openlayers 根据起点和终点半径画圆弧

Posted 2022-05-27 no_shutdown_

原理：

1.先计算出圆心（结果有两个值一个在两个点的连线上面一个在两个点的下面）

2.选择圆心，计算两点之间指定个数点的坐标

3.将坐标集合以线的形式绘制出来（点越多越接近圆）

图上根据选择的圆心绘制了两个反方向的圆弧，中间横线为两点之间的连线用来验证坐标计算的

计算坐标的代码tool.js

function get_circle(point_s, point_e, radius, flag = true) 
    //返回结果变量    let x0, y0, x1, y1
    //计算过程    if (point_e[0] == point_s[0]) 
        y0 = y1 = (point_s[1] + point_e[1]) / 2        let delta_y = (y0 - point_s[1]) ** 2        let delta_x = .sqrt(radius ** 2 - delta_y)
        x0 = point_e[0] - delta_x
        x1 = point_e[0] + delta_x
     else 
        let C1 = (point_e[0] ** 2 + point_e[1] ** 2 - point_s[0] ** 2 - point_s[1] ** 2) / 2 / (point_e[0] - point_s[0])
        let C2 = (point_e[1] - point_s[1]) / (point_e[0] - point_s[0])
        let A = 1 + C2 ** 2        let B = 2 * (point_s[0] - C1) * C2 - 2 * point_s[1]
        let C = (point_s[0] - C1) ** 2 + point_s[1] ** 2 - radius ** 2        y0 = (-B + .sqrt(B * B - 4 * A * C)) / 2 / A
        y1 = (-B - .sqrt(B * B - 4 * A * C)) / 2 / A
        x0 = C1 - C2 * y0
        x1 = C1 - C2 * y1
    
    //判断返回哪个圆心    
    if (flag) 
        return [x1, y1]
     else 
        return [x0, y0]
    



var rs_points = function rs_points(point_s, point_e, radius, dot_count, flag = true) 
    let line = []
    let rs = get_circle(point_s, point_e, radius,flag)
    let a = rs[0], b = rs[1]
    let ad_len = 0    
    let y = 0    
    //计算x坐标的距离    
    ad_len =Math.abs(point_s[0] - point_e[0])
   
    //根据需要打散的点数 计算每个x坐标    
    for (let i = 0; i < dot_count; i++) 
        let x = point_s[0]
        if (point_s[1] > point_e[1]) 
            x = x + ad_len / dot_count * i
         else 
            x = x + ad_len / dot_count * i
        
        //计算指定x点的y坐标        
        if (flag) 
            //如果返回的圆心上面            
            y = b + Math.sqrt(radius * radius - (x - a) * (x - a))
         else 
            //如果返回的圆心下面            
            y = b - Math.sqrt(radius * radius - (x - a) * (x - a))
        
        line.push([x, y])
    
    line.push(point_e)
    //返回所有在两点之间的圆弧上的点的集合    
    return line


exports.rs_points = rs_points