所以我一直在为这个相对简单的算法而努力。我不确定我的代码有什么问题,但是我没有得到它们实际相交的交点。
我正在使用Unity3D,试图在x,z平面中找到两条线相交的点,尽管不在x,y平面中。我假设适用于x,y的算法应适用于x,z;
我的代码:
Vector3 thisPoint1 = thisCar.position + (2 * thisCar.forward);
Vector3 thisPoint2 = thisCar.position + (20 * thisCar.forward);
Debug.DrawLine(thisPoint1, thisPoint2, Color.white, 2);
Vector3 otherPoint1 = threateningCar.position + (2 * threateningCar.forward);
Vector3 otherPoint2 = threateningCar.position + (20 * threateningCar.forward);
Debug.DrawLine(otherPoint1, otherPoint2, Color.white, 2);
float A1 = thisPoint2.z - thisPoint1.z;
float B1 = thisPoint1.x - thisPoint2.x;
float C1 = A1 * thisPoint1.x + B1 * thisPoint1.z;
float A2 = otherPoint2.z - otherPoint1.z;
float B2 = otherPoint1.x - otherPoint2.x;
float C2 = A2 * otherPoint1.z + B2 * otherPoint1.z;
float det = A1 * B2 - A2 * B1;
float x = (B2 * C1 - B1 * C2) / det;
float z = (A1 * C2 - A2 * C1) / det;
return new Vector3(x, this.transform.position.y, z);
任何人都可以帮助指出我在做什么错吗?
thisCar.forward
和threateningCar.forward
通常是[0,0,1], [0,0,-1]
或[1,0,0], [-1,0,0]
最佳答案
找到了!!!
float A2 = otherPoint2.z - otherPoint1.z;
float B2 = otherPoint1.x - otherPoint2.x;
float C2 = A2 * otherPoint1.z + B2 * otherPoint1.z;
应该:
float A2 = otherPoint2.z - otherPoint1.z;
float B2 = otherPoint1.x - otherPoint2.x;
float C2 = A2 * otherPoint1
.x + B2 * otherPoint1.z
;浪费了很多时间:/。
无论如何,这将帮助任何想要做线相交的人。