Does anyone know the best way to go about getting the mid-point of a pair of latitude and longitude points?
I mm using d3.js to draw points on a map and need to draw
Apologies for the long script - it just seemed fun to draw stuff :-). I've marked off sections that are not required
// your latitude / longitude
var co2 = [70, 48];
var co1 = [-70, -28];
// NOT REQUIRED
var ctx = document.getElementById("myChart").getContext("2d");
function drawPoint(color, point) {
ctx.fillStyle = color;
ctx.beginPath();
ctx.arc(point.x, point.y, 5, 0, 2 * Math.PI, false);
ctx.fill();
}
function drawCircle(point, r) {
ctx.strokeStyle = 'gray';
ctx.setLineDash([5, 5]);
ctx.beginPath();
ctx.arc(point.x, point.y, r, 0, 2 * Math.PI, false);
ctx.stroke();
}
// REQUIRED
// convert to cartesian
var projection = d3.geo.equirectangular()
var cot1 = projection(co1);
var cot2 = projection(co2);
var p0 = { x: cot1[0], y: cot1[1] };
var p1 = { x: cot2[0], y: cot2[1] };
// NOT REQUIRED
drawPoint('green', p0);
drawPoint('green', p1);
// REQUIRED
function dfn(p0, p1) {
return Math.pow(Math.pow(p0.x - p1.x, 2) + Math.pow(p0.y - p1.y, 2), 0.5);
}
// from http://math.stackexchange.com/a/87374
var d = dfn(p0, p1);
var m = {
x: (p0.x + p1.x) / 2,
y: (p0.y + p1.y) / 2,
}
var u = (p1.x - p0.x) / d
var v = (p1.y - p0.y) / d;
// increase 1, if you want a larger curvature
var r = d * 1;
var h = Math.pow(Math.pow(r, 2) - Math.pow(d, 2) / 4, 0.5);
// 2 possible centers
var c1 = {
x: m.x - h * v,
y: m.y + h * u
}
var c2 = {
x: m.x + h * v,
y: m.y - h * u
}
// NOT REQUIRED
drawPoint('gray', c1)
drawPoint('gray', c2)
drawCircle(c1, r)
drawCircle(c2, r)
// REQUIRED
// from http://math.stackexchange.com/a/919423
function mfn(p0, p1, c) {
// the -c1 is for moving the center to 0 and back again
var mt1 = {
x: r * (p0.x + p1.x - c.x * 2) / Math.pow(Math.pow(p0.x + p1.x - c.x * 2, 2) + Math.pow(p0.y + p1.y - c.y * 2, 2), 0.5)
};
mt1.y = (p0.y + p1.y - c.y * 2) / (p0.x + p1.x - c.x * 2) * mt1.x;
var ma = {
x: mt1.x + c.x,
y: mt1.y + c.y,
}
var mb = {
x: -mt1.x + c.x,
y: -mt1.y + c.y,
}
return (dfn(ma, p0) < dfn(mb, p0)) ? ma : mb;
}
var m1 = mfn(p0, p1, c1);
var m2 = mfn(p0, p1, c2);
var mo1 = projection.invert([m1.x, m1.y]);
var mo2 = projection.invert([m2.x, m2.y]);
// NOT REQUIRED
drawPoint('blue', m1);
drawPoint('blue', m2);
// your final output (in lat long)
console.log(mo1);
console.log(mo2);
Fiddle - https://jsfiddle.net/srjuc2gd/
And here's just the relevant portion (most of it is just copy-pasta from the beginning of this answer)
var Q31428016 = (function () {
// adjust curvature
var CURVATURE = 1;
// project to convert from lat / long to cartesian
var projection = d3.geo.equirectangular();
// distance between p0 and p1
function dfn(p0, p1) {
return Math.pow(Math.pow(p0.x - p1.x, 2) + Math.pow(p0.y - p1.y, 2), 0.5);
}
// mid point between p0 and p1
function cfn(p0, p1) {
return {
x: (p0.x + p1.x) / 2,
y: (p0.y + p1.y) / 2,
}
}
// get arc midpoint given end points, center and radius - http://math.stackexchange.com/a/919423
function mfn(p0, p1, c, r) {
var m = cfn(p0, p1);
// the -c1 is for moving the center to 0 and back again
var mt1 = {
x: r * (m.x - c.x) / Math.pow(Math.pow(m.x - c.x, 2) + Math.pow(m.y - c.y, 2), 0.5)
};
mt1.y = (m.y - c.y) / (m.x - c.x) * mt1.x;
var ma = {
x: mt1.x + c.x,
y: mt1.y + c.y,
}
var mb = {
x: -mt1.x + c.x,
y: -mt1.y + c.y,
}
return (dfn(ma, p0) < dfn(mb, p0)) ? ma : mb;
}
var Q31428016 = {};
Q31428016.convert = function (co1, co2) {
// convert to cartesian
var cot1 = projection(co1);
var cot2 = projection(co2);
var p0 = { x: cot1[0], y: cot1[1] };
var p1 = { x: cot2[0], y: cot2[1] };
// get center - http://math.stackexchange.com/a/87374
var d = dfn(p0, p1);
var m = cfn(p0, p1);
var u = (p1.x - p0.x) / d
var v = (p1.y - p0.y) / d;
var r = d * CURVATURE;
var h = Math.pow(Math.pow(r, 2) - Math.pow(d, 2) / 4, 0.5);
// 2 possible centers
var c1 = {
x: m.x - h * v,
y: m.y + h * u
}
var c2 = {
x: m.x + h * v,
y: m.y - h * u
}
// get arc midpoints
var m1 = mfn(p0, p1, c1, r);
var m2 = mfn(p0, p1, c2, r);
// convert back to lat / long
var mo1 = projection.invert([m1.x, m1.y]);
var mo2 = projection.invert([m2.x, m2.y]);
return [mo1, mo2]
}
return Q31428016;
})();
// your latitude / longitude
var co1 = [-70, -28];
var co2 = [70, 48];
var mo = Q31428016.convert(co1, co2)
// your output
console.log(mo[0]);
console.log(mo[1]);