how to fit 3d points with a non linear curve?

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hello, I have these points ( first colum is x cordinate, second is y cordinate and third is z cordinate).
my goal is to fit these points with a 3d curve: they are sclera surface points (sclera is a part of the eyeglobe that can be modelled as a ellipsoid) so i'm expecting an ellipse function in the space or just a part of it. I want to calculate the arclength of this part of the curve, so, I would have a mathematical equation to do the integral..
I've tried using the fitting toolbox but nothing ,I ve already searched on the forum but nothing...
can anyone help me?
points = [
-9.2836 -8.2360 -5.1328
-9.2261 -8.1850 -0.9157
-9.1251 -8.0953 -0.1483
-9.1246 -8.0949 -3.0498
-8.9797 -7.9664 -6.0975
-8.4251 -7.4743 -7.0451
-7.9054 -7.0133 -7.9331
-7.7326 -6.8600 -7.8448
-7.2632 -6.4436 -7.6050
-7.2632 -6.4436 -7.6050
-6.8659 -6.0911 -8.4124
-6.8434 -6.0711 -8.4581
-6.6529 -5.9021 -8.6478
-6.4493 -5.7215 -8.8505
-6.2142 -5.5130 -8.9064
-5.7780 -5.1260 -9.3982
-5.4242 -4.8121 -9.7969
-5.3082 -4.7092 -9.9064
-5.2255 -4.6358 -9.9665
-4.1891 -3.7164 -11.3792
-4.1542 -3.6854 -11.4154
-4.1364 -3.6696 -11.4286
-4.1162 -3.6517 -11.4354
-3.4141 -3.0289 -11.4215
-3.2885 -2.9174 -11.4190
-2.6937 -2.3897 -11.4072
-2.2985 -2.0391 -11.3994
-1.9732 -1.7505 -11.3929
-1.4682 -1.3025 -11.6907
-1.1315 -1.0038 -11.8893
-1.1254 -0.9984 -11.8887
-1.1177 -0.9916 -11.8879
-0.4374 -0.3880 -11.7442
0.0000 0.0000 -11.7540];

Accepted Answer

Alex Sha
Alex Sha on 26 Aug 2019
How about the fit function fellow:
z = p1*exp(-0.5*(((x-p2)/p3)^2))+p4*exp(-0.5*(((y-p5)/p6)^2))+p7
Root of Mean Square Error (RMSE): 0.288723987265949
Sum of Squared Residual: 2.83429238797343
Correlation Coef. (R): 0.995496865084864
R-Square: 0.991014008393792
Adjusted R-Square: 0.990115409233171
Determination Coef. (DC): 0.991014008393783
Chi-Square: -0.161670470284513
F-Statistic: 496.279531952741
Parameter Best Estimate
---------- -------------
p1 -16.1160680408573
p2 -1.49490026614748
p3 -7.84324518363849
p4 1.92526452647942E18
p5 -8.14010212547392
p6 0.00498580635142014
p7 4.26212632161697tu41.jpg
  1 Comment
mmrsagar
mmrsagar on 22 Dec 2021
Hi Alex,
How did you get the function and how did you find the best estimate of the parameters? Sorry if my question is too naive but a little explaination would help me a lot. thanks

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More Answers (1)

Bhargavi Maganuru
Bhargavi Maganuru on 23 Aug 2019

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