[OpenLayers-Users] Determining polygon area? Trying to see through the confusion.

Pierre GIRAUD bluecarto at gmail.com
Sat Oct 18 04:49:57 EDT 2008


I think that there's something already done in OpenLayers, at least in
the examples.
Have a look at the measure example [1], there's kind of a conversion
thing using distVincenty.

Hope this helps,
Pierre

[1] http://www.openlayers.org/dev/examples/measure.html

On Fri, Oct 17, 2008 at 7:25 PM, Bill Thoen <bthoen at gisnet.com> wrote:
> plen wrote:
>> I see that there is a feature.geometry.getArea() function.  This returns
>> some value which I am unsure as to its meaning (square degrees?).  I have
>> seen previous posts that sort of get into determining polygon areas but
>> nothing seems clear.  Maybe this is a bigger type issue than can be handled
>> by the OL API, not sure.  I am hoping that someone could answer (or lead to
>> an answer) any of the following:
>>
>> 1) What unit of measurement is returned by the getArea() function?
>>
>> 2) How to convert that to something like square miles, meters, kilometers,
>> etc.
>>
>> 3) Determining the area for regular polygons vs irrlegular polygons.
>
> It looks like it returns an area in terms of the units of your
> coordinate system. If you are using decimal degrees, then the result is
> essentially meaningless, i.e 'square degrees'.  To get a reasonable
> "flat land" area use coordinate units in meters or feet and then apply
> the appropriate conversion factor to get square miles, hectares, etc.
> The method does appear to accommodate irregular polygons (ones with
> holes in them) by subtracting the area of the holes from the area of the
> outer ring.
>
> If you're up to the coding, there's a description of Girad's Algorithm
> for calculating the surface area of a spherical triangle at
> http://math.rice.edu/~pcmi/sphere/ and that can be extended to determine
> the area of a polygon on a spherical surface using decimal degrees. I
> think this would be a more appropriate method for calculating areas on
> the earth's surface.
>
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