相对于线段转换坐标 [英] translate coordinates relative to line segment

问题描述

我需要编写一个函数（使用Javascript），它接受一个线段的端点和一个附加点，并返回该点相对于起点的坐标。所以，基于这张图片：

<$ p $函数vertical_coords（x1，y1，x2，y2，xp，yp）

返回 [xp'，yp'] 其中xp'是沿着垂直于线段的线的距离（xp，yp），yp'是距离从（x1，y1）到垂直线与线段相交点。

到目前为止我尝试过的方法是：

 函数rotateRad（cx，cy，x，y，radians）{
 var cos = Math.cos（弧度），
 sin =数学.sin（弧度），
 nx =（cos *（x-cx））+（sin *（y -cy））+ cx，
 ny = （sin *（x-cx））+ cy; 
 return [nx，ny]; 
} 
 
 [xp'，yp'] = rotateRad（x1，y1，xp，yp，Math.atan2（y2-y1，x2-x1））; 
  

我没有写这个函数;从 https://stackoverflow.com/a/17411276/1368860 获得它

1. yp'我使用这里的答案找到相交点：垂直于从给定点开始


2. 使用以下等式计算（xp，yp）和直线之间的距离： b
   $ b

   不是最优雅的解决方案，但它似乎可行。

   
   
   

   结果代码 https://jsfiddle.net/qke0m4mb/

    函数的n vertical_coords（x1，y1，x2，y2，xp，yp）{
    var dx = x2  -  x1，
    dy = y2  -  y1; 
    
    //找到交点
    var k =（dy *（xp-x1） -  dx *（yp-y1））/（dy * dy + dx * dx）; 
    var x4 = xp  -  k * dy; 
    var y4 = yp + k * dx; （y4-y1）*（y4-y1）+（x4-x1）*（x4-x1））;（b4 b1 b 
    
    var ypt = Math.sqrt 
    var xpt = distance（x1，y1，x2，y2，xp，yp）; 
    return [xpt，ypt]; 
   } 
    
    //点距线的距离
   函数距离（x1，y1，x2，y2，xp，yp）{
    var dx = x2  -  x1 ; 
    var dy = y2  -  y1; 
    
    return Math.abs（dy * xp  -  dx * yp + x2 * y1  -  y2 * x1）/ Math.sqrt（dy * dy + dx * dx）; 
   } 
     

   
   

   I need to write a function (in Javascript) that accepts the endpoints of a line segment and an additional point and returns coordinates for the point relative to the start point. So, based on this picture:

   function perpendicular_coords(x1,y1, x2,y2, xp,yp)
   

   returns [xp',yp'] where xp' is the distance from (xp,yp) along a line perpendicular to the line segment, and yp' is the distance from (x1,y1) to the point where that perpendicular line intersects the line segment.

   What I've tried so far:

   function rotateRad(cx, cy, x, y, radians) {
       var cos = Math.cos(radians),
           sin = Math.sin(radians),
           nx = (cos * (x - cx)) + (sin * (y - cy)) + cx,
           ny = (cos * (y - cy)) - (sin * (x - cx)) + cy;
       return [nx, ny];
   }
   
   [xp', yp'] = rotateRad(x1,y1, xp, yp, Math.atan2(y2-y1,x2-x1));
   

   I didn't write the function; got it from https://stackoverflow.com/a/17411276/1368860

   解决方案

   I took a different approach, by combining two functions:

   1. To get yp' I find the intersection point using the answer from here: Perpendicular on a line from a given point
   2. To get xp' I calculate the distance between (xp, yp) and the line, using the equation from here: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line

   Not the most elegant solution, but it seems to work.

   Resulting code https://jsfiddle.net/qke0m4mb/

   function perpendicular_coords(x1, y1, x2, y2, xp, yp) {
     var dx = x2 - x1,
         dy = y2 - y1;
   
     //find intersection point    
     var k = (dy * (xp-x1) - dx * (yp-y1)) / (dy*dy + dx*dx);
     var x4 = xp - k * dy;
     var y4 = yp + k * dx;
   
     var ypt = Math.sqrt((y4-y1)*(y4-y1)+(x4-x1)*(x4-x1));
     var xpt = distance(x1, y1, x2, y2, xp, yp);
     return [xpt, ypt];
   }
   
   // Distance of point from line
   function distance(x1, y1, x2, y2, xp, yp) {
     var dx = x2 - x1;
     var dy = y2 - y1;
   
     return Math.abs(dy*xp - dx*yp + x2*y1 - y2*x1) / Math.sqrt(dy*dy + dx*dx);
   }
   

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