Areaquad, latitude and longitude angles do not give the same answer when exchanged

1 Ansicht (letzte 30 Tage)
Dear people,
I'm trying to find the area of patch defined by a rectangle projected onto a sphere. I do this by finding the angles between the origin of the sphere and the edges of the rectangle. These angles can be used as the latitude and longitudes for defining the dimensions of the patch on the sphere.
Now the maximum angles for this in both the X and Y direction are therefor -+90. This would give you half a sphere and thus the function areaquad would result in 0.5, which it does. Now the issue is that if I swap the angles for latitude and longitude or X and Y, I do not get the same area. -+63.43 and -+75.96, does not give the same as -+75.96 and -+63.43 (X,Y respectively), the first one gives an areaquad of 0.3419 and the second of 0.3775.
However if I plot these it can also be seen that they should have the same area, see code and images below.
%for ii=1:points
%set latitudes and longitude limits
lat1=-beta(2); %-63.43 or -75.96
lat2=beta(1); %63.43 or 75.96
long1=-alpha(2); %-75.96 or -63.43
long2=alpha(1); %75.96 or 63.43
lat=[lat1, lat1, lat2, lat2, lat1];
long=[long1, long2, long2, long1, long1];
latm=[-90, 90, 90, -90, -90];
longm=[-90, -90, 90, 90, -90];
axis equal
overlap(ii)=areaquad(long1,lat1, long2, lat2);
please explain me what I'm doing wrong or what's going on here.

Antworten (1)

jean-luc martin
jean-luc martin am 15 Jul. 2017
Sorry for reading your question so late, hope you already find an answer. The area of the quadrangle is spherical (not rectangle) and delimited by a "lune" and its intersection by 2 parallels. So that the both quadrangles are not equivalent if you swap lat and lon values. Try for example with lats (-2,+2) and lons (-85,+85) : a thin band almost rectangular around equator, then opposite (-85,+85) and (-2,+2): a thin lune around greenwich meridian. You may also wish to use Google Earth Pro to visualize on the globe.
Hope this helps,

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