仅在象限I和IV中均匀分布圆周上的点 [英] Distribute points evenly on circle circumference in quadrants I and IV only

问题描述

我想仅在象限I和IV中均匀分布 n 点数。

参数，我有点数 n ，圈子坐标的中心 cx 和 cy 和半径 r 。

我可以在整个圆周上分配点使用下面的公式，但我正在寻找的公式只能在象限I和IV中分布它们

  var n = 5; 
 var cx = 1; 
 var cy = 1; 
 var r = 2; 
 
 //我将这个数组中每个点的坐标存储在
下面的数组中var coordinates = []; 
 
 for（var i = 0; i< n; i ++）{
 //定义点的角度与辐射的圆的中心
 //这个角度分布在所有4个象限上均匀分布
 var angle =（（360 / n）* i）*（Math.PI / 180）; 
 
 //计算圆周上的点坐标
 var pointX = cx + r * Math.cos（angle）; 
 var pointY = cx + r * Math.sin（angle）; 
 
 //存储点的坐标
 coordinates.push（[pointX，pointY]）; 
} 
  

解决方案

'需要解决这个问题：

1. 找到角度。每个点 var incrementBy = 180 / n


2. 开始角度为270º，结束角度为90º
   
3. 通过
   

代码进行迭代

  var increment = 180 / n 
 var startAngle = 270 
 for（var i = 0; i< n; i ++）
 {
 var angle = startAngle + increment * i 
 var rads = angle * Math.pi / 180 
 
 var tx = cx + r * Math.cos（angle）
 var ty = cy + r * Math.sin（angle）
 
 coords.push（[tx，ty]）
} 
  

note

我没有为传统象限转换（vs JS的y轴向下移动）。如果需要，那么在计算之后，只需反转 ty 值。当您增加到四分之一时，我也不会在超过360º时减小角度值。

I want to distribute n points evenly on a circle circumference in quadrants I and IV only.

As parameters, I have the numbers of point n, the center of circle coordiantes cx and cy and the radius r.

I can distribute the points over the whole circumference like using this formula below, but I am looking for the formula to spread them only in quadrants I and IV

var n = 5;
var cx = 1;
var cy = 1;
var r = 2;

//I store each point's coordinates in this array below
var coordinates = [];

for (var i=0; i < n; i++) {
    //defining point's angle with the center of the circle in radiant
    //this angle distribute the points evenly over all 4 quadrants
    var angle = ((360/n) * i) * (Math.PI/180);

    //calculating the point's coordinates on the circle circumference
    var pointX = cx + r * Math.cos(angle);
    var pointY = cx + r * Math.sin(angle);

    //storing the point's coordinates
    coordinates.push([pointX, pointY]);
}

解决方案

Here would be the steps I'd take to solve this:

1. find the angle betw. each point var incrementBy = 180 / n
2. start angle will be 270º and end angle will be 90º
3. iterate through via

code

 var increment = 180 / n
 var startAngle = 270
 for (var i = 0; i < n; i++)
 {
     var angle = startAngle + increment * i
     var rads = angle * Math.pi / 180

     var tx = cx + r * Math.cos(angle)
     var ty = cy + r * Math.sin(angle)

     coords.push([tx, ty])
 }

note

I didn't bother to convert for traditional quadrants (vs JS's y-axis moving downwards). If that is needed then, after your calculations, just invert the ty value. I also didn't bother to reduce the angle value when it exceeds 360º when you're incrementing back into Quad I.

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