We Keep Coding

How to generate a set of geospacial coordinates?

I tried using this:
function getRandomInRange(from, to, fixed) {
return parseFloat((Math.random() * (to - from) + from).toFixed(fixed));
}
var latLongPairs = 4;
for(var i =0; i<latLongPairs; i++) {
console.log(`${getRandomInRange(-180, 180, 3)}, ${getRandomInRange(-180, 180, 3)}`);
}
That's ok for random numbers like 39.21988,9.124741 but they might end up not valid coordinates. And I need 100 of them.

Your latitude range is incorrect - it should be ±90.
function getRandomInRange(from, to) {
return Math.random() * (to - from) + from;
}
const latLongPairs = 4;
for (let i = 0; i < latLongPairs; ++i) {
let lat = getRandomInRange(-90, 90);
let lon = getRandomInRange(-180, 180);
console.log(`${lat.toFixed(3)}, ${lon.toFixed(3)}`);
}

You might need to provide some initial Latitude and longitude with some radius (in meters)
var getRandomLocation = function(latitude, longitude, radiusInMeters) {
var getRandomCoordinates = function(radius, uniform) {
// Generate two random numbers
var a = Math.random(),
b = Math.random();
// Flip for more uniformity.
if (uniform) {
if (b < a) {
var c = b;
b = a;
a = c;
}
}
// It's all triangles.
return [
b * radius * Math.cos(2 * Math.PI * a / b),
b * radius * Math.sin(2 * Math.PI * a / b)
];
};
var randomCoordinates = getRandomCoordinates(radiusInMeters, true);
// Earths radius in meters via WGS 84 model.
var earth = 6378137;
// Offsets in meters.
var northOffset = randomCoordinates[0],
eastOffset = randomCoordinates[1];
// Offset coordinates in radians.
var offsetLatitude = northOffset / earth,
offsetLongitude = eastOffset / (earth * Math.cos(Math.PI * (latitude / 180)));
return `${latitude + (offsetLatitude * (180 / Math.PI))}, ${longitude + (offsetLongitude * (180 / Math.PI))}`
};
for (i = 0; i < 182; i++) {
console.log(getRandomLocation(41.8819, -87.6278, 50))
}

Plotting Points on Concentric Circles

I have an array of items that I would like to place on concentric circles. See diagram. I am having trouble with the math.
I can calculate the "steps" variable (each item is 40x40 so the steps are the circumference divided by the width plus margin). And I can calculate the points given the radius and the steps, but I don't know how to calculate the radius as a function of the current item index.
for(var i = 0; i < items.length; i++) {
var radius = functionOf(i)??;
var steps = Math.floor((2*radius*Math.PI)/60);
var x = Math.floor(0 + radius * Math.cos(2 * Math.PI * index / steps));
var y = Math.floor(0 + radius * Math.sin(2 * Math.PI * index / steps));
//draw item at x,y
}
Thoughts on how to calculate radius as a function of i?

You can calculate radius for every item using sum of arithmetic progression (for equidistant circles number of items form that progression), but there is simpler and faster approach - change radius after circle filling (pseudocode)
var radius = 0;
var i = 0;
while (i < items.length) {
var steps = Math.floor((2*radius*Math.PI)/60);
for(var index = 0; index < steps; index++) {
var x = Math.floor(0 + radius * Math.cos(2 * Math.PI * index / steps));
var y = Math.floor(0 + radius * Math.sin(2 * Math.PI * index / steps));
//draw item at x,y
i++;
if (i == items.length)
break;
}
radius = radius + 60; //start next circle
}
About arithmetic progression:
for progression with the first item a0 and difference d the sum of first n members is
S = n * (2 * a0 + d * (n - 1)) / 2
so to find what circle some index S belongs to, we have to solve quadratic inequality (find max integer x to fulfill the condition)
x^2 * d + x * (2 * a0 - d) - 2 * S <= 0
checked solution in Delphi:
function GetCircle(a0, d, i: Integer): Integer;
var
Discr: Double;
begin
Discr := (2 * a0 - d) * (2 * a0 - d) + 8 * i * d;
if Discr < 0 then
Exit(0);
Result := Floor((d - 2 * a0 + Sqrt(Discr)) / (2 * d));
end;
for case a0=1, d=6 (your picture differs a bit - progression is not exact) it gives
i=0: 0
i=1..7: 1
i=8..20: 2
i=21..39: 3
and so on
This means:
0th item is at the circle with radius 0
3rd item at the circle with radius 1 * R
11th item at the circle with radius 2 * R
...

Given your sample image, an easy approach is to add 6 elements for every outer circle, so that we have the sequence: 1, 6, 12, 18...
Then the radius of each circle can be enlarged by a constant amount, bigger then the element size:
// center of the circles
var centerx = 350;
var centery = 350;
// start drawing the central (first) element
if ( items.length > 0 ) {
// draw item at centerx, centery
}
var k = 1;
var i = 1;
while ( i < items.length ) {
// number of elements on this circle
var steps = k * 6;
// angular distance between elements
var angle_range = 2 * Math.PI / steps;
// every circle is bigger then the previuos of the same amount
var radius = k * 60;
var j = 0;
while ( j < steps && i < items.length ) {
var angle = j * angle_range;
var x = Math.round(centerx + radius * Math.cos(angle));
var y = Math.round(centery + radius * Math.sin(angle));
//draw item at x,y
i++;
j++;
}
k++;
}
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8"/>
<title>Circles</title>
<link href="stile.css"/>
</head>
<body>
<canvas id="myCanvas" width="700" height="700" style="border:1px solid #000000;">
</canvas>
<script>
// dummy draw function
function draw_item( itm, cntx, x, y) {
cntx.fillStyle = "#0022FF";
cntx.fillRect(x - 20, y - 20, 40, 40);
cntx.font = "14px Arial";
cntx.fillStyle = "#FFFFFF";
cntx.fillText(itm, x - 7, y + 5);
}
var c = document.getElementById("myCanvas");
var ctx = c.getContext("2d");
// dummy array
var items = [];
for ( var i = 0; i < 91; i++ ) {
items.push(i);
}
// center of the circles
var centerx = 350;
var centery = 350;
// start drawing the central (first) element
if ( items.length > 0 ) {
draw_item(items[0], ctx, centerx, centery);
}
var k = 1;
var i = 1;
while ( i < items.length ) {
// number of elements on this circle
var steps = k * 6;
// angular distance between elements
var angle_range = 2 * Math.PI / steps;
// every circle is bigger then the previuos of the same amount
var radius = k * 60;
var j = 0;
while ( j < steps && i < items.length ) {
var angle = j * angle_range;
var x = Math.round(centerx + radius * Math.cos(angle));
var y = Math.round(centery + radius * Math.sin(angle));
//draw item at x,y
draw_item(items[i], ctx, x, y);
i++;
j++;
}
k++;
}
</script>
</body>
</html>

Distribute points evenly on circle circumference in quadrants I and IV only

I want to distribute n points evenly on a circle circumference in quadrants I and IV only.
As parameters, I have the numbers of point n, the center of circle coordiantes cx and cy and the radius r.
I can distribute the points over the whole circumference like using this formula below, but I am looking for the formula to spread them only in quadrants I and IV
var n = 5;
var cx = 1;
var cy = 1;
var r = 2;
//I store each point's coordinates in this array below
var coordinates = [];
for (var i=0; i < n; i++) {
//defining point's angle with the center of the circle in radiant
//this angle distribute the points evenly over all 4 quadrants
var angle = ((360/n) * i) * (Math.PI/180);
//calculating the point's coordinates on the circle circumference
var pointX = cx + r * Math.cos(angle);
var pointY = cx + r * Math.sin(angle);
//storing the point's coordinates
coordinates.push([pointX, pointY]);
}

Here would be the steps I'd take to solve this:
find the angle betw. each point var incrementBy = 180 / n
start angle will be 270º and end angle will be 90º
iterate through via
code
var increment = 180 / n
var startAngle = 270
for (var i = 0; i < n; i++)
{
var angle = startAngle + increment * i
var rads = angle * Math.pi / 180
var tx = cx + r * Math.cos(rads)
var ty = cy + r * Math.sin(rads)
coords.push([tx, ty])
}
note
I didn't bother to convert for traditional quadrants (vs JS's y-axis moving downwards). If that is needed then, after your calculations, just invert the ty value. I also didn't bother to reduce the angle value when it exceeds 360º when you're incrementing back into Quad I.

like this?
var n = 5;
var r = 2;
var cx = 1;
var cy = 1;
var coordinates = [];
for(var i=0; i<n; ++i){
var a = (i+.5) / n * Math.PI;
coordinates.push([
cx + Math.sin(a) * r,
cy - Math.cos(a) * r
]);
}

Here is the way to equally distribute objects
var boxes = []
let count = 10;
for (let i = 0; i < count; i++) {
let size = 0.8;
let radius = 3;
let box = new THREE.Mesh(new THREE.BoxGeometry( size, size, size ),new THREE.MeshPhysicalMaterial(0x333333))
let gap = 0.5;
let angle = i * ((Math.PI * 2) / count);
let x = radius * Math.cos(angle);
let y = 0;
let z = radius * Math.sin(angle);
box.position.set(x,y,z);
boxes.push(box);`enter code here`
scene.add(box)
}
here is how it looks for 10 blocks
here is how it looks for 5 blocks

var n = 5;
var cx = 1;
var cy = 1;
var r = 2;
//I store each point's coordinates in this array below
var coordinates = [];
for (var i=0; i < n; i++) {
//defining the angle of the point with the center of the circle in radiant
var angle = ((360/n) * i) * (Math.PI/180);
//calculating the coordinates of the point on the circle circumference
var pointX = cx + r * Math.cos(angle);
var pointY = cx + r * Math.sin(angle);
// Here, we are going to use a boolean expression to determine if
// [pointX, pointY] is within quadrant 1 or 4.
// We can start with this boolean equation:
// (pointX >= cx && pointY >= cy) || (pointX >= cx && pointY <= cy)
// But this problem can be simplified to only pointX >= cx
if(pointX >= cx){
//storing the point's coordinates
coordinates.push([pointX, pointY]);
}
}

How do I move the starting point of X,Y coordinates plotted on a circle perimeter?

I want to plot a range of points on the lower left section (6 to 9 o'clock) of the perimeter of a circle. However, the starting point of rendering X,Y coordinates always begins at 3 o'clock.
!https://dl.dropboxusercontent.com/u/55849501/plotting-xy.png
Here is the rendering portion of my code:
var items = 5;
for(var i = 0; i < items; i++) {
var x = 96 + 100 * Math.cos(0.665 * Math.PI * i / items);
var y = 96 + 100 * Math.sin(0.665 * Math.PI * i / items);
$("#center").append("<div class='point' style='left:"+ x +"px;top:"+ y +"px'></div>");
}
And here is a jsfiddle of the code in action: http://jsfiddle.net/jE26S/198/
In summary:
I want the points to render starting at the 6 o'clock position instead of the 3 o'clock position.

What you are really doing here is interpolating between two values of theta. In your case, you want to start at Pi/2 and end at Pi. I took the liberty of re-writing your snippet using this interpolation paradigm. Also, you can adjust how far you want the dots/items away from the circle using outerCircleRadius.
var items = 5;
var startTheta = .5 * Math.PI;
var endTheta = 1 * Math.PI;
var outerCircleRadius = 112;
var cx = 90;
var cy = 90;
for(var i = 0; i < items; i++) {
var theta = startTheta + (endTheta - startTheta) * i / (items - 1)
var x = cx + outerCircleRadius * Math.cos(theta);
var y = cy + outerCircleRadius * Math.sin(theta);
$("#center").append("<div class='point' style='left:"+ x +"px;top:"+ y +"px'></div>");
}

Something like this?
var x = 86 + 100 * Math.cos(0.665 * Math.PI * (items-1+i-0.5) / items);
var y = 96 + 100 * Math.sin(0.665 * Math.PI * (items-1+i-0.5) / items);

Everything in Richard Shurtz's answer, except for the "items - 1" in the first line in the loop.
This worked for any number of items:
var items = 5;
var startTheta = .5 * Math.PI;
var endTheta = 1 * Math.PI;
var outerCircleRadius = 112;
var cx = 90;
var cy = 90;
for(var i = 0; i < items; i++) {
var theta = startTheta + (endTheta - startTheta) * i / items
var x = cx + outerCircleRadius * Math.cos(theta);
var y = cy + outerCircleRadius * Math.sin(theta);
$("#center").append("<div class='point' style='left:"+ x +"px;top:"+ y +"px'></div>");
}

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

i have a polyine which i have drawn with latlngs obtained from google maps directions service.
Now i want to find a point on the polyline that is closest to a given point.
The obvious way (to me) is to kind of loop through all the points in the polyline and find the distance between them and the given point, however this is inefficient because the points on the polyline can potentially be large.
I would be glad to hear any alternatives of doing this.
Thanks in advance.

I needed a cleaner version that was ported to V3, so here it is:
/**
* Snap marker to closest point on a line.
*
* Based on Distance to line example by
* Marcelo, maps.forum.nu - http://maps.forum.nu/gm_mouse_dist_to_line.html
* Then
* # work of Björn Brala - Swis BV who wrapped the algorithm in a class operating on GMap Objects
* And now
* Bill Chadwick, who factored the basic algorithm out of the class (removing much intermediate storage of results)
* and added distance along line to nearest point calculation
* Followed by
* Robert Crowe, who ported it to v3 of the Google Maps API and factored out the marker to make it more general.
*
* Usage:
*
* Create the class
* var oSnap = new cSnapToRoute();
*
* Initialize the subjects
* oSnap.init(oMap, oPolyline);
*
**/
function cSnapToRoute() {
this.routePoints = Array();
this.routePixels = Array();
this._oMap;
this._oPolyline;
/**
* #desc Initialize the objects.
* #param Map object
* #param GPolyline object - the 'route'
**/
this.init = function (oMap, oPolyline) {
this._oMap = oMap;
this._oPolyline = oPolyline;
this.loadRouteData(); // Load needed data for point calculations
}
/**
* #desc internal use only, Load route points into RoutePixel array for calculations, do this whenever zoom changes
**/
this.loadRouteData = function () {
this.routePixels = new Array();
var proj = this._oMap.getProjection();
for (var i = 0; i < this._oPolyline.getPath().getLength(); i++) {
var Px = proj.fromLatLngToPoint(this._oPolyline.getPath().getAt(i));
this.routePixels.push(Px);
}
}
/**
* #desc Get closest point on route to test point
* #param GLatLng() the test point
* #return new GLatLng();
**/
this.getClosestLatLng = function (latlng) {
var r = this.distanceToLines(latlng);
var proj = this._oMap.getProjection();
return proj.fromPointToLatLng(new google.maps.Point(r.x, r.y));
}
/**
* #desc Get distance along route in meters of closest point on route to test point
* #param GLatLng() the test point
* #return distance in meters;
**/
this.getDistAlongRoute = function (latlng) {
var r = this.distanceToLines(latlng);
return this.getDistToLine(r.i, r.fTo);
}
/**
* #desc internal use only, gets test point xy and then calls fundamental algorithm
**/
this.distanceToLines = function (thisLatLng) {
var tm = this._oMap;
var proj = this._oMap.getProjection();
var thisPx = proj.fromLatLngToPoint(thisLatLng);
var routePixels = this.routePixels;
return getClosestPointOnLines(thisPx, routePixels);
}
/**
* #desc internal use only, find distance along route to point nearest test point
**/
this.getDistToLine = function (iLine, fTo) {
var routeOverlay = this._oPolyline;
var d = 0;
for (var n = 1 ; n < iLine ; n++) {
d += routeOverlay.getPath().getAt(n - 1).distanceFrom(routeOverlay.getPath().getAt(n));
}
d += routeOverlay.getPath().getAt(iLine - 1).distanceFrom(routeOverlay.getPath().getAt(iLine)) * fTo;
return d;
}
}
/* 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 };
}

See Bill Chadwick's example here:
http://www.bdcc.co.uk/Gmaps/BdccGmapBits.htm
above example ported to v3 (code at bottom of this answer)
on his page under:
DISTANCE POINT TO POLYLINE OR POLYGON
from that post:
There is a similar, better demo here http://wtp2.appspot.com/cSnapToRouteDemo.html
It is finding the closest point on the line to the mouse. Also note that it is a Google Maps API v2 example (but the principle with v3 would be the same).
// Code to find the distance in metres between a lat/lng point and a polyline of lat/lng points
// All in WGS84. Free for any use.
//
// Bill Chadwick 2007
// updated to Google Maps API v3, Lawrence Ross 2014
// Construct a bdccGeo from its latitude and longitude in degrees
function bdccGeo(lat, lon)
{
var theta = (lon * Math.PI / 180.0);
var rlat = bdccGeoGeocentricLatitude(lat * Math.PI / 180.0);
var c = Math.cos(rlat);
this.x = c * Math.cos(theta);
this.y = c * Math.sin(theta);
this.z = Math.sin(rlat);
}
bdccGeo.prototype = new bdccGeo();
// internal helper functions =========================================
// Convert from geographic to geocentric latitude (radians).
function bdccGeoGeocentricLatitude(geographicLatitude)
{
var flattening = 1.0 / 298.257223563;//WGS84
var f = (1.0 - flattening) * (1.0 - flattening);
return Math.atan((Math.tan(geographicLatitude) * f));
}
// Returns the two antipodal points of intersection of two great
// circles defined by the arcs geo1 to geo2 and
// geo3 to geo4. Returns a point as a Geo, use .antipode to get the other point
function bdccGeoGetIntersection( geo1, geo2, geo3, geo4)
{
var geoCross1 = geo1.crossNormalize(geo2);
var geoCross2 = geo3.crossNormalize(geo4);
return geoCross1.crossNormalize(geoCross2);
}
//from Radians to Meters
function bdccGeoRadiansToMeters(rad)
{
return rad * 6378137.0; // WGS84 Equatorial Radius in Meters
}
//from Meters to Radians
function bdccGeoMetersToRadians(m)
{
return m / 6378137.0; // WGS84 Equatorial Radius in Meters
}
// properties =================================================
bdccGeo.prototype.getLatitudeRadians = function()
{
return (bdccGeoGeographicLatitude(Math.atan2(this.z,
Math.sqrt((this.x * this.x) + (this.y * this.y)))));
}
bdccGeo.prototype.getLongitudeRadians = function()
{
return (Math.atan2(this.y, this.x));
}
bdccGeo.prototype.getLatitude = function()
{
return this.getLatitudeRadians() * 180.0 / Math.PI;
}
bdccGeo.prototype.getLongitude = function()
{
return this.getLongitudeRadians() * 180.0 / Math.PI ;
}
// Methods =================================================
//Maths
bdccGeo.prototype.dot = function( b)
{
return ((this.x * b.x) + (this.y * b.y) + (this.z * b.z));
}
//More Maths
bdccGeo.prototype.crossLength = function( b)
{
var x = (this.y * b.z) - (this.z * b.y);
var y = (this.z * b.x) - (this.x * b.z);
var z = (this.x * b.y) - (this.y * b.x);
return Math.sqrt((x * x) + (y * y) + (z * z));
}
//More Maths
bdccGeo.prototype.scale = function( s)
{
var r = new bdccGeo(0,0);
r.x = this.x * s;
r.y = this.y * s;
r.z = this.z * s;
return r;
}
// More Maths
bdccGeo.prototype.crossNormalize = function( b)
{
var x = (this.y * b.z) - (this.z * b.y);
var y = (this.z * b.x) - (this.x * b.z);
var z = (this.x * b.y) - (this.y * b.x);
var L = Math.sqrt((x * x) + (y * y) + (z * z));
var r = new bdccGeo(0,0);
r.x = x / L;
r.y = y / L;
r.z = z / L;
return r;
}
// point on opposite side of the world to this point
bdccGeo.prototype.antipode = function()
{
return this.scale(-1.0);
}
//distance in radians from this point to point v2
bdccGeo.prototype.distance = function( v2)
{
return Math.atan2(v2.crossLength(this), v2.dot(this));
}
//returns in meters the minimum of the perpendicular distance of this point from the line segment geo1-geo2
//and the distance from this point to the line segment ends in geo1 and geo2
bdccGeo.prototype.distanceToLineSegMtrs = function(geo1, geo2)
{
//point on unit sphere above origin and normal to plane of geo1,geo2
//could be either side of the plane
var p2 = geo1.crossNormalize(geo2);
// intersection of GC normal to geo1/geo2 passing through p with GC geo1/geo2
var ip = bdccGeoGetIntersection(geo1,geo2,this,p2);
//need to check that ip or its antipode is between p1 and p2
var d = geo1.distance(geo2);
var d1p = geo1.distance(ip);
var d2p = geo2.distance(ip);
//window.status = d + ", " + d1p + ", " + d2p;
if ((d >= d1p) && (d >= d2p))
return bdccGeoRadiansToMeters(this.distance(ip));
else
{
ip = ip.antipode();
d1p = geo1.distance(ip);
d2p = geo2.distance(ip);
}
if ((d >= d1p) && (d >= d2p))
return bdccGeoRadiansToMeters(this.distance(ip));
else
return bdccGeoRadiansToMeters(Math.min(geo1.distance(this),geo2.distance(this)));
}
// distance in meters from GLatLng point to GPolyline or GPolygon poly
function bdccGeoDistanceToPolyMtrs(poly, point)
{
var d = 999999999;
var i;
var p = new bdccGeo(point.lat(),point.lng());
for(i=0; i<(poly.getPath().getLength()-1); i++)
{
var p1 = poly.getPath().getAt(i);
var l1 = new bdccGeo(p1.lat(),p1.lng());
var p2 = poly.getPath().getAt(i+1);
var l2 = new bdccGeo(p2.lat(),p2.lng());
var dp = p.distanceToLineSegMtrs(l1,l2);
if(dp < d)
d = dp;
}
return d;
}
// get a new GLatLng distanceMeters away on the compass bearing azimuthDegrees
// from the GLatLng point - accurate to better than 200m in 140km (20m in 14km) in the UK
function bdccGeoPointAtRangeAndBearing (point, distanceMeters, azimuthDegrees)
{
var latr = point.lat() * Math.PI / 180.0;
var lonr = point.lng() * Math.PI / 180.0;
var coslat = Math.cos(latr);
var sinlat = Math.sin(latr);
var az = azimuthDegrees* Math.PI / 180.0;
var cosaz = Math.cos(az);
var sinaz = Math.sin(az);
var dr = distanceMeters / 6378137.0; // distance in radians using WGS84 Equatorial Radius
var sind = Math.sin(dr);
var cosd = Math.cos(dr);
return new google.maps.LatLng(Math.asin((sinlat * cosd) + (coslat * sind * cosaz)) * 180.0 / Math.PI,
(Math.atan2((sind * sinaz), (coslat * cosd) - (sinlat * sind * cosaz)) + lonr) * 180.0 / Math.PI);
}

I do not think you can avoid checking all the points.
What if the not checked point is the nearest one?
If you have to do this operation many times, you can choose a data structure that is optimized for such a search, quadtree for example.
Note that you should not use lat lng as Descartes coordinates.
See also Finding nearest point in an efficient way
That is for the 2D plane, and not for lat lng, but you can approximate: https://stackoverflow.com/a/16271669/59019

Inspired by jmihalicza's answer, i came up with this function to find the closest point in an array of LatLngs to a given LatLng.
function closest takes a LatLng(llng) and an array of LatLngs (listData) and finds the distance between each latlng in the array and the given latlng, it then finds the least distance and returns the Latlng from the list which provided that distance.
function closest(llng, listData) {
var arr = listData;
var pnt = llng;
var distArr = [];
var dist = google.maps.geometry.spherical.computeDistanceBetween;
for (index in arr)
distArr.push([arr[index], dist(pnt, arr[index])]);
return distArr.sort(function(a,b){
return a[1]-b[1];
})[0][0];
}
EDIT
If you don't have access to the array of LatLngs which make up the polyline, but have access to the polyline itself, you can use polyline's getPath method to get the path which is an MVC array so you can use .getArray() to return an array of LatLngs to use with the above function (closest).