Tag Archives: matlab
I am setting up the default case in Virtual Vehicle composer app and while running the performance test on the motor for 1EM.I am getting an oscillating graph for Trqs.
I am setting up the default case in Virtual Vehicle composer app and while running the performance test on the motor for 1EM.I am getting an oscillating graph for Trqs. Please xplain why this is an oscillating graph.I am setting up the default case in Virtual Vehicle composer app and while running the performance test on the motor for 1EM.I am getting an oscillating graph for Trqs. Please xplain why this is an oscillating graph. I am setting up the default case in Virtual Vehicle composer app and while running the performance test on the motor for 1EM.I am getting an oscillating graph for Trqs. Please xplain why this is an oscillating graph. virtual vehicle composer MATLAB Answers — New Questions
HOW CAN I TURN AN EQUATION INTO COMPONENT IN A LIBRARY?
component CantileverBeamDeflection
% Define the nodes
nodes
P = foundation.mechanical.translational.translational; % Translational node P
C = foundation.mechanical.translational.translational; % Translational node C
end
% Define the parameters
parameters
L = {1, ‘m’}; % Length of the beam
E = {210e9, ‘Pa’}; % Young’s modulus
I = {1e-6, ‘m^4’}; % Second moment of area
w = {100, ‘N/m’}; % Uniform load per unit length
x = {0.5, ‘m’}; % Distance from the fixed end
end
% Define the variables
variables
delta = {0, ‘m’}; % Deflection
end
% Define the equations
equations
% Deflection equation for a fixed-free beam under uniform load
delta == (w*x^2)/(24*E*I) * (L^3 – 2*L*x + x^2);
% Force equilibrium at nodes
P.f == -w * x;
C.f == w * x;
end
endcomponent CantileverBeamDeflection
% Define the nodes
nodes
P = foundation.mechanical.translational.translational; % Translational node P
C = foundation.mechanical.translational.translational; % Translational node C
end
% Define the parameters
parameters
L = {1, ‘m’}; % Length of the beam
E = {210e9, ‘Pa’}; % Young’s modulus
I = {1e-6, ‘m^4’}; % Second moment of area
w = {100, ‘N/m’}; % Uniform load per unit length
x = {0.5, ‘m’}; % Distance from the fixed end
end
% Define the variables
variables
delta = {0, ‘m’}; % Deflection
end
% Define the equations
equations
% Deflection equation for a fixed-free beam under uniform load
delta == (w*x^2)/(24*E*I) * (L^3 – 2*L*x + x^2);
% Force equilibrium at nodes
P.f == -w * x;
C.f == w * x;
end
end component CantileverBeamDeflection
% Define the nodes
nodes
P = foundation.mechanical.translational.translational; % Translational node P
C = foundation.mechanical.translational.translational; % Translational node C
end
% Define the parameters
parameters
L = {1, ‘m’}; % Length of the beam
E = {210e9, ‘Pa’}; % Young’s modulus
I = {1e-6, ‘m^4’}; % Second moment of area
w = {100, ‘N/m’}; % Uniform load per unit length
x = {0.5, ‘m’}; % Distance from the fixed end
end
% Define the variables
variables
delta = {0, ‘m’}; % Deflection
end
% Define the equations
equations
% Deflection equation for a fixed-free beam under uniform load
delta == (w*x^2)/(24*E*I) * (L^3 – 2*L*x + x^2);
% Force equilibrium at nodes
P.f == -w * x;
C.f == w * x;
end
end crete component in library, new custom component, equation into code MATLAB Answers — New Questions
How can I create my desired waveform with the DAC of Arduino Due through MATLAB?
Arduino due has 2 pins for DAC but presumably there is no analogWrite command in MATLAB for Arduino. All I want to know is that if it is possible to use the 2 DAC pins with MATLAB and how.Arduino due has 2 pins for DAC but presumably there is no analogWrite command in MATLAB for Arduino. All I want to know is that if it is possible to use the 2 DAC pins with MATLAB and how. Arduino due has 2 pins for DAC but presumably there is no analogWrite command in MATLAB for Arduino. All I want to know is that if it is possible to use the 2 DAC pins with MATLAB and how. arduinodue dac analogwrite MATLAB Answers — New Questions
Linking line plot and map (geoshow) together MATLAB
Hello all:
I wish to combine the MATLAB spatial plot generated through mapping function and the line plot on the same axis scale. However, the linkprop function is not working in this case. Please suggest how to make both axis similar?
Here is my code:
figure(1);
AX1 = subplot(1,2,1);
ax=axesm(‘mercator’,’MapLatLimit’,[-65 65],’MapLonLimit’,[-180 180]);
f=worldmap([-60 80],[-180 180]);
h = geoshow(coast1, ‘DisplayType’, ‘polygon’,’facecolor’,’w’);
oceanColor = [.7 .8 .9];%[.5 .7 .9];
setm(ax,’FFaceColor’,oceanColor);
setm(ax, ‘meridianlabel’, ‘on’, ‘parallellabel’, ‘on’);
setm(gca,’mlabelparallel’,-90);
hold on;
%Define coordinates for tropical lines
yvec = -180:180;
xvec = ones(size(yvec));
geoshow(23.5*xvec,yvec,’DisplayType’,’Line’,’LineWidth’,1,’LineStyle’,’–‘,’Color’,’k’); hold on;
geoshow(-23.5*xvec,yvec,’DisplayType’,’Line’,’LineWidth’,1,’LineStyle’,’–‘,’Color’,’k’); hold on;
geoshow(35*xvec,yvec,’DisplayType’,’Line’,’LineWidth’,0.8,’LineStyle’,’–‘,’Color’,’r’); hold on;
geoshow(-35*xvec,yvec,’DisplayType’,’Line’,’LineWidth’,0.8,’LineStyle’,’–‘,’Color’,’r’); hold on;
AX2 = subplot(1,2,2);
h1 = plot(B1,X_dist(:,2),’r–‘,’linew’,0.8); hold on;
h2 = plot(B,X_dist(:,2),’r-‘,’linew’,1.8); hold on;
h3 = plot(B2,X_dist(:,2),’r–‘,’linew’,2); hold on;
set(gca,’fontsize’,14,’fontname’,’arial’);
degreetick ‘y’;
linkprop(AX1,’YLim’);
linkprop(AX2,’YLim’);
However, I am unable to adjust the size and Y-labels of both subplots. The latitudes of both subplots should match.Hello all:
I wish to combine the MATLAB spatial plot generated through mapping function and the line plot on the same axis scale. However, the linkprop function is not working in this case. Please suggest how to make both axis similar?
Here is my code:
figure(1);
AX1 = subplot(1,2,1);
ax=axesm(‘mercator’,’MapLatLimit’,[-65 65],’MapLonLimit’,[-180 180]);
f=worldmap([-60 80],[-180 180]);
h = geoshow(coast1, ‘DisplayType’, ‘polygon’,’facecolor’,’w’);
oceanColor = [.7 .8 .9];%[.5 .7 .9];
setm(ax,’FFaceColor’,oceanColor);
setm(ax, ‘meridianlabel’, ‘on’, ‘parallellabel’, ‘on’);
setm(gca,’mlabelparallel’,-90);
hold on;
%Define coordinates for tropical lines
yvec = -180:180;
xvec = ones(size(yvec));
geoshow(23.5*xvec,yvec,’DisplayType’,’Line’,’LineWidth’,1,’LineStyle’,’–‘,’Color’,’k’); hold on;
geoshow(-23.5*xvec,yvec,’DisplayType’,’Line’,’LineWidth’,1,’LineStyle’,’–‘,’Color’,’k’); hold on;
geoshow(35*xvec,yvec,’DisplayType’,’Line’,’LineWidth’,0.8,’LineStyle’,’–‘,’Color’,’r’); hold on;
geoshow(-35*xvec,yvec,’DisplayType’,’Line’,’LineWidth’,0.8,’LineStyle’,’–‘,’Color’,’r’); hold on;
AX2 = subplot(1,2,2);
h1 = plot(B1,X_dist(:,2),’r–‘,’linew’,0.8); hold on;
h2 = plot(B,X_dist(:,2),’r-‘,’linew’,1.8); hold on;
h3 = plot(B2,X_dist(:,2),’r–‘,’linew’,2); hold on;
set(gca,’fontsize’,14,’fontname’,’arial’);
degreetick ‘y’;
linkprop(AX1,’YLim’);
linkprop(AX2,’YLim’);
However, I am unable to adjust the size and Y-labels of both subplots. The latitudes of both subplots should match. Hello all:
I wish to combine the MATLAB spatial plot generated through mapping function and the line plot on the same axis scale. However, the linkprop function is not working in this case. Please suggest how to make both axis similar?
Here is my code:
figure(1);
AX1 = subplot(1,2,1);
ax=axesm(‘mercator’,’MapLatLimit’,[-65 65],’MapLonLimit’,[-180 180]);
f=worldmap([-60 80],[-180 180]);
h = geoshow(coast1, ‘DisplayType’, ‘polygon’,’facecolor’,’w’);
oceanColor = [.7 .8 .9];%[.5 .7 .9];
setm(ax,’FFaceColor’,oceanColor);
setm(ax, ‘meridianlabel’, ‘on’, ‘parallellabel’, ‘on’);
setm(gca,’mlabelparallel’,-90);
hold on;
%Define coordinates for tropical lines
yvec = -180:180;
xvec = ones(size(yvec));
geoshow(23.5*xvec,yvec,’DisplayType’,’Line’,’LineWidth’,1,’LineStyle’,’–‘,’Color’,’k’); hold on;
geoshow(-23.5*xvec,yvec,’DisplayType’,’Line’,’LineWidth’,1,’LineStyle’,’–‘,’Color’,’k’); hold on;
geoshow(35*xvec,yvec,’DisplayType’,’Line’,’LineWidth’,0.8,’LineStyle’,’–‘,’Color’,’r’); hold on;
geoshow(-35*xvec,yvec,’DisplayType’,’Line’,’LineWidth’,0.8,’LineStyle’,’–‘,’Color’,’r’); hold on;
AX2 = subplot(1,2,2);
h1 = plot(B1,X_dist(:,2),’r–‘,’linew’,0.8); hold on;
h2 = plot(B,X_dist(:,2),’r-‘,’linew’,1.8); hold on;
h3 = plot(B2,X_dist(:,2),’r–‘,’linew’,2); hold on;
set(gca,’fontsize’,14,’fontname’,’arial’);
degreetick ‘y’;
linkprop(AX1,’YLim’);
linkprop(AX2,’YLim’);
However, I am unable to adjust the size and Y-labels of both subplots. The latitudes of both subplots should match. mapping, linkprop MATLAB Answers — New Questions
interpolation of a sinusoidal tunnel data
hallo eveyone, i have fluent data of a flow instive a wavy tunnel as present here. I export the data as ACII file and i want to crate matrixs so i can work on the data preaty easely. the problem is that all the scatterd interpolation mathods do not work smothly on a wavy wall, wich asagnifetly decline. i need bater mathod to work with that interpolat on the wall in between those data points and not in the normal x y deraction? any segestions ??? may data is simpal scatterd data of x,y,and uhallo eveyone, i have fluent data of a flow instive a wavy tunnel as present here. I export the data as ACII file and i want to crate matrixs so i can work on the data preaty easely. the problem is that all the scatterd interpolation mathods do not work smothly on a wavy wall, wich asagnifetly decline. i need bater mathod to work with that interpolat on the wall in between those data points and not in the normal x y deraction? any segestions ??? may data is simpal scatterd data of x,y,and u hallo eveyone, i have fluent data of a flow instive a wavy tunnel as present here. I export the data as ACII file and i want to crate matrixs so i can work on the data preaty easely. the problem is that all the scatterd interpolation mathods do not work smothly on a wavy wall, wich asagnifetly decline. i need bater mathod to work with that interpolat on the wall in between those data points and not in the normal x y deraction? any segestions ??? may data is simpal scatterd data of x,y,and u interpolation, fluent MATLAB Answers — New Questions
Surf on a cylinder
How can I plot this surf
on a cylindrical surface, like this?How can I plot this surf
on a cylindrical surface, like this? How can I plot this surf
on a cylindrical surface, like this? surf, cylindrical MATLAB Answers — New Questions
Equations for predicting outputs under SVM regression (RBF or polynomial)
Hello everyone,
I am trying to calculate output (Y) for new input data (X) using a pretrained SVM (trained with kernel function RBF or polynomial).
I know the equations for linear SVM regression:
ex) Y = inputdata * Beta + Bias
However, I am trying to find the equations to calculate the Y response under ‘RBF’ or ‘Polynomial’ SVM regression.
Please help
Thank you.Hello everyone,
I am trying to calculate output (Y) for new input data (X) using a pretrained SVM (trained with kernel function RBF or polynomial).
I know the equations for linear SVM regression:
ex) Y = inputdata * Beta + Bias
However, I am trying to find the equations to calculate the Y response under ‘RBF’ or ‘Polynomial’ SVM regression.
Please help
Thank you. Hello everyone,
I am trying to calculate output (Y) for new input data (X) using a pretrained SVM (trained with kernel function RBF or polynomial).
I know the equations for linear SVM regression:
ex) Y = inputdata * Beta + Bias
However, I am trying to find the equations to calculate the Y response under ‘RBF’ or ‘Polynomial’ SVM regression.
Please help
Thank you. svm, rbf, polynomial, regression MATLAB Answers — New Questions
save fit data from ARX reports
I have a lot of time series data (312 experiments) and several models. I want to make sure that the models outperform an AR model with the same number of parameters. However, due to the read-only mode of the AR report.fit, I have to fit each experiment separately because the Report.Fit struct is read-only. Is there any way I can extract the Sys.Report.Fit data so that I can save all 312 experiments using flist?I have a lot of time series data (312 experiments) and several models. I want to make sure that the models outperform an AR model with the same number of parameters. However, due to the read-only mode of the AR report.fit, I have to fit each experiment separately because the Report.Fit struct is read-only. Is there any way I can extract the Sys.Report.Fit data so that I can save all 312 experiments using flist? I have a lot of time series data (312 experiments) and several models. I want to make sure that the models outperform an AR model with the same number of parameters. However, due to the read-only mode of the AR report.fit, I have to fit each experiment separately because the Report.Fit struct is read-only. Is there any way I can extract the Sys.Report.Fit data so that I can save all 312 experiments using flist? read only mode for sys.report. fit in ar models MATLAB Answers — New Questions
Unable to find a valid PropName for waitfor with a Matlab app
Using app designer, I have a main app that calls a form. From the main app, I instantiate the form
app.FormApp = FormApp(app);
I can halt execution by waiting for the form to be deleted
waitfor(app.formApp);
My desire is to keep the form open and repopulate from a loop in the main app. However, setting a public property on FormApp and calling it fails. I.e.
waitfor(app.formApp, app.formApp.pauseLoop, false)
throws "Error using waitfor
Invalid property." I’ve tried lots of combinations trying to get a property of the FormApp that waitfor will use. Any help appreciated.
I’m using R2023b Update 5 on Windows 10.Using app designer, I have a main app that calls a form. From the main app, I instantiate the form
app.FormApp = FormApp(app);
I can halt execution by waiting for the form to be deleted
waitfor(app.formApp);
My desire is to keep the form open and repopulate from a loop in the main app. However, setting a public property on FormApp and calling it fails. I.e.
waitfor(app.formApp, app.formApp.pauseLoop, false)
throws "Error using waitfor
Invalid property." I’ve tried lots of combinations trying to get a property of the FormApp that waitfor will use. Any help appreciated.
I’m using R2023b Update 5 on Windows 10. Using app designer, I have a main app that calls a form. From the main app, I instantiate the form
app.FormApp = FormApp(app);
I can halt execution by waiting for the form to be deleted
waitfor(app.formApp);
My desire is to keep the form open and repopulate from a loop in the main app. However, setting a public property on FormApp and calling it fails. I.e.
waitfor(app.formApp, app.formApp.pauseLoop, false)
throws "Error using waitfor
Invalid property." I’ve tried lots of combinations trying to get a property of the FormApp that waitfor will use. Any help appreciated.
I’m using R2023b Update 5 on Windows 10. app designer, uifigure, waitfor MATLAB Answers — New Questions
“Try the New Desktop” icon persists even after uninstalling the add-on (New Desktop for MATLAB (Beta))
I recently installed the New Desktop for MATLAB from MATLAB FEX (by reading this article of Matlab Blog) to enable dark mode. However, I didn’t prefer it, so I uninstalled it through Add-Ons > Manage Add-Ons. While it’s no longer listed in my add-ons, the "Try the New Desktop" icon remains at the top of my MATLAB environment. I’ve even restarted my computer, but the icon persists.
Could you please advise on how to remove this icon permanently?I recently installed the New Desktop for MATLAB from MATLAB FEX (by reading this article of Matlab Blog) to enable dark mode. However, I didn’t prefer it, so I uninstalled it through Add-Ons > Manage Add-Ons. While it’s no longer listed in my add-ons, the "Try the New Desktop" icon remains at the top of my MATLAB environment. I’ve even restarted my computer, but the icon persists.
Could you please advise on how to remove this icon permanently? I recently installed the New Desktop for MATLAB from MATLAB FEX (by reading this article of Matlab Blog) to enable dark mode. However, I didn’t prefer it, so I uninstalled it through Add-Ons > Manage Add-Ons. While it’s no longer listed in my add-ons, the "Try the New Desktop" icon remains at the top of my MATLAB environment. I’ve even restarted my computer, but the icon persists.
Could you please advise on how to remove this icon permanently? matlab, file exchange, new desktop for matlab (beta) MATLAB Answers — New Questions
Not able to plot a proper graph for the equation.
y = (1*1*((25)-(x.^2)))/((x.^4)-((x.^2)*((1*(1.78))+(25)+1))+(1*(25)));
fplot(y);
i used this code but am only getting a straight line instead of the 2 peak FRF that I want.y = (1*1*((25)-(x.^2)))/((x.^4)-((x.^2)*((1*(1.78))+(25)+1))+(1*(25)));
fplot(y);
i used this code but am only getting a straight line instead of the 2 peak FRF that I want. y = (1*1*((25)-(x.^2)))/((x.^4)-((x.^2)*((1*(1.78))+(25)+1))+(1*(25)));
fplot(y);
i used this code but am only getting a straight line instead of the 2 peak FRF that I want. mathematics, graph, plot MATLAB Answers — New Questions
Problems with the use of streamline for a magnetic field
Hi,
I’m trying to plot the streamlines of a magnetic field generated by a magnet and a coil. I have my Field generated in 6 vectors, XYZ for the coordinates et UVW for the components. I managed to reorganize those data in 6 3D-arrays. I ploted the quiver and the plot seems satisfying. Now I wanted to plot the streamlines in order to observe the loopback. Unfortunately, the streamlines are not really good and does not follow the vectors ploted by quiver. Here is the figure :
when i use the following xyz 3D arrays and i have the error sample points must be unique and i don’t understand why becuase all the couples are unique.
x=
val(:,:,1) =
-2.000000000000000e-01 -2.000000000000000e-01 -2.000000000000000e-01
-1.387778780781446e-17 -1.387778780781446e-17 -1.387778780781446e-17
2.000000000000000e-01 2.000000000000000e-01 2.000000000000000e-01
val(:,:,2) =
-2.000000000000000e-01 -2.000000000000000e-01 -2.000000000000000e-01
-1.387778780781446e-17 -1.387778780781446e-17 -1.387778780781446e-17
2.000000000000000e-01 2.000000000000000e-01 2.000000000000000e-01
val(:,:,3) =
-2.000000000000000e-01 -2.000000000000000e-01 -2.000000000000000e-01
-1.387778780781446e-17 -1.387778780781446e-17 -1.387778780781446e-17
2.000000000000000e-01 2.000000000000000e-01 2.000000000000000e-01
y=
val(:,:,1) =
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
val(:,:,2) =
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
val(:,:,3) =
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
z=
val(:,:,1) =
-5.000000000000000e-02 -5.000000000000000e-02 -5.000000000000000e-02
-5.000000000000000e-02 -5.000000000000000e-02 -5.000000000000000e-02
-5.000000000000000e-02 -5.000000000000000e-02 -5.000000000000000e-02
val(:,:,2) =
1.500000000000000e-01 1.500000000000000e-01 1.500000000000000e-01
1.500000000000000e-01 1.500000000000000e-01 1.500000000000000e-01
1.500000000000000e-01 1.500000000000000e-01 1.500000000000000e-01
val(:,:,3) =
3.500000000000000e-01 3.500000000000000e-01 3.500000000000000e-01
3.500000000000000e-01 3.500000000000000e-01 3.500000000000000e-01
3.500000000000000e-01 3.500000000000000e-01 3.500000000000000e-01Hi,
I’m trying to plot the streamlines of a magnetic field generated by a magnet and a coil. I have my Field generated in 6 vectors, XYZ for the coordinates et UVW for the components. I managed to reorganize those data in 6 3D-arrays. I ploted the quiver and the plot seems satisfying. Now I wanted to plot the streamlines in order to observe the loopback. Unfortunately, the streamlines are not really good and does not follow the vectors ploted by quiver. Here is the figure :
when i use the following xyz 3D arrays and i have the error sample points must be unique and i don’t understand why becuase all the couples are unique.
x=
val(:,:,1) =
-2.000000000000000e-01 -2.000000000000000e-01 -2.000000000000000e-01
-1.387778780781446e-17 -1.387778780781446e-17 -1.387778780781446e-17
2.000000000000000e-01 2.000000000000000e-01 2.000000000000000e-01
val(:,:,2) =
-2.000000000000000e-01 -2.000000000000000e-01 -2.000000000000000e-01
-1.387778780781446e-17 -1.387778780781446e-17 -1.387778780781446e-17
2.000000000000000e-01 2.000000000000000e-01 2.000000000000000e-01
val(:,:,3) =
-2.000000000000000e-01 -2.000000000000000e-01 -2.000000000000000e-01
-1.387778780781446e-17 -1.387778780781446e-17 -1.387778780781446e-17
2.000000000000000e-01 2.000000000000000e-01 2.000000000000000e-01
y=
val(:,:,1) =
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
val(:,:,2) =
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
val(:,:,3) =
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
z=
val(:,:,1) =
-5.000000000000000e-02 -5.000000000000000e-02 -5.000000000000000e-02
-5.000000000000000e-02 -5.000000000000000e-02 -5.000000000000000e-02
-5.000000000000000e-02 -5.000000000000000e-02 -5.000000000000000e-02
val(:,:,2) =
1.500000000000000e-01 1.500000000000000e-01 1.500000000000000e-01
1.500000000000000e-01 1.500000000000000e-01 1.500000000000000e-01
1.500000000000000e-01 1.500000000000000e-01 1.500000000000000e-01
val(:,:,3) =
3.500000000000000e-01 3.500000000000000e-01 3.500000000000000e-01
3.500000000000000e-01 3.500000000000000e-01 3.500000000000000e-01
3.500000000000000e-01 3.500000000000000e-01 3.500000000000000e-01 Hi,
I’m trying to plot the streamlines of a magnetic field generated by a magnet and a coil. I have my Field generated in 6 vectors, XYZ for the coordinates et UVW for the components. I managed to reorganize those data in 6 3D-arrays. I ploted the quiver and the plot seems satisfying. Now I wanted to plot the streamlines in order to observe the loopback. Unfortunately, the streamlines are not really good and does not follow the vectors ploted by quiver. Here is the figure :
when i use the following xyz 3D arrays and i have the error sample points must be unique and i don’t understand why becuase all the couples are unique.
x=
val(:,:,1) =
-2.000000000000000e-01 -2.000000000000000e-01 -2.000000000000000e-01
-1.387778780781446e-17 -1.387778780781446e-17 -1.387778780781446e-17
2.000000000000000e-01 2.000000000000000e-01 2.000000000000000e-01
val(:,:,2) =
-2.000000000000000e-01 -2.000000000000000e-01 -2.000000000000000e-01
-1.387778780781446e-17 -1.387778780781446e-17 -1.387778780781446e-17
2.000000000000000e-01 2.000000000000000e-01 2.000000000000000e-01
val(:,:,3) =
-2.000000000000000e-01 -2.000000000000000e-01 -2.000000000000000e-01
-1.387778780781446e-17 -1.387778780781446e-17 -1.387778780781446e-17
2.000000000000000e-01 2.000000000000000e-01 2.000000000000000e-01
y=
val(:,:,1) =
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
val(:,:,2) =
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
val(:,:,3) =
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
-2.000000000000000e-01 -1.387778780781446e-17 2.000000000000000e-01
z=
val(:,:,1) =
-5.000000000000000e-02 -5.000000000000000e-02 -5.000000000000000e-02
-5.000000000000000e-02 -5.000000000000000e-02 -5.000000000000000e-02
-5.000000000000000e-02 -5.000000000000000e-02 -5.000000000000000e-02
val(:,:,2) =
1.500000000000000e-01 1.500000000000000e-01 1.500000000000000e-01
1.500000000000000e-01 1.500000000000000e-01 1.500000000000000e-01
1.500000000000000e-01 1.500000000000000e-01 1.500000000000000e-01
val(:,:,3) =
3.500000000000000e-01 3.500000000000000e-01 3.500000000000000e-01
3.500000000000000e-01 3.500000000000000e-01 3.500000000000000e-01
3.500000000000000e-01 3.500000000000000e-01 3.500000000000000e-01 streamlines MATLAB Answers — New Questions
How to find solution to system of 2 circle equations in Matlab?
x^2+y^2=25;
x^2+y^2=36;
%%how to find values of x and y??
%Please help someonex^2+y^2=25;
x^2+y^2=36;
%%how to find values of x and y??
%Please help someone x^2+y^2=25;
x^2+y^2=36;
%%how to find values of x and y??
%Please help someone #cicle, #equations MATLAB Answers — New Questions
Global Stiffness Matrix 8×8
% Declaring all the variables
E1 = 200e9; % Young’s modulus for elements 1 and 2 in Pa
A1 = 0.000006; % Cross-sectional area for elements 1 and 2 in m^2
E3 = 200e9; % Young’s modulus for element 3 in Pa
A3 = 0.000006; % Cross-sectional area for element 3 in m^2
L = 0.5; % Length of each element in m
F = 5000; % Applied force at node 2 in N
disp(‘E1(Pa) = ‘),disp(E1);
disp(‘E3(Pa) = ‘),disp(E3);
disp(‘A1(m^2) = ‘),disp(A1);
disp(‘A3(m^2) = ‘),disp(A3);
disp(‘L(m) = ‘),disp(L);
disp(‘θ(°) = 5 ‘);
C=cos(5*pi/180);
S=sin(5*pi/180);
disp(‘ Local Stiffness Matrixs :’)
k1=E1*A1*[C*C C*S -C*C -C*S
C*S S*S -C*S -S*S
-C*C -C*S C*C C*S
-C*S -S*S C*S S*S]/0.25
k2=E1*A1*[0 0 0 0
0 1 0 -1
0 0 0 0
0 -1 0 1]/0.0218
k3=E3*A3*[C*C C*S -C*C -C*S
C*S S*S -C*S -S*S
-C*C -C*S C*C C*S
-C*S -S*S C*S S*S]/0.25
disp(‘ Global Stiffness Matrix :’)
K=[k1(1,1) k1(1,2) k1(1,3) k1(1,4) 0 0 0 0
k1(2,1) k1(2,2) k1(2,3) k1(2,4) 0 0 0 0
k1(3,1) k1(3,2) k1(3,3)+k2(1,1) k1(3,4)+k2(1,2) k2(1,3) k2(1,4) 0 0
k1(4,1) k1(4,2) k1(4,3)+k2(2,1) k1(4,4)+k2(2,2) k2(2,3) k2(2,4) 0 0
0 0 k2(3,1) k2(3,2) k2(3,3)+k3(1,1) k2(3,4)+k3(1,2) k3(1,3) k3(1,4)
0 0 k2(4,1) k2(4,2) k2(4,3)+k3(2,1) k2(4,4)+k3(2,2) k3(2,3) k3(2,4)
0 0 0 0 k3(3,1) k3(3,2) k3(3,3) k3(3,4)
0 0 0 0 k3(4,1) k3(4,2) k3(4,3) k3(4,4) ]
disp(‘ Transformation matrix :’)
T= [ C S 0 0 0 0 0 0
-S C 0 0 0 0 0 0
0 0 C S 0 0 0 0
0 0 -S C 0 0 0 0
0 0 0 0 C S 0 0
0 0 0 0 -S C 0 0
0 0 0 0 0 0 C S
0 0 0 0 0 0 -S C ]
disp(‘ Inverse of Transformation matrix :’)
TT= inv(T)
disp(‘ [T][K][TT] :’)
M=T*K*TT
disp(‘ Applying the boundary condition’)
U1=0
V1=0
U3=0
V3=0
V4=0
disp(‘Force at node 2 :’);
Fx2=[0,5000];
IK=[M(3,3), M(5,3); M(5,3), M(5,5)];
sol=Fx2/IK;
disp (5000);
disp(‘Displacement of node 2 and 3 at local coordinate(m) (0°):’);
U2=sol(1);
U3=sol(2);
t=[C,S;-S,C];
u=[U2; U3];
disp(‘d2x = ‘),disp(U2);
disp(‘d3x = ‘),disp(U3);
disp(‘Displacement of node 2 and 3 at Global coordinate(m) (5°):’);
t=[C,S;-S,C];
u=[U2; U3];
solu=tu;
u2=solu(1);
u3=solu(2);
disp(‘d2x = ‘),disp(u2);
disp(‘d3x = ‘),disp(u3);
I don’t know if the Global Stiffness Matrix arrangement is correct can someone help me?% Declaring all the variables
E1 = 200e9; % Young’s modulus for elements 1 and 2 in Pa
A1 = 0.000006; % Cross-sectional area for elements 1 and 2 in m^2
E3 = 200e9; % Young’s modulus for element 3 in Pa
A3 = 0.000006; % Cross-sectional area for element 3 in m^2
L = 0.5; % Length of each element in m
F = 5000; % Applied force at node 2 in N
disp(‘E1(Pa) = ‘),disp(E1);
disp(‘E3(Pa) = ‘),disp(E3);
disp(‘A1(m^2) = ‘),disp(A1);
disp(‘A3(m^2) = ‘),disp(A3);
disp(‘L(m) = ‘),disp(L);
disp(‘θ(°) = 5 ‘);
C=cos(5*pi/180);
S=sin(5*pi/180);
disp(‘ Local Stiffness Matrixs :’)
k1=E1*A1*[C*C C*S -C*C -C*S
C*S S*S -C*S -S*S
-C*C -C*S C*C C*S
-C*S -S*S C*S S*S]/0.25
k2=E1*A1*[0 0 0 0
0 1 0 -1
0 0 0 0
0 -1 0 1]/0.0218
k3=E3*A3*[C*C C*S -C*C -C*S
C*S S*S -C*S -S*S
-C*C -C*S C*C C*S
-C*S -S*S C*S S*S]/0.25
disp(‘ Global Stiffness Matrix :’)
K=[k1(1,1) k1(1,2) k1(1,3) k1(1,4) 0 0 0 0
k1(2,1) k1(2,2) k1(2,3) k1(2,4) 0 0 0 0
k1(3,1) k1(3,2) k1(3,3)+k2(1,1) k1(3,4)+k2(1,2) k2(1,3) k2(1,4) 0 0
k1(4,1) k1(4,2) k1(4,3)+k2(2,1) k1(4,4)+k2(2,2) k2(2,3) k2(2,4) 0 0
0 0 k2(3,1) k2(3,2) k2(3,3)+k3(1,1) k2(3,4)+k3(1,2) k3(1,3) k3(1,4)
0 0 k2(4,1) k2(4,2) k2(4,3)+k3(2,1) k2(4,4)+k3(2,2) k3(2,3) k3(2,4)
0 0 0 0 k3(3,1) k3(3,2) k3(3,3) k3(3,4)
0 0 0 0 k3(4,1) k3(4,2) k3(4,3) k3(4,4) ]
disp(‘ Transformation matrix :’)
T= [ C S 0 0 0 0 0 0
-S C 0 0 0 0 0 0
0 0 C S 0 0 0 0
0 0 -S C 0 0 0 0
0 0 0 0 C S 0 0
0 0 0 0 -S C 0 0
0 0 0 0 0 0 C S
0 0 0 0 0 0 -S C ]
disp(‘ Inverse of Transformation matrix :’)
TT= inv(T)
disp(‘ [T][K][TT] :’)
M=T*K*TT
disp(‘ Applying the boundary condition’)
U1=0
V1=0
U3=0
V3=0
V4=0
disp(‘Force at node 2 :’);
Fx2=[0,5000];
IK=[M(3,3), M(5,3); M(5,3), M(5,5)];
sol=Fx2/IK;
disp (5000);
disp(‘Displacement of node 2 and 3 at local coordinate(m) (0°):’);
U2=sol(1);
U3=sol(2);
t=[C,S;-S,C];
u=[U2; U3];
disp(‘d2x = ‘),disp(U2);
disp(‘d3x = ‘),disp(U3);
disp(‘Displacement of node 2 and 3 at Global coordinate(m) (5°):’);
t=[C,S;-S,C];
u=[U2; U3];
solu=tu;
u2=solu(1);
u3=solu(2);
disp(‘d2x = ‘),disp(u2);
disp(‘d3x = ‘),disp(u3);
I don’t know if the Global Stiffness Matrix arrangement is correct can someone help me? % Declaring all the variables
E1 = 200e9; % Young’s modulus for elements 1 and 2 in Pa
A1 = 0.000006; % Cross-sectional area for elements 1 and 2 in m^2
E3 = 200e9; % Young’s modulus for element 3 in Pa
A3 = 0.000006; % Cross-sectional area for element 3 in m^2
L = 0.5; % Length of each element in m
F = 5000; % Applied force at node 2 in N
disp(‘E1(Pa) = ‘),disp(E1);
disp(‘E3(Pa) = ‘),disp(E3);
disp(‘A1(m^2) = ‘),disp(A1);
disp(‘A3(m^2) = ‘),disp(A3);
disp(‘L(m) = ‘),disp(L);
disp(‘θ(°) = 5 ‘);
C=cos(5*pi/180);
S=sin(5*pi/180);
disp(‘ Local Stiffness Matrixs :’)
k1=E1*A1*[C*C C*S -C*C -C*S
C*S S*S -C*S -S*S
-C*C -C*S C*C C*S
-C*S -S*S C*S S*S]/0.25
k2=E1*A1*[0 0 0 0
0 1 0 -1
0 0 0 0
0 -1 0 1]/0.0218
k3=E3*A3*[C*C C*S -C*C -C*S
C*S S*S -C*S -S*S
-C*C -C*S C*C C*S
-C*S -S*S C*S S*S]/0.25
disp(‘ Global Stiffness Matrix :’)
K=[k1(1,1) k1(1,2) k1(1,3) k1(1,4) 0 0 0 0
k1(2,1) k1(2,2) k1(2,3) k1(2,4) 0 0 0 0
k1(3,1) k1(3,2) k1(3,3)+k2(1,1) k1(3,4)+k2(1,2) k2(1,3) k2(1,4) 0 0
k1(4,1) k1(4,2) k1(4,3)+k2(2,1) k1(4,4)+k2(2,2) k2(2,3) k2(2,4) 0 0
0 0 k2(3,1) k2(3,2) k2(3,3)+k3(1,1) k2(3,4)+k3(1,2) k3(1,3) k3(1,4)
0 0 k2(4,1) k2(4,2) k2(4,3)+k3(2,1) k2(4,4)+k3(2,2) k3(2,3) k3(2,4)
0 0 0 0 k3(3,1) k3(3,2) k3(3,3) k3(3,4)
0 0 0 0 k3(4,1) k3(4,2) k3(4,3) k3(4,4) ]
disp(‘ Transformation matrix :’)
T= [ C S 0 0 0 0 0 0
-S C 0 0 0 0 0 0
0 0 C S 0 0 0 0
0 0 -S C 0 0 0 0
0 0 0 0 C S 0 0
0 0 0 0 -S C 0 0
0 0 0 0 0 0 C S
0 0 0 0 0 0 -S C ]
disp(‘ Inverse of Transformation matrix :’)
TT= inv(T)
disp(‘ [T][K][TT] :’)
M=T*K*TT
disp(‘ Applying the boundary condition’)
U1=0
V1=0
U3=0
V3=0
V4=0
disp(‘Force at node 2 :’);
Fx2=[0,5000];
IK=[M(3,3), M(5,3); M(5,3), M(5,5)];
sol=Fx2/IK;
disp (5000);
disp(‘Displacement of node 2 and 3 at local coordinate(m) (0°):’);
U2=sol(1);
U3=sol(2);
t=[C,S;-S,C];
u=[U2; U3];
disp(‘d2x = ‘),disp(U2);
disp(‘d3x = ‘),disp(U3);
disp(‘Displacement of node 2 and 3 at Global coordinate(m) (5°):’);
t=[C,S;-S,C];
u=[U2; U3];
solu=tu;
u2=solu(1);
u3=solu(2);
disp(‘d2x = ‘),disp(u2);
disp(‘d3x = ‘),disp(u3);
I don’t know if the Global Stiffness Matrix arrangement is correct can someone help me? stiffness matrix MATLAB Answers — New Questions
Please can anyone help me to correct this code !!!!
% Définition des matrices de système
A = [0 0 1 -1 0 0; 0 1 0 0 1 0; 0 0 0 0 0 1];
B1 = [0 1 0; 0 0 0; 0 0 0];
B2 = [0 0 1; 0 0 0; 0 0 0];
C1 = [1 0 0; 0 1 0; 0 0 1];
C2 = [0 0 1; 0 0 0; 0 0 0];
D1 = [1 0 0; 0 1 0; 0 0 1];
% Définition des fonctions d’appartenance floues
f1 = @(t) (t – 0.5)^2;
f2 = @(t) t;
% Définition des règles floues
rule1 = ‘IF f1(t) is "Long" and f2(t) is "Extended", THEN _x(t) = A1*x(t) + B1*x(t) + B2*sat(u(t) – s(t))’;
rule2 = ‘IF f1(t) is "Short" and f2(t) is "Compressed", THEN _x(t) = A2*x(t) + B1*x(t) + B2*sat(u(t) – s(t))’;
rule3 = ‘IF f1(t) is "Long" and f2(t) is "Compressed", THEN _x(t) = A3*x(t) + B1*x(t) + B2*sat(u(t) – s(t))’;
rule4 = ‘IF f1(t) is "Short" and f2(t) is "Extended", THEN _x(t) = A4*x(t) + B1*x(t) + B2*sat(u(t) – s(t))’;
% Définition des gains de contrôle
P1 = [1 0 0; 0 1 0; 0 0 1];
P2 = [0 1 0; 0 0 1; 0 0 0];
P3 = [0 0 1; 0 0 0; 0 0 1];
P4 = [1 0 0; 0 1 0; 0 0 1];
% Définition de la fonction de contrôle
control = @(t, x) P1*x + P2*x + P3*sat(u(t) – s(t));
% Simulation
t = 0:0.01:10;
x0 = [1; 0; 0];
u = zeros(size(t));
s = 0.5*t;
z1 = zeros(size(t));
z2 = zeros(size(t));
for i = 1:length(t)
x = x0;
for j = 1:length(t)
if f1(t(j)) > 0.5 && f2(t(j)) > 0.5
x = A1*x + B1*x + B2*sat(u(j) – s(j));
elseif f1(t(j)) < 0.5 && f2(t(j)) < 0.5
x = A2*x + B1*x + B2*sat(u(j) – s(j));
elseif f1(t(j)) > 0.5 && f2(t(j)) < 0.5
x = A3*x + B1*x + B2*sat(u(j) – s(j));
else
x = A4*x + B1*x + B2*sat(u(j) – s(j));
end
end
z1(i) = C1*x;
z2(i) = C2*x;
u(i) = control(t(i), x);
end
% Affichage des résultats
plot(t, z1, t, z2);
xlabel(‘Time (s)’);
ylabel(‘State’);
title(‘Active Suspension System’);
And this the error message ‘ paper1
Error using *
Incorrect dimensions for matrix multiplication. Check that the number of columns in the first matrix matches the number of
rows in the second matrix. To perform elementwise multiplication, use ‘.*’.
Error in paper1 (line 42)
x = A2*x + B1*x + B2*sat(u(j) – s(j)); ‘% Définition des matrices de système
A = [0 0 1 -1 0 0; 0 1 0 0 1 0; 0 0 0 0 0 1];
B1 = [0 1 0; 0 0 0; 0 0 0];
B2 = [0 0 1; 0 0 0; 0 0 0];
C1 = [1 0 0; 0 1 0; 0 0 1];
C2 = [0 0 1; 0 0 0; 0 0 0];
D1 = [1 0 0; 0 1 0; 0 0 1];
% Définition des fonctions d’appartenance floues
f1 = @(t) (t – 0.5)^2;
f2 = @(t) t;
% Définition des règles floues
rule1 = ‘IF f1(t) is "Long" and f2(t) is "Extended", THEN _x(t) = A1*x(t) + B1*x(t) + B2*sat(u(t) – s(t))’;
rule2 = ‘IF f1(t) is "Short" and f2(t) is "Compressed", THEN _x(t) = A2*x(t) + B1*x(t) + B2*sat(u(t) – s(t))’;
rule3 = ‘IF f1(t) is "Long" and f2(t) is "Compressed", THEN _x(t) = A3*x(t) + B1*x(t) + B2*sat(u(t) – s(t))’;
rule4 = ‘IF f1(t) is "Short" and f2(t) is "Extended", THEN _x(t) = A4*x(t) + B1*x(t) + B2*sat(u(t) – s(t))’;
% Définition des gains de contrôle
P1 = [1 0 0; 0 1 0; 0 0 1];
P2 = [0 1 0; 0 0 1; 0 0 0];
P3 = [0 0 1; 0 0 0; 0 0 1];
P4 = [1 0 0; 0 1 0; 0 0 1];
% Définition de la fonction de contrôle
control = @(t, x) P1*x + P2*x + P3*sat(u(t) – s(t));
% Simulation
t = 0:0.01:10;
x0 = [1; 0; 0];
u = zeros(size(t));
s = 0.5*t;
z1 = zeros(size(t));
z2 = zeros(size(t));
for i = 1:length(t)
x = x0;
for j = 1:length(t)
if f1(t(j)) > 0.5 && f2(t(j)) > 0.5
x = A1*x + B1*x + B2*sat(u(j) – s(j));
elseif f1(t(j)) < 0.5 && f2(t(j)) < 0.5
x = A2*x + B1*x + B2*sat(u(j) – s(j));
elseif f1(t(j)) > 0.5 && f2(t(j)) < 0.5
x = A3*x + B1*x + B2*sat(u(j) – s(j));
else
x = A4*x + B1*x + B2*sat(u(j) – s(j));
end
end
z1(i) = C1*x;
z2(i) = C2*x;
u(i) = control(t(i), x);
end
% Affichage des résultats
plot(t, z1, t, z2);
xlabel(‘Time (s)’);
ylabel(‘State’);
title(‘Active Suspension System’);
And this the error message ‘ paper1
Error using *
Incorrect dimensions for matrix multiplication. Check that the number of columns in the first matrix matches the number of
rows in the second matrix. To perform elementwise multiplication, use ‘.*’.
Error in paper1 (line 42)
x = A2*x + B1*x + B2*sat(u(j) – s(j)); ‘ % Définition des matrices de système
A = [0 0 1 -1 0 0; 0 1 0 0 1 0; 0 0 0 0 0 1];
B1 = [0 1 0; 0 0 0; 0 0 0];
B2 = [0 0 1; 0 0 0; 0 0 0];
C1 = [1 0 0; 0 1 0; 0 0 1];
C2 = [0 0 1; 0 0 0; 0 0 0];
D1 = [1 0 0; 0 1 0; 0 0 1];
% Définition des fonctions d’appartenance floues
f1 = @(t) (t – 0.5)^2;
f2 = @(t) t;
% Définition des règles floues
rule1 = ‘IF f1(t) is "Long" and f2(t) is "Extended", THEN _x(t) = A1*x(t) + B1*x(t) + B2*sat(u(t) – s(t))’;
rule2 = ‘IF f1(t) is "Short" and f2(t) is "Compressed", THEN _x(t) = A2*x(t) + B1*x(t) + B2*sat(u(t) – s(t))’;
rule3 = ‘IF f1(t) is "Long" and f2(t) is "Compressed", THEN _x(t) = A3*x(t) + B1*x(t) + B2*sat(u(t) – s(t))’;
rule4 = ‘IF f1(t) is "Short" and f2(t) is "Extended", THEN _x(t) = A4*x(t) + B1*x(t) + B2*sat(u(t) – s(t))’;
% Définition des gains de contrôle
P1 = [1 0 0; 0 1 0; 0 0 1];
P2 = [0 1 0; 0 0 1; 0 0 0];
P3 = [0 0 1; 0 0 0; 0 0 1];
P4 = [1 0 0; 0 1 0; 0 0 1];
% Définition de la fonction de contrôle
control = @(t, x) P1*x + P2*x + P3*sat(u(t) – s(t));
% Simulation
t = 0:0.01:10;
x0 = [1; 0; 0];
u = zeros(size(t));
s = 0.5*t;
z1 = zeros(size(t));
z2 = zeros(size(t));
for i = 1:length(t)
x = x0;
for j = 1:length(t)
if f1(t(j)) > 0.5 && f2(t(j)) > 0.5
x = A1*x + B1*x + B2*sat(u(j) – s(j));
elseif f1(t(j)) < 0.5 && f2(t(j)) < 0.5
x = A2*x + B1*x + B2*sat(u(j) – s(j));
elseif f1(t(j)) > 0.5 && f2(t(j)) < 0.5
x = A3*x + B1*x + B2*sat(u(j) – s(j));
else
x = A4*x + B1*x + B2*sat(u(j) – s(j));
end
end
z1(i) = C1*x;
z2(i) = C2*x;
u(i) = control(t(i), x);
end
% Affichage des résultats
plot(t, z1, t, z2);
xlabel(‘Time (s)’);
ylabel(‘State’);
title(‘Active Suspension System’);
And this the error message ‘ paper1
Error using *
Incorrect dimensions for matrix multiplication. Check that the number of columns in the first matrix matches the number of
rows in the second matrix. To perform elementwise multiplication, use ‘.*’.
Error in paper1 (line 42)
x = A2*x + B1*x + B2*sat(u(j) – s(j)); ‘ matlab, matrix, optimization MATLAB Answers — New Questions
a cylindirical annular fuel element consist of two regions, solve the temperature distribution by using finite difference method??
hi everyone i have a problem I dont know how to write code please help me if i do that i will graduate. I should use Finite Difference Method.
Consider a cylindirical annular fuel element consist of two regions, one is UO2 (fuel) other is Zircaloy-4 (cladding). Assume that heat is generated only in fuel region. There is internal and external cooling for the fuel element. You are asked to solve the temperature distribution from Rv to outer surface of the cladding by using finite difference method.
Fuel radius Rfo=0.705 cm
internal cavity radius: Rv=0.495 cm
thickness of clad, tc= 0.0654 cm
cald radius :Rclad=Rfo+tc
Water Bulk Temperature: Tb=316 C
Heat transfer coefficient of water: h_w=55 W/cm^2 K
1/r d/dr rk dT/dr= -q′′′
Volumetric Heat Generation Rate: q(r)’’’= q_1^”’+(q_2^”’ r^2)/(Rfo^2 )
q_1^”’=0.75q_0^”’
q_2^”’=0.50q_0^”’
Case1q_0’’’: 388 W/cm3
Case2q_0’’’: 388*1.5 W/cm3
Case3q_0’’’: 388*2 W/cm3
A= 4.52
B=2.46*〖10〗^(-2)
E= 3.5*〖10〗^7
F=16361
Conductivity are temperature dependent: kf=1/(A+BT)+(E/T^2 )exp(-F/T)
Conductivity of Clad: kc= 0.113+2.25*(10^-5*)T + (0.725*10^8)T^2 𝑊/c𝑚𝐾
Boundary condition: -kf dT/dr= h_w (T(Rv)-Tb) @r=Rv
Boundary condition: -kc dT/dr= h_w (T(Rclad)-Tb) @r=Rclad
1)Plot the conductivity distributions for the three cases in a single graph.
2) Plot the Temperature distributions for the three cases in a single graph.
I had to create a matrix for k, start from a T value and iterate until I found the appropriate value. But I can’t do it, it’s so hard for me, help me ??hi everyone i have a problem I dont know how to write code please help me if i do that i will graduate. I should use Finite Difference Method.
Consider a cylindirical annular fuel element consist of two regions, one is UO2 (fuel) other is Zircaloy-4 (cladding). Assume that heat is generated only in fuel region. There is internal and external cooling for the fuel element. You are asked to solve the temperature distribution from Rv to outer surface of the cladding by using finite difference method.
Fuel radius Rfo=0.705 cm
internal cavity radius: Rv=0.495 cm
thickness of clad, tc= 0.0654 cm
cald radius :Rclad=Rfo+tc
Water Bulk Temperature: Tb=316 C
Heat transfer coefficient of water: h_w=55 W/cm^2 K
1/r d/dr rk dT/dr= -q′′′
Volumetric Heat Generation Rate: q(r)’’’= q_1^”’+(q_2^”’ r^2)/(Rfo^2 )
q_1^”’=0.75q_0^”’
q_2^”’=0.50q_0^”’
Case1q_0’’’: 388 W/cm3
Case2q_0’’’: 388*1.5 W/cm3
Case3q_0’’’: 388*2 W/cm3
A= 4.52
B=2.46*〖10〗^(-2)
E= 3.5*〖10〗^7
F=16361
Conductivity are temperature dependent: kf=1/(A+BT)+(E/T^2 )exp(-F/T)
Conductivity of Clad: kc= 0.113+2.25*(10^-5*)T + (0.725*10^8)T^2 𝑊/c𝑚𝐾
Boundary condition: -kf dT/dr= h_w (T(Rv)-Tb) @r=Rv
Boundary condition: -kc dT/dr= h_w (T(Rclad)-Tb) @r=Rclad
1)Plot the conductivity distributions for the three cases in a single graph.
2) Plot the Temperature distributions for the three cases in a single graph.
I had to create a matrix for k, start from a T value and iterate until I found the appropriate value. But I can’t do it, it’s so hard for me, help me ?? hi everyone i have a problem I dont know how to write code please help me if i do that i will graduate. I should use Finite Difference Method.
Consider a cylindirical annular fuel element consist of two regions, one is UO2 (fuel) other is Zircaloy-4 (cladding). Assume that heat is generated only in fuel region. There is internal and external cooling for the fuel element. You are asked to solve the temperature distribution from Rv to outer surface of the cladding by using finite difference method.
Fuel radius Rfo=0.705 cm
internal cavity radius: Rv=0.495 cm
thickness of clad, tc= 0.0654 cm
cald radius :Rclad=Rfo+tc
Water Bulk Temperature: Tb=316 C
Heat transfer coefficient of water: h_w=55 W/cm^2 K
1/r d/dr rk dT/dr= -q′′′
Volumetric Heat Generation Rate: q(r)’’’= q_1^”’+(q_2^”’ r^2)/(Rfo^2 )
q_1^”’=0.75q_0^”’
q_2^”’=0.50q_0^”’
Case1q_0’’’: 388 W/cm3
Case2q_0’’’: 388*1.5 W/cm3
Case3q_0’’’: 388*2 W/cm3
A= 4.52
B=2.46*〖10〗^(-2)
E= 3.5*〖10〗^7
F=16361
Conductivity are temperature dependent: kf=1/(A+BT)+(E/T^2 )exp(-F/T)
Conductivity of Clad: kc= 0.113+2.25*(10^-5*)T + (0.725*10^8)T^2 𝑊/c𝑚𝐾
Boundary condition: -kf dT/dr= h_w (T(Rv)-Tb) @r=Rv
Boundary condition: -kc dT/dr= h_w (T(Rclad)-Tb) @r=Rclad
1)Plot the conductivity distributions for the three cases in a single graph.
2) Plot the Temperature distributions for the three cases in a single graph.
I had to create a matrix for k, start from a T value and iterate until I found the appropriate value. But I can’t do it, it’s so hard for me, help me ?? finite difference method, temperature distribution, annular pellet, conductivity, cladding, heat generation, homework, temperature, external cooling, uo2 (fuel), matlab, plot, matrix, graph, function, differential equations, matrices, iteration MATLAB Answers — New Questions
Connecting Bluetooth Module and Arduino without using arduinosetup command
I’m connecting IMU9250 sensor and Bluetooth module HC-05 to my arduino and I want to get the data of the IMU using the Bluetooth module instead of using the Arduino cable.
I’m trying to use arduinosetup command but it fails at last step while testing the connection between Arduino and Bluetooth module.
I’ve tested the connection of the Bluetooth module only usning blutooth() function and it connects fine!
My question is: Is there a way to establish the connection between MATLAB and the HC-05 Bluetooth module connected to my Arduino Uno without using arduinosetup. Just by using objects and functions step by step manually.I’m connecting IMU9250 sensor and Bluetooth module HC-05 to my arduino and I want to get the data of the IMU using the Bluetooth module instead of using the Arduino cable.
I’m trying to use arduinosetup command but it fails at last step while testing the connection between Arduino and Bluetooth module.
I’ve tested the connection of the Bluetooth module only usning blutooth() function and it connects fine!
My question is: Is there a way to establish the connection between MATLAB and the HC-05 Bluetooth module connected to my Arduino Uno without using arduinosetup. Just by using objects and functions step by step manually. I’m connecting IMU9250 sensor and Bluetooth module HC-05 to my arduino and I want to get the data of the IMU using the Bluetooth module instead of using the Arduino cable.
I’m trying to use arduinosetup command but it fails at last step while testing the connection between Arduino and Bluetooth module.
I’ve tested the connection of the Bluetooth module only usning blutooth() function and it connects fine!
My question is: Is there a way to establish the connection between MATLAB and the HC-05 Bluetooth module connected to my Arduino Uno without using arduinosetup. Just by using objects and functions step by step manually. arduino, bluetooth, hc-05, imu, robotics, serial MATLAB Answers — New Questions
Why MATLAB crashed while running PV based boost converter simulink model
MATLAB simulink crashed while running model based on PV array system with boost converter. When I changed irradiation of PV array it crash simulink model and also closed MATLAB. Initially i used stair function then tried with manually constant block during running model.MATLAB simulink crashed while running model based on PV array system with boost converter. When I changed irradiation of PV array it crash simulink model and also closed MATLAB. Initially i used stair function then tried with manually constant block during running model. MATLAB simulink crashed while running model based on PV array system with boost converter. When I changed irradiation of PV array it crash simulink model and also closed MATLAB. Initially i used stair function then tried with manually constant block during running model. pv array, boost converter, simulink, simscape electrical MATLAB Answers — New Questions
Why are the labels rotated?
I am working on code in which the program must detect whether the spring is inserted correctly in the handle or not. I labeled all the pictures. when I run the code, I see that the label is rotated and not like the original picture. I also noticed that it doesn’t pull the Spring_incorrect label into the image I’m also interested in the classNames part. if I write something completely different inside, it still pulls the label.
PS: how can i see all the pictures? it jsut pulls the first one.
dataDir = fullfile(‘C:UsersandraDesktopslike vrc’);
imDir = fullfile(dataDir,’nepravilno’);
pxDir = fullfile(‘C:UsersandraDesktoptestLabelingProjectGroundTruthProjectPixelLabelData’);
imds = imageDatastore(imDir);
classNames = ["Correct" "Spring" "incorrect" "Spring_incorrect"];
pixelLabelID = [1 2 3 4];
pxds = pixelLabelDatastore(pxDir,classNames,pixelLabelID);
I = read(imds);
C = read(pxds);
categories(C{1})
B = labeloverlay(I,C{1});
figure
imshow(B)I am working on code in which the program must detect whether the spring is inserted correctly in the handle or not. I labeled all the pictures. when I run the code, I see that the label is rotated and not like the original picture. I also noticed that it doesn’t pull the Spring_incorrect label into the image I’m also interested in the classNames part. if I write something completely different inside, it still pulls the label.
PS: how can i see all the pictures? it jsut pulls the first one.
dataDir = fullfile(‘C:UsersandraDesktopslike vrc’);
imDir = fullfile(dataDir,’nepravilno’);
pxDir = fullfile(‘C:UsersandraDesktoptestLabelingProjectGroundTruthProjectPixelLabelData’);
imds = imageDatastore(imDir);
classNames = ["Correct" "Spring" "incorrect" "Spring_incorrect"];
pixelLabelID = [1 2 3 4];
pxds = pixelLabelDatastore(pxDir,classNames,pixelLabelID);
I = read(imds);
C = read(pxds);
categories(C{1})
B = labeloverlay(I,C{1});
figure
imshow(B) I am working on code in which the program must detect whether the spring is inserted correctly in the handle or not. I labeled all the pictures. when I run the code, I see that the label is rotated and not like the original picture. I also noticed that it doesn’t pull the Spring_incorrect label into the image I’m also interested in the classNames part. if I write something completely different inside, it still pulls the label.
PS: how can i see all the pictures? it jsut pulls the first one.
dataDir = fullfile(‘C:UsersandraDesktopslike vrc’);
imDir = fullfile(dataDir,’nepravilno’);
pxDir = fullfile(‘C:UsersandraDesktoptestLabelingProjectGroundTruthProjectPixelLabelData’);
imds = imageDatastore(imDir);
classNames = ["Correct" "Spring" "incorrect" "Spring_incorrect"];
pixelLabelID = [1 2 3 4];
pxds = pixelLabelDatastore(pxDir,classNames,pixelLabelID);
I = read(imds);
C = read(pxds);
categories(C{1})
B = labeloverlay(I,C{1});
figure
imshow(B) image processing, matlab MATLAB Answers — New Questions
Open a 2nd Matlab session and read model-file open in the 1st session (programatically)
I have a unique situation where I need to programatically open a 2nd session of Matlab and run a script. That is easy enough. However, I need this script to check for some info in the 1st session. For example, return the current path in the first session, return a list of all the Simulink models open in the first session, maybe even rename some signals in one of the models in session 1. This I have no idea how to do.
I’ve read a bit about UDP, doesn’t look like that would work for my use-case. Any suggestions welcome. Thanks.
JI have a unique situation where I need to programatically open a 2nd session of Matlab and run a script. That is easy enough. However, I need this script to check for some info in the 1st session. For example, return the current path in the first session, return a list of all the Simulink models open in the first session, maybe even rename some signals in one of the models in session 1. This I have no idea how to do.
I’ve read a bit about UDP, doesn’t look like that would work for my use-case. Any suggestions welcome. Thanks.
J I have a unique situation where I need to programatically open a 2nd session of Matlab and run a script. That is easy enough. However, I need this script to check for some info in the 1st session. For example, return the current path in the first session, return a list of all the Simulink models open in the first session, maybe even rename some signals in one of the models in session 1. This I have no idea how to do.
I’ve read a bit about UDP, doesn’t look like that would work for my use-case. Any suggestions welcome. Thanks.
J multiple sessions, matlab, models, communicate MATLAB Answers — New Questions
