MATLAB: How to create two rectangles with the same shape, one stay at specific position that created by using matrix, and the other change its position from 0 to 90 degree

MATLAB

How can I create two rectangles with the same shape, one stay at specific position that created by using matrix, and the other change its position from 0 to 90 degree? I got the code below just for the changing one. Let me know if somgthing I do wrong. Thank you.

%% Main Code
% Initialise the graph
hSquare = plot (NaN, NaN , '-b'); %Open empty graph. NaN - Not a number.
axis([-10 10 -10 10]); grid on;
% Main
X       = [2 6 6 2 2];
Y       = [2 2 4 4 2];
x1 = [get(hSquare, 'XData') X];
y1= [get(hSquare, 'YData') Y];
 set(hSquare, 'XData', x1, 'YData', y1);
 drawnow;
 for thetad =1:90
     R = [cosd(thetad) -sind(thetad); sind(thetad) cosd(thetad)];
     x2 = X(1:4);
     y2 = Y(1:4);
     V = R*[X;Y];% do the rotation 
    
     Vt = V(1,:);
    
     Vb = V(2,:);
    
     plot(Vt,Vb,'r-');
     axis([-10 10 -10 10]); grid on;
     drawnow;
     
  
     pause(0.05);
 end

Best Answer

%% Main Code
% Initialise the graph
hSquare = plot (NaN, NaN , '-b'); %Open empty graph. NaN - Not a number.
axis([-10 10 -10 10]); grid on;
% Main
X       = [2 6 6 2 2];
Y       = [2 2 4 4 2];
Xm = mean(X) ; 
Ym = mean(Y) ;
x1 = [get(hSquare, 'XData') X];
y1= [get(hSquare, 'YData') Y];
 set(hSquare, 'XData', x1, 'YData', y1);
 drawnow;
 for thetad =1:90
     R = [cosd(thetad) -sind(thetad); sind(thetad) cosd(thetad)];
     x2 = X(1:4);
     y2 = Y(1:4);
     V = R*([X;Y]-[Xm;Ym])+[Xm;Ym];% do the rotation relative to the centre of the square
    
     Vt = V(1,:);
    
     Vb = V(2,:);
    
     plot(X,Y,'b',Vt,Vb,'r-');
     axis([-10 10 -10 10]); grid on;
     drawnow;
     
  
     pause(0.05);
 end

Related Solutions

MATLAB: Rotating a 2d shape

Shaan - see rotation matrix which you can construct to rotate your (x,y) points (of your 2D shape) so that you can rotate them counter-clockwise.

For example, if you know the four vertices of the square that you wish to draw, then you can use the MATLAB fill function to create it

 X = [-2 -2 2 2];
 Y = [-2 2 2 -2];
 hSquare = fill(X,Y,'k');

where X and Y are the x and y coordinates of the vertices, and hSquare is the handle to the square (polygon) that you have drawn.

Suppose you now want to rotate the square by one degree in a counter-clockwise direction. The rotation matrix is the usual

 thetad = 1;
 R      = [cosd(thetad) -sind(thetad); sind(thetad) cosd(thetad)];

We can get the vertices from the drawn square since we have its handle

 V = get(hSquare,'Vertices')';

We could multiply V by R to get the rotated set of vertices, but this would only be valid if the centre of the square is (0,0). As that will not always be the case, then you will need to translate the square from its current coordinate system to one centred at (0,0). We can do this and the rotation as follows

 V = R*(V - C) + C;

where C is an appropriately sized matrix with the centre of the square (use repmat to create this C given the (x,y) coordinates for the centre). So we translate to the coordinate system centred at (0,0) (the subtraction), rotate (the multiplication), and then translate back to the original coordinate system whose origin is the centre of the square (the addition). We can then update the square with the new vertices as

 set(hSquare,'Vertices',V');

and the rotation by one degree is complete. Putting this all together to rotate the square over 360 degrees, we would do

 X       = [-2 -2 2 2];
 Y       = [-2 2 2 -2];
 hSquare = fill(X,Y,'k');
 thetad  = 1;
 R       = [cosd(thetad) -sind(thetad); sind(thetad) cosd(thetad)];
 C       = repmat([0 0], 4, 1)';
 axis([-10 10 -10 10])
 for k=1:360
     V = get(hSquare,'Vertices')';  % get the current set of vertices
     V = R*(V - C) + C;             % do the rotation relative to the centre of the square
     set(hSquare,'Vertices',V');    % update the vertices
     pause(0.01);
 end

MATLAB: Creating a shape with varying arcs

Note that 'smoothly' might not mean what you think it means: depending on your relative radii, you might get a change in concavity even with a "smooth" (continuous derivative) change in radius as a function of angle.

See if the below helps:

r1 = 2;
r2 = 3;
x1 = 4*pi/6; % r = r1
x2 = 2*pi/6; % r 'smoothly' increases
x3 = 3*pi/6; % r = r2
x4 = 2*pi - (x1 + x2 + x3); % r 'smoothly' decreases
t_table = cumsum([ 0, x1, x2, x3, x4, x1]); % note extra wrap-around entry to keep pchip smooth
r_table =        [r1, r1, r2, r2, r1, r1];
n = 1000;
method = {'linear', 'pchip', 'spline'};
theta = linspace(0, 2*pi, n + 1);
%radius = interp1(t_table, r_table, theta, 'linear');
for k = 1:numel(method)
radius = interp1(t_table, r_table, theta, method{k});
hf{k} = figure('WindowStyle', 'Docked');
ha{k}(1) = subplot(2, 1, 1);
plot(theta, radius);
xlabel('\theta [rad]')
ylabel('r [u]')
axis([0, 2*pi, 0, r2*1.1])
grid('on')
ha{k}(2) = subplot(2, 1, 2);
plot(radius.*cos(theta), radius.*sin(theta));
title(['Cam Shape (', method{k}, ')'])
axis('equal')
grid('on')
axis([-1, 1, -1, 1]*r2*1.1)
end

(I think keeping a convex shape will depend on if your second region of [larger] constant radius lies within the cone/triangle defined by the two tangents at either end of your first region of [smaller] constant radius.)

Best Answer

Related Solutions

MATLAB: Rotating a 2d shape

MATLAB: Creating a shape with varying arcs

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