正确的三角用于旋转点围绕原点 [英] Proper Trigonometry For Rotating A Point Around The Origin

问题描述

执行下列任一下面的方法用于旋转点的正确数学？如果是这样，哪一个是正确的？

  POINT rotate_point（浮点CX，CY浮动，浮动角度，点P）
{
  浮S =罪（角）;
  浮C = COS（角度）;  //回翻译点原点：
  p.x  -  = CX;
  p.y  -  = CY;  // 哪一个是对的：
  // 这个？
  浮xnew = p.x * C  -  p.y * S;
  浮ynew = p.x * S + p.y * C;
  // 或这个？
  浮xnew = p.x * C + p.y * S;
  浮ynew = -p.x * S + p.y * C;  //翻译点回：
  p.x = xnew + CX;
  p.y = ynew + CY;
}
 

解决方案

这要看你如何定义角度。如果它是逆时针方向测量（这是数学约定），则正确的旋转是您的第一个：

  //这个？
浮xnew = p.x * C  -  p.y * S;
浮ynew = p.x * S + p.y * C;
 

但如果是顺时针测量，然后第二个是正确的：

  //或那样吗？
浮xnew = p.x * C + p.y * S;
浮ynew = -p.x * S + p.y * C;
 

Do either of the below approaches use the correct mathematics for rotating a point? If so, which one is correct?

POINT rotate_point(float cx,float cy,float angle,POINT p)
{
  float s = sin(angle);
  float c = cos(angle);

  // translate point back to origin:
  p.x -= cx;
  p.y -= cy;

  // Which One Is Correct:
  // This?
  float xnew = p.x * c - p.y * s;
  float ynew = p.x * s + p.y * c;
  // Or This?
  float xnew = p.x * c + p.y * s;
  float ynew = -p.x * s + p.y * c;

  // translate point back:
  p.x = xnew + cx;
  p.y = ynew + cy;
}

解决方案

It depends on how you define angle. If it is measured counterclockwise (which is the mathematical convention) then the correct rotation is your first one:

// This?
float xnew = p.x * c - p.y * s;
float ynew = p.x * s + p.y * c;

But if it is measured clockwise, then the second is correct:

// Or This?
float xnew = p.x * c + p.y * s;
float ynew = -p.x * s + p.y * c;

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