Help generating a triangle using arrays, rotating triangle

The assignment asks me to generate a triangle: Triangle dimensions:

Side 1: y = 0 for x = 0 to 2

Side 2: x = 0 for y = 0 to 1

Hypotenuse: y = 1 - 0.5x for x = 0 to 2

generate a 1 x n array 'x' containing the values of 0 to 2 in steps of 0.1. Then, using (0.1), generate the corresponding array 'y'. This gives you the hypotenuse. I am then asked to generate a 2 x (n + 2) array 'Triangle' that contains the points necessary to plot the triangle, with x-coordinates in the first row and y-coordinates in the second row. There is more to the problem, but I am having trouble with this part and I believe once I get this down I will be able to complete the rest of the assignment (also I can't go on unless I have the first part completed). However, I have never had linear algebra before so I am a bit rusty when it comes to matrices and arrays, this is the code I have thus far:

 clc
clear
line = ['r', 'g', 'k', 'b']; %Line colors
x = linspace(0,2,0.1);
y = linspace(0,1,0.1);
Triangle = [x; y]
plot(Triangle(1,:), Triangle(2,:), 'bs-', 'LineWidth',2)
axis([0 2 0 2])
axis('square')
axis off
figure(1)
end

However, when I run the code it says I have an empty matrix, 2 by 0. If someone could give me some advice on what to do here and why my code isn't working I would greatly appreciate it. Thanks.

Answer 1

[Orion]

You misused linspace.

x = linspace(0,2,0.1);

will return an empty matrix, same for y. and then you concatenate two empty matrix., so you create a 2-by-0 Empty matrix

see the doc of linspace :

y = linspace(a,b,n) generates a row vector y of n points linearly spaced between and including a and b. For n = 1, linspace returns b.

in your code you want to create a vector starting at 0, ending at 2 and incrementing of 0.1 :

x = 0:0.1:2;

and for y

y = 1-0.5*x;

so your code should be more like :

clc
clear
line = ['r', 'g', 'k', 'b']; %Line colors
x = 0:0.1:2;
y = 1-0.5*x;
plot(x,y, 'bs-', 'LineWidth',2)
axis([0 2 0 2])
axis('square')
axis off

[Colin]

Thank you very much, but this is giving me only a line segment, any advice on how I would plot the entire triangle?

[Orion]

maybe this

clc
clear
line = ['r', 'g', 'k', 'b']; %Line colors
x = 0:0.1:2;
y = 1-0.5*x;
plot(x,y, 'bs-', 'LineWidth',2)
axis([0 2 0 2])
axis('square')
axis off;

hold on;
% plot the x axis.
plot(x,zeros(size(x)), 'bs-', 'LineWidth',2);

% plot the y axis
plot(zeros(size(x)),y, 'bs-', 'LineWidth',2);

In this two extra plots, i used the same number of elements as in x.

Then, it's up to you to draw the number of point you need/want

[Colin]

Thanks again, I have one last question: the code you just posted seems to me like it may be hard to rotate, since ultimately this assignment requires me to rotate the triangle. Therefore, I was thinking it may be easier just to do something like:

if 
clc
clear
line = ['r', 'g', 'k', 'b']; %Line colors
Triangle = [0 2 0 0;0 0 1 0];
plot(Triangle(1,:, Triangle(2,:, 'b', 'LineWidth',2)
axis([0 2 0 2])
title('Basic Triangle')
axis('square')
axis off
figure(1)
end

end

Would you agree?

Answer 2

[Youssef Khmou]

Colin, you can try using the function line as the following :

 x1=[0 2];
 y1=[0 0];

 x2=[0 0];
y2=[0 1];
x3=[0 2];
y3=[1 0];

 figure;
line(x1,y1); hold on
line(x2,y2);
line(x3,y3);
hold off
axis([-2 2 -2 2])

[Colin]

Thank you, but in the assignment it says I need an array Triangle, could I define Triangle using the variables you listed?

[Youssef Khmou]

you mean you need to get a couple (x,y) that gives an array? it is possible to concatenate the answer above :

 X=[0 2 0 0 0 2];
 Y=[0 0 0 1 1 0];
 figure; plot(X,Y)
 axis([-2 2 -2 2])

[Colin]

I am supposed to generate a 1xn array 'x' containing the values of x=0 to 2 in steps of 0.1. Then I am to generate another array 'y' using the equation for the hypotenuse. They say that these two arrays will give me the hypotenuse of my triangle. Then, I am to generate a 2x(n+2) array called 'Triangle' that contains the other points necessary to plot the triangle. After I do all of that I am to find the centroid (which won't be a problem) and rotate the triangle in a full circle. Ultimately the output should look something like what this guy has done: http://www.youtube.com/watch?v=OAMU0kT7fBc

Answer 3

[Colin]

This is my updated code:

clc
clear
line = ['r', 'g', 'k', 'b']; %Line colors
x = 0:0.1:2;
y = 1-0.5*x;
Triangle = [0 2 0 x 0 0; 0 0 1 y 0 1]
plot(Triangle(1,:),Triangle(2,:), 'bs-','LineWidth',2)
axis([0 2 0 2])
axis('square')
axis off
figure(1)

i 
figure(2)
%Triangle is assumed to be a planar lamina
f = @(x) 1 - (x/2);
mass = integral(f, 0, 2) %Multiplied by rho, measured in mass per unit area
g = @(x) x.*(1-(x/2));
y_moment = integral(g, 0, 2)
x_centroid = y_moment/mass
end

end

Answer 4

[Orion]

I haven't done things like that for a long time. always fun.

see the attached file and adapt it.

[Colin]

Thanks man, here is the code I have so far, its almost done but I cant get it to rotate in a full circle yet, anyway thanks for the help

if clc
clear
line = ['r', 'g', 'k', 'b']; %Line colors
x = 0:0.1:2;  %X coordinates
y = 1-0.5*x;  %Y coordinates
Triangle = [0 2 0 x 0 0; 0 0 1 y 0 1];  %Triangle x, y coordinates
plot(Triangle(1,:),Triangle(2,:), 'bs-','LineWidth',2) %Plot the triangle
axis([0 2 0 2])
axis('square')
axis off
figure(1)
%Triangle is assumed to be a planar lamina
f = @(x) 1 - (x/2); %Function representing the hypotenuse of the triangle
mass = integral(f, 0, 2); %Multiplied by rho, measured in mass per unit area
g = @(x) x.*(1-(x/2)); %Function to determine the first moment about y
y_moment = integral(g, 0, 2); %Multiplied by rho
x_centroid = y_moment/mass;   %Compute the x coordinate of the centroid
fprintf('The x component of the centroid is %0.2f \n',x_centroid)  %Output centroid
y = 0:0.05:1;   %Y coordinates
x = 2-(2*y);    %X coordinates
Triangle_1 = [0 2 0 x 0 0; 0 0 1 y 0 1];  %Triangle x,y coordinates, same triangle as above
plot(Triangle_1(1,:),Triangle_1(2,:),'bs-','LineWidth',2)  %Plot the triangle
axis([0 2 0 2])
axis('square')
axis off
figure(1)
h = @(x) ((1-(x/2))/2).*(1-(x/2))
x_moment = integral(h,0,2)
y_centroid = x_moment/mass;
fprintf('The y component of the centroid is %0.2f \n',y_centroid)
%Using the first triangle for rotation
x = 0:0.1:2;  %X coordinates
y = 1-0.5*x;  %Y coordinates
Triangle = [0 2 0 x 0 0; 0 0 1 y 0 1];  %Triangle x, y coordinates
plot(Triangle(1,:),Triangle(2,:), 'bs-','LineWidth',2) %Plot the triangle
axis([0 2 0 2])
axis('square')
axis off
figure(1)
figure(2) %New window
for k=0:64
  phi = k*pi/64;
  A = [cos(phi) -sin(phi); sin(phi) cos(phi)];
  TriangleRotated = A*Triangle;
  plot(TriangleRotated(1,:),TriangleRotated(2,:),line(mod(k,4)+1));
  if k==0
      axis([0 2 0 2]);
      axis('square');
      axis off
      hold on
  end;
  figure(2);
end
hold off;
figure(2);

end

[Orion]

one last, just for fun

run the file

[Colin]

Awesome, here is my code (I got the full rotation down)

clc
clear
line = ['r', 'g', 'k', 'b']; %Line colors
x = 0:0.1:2;  %X coordinates
y = 1-0.5*x;  %Y coordinates
Triangle = [0 2 0 x 0 0; 0 0 1 y 0 1];  %Triangle x, y coordinates
plot(Triangle(1,:),Triangle(2,:), 'bs-','LineWidth',2) %Plot the triangle
axis([0 2 0 2])
axis('square')
axis off
figure(1)
%Triangle is assumed to be a planar lamina
f = @(x) 1 - (x/2); %Function representing the hypotenuse of the triangle
mass = integral(f, 0, 2); %Multiplied by rho, measured in mass per unit area
g = @(x) x.*(1-(x/2)); %Function to determine the first moment about y
y_moment = integral(g, 0, 2); %Multiplied by rho
x_centroid = y_moment/mass;   %Compute the x coordinate of the centroid
fprintf('The x component of the centroid is %0.2f \n',x_centroid)  %Output centroid
y = 0:0.05:1;   %Y coordinates
x = 2-(2*y);    %X coordinates
Triangle_1 = [0 2 0 x 0 0; 0 0 1 y 0 1];  %Triangle x,y coordinates, same triangle as above
plot(Triangle_1(1,:),Triangle_1(2,:),'bs-','LineWidth',2)  %Plot the triangle
axis([0 2 0 2])
axis('square')
axis off
figure(1)
%Again, triangle is a planar lamina
h = @(x) ((1-(x/2))/2).*(1-(x/2)); %Function to determine first moment about x
x_moment = integral(h,0,2);  %Multiplied by rho
y_centroid = x_moment/mass;  %Compute the y component of the centroid
fprintf('The y component of the centroid is %0.2f \n',y_centroid)  %Output centroid component
%Using the first triangle for rotation
x = 0:0.1:2;  %X coordinates
y = 1-0.5*x;  %Y coordinates
Triangle = [0 2 0 x 0 0; 0 0 1 y 0 1];  %Triangle x, y coordinates
plot(Triangle(1,:),Triangle(2,:), 'bs-','LineWidth',2) %Plot the triangle
axis([0 2 0 2])
axis('square')
axis off
figure(1)
figure(2) %New window
for k=0:64 %64 different positions
  phi = k*(2*pi)/64;  %Angle of rotation, 64 rotations around a full circle (2 pi radians)
  A = [cos(phi) -sin(phi); sin(phi) cos(phi)];  %General form of 2-D rotation
  TriangleRotated = A*Triangle;   %To rotate the triangle
  plot(TriangleRotated(1,:),TriangleRotated(2,:),line(mod(k,4)+1));
  %Plot scaling for first iteration of the loop 
  if k==0
      axis([-2 2 -2 2]);
      axis('square');
      axis off
      hold on
  end;
  figure(2);  %Displays the plot
end %End of the first iteration
hold off;  %Release plot
figure(2); %Display plot 

end