# Calculating surface using Lambert correction?

i'm using lambert conformal conic , so we have distortions along any given parallel , i would like to get the area used to solve this distortion.

I have calculated the area using Shoelace formula , surprisingly it gave me the same area exactly as the area calculated by Arcgis, that means that Arcgis don't calculate Lambert Correction, this correction is necessary to correct distortions.

Please correct me, if this is wrong.

i would like to calculate the surface using Lambert correction, if you can help me to find the formula of Lambert Correction?

i'm using Arcgis 10.1 , i would like to solve the problem using ArcPy.

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Can you provide an example of when the "graphical" area is different from the "projected" area? Some links or further exposition would be nice. – blah238 Nov 10 '12 at 12:55
i'm using Lambert Conformal Conic but it portrays shape more accurately than area. this post could explain the difference forums.esri.com/Thread.asp?c=93&f=982&t=271682 – geogeek Nov 10 '12 at 13:07
That discussion forum post made no mention of Graphical Area so I too would like to see your Question revised to further explain what you (or others) mean by the term. – PolyGeo Nov 10 '12 at 20:42

The magnitude of the "Lambert Distortion" relies on a number of factors; your starting datum/ellipsoid, the relative size of the area you're calculating, the two standard parallels of your projection, etc. There's no universal "Lambert Correction" formula we can give you to plug into Arcpy. I believe there is a Lambert correction formula assuming a perfect sphere, but I'm willing to bet your projection probably isn't sphere-based.

The best way to "very accurately" calculate the area is to reproject out of Lambert Conformal Conic and into an Equal Area projection, and then calculate the area.

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to reproject into an Equal Area projection, is a really smart idea, once i test it i will you give my feedback. concerning the Equal Area projection, i can test with any Equal Area projection, the distance to the datum or any other parameters will not matter. – geogeek Nov 15 '12 at 7:29
your answer seems to be the closest to the solution, thanks – geogeek Nov 20 '12 at 23:07

Found this article, basically, you can call the Calculate Field tool from python and get the geodesic area -- if you are in geographic. When projected, it sounds like you can't get the geodesic area or distance? It seems you can only choose a more area-preserving projection.

http://blogs.esri.com/esri/arcgis/2011/07/21/calculating_geodesic/

"The field calculate actually calls the Calculate Field geoprocessing tool (you can see your calculations in the Result window), so the functionality the same between the two methods. For your second question: this will only work for data in a geographic coordinate system. We are working on enabling geodesic measurement for a number of tools, and in both projected and geographic coordinate systems, but this won’t be available until after 10.1."

See my post on geodesic calculations in javascript API -- seems better than desktop in some ways -- http://www.spatialexception.org/posts/arcgis-server-spatial-reference-faqs

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Not entirely true; in the underlying implementation if it's a projected spatial ref it will query the representative GCS it wraps and use that. – Jason Scheirer Nov 14 '12 at 8:15
If he needs to calculate the area "very accurately" then a geodesic area calculation is probably not the best approach. – Mintx Nov 14 '12 at 17:05
Of note, looking at the flex api for geodesic length and area. It says "The length [or area] will be calculated using a custom cylindrical equal-area projection". This is not mentioned in the javascript api. resources.arcgis.com/en/help/flex-api/apiref/com/esri/ags/utils/… When the term geodesic is used, I've been thinking, oh, cos sin tan on a ellipsoid are being used. Maybe not... – awesomo Nov 14 '12 at 21:28
the geodetic area could be very accurate, but i should use Lambert conformal conic. – geogeek Nov 15 '12 at 7:24
The geodetic area is based on your datum. It does not have anything to do with the Lambert conformal conic projection. – Mintx Nov 20 '12 at 16:14