java - 使用投影 vector 的基本 AABB 碰撞

Question

过去几天我一直在四处寻找，并研究 vector ，但仍然无法完全理解数学......

我有两个AABB's 。在发生碰撞时，我希望我的方法返回一个 vector ，然后我可以将其添加到位置 vector 以使我的对象回到边界内。

这是我当前的代码:

(位置 vector 为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);
}

(不言自明，但这里也是 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有效。但是与“其他”碰撞的(这个)对象被推出并朝向“其他”最近的角落。

我想我已经很接近了。当然，对于不同的人来说，这将是显而易见的事情，但我无法完全弄清楚。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，只是没有我想要的那么整洁。此外，当仅沿一个轴移动到“另一个”主体时，它工作得很好，但如果你以一定角度插入主体，你就会进入形状内部，它会随机地将你抛出去。

这可能只是因为我每步移动了太多像素，但我应该扫描测试我的碰撞

(此代码查看投影 vector 以查看哪个分量更大，然后 0 表示最大分量​​。这意味着我仅沿着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

随着伊什塔尔答案的实现，事情看起来不错。但我发现，如果我将一个小物体与一个宽物体碰撞，它会准确地修复中心附近的碰撞，但当你离开角落附近时，你就会陷入形状中。

像这样:

         _
 _______l_l_________
|                   |
|______OK___________|


 _--________________
| --                |
|_____SINKS IN______|

编辑三个

当前碰撞代码:

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
}

Answer 1

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);

返回的平移 vector 将使 ox(x 中重叠)和 oy(y 中重叠)都为零。但这不是您需要的。如果 ox 为 0，则 x 方向上没有重叠，因此根本没有重叠。如果 oy 为 0，同上。因此，您需要找到仅使 ox 或 oy 为零的 vector 。

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.