Mismatch between AREAINT and AREAQUAD around (0,0).

Dear all,

 E_earth = referenceEllipsoid( 'earth' ) ;

 lat = [-1, 1] + 0 ;  % +0 is placeholder for easy manual shift.
 lon = [-1, 1] + 0 ;

 areaquad( min(lat), min(lon), max(lat), max(lon), E_earth )
 areaint( [min(lat), min(lat), max(lat), max(lat), min(lat), NaN], ...
     [max(lon), min(lon), min(lon), max(lon), max(lon), NaN], E_earth )

This outputs:

 ans =
   4.9234e+10

 ans =
   7.7333e+10

Their ratio is close to pi/2, and the output of AREAQUAD matches what I get with ArcGIS for example.

Thanks!

Cedric

[Marcus Adkins]

Comparing the results of a 0.02x0.02 (lat/lon) area against STK's computations, you get agreement to 5 significant digits until you hit +/- 1 deg latitude. In that range the Matlab areaInt function is off by as much as 50%

Answer 1

[Amy Haskins]

The doc for areaint hints that the accuracy of the function improves with denser sampling. In your you example, you only have the corner points of the quadrangle in the polygon passed to areaint. The function outlinegeoquad can be used to get a more densely sampled boundary of the quadrangle. Using this function, I get answers which are increasing close to that returned by areaquad.

>> E_earth = referenceEllipsoid( 'earth' ) ;
>> [latq,lonq] = outlinegeoquad(lat,lon,0.1,0.1);
>> areaint( latq, lonq, E_earth )

ans =

     4.9304e+10

>> [latq,lonq] = outlinegeoquad(lat,lon,0.01,0.01);
>> areaint( latq, lonq, E_earth )

ans =

     4.9235e+10

[Cedric]

I see now

"Accuracy of the integration method is inversely proportional to the distance between lat/lon points"

and I hadn't realized that a 2deg difference around the origin would be troublesome when it isn't elsewhere.

Thank you very much!