areaint has a singularity?

In MATLAB17a I am computing surface integrals for rectangular polygons on a sphere ("earth") using areaint and wgs84Ellipsoid.

I get annomalously large surface areas in meter-squared when my rectangular polygons cover latitude 0 and longitude 0, 180, or 360 degrees. Away from these points the surface area is what you would expect. Is this a singularity (an error), or am I doing something wrong?

% example code
dy  = 2.5;
E = wgs84Ellipsoid;

% declare longitudes and latitudes
%lonc = -5:1:5;
lonc = 175:1:185;
latc = -5:1:5;

AA = ones(length(latc),length(lonc)); %declare area
for i=1:length(lonc)
    disp(num2str(i))
    for j=1:length(latc)
        % make rectangular polygons centered on lonc and latc
        lonp = [lonc(i)-dy/2   lonc(i)+dy/2   lonc(i)+dy/2  lonc(i)-dy/2  lonc(i)-dy/2];
        latp = [latc(j)-dy/2   latc(j)-dy/2   latc(j)+dy/2  latc(j)+dy/2  latc(j)-dy/2];
        AA(j,i) = areaint(latp,lonp,E);
    end
end

% plot surface area 
figure
pcolor(lonc,latc,AA); 
colorbar

[Marcus Adkins]

I've been comparing small target sizes against the same targets (+/- 0.02 deg in lat/lon) against area targets created using Satellite Took Kit. The area size agreement is within 5 significant digits until you get within about +/- 1 degree in latitude around the equator - including targets that cross the equator. The areas computed by areaInt can be as much as 50% larger than they should be. While you can use functions like outlinegeoquad to make more interior points which buys back a lot of this - it only applies if your original target edges are parallel to the lat/lon axes. So yes, I would call this a bug.

Answer 1

[Harsha Priya Daggubati]

Hi,

I suspect this discrepancy is due to the dy value you chose.