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.
https://www.spaceroots.org/documents/circle/CircleFitter.java
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
<reudenb...@uni-marburg.de> wrote:
Because it seems to be an arc and not a circle issue that you can solve the
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
arc.
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....
Barry
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