fit a 3d curve
199 views (last 30 days)
Show older comments
Dear all,
I am a beginner in Matlab and I need your help. Here is my problem: I have a cloud of data obtained by measurement. Thanks to those datas I have made a matrix(49x49) which allowed me to plot a paraboloid. I would like to fit this 3d curve based on data, but I don't know how to start. Could you please help me to find a way to solve this problem?
Sincerely,
Elza
Accepted Answer
More Answers (4)
Jarrod Rivituso
on 14 Apr 2011
Could you cast it as a linear regression problem and then solve the equations using backslash? Some details:
Say we are trying to fit to an equation of the form
z = c1*f1(x,y) + c2*f2(x,y) + ...
where f1 and f2 are the basis functions of the linear equation, and c1,c2,etc. are the coefficients. You could write this as a matrix equation...
z = [f1(x,y) , f2(x,y) , ...] * [c1 ; c2 ; ...]
Here's an example in MATLAB with some data I made up...
%Define grid of x,y values
x = -1:0.01:1;
y = -1:0.01:1;
[X,Y] = meshgrid(x,y);
%Make up some data
data = X.^2 + 3*Y.^2 + 0.01*rand(size(X));
%Set the basis functions
f1 = ones(size(X));
f2 = Y.^2;
f3 = X.^2;
%Write as matrix equation
A = [f1(:),f2(:),f3(:)];
y = data(:);
%Solve for coefficients
coeffs = A\y
0 Comments
Sonali
on 27 Aug 2023
Rt = 30.5;
area_ratio = 31.4754;
theta = 50;
theta_exit = 16.5;
theta_n = theta*pi/180;
k = 0.8;
Ln = k*(sqrt(area_ratio)-1)*Rt/tand(theta_exit);
% FIRST CURVE (FC)
% Angles for First Curve
Angle_FC = -(pi+(45*pi/180));
FC_step = (-pi/2-Angle_FC)/100;
theta_FC = (-3*pi/4):FC_step:(-pi/2);
% Coordinates for First Curve
x_FC = cos(theta_FC)*1.5*Rt;
y_FC = sin(theta_FC)*1.5*Rt+(1.5*Rt+Rt);
x_FC1 = x_FC';
y_FC1 = y_FC';
% SECOND CURVE (SC)
% Angle for Second Curve
Angle_SC = -pi/2;
SC_step = ((theta_n-pi/2)-Angle_SC)/100;
theta_SC = -pi/2:SC_step:(theta_n-pi/2);
% Coordinates for Second Curve
x_SC = cos(theta_SC)*0.382*Rt;
y_SC = sin(theta_SC)*0.382*Rt+(0.382*Rt+Rt);
x_SC1 = x_SC';
y_SC1 = y_SC';
% THIRD CURVE (TC)
% Coordinates for Third Curve
x_TC = cos(theta_n-pi/2)*0.382*Rt;
y_TC = sin(theta_n-pi/2)*0.382*Rt+(0.382*Rt+Rt);
% Exit Coordinates
y_exit = sqrt(area_ratio)*Rt;
matrix_y = [y_TC^2 y_TC 1; y_exit^2 y_exit 1; 2*y_TC 1 0];
matrix_x = [x_TC; Ln ;1/tan(theta_n)];
x_exit = matrix_y^-1*matrix_x;
a = x_exit(1,1);
b = x_exit(2,1);
c = x_exit(3,1);
y = y_TC:0.8:y_exit;
x = a*y.^2+b*y+c;
% Complete Coordinate
Rao_x = [x_FC x_SC x];
Rao_y = [y_FC y_SC y];
% Lower Geometry
Rao_y1 = -1*Rao_y;
%Coordinates
R_X = Rao_x';
R_Y = Rao_y';
Rao = [R_X R_Y];
% Figures
figure(1)
plot (Rao_x,Rao_y,'r','LineWidth',3)
hold on
plot (Rao_x,Rao_y1,'r','LineWidth',3)
hold of
I need this code To be generate in 3D model
can you please provide
0 Comments
See Also
Categories
Find more on Get Started with Curve Fitting Toolbox in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!