Thread: land area calculations

http://groups.google.com/group/sci.e...a916915401100a have the right formula?
To quote:
You can use surveyors formula to calculate area of a land parcel.
A = (1/2)[Determinant(x1,x2,y1,y2)+Determinant(x2,x3,y2,y3)+ ...
+Determinant(xn,x1,yn,y1)],
where Determinant(x1,x2,y1,y2) = x1*y2-x2*y1

This came up before and myself and RobA worked out a real easy way to do it using the histogram window.

Get to the bottom because it gets easier and easier as we found out better ways to do it.

Looking For Free Area Calc Software?!

By the end its just multiply two numbers together and the histo windows will tell you outright the hard one of the two. If you get stuck, post the map or a map and ill show you how to do it. It does need to be a map with rectangular axes not some curved axes projection based type tho.

waldronate - I do believe that that is the one

have some rep my friend

Redrobes - thanks for that thread, it is quite interesting and good to know

Yes, a good thread

If you are worried about precision though, remember pixels do not have equal height and width.

Originally Posted by Sigurd

If you are worried about precision though, remember pixels do not have equal height and width.

If you're talking about determining area by summing pixel areas, then the problem is the pixel size (relative to the size of the region), not shape.

Pixels which are entirely within the boundary of the region are making accurate contributions to the sum, but those which overlap the boundary are contributing more area than they should -- the parts of the pixels which are outside the boundary shouldn't count. As the size of the pixels approaches zero, so does the amount of error, and this would work with pixels of any shape as long as they tile the region.

This is similar to how a Riemann sum works in calculus, where you find the area under a curve by adding the areas of rectangles drawn under it and looking at what happens to that sum as the width of the rectangles goes to zero. As in that method, using any finite number of sub-areas will generally give only an approximate result.

Square or rectangular, however, the pixels are probably all small enough that the error isn't worth worrying about.

Here are some examples using square and rectangular "pixels". In each example the figure on the right, using smaller pixels than the one on the left, covers an area closer to the actual area bounded by the curve -- replacing a large boundary rectangle by smaller ones allows you to remove some of the smaller ones, and their areas are subtracted from the overestimate.