The model:

<</matlabcentral/answers/uploaded_files/27438/3.PNG>>

The simMechanics system is as follow:

<</matlabcentral/answers/uploaded_files/27436/1.PNG>>

simin is input from workspace and is the motion of Revolute Joint as the time goes on.

<</matlabcentral/answers/uploaded_files/27437/2.PNG>>

But it shows some kind of error:
In the dynamically coupled component containing Revolute Joint Revolute_Joint, there are fewer joint primitive degrees of freedom with automatically computed force or torque (0) than with motion from inputs (1). The prescribed motion trajectories in this component may not be achievable. Solve this problem by reducing the number of joint primitives with motion from inputs or increasing the number of joint primitives with automatically computed force or torque. Resolve this issue in order to simulate the model.

Does anybody knows what’s wrong and what should I do to solve the problem.The model:

<</matlabcentral/answers/uploaded_files/27438/3.PNG>>

The simMechanics system is as follow:

<</matlabcentral/answers/uploaded_files/27436/1.PNG>>

simin is input from workspace and is the motion of Revolute Joint as the time goes on.

<</matlabcentral/answers/uploaded_files/27437/2.PNG>>

But it shows some kind of error:
In the dynamically coupled component containing Revolute Joint Revolute_Joint, there are fewer joint primitive degrees of freedom with automatically computed force or torque (0) than with motion from inputs (1). The prescribed motion trajectories in this component may not be achievable. Solve this problem by reducing the number of joint primitives with motion from inputs or increasing the number of joint primitives with automatically computed force or torque. Resolve this issue in order to simulate the model.

Does anybody knows what’s wrong and what should I do to solve the problem. The model:

<</matlabcentral/answers/uploaded_files/27438/3.PNG>>

The simMechanics system is as follow:

<</matlabcentral/answers/uploaded_files/27436/1.PNG>>

simin is input from workspace and is the motion of Revolute Joint as the time goes on.

<</matlabcentral/answers/uploaded_files/27437/2.PNG>>

But it shows some kind of error:
In the dynamically coupled component containing Revolute Joint Revolute_Joint, there are fewer joint primitive degrees of freedom with automatically computed force or torque (0) than with motion from inputs (1). The prescribed motion trajectories in this component may not be achievable. Solve this problem by reducing the number of joint primitives with motion from inputs or increasing the number of joint primitives with automatically computed force or torque. Resolve this issue in order to simulate the model.

Does anybody knows what’s wrong and what should I do to solve the problem. simulink, simmechanics MATLAB Answers — New Questions

​

I encountered a nonlinear fitting problem, and the fitting formula is shown in Equation (1), which includes two infinite integrals (in practice, the integration range can be set from 0.01E-6 to 200E-6).

In these formulas, except for x and y being vectors, all other variables are scalars, and Rmedian and sigma are the parameters to be fitted.

I found a related post and tried to write the code based on it, but it keeps reporting errors. The error message seems to indicate that the vector dimensions are inconsistent, preventing the operation from proceeding. However, these functions are all calculations for individual scalars.
Error using /
Matrix dimensions must agree.

Error in dsdmain>@(r)1/(2*r*sigma*sqrt(2*pi))*exp(-(log(2*r)-log(2*Rmean)^2)/(2*sigma^2)) (line 13)
gauss = @(r) 1/(2*r*sigma*sqrt(2*pi))* exp( -(log(2*r)-log(2*Rmean)^2)/(2*sigma^2) );
My question is: Can I refer to the content of this post to solve my problem? If yes, what does this error message mean? If not, how should I resolve my problem? (Note: The range of Rmedian is 1E-6 to 5E-6)

After modifying the code according to Walter Roberson and Torsten’s suggestion, the program no longer throws errors. But no matter how I set the initial values on my end, it always prompts:
Initial point is a local minimum.
Optimization completed because the size of the gradient at the initial point
is less than the default value of the optimality tolerance.
<stopping criteria details>
theta = initial point
1.0e-05 *
0.2000
0.1000
Optimization completed: The final point is the initial point.
The first-order optimality measure, 0.000000e+00, is less than
options.OptimalityTolerance = 1.000000e-06.
Optimization Metric Options
relative first-order optimality = 0.00e+00 OptimalityTolerance = 1e-06 (default)

I have checked all the formulas and the units of the variables, and I didn’t find any problems.

————————————– Beblow is my code for issue reproduction ———————————————————-
function testmain()
clc
function Kvec = model(param,xdata)
% Vector of Kd for every xdata:
Kvec = zeros(size(xdata));
Rmean = param(1);
Rstd = param(2);

for i = 1:length(xdata)
model = @(r) unified(xdata(i),r,Rmean,Rstd,delta,Delta,D,lambda);
Kvec(i) = integral(model,0.01E-6,200E-6);
end
end
function s = unified(g,R,Rmean,Rstd,delta,Delta,D,lambda)
%unified Unified fitting model for DSD
% exponentional combined

factor = 1./(2*R*Rstd*sqrt(2*pi)) *2 ; % int(P(r)) = 0.5,1/0.5=2
p1 = -(log(2*R)-log(2*Rmean)).^2/(2*Rstd^2);
c = -2*2.675E8^2*g.^2/D;

tmp = 0;
for il = 1:length(lambda)
a2 = (lambda(il)./R).^2;
an4 = (R/lambda(il)).^4;
Psi = 2+exp(-a2*D*(Delta-delta))-2*exp(-a2*D*delta)-2*exp(-a2*D*Delta)+exp(-a2*D*(Delta+delta));
tmp = tmp+an4./(a2.*R.*R-2).*(2*delta-Psi./(a2*D));
end
p2 = c*tmp;

s = factor.*exp(p1+p2);
end
Delta = 0.075;
delta = 0.002;
D = 0.098E-9;
lambda = [2.0816 5.9404 9.2058 12.4044 15.5792 18.7426 21.8997 25.0528 28.2034 31.3521];

g = [ 0.300616, 0.53884, 0.771392, 1.009616, 1.24784, 1.480392, 1.718616, 1.95684, 2.189392, 2.427616, 2.66584, 2.898392 ];
xdata = g;
ydata = [100, 91.16805426, 80.52955192, 67.97705378, 55.1009735,41.87307917, 30.39638776, 21.13515607, 13.7125649, 8.33083767, 5.146756077, 2.79768609];
ydata = ydata/ydata(1); % normalize

% Inital guess for parameters:
Rmean0 = 2E-6;
Rstd0 = 1E-6;
p0 = [Rmean0;Rstd0];
% lsqcurvefit is in the optimization toolbox.
% fit, from the curve fitting toolbox may be an alternative
theta = lsqcurvefit(@model,p0,xdata,ydata,[0.01E-6;0.1E-6],[10E-6,2E-6])
endI encountered a nonlinear fitting problem, and the fitting formula is shown in Equation (1), which includes two infinite integrals (in practice, the integration range can be set from 0.01E-6 to 200E-6).

In these formulas, except for x and y being vectors, all other variables are scalars, and Rmedian and sigma are the parameters to be fitted.

I found a related post and tried to write the code based on it, but it keeps reporting errors. The error message seems to indicate that the vector dimensions are inconsistent, preventing the operation from proceeding. However, these functions are all calculations for individual scalars.
Error using /
Matrix dimensions must agree.

Error in dsdmain>@(r)1/(2*r*sigma*sqrt(2*pi))*exp(-(log(2*r)-log(2*Rmean)^2)/(2*sigma^2)) (line 13)
gauss = @(r) 1/(2*r*sigma*sqrt(2*pi))* exp( -(log(2*r)-log(2*Rmean)^2)/(2*sigma^2) );
My question is: Can I refer to the content of this post to solve my problem? If yes, what does this error message mean? If not, how should I resolve my problem? (Note: The range of Rmedian is 1E-6 to 5E-6)

After modifying the code according to Walter Roberson and Torsten’s suggestion, the program no longer throws errors. But no matter how I set the initial values on my end, it always prompts:
Initial point is a local minimum.
Optimization completed because the size of the gradient at the initial point
is less than the default value of the optimality tolerance.
<stopping criteria details>
theta = initial point
1.0e-05 *
0.2000
0.1000
Optimization completed: The final point is the initial point.
The first-order optimality measure, 0.000000e+00, is less than
options.OptimalityTolerance = 1.000000e-06.
Optimization Metric Options
relative first-order optimality = 0.00e+00 OptimalityTolerance = 1e-06 (default)

I have checked all the formulas and the units of the variables, and I didn’t find any problems.

————————————– Beblow is my code for issue reproduction ———————————————————-
function testmain()
clc
function Kvec = model(param,xdata)
% Vector of Kd for every xdata:
Kvec = zeros(size(xdata));
Rmean = param(1);
Rstd = param(2);

for i = 1:length(xdata)
model = @(r) unified(xdata(i),r,Rmean,Rstd,delta,Delta,D,lambda);
Kvec(i) = integral(model,0.01E-6,200E-6);
end
end
function s = unified(g,R,Rmean,Rstd,delta,Delta,D,lambda)
%unified Unified fitting model for DSD
% exponentional combined

factor = 1./(2*R*Rstd*sqrt(2*pi)) *2 ; % int(P(r)) = 0.5,1/0.5=2
p1 = -(log(2*R)-log(2*Rmean)).^2/(2*Rstd^2);
c = -2*2.675E8^2*g.^2/D;

tmp = 0;
for il = 1:length(lambda)
a2 = (lambda(il)./R).^2;
an4 = (R/lambda(il)).^4;
Psi = 2+exp(-a2*D*(Delta-delta))-2*exp(-a2*D*delta)-2*exp(-a2*D*Delta)+exp(-a2*D*(Delta+delta));
tmp = tmp+an4./(a2.*R.*R-2).*(2*delta-Psi./(a2*D));
end
p2 = c*tmp;

s = factor.*exp(p1+p2);
end
Delta = 0.075;
delta = 0.002;
D = 0.098E-9;
lambda = [2.0816 5.9404 9.2058 12.4044 15.5792 18.7426 21.8997 25.0528 28.2034 31.3521];

g = [ 0.300616, 0.53884, 0.771392, 1.009616, 1.24784, 1.480392, 1.718616, 1.95684, 2.189392, 2.427616, 2.66584, 2.898392 ];
xdata = g;
ydata = [100, 91.16805426, 80.52955192, 67.97705378, 55.1009735,41.87307917, 30.39638776, 21.13515607, 13.7125649, 8.33083767, 5.146756077, 2.79768609];
ydata = ydata/ydata(1); % normalize

% Inital guess for parameters:
Rmean0 = 2E-6;
Rstd0 = 1E-6;
p0 = [Rmean0;Rstd0];
% lsqcurvefit is in the optimization toolbox.
% fit, from the curve fitting toolbox may be an alternative
theta = lsqcurvefit(@model,p0,xdata,ydata,[0.01E-6;0.1E-6],[10E-6,2E-6])
end I encountered a nonlinear fitting problem, and the fitting formula is shown in Equation (1), which includes two infinite integrals (in practice, the integration range can be set from 0.01E-6 to 200E-6).

In these formulas, except for x and y being vectors, all other variables are scalars, and Rmedian and sigma are the parameters to be fitted.

I found a related post and tried to write the code based on it, but it keeps reporting errors. The error message seems to indicate that the vector dimensions are inconsistent, preventing the operation from proceeding. However, these functions are all calculations for individual scalars.
Error using /
Matrix dimensions must agree.

Error in dsdmain>@(r)1/(2*r*sigma*sqrt(2*pi))*exp(-(log(2*r)-log(2*Rmean)^2)/(2*sigma^2)) (line 13)
gauss = @(r) 1/(2*r*sigma*sqrt(2*pi))* exp( -(log(2*r)-log(2*Rmean)^2)/(2*sigma^2) );
My question is: Can I refer to the content of this post to solve my problem? If yes, what does this error message mean? If not, how should I resolve my problem? (Note: The range of Rmedian is 1E-6 to 5E-6)

After modifying the code according to Walter Roberson and Torsten’s suggestion, the program no longer throws errors. But no matter how I set the initial values on my end, it always prompts:
Initial point is a local minimum.
Optimization completed because the size of the gradient at the initial point
is less than the default value of the optimality tolerance.
<stopping criteria details>
theta = initial point
1.0e-05 *
0.2000
0.1000
Optimization completed: The final point is the initial point.
The first-order optimality measure, 0.000000e+00, is less than
options.OptimalityTolerance = 1.000000e-06.
Optimization Metric Options
relative first-order optimality = 0.00e+00 OptimalityTolerance = 1e-06 (default)

I have checked all the formulas and the units of the variables, and I didn’t find any problems.

————————————– Beblow is my code for issue reproduction ———————————————————-
function testmain()
clc
function Kvec = model(param,xdata)
% Vector of Kd for every xdata:
Kvec = zeros(size(xdata));
Rmean = param(1);
Rstd = param(2);

for i = 1:length(xdata)
model = @(r) unified(xdata(i),r,Rmean,Rstd,delta,Delta,D,lambda);
Kvec(i) = integral(model,0.01E-6,200E-6);
end
end
function s = unified(g,R,Rmean,Rstd,delta,Delta,D,lambda)
%unified Unified fitting model for DSD
% exponentional combined

factor = 1./(2*R*Rstd*sqrt(2*pi)) *2 ; % int(P(r)) = 0.5,1/0.5=2
p1 = -(log(2*R)-log(2*Rmean)).^2/(2*Rstd^2);
c = -2*2.675E8^2*g.^2/D;

tmp = 0;
for il = 1:length(lambda)
a2 = (lambda(il)./R).^2;
an4 = (R/lambda(il)).^4;
Psi = 2+exp(-a2*D*(Delta-delta))-2*exp(-a2*D*delta)-2*exp(-a2*D*Delta)+exp(-a2*D*(Delta+delta));
tmp = tmp+an4./(a2.*R.*R-2).*(2*delta-Psi./(a2*D));
end
p2 = c*tmp;

s = factor.*exp(p1+p2);
end
Delta = 0.075;
delta = 0.002;
D = 0.098E-9;
lambda = [2.0816 5.9404 9.2058 12.4044 15.5792 18.7426 21.8997 25.0528 28.2034 31.3521];

g = [ 0.300616, 0.53884, 0.771392, 1.009616, 1.24784, 1.480392, 1.718616, 1.95684, 2.189392, 2.427616, 2.66584, 2.898392 ];
xdata = g;
ydata = [100, 91.16805426, 80.52955192, 67.97705378, 55.1009735,41.87307917, 30.39638776, 21.13515607, 13.7125649, 8.33083767, 5.146756077, 2.79768609];
ydata = ydata/ydata(1); % normalize

% Inital guess for parameters:
Rmean0 = 2E-6;
Rstd0 = 1E-6;
p0 = [Rmean0;Rstd0];
% lsqcurvefit is in the optimization toolbox.
% fit, from the curve fitting toolbox may be an alternative
theta = lsqcurvefit(@model,p0,xdata,ydata,[0.01E-6;0.1E-6],[10E-6,2E-6])
end curve fitting, integral MATLAB Answers — New Questions

​

Hello,
I have a question regarding the relationship between Simulink sample rate and HDL Coder target frequency.
Is it okay if the Simulink sample rate is equal to the target frequency specified in HDL Coder?
I’ve often heard that the hardware clock rate (target frequency) should generally be at least 2× the sample rate to ensure proper timing and pipelining. But in Simulink and HDL Coder, I’m unsure how strict this rule is.
Specifically:
If I set my Simulink sample rate to 120 MSPS and also specify 120 MHz as the HDL Coder target frequency, is this configuration valid?
Or should the target frequency always be greater than the sample rate, even in fully pipelined or parallel architectures?
I’d really appreciate any clarification on how these two values interact and whether matching them could lead to timing issues during synthesis or implementation.
Thanks in advance!Hello,
I have a question regarding the relationship between Simulink sample rate and HDL Coder target frequency.
Is it okay if the Simulink sample rate is equal to the target frequency specified in HDL Coder?
I’ve often heard that the hardware clock rate (target frequency) should generally be at least 2× the sample rate to ensure proper timing and pipelining. But in Simulink and HDL Coder, I’m unsure how strict this rule is.
Specifically:
If I set my Simulink sample rate to 120 MSPS and also specify 120 MHz as the HDL Coder target frequency, is this configuration valid?
Or should the target frequency always be greater than the sample rate, even in fully pipelined or parallel architectures?
I’d really appreciate any clarification on how these two values interact and whether matching them could lead to timing issues during synthesis or implementation.
Thanks in advance! Hello,
I have a question regarding the relationship between Simulink sample rate and HDL Coder target frequency.
Is it okay if the Simulink sample rate is equal to the target frequency specified in HDL Coder?
I’ve often heard that the hardware clock rate (target frequency) should generally be at least 2× the sample rate to ensure proper timing and pipelining. But in Simulink and HDL Coder, I’m unsure how strict this rule is.
Specifically:
If I set my Simulink sample rate to 120 MSPS and also specify 120 MHz as the HDL Coder target frequency, is this configuration valid?
Or should the target frequency always be greater than the sample rate, even in fully pipelined or parallel architectures?
I’d really appreciate any clarification on how these two values interact and whether matching them could lead to timing issues during synthesis or implementation.
Thanks in advance! simulink, sample rate, hdl coder, target frequency MATLAB Answers — New Questions

​

Hi everyone,
I’m trying to generate a uniform grid where the red lines intersect at a 45-degree angle to the horizontal lines in blue using the finite difference method in MATLAB.
However, the resulting grid is slightly off, as shown in the attached image. Intesection angle is uniform but higher than 45-degree.
I set computational domain as and a grid is generated on physical domain with coordinates .
As I want to generate a mesh described above, I aimed to solve the following PDE :

where we have (horizontal lines). The boundary conditions are .
The above equation comes down to the following:

Based on the above equation, I discretized to finite difference form (central difference for and forward differece for ).
Plus, I found the grid collapses when I set different angles (for example 90-degree (replace )).
Am I missing something in implementation?

I need your help.
Here is my code.
% Parameter settings
eta_max = 10;
xi_max = 10;
Delta_eta = 0.1;
Delta_xi = 0.1;
tol = 1e-6;
max_iterations = 100000;

% Grid size
eta_steps = floor(eta_max / Delta_eta) + 1;
xi_steps = floor(xi_max / Delta_xi) + 1;

% Initial conditions
x = zeros(eta_steps, xi_steps);
y = repmat((0:eta_steps-1)’ * Delta_eta, 1, xi_steps);

% Boundary conditions
x(1, 🙂 = linspace(0, xi_max, xi_steps);

% Finite difference method
converged = false;
iteration = 0;

while ~converged && iteration < max_iterations
iteration = iteration + 1;
max_error = 0;

for n = 1:eta_steps – 1
for m = 2:xi_steps – 1
x_xi = (x(n, m+1) – x(n, m-1)) / (2 * Delta_xi);
y_xi = (y(n, m+1) – y(n, m-1)) / (2 * Delta_xi);

new_x = x(n, m) + Delta_eta * ((pi/4) – y_xi)/(x_xi);
error_x = abs(new_x – x(n, m));
max_error = max([max_error, error_x]);

x(n + 1, m) = new_x;
end

x(n+1, xi_steps) = x(n+1, xi_steps-1) + Delta_xi;
x(n+1, 1) = x(n+1, 2) – Delta_xi;
end

if max_error < tol
converged = true;
end
end

% Plot the results
figure; hold on; axis equal; grid on;

for m = 1:xi_steps
plot(x(:, m), y(:, m), ‘r’);
end

for n = 1:eta_steps
plot(x(n, :), y(n, :), ‘b’);
end

title([‘Iterations until convergence: ‘, num2str(iteration)]);
xlabel(‘x’); ylabel(‘y’);Hi everyone,
I’m trying to generate a uniform grid where the red lines intersect at a 45-degree angle to the horizontal lines in blue using the finite difference method in MATLAB.
However, the resulting grid is slightly off, as shown in the attached image. Intesection angle is uniform but higher than 45-degree.
I set computational domain as and a grid is generated on physical domain with coordinates .
As I want to generate a mesh described above, I aimed to solve the following PDE :

where we have (horizontal lines). The boundary conditions are .
The above equation comes down to the following:

Based on the above equation, I discretized to finite difference form (central difference for and forward differece for ).
Plus, I found the grid collapses when I set different angles (for example 90-degree (replace )).
Am I missing something in implementation?

I need your help.
Here is my code.
% Parameter settings
eta_max = 10;
xi_max = 10;
Delta_eta = 0.1;
Delta_xi = 0.1;
tol = 1e-6;
max_iterations = 100000;

% Grid size
eta_steps = floor(eta_max / Delta_eta) + 1;
xi_steps = floor(xi_max / Delta_xi) + 1;

% Initial conditions
x = zeros(eta_steps, xi_steps);
y = repmat((0:eta_steps-1)’ * Delta_eta, 1, xi_steps);

% Boundary conditions
x(1, 🙂 = linspace(0, xi_max, xi_steps);

% Finite difference method
converged = false;
iteration = 0;

while ~converged && iteration < max_iterations
iteration = iteration + 1;
max_error = 0;

for n = 1:eta_steps – 1
for m = 2:xi_steps – 1
x_xi = (x(n, m+1) – x(n, m-1)) / (2 * Delta_xi);
y_xi = (y(n, m+1) – y(n, m-1)) / (2 * Delta_xi);

new_x = x(n, m) + Delta_eta * ((pi/4) – y_xi)/(x_xi);
error_x = abs(new_x – x(n, m));
max_error = max([max_error, error_x]);

x(n + 1, m) = new_x;
end

x(n+1, xi_steps) = x(n+1, xi_steps-1) + Delta_xi;
x(n+1, 1) = x(n+1, 2) – Delta_xi;
end

if max_error < tol
converged = true;
end
end

% Plot the results
figure; hold on; axis equal; grid on;

for m = 1:xi_steps
plot(x(:, m), y(:, m), ‘r’);
end

for n = 1:eta_steps
plot(x(n, :), y(n, :), ‘b’);
end

title([‘Iterations until convergence: ‘, num2str(iteration)]);
xlabel(‘x’); ylabel(‘y’); Hi everyone,
I’m trying to generate a uniform grid where the red lines intersect at a 45-degree angle to the horizontal lines in blue using the finite difference method in MATLAB.
However, the resulting grid is slightly off, as shown in the attached image. Intesection angle is uniform but higher than 45-degree.
I set computational domain as and a grid is generated on physical domain with coordinates .
As I want to generate a mesh described above, I aimed to solve the following PDE :

where we have (horizontal lines). The boundary conditions are .
The above equation comes down to the following:

Based on the above equation, I discretized to finite difference form (central difference for and forward differece for ).
Plus, I found the grid collapses when I set different angles (for example 90-degree (replace )).
Am I missing something in implementation?

I need your help.
Here is my code.
% Parameter settings
eta_max = 10;
xi_max = 10;
Delta_eta = 0.1;
Delta_xi = 0.1;
tol = 1e-6;
max_iterations = 100000;

% Grid size
eta_steps = floor(eta_max / Delta_eta) + 1;
xi_steps = floor(xi_max / Delta_xi) + 1;

% Initial conditions
x = zeros(eta_steps, xi_steps);
y = repmat((0:eta_steps-1)’ * Delta_eta, 1, xi_steps);

% Boundary conditions
x(1, 🙂 = linspace(0, xi_max, xi_steps);

% Finite difference method
converged = false;
iteration = 0;

while ~converged && iteration < max_iterations
iteration = iteration + 1;
max_error = 0;

for n = 1:eta_steps – 1
for m = 2:xi_steps – 1
x_xi = (x(n, m+1) – x(n, m-1)) / (2 * Delta_xi);
y_xi = (y(n, m+1) – y(n, m-1)) / (2 * Delta_xi);

new_x = x(n, m) + Delta_eta * ((pi/4) – y_xi)/(x_xi);
error_x = abs(new_x – x(n, m));
max_error = max([max_error, error_x]);

x(n + 1, m) = new_x;
end

x(n+1, xi_steps) = x(n+1, xi_steps-1) + Delta_xi;
x(n+1, 1) = x(n+1, 2) – Delta_xi;
end

if max_error < tol
converged = true;
end
end

% Plot the results
figure; hold on; axis equal; grid on;

for m = 1:xi_steps
plot(x(:, m), y(:, m), ‘r’);
end

for n = 1:eta_steps
plot(x(n, :), y(n, :), ‘b’);
end

title([‘Iterations until convergence: ‘, num2str(iteration)]);
xlabel(‘x’); ylabel(‘y’); finite difference method, grid generation, mesh generation, numerical analysis, differential equations MATLAB Answers — New Questions

​

Why numerical solutions is not consistent with the exact solutions? Also the error is too large. How to modify the code to get consistent results? See the attached image for the equation of motions and their exact solutions.
clear; clc;
tic;
c1 = 0; % real constant
c2 = 0; % real constant
a = 1; % real constant
b=0.1; % two values b=0.1, b=0.01
lambda = -1./(2*(1 + 1i)); % complex or imaginary
N = 1000;
X = linspace(-10, 10, N);
T = linspace(0, 10, N);

% Calculate initial conditions
[x, ~] = meshgrid(X, 0); % t = 0 for initial conditions
%psi_init = A * exp(-1i * lambda * x);
%phi_init = B * exp(1i * lambda * x);
nq_init = -a*(b.^5*(1+(0 + c2).^2).^2 – 8*a*(0+c2).*(x+c1) – 4*a*b.^4*(0+c2).*(1+(0+c2).^2).*(x+c1) – 4*a*b.^2*(0+c2).*(x+c1).*(3+(0+c2).^2 + a.^2*(x+c1).^2) + b*(-3+(0+c2).^4 + 6*a.^2*(x+c1).^2 + a.^4*(x+c1).^4 + 2*(0+c2).^2.*(3 + a.^2*(x+c1).^2)) + 2*b.^3*(-1 + (0+c2).^4 + a.^2*(x+c1).^2 + (0+c2).^2*(4 + 3*a.^2*(x+c1).^2)));
dq_init = 1 + (0 + c2).^2 + b.^2 + (b*(0 + c2) + a*(x + c1)).^2;
q_init = (nq_init ./ (dq_init .^ 2));
ns_init = -(-3 + (0 + c2).^2 + b.^2*(-3 + (0 + c2).^2) + 4*1i*a*(x+c1) – 2*a*b*(0 + c2).*(x+c1) + a.^2*(x+c1).^2).*(exp(1i*(a*x + b*0)));
s_init = ns_init ./ dq_init;

initial_conditions = [q_init(:); s_init(:)];

t_span = T;

% Solve the differential equation numerically
options = odeset(‘RelTol’, 1e-4, ‘AbsTol’, 1e-8);
[t, u] = ode45(@(t, u) dydt(t, u, N), t_span, initial_conditions, options);

% Extract q and s from the solution
q = reshape(u(:, 1:N), [length(t), N]);
s = reshape(u(:, N+1:end), [length(t), N]);

% Calculate the exact solution over the mesh grid
[x, t] = meshgrid(X, T);
nq = -a*(b.^5*(1+(t + c2).^2).^2 – 8*a*(t+c2).*(x+c1) – 4*a*b.^4*(t+c2).*(1+(t+c2).^2).*(x+c1) – 4*a*b.^2*(t+c2).*(x+c1).*(3+(t+c2).^2 + a.^2*(x+c1).^2) + b*(-3+(t+c2).^4 + 6*a.^2*(x+c1).^2 + a.^4*(x+c1).^4 + 2*(t+c2).^2.*(3 + a.^2*(x+c1).^2)) + 2*b.^3*(-1 + (t+c2).^4 + a.^2*(x+c1).^2 + (t+c2).^2*(4 + 3*a.^2*(x+c1).^2)));
dq = 1 + (t + c2).^2 + b.^2 + (b*(t + c2) + a*(x + c1)).^2;
ns = -(-3 + (t + c2).^2 + b.^2*(-3 + (t + c2).^2) + 4*1i*a*(x+c1) – 2*a*b*(t + c2).*(x+c1) + a.^2*(x+c1).^2).*(exp(1i*(a*x + b*0)));
exact_q = (nq ./ (dq .^ 2));
exact_s = ns ./ dq;

% Compute the error
error_q = abs(q’ – exact_q);
error_s = abs(s’ – exact_s);

% Plotting the numerical solution
figure;
subplot(2, 1, 1);
surf(t, x, abs(q’));
title(‘Numerical Solution for |q_{n+1} – q_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘|q_{n+1} – q_n|’);
shading interp;

subplot(2, 1, 2);
surf(t, x, abs(s’));
title(‘Numerical Solution for |s_{n+1} – s_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘|s_{n+1} – s_n|’);
shading interp;

% 3D surface plot for exact solution
figure;
subplot(2, 1, 1);
surf(t, x, abs(exact_q’));
title(‘Exact Solution for |q_{n+1} – q_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘|q_{n+1} – q_n|’);
shading interp;

subplot(2, 1, 2);
surf(t, x, abs(exact_s’));
title(‘Exact Solution for |s_{n+1} – s_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘|s_{n+1} – s_n|’);
shading interp;

% Plot the error
figure;
subplot(2, 1, 1);
surf(t, x, error_q’);
title(‘Error for |q_{n+1} – q_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘Error’);
shading interp;

subplot(2, 1, 2);
surf(t, x, error_s’);
title(‘Error for |s_{n+1} – s_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘Error’);
shading interp;

% 2D error plot at specific times
figure;
times_to_plot = [0, 5, 10]; % Example times to plot error
for i = 1:length(times_to_plot)
time_idx = find(t_span >= times_to_plot(i), 1);
subplot(length(times_to_plot), 1, i);
plot(X, error_q(:, time_idx), ‘r’, X, error_s(:, time_idx), ‘b’);
title([‘Error at t = ‘, num2str(times_to_plot(i))]);
xlabel(‘x’);
ylabel(‘Error’);
legend(‘|q_{n+1} – q_n| Error’, ‘|s_{n+1} – s_n| Error’);
end

% Generate error table for fixed times from 0 to 10 in increments of 0.5
fixed_times = 0:0.5:10;
error_table_q = zeros(length(fixed_times), 1);
error_table_s = zeros(length(fixed_times), 1);

for i = 1:length(fixed_times)
time_idx = find(t_span >= fixed_times(i), 1);
error_table_q(i) = mean(error_q(:, time_idx));
error_table_s(i) = mean(error_s(:, time_idx));
end

% Display the error table
error_table = table(fixed_times’, error_table_q, error_table_s, …
‘VariableNames’, {‘Time’, ‘Error_q’, ‘Error_s’});
disp(error_table);

toc;

function dydt = dydt(~, u, N)
dydt = zeros(2*N, 1);
q = u(1:N);
s = u(N+1:2*N);

% Handle boundary conditions
for n = 3:N-2
dydt(n) = 1/2 * s(n) * conj(s(n)) – 1/2 * s(n+1) * conj(s(n+1));
dydt(N+n) = 1/2 * (q(n+1) – q(n)) * (s(n+1) – s(n));
end
endWhy numerical solutions is not consistent with the exact solutions? Also the error is too large. How to modify the code to get consistent results? See the attached image for the equation of motions and their exact solutions.
clear; clc;
tic;
c1 = 0; % real constant
c2 = 0; % real constant
a = 1; % real constant
b=0.1; % two values b=0.1, b=0.01
lambda = -1./(2*(1 + 1i)); % complex or imaginary
N = 1000;
X = linspace(-10, 10, N);
T = linspace(0, 10, N);

% Calculate initial conditions
[x, ~] = meshgrid(X, 0); % t = 0 for initial conditions
%psi_init = A * exp(-1i * lambda * x);
%phi_init = B * exp(1i * lambda * x);
nq_init = -a*(b.^5*(1+(0 + c2).^2).^2 – 8*a*(0+c2).*(x+c1) – 4*a*b.^4*(0+c2).*(1+(0+c2).^2).*(x+c1) – 4*a*b.^2*(0+c2).*(x+c1).*(3+(0+c2).^2 + a.^2*(x+c1).^2) + b*(-3+(0+c2).^4 + 6*a.^2*(x+c1).^2 + a.^4*(x+c1).^4 + 2*(0+c2).^2.*(3 + a.^2*(x+c1).^2)) + 2*b.^3*(-1 + (0+c2).^4 + a.^2*(x+c1).^2 + (0+c2).^2*(4 + 3*a.^2*(x+c1).^2)));
dq_init = 1 + (0 + c2).^2 + b.^2 + (b*(0 + c2) + a*(x + c1)).^2;
q_init = (nq_init ./ (dq_init .^ 2));
ns_init = -(-3 + (0 + c2).^2 + b.^2*(-3 + (0 + c2).^2) + 4*1i*a*(x+c1) – 2*a*b*(0 + c2).*(x+c1) + a.^2*(x+c1).^2).*(exp(1i*(a*x + b*0)));
s_init = ns_init ./ dq_init;

initial_conditions = [q_init(:); s_init(:)];

t_span = T;

% Solve the differential equation numerically
options = odeset(‘RelTol’, 1e-4, ‘AbsTol’, 1e-8);
[t, u] = ode45(@(t, u) dydt(t, u, N), t_span, initial_conditions, options);

% Extract q and s from the solution
q = reshape(u(:, 1:N), [length(t), N]);
s = reshape(u(:, N+1:end), [length(t), N]);

% Calculate the exact solution over the mesh grid
[x, t] = meshgrid(X, T);
nq = -a*(b.^5*(1+(t + c2).^2).^2 – 8*a*(t+c2).*(x+c1) – 4*a*b.^4*(t+c2).*(1+(t+c2).^2).*(x+c1) – 4*a*b.^2*(t+c2).*(x+c1).*(3+(t+c2).^2 + a.^2*(x+c1).^2) + b*(-3+(t+c2).^4 + 6*a.^2*(x+c1).^2 + a.^4*(x+c1).^4 + 2*(t+c2).^2.*(3 + a.^2*(x+c1).^2)) + 2*b.^3*(-1 + (t+c2).^4 + a.^2*(x+c1).^2 + (t+c2).^2*(4 + 3*a.^2*(x+c1).^2)));
dq = 1 + (t + c2).^2 + b.^2 + (b*(t + c2) + a*(x + c1)).^2;
ns = -(-3 + (t + c2).^2 + b.^2*(-3 + (t + c2).^2) + 4*1i*a*(x+c1) – 2*a*b*(t + c2).*(x+c1) + a.^2*(x+c1).^2).*(exp(1i*(a*x + b*0)));
exact_q = (nq ./ (dq .^ 2));
exact_s = ns ./ dq;

% Compute the error
error_q = abs(q’ – exact_q);
error_s = abs(s’ – exact_s);

% Plotting the numerical solution
figure;
subplot(2, 1, 1);
surf(t, x, abs(q’));
title(‘Numerical Solution for |q_{n+1} – q_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘|q_{n+1} – q_n|’);
shading interp;

subplot(2, 1, 2);
surf(t, x, abs(s’));
title(‘Numerical Solution for |s_{n+1} – s_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘|s_{n+1} – s_n|’);
shading interp;

% 3D surface plot for exact solution
figure;
subplot(2, 1, 1);
surf(t, x, abs(exact_q’));
title(‘Exact Solution for |q_{n+1} – q_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘|q_{n+1} – q_n|’);
shading interp;

subplot(2, 1, 2);
surf(t, x, abs(exact_s’));
title(‘Exact Solution for |s_{n+1} – s_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘|s_{n+1} – s_n|’);
shading interp;

% Plot the error
figure;
subplot(2, 1, 1);
surf(t, x, error_q’);
title(‘Error for |q_{n+1} – q_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘Error’);
shading interp;

subplot(2, 1, 2);
surf(t, x, error_s’);
title(‘Error for |s_{n+1} – s_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘Error’);
shading interp;

% 2D error plot at specific times
figure;
times_to_plot = [0, 5, 10]; % Example times to plot error
for i = 1:length(times_to_plot)
time_idx = find(t_span >= times_to_plot(i), 1);
subplot(length(times_to_plot), 1, i);
plot(X, error_q(:, time_idx), ‘r’, X, error_s(:, time_idx), ‘b’);
title([‘Error at t = ‘, num2str(times_to_plot(i))]);
xlabel(‘x’);
ylabel(‘Error’);
legend(‘|q_{n+1} – q_n| Error’, ‘|s_{n+1} – s_n| Error’);
end

% Generate error table for fixed times from 0 to 10 in increments of 0.5
fixed_times = 0:0.5:10;
error_table_q = zeros(length(fixed_times), 1);
error_table_s = zeros(length(fixed_times), 1);

for i = 1:length(fixed_times)
time_idx = find(t_span >= fixed_times(i), 1);
error_table_q(i) = mean(error_q(:, time_idx));
error_table_s(i) = mean(error_s(:, time_idx));
end

% Display the error table
error_table = table(fixed_times’, error_table_q, error_table_s, …
‘VariableNames’, {‘Time’, ‘Error_q’, ‘Error_s’});
disp(error_table);

toc;

function dydt = dydt(~, u, N)
dydt = zeros(2*N, 1);
q = u(1:N);
s = u(N+1:2*N);

% Handle boundary conditions
for n = 3:N-2
dydt(n) = 1/2 * s(n) * conj(s(n)) – 1/2 * s(n+1) * conj(s(n+1));
dydt(N+n) = 1/2 * (q(n+1) – q(n)) * (s(n+1) – s(n));
end
end Why numerical solutions is not consistent with the exact solutions? Also the error is too large. How to modify the code to get consistent results? See the attached image for the equation of motions and their exact solutions.
clear; clc;
tic;
c1 = 0; % real constant
c2 = 0; % real constant
a = 1; % real constant
b=0.1; % two values b=0.1, b=0.01
lambda = -1./(2*(1 + 1i)); % complex or imaginary
N = 1000;
X = linspace(-10, 10, N);
T = linspace(0, 10, N);

% Calculate initial conditions
[x, ~] = meshgrid(X, 0); % t = 0 for initial conditions
%psi_init = A * exp(-1i * lambda * x);
%phi_init = B * exp(1i * lambda * x);
nq_init = -a*(b.^5*(1+(0 + c2).^2).^2 – 8*a*(0+c2).*(x+c1) – 4*a*b.^4*(0+c2).*(1+(0+c2).^2).*(x+c1) – 4*a*b.^2*(0+c2).*(x+c1).*(3+(0+c2).^2 + a.^2*(x+c1).^2) + b*(-3+(0+c2).^4 + 6*a.^2*(x+c1).^2 + a.^4*(x+c1).^4 + 2*(0+c2).^2.*(3 + a.^2*(x+c1).^2)) + 2*b.^3*(-1 + (0+c2).^4 + a.^2*(x+c1).^2 + (0+c2).^2*(4 + 3*a.^2*(x+c1).^2)));
dq_init = 1 + (0 + c2).^2 + b.^2 + (b*(0 + c2) + a*(x + c1)).^2;
q_init = (nq_init ./ (dq_init .^ 2));
ns_init = -(-3 + (0 + c2).^2 + b.^2*(-3 + (0 + c2).^2) + 4*1i*a*(x+c1) – 2*a*b*(0 + c2).*(x+c1) + a.^2*(x+c1).^2).*(exp(1i*(a*x + b*0)));
s_init = ns_init ./ dq_init;

initial_conditions = [q_init(:); s_init(:)];

t_span = T;

% Solve the differential equation numerically
options = odeset(‘RelTol’, 1e-4, ‘AbsTol’, 1e-8);
[t, u] = ode45(@(t, u) dydt(t, u, N), t_span, initial_conditions, options);

% Extract q and s from the solution
q = reshape(u(:, 1:N), [length(t), N]);
s = reshape(u(:, N+1:end), [length(t), N]);

% Calculate the exact solution over the mesh grid
[x, t] = meshgrid(X, T);
nq = -a*(b.^5*(1+(t + c2).^2).^2 – 8*a*(t+c2).*(x+c1) – 4*a*b.^4*(t+c2).*(1+(t+c2).^2).*(x+c1) – 4*a*b.^2*(t+c2).*(x+c1).*(3+(t+c2).^2 + a.^2*(x+c1).^2) + b*(-3+(t+c2).^4 + 6*a.^2*(x+c1).^2 + a.^4*(x+c1).^4 + 2*(t+c2).^2.*(3 + a.^2*(x+c1).^2)) + 2*b.^3*(-1 + (t+c2).^4 + a.^2*(x+c1).^2 + (t+c2).^2*(4 + 3*a.^2*(x+c1).^2)));
dq = 1 + (t + c2).^2 + b.^2 + (b*(t + c2) + a*(x + c1)).^2;
ns = -(-3 + (t + c2).^2 + b.^2*(-3 + (t + c2).^2) + 4*1i*a*(x+c1) – 2*a*b*(t + c2).*(x+c1) + a.^2*(x+c1).^2).*(exp(1i*(a*x + b*0)));
exact_q = (nq ./ (dq .^ 2));
exact_s = ns ./ dq;

% Compute the error
error_q = abs(q’ – exact_q);
error_s = abs(s’ – exact_s);

% Plotting the numerical solution
figure;
subplot(2, 1, 1);
surf(t, x, abs(q’));
title(‘Numerical Solution for |q_{n+1} – q_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘|q_{n+1} – q_n|’);
shading interp;

subplot(2, 1, 2);
surf(t, x, abs(s’));
title(‘Numerical Solution for |s_{n+1} – s_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘|s_{n+1} – s_n|’);
shading interp;

% 3D surface plot for exact solution
figure;
subplot(2, 1, 1);
surf(t, x, abs(exact_q’));
title(‘Exact Solution for |q_{n+1} – q_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘|q_{n+1} – q_n|’);
shading interp;

subplot(2, 1, 2);
surf(t, x, abs(exact_s’));
title(‘Exact Solution for |s_{n+1} – s_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘|s_{n+1} – s_n|’);
shading interp;

% Plot the error
figure;
subplot(2, 1, 1);
surf(t, x, error_q’);
title(‘Error for |q_{n+1} – q_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘Error’);
shading interp;

subplot(2, 1, 2);
surf(t, x, error_s’);
title(‘Error for |s_{n+1} – s_n|’);
xlabel(‘t’);
ylabel(‘x’);
zlabel(‘Error’);
shading interp;

% 2D error plot at specific times
figure;
times_to_plot = [0, 5, 10]; % Example times to plot error
for i = 1:length(times_to_plot)
time_idx = find(t_span >= times_to_plot(i), 1);
subplot(length(times_to_plot), 1, i);
plot(X, error_q(:, time_idx), ‘r’, X, error_s(:, time_idx), ‘b’);
title([‘Error at t = ‘, num2str(times_to_plot(i))]);
xlabel(‘x’);
ylabel(‘Error’);
legend(‘|q_{n+1} – q_n| Error’, ‘|s_{n+1} – s_n| Error’);
end

% Generate error table for fixed times from 0 to 10 in increments of 0.5
fixed_times = 0:0.5:10;
error_table_q = zeros(length(fixed_times), 1);
error_table_s = zeros(length(fixed_times), 1);

for i = 1:length(fixed_times)
time_idx = find(t_span >= fixed_times(i), 1);
error_table_q(i) = mean(error_q(:, time_idx));
error_table_s(i) = mean(error_s(:, time_idx));
end

% Display the error table
error_table = table(fixed_times’, error_table_q, error_table_s, …
‘VariableNames’, {‘Time’, ‘Error_q’, ‘Error_s’});
disp(error_table);

toc;

function dydt = dydt(~, u, N)
dydt = zeros(2*N, 1);
q = u(1:N);
s = u(N+1:2*N);

% Handle boundary conditions
for n = 3:N-2
dydt(n) = 1/2 * s(n) * conj(s(n)) – 1/2 * s(n+1) * conj(s(n+1));
dydt(N+n) = 1/2 * (q(n+1) – q(n)) * (s(n+1) – s(n));
end
end ode45, numerical integration, differential equations MATLAB Answers — New Questions

​

Hello, I have an image and I want to get the median of data at each radius value. This is shown by the yellow circles – so each circle (take the median of all pixels on it).

However, in my attemy, I missing all the corners of data that are actually at a larger radius than the image size, but not full circles. Any suggestions how to include this data.
This was my attempt:
% 1st draw circle of radius R on the flog2 plot

theta = linspace(0,360); %use n=121 for every 3 degrees (reduce from every degree for speed
centre = [cx,cy]; %circle centre
% I= Image

[X,Y]=ndgrid(1:size(I,1),1:size(I,2));

X=X-cx;Y=Y-cy;%shift coordinate grid
hold(ax1,"on");

data=[];
for i=1:cy-1
radius=i; %circle radius
data(i,1)=i;
y = radius*sind(theta)+centre(2);
x = radius*cosd(theta)+centre(1);

%Get Data on radius
L=sqrt(X.^2+Y.^2)==radius;
data(i,2) = median(I(L));
if mod(i,5)==0
plot(ax1,x,y,’y’); %draw circle
end

% Now use interp so use all pixels on circle. This is
% actually what I finally use now
[xx,yy]=pol2cart(theta,radius); % this is for the interp2 approach
x=xx+cx; y=yy+cy;
vals=median(interp2(I,x,y));
data(i,3)=vals;
end

% Now plot data
hold(ax2,"on");
plot(ax2,data(:,1),data(:,3),’y-‘,’LineWidth’,3);Hello, I have an image and I want to get the median of data at each radius value. This is shown by the yellow circles – so each circle (take the median of all pixels on it).

However, in my attemy, I missing all the corners of data that are actually at a larger radius than the image size, but not full circles. Any suggestions how to include this data.
This was my attempt:
% 1st draw circle of radius R on the flog2 plot

theta = linspace(0,360); %use n=121 for every 3 degrees (reduce from every degree for speed
centre = [cx,cy]; %circle centre
% I= Image

[X,Y]=ndgrid(1:size(I,1),1:size(I,2));

X=X-cx;Y=Y-cy;%shift coordinate grid
hold(ax1,"on");

data=[];
for i=1:cy-1
radius=i; %circle radius
data(i,1)=i;
y = radius*sind(theta)+centre(2);
x = radius*cosd(theta)+centre(1);

%Get Data on radius
L=sqrt(X.^2+Y.^2)==radius;
data(i,2) = median(I(L));
if mod(i,5)==0
plot(ax1,x,y,’y’); %draw circle
end

% Now use interp so use all pixels on circle. This is
% actually what I finally use now
[xx,yy]=pol2cart(theta,radius); % this is for the interp2 approach
x=xx+cx; y=yy+cy;
vals=median(interp2(I,x,y));
data(i,3)=vals;
end

% Now plot data
hold(ax2,"on");
plot(ax2,data(:,1),data(:,3),’y-‘,’LineWidth’,3); Hello, I have an image and I want to get the median of data at each radius value. This is shown by the yellow circles – so each circle (take the median of all pixels on it).

However, in my attemy, I missing all the corners of data that are actually at a larger radius than the image size, but not full circles. Any suggestions how to include this data.
This was my attempt:
% 1st draw circle of radius R on the flog2 plot

theta = linspace(0,360); %use n=121 for every 3 degrees (reduce from every degree for speed
centre = [cx,cy]; %circle centre
% I= Image

[X,Y]=ndgrid(1:size(I,1),1:size(I,2));

X=X-cx;Y=Y-cy;%shift coordinate grid
hold(ax1,"on");

data=[];
for i=1:cy-1
radius=i; %circle radius
data(i,1)=i;
y = radius*sind(theta)+centre(2);
x = radius*cosd(theta)+centre(1);

%Get Data on radius
L=sqrt(X.^2+Y.^2)==radius;
data(i,2) = median(I(L));
if mod(i,5)==0
plot(ax1,x,y,’y’); %draw circle
end

% Now use interp so use all pixels on circle. This is
% actually what I finally use now
[xx,yy]=pol2cart(theta,radius); % this is for the interp2 approach
x=xx+cx; y=yy+cy;
vals=median(interp2(I,x,y));
data(i,3)=vals;
end

% Now plot data
hold(ax2,"on");
plot(ax2,data(:,1),data(:,3),’y-‘,’LineWidth’,3); image, ngrid, pol2cart MATLAB Answers — New Questions

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Hello, i have 3D image but when I view the image in the 3D software to be printed, the image does not extrude. here i attached my code and images. can anyone check on my code. Many thanks.

a = imread (‘stomachgray.tif’);
mask = zeros(size(a));
mask(100:end-100,100:end-100) = 1;
bw = activecontour(a,mask,1000);

c = im2double(bw);
shading flat

d = imgaussfilt3 (c,4);

colormap(bone)

h = hgtransform;

mesh(d*100, ‘Parent’, h, ‘FaceColor’, ‘r’ )
view(3)
lighting gouraud
camlight right
% Make it taller
set (gca, ‘units’, ‘cent’)
set(h, ‘Matrix’, makehgtform(‘scale’, [10 10 500]))
[X,Y] = meshgrid(1:length(h));
surf2stl(‘stomachSurf7.stl’,X,Y,d);

end

<</matlabcentral/answers/uploaded_files/64209/3d%20mtlb.JPG>>

<</matlabcentral/answers/uploaded_files/64210/3d%20view.JPG>>Hello, i have 3D image but when I view the image in the 3D software to be printed, the image does not extrude. here i attached my code and images. can anyone check on my code. Many thanks.

a = imread (‘stomachgray.tif’);
mask = zeros(size(a));
mask(100:end-100,100:end-100) = 1;
bw = activecontour(a,mask,1000);

c = im2double(bw);
shading flat

d = imgaussfilt3 (c,4);

colormap(bone)

h = hgtransform;

mesh(d*100, ‘Parent’, h, ‘FaceColor’, ‘r’ )
view(3)
lighting gouraud
camlight right
% Make it taller
set (gca, ‘units’, ‘cent’)
set(h, ‘Matrix’, makehgtform(‘scale’, [10 10 500]))
[X,Y] = meshgrid(1:length(h));
surf2stl(‘stomachSurf7.stl’,X,Y,d);

end

<</matlabcentral/answers/uploaded_files/64209/3d%20mtlb.JPG>>

<</matlabcentral/answers/uploaded_files/64210/3d%20view.JPG>> Hello, i have 3D image but when I view the image in the 3D software to be printed, the image does not extrude. here i attached my code and images. can anyone check on my code. Many thanks.

a = imread (‘stomachgray.tif’);
mask = zeros(size(a));
mask(100:end-100,100:end-100) = 1;
bw = activecontour(a,mask,1000);

c = im2double(bw);
shading flat

d = imgaussfilt3 (c,4);

colormap(bone)

h = hgtransform;

mesh(d*100, ‘Parent’, h, ‘FaceColor’, ‘r’ )
view(3)
lighting gouraud
camlight right
% Make it taller
set (gca, ‘units’, ‘cent’)
set(h, ‘Matrix’, makehgtform(‘scale’, [10 10 500]))
[X,Y] = meshgrid(1:length(h));
surf2stl(‘stomachSurf7.stl’,X,Y,d);

end

<</matlabcentral/answers/uploaded_files/64209/3d%20mtlb.JPG>>

<</matlabcentral/answers/uploaded_files/64210/3d%20view.JPG>> mesh, extrude, matlab, 3d image MATLAB Answers — New Questions

​

I have particle size data for differrent De(parameter) values.
I have plotted 2Dbar graphs, I want stack all figures into one 3D figure window.
I stacked everything, that looks like the image below

But I want to rotate axis in such a way that: Particle_density(Y-axis) in vertical, Particle size(X-axis) in horizontal and De (z-axis) per pendicular to the screen. It should look like this image below

I have attached my matlab code to this thread.
Please guide me, how to do this. Thanks in advanceI have particle size data for differrent De(parameter) values.
I have plotted 2Dbar graphs, I want stack all figures into one 3D figure window.
I stacked everything, that looks like the image below

But I want to rotate axis in such a way that: Particle_density(Y-axis) in vertical, Particle size(X-axis) in horizontal and De (z-axis) per pendicular to the screen. It should look like this image below

I have attached my matlab code to this thread.
Please guide me, how to do this. Thanks in advance I have particle size data for differrent De(parameter) values.
I have plotted 2Dbar graphs, I want stack all figures into one 3D figure window.
I stacked everything, that looks like the image below

But I want to rotate axis in such a way that: Particle_density(Y-axis) in vertical, Particle size(X-axis) in horizontal and De (z-axis) per pendicular to the screen. It should look like this image below

I have attached my matlab code to this thread.
Please guide me, how to do this. Thanks in advance rotation, bar, plot3 MATLAB Answers — New Questions

​

Hi guys, i have to make a triangulation 3D of a solid of which i have the coordinates (x,y,z) of 20ooo points and i need to export the result (so the tetrahedra) to an stl file.
How can I use properly the stlwrite function, in particular the option indicated with TRIANGULATION?

thanks a lot.Hi guys, i have to make a triangulation 3D of a solid of which i have the coordinates (x,y,z) of 20ooo points and i need to export the result (so the tetrahedra) to an stl file.
How can I use properly the stlwrite function, in particular the option indicated with TRIANGULATION?

thanks a lot. Hi guys, i have to make a triangulation 3D of a solid of which i have the coordinates (x,y,z) of 20ooo points and i need to export the result (so the tetrahedra) to an stl file.
How can I use properly the stlwrite function, in particular the option indicated with TRIANGULATION?

thanks a lot. triangulation delaunay stl .stl stlwrite MATLAB Answers — New Questions

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