Re: fitting circle to data

[macy]

>From my naive approach:

Given a circle or radius, R, with origin at x0 and y0:
the error for any data point is:
= R^2-(xi-x0)^2-(yi-y))^2

now you must minimize the sum of the squares:
sum( (R^2-(xi-x0)^2-(yi-y))^2)^2 );

Doing it without using sqrt function should preserve your final answers
completely in real space, right?


> Hello,
>
> i need to extract the radius and centre points of a circle fitting a
> set of datas. I was trying with this way
>
> xt = %data
> yt = %data
>
> F = @(x,p) sqrt( p(3).^2 -(x-p(1)).^2)+p(2)
> pin = [1;1;1]
>
> [f, p] = leasqr(xt,yt,pin,f);
>
> testing it with xt = 2+4*cos(thetas) and yt = 3+4*sin(thetas) works well,
> i get
> p =
>
>    2.0000 + 0.0000i
>    3.0000 + 0.0000i
>    4.0000 - 0.0000i
>
> but using the real data i get
> p =
>
>    -2.23630 -  0.62042i
>    12.24122 - 64.02140i
>     0.34923 + 59.63389i
>
> and should be aprox. p = [0, 300, 290]
>
> If i use pin =  [0; 300; 290] get
> p =
>
>     -2.2316 -   1.5828i
>     12.2604 - 162.9057i
>      0.3318 + 161.1838i
>
> is that right? how souhld i use this comples result?
> Thank you
>
> PD: here a few xt and yt datas
> xt =
>
>    29.94340
>    28.45820
>    27.19490
>    25.85900
>    24.73510
>    23.54580
>    21.88660
>    19.90880
>    17.77030
>    15.64780
>    13.04540
>    10.67190
>     8.43594
>     6.35659
>     4.31690
>     2.28645
>     0.19964
>    -1.74150
>    -3.66550
>    -5.75025
>    -7.75397
> ...
>  yt =
>
>    13.277
>    13.079
>    12.952
>    12.747
>    12.670
>    12.638
>    12.475
>    12.225
>    12.059
>    11.977
>    11.773
>    11.676
>    11.556
>    11.526
>    11.454
>    11.415
>    11.394
>    11.449
>    11.444
>    11.491
>    11.556
> ...
>
>
>
> --
> Gastón Araguás
> ______________________________________________________
> CIII - Centro de Investigación en Informática para la Ingeniería
> Univ. Tecnológica Nacional Facultad Regional Córdoba. Argentina