Calculating an average polygon

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07-01-2013 02:18 PM
SteveMayall
New Contributor
Hello everybody,

I found this thread in the old ESRI forums with the same type of problem I am facing today. The solution to the problem is explained in little detail and since I am not an expert, I was wondering if somebody could describe it to me in little more detail.

http://forums.esri.com/Thread.asp?t=248519&c=93&f=995&mc=7&ESRISessionID=eiVtjzEylR-m_pw7WZEffISQch5...

To simplify, I have multiple polygons and based on their size, shape and location I want to create a single new polygon which would best represent the average size, shape and location of all these input polygons.

If somebody would be kind enough to explain me the process it would be much appreciated.

Thank you!
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ShaunWalbridge
Esri Regular Contributor
I don't have an exact answer, but my understanding is, this is one of those questions which is easy to ask, but in turn generates many related questions before you can figure out what specifically you'd like. See, for example, the answer on finding the center of geometry of an object, and computing an average polygon width, in addition to the thread you'd linked to. Can you provide more details on the kind of analysis you'd like to perform, and what goals you'd like to accomplish? That'll probably help as there are likely many subjective alternatives to compute these averages.

cheers,
Shaun
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SteveMayall
New Contributor
Thanks for your help, I hope this will clarify things a little bit.

We are doing a mental map analysis. We have let's say 50 polygons of a certain city square or park which are all different in shape, size and location. Our goal is to create a single map which would represent the answers from our whole specimen of over 50 people. So basically, we need a method for generating a single polygon out of these 50 which would be an "average" polygon in size, shape and location. I hope I made my point and managed to explain what we need to do.

Thanks again and if it is still not clear, I will gladly explain it further :)
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