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388

answers:

2

To draw a circle on map I have a center GLatLng (A) and a radius (r) in meters.

Here's a diagram:

           -----------
        --/           \--
      -/                 \-
     /                     \
    /                       \
   /                   r     \
   |            *-------------*
   \             A           / B
    \                       /
     \                     /
      -\                 /-
        --\           /--
           -----------

How to calculate the GLatLng at position B? Assuming that r is parallel to the equator.

Getting the radius when A and B is given is trivial using the GLatLng.distanceFrom() method - but doing it the other way around not so. Seems that I need to do some heavier math.

+1  A: 

The answer to this question and more can be found here: http://williams.best.vwh.net/avform.htm

Paul Tomblin
I do see a lot of acos, sin, tan, etc on that page. But I'll stick with the answer from Daniel. Thanks for helping.
Rene Saarsoo
+6  A: 

We will need a method that returns the destination point when given a bearing and the distance travelled from a source point. Luckily, there is a very good JavaScript implementation by Chris Veness at Calculate distance, bearing and more between Latitude/Longitude points.

The following has been adapted to work with the google.maps.LatLng class:

Number.prototype.toRad = function() {
   return this * Math.PI / 180;
}

Number.prototype.toDeg = function() {
   return this * 180 / Math.PI;
}

google.maps.LatLng.prototype.destinationPoint = function(brng, dist) {
   dist = dist / 6371;  
   brng = brng.toRad();  

   var lat1 = this.lat().toRad(), lon1 = this.lng().toRad();

   var lat2 = Math.asin(Math.sin(lat1) * Math.cos(dist) + 
                        Math.cos(lat1) * Math.sin(dist) * Math.cos(brng));

   var lon2 = lon1 + Math.atan2(Math.sin(brng) * Math.sin(dist) *
                                Math.cos(lat1), 
                                Math.cos(dist) - Math.sin(lat1) *
                                Math.sin(lat2));

   if (isNaN(lat2) || isNaN(lon2)) return null;

   return new google.maps.LatLng(lat2.toDeg(), lon2.toDeg());
}

You would simply use it as follows:

var pointA = new google.maps.LatLng(25.48, -71.26); 
var radiusInKm = 10;

var pointB = pointA.destinationPoint(90, radiusInKm);

Here is a complete example using Google Maps API v3:

<!DOCTYPE html>
<html> 
<head> 
   <meta http-equiv="content-type" content="text/html; charset=UTF-8"/> 
   <title>Google Maps Geometry</title> 
   <script src="http://maps.google.com/maps/api/js?sensor=false" 
           type="text/javascript"></script> 
</head> 
<body> 
   <div id="map" style="width: 400px; height: 300px"></div> 

   <script type="text/javascript"> 
      Number.prototype.toRad = function() {
         return this * Math.PI / 180;
      }

      Number.prototype.toDeg = function() {
         return this * 180 / Math.PI;
      }

      google.maps.LatLng.prototype.destinationPoint = function(brng, dist) {
         dist = dist / 6371;  
         brng = brng.toRad();  

         var lat1 = this.lat().toRad(), lon1 = this.lng().toRad();

         var lat2 = Math.asin(Math.sin(lat1) * Math.cos(dist) + 
                              Math.cos(lat1) * Math.sin(dist) * Math.cos(brng));

         var lon2 = lon1 + Math.atan2(Math.sin(brng) * Math.sin(dist) *
                                      Math.cos(lat1), 
                                      Math.cos(dist) - Math.sin(lat1) *
                                      Math.sin(lat2));

         if (isNaN(lat2) || isNaN(lon2)) return null;

         return new google.maps.LatLng(lat2.toDeg(), lon2.toDeg());
      }

      var pointA = new google.maps.LatLng(40.70, -74.00);   // Circle center
      var radius = 10;                                      // 10km

      var mapOpt = { 
         mapTypeId: google.maps.MapTypeId.TERRAIN,
         center: pointA,
         zoom: 10
      };

      var map = new google.maps.Map(document.getElementById("map"), mapOpt);

      // Draw the circle
      new google.maps.Circle({
         center: pointA,
         radius: radius * 1000,       // Convert to meters
         fillColor: '#FF0000',
         fillOpacity: 0.2,
         map: map
      });

      // Show marker at circle center
      new google.maps.Marker({
         position: pointA,
         map: map
      });

      // Show marker at destination point
      new google.maps.Marker({
         position: pointA.destinationPoint(90, radius),
         map: map
      });
   </script> 
</body> 
</html>

Screenshot:

Google Maps Geometry

UPDATE:

In reply to Paul's comment below, this is what happens when the circle wraps around one of the poles.

Plotting pointA near the north pole, with a radius of 1,000km:

  var pointA = new google.maps.LatLng(85, 0);   // Close to north pole
  var radius = 1000;                            // 1000km

Screenshot for pointA.destinationPoint(90, radius):

Close to north pole

Daniel Vassallo
So does this always give the destination point lying in the east?
Nirmal
@Nirmal: No it depends on the first paramater you pass to `destinationPoint()`. 90 is East, but you can use any bearing, starting from 0 = North, moving clockwise.
Daniel Vassallo
Thanks, that's perfect.
Rene Saarsoo
Will this work near the poles? I.e. will it produce something that doesn't look circular on the map, but which is circular on the globe?
Paul Tomblin
@Paul: Interesting observation. I've updated the answer with an example. Google Maps uses the Mercator projection (http://en.wikipedia.org/wiki/Mercator_projection), which distorts the size and shape of large objects, as the scale increases from the Equator to the poles, where it becomes infinite. Using the above example, with `var pointA = new google.maps.LatLng(85, 0);` and a radius of 1000km `var radius = 1000;` appears to work. The circle cap wraps around the north pole, so on the 2D projection it does not have the shape of a circle anymore. The destination point appears to be right.
Daniel Vassallo
Can i find a detailed description of the geometry in destinationPoint ?
The Machine
@Daniel, that looks like the correct result. Thanks.
Paul Tomblin
@The Machine: Did you check http://www.movable-type.co.uk/scripts/latlong.html?... It links to some external articles with additional details.
Daniel Vassallo
@Daniel, That was a quality answer one could get for the original question. Please keep up the good work!
Nirmal