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

Question

luca antonioli 2019-8-20

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/476878-how-to-fit-3d-points-with-a-non-linear-curve

评论： mmrsagar 2021-12-22

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];
    

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Answer 1

Alex Sha 2019-8-26

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https://ww2.mathworks.cn/matlabcentral/answers/476878-how-to-fit-3d-points-with-a-non-linear-curve#answer_389107

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 4.26212632161697

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mmrsagar 2021-12-22

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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