Basics of MATLAB and Beyond

x = [0 0 1 1 0 .41 1 1 0 0 .41 .6 .6 .4 .4 .60 .4 .6 .6 .4 .6];y = [0 1 1 0 0 .40 1 0 0 1 .61 .6 .4 .4 .6 .61 .6 .6 .4 .4 .4];z = [0 0 0 0 0 10 0 0 0 0 10 1 1 1 1 10 1 1 1 1 1];patch(x,y,z,’y’)view(3);xyz,boxz10.51 00.5y0 00.5x1Exercise 22 (Page 171)The faces are coloured according to the colours of the vertex. We havevertices defined for z = 0 and z = 1, but the heat source is located atz =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 onlyhave two z values. To produce a better display we simply need to addvertices at a range of z values. The following code does the job:N = 100; % Number of points around the circleM = 30; % Number of circles in the cylinderdt = 2*pi/N;t = (0:dt:(N-1)*dt)’;h = linspace(0,1,M); % vector of heightsxv = 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