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.

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.

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

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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

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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.

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.

<p>For a point-and-click solution, Meander ( http://peacockmedia.co.uk/meander ) allows you to trace the perimeter of any shape you can see on your screen and read off the perimeter and area</p>

<p>Hi shieladixon. Does this measure projected area (I.e. treat the area as a flat plane), or does it take into account changes in elevation? <br>I come from a farming background in a hilly area. Topo maps and surveyed areas give you the first, but on the ground, a bit of undulation can take a 2ha paddock up to a 2.5ha paddock. Important if you're calculating seed/fertilizer quantities, etc.</p>

<p>Hi, thanks for your interest. At present it allows you to add in a hilliness factor for the line (eg walking route) but doesn't take this into account when calculating the area. Now that you've given me a practical application I'll update the app to do this.</p>

This is OK for very simple approximations, but Green's Theorem and Surface Internals would be far more effective!

Yeah yeah. :D<br>The thought of keeping track of the line integral x*dy - y*dx as you walk around a paddock sounds reeeeellllllll fun! :P

'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.

This would definitely be a method you'd keep for moderate size areas. :D<br> I'd imagine this would take 4 hours on a large paddock, which tend to be straight edged any case. ;)<br> <br> In the situations where this is necessary (E.g. measuring a curved lawn), the various forms of <a href="http://en.wikipedia.org/wiki/Simpson%27s_method">Simpson's Method</a> may be a more accurate form of this approach. It gives approximately twice as much accuracy, allowing you to make the strips twice as big.<br> <br> For gardens/areas that are almost circular, using Simpson's Method from the centre to measure out arcs (Keep the width of the ends constant) and divide the total by two is also a good approach.<br>

I'm afraid my math education is far less comprehensive (and perhaps more ancient) than yours. <br> <br>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. <br> <br>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. <br> <br>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.

The simplest version is the 2/4 rule. Split your area up into an even number of sectors of equal widths. This will give you an odd number of measurements to make.<br>E.g. assume we split the area into six sectors, and seven measurements, which are [1,2,1,3,1,2,2].<br><br>Take the first and last value and add them together.<br>1 + 2 = 3<br>Now take each even value, add them together and multiply by four.<br>2 + 3 + 2 = 7<br>7 * 4 = 28<br>Take the remaining odd values, add them together and multiply by two.<br>1 + 1 = 2<br>2 * 2 = 4<br><br>Add all these results together and divide by 3.<br>3 + 28 + 4 = 35<br>35 / 3 = 11 2/3<br><br>Simpson's rule is based on taking polynomial approximations of each section.

Perhaps someone could explain? <br>I thought that you find the area through: <br>area = (width * height) / 2 <br> <br>Is that any different? <br>Great job, by the way!

This saves you ensuring that the height is at a perfect right angle. All you need to do is go for a walk around the paddock with a trundle wheel.

Thank you for the information and this method.<br><br>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 <br><br>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.<br><br>otherwise it is a great job you've done in there..<br><br>thanks.

Hi sonogo. I've uploaded some password-less, protected versions of the spreadsheet. The issue being that the version I've created has formulas that only go up to 30 triangles. I've left the unprotected versions in as well, if people want to adapt them.

Thank You very much for the very handy and useful method and file you gave in this instructable.<br><br><br>keep the good job.<br><br><br>

Thank you for sharing this. Just I know this so useful formula.

I enjoyed reading it. I'm sure I'll be using it some day. Thanks, well done.

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!

I used geogebra (<a href="http://www.geogebra.org/cms/"> http://www.geogebra.org/cms/</a> ) for the diagrams, and I'm not entirely certain how to do non-Euclidean corrections. Thank you for the featuring.