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Measuring Area of Irregular Shapes

Measuring Area of Irregular Shapes
This instructable combines a few mathematical tricks to enable you to calculate the area of irregular shapes.

I come from a farming background. One of the things we often had to do, was measure the area of a section of land in order to calculate how much fertiliser  was required.
This can be done from maps, (E.g. using Google Maps measure tools) but this doesn't take into account variations in height. Our farm was in a moderately hilly area, and often the real area of a paddock was 20% more than what the maps said.

This instructable could also be used to measure area of irregular plane shapes, as well as surface area of 3D shapes.
 
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Step 1Heron's Formula - Calculating the Area of a Single Triangle.

Heron\
Heron's formula is a way to calculate the area of ANY triangle, knowing the lengths of it's three sides, a, b and c. More information can be found at http://en.wikipedia.org/wiki/Heron%27s_formula

Firstly, calculate the semiperimeter. This is just half of the triangle's perimeter.

s = (a + b + c) / 2
In our example,
s = (6.56 + 6.01 + 5.76)/2
s = 9.165
-----------------------------------------------------------------
Then, calculate the following differences:
s - a
s - b
s - c

E.g.
s - a = 9.165 - 6.56 = 2.605
s - b = 9.165 - 6.01 = 3.155
s - c = 9.165 - 5.76 =  3.405
------------------------------------------------------------------
Finally, substitute your answers into the following equation:
A = sqrt( s * (s-a) * (s-b) * (s-c) )
(Multiply the three differences together, along with the semiperimeter, and then take the square root.)

A = sqrt( 9.165 * 2.605 * 3.155 * 3.405)
A = sqrt( 256.482)
A = 16.015
This gives the area of a single triangle.
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15 comments
Oct 22, 2010. 1:32 PMXellers says:
This is OK for very simple approximations, but Green's Theorem and Surface Internals would be far more effective!
Oct 18, 2010. 7:48 AMmikchil says:
'nother way is to divide the area into horizontal strips of equal width & measure the lengths of the strip centerlines from edge to edge of shape in question. Add the lengths & multiply by the strip width and you have a pretty good approximation of the area. The narrower the strips, the greater tha accuracy. It's sorta like integration iin calculus if memory serves. This works well for any kind of shape including those with curved edges. If there are holes in the middle, leave them out of the measurements.
Oct 23, 2010. 5:49 PMmikchil says:
I'm afraid my math education is far less comprehensive (and perhaps more ancient) than yours.

I looked up Simpson's rule and it made my head hurt. Didn't do very well in the calculus classes I attempted because of those lousey headaches.

My thoughts run more along the lines of the rectangular method using midpoint approximation. I often have to estimate irregular shapes on plans measuring from a few hundred square feet to hundreds of acres. After using this method for a number of years, I find that if I choose the width of the rectangles correctly, my results are close enough the output of high end digitizers or autocad and don't take much time to do using a simple measuring wheel and a straight edge. If the margins are made up of straight lines, subdividing into simple straight sided shapes can be a quicker approach if the original shape is regular enough but that's a judgement call.

I'm sure that the math approach will provide better accuracy if you have the right information but I don't find the error in the approach that I use to be significant in my world.
Oct 20, 2010. 5:29 PMSe1f_Destruct says:
Perhaps someone could explain?
I thought that you find the area through:
area = (width * height) / 2

Is that any different?
Great job, by the way!
Oct 17, 2010. 2:29 AMsonogo says:
Thank you for the information and this method.

I would like to comment on the xls file you made. it is okay but I think you should have put some protection on the cells other than the value of a,b,and c

if some one write on the cell of S or a-S for example it would damage the file. some people doesn't know how to drag the formula to the other cells.

otherwise it is a great job you've done in there..

thanks.
Oct 18, 2010. 9:19 PMsonogo says:
Thank You very much for the very handy and useful method and file you gave in this instructable.


keep the good job.


Oct 16, 2010. 7:14 PMrimar2000 says:
Thank you for sharing this. Just I know this so useful formula.
Oct 16, 2010. 1:58 AMgserrano701 says:
I enjoyed reading it. I'm sure I'll be using it some day. Thanks, well done.
Oct 15, 2010. 9:44 PMkelseymh says:
This is great! Well written, citations where appropriate, and clear diagrams. And it's got math :-) Too bad you didn't go in for non-Euclidean corrections, but I guess you're not trying to measure the area of a hyperboloid of revolution ;-> Featured and rated!

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