View Full Version : land area calculations

12-12-2008, 12:32 PM
Any old time land surveyors out there?

I am looking for the method for calculating the area of an iregular land mass.

I am asking for the oldtimers 'cause back in the eighties, I started to take a survey course, but had to leave before it got too far into it. Just before I had to leave, might even have been the last day, the teacher gave us a method to calculate this type of area using the xy coordinates. Stupid me, I didn`t write down his method or the setup for the coords, just his proof. I was looking through my notes recently (yes I am anal enough to have kept some partial course notes for more than 25 years) and found the proof. For the very life of me, I can not figure out the method, and I really have tried to solve it.


12-12-2008, 03:15 PM
http://groups.google.com/group/sci.engr.surveying/browse_thread/thread/eaa916915401100a 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)+ ...
where Determinant(x1,x2,y1,y2) = x1*y2-x2*y1

12-12-2008, 05:32 PM
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.


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.

12-12-2008, 09:06 PM
waldronate - I do believe that that is the one :)

have some rep my friend 8)

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

12-14-2008, 12:31 PM
Yes, a good thread :)

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

12-14-2008, 03:49 PM
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 (http://en.wikipedia.org/wiki/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.