javascript - 在最接近第三点javascript的线上找到一个点

Question

我试图在一条直线上找到离直线第三个点最近的点。这些点是纬度/经度。

简单的图形显示了我正在努力实现的目标。我将它用于 javascript，但任何语言或公式仍然有效。我知道这是基本的几何学，但我仍然无法在 google 上找到公式 :S lol...留在学校!

var a = '48,-90';
var b = '49,-92';
var c = '48.25,-91.8';
var d = 'calculated point on line';

Answer 1

RCrowe @ Find a point in a polyline which is closest to a latlng

/* desc Static function. Find point on lines nearest test point
   test point pXy with properties .x and .y
   lines defined by array aXys with nodes having properties .x and .y 
   return is object with .x and .y properties and property i indicating nearest segment in aXys 
   and property fFrom the fractional distance of the returned point from aXy[i-1]
   and property fTo the fractional distance of the returned point from aXy[i]   */


function getClosestPointOnLines(pXy, aXys) {

    var minDist;
    var fTo;
    var fFrom;
    var x;
    var y;
    var i;
    var dist;

    if (aXys.length > 1) {

        for (var n = 1 ; n < aXys.length ; n++) {

            if (aXys[n].x != aXys[n - 1].x) {
                var a = (aXys[n].y - aXys[n - 1].y) / (aXys[n].x - aXys[n - 1].x);
                var b = aXys[n].y - a * aXys[n].x;
                dist = Math.abs(a * pXy.x + b - pXy.y) / Math.sqrt(a * a + 1);
            }
            else
                dist = Math.abs(pXy.x - aXys[n].x)

            // length^2 of line segment 
            var rl2 = Math.pow(aXys[n].y - aXys[n - 1].y, 2) + Math.pow(aXys[n].x - aXys[n - 1].x, 2);

            // distance^2 of pt to end line segment
            var ln2 = Math.pow(aXys[n].y - pXy.y, 2) + Math.pow(aXys[n].x - pXy.x, 2);

            // distance^2 of pt to begin line segment
            var lnm12 = Math.pow(aXys[n - 1].y - pXy.y, 2) + Math.pow(aXys[n - 1].x - pXy.x, 2);

            // minimum distance^2 of pt to infinite line
            var dist2 = Math.pow(dist, 2);

            // calculated length^2 of line segment
            var calcrl2 = ln2 - dist2 + lnm12 - dist2;

            // redefine minimum distance to line segment (not infinite line) if necessary
            if (calcrl2 > rl2)
                dist = Math.sqrt(Math.min(ln2, lnm12));

            if ((minDist == null) || (minDist > dist)) {
                if (calcrl2 > rl2) {
                    if (lnm12 < ln2) {
                        fTo = 0;//nearer to previous point
                        fFrom = 1;
                    }
                    else {
                        fFrom = 0;//nearer to current point
                        fTo = 1;
                    }
                }
                else {
                    // perpendicular from point intersects line segment
                    fTo = ((Math.sqrt(lnm12 - dist2)) / Math.sqrt(rl2));
                    fFrom = ((Math.sqrt(ln2 - dist2)) / Math.sqrt(rl2));
                }
                minDist = dist;
                i = n;
            }
        }

        var dx = aXys[i - 1].x - aXys[i].x;
        var dy = aXys[i - 1].y - aXys[i].y;

        x = aXys[i - 1].x - (dx * fTo);
        y = aXys[i - 1].y - (dy * fTo);

    }

    return { 'x': x, 'y': y, 'i': i, 'fTo': fTo, 'fFrom': fFrom };
}