11

I need to calculate areas and intersection areas for polygons (some real geographical objects like lake, city, country, e.t.c.). Polygons located in California, New Zealand, Russia.Anadyr, Sweden

All polygons are in WGS84.

What I did using GeoTool java API:

  1. Project all polygons using EPSG:3488, EPSG:NAD83(NSRS2007) / California Albers and calculated areas and overlap areas.
  2. Did the same using World_Mollweide and World_Eckert_IV
  3. Picked "local specific projections" for polygons from California, New Zealand, e.t.c.

I assume that #3 is the most accurate result, since I pick projection which covers polygon area

Result:

'#2 showed the worst result comparing to #3

'#1 and #3 areas and intersection areas difference is less than 0.1%

Why? I pick absolutely wrong projection EPSG:3488 (California) for polygons from Sweden and get pretty the same areas and intersection areas?

UPD: Looks like I didn't explain my confusion correctly. Here is sample output with explanation

#area_from_new_zealand_1
EPSG_27200 area[11733479] CRS[World_Mollweide] area[11736023] diff[2544] [0.0%]
EPSG_27200 area[11733479] CRS[World_Eckert_IV] area[11736033] diff[2554] [0.0%] 
EPSG_27200 area[11733479] CRS[EPSG:NAD83(NSRS2007) / California Albers] area[11736034] diff[2555] [0.0%] 

#area_from_new_zealand_2
EPSG_27200 area[2952725]  CRS[World_Mollweide] area[2953281] diff[556] [0.0%] 
EPSG_27200 area[2952725]  CRS[World_Eckert_IV] area[2953342] diff[617] [0.0%] 
EPSG_27200 area[2952725]  CRS[EPSG:NAD83(NSRS2007) / California Albers] area[2953467] diff[743] [0.0%] 

#intersection_area_between_two_new_zealand_areas
EPSG_27200 intersection area[1001857] CRS[World_Mollweide]                          area[1002082] diff[225] [0.0%] 
EPSG_27200 intersection area[1001857] CRS[World_Eckert_IV]                          area[1002082] diff[225] [0.0%] 
EPSG_27200 intersection area[1001857] CRS[EPSG:NAD83(NSRS2007) / California Albers] area[1002096] diff[239] [0.0%] 


#area_from_alaska_1
EPSG_3338 area[56278347]    CRS[World_Mollweide] area[56041510] diff[236837] [0.4%] 
EPSG_3338 area[56278347]    CRS[World_Eckert_IV] area[56041585] diff[236763] [0.4%] 
EPSG_3338 area[56278347]    CRS[EPSG:NAD83(NSRS2007) / California Albers] area[56278426] diff[79] [0.0%] 

#area_from_alaska_2
EPSG_3338 area[17564799282] CRS[World_Mollweide] area[17486015889] diff[78783393] [0.4%] 
EPSG_3338 area[17564799282] CRS[World_Eckert_IV] area[17486869816] diff[77929466] [0.4%]
EPSG_3338 area[17564799282] CRS[EPSG:NAD83(NSRS2007) / California Albers] area[17566197286] diff[1398004] [0.0%] 
 
 #intersection_area_between_two_alaska_areas 
EPSG_3338 intersection area[43808167] CRS[World_Mollweide] area[45066901] diff[1258734] [2.8%] 
EPSG_3338 intersection area[43808167] CRS[World_Eckert_IV] area[45163183] diff[1355016] [3.0%] 
EPSG_3338 intersection area[43808167] CRS[EPSG:NAD83(NSRS2007) / California Albers] area[43885182] diff[77015] [0.2%]

My confusion is: EPSG:3488 designed to be used in California

I pick "wrong" projection EPSG:3488 for Alaska, New Zealand areas and see that resulting calculations don't differ "significantly" from correct projections. EPSG:3488 even performs better than Mollweide, Eckert_IV projections designed to be used around the world.

Improve this question
4
  • I have also found that there is close to no observable difference between these two projections, however the difference still exists. In ArcGIS, you cannot create a "feature dataset" unless your data is in the same projection even with such a small difference as found between WGS84 and NAD83. The following webpage was very informative to me and I hope you find it to be useful as well. differencebetween.net/technology/… I would have put this as a comment but I do not have 50 rep :(
    – Justin Q
    Aug 13, 2015 at 16:02
  • what are you comparing the results to?
    – Ian Turton ♦
    Aug 14, 2015 at 9:07
  • @iant please see updated question. I added comparison output.
    – Capacytron
    Aug 14, 2015 at 9:37
  • You could try the AUTO projections (UTM centred on a user supplied point) - construct CRS as String code = "AUTO:42001," + x + "," + y; // System.out.println(code); CoordinateReferenceSystem auto = CRS.decode(code);
    – Ian Turton ♦
    Aug 14, 2015 at 9:50

2 Answers 2

Reset to default
9

"EPSG:3488, EPSG:NAD83(NSRS2007) / California Albers" is an equal-area projection. It is based on the Albers Conic, which is defined for the northern hemisphere. Because Sweden is within its range of definition, it is equal-area in Sweden. This means that (up to floating point rounding error) it will give absolutely correct areas.

Neither the Mollweide nor the Eckert are exactly equal-area, but (as M. Kennedy kindly points out in a comment) they are approximately so. The distortions they introduce will be comparable to the differences between the sphere and the ellipsoid, which are limited to about one part in 300 (0.3%).

Improve this answer
10
  • 1
    Bill, Eckert IV and Mollweide are equal-area projections, but only have spherical algorithms.
    – mkennedy
    Aug 13, 2015 at 16:41
  • 1
    @mkennedy Oops--I should have checked. Thank you for the correction, Melita. I'll fix that comment.
    – whuber
    Aug 13, 2015 at 16:41
  • 1
    @mkennedy so ESRI:53009 and ESRI:54009 are actually identical? I see that Snyder does not give formulas on the ellispoid, so what's the sense of those World_xxx projections from ESRI based on WGS84?
    – AndreJ
    Aug 13, 2015 at 17:14
  • 1
    Esri:53009 (and the other 53xxx entries) use a sphere-based GeoCRS with R=6371000.0 m. The Esri:54xxx range uses a WGS84 geoCRS so the actual radius used is the semimajor axis, 6378137.0. They were both added just as test/sample definitions.
    – mkennedy
    Aug 13, 2015 at 17:25
  • 2
    The South Pole is tens of thousands of kilometers from Sweden: it's in Antarctica. The North pole is only a few thousand kilometers from Sweden. But it doesn't matter: if you use an equal area projection and it is capable of projecting a region whose area you want to compute, then it will compute the area correctly (for the datum on which it is based). Some equal-area projections are capable of projecting the entire world except for a single point (which depends on the projection).
    – whuber
    Aug 13, 2015 at 22:12
1

@whuber's assertion that an equal-area projection "will give absolutely correct areas" comes with an asterisk, namely, assuming that the edges of the polygon are straight lines in said projection. This is often a good approximation, particularly if the edges are short; but it is rarely strictly true.

If, on the other hand, the edges of your polygon are geodesics or rhumb lines, other techniques can be used to determine the area accurate to round-off. My online planimeter implements these. Give it a try.

Improve this answer
7
  • Hi! Thanks for the all input. So want could be the summary? EPSG:3488 uses Albers equal-area projection, that is why it correctly calculates areas all around the world, even on South pole?
    – Capacytron
    Aug 18, 2015 at 8:54
  • Probably Albers equal-area is going to provide a decent enough result for your application. However blindly using this projection to calculate the area of Antarctica (which encircles the pole) will give a nonsense result. So for general use, I would recommend calculating the geodesic area so you don't have to worry about the limitations of each particular equal-area projection.
    – cffk
    Aug 18, 2015 at 9:49
  • Thanks for the reply. Unfortunately, we need square area in meters.
    – Capacytron
    Aug 18, 2015 at 9:56
  • The Planimeter tool gives you the area in square meters.
    – cffk
    Aug 18, 2015 at 10:58
  • The same functionality is available as a Java package, Geographiclib-Java. The documentation includes code to compute the area of a polygon specified as a set of latitude/longitude points with the result given in square meters.
    – cffk
    Aug 18, 2015 at 11:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.