Tag Archives: matlab
4-class perceptron classification
Good evening, I hope everyone is well. I have a single-layer perceptron that is meant to take in two inputs and provide an output as one of four classifications. I then need to plot the inputs and the hyperplanes dividing the four classes. I’m getting errors when I try to run the code that I suspect are related to the number of outputs I’m trying to get. Here’s the code I’m working with:
if true
% Initialize the Input Space (extended input matrix)
x = [1 1 1 2 2 -1 -2 -1 -2; 1 1 2 -1 0 2 1 -1 -2];
% Initialize the extended target vector
t = [0 1 1 2 2 3 3 4 4];
% Plot the input locations
figure
hold on
for i=1:length(x)
if (t(i)==1)
scatter(x(1,i), x(2,i),’k ‘, ‘filled’);
elseif (t(i)==2)
scatter(x(1,i), x(2,i),’r ‘, ‘filled’);
elseif (t(i)==3)
scatter(x(1,i), x(2,i),’b ‘, ‘filled’);
elseif (t(i)==4)
scatter(x(1,i), x(2,i),’g ‘, ‘filled’);
end
end
grid on
line([0 0], ylim, ‘linewidth’, 1); %y-axis
line(xlim, [0 0], ‘linewidth’, 1); %x-axis
legend(‘class_1’, ‘class_2’, ‘class_3’, ‘class_4’)
net = perceptron;
net.trainparam.epochs = 100; % Set # of training epochs
net.trainparam.goal = 1e-2; % Set desired max error
net.trainparam.lr = 0.1; % Set desired learning rate
train(net, x, t); % Train the perceptron
predictions = net(x); % Get data predictions
net.IW{:}; % Learned weights
net.b{:}; % Learned biases
% Equation of a line: w1*x1 + w2*x2 + b = 0
% Plot the lines
plot([-net.b{1}/net.IW{:}(1,1),0],[0,-net.b{1}/net.IW{:}(1,2)])
plot([-net.b{2}/net.IW{:}(2,1),0],[0,-net.b{2}/net.IW{:}(2,2)])
hold off;
end
The trainparam epochs, goal and lr are initial conditions an can change but I don’t think that’s where the problem is.
I’d appreciate any help you’re able to provide. Best regards.Good evening, I hope everyone is well. I have a single-layer perceptron that is meant to take in two inputs and provide an output as one of four classifications. I then need to plot the inputs and the hyperplanes dividing the four classes. I’m getting errors when I try to run the code that I suspect are related to the number of outputs I’m trying to get. Here’s the code I’m working with:
if true
% Initialize the Input Space (extended input matrix)
x = [1 1 1 2 2 -1 -2 -1 -2; 1 1 2 -1 0 2 1 -1 -2];
% Initialize the extended target vector
t = [0 1 1 2 2 3 3 4 4];
% Plot the input locations
figure
hold on
for i=1:length(x)
if (t(i)==1)
scatter(x(1,i), x(2,i),’k ‘, ‘filled’);
elseif (t(i)==2)
scatter(x(1,i), x(2,i),’r ‘, ‘filled’);
elseif (t(i)==3)
scatter(x(1,i), x(2,i),’b ‘, ‘filled’);
elseif (t(i)==4)
scatter(x(1,i), x(2,i),’g ‘, ‘filled’);
end
end
grid on
line([0 0], ylim, ‘linewidth’, 1); %y-axis
line(xlim, [0 0], ‘linewidth’, 1); %x-axis
legend(‘class_1’, ‘class_2’, ‘class_3’, ‘class_4’)
net = perceptron;
net.trainparam.epochs = 100; % Set # of training epochs
net.trainparam.goal = 1e-2; % Set desired max error
net.trainparam.lr = 0.1; % Set desired learning rate
train(net, x, t); % Train the perceptron
predictions = net(x); % Get data predictions
net.IW{:}; % Learned weights
net.b{:}; % Learned biases
% Equation of a line: w1*x1 + w2*x2 + b = 0
% Plot the lines
plot([-net.b{1}/net.IW{:}(1,1),0],[0,-net.b{1}/net.IW{:}(1,2)])
plot([-net.b{2}/net.IW{:}(2,1),0],[0,-net.b{2}/net.IW{:}(2,2)])
hold off;
end
The trainparam epochs, goal and lr are initial conditions an can change but I don’t think that’s where the problem is.
I’d appreciate any help you’re able to provide. Best regards. Good evening, I hope everyone is well. I have a single-layer perceptron that is meant to take in two inputs and provide an output as one of four classifications. I then need to plot the inputs and the hyperplanes dividing the four classes. I’m getting errors when I try to run the code that I suspect are related to the number of outputs I’m trying to get. Here’s the code I’m working with:
if true
% Initialize the Input Space (extended input matrix)
x = [1 1 1 2 2 -1 -2 -1 -2; 1 1 2 -1 0 2 1 -1 -2];
% Initialize the extended target vector
t = [0 1 1 2 2 3 3 4 4];
% Plot the input locations
figure
hold on
for i=1:length(x)
if (t(i)==1)
scatter(x(1,i), x(2,i),’k ‘, ‘filled’);
elseif (t(i)==2)
scatter(x(1,i), x(2,i),’r ‘, ‘filled’);
elseif (t(i)==3)
scatter(x(1,i), x(2,i),’b ‘, ‘filled’);
elseif (t(i)==4)
scatter(x(1,i), x(2,i),’g ‘, ‘filled’);
end
end
grid on
line([0 0], ylim, ‘linewidth’, 1); %y-axis
line(xlim, [0 0], ‘linewidth’, 1); %x-axis
legend(‘class_1’, ‘class_2’, ‘class_3’, ‘class_4’)
net = perceptron;
net.trainparam.epochs = 100; % Set # of training epochs
net.trainparam.goal = 1e-2; % Set desired max error
net.trainparam.lr = 0.1; % Set desired learning rate
train(net, x, t); % Train the perceptron
predictions = net(x); % Get data predictions
net.IW{:}; % Learned weights
net.b{:}; % Learned biases
% Equation of a line: w1*x1 + w2*x2 + b = 0
% Plot the lines
plot([-net.b{1}/net.IW{:}(1,1),0],[0,-net.b{1}/net.IW{:}(1,2)])
plot([-net.b{2}/net.IW{:}(2,1),0],[0,-net.b{2}/net.IW{:}(2,2)])
hold off;
end
The trainparam epochs, goal and lr are initial conditions an can change but I don’t think that’s where the problem is.
I’d appreciate any help you’re able to provide. Best regards. perceptron, classification, neural network MATLAB Answers — New Questions
How to setup MATLAB-Compiler to use MSVC2017 Build Tools
I was able to sucessfully setup a system using the MATLAB 2018a compiler with Visual Studio 2017 Professional. However on another machine I only run the Visual Stiudio 2017 Build Tools which have all the files needed to compile but only lack the IDE.
It seems this compiler is not yet supported by the MATLAB compiler.
Is there a way / patch that allows usage of the build tools?
Thanks in advance!I was able to sucessfully setup a system using the MATLAB 2018a compiler with Visual Studio 2017 Professional. However on another machine I only run the Visual Stiudio 2017 Build Tools which have all the files needed to compile but only lack the IDE.
It seems this compiler is not yet supported by the MATLAB compiler.
Is there a way / patch that allows usage of the build tools?
Thanks in advance! I was able to sucessfully setup a system using the MATLAB 2018a compiler with Visual Studio 2017 Professional. However on another machine I only run the Visual Stiudio 2017 Build Tools which have all the files needed to compile but only lack the IDE.
It seems this compiler is not yet supported by the MATLAB compiler.
Is there a way / patch that allows usage of the build tools?
Thanks in advance! compiler, visual studio MATLAB Answers — New Questions
loop Will only open the files in starting folder
I want to create a script that opens a folder and goes in to every subfolder and pulls out .mat files. The following code only pulls out the .mat files that are located in the starting fold even though it gathers all sub folder directories. Any help would be greatly appreciated!
startingFolder = pwd;
root = uigetdir(startingFolder);
allSubfolders = genpath(root);
subFolders = regexp(allSubfolders, ‘;’, ‘split’);
for subFolderCount = 3 : length(subFolders)
% Get this subfolder.
thisSubFolder = subFolders{subFolderCount};
% Get a list of MAT files in this subfolder.
matFiles = dir(fullfile(thisSubFolder, ‘*.mat’));
endI want to create a script that opens a folder and goes in to every subfolder and pulls out .mat files. The following code only pulls out the .mat files that are located in the starting fold even though it gathers all sub folder directories. Any help would be greatly appreciated!
startingFolder = pwd;
root = uigetdir(startingFolder);
allSubfolders = genpath(root);
subFolders = regexp(allSubfolders, ‘;’, ‘split’);
for subFolderCount = 3 : length(subFolders)
% Get this subfolder.
thisSubFolder = subFolders{subFolderCount};
% Get a list of MAT files in this subfolder.
matFiles = dir(fullfile(thisSubFolder, ‘*.mat’));
end I want to create a script that opens a folder and goes in to every subfolder and pulls out .mat files. The following code only pulls out the .mat files that are located in the starting fold even though it gathers all sub folder directories. Any help would be greatly appreciated!
startingFolder = pwd;
root = uigetdir(startingFolder);
allSubfolders = genpath(root);
subFolders = regexp(allSubfolders, ‘;’, ‘split’);
for subFolderCount = 3 : length(subFolders)
% Get this subfolder.
thisSubFolder = subFolders{subFolderCount};
% Get a list of MAT files in this subfolder.
matFiles = dir(fullfile(thisSubFolder, ‘*.mat’));
end open files MATLAB Answers — New Questions
Identifying an object and its centroid in an image and then cropping the original image based on this centroid
Hello,
I need to detect a particular objects out of many objects in an image. Then I need to find its centroid and then crop the image around that particular tracked object using its centroid as the centre of that rectangular crop.
For example, let’s say I need to idenfity the marked watermelon piece (as in the attached picture) and then crop the image around this piece of watermelon.
Any help in this regard will be greatly appreciated. Thank you very much!Hello,
I need to detect a particular objects out of many objects in an image. Then I need to find its centroid and then crop the image around that particular tracked object using its centroid as the centre of that rectangular crop.
For example, let’s say I need to idenfity the marked watermelon piece (as in the attached picture) and then crop the image around this piece of watermelon.
Any help in this regard will be greatly appreciated. Thank you very much! Hello,
I need to detect a particular objects out of many objects in an image. Then I need to find its centroid and then crop the image around that particular tracked object using its centroid as the centre of that rectangular crop.
For example, let’s say I need to idenfity the marked watermelon piece (as in the attached picture) and then crop the image around this piece of watermelon.
Any help in this regard will be greatly appreciated. Thank you very much! image processing, image, image analysis MATLAB Answers — New Questions
Whats wrong with tester_app_3?
I am not sending the functions that are needed to run the app and neither the other app required to run the app since they are many functions and the problem is in tester_app_3.
Do I need a while loop somewhere?
Seems the values for the table osmotic data aren’t saved in my table columns when I run the app?
I think there is something missing in tester_app_3 which I can’t see.
Do I need a loop to accumulate the column values in osmotic data, I just get the last value printed. It also seems that the program doesn’t start on zone nr 1 when I start the app again though I have deleted the previous files by writing the syntax delete(file).
I have matlab 2018I am not sending the functions that are needed to run the app and neither the other app required to run the app since they are many functions and the problem is in tester_app_3.
Do I need a while loop somewhere?
Seems the values for the table osmotic data aren’t saved in my table columns when I run the app?
I think there is something missing in tester_app_3 which I can’t see.
Do I need a loop to accumulate the column values in osmotic data, I just get the last value printed. It also seems that the program doesn’t start on zone nr 1 when I start the app again though I have deleted the previous files by writing the syntax delete(file).
I have matlab 2018 I am not sending the functions that are needed to run the app and neither the other app required to run the app since they are many functions and the problem is in tester_app_3.
Do I need a while loop somewhere?
Seems the values for the table osmotic data aren’t saved in my table columns when I run the app?
I think there is something missing in tester_app_3 which I can’t see.
Do I need a loop to accumulate the column values in osmotic data, I just get the last value printed. It also seems that the program doesn’t start on zone nr 1 when I start the app again though I have deleted the previous files by writing the syntax delete(file).
I have matlab 2018 app designer, tables, while loop, overwriting files MATLAB Answers — New Questions
Returning data from uibutton callback
Hi,
I want to create a script that lets me add rows to a uitable with a ui buttom. My script mostly does what I want, except that it doesn’t store the inputed values when I edit the table, nor does it return the eddited table back to memory to be used later. Is anyone able to point me in the right direction?
clc
Support = [0];
Type = {false};
t = table(Support,Type);
fig = uifigure;
uit = uitable(fig,"Data",t);
btn = uibutton(fig,"Text","Add Row","Position", [400 100 100 50],"ButtonPushedFcn",@(src,event) buttonCallback(t,fig));
uit.ColumnEditable = true;
function buttonCallback(t,fig)
Support = 0;
Type = {false};
addrow = table(Support,Type);
t = [t;addrow];
uit = uitable(fig,"Data",t);
uit.ColumnEditable = true;
btn = uibutton(fig,"Text","Add Row","Position", [400 100 100 50],"ButtonPushedFcn",@(src,event) buttonCallback(t,fig));
endHi,
I want to create a script that lets me add rows to a uitable with a ui buttom. My script mostly does what I want, except that it doesn’t store the inputed values when I edit the table, nor does it return the eddited table back to memory to be used later. Is anyone able to point me in the right direction?
clc
Support = [0];
Type = {false};
t = table(Support,Type);
fig = uifigure;
uit = uitable(fig,"Data",t);
btn = uibutton(fig,"Text","Add Row","Position", [400 100 100 50],"ButtonPushedFcn",@(src,event) buttonCallback(t,fig));
uit.ColumnEditable = true;
function buttonCallback(t,fig)
Support = 0;
Type = {false};
addrow = table(Support,Type);
t = [t;addrow];
uit = uitable(fig,"Data",t);
uit.ColumnEditable = true;
btn = uibutton(fig,"Text","Add Row","Position", [400 100 100 50],"ButtonPushedFcn",@(src,event) buttonCallback(t,fig));
end Hi,
I want to create a script that lets me add rows to a uitable with a ui buttom. My script mostly does what I want, except that it doesn’t store the inputed values when I edit the table, nor does it return the eddited table back to memory to be used later. Is anyone able to point me in the right direction?
clc
Support = [0];
Type = {false};
t = table(Support,Type);
fig = uifigure;
uit = uitable(fig,"Data",t);
btn = uibutton(fig,"Text","Add Row","Position", [400 100 100 50],"ButtonPushedFcn",@(src,event) buttonCallback(t,fig));
uit.ColumnEditable = true;
function buttonCallback(t,fig)
Support = 0;
Type = {false};
addrow = table(Support,Type);
t = [t;addrow];
uit = uitable(fig,"Data",t);
uit.ColumnEditable = true;
btn = uibutton(fig,"Text","Add Row","Position", [400 100 100 50],"ButtonPushedFcn",@(src,event) buttonCallback(t,fig));
end uitable, uibutton MATLAB Answers — New Questions
Particle Weight update does not seem to work
Hello,
I have implemented a 2D occupancy map and a robot model. The robot will move through the map. It is equipped with a simulated Lidar sensor. If I let the Monte Carlo Localization function of Matlab run, my code works. The particles converge to the robot’s position. However, if I check the particle weights during the process, they are all the same. If I have 2000 particles, all 2000 particles always have the same weights. Does someone know why this is or what the problem might be?
I am assuming that the code works correctly since it does converge. So the weights have to be changing in the background without my knowledge.
The main part of the code is this in a loop for each time step. At the end Vehiclepose will be updated for the next time step.
[ranges,angles] = lidar(Vehiclepose,map);
scan = lidarScan(ranges,angles);
[isUpdated,estimatedPose, estimatedCovariance] = mcl(Vehiclepose, scan);
[particles,weights] = getParticles(mcl);
Thank you.Hello,
I have implemented a 2D occupancy map and a robot model. The robot will move through the map. It is equipped with a simulated Lidar sensor. If I let the Monte Carlo Localization function of Matlab run, my code works. The particles converge to the robot’s position. However, if I check the particle weights during the process, they are all the same. If I have 2000 particles, all 2000 particles always have the same weights. Does someone know why this is or what the problem might be?
I am assuming that the code works correctly since it does converge. So the weights have to be changing in the background without my knowledge.
The main part of the code is this in a loop for each time step. At the end Vehiclepose will be updated for the next time step.
[ranges,angles] = lidar(Vehiclepose,map);
scan = lidarScan(ranges,angles);
[isUpdated,estimatedPose, estimatedCovariance] = mcl(Vehiclepose, scan);
[particles,weights] = getParticles(mcl);
Thank you. Hello,
I have implemented a 2D occupancy map and a robot model. The robot will move through the map. It is equipped with a simulated Lidar sensor. If I let the Monte Carlo Localization function of Matlab run, my code works. The particles converge to the robot’s position. However, if I check the particle weights during the process, they are all the same. If I have 2000 particles, all 2000 particles always have the same weights. Does someone know why this is or what the problem might be?
I am assuming that the code works correctly since it does converge. So the weights have to be changing in the background without my knowledge.
The main part of the code is this in a loop for each time step. At the end Vehiclepose will be updated for the next time step.
[ranges,angles] = lidar(Vehiclepose,map);
scan = lidarScan(ranges,angles);
[isUpdated,estimatedPose, estimatedCovariance] = mcl(Vehiclepose, scan);
[particles,weights] = getParticles(mcl);
Thank you. montecarlolocalization, particlefilter, weights, lidar MATLAB Answers — New Questions
Undefined function ‘ ‘ for input arguments of type ‘gpuArray’. – Why ?
So Im using Paralell Computing Toolbox v. 23.2, on R2023b and having this problem when passing an array from one function to another, and can’t seem to understand what is causing this.
My nested loop looks like this, everything is initiated correctly in its running, except for when I add the nu_t term.
for t_n = 1:t-1
for i = 2:length(X) – 1
for j = 2:length(Y) – 1
nu_t = addTurbulentViscotiy(U_temp, C_s, Delta)
end
end
U_temp = applyNoSlipBoundary(U_temp, t_n);
P_temp = applyNeumannBoundary(P_temp, t_n);
end
And the function that doesnt work looks like this:
function [nu_t] = addTurbulentViscotiy(U, C_s, Delta)
% Computing the strain rate tensor components for S
S_11 = (U(t_n, i+1, j, 1) – U(t_n, i-1, j, 1)) / (2 * Delta);
S_22 = (U(t_n, i, j+1, 2) – U(t_n, i, j-1, 2)) / (2 * Delta);
S_12 = 0.5 * ((U(t_n, i, j+1, 1) – U(t_n, i, j-1, 1)) / (2 * Delta) …
+ (U(t_n, i+1, j, 2) – U(t_n, i-1, j, 2)) / (2 * Delta));
S_21 = S_12;
% Calculating the magnitude of S
S_mag = sqrt(S_11.^2 + S_22.^2 + 2* S12.^2);
% Calculating the turbulent Viscocity
nu_t = (C_s*Delta).^2*S_mag
end
However, these function work. Its the same values being passed:
function [U_bound] = applyNoSlipBoundary(U_field, t_n)
U_field(t_n + 1, 1, :, 🙂 = 0; % Top boundary
U_field(t_n + 1, end, :, 🙂 = 0; % Bottom boundary
U_field(t_n + 1, :, 1, 🙂 = 0; % Left boundary
U_field(t_n + 1, :, end, 🙂 = 0; % Right boundary
U_bound = U_field;
end
% Apply Neumann boundary conditions (zero gradient)
function [P_bound] = applyNeumannBoundary(P_temp, t_n)
P_temp(t_n + 1, 1, 🙂 = P_temp(t_n + 1, 2, :); % Top boundary
P_temp(t_n + 1, end, 🙂 = P_temp(t_n + 1, end-1, :); % Bottom boundary
P_temp(t_n + 1, :, 1) = P_temp(t_n + 1, :, 2); % Left boundary
P_temp(t_n + 1, :, end) = P_temp(t_n + 1, :, end-1); % Right boundary
P_bound = P_temp;
end
The error gotten is
Undefined function ‘NSsolverTest2’ for input arguments of type ‘gpuArray’.
The functions are in the same file, altough not nested.
Why do they work in one function but not the other? What is causing this I looked through the documentations but can’t find anything there. The only difference is where in the loop theyre being called. Anybody that could help me figure this out?
Thanks in advance!So Im using Paralell Computing Toolbox v. 23.2, on R2023b and having this problem when passing an array from one function to another, and can’t seem to understand what is causing this.
My nested loop looks like this, everything is initiated correctly in its running, except for when I add the nu_t term.
for t_n = 1:t-1
for i = 2:length(X) – 1
for j = 2:length(Y) – 1
nu_t = addTurbulentViscotiy(U_temp, C_s, Delta)
end
end
U_temp = applyNoSlipBoundary(U_temp, t_n);
P_temp = applyNeumannBoundary(P_temp, t_n);
end
And the function that doesnt work looks like this:
function [nu_t] = addTurbulentViscotiy(U, C_s, Delta)
% Computing the strain rate tensor components for S
S_11 = (U(t_n, i+1, j, 1) – U(t_n, i-1, j, 1)) / (2 * Delta);
S_22 = (U(t_n, i, j+1, 2) – U(t_n, i, j-1, 2)) / (2 * Delta);
S_12 = 0.5 * ((U(t_n, i, j+1, 1) – U(t_n, i, j-1, 1)) / (2 * Delta) …
+ (U(t_n, i+1, j, 2) – U(t_n, i-1, j, 2)) / (2 * Delta));
S_21 = S_12;
% Calculating the magnitude of S
S_mag = sqrt(S_11.^2 + S_22.^2 + 2* S12.^2);
% Calculating the turbulent Viscocity
nu_t = (C_s*Delta).^2*S_mag
end
However, these function work. Its the same values being passed:
function [U_bound] = applyNoSlipBoundary(U_field, t_n)
U_field(t_n + 1, 1, :, 🙂 = 0; % Top boundary
U_field(t_n + 1, end, :, 🙂 = 0; % Bottom boundary
U_field(t_n + 1, :, 1, 🙂 = 0; % Left boundary
U_field(t_n + 1, :, end, 🙂 = 0; % Right boundary
U_bound = U_field;
end
% Apply Neumann boundary conditions (zero gradient)
function [P_bound] = applyNeumannBoundary(P_temp, t_n)
P_temp(t_n + 1, 1, 🙂 = P_temp(t_n + 1, 2, :); % Top boundary
P_temp(t_n + 1, end, 🙂 = P_temp(t_n + 1, end-1, :); % Bottom boundary
P_temp(t_n + 1, :, 1) = P_temp(t_n + 1, :, 2); % Left boundary
P_temp(t_n + 1, :, end) = P_temp(t_n + 1, :, end-1); % Right boundary
P_bound = P_temp;
end
The error gotten is
Undefined function ‘NSsolverTest2’ for input arguments of type ‘gpuArray’.
The functions are in the same file, altough not nested.
Why do they work in one function but not the other? What is causing this I looked through the documentations but can’t find anything there. The only difference is where in the loop theyre being called. Anybody that could help me figure this out?
Thanks in advance! So Im using Paralell Computing Toolbox v. 23.2, on R2023b and having this problem when passing an array from one function to another, and can’t seem to understand what is causing this.
My nested loop looks like this, everything is initiated correctly in its running, except for when I add the nu_t term.
for t_n = 1:t-1
for i = 2:length(X) – 1
for j = 2:length(Y) – 1
nu_t = addTurbulentViscotiy(U_temp, C_s, Delta)
end
end
U_temp = applyNoSlipBoundary(U_temp, t_n);
P_temp = applyNeumannBoundary(P_temp, t_n);
end
And the function that doesnt work looks like this:
function [nu_t] = addTurbulentViscotiy(U, C_s, Delta)
% Computing the strain rate tensor components for S
S_11 = (U(t_n, i+1, j, 1) – U(t_n, i-1, j, 1)) / (2 * Delta);
S_22 = (U(t_n, i, j+1, 2) – U(t_n, i, j-1, 2)) / (2 * Delta);
S_12 = 0.5 * ((U(t_n, i, j+1, 1) – U(t_n, i, j-1, 1)) / (2 * Delta) …
+ (U(t_n, i+1, j, 2) – U(t_n, i-1, j, 2)) / (2 * Delta));
S_21 = S_12;
% Calculating the magnitude of S
S_mag = sqrt(S_11.^2 + S_22.^2 + 2* S12.^2);
% Calculating the turbulent Viscocity
nu_t = (C_s*Delta).^2*S_mag
end
However, these function work. Its the same values being passed:
function [U_bound] = applyNoSlipBoundary(U_field, t_n)
U_field(t_n + 1, 1, :, 🙂 = 0; % Top boundary
U_field(t_n + 1, end, :, 🙂 = 0; % Bottom boundary
U_field(t_n + 1, :, 1, 🙂 = 0; % Left boundary
U_field(t_n + 1, :, end, 🙂 = 0; % Right boundary
U_bound = U_field;
end
% Apply Neumann boundary conditions (zero gradient)
function [P_bound] = applyNeumannBoundary(P_temp, t_n)
P_temp(t_n + 1, 1, 🙂 = P_temp(t_n + 1, 2, :); % Top boundary
P_temp(t_n + 1, end, 🙂 = P_temp(t_n + 1, end-1, :); % Bottom boundary
P_temp(t_n + 1, :, 1) = P_temp(t_n + 1, :, 2); % Left boundary
P_temp(t_n + 1, :, end) = P_temp(t_n + 1, :, end-1); % Right boundary
P_bound = P_temp;
end
The error gotten is
Undefined function ‘NSsolverTest2’ for input arguments of type ‘gpuArray’.
The functions are in the same file, altough not nested.
Why do they work in one function but not the other? What is causing this I looked through the documentations but can’t find anything there. The only difference is where in the loop theyre being called. Anybody that could help me figure this out?
Thanks in advance! parallel computing toolbox, gpu, arrays, function, embedded matlab function MATLAB Answers — New Questions
plot of riemann surface
I am plotting the Riemann surface and I do expect symmetry with respect to the x and y axis, but it seems that I get only the one half of the surface
those lines give the surface
w01 = sqrt(r).*exp(1i*theta/2); % first branch
w02 = sqrt(r).*exp(1i*(theta+2*pi)/2); % second branch
(with z = re + 1j*im;
theta = angle(z); % atan2(imag(z), real(z));
r = 2*abs(z);)I am plotting the Riemann surface and I do expect symmetry with respect to the x and y axis, but it seems that I get only the one half of the surface
those lines give the surface
w01 = sqrt(r).*exp(1i*theta/2); % first branch
w02 = sqrt(r).*exp(1i*(theta+2*pi)/2); % second branch
(with z = re + 1j*im;
theta = angle(z); % atan2(imag(z), real(z));
r = 2*abs(z);) I am plotting the Riemann surface and I do expect symmetry with respect to the x and y axis, but it seems that I get only the one half of the surface
those lines give the surface
w01 = sqrt(r).*exp(1i*theta/2); % first branch
w02 = sqrt(r).*exp(1i*(theta+2*pi)/2); % second branch
(with z = re + 1j*im;
theta = angle(z); % atan2(imag(z), real(z));
r = 2*abs(z);) riemann surface, plot, symmetry MATLAB Answers — New Questions
Ayers ‘Differential Equations Problem 4
Could you help me do this in MatLab , I have figured it out with paper and pencil, Ayer’s Schaums Outline Problem 4 , I’ve tried putting the differential into dsolve but cannot prove the primitive or find the solution Ayers gets.
4. Show that (y-C)^2=Cx is the primitive of the differential equation 4x (dy/dx)^2+2x dy/dx-y = 0 and find the equations of the integral curves through the points x=1 y=2
Here 2(y-C)*dy/dx = C and dy/dx=C/(2(y-C))
Then (4xC^2)/(4(y-C)^2) + (2xC)/(2(y-C)) – y = (C^2x+Cx(y-C)-y(y-C)^2)/(y-C)^2 = (y[Cx-(y-C)^2])/(y-C)^2 = 0
When x=1,y=2: (2-C)^2 = C and C = 1,4
The equations of the integral curves through (1,2) are (y-1)^2 = x and (y-4)^2 = 4x
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%% My own commentary
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The primitive is given as (y-C)^2=Cx
derivative of (y-C)^2 = -2C+2y = 2(y-C), derivative of Cx = C*1 =C
He puts the derivative of the unknown function which is dy/dx, as a factor in the primitive
derivative of (y-C)^2 * dy/dx = derivative of Cx thus: 2(y-C) * dy/dx = C
Now solve for dy/dx : divide each side by 2(y-C) which cancels 2(y-C) on the right :
dy/dx = C/2(y-C) then (dy/dx)^2 = C^2/(2(y-C))^2 = C^2/(4( y-C)^2)
Now plug the coordinates x=1 y=2 into (y-C)^2 = Cx : (2-C)^2 = C*1 which is
(2-C)^2 = C , (2-C)(2-C)=C : 4-2C-2C+C^2 = C : 4-4C+C^2=C :
C^2-4C+4 =C : -4C and 4 to the right C^2=C+4C-4 : C to the left
C^2-C=4C-4, 4C to the left C^2-4C-C=-4 collect common terms left C^2-5C = -4
which is quadratic, complete the square :
C^2-5C+2.5^2 = -4+2.5^2
(C-2.5)^2 = -4+6.25
(C-2.5)^2 = 2.25 , square root each side
C-2.5 = + – SquareRoot(2.25)
C = +1.5+2.5 or -1.5+2.5
C = 4 or 1
The Differential equation was given as : 4x * C^2/4(y-C)^2 + 2x * C/2(y-C) -y = 0
combine 4x and 2x with respective numerators (4x*C^2)/(4(y-C)^2) + (2x*C)/2(y-C) – y = 0
cancel 4 and 2 1st and 2nd fractions left side xC^2/(y-C)^2 + xC/(y-C) – y = 0
common denominator (y-C)^2
C^2x/(y-C)^2 + Cx(y-C)/(y-C)^2 – y(y-C)^2/(y-C)^2 = 0
Combine numerators (C^2x+Cx(y-C)-y(y-C)^2)/(y-C)^2 = 0
The Cx(y-C) term in numerator can be factored Cxy-CxC
(C^2x+Cxy-CxC-y(y-C)^2)/(y-C)^2 = (C^2x+Cxy-C^2x-y(y-C)^2)/(y-C)^2 = 0 cancel C^2x
(Cxy-y(y-C)^2)/((y-C)^2) factor Cxy and y(y-C)^2 = (y(Cx-(y-C)^2)/(y-C)^2Could you help me do this in MatLab , I have figured it out with paper and pencil, Ayer’s Schaums Outline Problem 4 , I’ve tried putting the differential into dsolve but cannot prove the primitive or find the solution Ayers gets.
4. Show that (y-C)^2=Cx is the primitive of the differential equation 4x (dy/dx)^2+2x dy/dx-y = 0 and find the equations of the integral curves through the points x=1 y=2
Here 2(y-C)*dy/dx = C and dy/dx=C/(2(y-C))
Then (4xC^2)/(4(y-C)^2) + (2xC)/(2(y-C)) – y = (C^2x+Cx(y-C)-y(y-C)^2)/(y-C)^2 = (y[Cx-(y-C)^2])/(y-C)^2 = 0
When x=1,y=2: (2-C)^2 = C and C = 1,4
The equations of the integral curves through (1,2) are (y-1)^2 = x and (y-4)^2 = 4x
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%% My own commentary
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The primitive is given as (y-C)^2=Cx
derivative of (y-C)^2 = -2C+2y = 2(y-C), derivative of Cx = C*1 =C
He puts the derivative of the unknown function which is dy/dx, as a factor in the primitive
derivative of (y-C)^2 * dy/dx = derivative of Cx thus: 2(y-C) * dy/dx = C
Now solve for dy/dx : divide each side by 2(y-C) which cancels 2(y-C) on the right :
dy/dx = C/2(y-C) then (dy/dx)^2 = C^2/(2(y-C))^2 = C^2/(4( y-C)^2)
Now plug the coordinates x=1 y=2 into (y-C)^2 = Cx : (2-C)^2 = C*1 which is
(2-C)^2 = C , (2-C)(2-C)=C : 4-2C-2C+C^2 = C : 4-4C+C^2=C :
C^2-4C+4 =C : -4C and 4 to the right C^2=C+4C-4 : C to the left
C^2-C=4C-4, 4C to the left C^2-4C-C=-4 collect common terms left C^2-5C = -4
which is quadratic, complete the square :
C^2-5C+2.5^2 = -4+2.5^2
(C-2.5)^2 = -4+6.25
(C-2.5)^2 = 2.25 , square root each side
C-2.5 = + – SquareRoot(2.25)
C = +1.5+2.5 or -1.5+2.5
C = 4 or 1
The Differential equation was given as : 4x * C^2/4(y-C)^2 + 2x * C/2(y-C) -y = 0
combine 4x and 2x with respective numerators (4x*C^2)/(4(y-C)^2) + (2x*C)/2(y-C) – y = 0
cancel 4 and 2 1st and 2nd fractions left side xC^2/(y-C)^2 + xC/(y-C) – y = 0
common denominator (y-C)^2
C^2x/(y-C)^2 + Cx(y-C)/(y-C)^2 – y(y-C)^2/(y-C)^2 = 0
Combine numerators (C^2x+Cx(y-C)-y(y-C)^2)/(y-C)^2 = 0
The Cx(y-C) term in numerator can be factored Cxy-CxC
(C^2x+Cxy-CxC-y(y-C)^2)/(y-C)^2 = (C^2x+Cxy-C^2x-y(y-C)^2)/(y-C)^2 = 0 cancel C^2x
(Cxy-y(y-C)^2)/((y-C)^2) factor Cxy and y(y-C)^2 = (y(Cx-(y-C)^2)/(y-C)^2 Could you help me do this in MatLab , I have figured it out with paper and pencil, Ayer’s Schaums Outline Problem 4 , I’ve tried putting the differential into dsolve but cannot prove the primitive or find the solution Ayers gets.
4. Show that (y-C)^2=Cx is the primitive of the differential equation 4x (dy/dx)^2+2x dy/dx-y = 0 and find the equations of the integral curves through the points x=1 y=2
Here 2(y-C)*dy/dx = C and dy/dx=C/(2(y-C))
Then (4xC^2)/(4(y-C)^2) + (2xC)/(2(y-C)) – y = (C^2x+Cx(y-C)-y(y-C)^2)/(y-C)^2 = (y[Cx-(y-C)^2])/(y-C)^2 = 0
When x=1,y=2: (2-C)^2 = C and C = 1,4
The equations of the integral curves through (1,2) are (y-1)^2 = x and (y-4)^2 = 4x
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%% My own commentary
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The primitive is given as (y-C)^2=Cx
derivative of (y-C)^2 = -2C+2y = 2(y-C), derivative of Cx = C*1 =C
He puts the derivative of the unknown function which is dy/dx, as a factor in the primitive
derivative of (y-C)^2 * dy/dx = derivative of Cx thus: 2(y-C) * dy/dx = C
Now solve for dy/dx : divide each side by 2(y-C) which cancels 2(y-C) on the right :
dy/dx = C/2(y-C) then (dy/dx)^2 = C^2/(2(y-C))^2 = C^2/(4( y-C)^2)
Now plug the coordinates x=1 y=2 into (y-C)^2 = Cx : (2-C)^2 = C*1 which is
(2-C)^2 = C , (2-C)(2-C)=C : 4-2C-2C+C^2 = C : 4-4C+C^2=C :
C^2-4C+4 =C : -4C and 4 to the right C^2=C+4C-4 : C to the left
C^2-C=4C-4, 4C to the left C^2-4C-C=-4 collect common terms left C^2-5C = -4
which is quadratic, complete the square :
C^2-5C+2.5^2 = -4+2.5^2
(C-2.5)^2 = -4+6.25
(C-2.5)^2 = 2.25 , square root each side
C-2.5 = + – SquareRoot(2.25)
C = +1.5+2.5 or -1.5+2.5
C = 4 or 1
The Differential equation was given as : 4x * C^2/4(y-C)^2 + 2x * C/2(y-C) -y = 0
combine 4x and 2x with respective numerators (4x*C^2)/(4(y-C)^2) + (2x*C)/2(y-C) – y = 0
cancel 4 and 2 1st and 2nd fractions left side xC^2/(y-C)^2 + xC/(y-C) – y = 0
common denominator (y-C)^2
C^2x/(y-C)^2 + Cx(y-C)/(y-C)^2 – y(y-C)^2/(y-C)^2 = 0
Combine numerators (C^2x+Cx(y-C)-y(y-C)^2)/(y-C)^2 = 0
The Cx(y-C) term in numerator can be factored Cxy-CxC
(C^2x+Cxy-CxC-y(y-C)^2)/(y-C)^2 = (C^2x+Cxy-C^2x-y(y-C)^2)/(y-C)^2 = 0 cancel C^2x
(Cxy-y(y-C)^2)/((y-C)^2) factor Cxy and y(y-C)^2 = (y(Cx-(y-C)^2)/(y-C)^2 ayers problem4 MATLAB Answers — New Questions
Google Earth Path Matlab
***This is the code. I wrote the parameters as a matrix, but when I upload it to google earth, a straight line appears. Normally it should rise. How can I solve this?
% Assuming stage 1 descent
% latitude
lat_descent1 = [
6
6
6
6
6.00001697808819
6.00001537579460
6.00000819533646
5.99997308778827
5.99993743652977
5.99986643730126
5.99977748747485
5.99967127631096
5.99954042243161
5.99938373713836
5.99919807860050
5.99897890584989
5.99872744189481
5.99843907725965
5.99811116804459
5.99774107034979
5.99730870592835
5.99684630397024
5.99633378813147
5.99576851607858
5.99514784547808
5.99446913399650
5.99372973930037
5.99292701905621
5.99205833093055
5.99112103258991
5.99011248170083
5.98905626627310
5.98789786285266
5.98666076970945
5.98534241293727
5.98394021862993
5.98245161288123
5.98087402178496
5.97920487143495
5.97744158792498
5.97558159734886
5.97373789310211
5.97167615820284
5.96950923763312
5.96723444408868
5.96484909026525
5.96235048885854
5.95973595256429
5.95700279407821
5.95414832609604
5.95116986131349
5.94806471242629
5.94492652765109
5.94155625671142
5.93804789054305
5.93439800833756
5.93060313256953
5.92665972899659
5.92256420665941
5.91831291788169
5.91390215827018
5.90932816671465
5.90454689863020
5.89963276058423
5.89454094174210
5.88926679244302
5.88380555315758
5.87815235448773
5.87230221716680
5.86625005205950
5.85999066016193
5.85351873260155
5.84682885063721
5.84063471984696
5.83349204278013
5.82611432628191
5.81849567515460
5.81063008147944
5.80251142461667
5.79413347120547
5.78548987516401
5.77657417768940
5.76737980725774
5.75790007962409
5.74812819782249
5.73805725216592
5.72768022024636
5.71698996693473
5.70597924438095
5.69464069201386
5.68296683654132
5.67095009195012
5.65858275950603
5.63984071719353
5.62678297994443
5.61342475612478
5.59976883512710
];
% longitude
lon_descent1 = [
50
50
50
50
50.0000021566457
50.0000019531139
50.0000010410146
49.9999964117640
49.9999918695666
49.9999825682685
49.9999707308957
49.9999565505254
49.9999389636032
49.9999177405055
49.9998924374550
49.9998626677352
49.9998280196824
49.9997880927597
49.9997424870414
49.9996908026018
49.9996342138181
49.9995691475952
49.9994967676766
49.9994166653836
49.9993284320374
49.9992316589594
49.9991259374707
49.9990108588927
49.9988860145467
49.9987509957538
49.9986053938355
49.9984565540059
49.9982885556130
49.9981087445194
49.9979167115520
49.9977120475376
49.9974943433032
49.9972631896756
49.9970181774818
49.9967588975486
49.9964849407028
49.9962155183357
49.9959108455819
49.9955901010110
49.9952528503718
49.9948986594133
49.9945270938844
49.9941377195339
49.9937301021107
49.9933038073638
49.9928584010420
49.9923934488942
49.9919216815966
49.9914157329126
49.9908883894136
49.9903390972186
49.9897672931963
49.9891724049652
49.9885538508935
49.9879110400996
49.9872433724513
49.9865502385665
49.9858240078422
49.9850777367560
49.9843036741592
49.9835010702375
49.9826691576180
49.9818071513690
49.9809142490003
49.9799896304630
49.9790324581497
49.9780418768942
49.9770170139718
49.9760604246596
49.9749642399754
49.9738309877956
49.9726597115747
49.9714494361610
49.9701991677962
49.9689078941154
49.9675745841473
49.9661981883141
49.9647776384314
49.9633118477084
49.9617997107476
49.9602401035451
49.9586318834906
49.9569738893672
49.9552649413515
49.9535038410134
49.9516893713166
49.9498202966180
49.9478953626683
49.9449683556856
49.9429332310077
49.9408499214220
49.9387188083541 ];
%Altitude
alt_descent1 = [
0
0.00396034764708020
0.0159241062374349
0.0360139180911574
0.0643925531312561
0.101090025254052
0.146254621966364
0.199759321470992
0.262188654018246
0.333151727883887
0.412853926070056
0.501525064427369
0.599121879851737
0.705703840632850
0.821315247307211
0.945973591174152
1.07974164112784
1.22261981231441
1.37463464222947
1.53581266836773
1.70379693415271
1.88328144395837
2.07186615782666
2.26954214971465
2.47630049357667
2.69213226336979
2.91702853304741
3.15098037656844
3.39397886788720
3.64601508095893
3.90708008973979
4.16754940449846
4.44644695869465
4.73417896234969
5.03071308064045
5.33601697874747
5.65005832184943
5.97280477512322
6.30422400374846
6.64428367290384
6.99295144776897
7.33220089763199
7.69810250038245
8.07265778268538
8.45585577107613
8.84768549209184
9.24813597226694
9.65719623813857
10.0748553162439
10.5011022331173
10.9359260152960
11.3793156893153
11.8224693239281
12.2834346251830
12.7533654041881
13.2323454098369
13.7204657152670
14.2178247178590
14.7245281392416
15.2406890252851
15.7664277461090
16.3018719960746
16.8532537211540
17.4086308385522
17.9742828265944
18.5504000071342
19.1371827036437
19.7348412412148
20.3435959465605
20.9636771480064
21.5953251755045
22.2387903606195
22.8943330365400
23.5619369419837
24.2422603239993
24.9353284452327
25.6414037542454
26.3607557786936
27.0936611253301
27.8404034800042
28.6012736076609
29.3765693523374
30.1665956371744
30.9716644643995
31.7920949153431
32.6282131504277
33.4803524091722
34.3488530101931
35.2340623511991
36.1363349089997
37.0560322394967
37.9935229776866
38.9491828376649
40.3465964075785
41.3373209459624
42.3421554417855
43.3607296633954
];
kmlStr = ge_plot3(lon_descent1, lat_descent1, alt_descent1,’lineColor’, ‘FFFF0000′,’lineWidth’, 5,’name’, ‘Stage 1 Descent’);
ge_output(‘rocket_trajectory13.kml’, kmlStr);***This is the code. I wrote the parameters as a matrix, but when I upload it to google earth, a straight line appears. Normally it should rise. How can I solve this?
% Assuming stage 1 descent
% latitude
lat_descent1 = [
6
6
6
6
6.00001697808819
6.00001537579460
6.00000819533646
5.99997308778827
5.99993743652977
5.99986643730126
5.99977748747485
5.99967127631096
5.99954042243161
5.99938373713836
5.99919807860050
5.99897890584989
5.99872744189481
5.99843907725965
5.99811116804459
5.99774107034979
5.99730870592835
5.99684630397024
5.99633378813147
5.99576851607858
5.99514784547808
5.99446913399650
5.99372973930037
5.99292701905621
5.99205833093055
5.99112103258991
5.99011248170083
5.98905626627310
5.98789786285266
5.98666076970945
5.98534241293727
5.98394021862993
5.98245161288123
5.98087402178496
5.97920487143495
5.97744158792498
5.97558159734886
5.97373789310211
5.97167615820284
5.96950923763312
5.96723444408868
5.96484909026525
5.96235048885854
5.95973595256429
5.95700279407821
5.95414832609604
5.95116986131349
5.94806471242629
5.94492652765109
5.94155625671142
5.93804789054305
5.93439800833756
5.93060313256953
5.92665972899659
5.92256420665941
5.91831291788169
5.91390215827018
5.90932816671465
5.90454689863020
5.89963276058423
5.89454094174210
5.88926679244302
5.88380555315758
5.87815235448773
5.87230221716680
5.86625005205950
5.85999066016193
5.85351873260155
5.84682885063721
5.84063471984696
5.83349204278013
5.82611432628191
5.81849567515460
5.81063008147944
5.80251142461667
5.79413347120547
5.78548987516401
5.77657417768940
5.76737980725774
5.75790007962409
5.74812819782249
5.73805725216592
5.72768022024636
5.71698996693473
5.70597924438095
5.69464069201386
5.68296683654132
5.67095009195012
5.65858275950603
5.63984071719353
5.62678297994443
5.61342475612478
5.59976883512710
];
% longitude
lon_descent1 = [
50
50
50
50
50.0000021566457
50.0000019531139
50.0000010410146
49.9999964117640
49.9999918695666
49.9999825682685
49.9999707308957
49.9999565505254
49.9999389636032
49.9999177405055
49.9998924374550
49.9998626677352
49.9998280196824
49.9997880927597
49.9997424870414
49.9996908026018
49.9996342138181
49.9995691475952
49.9994967676766
49.9994166653836
49.9993284320374
49.9992316589594
49.9991259374707
49.9990108588927
49.9988860145467
49.9987509957538
49.9986053938355
49.9984565540059
49.9982885556130
49.9981087445194
49.9979167115520
49.9977120475376
49.9974943433032
49.9972631896756
49.9970181774818
49.9967588975486
49.9964849407028
49.9962155183357
49.9959108455819
49.9955901010110
49.9952528503718
49.9948986594133
49.9945270938844
49.9941377195339
49.9937301021107
49.9933038073638
49.9928584010420
49.9923934488942
49.9919216815966
49.9914157329126
49.9908883894136
49.9903390972186
49.9897672931963
49.9891724049652
49.9885538508935
49.9879110400996
49.9872433724513
49.9865502385665
49.9858240078422
49.9850777367560
49.9843036741592
49.9835010702375
49.9826691576180
49.9818071513690
49.9809142490003
49.9799896304630
49.9790324581497
49.9780418768942
49.9770170139718
49.9760604246596
49.9749642399754
49.9738309877956
49.9726597115747
49.9714494361610
49.9701991677962
49.9689078941154
49.9675745841473
49.9661981883141
49.9647776384314
49.9633118477084
49.9617997107476
49.9602401035451
49.9586318834906
49.9569738893672
49.9552649413515
49.9535038410134
49.9516893713166
49.9498202966180
49.9478953626683
49.9449683556856
49.9429332310077
49.9408499214220
49.9387188083541 ];
%Altitude
alt_descent1 = [
0
0.00396034764708020
0.0159241062374349
0.0360139180911574
0.0643925531312561
0.101090025254052
0.146254621966364
0.199759321470992
0.262188654018246
0.333151727883887
0.412853926070056
0.501525064427369
0.599121879851737
0.705703840632850
0.821315247307211
0.945973591174152
1.07974164112784
1.22261981231441
1.37463464222947
1.53581266836773
1.70379693415271
1.88328144395837
2.07186615782666
2.26954214971465
2.47630049357667
2.69213226336979
2.91702853304741
3.15098037656844
3.39397886788720
3.64601508095893
3.90708008973979
4.16754940449846
4.44644695869465
4.73417896234969
5.03071308064045
5.33601697874747
5.65005832184943
5.97280477512322
6.30422400374846
6.64428367290384
6.99295144776897
7.33220089763199
7.69810250038245
8.07265778268538
8.45585577107613
8.84768549209184
9.24813597226694
9.65719623813857
10.0748553162439
10.5011022331173
10.9359260152960
11.3793156893153
11.8224693239281
12.2834346251830
12.7533654041881
13.2323454098369
13.7204657152670
14.2178247178590
14.7245281392416
15.2406890252851
15.7664277461090
16.3018719960746
16.8532537211540
17.4086308385522
17.9742828265944
18.5504000071342
19.1371827036437
19.7348412412148
20.3435959465605
20.9636771480064
21.5953251755045
22.2387903606195
22.8943330365400
23.5619369419837
24.2422603239993
24.9353284452327
25.6414037542454
26.3607557786936
27.0936611253301
27.8404034800042
28.6012736076609
29.3765693523374
30.1665956371744
30.9716644643995
31.7920949153431
32.6282131504277
33.4803524091722
34.3488530101931
35.2340623511991
36.1363349089997
37.0560322394967
37.9935229776866
38.9491828376649
40.3465964075785
41.3373209459624
42.3421554417855
43.3607296633954
];
kmlStr = ge_plot3(lon_descent1, lat_descent1, alt_descent1,’lineColor’, ‘FFFF0000′,’lineWidth’, 5,’name’, ‘Stage 1 Descent’);
ge_output(‘rocket_trajectory13.kml’, kmlStr); ***This is the code. I wrote the parameters as a matrix, but when I upload it to google earth, a straight line appears. Normally it should rise. How can I solve this?
% Assuming stage 1 descent
% latitude
lat_descent1 = [
6
6
6
6
6.00001697808819
6.00001537579460
6.00000819533646
5.99997308778827
5.99993743652977
5.99986643730126
5.99977748747485
5.99967127631096
5.99954042243161
5.99938373713836
5.99919807860050
5.99897890584989
5.99872744189481
5.99843907725965
5.99811116804459
5.99774107034979
5.99730870592835
5.99684630397024
5.99633378813147
5.99576851607858
5.99514784547808
5.99446913399650
5.99372973930037
5.99292701905621
5.99205833093055
5.99112103258991
5.99011248170083
5.98905626627310
5.98789786285266
5.98666076970945
5.98534241293727
5.98394021862993
5.98245161288123
5.98087402178496
5.97920487143495
5.97744158792498
5.97558159734886
5.97373789310211
5.97167615820284
5.96950923763312
5.96723444408868
5.96484909026525
5.96235048885854
5.95973595256429
5.95700279407821
5.95414832609604
5.95116986131349
5.94806471242629
5.94492652765109
5.94155625671142
5.93804789054305
5.93439800833756
5.93060313256953
5.92665972899659
5.92256420665941
5.91831291788169
5.91390215827018
5.90932816671465
5.90454689863020
5.89963276058423
5.89454094174210
5.88926679244302
5.88380555315758
5.87815235448773
5.87230221716680
5.86625005205950
5.85999066016193
5.85351873260155
5.84682885063721
5.84063471984696
5.83349204278013
5.82611432628191
5.81849567515460
5.81063008147944
5.80251142461667
5.79413347120547
5.78548987516401
5.77657417768940
5.76737980725774
5.75790007962409
5.74812819782249
5.73805725216592
5.72768022024636
5.71698996693473
5.70597924438095
5.69464069201386
5.68296683654132
5.67095009195012
5.65858275950603
5.63984071719353
5.62678297994443
5.61342475612478
5.59976883512710
];
% longitude
lon_descent1 = [
50
50
50
50
50.0000021566457
50.0000019531139
50.0000010410146
49.9999964117640
49.9999918695666
49.9999825682685
49.9999707308957
49.9999565505254
49.9999389636032
49.9999177405055
49.9998924374550
49.9998626677352
49.9998280196824
49.9997880927597
49.9997424870414
49.9996908026018
49.9996342138181
49.9995691475952
49.9994967676766
49.9994166653836
49.9993284320374
49.9992316589594
49.9991259374707
49.9990108588927
49.9988860145467
49.9987509957538
49.9986053938355
49.9984565540059
49.9982885556130
49.9981087445194
49.9979167115520
49.9977120475376
49.9974943433032
49.9972631896756
49.9970181774818
49.9967588975486
49.9964849407028
49.9962155183357
49.9959108455819
49.9955901010110
49.9952528503718
49.9948986594133
49.9945270938844
49.9941377195339
49.9937301021107
49.9933038073638
49.9928584010420
49.9923934488942
49.9919216815966
49.9914157329126
49.9908883894136
49.9903390972186
49.9897672931963
49.9891724049652
49.9885538508935
49.9879110400996
49.9872433724513
49.9865502385665
49.9858240078422
49.9850777367560
49.9843036741592
49.9835010702375
49.9826691576180
49.9818071513690
49.9809142490003
49.9799896304630
49.9790324581497
49.9780418768942
49.9770170139718
49.9760604246596
49.9749642399754
49.9738309877956
49.9726597115747
49.9714494361610
49.9701991677962
49.9689078941154
49.9675745841473
49.9661981883141
49.9647776384314
49.9633118477084
49.9617997107476
49.9602401035451
49.9586318834906
49.9569738893672
49.9552649413515
49.9535038410134
49.9516893713166
49.9498202966180
49.9478953626683
49.9449683556856
49.9429332310077
49.9408499214220
49.9387188083541 ];
%Altitude
alt_descent1 = [
0
0.00396034764708020
0.0159241062374349
0.0360139180911574
0.0643925531312561
0.101090025254052
0.146254621966364
0.199759321470992
0.262188654018246
0.333151727883887
0.412853926070056
0.501525064427369
0.599121879851737
0.705703840632850
0.821315247307211
0.945973591174152
1.07974164112784
1.22261981231441
1.37463464222947
1.53581266836773
1.70379693415271
1.88328144395837
2.07186615782666
2.26954214971465
2.47630049357667
2.69213226336979
2.91702853304741
3.15098037656844
3.39397886788720
3.64601508095893
3.90708008973979
4.16754940449846
4.44644695869465
4.73417896234969
5.03071308064045
5.33601697874747
5.65005832184943
5.97280477512322
6.30422400374846
6.64428367290384
6.99295144776897
7.33220089763199
7.69810250038245
8.07265778268538
8.45585577107613
8.84768549209184
9.24813597226694
9.65719623813857
10.0748553162439
10.5011022331173
10.9359260152960
11.3793156893153
11.8224693239281
12.2834346251830
12.7533654041881
13.2323454098369
13.7204657152670
14.2178247178590
14.7245281392416
15.2406890252851
15.7664277461090
16.3018719960746
16.8532537211540
17.4086308385522
17.9742828265944
18.5504000071342
19.1371827036437
19.7348412412148
20.3435959465605
20.9636771480064
21.5953251755045
22.2387903606195
22.8943330365400
23.5619369419837
24.2422603239993
24.9353284452327
25.6414037542454
26.3607557786936
27.0936611253301
27.8404034800042
28.6012736076609
29.3765693523374
30.1665956371744
30.9716644643995
31.7920949153431
32.6282131504277
33.4803524091722
34.3488530101931
35.2340623511991
36.1363349089997
37.0560322394967
37.9935229776866
38.9491828376649
40.3465964075785
41.3373209459624
42.3421554417855
43.3607296633954
];
kmlStr = ge_plot3(lon_descent1, lat_descent1, alt_descent1,’lineColor’, ‘FFFF0000′,’lineWidth’, 5,’name’, ‘Stage 1 Descent’);
ge_output(‘rocket_trajectory13.kml’, kmlStr); kml, kmz, google, earth, simulation, trajectory, path, aerospace MATLAB Answers — New Questions
Is Entity Gate needed?
Trying to simulate a certain section of block diagram, specifcally the feedback signal it passes through what looks to be a gate that only closes based on a certain trim condition of the aircraft. My question is would I need to use a Entity Gate Block to accurately simulate the block diagram or would it be more accurate to use a Switch Block? Thanks for any input/suggestion.Trying to simulate a certain section of block diagram, specifcally the feedback signal it passes through what looks to be a gate that only closes based on a certain trim condition of the aircraft. My question is would I need to use a Entity Gate Block to accurately simulate the block diagram or would it be more accurate to use a Switch Block? Thanks for any input/suggestion. Trying to simulate a certain section of block diagram, specifcally the feedback signal it passes through what looks to be a gate that only closes based on a certain trim condition of the aircraft. My question is would I need to use a Entity Gate Block to accurately simulate the block diagram or would it be more accurate to use a Switch Block? Thanks for any input/suggestion. switch, entity gate MATLAB Answers — New Questions
Custom colorbar labeling centered on colors
I have a figure for which I want to create custom colorbar labeling. Below is the sample code:
figure;
set(gca,’xtick’,[],’xticklabel’,[],’ytick’,[],’yticklabel’,[],’color’,’w’);
set(gcf,’units’,’pixel’,’position’,[70,70,600,600],’papersize’,[600,600],’color’,’w’);
ax=gca;
colormap(flipud(jet(11)));
caxis([0 11]);
hcb=colorbar(‘SouthOutside’);
axPos = ax.Position;
colorbarpos=hcb.Position;
hcb.Ruler.TickLabelRotation=0;
set(hcb,’YTick’,[0:1:11],’TickLength’,colorbarpos(4)/colorbarpos(3));
This gives a standard southoutside, reverse jet colorbar like below:
Instead, I’d like to create custom labels centered on each color…something like below:
Any ideas on how or if this can be done?I have a figure for which I want to create custom colorbar labeling. Below is the sample code:
figure;
set(gca,’xtick’,[],’xticklabel’,[],’ytick’,[],’yticklabel’,[],’color’,’w’);
set(gcf,’units’,’pixel’,’position’,[70,70,600,600],’papersize’,[600,600],’color’,’w’);
ax=gca;
colormap(flipud(jet(11)));
caxis([0 11]);
hcb=colorbar(‘SouthOutside’);
axPos = ax.Position;
colorbarpos=hcb.Position;
hcb.Ruler.TickLabelRotation=0;
set(hcb,’YTick’,[0:1:11],’TickLength’,colorbarpos(4)/colorbarpos(3));
This gives a standard southoutside, reverse jet colorbar like below:
Instead, I’d like to create custom labels centered on each color…something like below:
Any ideas on how or if this can be done? I have a figure for which I want to create custom colorbar labeling. Below is the sample code:
figure;
set(gca,’xtick’,[],’xticklabel’,[],’ytick’,[],’yticklabel’,[],’color’,’w’);
set(gcf,’units’,’pixel’,’position’,[70,70,600,600],’papersize’,[600,600],’color’,’w’);
ax=gca;
colormap(flipud(jet(11)));
caxis([0 11]);
hcb=colorbar(‘SouthOutside’);
axPos = ax.Position;
colorbarpos=hcb.Position;
hcb.Ruler.TickLabelRotation=0;
set(hcb,’YTick’,[0:1:11],’TickLength’,colorbarpos(4)/colorbarpos(3));
This gives a standard southoutside, reverse jet colorbar like below:
Instead, I’d like to create custom labels centered on each color…something like below:
Any ideas on how or if this can be done? colorbar, custom, colormap MATLAB Answers — New Questions
Population Growth Model Development
Research in cell and tissue engineering often involves growing cells in the lab in a dish. Imagine
having a single 120 cm2 dish that has been seeded with 1500 cells/cm2. The dish can only
sustain 9×107 cells. With this seeding of the dish, the maximum that the dish can sustain
(carrying capacity) is reached in about 20-25 days. The growth rate, r, of this cell type is known
to be 0.75 cells per day (𝑏𝑖𝑟𝑡ℎ 𝑟𝑎𝑡𝑒 ― 𝑑𝑒𝑎𝑡ℎ 𝑟𝑎𝑡𝑒).
This cell population follows a logistic population growth model:
𝑑𝑃(𝑡)
𝑑𝑡 = 𝑟𝑃(𝑡)(1 ― 𝑃(𝑡)/𝐾 ),
where P(t) is the size of the population at time, t, K is a constant corresponding to the
saturation level (carrying capacity) and r > 0 is the birth rate.
1. Write a MATLAB script for the numerical solution of this cell population problem
utilizing the Euler differential equation solver as demonstrated in class.Research in cell and tissue engineering often involves growing cells in the lab in a dish. Imagine
having a single 120 cm2 dish that has been seeded with 1500 cells/cm2. The dish can only
sustain 9×107 cells. With this seeding of the dish, the maximum that the dish can sustain
(carrying capacity) is reached in about 20-25 days. The growth rate, r, of this cell type is known
to be 0.75 cells per day (𝑏𝑖𝑟𝑡ℎ 𝑟𝑎𝑡𝑒 ― 𝑑𝑒𝑎𝑡ℎ 𝑟𝑎𝑡𝑒).
This cell population follows a logistic population growth model:
𝑑𝑃(𝑡)
𝑑𝑡 = 𝑟𝑃(𝑡)(1 ― 𝑃(𝑡)/𝐾 ),
where P(t) is the size of the population at time, t, K is a constant corresponding to the
saturation level (carrying capacity) and r > 0 is the birth rate.
1. Write a MATLAB script for the numerical solution of this cell population problem
utilizing the Euler differential equation solver as demonstrated in class. Research in cell and tissue engineering often involves growing cells in the lab in a dish. Imagine
having a single 120 cm2 dish that has been seeded with 1500 cells/cm2. The dish can only
sustain 9×107 cells. With this seeding of the dish, the maximum that the dish can sustain
(carrying capacity) is reached in about 20-25 days. The growth rate, r, of this cell type is known
to be 0.75 cells per day (𝑏𝑖𝑟𝑡ℎ 𝑟𝑎𝑡𝑒 ― 𝑑𝑒𝑎𝑡ℎ 𝑟𝑎𝑡𝑒).
This cell population follows a logistic population growth model:
𝑑𝑃(𝑡)
𝑑𝑡 = 𝑟𝑃(𝑡)(1 ― 𝑃(𝑡)/𝐾 ),
where P(t) is the size of the population at time, t, K is a constant corresponding to the
saturation level (carrying capacity) and r > 0 is the birth rate.
1. Write a MATLAB script for the numerical solution of this cell population problem
utilizing the Euler differential equation solver as demonstrated in class. matlab MATLAB Answers — New Questions
trainYOLOv4ObjectDetector>iParseInputsYOLOv4 The class names specified in the detector must match the class names in training data.
Hi.When I try to train a YOLOv4,he said that Error using trainYOLOv4ObjectDetector>iParseInputsYOLOv4
The class names specified in the detector must match the class names in training data.
But I checked the content of the label and it’s the same as the detector.And I didn’t have this problem when training FasterrCNN.Here is my code:
load(‘training_data.mat’)
>> trainingData = objectDetectorTrainingData(gTruth);
% Create a partition of the data into a 70/30 train/test split
cvp = cvpartition(size(trainingData, 1), ‘HoldOut’, 0.2);
% Get the indices of the training and testing sets
trainIdx = cvp.training;
testIdx = cvp.test;
% Extract the training and testing data using the indices
trainData = trainingData(trainIdx, :);
testData = trainingData(testIdx, :);
imdsTrain = imageDatastore(trainData.imageFilename);
imdsTest = imageDatastore(testData.imageFilename);
bldsTrain = boxLabelDatastore(trainData(:, 2:end));
bldsTest = boxLabelDatastore(testData(:, 2:end));
%imdsValidationData = imageDatastore(validationData.imageFilename);
%bldsValidationData = boxLabelDatastore(validationData(:, 2:end));
ds = combine(imdsTrain,bldsTrain);
inputSize = [2880 2880 3];
trainingDataForEstimation = transform(ds,@(data)preprocessData(data,inputSize));
numAnchors = 6;
[anchors, meanIoU] = estimateAnchorBoxes(trainingDataForEstimation,numAnchors);
area = anchors(:,1).*anchors(:,2);
[~,idx] = sort(area,"descend");
anchors = anchors(idx,:);
anchorBoxes = {anchors(1:3,:);anchors(4:6,:)};
classes =gTruth.LabelDefinitions.Name’;
detector = yolov4ObjectDetector("tiny-yolov4-coco",classes,anchorBoxes,InputSize=inputSize);
options = trainingOptions("sgdm", …
InitialLearnRate=0.001, …
MiniBatchSize=16,…
MaxEpochs=40, …
BatchNormalizationStatistics="moving",…
ResetInputNormalization=false,…
VerboseFrequency=30);
trainedDetector = trainYOLOv4ObjectDetector(ds,detector,options);
Error using trainYOLOv4ObjectDetector>iParseInputsYOLOv4
The class names specified in the detector must match the class names in training data.
Error in trainYOLOv4ObjectDetector (line 119)
[trainingData, params] = iParseInputsYOLOv4(trainingData,detector,options,mfilename,varargin{:});
detector.ClassNames
ans =
12×1 cell array
{‘lozenge’ }
{‘hexagon’ }
{‘trapezium’ }
{‘heart’ }
{‘semicircle’}
{‘pentagon’ }
{‘star’ }
{‘elliptical’}
{‘rectangle’ }
{‘square’ }
{‘triangle’ }
{‘circle’ }
gTruth.LabelDefinitions.Name
ans =
12×1 cell array
{‘lozenge’ }
{‘hexagon’ }
{‘trapezium’ }
{‘heart’ }
{‘semicircle’}
{‘pentagon’ }
{‘star’ }
{‘elliptical’}
{‘rectangle’ }
{‘square’ }
{‘triangle’ }
{‘circle’ }Hi.When I try to train a YOLOv4,he said that Error using trainYOLOv4ObjectDetector>iParseInputsYOLOv4
The class names specified in the detector must match the class names in training data.
But I checked the content of the label and it’s the same as the detector.And I didn’t have this problem when training FasterrCNN.Here is my code:
load(‘training_data.mat’)
>> trainingData = objectDetectorTrainingData(gTruth);
% Create a partition of the data into a 70/30 train/test split
cvp = cvpartition(size(trainingData, 1), ‘HoldOut’, 0.2);
% Get the indices of the training and testing sets
trainIdx = cvp.training;
testIdx = cvp.test;
% Extract the training and testing data using the indices
trainData = trainingData(trainIdx, :);
testData = trainingData(testIdx, :);
imdsTrain = imageDatastore(trainData.imageFilename);
imdsTest = imageDatastore(testData.imageFilename);
bldsTrain = boxLabelDatastore(trainData(:, 2:end));
bldsTest = boxLabelDatastore(testData(:, 2:end));
%imdsValidationData = imageDatastore(validationData.imageFilename);
%bldsValidationData = boxLabelDatastore(validationData(:, 2:end));
ds = combine(imdsTrain,bldsTrain);
inputSize = [2880 2880 3];
trainingDataForEstimation = transform(ds,@(data)preprocessData(data,inputSize));
numAnchors = 6;
[anchors, meanIoU] = estimateAnchorBoxes(trainingDataForEstimation,numAnchors);
area = anchors(:,1).*anchors(:,2);
[~,idx] = sort(area,"descend");
anchors = anchors(idx,:);
anchorBoxes = {anchors(1:3,:);anchors(4:6,:)};
classes =gTruth.LabelDefinitions.Name’;
detector = yolov4ObjectDetector("tiny-yolov4-coco",classes,anchorBoxes,InputSize=inputSize);
options = trainingOptions("sgdm", …
InitialLearnRate=0.001, …
MiniBatchSize=16,…
MaxEpochs=40, …
BatchNormalizationStatistics="moving",…
ResetInputNormalization=false,…
VerboseFrequency=30);
trainedDetector = trainYOLOv4ObjectDetector(ds,detector,options);
Error using trainYOLOv4ObjectDetector>iParseInputsYOLOv4
The class names specified in the detector must match the class names in training data.
Error in trainYOLOv4ObjectDetector (line 119)
[trainingData, params] = iParseInputsYOLOv4(trainingData,detector,options,mfilename,varargin{:});
detector.ClassNames
ans =
12×1 cell array
{‘lozenge’ }
{‘hexagon’ }
{‘trapezium’ }
{‘heart’ }
{‘semicircle’}
{‘pentagon’ }
{‘star’ }
{‘elliptical’}
{‘rectangle’ }
{‘square’ }
{‘triangle’ }
{‘circle’ }
gTruth.LabelDefinitions.Name
ans =
12×1 cell array
{‘lozenge’ }
{‘hexagon’ }
{‘trapezium’ }
{‘heart’ }
{‘semicircle’}
{‘pentagon’ }
{‘star’ }
{‘elliptical’}
{‘rectangle’ }
{‘square’ }
{‘triangle’ }
{‘circle’ } Hi.When I try to train a YOLOv4,he said that Error using trainYOLOv4ObjectDetector>iParseInputsYOLOv4
The class names specified in the detector must match the class names in training data.
But I checked the content of the label and it’s the same as the detector.And I didn’t have this problem when training FasterrCNN.Here is my code:
load(‘training_data.mat’)
>> trainingData = objectDetectorTrainingData(gTruth);
% Create a partition of the data into a 70/30 train/test split
cvp = cvpartition(size(trainingData, 1), ‘HoldOut’, 0.2);
% Get the indices of the training and testing sets
trainIdx = cvp.training;
testIdx = cvp.test;
% Extract the training and testing data using the indices
trainData = trainingData(trainIdx, :);
testData = trainingData(testIdx, :);
imdsTrain = imageDatastore(trainData.imageFilename);
imdsTest = imageDatastore(testData.imageFilename);
bldsTrain = boxLabelDatastore(trainData(:, 2:end));
bldsTest = boxLabelDatastore(testData(:, 2:end));
%imdsValidationData = imageDatastore(validationData.imageFilename);
%bldsValidationData = boxLabelDatastore(validationData(:, 2:end));
ds = combine(imdsTrain,bldsTrain);
inputSize = [2880 2880 3];
trainingDataForEstimation = transform(ds,@(data)preprocessData(data,inputSize));
numAnchors = 6;
[anchors, meanIoU] = estimateAnchorBoxes(trainingDataForEstimation,numAnchors);
area = anchors(:,1).*anchors(:,2);
[~,idx] = sort(area,"descend");
anchors = anchors(idx,:);
anchorBoxes = {anchors(1:3,:);anchors(4:6,:)};
classes =gTruth.LabelDefinitions.Name’;
detector = yolov4ObjectDetector("tiny-yolov4-coco",classes,anchorBoxes,InputSize=inputSize);
options = trainingOptions("sgdm", …
InitialLearnRate=0.001, …
MiniBatchSize=16,…
MaxEpochs=40, …
BatchNormalizationStatistics="moving",…
ResetInputNormalization=false,…
VerboseFrequency=30);
trainedDetector = trainYOLOv4ObjectDetector(ds,detector,options);
Error using trainYOLOv4ObjectDetector>iParseInputsYOLOv4
The class names specified in the detector must match the class names in training data.
Error in trainYOLOv4ObjectDetector (line 119)
[trainingData, params] = iParseInputsYOLOv4(trainingData,detector,options,mfilename,varargin{:});
detector.ClassNames
ans =
12×1 cell array
{‘lozenge’ }
{‘hexagon’ }
{‘trapezium’ }
{‘heart’ }
{‘semicircle’}
{‘pentagon’ }
{‘star’ }
{‘elliptical’}
{‘rectangle’ }
{‘square’ }
{‘triangle’ }
{‘circle’ }
gTruth.LabelDefinitions.Name
ans =
12×1 cell array
{‘lozenge’ }
{‘hexagon’ }
{‘trapezium’ }
{‘heart’ }
{‘semicircle’}
{‘pentagon’ }
{‘star’ }
{‘elliptical’}
{‘rectangle’ }
{‘square’ }
{‘triangle’ }
{‘circle’ } yolov4, classnames MATLAB Answers — New Questions
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I am working on a bit of code to generate a the probability of an item dropping. I have it set so that its going through an if loop chain, where if it isnt one thing it goes to the next thing and if it isnt that it goes to the next and so forth, but when I try to run it with a for loop to dictate how many times it needs to run it just gives all of the attempts to a singular output giving that one output a probability of 100% each run, which isnt whats supposed to happen.
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none = 0;
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This is the code I’ve been working on.I am working on a bit of code to generate a the probability of an item dropping. I have it set so that its going through an if loop chain, where if it isnt one thing it goes to the next thing and if it isnt that it goes to the next and so forth, but when I try to run it with a for loop to dictate how many times it needs to run it just gives all of the attempts to a singular output giving that one output a probability of 100% each run, which isnt whats supposed to happen.
numOfSims = 100;
potionOfStrength = 0;
braceletOfStrength = 0;
hammerOfThunder = 0;
none = 0;
X = rand;
X2 = rand;
X3 = rand;
for iteration_num = 1:numOfSims
if X<0.4
potionOfStrength = potionOfStrength +1;
else
if X2<0.2
braceletOfStrength = braceletOfStrength +1;
else
if X3<0.2
hammerOfThunder = hammerOfThunder +1;
else
none = none +1;
end
end
end
This is the code I’ve been working on. I am working on a bit of code to generate a the probability of an item dropping. I have it set so that its going through an if loop chain, where if it isnt one thing it goes to the next thing and if it isnt that it goes to the next and so forth, but when I try to run it with a for loop to dictate how many times it needs to run it just gives all of the attempts to a singular output giving that one output a probability of 100% each run, which isnt whats supposed to happen.
numOfSims = 100;
potionOfStrength = 0;
braceletOfStrength = 0;
hammerOfThunder = 0;
none = 0;
X = rand;
X2 = rand;
X3 = rand;
for iteration_num = 1:numOfSims
if X<0.4
potionOfStrength = potionOfStrength +1;
else
if X2<0.2
braceletOfStrength = braceletOfStrength +1;
else
if X3<0.2
hammerOfThunder = hammerOfThunder +1;
else
none = none +1;
end
end
end
This is the code I’ve been working on. probability, for loop, if statement MATLAB Answers — New Questions
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