How to use pdeplot with appdesigner?

6 Ansichten (letzte 30 Tage)
JClarcq
JClarcq am 5 Apr. 2018
Hello,
I try to display a gradient contour from a thermal pde solution.
pdeplot(app.data.thermalModelT,'XYData',app.data.temperature(:,end),'Contour','on');
as expected it displays a new figure. Thus I tried
pdeplot(app.UIAxes, app.data.thermalModelT,'XYData',app.data.temperature(:,end),'Contour','on');
like I would do for normal plot. But that is not supported. Is there a solution?
Thanks,
  1 Kommentar
Mohammad reza Nejati
Mohammad reza Nejati am 4 Mär. 2020
I have the same problem. could anyone help please ?

Melden Sie sich an, um zu kommentieren.

Antworten (2)

Ruben Gavín Mulelro
Ruben Gavín Mulelro am 29 Mai 2021
Same problem.. Have you found any solution?

Angelo Hafner
Angelo Hafner am 14 Jul. 2019
Today I was working all day looking for an answer. It folows my code...
nt = 21;
r = linspace(r1,r2,nt);
th = linspace(0,2*pi,nt);
[R,TH] = meshgrid(r,th);
% polar to cartezian (may be not necessary in your case
X = R .* cos(TH);
Y = R .* sin(TH);
The important thing here is to do the meshgrid
querypoints = [X(:),Y(:)]';
uintrp = interpolateSolution(results,querypoints);
and here is reshape
uintrp = reshape(uintrp,size(X));
mesh(X,Y,uintrp)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by