How to add two different surface curves in a single plot?

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Deepshikha Deo 2024-3-15

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https://ww2.mathworks.cn/matlabcentral/answers/2094776-how-to-add-two-different-surface-curves-in-a-single-plot

评论： Star Strider 2024-3-20

采纳的回答： Star Strider

I have a bell shaped curve in 3D (generated from curve fitter app) as shown in the figures.

Both are the same curve the only difference is one is without point data and the other has point data.

Problem: I want to add another surface to the same plot for the same data which is at z=1 and parallel to x- and y-axis. Also, is it possible to find the values of major and minor axes of the ellipse formed from the intersection of both the surfaces. Or in other words, the values between the intersection of x and z values of both the curves and the y and z values of both the curves.

I hope I am able to express myself clearly.

Thank you

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Answer 1

Star Strider 2024-3-15

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https://ww2.mathworks.cn/matlabcentral/answers/2094776-how-to-add-two-different-surface-curves-in-a-single-plot#answer_1425911

在 MATLAB Online 中打开

Try this —

‘I want to add another surface to the same plot for the same data which is at z=1 and parallel to x- and y-axis.’

[X,Y] = ndgrid(-3:0.1:3);

f = @(x,y) exp(-(x.^2+(2*y).^2)*0.5);

Z = f(X,Y)*3;

figure

surf(X, Y, Z)

hold on

surf(X, Y, ones(size(Z)), 'FaceColor','r', 'FaceAlpha',0.5, 'EdgeColor','none')

hold off

colormap(turbo)

‘Also, is it possible to find the values of major and minor axes of the ellipse formed from the intersection of both the surfaces.’

figure

[c,h] = contour(X, Y, Z, [1 1]);

axis('equal')

grid

Ax = gca;

Ax.XAxisLocation = 'origin';

Ax.YAxisLocation = 'origin';

elpsfcn = @(b,xy) xy(1,:).^2/b(1)^2 + xy(2,:).^2/b(2)^2 - b(3);

opts = optimoptions('fminunc', 'MaxFunctionEvaluations', 5E+3, 'MaxIterations',1E+4);

[B, fv] = fminunc(@(b) norm(elpsfcn(b,c(:,2:end))), rand(3,1), opts)

Local minimum possible. fminunc stopped because it cannot decrease the objective function along the current search direction.

B = 3×1

52.8204 26.4510 0.0008

fv = 1.2653e-05

fprintf('Semimajor Axis = %.4f\nSemiminor Axis = %.4f\nConstant = %.4f\n', B)

Semimajor Axis = 52.8204 Semiminor Axis = 26.4510 Constant = 0.0008

text(-2.5, 2.5, sprintf('$\\frac{x^2}{%.2f^2} + \\frac{y^2}{%.2f^2} = %.4f$',B), 'Interpreter','latex', 'FontSize',16)

.

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Deepshikha Deo 2024-3-18

BG_Apr03.zip

Thanks a lot for taking your time and answering the question with an example. I tried the sample code and it works well but I am having the trouble with replacing it with my data, I hope you can help me.

I have used curve fitter app and exported the output to the workspace. Thus a sfit of name 'fittedmodel' has been created in the workspace.

I replaced sixth line of your code (which is surf(X, Y, Z) ) to plot(fittedmodel,[x,y],z)

This created the following image

After that I followed the code as it is and it showed an error that stated:

Error using surf

Z must be a matrix, not a scalar or vector.

In an another trial, I added the function in attempts to create a matrix as follows:

f=@(x,y) a0*exp(-(((x-x0).*cos(phi)+(y-y0).*sin(phi)).^2/0.5*ax^2)-(((y-y0).*cos(phi)-(x-x0).*sin(phi)).^2/0.5*ay^2))

z=f(X,Y)*3;

But it also showed an error stating:

Unrecognized function or variable 'a0'.

Error in @(x,y)a0*exp(-(((x-x0).*cos(phi)+(y-y0).*sin(phi)).^2/0.5*ax^2)-(((y-y0).*cos(phi)-(x-x0).*sin(phi)).^2/0.5*ay^2))

I request you to help me with the problem. I have also attached the zip file which contains the data in excel and saved sfit file.

Thank you

Regards

Deepshikha Deo 2024-3-19

Thanks a lot for putting so much efforts. I appreciate the time you have given to my problem.

Yes, the curve fitter app is needed to open the sfit file (the second fit named 'fp' is the one). The grid_code in excel file was used to calculate the values for z column in excel sheet, thus it is not needed further but could not be removed too. I have attached the code which contains the equation (function) with the brief description of its unknown parameters. This function was called in curve fitter app as a custom equation since the real equation was too long to type.

function[zVal]=gaussianFit(x,y,a0,ax,ay,phi,x0,y0)

% no constants in the equation

% the meaning of unknowns is:

% x0 and y0 are the center location of the peak value. Values should be within the range of x and y data

% a0 is the peak value or height of Gaussian curve. Its values should be greater than zero and not more than the maximum z value

% phi is the angle of rotation of ellipse formed.

% ax is the major axis of ellipse or the standard deviation in x-direction. It should be greater than zero

% ay is the minor axis of ellipse or the standard deviation in y-direction. It should be greater than zero

% x and y are the x & y coordinates values

% zVal is the UHI values (z values)

zVal=a0*exp(-(((x-x0).*cos(phi)+(y-y0).*sin(phi)).^2/0.5*ax^2)-(((y-y0).*cos(phi)-(x-x0).*sin(phi)).^2/0.5*ay^2));

end

There is also an onramp course with the name "curve fitting onramp" (https://matlabacademy.mathworks.com/details/curve-fitting-onramp/orcf?s_tid=srchtitle_site_search_1_curve%20fitter%20course) for its understanding.

In the curve fitter app, after adding the data, I selected "custom equation" as fit type and typed the function. After manipulating the lower and upper bounds for six parameters, I hit the 'Fit' which generated the surface fit. The following code is generated by export generate code.

Note- The variables' names may wary.

function [fitresult, gof] = createFit2(POINT_X, POINT_Y, UHI)

%CREATEFIT2(POINT_X,POINT_Y,UHI)

% Create a fit.

%

% Data for 'fp' fit:

% X Input: POINT_X from BG_B_Apr2003

% Y Input: POINT_Y from BG_B_Apr2003

% Z Output: UHI from BG_B_Apr2003

% Output:

% fitresult : a fit object representing the fit.

% gof : structure with goodness-of fit info.

%

% See also FIT, CFIT, SFIT.

% Auto-generated by MATLAB on 19-Mar-2024 05:28:55

%% Fit: 'fp'.

[xData, yData, zData] = prepareSurfaceData( POINT_X, POINT_Y, UHI );

% Set up fittype and options.

ft = fittype( 'gaussianFit(x,y,a0,ax,ay,phi,x0,y0)', 'independent', {'x', 'y'}, 'dependent', 'z' );

opts = fitoptions( 'Method', 'NonlinearLeastSquares' );

opts.Display = 'Off';

opts.Lower = [0 0 0 0 77 12];

opts.StartPoint = [0.754686681982361 0.276025076998578 0.679702676853675 0.655098003973841 0.162611735194631 0.118997681558377];

opts.Upper = [Inf Inf Inf 360 78 14];

% Fit model to data.

[fitresult, gof] = fit( [xData, yData], zData, ft, opts );

% Plot fit with data.

figure( 'Name', 'fp' );

h = plot( fitresult, [xData, yData], zData );

legend( h, 'fp', 'UHI vs. POINT_X, POINT_Y', 'Location', 'NorthEast', 'Interpreter', 'none' );

% Label axes

xlabel( 'POINT_X', 'Interpreter', 'none' );

ylabel( 'POINT_Y', 'Interpreter', 'none' );

zlabel( 'UHI', 'Interpreter', 'none' );

grid on

By using export, the output can be exported to the workspace. The following code is used to plot the exported output surface.

plot(fittedmodel) % fittedmodel is the name of the output surface. This will create plot without the point data values

plot(fittedmodel,[x,y],z) % This will plot the surface with point data values. (the name of x, y, and z may vary)

The surface was generated from zero thus giving me the values for the parameters from zero. This is fine for all values except ax and ay. The concept (suggested in research papers), I followed have shown the values for ax and ay from the value 1 of z-axis. I was not able to find the values when z=1 using curve fitter app as it not should exceed the values of ax and ay at 0 of z-axis. After that I came up with the idea I put in the question which I again failed to solve by myself.

I attached this image as an example for better understanding. It shows the 2-D image of the surface (taked from one of the research papers). In this image, the line ax(1K) represents the standard deviation in x-direction whereas the curve is starting from zero.

I know, this can also be done using optimization too but my lack of knowledge became a hinderance to do so.

Once again thanks a lot for the time you have given to my probelm.

Best Regards

Star Strider 2024-3-19

The semimajor and semimnor axis values are with respect to the data provided. Since the data values are relatively large in magnitude (77 for ‘X’ and 13 for ‘Y’), the corresponding values are large.

Running it a few more times, and adding a second ellipse parameter estimation with offsets for ‘X’ and ‘Y’, and the objective function now defined as:

elpsfcn = @(b,xy) (xy(1,:)-b(4)).^2/b(1)^2 + (xy(2,:)-b(5)).^2/b(2)^2 - b(3);

with the ‘b(4)’ and ‘b(5)’ offsets themselves (77.5579, 12.9724) appropriately estimated, gives a much better result:

Semimajor Axis = 1.7444

Semiminor Axis = 1.7627

Constant = 0.0026

This ratio is now 0.99, and more representative of the observed ellipse shape. (See the second ‘Contour With Fitted Function’ plot.)

‘Please, tell me how to create that grid to create surface if I already have x and y values as (n x 1) size variable.’

I am not certain what you want, however I probably already provided that in my code, specifically:

xv = linspace(Xu, Xd, 100);

yv = linspace(Yd, Yu, 100);

[Xm,Ym] = ndgrid(xv, yv);

ZF = scatteredInterpolant(T1.X, T1.Y, T1.Z);

Zm = ZF(Xm, Ym);

These are excerpted from the code that created the first surf plot. They interpolate the existing data and create it as the surface.

That is likely the best that I can do with these data.

.

Deepshikha Deo 2024-3-20

Thank you for your efforts.

Regards

Star Strider 2024-3-20

As always, my pleasure!

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How to add two different surface curves in a single plot?

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