Caleb Pitman Caleb Pitman - 1 month ago 15
Javascript Question

find a point on a line closest to a third point javascript

I'm trying to find a point on a line closest to a third point off of the line. The points are latitude/longitude.

The simple graphic shows what I'm trying to achieve. I'm using it for javascript, but any language or formula would still work. I know this is basic geometry, but I'm still having trouble finding a formula on google :S lol... stay in school!

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


enter image description here

Answer

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 };
}