Geometry.getarea() and Polygon.getarea() what are the units?

Question

So I am working with a set of images where the the corners are the lat long on the map. I am creating Polygons to check if there is overlap of the given images. I need to know the units associated with the .getarea() methods for both Polygon and geometry. I'm using the following objects to create my polygons and geometry

http://tsusiatsoftware.net/jts/javadoc/com/vividsolutions/jts/geom/Geometry.html#getArea() http://www.vividsolutions.com/jts/javadoc/com/vividsolutions/jts/geom/Polygon.html

When using both of the above objects I get a number back however I have found no indication on what the units associated with the number are. So are we talking meters, kilometers, miles?

a sample of cordinates i'm using are +30.658739 -086.345670, +30.659997 -086.34002, +30.664041 -086.345082, +30.662783 -086.342750 I'm looking for the area between these 4 points.
the value i get back from the .getArea() is 1.31039680000139E-5 The points are realitively close so i'm thinking its in meters which would be 1310.4 meters

Answer 1

This is difficult to tell without knowing what the underlying coordinate system is. After taking a quick glance at the Javadoc API you linked to, I suspect that your geometry package is probably dealing in raw cartesian coordinates and is agnostic about the units of measurement being used. If this is the case, you are on a very slippery slope. Here's the problem:

Not All Degrees Are Created Equal

Degrees in latitude all have the same spatial resolution. Each degree of latitude corresponds to roughly 111km of distance. This is not the case for degrees of longitude. At the equator, degrees of longitude correspond to 111km of distance, but at the poles, 1 degree of longitude has ZERO km of distance. In other words, if you have a Lat/Lon box with the upper left at 10:0 and lower-right at 0:10, it will have more surface area than a Lat/Lon box with upper left at 20:0 and lower-right at 10:10, even though the sides of those boxes are all 10 degrees long.

A second issue is the curvature of the earth. Because of the earth's curvature, a 100km by 100km square on the earth's surface will have more surface area than 10000 km^2, because the shortest distance from one point on the earth's surface to another is not actually straight line, but an arc.

A third and often-overlooked but less-important issue is that the Earth is not actually a sphere, but an ellipsoid. It tends to bulge near the equator, which breaks our assumption that a degree of latitude anywhere on earth is the same distance as another degree of latitude anywhere else. However, this issue does not introduce as much error into our surface area estimates as the first two.

In other words, spherical (or in reality, ellipsoidal) surface area isn't an easy problem to solve, at least not as easy as mapping to cartesian coordinates and using them to find euclidean measurements of surface area. You can get away with it if the surface area you are dealing with spans a very small angular distance, but the larger your lat/lon box is, the more distortion you get.

Possible Solution

This only works if your images are rectangular and top/bottom of the image have constant latitude from left-to right. This still introduces more error with wider ranges of latitude because it still ignores the curvature of the earth, but does a better job than assuming all degrees are created equal in a cartesian coordinate system. If this is the case, then the intersection of your image will be bounded by the following coordinates:

topLat:*leftLon*, bottomLat:*rightLon*

Calculate the average latitude, then use it to find distance per degree of longitude at that average latitude, kmPerLon. We now have the following equation:

area = ((topLat - bottomLat) * 111km) * (rightLon - leftLon) * kmPerLon)

The area you get from this will be measured in square kilometers, but again, I'd like to reiterate that this only works reasonably well if your images are rectangularly aligned to parallel latitudes and do not span much angular distance.