试图计算圆上两点之间的角度？ [英] Trying to calculate the angle between two points on a circle?

问题描述

我试图计算两个点之间的角度，一个是在数学上圆圈，两点是用户在一定时间内拖动的距离。

这是我到目前为止：

  intervalId = setInterval （函数（）{
 
 if（p1x！= undefined）{
 p2x = Xpos; 
 
} 
 
 if（p1y！ = $ p 
 
 
 $ b （p1a * p1a）+（p1b * p1b））; 
 
 p2a = p2x -  wheelMiddleVer; 
 p2b = p2y  -  wheelMiddleHor; 
b = Math.sqrt（（p2a * p2a）+（p2b * p2b））; 
 
u = p1x  -  p2x; 
v = p1y  -  p2y; 
c = Math.sqrt（（u * u）+（v * v））; 
 
 
} 
 
 p1x = Xpos; 
 p1y = Ypos; 
 
 
}，1000）; 
  

我不知道如何完成它。我尝试过使用cos A =（b ^ 2 + c ^ 2 - a ^ 2）/ 2bc公式，但对我而言并不成功。我会很感激你对此的投入。如果我能让问题更清楚，让我知道。

解决方案

首先，请记住，大多数语言都有一个Y轴反转。意思是比例越高 - > Y值越小。

至于你的问题，你应该使用arctan：

  var angle = Math.atan（（p1y  -  p2y）/（p2x  -  p1x））*（180 / Math.PI）; 
  

First and foremost let me say I am an absolutely atrocious at maths, please bear with me.

I'm trying to calculate the angle between two points on a circle, the two points being the distance the user has dragged in a certain amount of time.

This is what I have so far:

intervalId = setInterval(function(){

      if(p1x != undefined){
        p2x = Xpos;

      }

      if(p1y != undefined){
        p2y = Ypos;
      }


      if( p1x != p2x || p1y != p2y ){

        p1a = p1x - wheelMiddleVer;
        p1b = p1y - wheelMiddleHor;
        a = Math.sqrt((p1a * p1a) + (p1b * p1b));

        p2a = p2x - wheelMiddleVer;
        p2b = p2y - wheelMiddleHor;
        b = Math.sqrt((p2a * p2a) + (p2b * p2b));

        u = p1x - p2x;
        v = p1y - p2y;
        c = Math.sqrt((u * u) + (v * v));


      }

      p1x = Xpos;
      p1y = Ypos;


  }, 1000);

I'm not sure how to finish it. I have tried using the cos A = (b^2 + c^2 - a^2)/2bc formula but it's not worked out for me. I will appreciate your input on this. If I can make the question clearer let me know.

解决方案

First of all, keep in mind that most of the languages have an invert Y axis. Meaning the higher the point on the scale -> the smaller it Y value is.

As for your question, you should use arctan:

var angle = Math.atan((p1y - p2y) / (p2x - p1x)) * (180 / Math.PI);

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