Plotting Cylinder Surface Using fill3

I am attempting to plot a cylinder (or any abitrary) surface using the fill3 or patch function.

MATLAB, however, does not produce the cylinder surface using the minmal working example below, unless the angle theta is restricted to less than pi.

Any help explaining this issue and finding a solution is greatly appreciated!

figure()
theta = 0:0.5:2*pi;
xs = [cos(theta) flip(cos(theta))];
ys = [sin(theta) flip(sin(theta))];
fill3(xs, ys, [zeros(size(theta)) ones(size(theta))], 'k', 'FaceAlpha', 0.2)

Answer 1

[Star Strider]

It is easier to use the surf function for this —

% figure()
% theta = 0:0.5:2*pi;
theta = linspace(0, 2*pi, 11);
xs = [cos(theta); flip(sin(theta))];
ys = [sin(theta); flip(cos(theta))];

figure
surf(xs, ys, [zeros(size(theta)); ones(size(theta))],  'FaceAlpha', 0.2)
view(-27,30)

figure
surf([cos(theta); cos(theta)].', [sin(theta); sin(theta)].', [ones(size(theta)); zeros(size(theta))].', 'FaceColor','c', 'FaceAlpha',0.5)
view(-27,30)

It can probably be done with patch, however I have always used surf for these sorts of problems.

.

[Florian Radack]

Thanks for the answer, this solution works. However, it does not explain the behaviour of the surf or fill3 functions.

[Star Strider]

As always, my pleasure!

I can almost always get patch to work in 2-D plots, however getting it to work in 3-D plots requires some sort of magic that I’ve not yet discovered.

Actually, I just now figured out how to get patch to work here, however it requires a loop for each segment of the circle —

theta = linspace(0, 2*pi, 11);
xs = cos(theta);
ys = sin(theta);

figure
hold on
for k = 1: numel(xs)-1
patch([xs([k k+1]) flip(xs([k k+1]))], [ys([k k+1]) flip(ys([k k+1]))], [0 0 1 1], 'g', 'FaceAlpha',0.5)
end
hold off
grid on
view(-27,30)

This is similar to the approach with the entire circle, however for whatever reason (perhaps that the beginnings and ends are essentially the same point), it fails with the complete circle.

It is then straightforward to add the top an bottom surfaces, closing the cylinder —

figure
hold on
for k = 1: numel(xs)-1
patch([xs([k k+1]) flip(xs([k k+1]))], [ys([k k+1]) flip(ys([k k+1]))], [0 0 1 1], 'g', 'FaceAlpha',0.5)
end
patch(xs,  ys, zeros(size(theta)), 'r', 'FaceAlpha',0.75)
patch(xs,  ys, ones(size(theta)), 'b', 'FaceAlpha',0.5)
hold off
grid on
view(-27,30)

This seems to ‘sort of’ solve it, without solving the mystery of the complete circle failing to work with essentially the same sort of approach.

.

Answer 2

[Wick]

I realize this is an old question, but here's the solution for making the cylinder side with patch objects using direct manipulation of the vertices and faces.

clearvars

theta = 0:0.5:2*pi;

N = length(theta);

% allowing that the (x,y) at the top and bottom might not be the same if the cylinder isn't aligned along the z axis

x1 = cos(theta');

x2 = x1;

y1 = sin(theta');

y2 = y1;

z1 = zeros(size(x1));

z2 = ones(size(x2));

vertices = [[x1;x2] [y1;y2] [z1;z2]];

% building faces that go down each face then back up. Each edge repeats on adjacent faces

a = (1:N)';

d = [2:N 1]';

b = a + N;

c = d + N;

faces = [a b c d];

% You can remove the last line of 'faces' if your angle vector goes all the

% way back to exactly 2*pi. The original post only approximated that so the

% last line closes the outer curve by repeating the very first edge.

figure()

patch('Vertices',vertices,'Faces',faces,'FaceColor','r')

view(-52,56)