Question

我有一点跟随我在Processing中制作的鼠标。

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我怎么能改变这个或添加到这个以使得该点具有有限的转弯率？我的想法与这场比赛中的导弹相似。 https://scratch.mit.edu/projects/181364872/ 我不知道我是怎么开始限制转折点的。任何帮助将不胜感激。

（我也标记了java，因为虽然这是在Processing中，但很多时候，Processing几乎都是带有内置方法的Java。）

Answer 1

执行此操作的一种方法是保持对象的当前方向。然后，您可以使用向量与鼠标的交叉积，以及沿对象方向的向量，以找到需要转向指向鼠标的角度。然后，您可以限制转弯并添加更改以获得新方向。

double direction = ?; // the current direction of object in radians
double x = ?;  // the current position
double y = ?;
double maxTurn = ?; // Max turn speed in radians
double speed = ?;
void move() {
   double mvx = mouseX - x;   // get vector to mouse
   double mvy = mouseY - y;
   double dist = Math.sqrt(mvx * mvx + mvy * mvy);  // get length of vector
   if(dist > 0){  // must be a distance from mouse
       mvx /= dist;  // normalize vector
       mvy /= dist; 
       double ovx = Math.cos(direction); // get direction vector
       double ovx = Math.sin(direction);
       double angle = Math.asin(mvx * ovy - mvy * ovx); // get angle between vectors
       if(-mvy * ovy - mvx * ovx < 0){ // is moving away
          angle = Math.sign(angle) * Math.PI - angle;  // correct angle
       }
       // angle in radians is now in the range -PI to PI
       // limit turn angle to maxTurn
       if(angle < 0){
          if(angle < -maxTurn){
             angle = -maxTurn;
          }
       }else{
          if(angle > maxTurn){
             angle = maxTurn;
          }
       }
       direction += angle; // turn 
       // move along new direction
       x += Math.cos(direction) * speed;
       y += Math.sin(direction) * speed;
    }
}

我有一个跟随鼠标的点，我怎么能给它一个有限的转弯率？

1 个答案: