英文:
How to find smallest Transition between two angles
问题
我有一个Java程序，在其中一个对象需要旋转到另一个对象。为此，我计算了两个向量的角度，然后计算它们之间的差异。现在我需要找出它们之间的最小旋转过渡。例如，如果第一个向量的角度是 -179°，而到第二个向量的角度是 179°，最小的旋转角度将是 2°，而旋转不需要是整个圆周旋转：-179° - 2° = -181° = 179°。
我希望解释没有问题。
那么我如何可以在Java中轻松计算这个呢？
我在思维上有一点障碍，无法想出解决方法。
编辑：角度以弧度表示
angle_Knight = atan2(knightNode.getLocalTranslation().y, knightNode.getLocalTranslation().x);
angle_follow = atan2(followKnight.getLocalTranslation().y, followKnight.getLocalTranslation().x);
angle = angle_follow - angle_Knight;
谢谢
英文:
I have a Java program in which one object needs to rotate to another object. For this I calculate the angles of the two Vectors and then the difference between them. Now I need to find out which will be the smallest transition between them. For example if the angle of the first vector -179° and the angle to the second vector is 179° the smallest rotation angle would be 2° and the rotation don't needs to be a whole round rotation : -179° -2° = - 181° = 179.
I hope the explanation is ok.
So how can I easy calculate this in Java?
I have a little bloackade in my head and can't figure this out.
Edit: the angles are in radians
angle_Knight = atan2(knightNode.getLocalTranslation().y,knightNode.getLocalTranslation().x));
angle_follow = atan2(followKnight.getLocalTranslation().y,followKnight.getLocalTranslation().x));
angle = angle_follow - angle_Knight;
Thanks
答案1
得分: 1
你离目标很近了。当得到的角度在 (-π, π) 范围之外时，只需添加或减去一个完整的圆周角（360度或2π弧度）。
角度 = 跟随角度 - 骑士角度;
如果 (角度 < -Math.PI) {
角度 = 角度 + 2*Math.PI;
} else 如果 (角度 > Math.PI) {
角度 = 角度 - 2*Math.PI;
}
英文:
You're almost there. Just add or subtract a full turn (360 degrees or 2pi radians) when the angle resulting angle is outside the (-pi, pi) range.
angle = angle_follow - angle_Knight;
if (angle < -Math.PI) {
angle = angle + 2*Math.PI;
} else if (angle > Math.PI) {
angle = angle - 2*Math.PI;
}
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