Java 2d多边形碰撞响应

Question

我制作了一个使用凸多边形的游戏。 我的计划是让这些多边形根据其质量和速度发生碰撞和反弹。 但是首先我需要确保它们没有重叠。 此代码检查多边形A的每个边缘，以查看多边形B的任何顶点在垂直于该边缘的轴上是否重叠。 整个方法返回修复多边形A所需的结果Vector：

    /**Calculates adjustment vector for EntityPolygon A*/
public Vector calculateCollision(EntityPolygon A, EntityPolygon B) {

    //this is a large number so the first comparison of overlap is true
    double overlap = 10000;

            //this is the angle of the axis to apply the overlap vector
    double angle = 0;

            //I ran a for loop for every edge of the polygon
    for(int x = 0; x <= A.numPoint - 1; x++) {

                    //create variables
        Vector edge;
        Vector axis;
        double centerA;
        double centerB;
        double maxA;
        double maxB;
        double minA;
        double minB;

                    //this if statement finds this point and the next point
                    //to make a Vector of the edge
        if(x != A.numPoint - 1) {
            edge = new Vector(A.point[x], A.point[x + 1]);
        } else {
            edge = new Vector(A.point[x], A.point[0]);
        }

                    //this finds the axis perpendicular axis of the edge
        axis = edge.getRightNormal();

        //finds the location of both polygon's centers when projected onto
                    //the velocity(projectionOnVelocity() projects the point on the                                                           
                    //new axis)
        centerA = A.getLocation().getProjectionOnVelocity(axis);
        centerB = B.getLocation().getProjectionOnVelocity(axis);

                    //finds the location of polygons A and B on the axis by
                    //setting the min and max of their highest and lowest points
        maxA = findMax(A, axis);
        maxB = findMax(B, axis);
        minA = findMin(A, axis);
        minB = findMin(B,axis);

        //final comparison to find overlapping vector.
        if(centerA > centerB) {//if A is above B on the axis
            if(maxB > minA) {//if the max point on B is above min on A
                double m = maxB - minA;
                if(m < overlap) {
                    overlap = m;
                    angle = axis.angle;
                }
            } else {
                                    //(0,0) vector
                return Vector.getDefault();
            }
        } else if(centerB > centerA) {//if B is above A on axis
            if(maxA > minB) {//if the max point on A is above min on B
                double m = maxA - minB;
                if(m < overlap) {
                    overlap = m;
                    angle = axis.angle + Math.PI;
                }
            } else {
                                    //(0,0) vector
                return Vector.getDefault();
            }
        }
    }
            //if the overlap value has been set by the edges of Polygon A
    if(overlap != 10000) {
                    //returns the adjustment vector along overlap edge axis
        return new Vector(angle, overlap, true);
    } else {
                    (0,0) vector
        return Vector.getDefault();
    }
}

这段代码有一个错误，当一个非常平坦的块位于一个正方形块之上时，该错误会导致问题，而该平坦块认为它比实际下降的要远得多。 结果，平板和正方形发生碰撞，将它们推到下图中的当前位置

这是初步阶段的实际游戏的图片。 甚至触摸之前，右侧的蓝色方块已被顶部的障碍物移动。 当正方形在屏幕的最左侧时，这会在程序开始时发生。

Answer 1

我同意Beta，您的算法看起来不正确（主要是因为我敢肯定多边形的交点不会那么容易。仅对于检查2个形状的交点（没有任何智能），您必须检查是否有任何边缘相交）具有其他形状的任何边缘，但这即使是中等简单的形状也很慢。

矩形相交是很容易做到的，三角形相交应该不会太难，如果您可以将形状拆分为其中的一个（或者实际上是一组具有相当简单公式的形状），它将使事情变得容易得多。

您也可以查看此问题 。

多边形相交是一个经过充分研究的问题，只是Google，您应该有很多选择。