[OpenLayers-Users] Function to create a geographically correct circle

[Michael]

I suppose it could work if we added the number of sides down in the line 
that provides "*  for (x = 0; x <= 360; x+=6)*".  Probably needs more 
tweaking to get squares and triangles to look like we expect.

I'm not sure how many folks besides me need this, but if there's a 
perceived need, I'm willing to tune it up.

oops.  Just noticed in the code that I use my shorthand definition of 
pt=new OpenLayers Point.  fixing code below

Michael

On 6/28/2011 6:33 AM, Janis Elmeris wrote:
> It would be a good idea to add this function to 
> OpenLayers.Geometry.Polygon along with createRegularPolygon that is 
> already there. Something like createGeodesicRegularPolygon as there is 
> getGeodesicArea.
>
> Janis
>
> On 2011.06.06. 8:20, Michael wrote:
>> Regular Polygons are fine for many applications.  I needed range 
>> rings for ocean races, which did not correctly reflect the projection 
>> of the map.  Although it projected like a nice round circle, the 
>> correct shape should have been oval.
>>
>> Did not find a geographic circle function in OL, so I adapted one 
>> from one found online.
>>
>> I present it here should anyone want it, or have a better idea, or 
>> want to point out that this was in OL all along.
>>
>> *function GeoCircle(latin, lonin, radius)// thanks to VE site for 
>> basic algo. Returns an OL Polygon of diameter in nautical miles
>>   {
>>     var locs = new Array();
>>     var lat1 = latin * Math.PI/180.0;
>>     var lon1 = lonin * Math.PI/180.0;
>>     var d = radius/3440.07;// change this value to the radius of the 
>> Earth in your units
>>     var x;
>>     for (x = 0; x <= 360; x+=6)  //gives 60 points.  More than enough 
>> for a smooth circle most likely
>>     {
>>         var tc = (x / 90)* Math.PI / 2;
>>         var lat = 
>> Math.asin(Math.sin(lat1)*Math.cos(d)+Math.cos(lat1)*Math.sin(d)*Math.cos(tc));
>>         lat = 180.0 * lat / Math.PI;
>>         var lon;
>>          if (Math.cos(lat1)==0)
>>         {
>>             lon=lonin; // endpoint a pole
>>         }
>>         else
>>         {
>>             lon = ((lon1 - Math.asin(Math.sin(tc) * 
>> Math.sin(d)/Math.cos(lat1)) + Math.PI) % (2 * Math.PI)) - Math.PI;
>>         }
>>         lon = 180.0 * lon / Math.PI;
>>         var loc = new OpenLayers.Point(lon,lat);
>>         locs.push(loc);
>>     }
>>     var poly = new OpenLayers.Feature.Vector(new 
>> OpenLayers.Geometry.Polygon(new OpenLayers.Geometry.LinearRing(locs)));
>>
>>     return poly;  //this can be added to a map
>> }
>> *
>>
>>
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>
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