Fitting a curve (not a surface) for set of 3d points

We have a set of 3d points but could only generate a best fit surface using curve fitting tool. Is there a way to get the best fit curve? After obtaining such curve, how does one extract the sample points between the end points of the original data?

[Riccardo Scorretti]

Can you share the data?

[PASUNURU SAI VINEETH]

x y z

-27.9887180593615 -1340.75546563430 -2467.65439789032

-28.1559650942894 -1264.59303658420 -2404.50824038076

-29.7919277205741 -1193.39703512010 -2300.00965848429

-31.3981715714723 -1120.17736318384 -2235.27618908221

-31.5451983267780 -1051.10973092073 -2140.30777056763

-32.8491580387443 -984.646322264197 -2063.56414843781

-33.2796553901372 -916.991538791753 -1996.11865926438

-35.8395784569732 -852.031242733576 -1911.15081465986

-37.1566640499132 -790.346389117523 -1815.82738997354

-38.3204915465027 -730.214519353841 -1746.18203218839

-40.1649399572649 -673.409308255575 -1674.26444109182

-41.3035654666637 -620.400594024487 -1587.01503697274

-42.1747052709711 -566.984092840253 -1500.55643509168

-46.1774120240010 -510.506804241280 -1356.63240367480

-46.8217316506543 -465.341776809379 -1312.21258689622

-48.1823821035027 -419.235048839213 -1236.80682112429

-46.8035473839173 -379.332398226134 -1160.38961081556

-50.4447955974995 -332.367954503435 -1049.62531161825

-51.6744777763725 -299.816833129747 -985.059960723105

-53.7621363830258 -263.954442877378 -898.073182985995

-57.1994027666078 -225.149444101340 -777.496668735717

-56.6340768082701 -199.990460428311 -708.719962466197

-58.8969572639191 -165.998414537714 -571.734657662908

-59.2922728658662 -147.613479302020 -529.946182294136

-63.0658812601543 -122.123656702184 -427.943227983156

-64.9755480437970 -102.744870160475 -328.893225669812

-64.7611060843518 -85.0805137022426 -223.268913127146

-66.8461254640729 -72.8327833085849 -156.365146230629

-68.9474184651645 -60.7799766280343 -50.0131596499198

Answer 1

[Star Strider]

I have no idea what model you want to fit to the data.

This does a linear regression —

A = [-27.9887180593615	-1340.75546563430	-2467.65439789032
    -28.1559650942894	-1264.59303658420	-2404.50824038076
    -29.7919277205741	-1193.39703512010	-2300.00965848429
    -31.3981715714723	-1120.17736318384	-2235.27618908221
    -31.5451983267780	-1051.10973092073	-2140.30777056763
    -32.8491580387443	-984.646322264197	-2063.56414843781
    -33.2796553901372	-916.991538791753	-1996.11865926438
    -35.8395784569732	-852.031242733576	-1911.15081465986
    -37.1566640499132	-790.346389117523	-1815.82738997354
    -38.3204915465027	-730.214519353841	-1746.18203218839
    -40.1649399572649	-673.409308255575	-1674.26444109182
    -41.3035654666637	-620.400594024487	-1587.01503697274
    -42.1747052709711	-566.984092840253	-1500.55643509168
    -46.1774120240010	-510.506804241280	-1356.63240367480
    -46.8217316506543	-465.341776809379	-1312.21258689622
    -48.1823821035027	-419.235048839213	-1236.80682112429
    -46.8035473839173	-379.332398226134	-1160.38961081556
    -50.4447955974995	-332.367954503435	-1049.62531161825
    -51.6744777763725	-299.816833129747	-985.059960723105
    -53.7621363830258	-263.954442877378	-898.073182985995
    -57.1994027666078	-225.149444101340	-777.496668735717
    -56.6340768082701	-199.990460428311	-708.719962466197
    -58.8969572639191	-165.998414537714	-571.734657662908
    -59.2922728658662	-147.613479302020	-529.946182294136
    -63.0658812601543	-122.123656702184	-427.943227983156
    -64.9755480437970	-102.744870160475	-328.893225669812
    -64.7611060843518	-85.0805137022426	-223.268913127146
    -66.8461254640729	-72.8327833085849	-156.365146230629
    -68.9474184651645	-60.7799766280343	-50.0131596499198];

figure
stem3(A(:,1), A(:,2), A(:,3), 'p')
grid on
xlabel('X')
ylabel('Y')
zlabel('Z')
title('Original Data')

DM = [A(:,[1 2]) ones(size(A(:,1)))];                           % Design Matrix
B = DM \ A(:,3);                                                % Linear Regression
fprintf('\nRegression Coefficients:\n\tB(1) = %10.3f\n\tB(2) = %10.3f\n\tB(3) = %10.3f\n',B)

Regression Coefficients:
	B(1) =    -47.407
	B(2) =      0.322
	B(3) =  -3333.979

Zhat = DM * B;                                                  % Fitted 'Z' Values
ResidNorm = norm(A(:,3)-Zhat)                                   % Norm Of Residuals

ResidNorm = 205.0220

figure
scatter3(A(:,1), A(:,2), A(:,3), 'p', 'filled')
hold on
plot3(A(:,1), A(:,2), Zhat, '-r')
hold off
grid on
legend('Data','Linear Regression', 'Location','best')
xlabel('X')
ylabel('Y')
zlabel('Z')
title('Data & Fitted Linear Regression')

The regress function and fitlm provide statistics on the parameters and linear fit. To do a nonlinear regression (with an appropriate model), use fitnlm.

.

[Torsten]

In principle, you again fit a surface to the data:

z = a*x + b*y + c

is a plane, not a curve.

[Alex Sha]

This model is simple, but with much better result than linear model:

z = 1/(p1*power(x/y,p2))+p3;

Sum Squared Error (SSE): 4673.16132730696
Root of Mean Square Error (RMSE): 12.6942307386099
Correlation Coef. (R): 0.999845884710939
R-Square: 0.999691793173401

Parameter	Best Estimate
----------	-------------
p1	-0.000423738956973825
p2	0.17801739938203
p3	2244.92292110732

[PASUNURU SAI VINEETH]

Thanks for the response. Here are 2 observations which I wanted to bring to your notice.

1) The suggested linear regression approach doesn't provide a smooth fit if observed from a different view than in the original response. Since it's of the form z=B1*x+B2*y+B3 where B1,B2 and B3 are constants for the whole curve, I expected the fit to be smooth.

2) Based on the physics of the problem (ball flight), a parabola would be a more appropriate fit. So, I went ahead with the following form for Design Matrix

DM = [ax.*ax ay.*ay ax.*ay ax ay ones(size(az))];

Data = readmatrix('data1.csv','NumHeaderLines',1);

x=0.001*Data(:,2);
y=0.001*Data(:,3);
z=0.001*Data(:,4);

[i]=find(y==min(y));
ax=x(i+1:end,:);
ay=y(i+1:end,:);
az=z(i+1:end,:);

DM = [ax.*ax ay.*ay ax.*ay ax ay ones(size(az))];     
B = DM \ az;  

Zhat = DM * B;                                                  % Fitted 'Z' Values
ResidNorm = norm(az-Zhat)  ;                                 % Norm Of Residuals

scatter3(ax,ay,az,'o')
hold on
grid on
plot3(ax, ay,Zhat, '-r')
xlabel('X')
ylabel('Y')
zlabel('Z')

Unfortunately, even this fit isn't smooth (despite being second degree in x and y). Can you please suggest a way to get a smooth non-linear fit and also the points sampling?

[Star Strider]

My linear regression was simply a demonstration because no model or other description of the data or problem was provided.

That appears to be essentially an exact fit.

A simpler model works as well:

DM = [ax.^2 ay.^2 ones(size(az))];

I doubt that it is possible to improve it, or to smooth it significantly, other than by interpolating ‘DM’ to a matrix with fewer points.

.