[R-sig-Geo] Find a circle center with spatial points
inacio.adrien at gmail.com
Mon Mar 21 15:00:07 CET 2016
Thanks for your help,
I try to find how I can use the java script but I don't really understand
"After download you should for standalone compilation delete line 37
compile it with:
*" javac CircleFitter.java -Xlint:uncheckedthen you can run it with:java
*"grab the output in a file using a OS depending pipe"*
Have I to download an application wich permit to work on java script, or
all can be done with R (I have never work with java so i'm a little bit
Thank you again.
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2016-03-18 23:28 GMT+01:00 Chris Reudenbach <reudenbach at uni-marburg.de>:
> Oops :-[ , Barry thanks for clarification , I got it now.
> I don't understand the math either but as a kind of "reparation"
> for not reading well enough I found the authors java code.
> After download you should for standalone compilation delete line 37
> "package org.spaceroots;"
> compile it with:
> javac CircleFitter.java -Xlint:unchecked
> then you can run it with:
> java CircleFitter input.file
> The input data should be formated like:
> # input
> # x y
> 0 5
> 1 4.5
> 2.5 4
> 3 3.5
> 4 2
> 5 0
> the above input yields
> initial circle: -001.69467803 -000.69446643 006.23578103
> converged after 7 iterations
> final circle: -001.34339845 -001.34426151 006.44308386
> with the format x,y,radius
> which at least make sense with respect to the data.
> You easily can run it from R by system() and grab the output in a file
> using a OS depending pipe
> or you may run it using the rJava package which could be more complex.
> cheers Chris
> if you want you can call it from R with javaR
> Am 18.03.2016 um 20:11 schrieb Barry Rowlingson:
>> On Fri, Mar 18, 2016 at 6:49 PM, Chris Reudenbach
>> <reudenbach at uni-marburg.de> wrote:
>>> Because it seems to be an arc and not a circle issue that you can solve
>>> problem by
>>> picking arbitrary two points of your assumed "arc" then construct
>>> (calculate) the perpendicular bisector of
>>> the line between them and do so for another arbitrary two points of the
>>> assumed "arc".
>>> The intersection of the perpendicular lines is the assumed center of the
>>> If you iterate over all points this should be a pretty good estimation of
>>> the real center.
>> This is the "sample 3 points and find the fitted circle" idea, you
>> are likely to get massive "outliers" and if you take the mean
>> coordinate it could fail horribly. See the paper I linked to for an
>> example. They use the median to get an initial "robust" estimate of
>> x,y,R, and then use some specialised optimisation to improve the
>> estimate - you can't just throw it into "optim"!
>> Not sure I understand the maths in it yet though....
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