如何旋转某个点周围的顶点？ [英] How to rotate a vertex around a certain point?

问题描述

假设你在二维空间中有两个点，你需要将这些点中的一个旋转X度，另一个点作为中心点。

  float distX = Math.abs（centerX-point2X）; 
 float distY = Math.abs（centerY-point2Y）; 
 
 float dist = FloatMath.sqrt（distX * distX + distY * distY）; 
  

到目前为止，我只是找到了两点之间的距离......任何想法我应该在哪里去那里？

解决方案

最简单的方法是编写三个转换：

1. 将点1带到原点的翻译
   
2. 围绕原点旋转所需角度​​
   
3. 带来点1的翻译回到原来的位置

当您完成所有工作时，您将得到以下转换结果：

  newX = centerX +（point2x-centerX）* Math.cos（x） - （point2y-centerY）* Math.sin（x）; 
 
 newY = centerY +（point2x-centerX）* Math.sin（x）+（point2y-centerY）* Math.cos（x）; 
  

请注意，这会假设角度 x 对于顺时针旋转是负值（所谓的标准或右手方向对于坐标系）。如果不是这种情况，那么您需要按照涉及 sin（x）的条款来颠倒标志。

Imagine you have two points in 2d space and you need to rotate one of these points by X degrees with the other point acting as a center.

float distX = Math.abs( centerX -point2X );
float distY = Math.abs( centerY -point2Y );

float dist = FloatMath.sqrt( distX*distX + distY*distY );

So far I just got to finding the distance between the two points... any ideas where should I go from that?

解决方案

The easiest approach is to compose three transformations:

1. A translation that brings point 1 to the origin
2. Rotation around the origin by the required angle
3. A translation that brings point 1 back to its original position

When you work this all out, you end up with the following transformation:

newX = centerX + (point2x-centerX)*Math.cos(x) - (point2y-centerY)*Math.sin(x);

newY = centerY + (point2x-centerX)*Math.sin(x) + (point2y-centerY)*Math.cos(x);

Note that this makes the assumption that the angle x is negative for clockwise rotation (the so-called standard or right-hand orientation for the coordinate system). If that's not the case, then you would need to reverse the sign on the terms involving sin(x).

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