should be lower mass)
* @param s2 : this stays where it is
* @param ox : the overlap along the x axis
* @param oy : the overlap along the y axis
*/
private static void fixCollision(Sprite s1, Sprite s2, float ox, float oy) {
//direction
Vector2f dir = new Vector2f(s1.getPosition());
dir.sub(s2.getPosition());
dir.normalise();
Vector2f projection = new Vector2f(dir.x * (ox), dir.y * (oy));
if(Math.abs(projection.x) > Math.abs(projection.y)) projection.x = 0;
else if(Math.abs(projection.y) > Math.abs(projection.x)) projection.y = 0;
if(ox > oy) s1.getPosition().add( new Vector2f(0, dir.y * oy) ); //overlap is bigger on x so project on y
else if(ox < oy) s1.getPosition().add( new Vector2f(dir.x * ox, 0)); //overlap is bigger on x so project on x
else s1.getPosition().add( new Vector2f(dir.x * ox, dir.y * oy)); //corner to corner
}
最佳答案
float ox = Math.abs(xAxis - cw); //overlap on x
float oy = Math.abs(yAxis - ch); //overlap on y
//direction
Vector2f dir = new Vector2f(this.position);
dir.sub(other.getPosition()); //subtract other.position from this.position
dir.normalise();
return new Vector2f(dir.x * ox, dir.y * oy);
返回的转换向量将使ox（x的重叠）和oy（y的重叠）均为零。但这不是您所需要的。如果ox为0，则在x方向上没有重叠，因此根本没有重叠。如果oy为0，则同上。因此，您需要找到仅使ox或oy为零的向量。
if (ox > oy )
return new Vector3f(0,dir.y*oy);//top-bottom collision
else if (ox < oy )
return new Vector3f(dir.x*ox,0);//left-right collision
else //if(ox == oy)
return new Vector3f(dir.x*ox,dir.y*oy); //corner-corner collision,
//unlikely with float's.