Page 205 - Basics of MATLAB and Beyond

x = [0 0 1 1 0 .4
1 1 1 0 0 .4
1.6.6.4.4.6
0 .4 .6 .6 .4 .6];
y = [0 1 1 0 0 .4
0 1 0 0 1 .6
1.6.4.4.6.6
1 .6 .6 .4 .4 .4];
z = [0 0 0 0 0 1
1
0 0 0 0 0 1
0 1 1 1 1 1 0.5
z
0 1 1 1 1 1];
0
1
patch(x,y,z,’y’) 1
0.5
0.5
view(3);xyz,box y 0 0 x
Exercise 22 (Page 171)
The faces are coloured according to the colours of the vertex. We have
vertices defined for z = 0 and z = 1, but the heat source is located at
z =0.25. The result should be symmetric about the z =0.25 plane,
but our result does not have this symmetry. The reason is that we only
have two z values. To produce a better display we simply need to add
vertices at a range of z values. The following code does the job:
N = 100; % Number of points around the circle
M = 30; % Number of circles in the cylinder
dt = 2*pi/N;
t = (0:dt:(N-1)*dt)’;
h = linspace(0,1,M); % vector of heights
xv = cos(t);
yv = sin(t);
% Reproduce the vertices at different heights:
x = repmat(xv,M,1);
y = repmat(yv,M,1);
z = ones(N,1)*h;
z = z(:);
vert = [x y z];

% These are the facets of a single ’layer’:
facets = zeros(N,4);
facets(1:N-1,1) = (1:N-1)’;
facets(1:N-1,2) = ((N+1):(2*N-1))’;
facets(1:N-1,3) = ((N+2):(2*N))’;
facets(1:N-1,4) = (2:N)’;
facets(N,:) = [N 2*N N+1 1];

c 2000 by CRC Press LLC