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 Circle Fit Parameter Initialisation Issue
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apg504

United Kingdom
1 Posts

Posted - 11/14/2013 :  1:53:15 PM  Show Profile  Edit Topic  Reply with Quote  View user's IP address  Delete Topic
I'm trying to perform a circle fit to some data points from impedance analysis measurements and although I seem to have compiled a workable function the parameter initialisation is returning no parameters. The equation is; y=yc+((r^2-(x-xc)^2))^0.5 with parameters; yc, xc, r. The data points are given by;

x ................ y
75.13571 -63.27658
76.25993 -114.3841
77.43791 -189.3655
79.08652 -304.7553
81.96098 -484.4481
87.57366 -766.7599
101.8244 -1208.366
116.2378 -1908.358
140.8182 -3009.153
184.4843 -4741.054
269.0154 -7466.838
412.0339 -11744.07
684.5799 -18475.14
1188.263 -29018.66
2188.201 -45496.34
4102.733 -71208.32
7954.601 -110888.1
14954.79 -171847.8
27267.98 -264488.6
48071.97 -405350.7
81058.66 -617535.1
134743.9 -940658.4
225447.5 -1.43007E6
384860.3 -2.16418E6
659018.9 -3.25763E6
1.17111E6 -4.85358E6
2.19878E6 -7.13682E6

I select the data as normal and go to non-linear curve fitting, choose the function, initialise the parameters which seems to do nothing and then press fit, which fails. Help!

greg

USA
1378 Posts

Posted - 11/14/2013 :  4:06:32 PM  Show Profile  Edit Reply  Reply with Quote  View user's IP address  Delete Reply
OriginPro 9 and 9.1 have Implicit fitting and come with the Circle function built in:
f = (x-xc)^2 + (y-yc)^2 - r^2
Without Implicit fitting, functions can only solve for one root, so any Circle function could only include either the positive root ( such as your y = yc + ((r^2-(x-xc)^2))^0.5 ) or the negative root ( y = yc - ((r^2-(x-xc)^2))^0.5 ).
Unfortunately, your data is definitely the negative part of a circle so you need to change your function accordingly.
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