Wrap part of a shape with a surface

I have the following code:

%Initially written by W. Rose 20251018
%Modified by poster rdemyan

clear

N=80;                   % number of angular steps in one rotation
w=7.908e-4;  R=6e-4;    % constants in equation for p(t) [units=length]
B=0.867;                % constant in equation for h(t): h(t)=B*p(t)  [dimensionless]
dt=2*pi/N;              % angle step size [radians]
t=(0:dt:2*pi)';         % angle of rotation (vector, length N+1) [radians]


p=((w/2)*cos(t)+(R^2*(sin(t).^2+0.5*cos(t).^2)-(w^2/2)*sin(t).^2).^0.5)./(sin(t).^2+0.5*cos(t).^2); 

% Compute h(t):
h=B*p;  

% Compute arrays of 3D points
pt=[p.*cos(t),p.*sin(t),h/2];       % x,y,z coords of points along top of shape
pb=[p.*cos(t),p.*sin(t),-h/2];      % x,y,z coords of points along bottom of shape

figure
       
for i=1:N+1

    view(40,60);
    hold on;
    axis equal; 

    % Draw HEIGHT LINES between top and bottom points 
    f=plot3([pt(i,1),pb(i,1)],...
          [pt(i,2),pb(i,2)],...
          [pt(i,3),pb(i,3)],LineWidth=4)
    f.Color = [0,0.7,0.7,0.7]

    drawnow;                        
end

With N = 80, I get the following shape.

With N = 2000, I get the following shape:

Instead of increasing N I would like to wrap the height lines with a surface. Increasing N does provide a passable figure, but there are lines that have an uneven transparency. Yes, if I increase FaceAlpha to 1, then there is no uneveness, but I want to have some transparancy. I think that a surface wrap might produce a much more uniform figure and also I would like to decrease the FaceAlpha value further from the value of 0.7. Further, for my own edification, I would like to know how to wrap the height lines in a surface to create the desired figure rather than just increasing the value of N. Finally, I do not want to have a top or bottom surface, just wrapped around as shown in Fig. 2. I want to be able to "see" inside as shown.

[Rik]

You have the bottom and top points in a resolution you can specify. That means you can generate a list of triangular faces (remember: every trapezoid can be split into 2 triangles). If you treat your list of points as list of vertices and define the faces as indices into that list of vertices, you are at the definition the patch function expects.

Did you try something like this yet?

I admit you would expect the surface function to do this (or actually surf), which is why my comment first referred to that function instead.

Answer 1

[Paul]

Hi Robert,

Perhaps something like this that draws a pathc between each height line? Salt to taste on the patch command.

Is the f.Color using an undocumented feature?

N=80;                   % number of angular steps in one rotation
w=7.908e-4;  R=6e-4;    % constants in equation for p(t) [units=length]
B=0.867;                % constant in equation for h(t): h(t)=B*p(t)  [dimensionless]
dt=2*pi/N;              % angle step size [radians]
t=(0:dt:2*pi)';         % angle of rotation (vector, length N+1) [radians]
p=((w/2)*cos(t)+(R^2*(sin(t).^2+0.5*cos(t).^2)-(w^2/2)*sin(t).^2).^0.5)./(sin(t).^2+0.5*cos(t).^2); 
% Compute h(t):
h=B*p;  
% Compute arrays of 3D points
pt=[p.*cos(t),p.*sin(t),h/2];       % x,y,z coords of points along top of shape
pb=[p.*cos(t),p.*sin(t),-h/2];      % x,y,z coords of points along bottom of shape
figure
view(40,60);
hold on;
axis equal; 
       
for i=1:N+1

    % Draw HEIGHT LINES between top and bottom points 
    f=plot3([pt(i,1),pb(i,1)],...
          [pt(i,2),pb(i,2)],...
          [pt(i,3),pb(i,3)],LineWidth=4);
    f.Color = [0,0.7,0.7,0.7];  % four-element color? 
    
    if i <= N
    patch(...
        [pt(i,1),pb(i,1),pb(i+1,1),pt(i+1,1)],...
        [pt(i,2),pb(i,2),pb(i+1,2),pt(i+1,2)],...
        [pt(i,3),pb(i,3),pb(i+1,3),pt(i+1,3)],...
        'r','FaceColor',[0,0.7,0.7],'EdgeColor','None','FaceAlpha',0.7);
    end
    
    drawnow;                        
end

●

[Robert Demyanovich]

Hi Paul:

This works very well. I simply removed the lines of code that define f and its color and this removes the lines. Now I have a nice smooth surface that is even in color even if FaceAlpha is equal to 0.3. With regards to f.Color, I found this line of code after an internet search; I don't remember exactly but I think it was from an example from the MatLab documentation. The last value is FaceAlpha, IIRC and not a color.

[Paul]

Just to make sure we're on the same page, the surface isn't really smooth because each patch is a plane. Viewing zoomed in from above makes the transitions at the vertices evident. But maybe it's smooth enough.

N=80;                   % number of angular steps in one rotation
w=7.908e-4;  R=6e-4;    % constants in equation for p(t) [units=length]
B=0.867;                % constant in equation for h(t): h(t)=B*p(t)  [dimensionless]
dt=2*pi/N;              % angle step size [radians]
t=(0:dt:2*pi)';         % angle of rotation (vector, length N+1) [radians]
p=((w/2)*cos(t)+(R^2*(sin(t).^2+0.5*cos(t).^2)-(w^2/2)*sin(t).^2).^0.5)./(sin(t).^2+0.5*cos(t).^2); 
% Compute h(t):
h=B*p;  
% Compute arrays of 3D points
pt=[p.*cos(t),p.*sin(t),h/2];       % x,y,z coords of points along top of shape
pb=[p.*cos(t),p.*sin(t),-h/2];      % x,y,z coords of points along bottom of shape
figure
view(40,60);
hold on;
axis equal; 
       
for i=1:N
   
    % change edge color for this view
    patch(...
        [pt(i,1),pb(i,1),pb(i+1,1),pt(i+1,1)],...
        [pt(i,2),pb(i,2),pb(i+1,2),pt(i+1,2)],...
        [pt(i,3),pb(i,3),pb(i+1,3),pt(i+1,3)],...
        'r','FaceColor',[0,0.7,0.7],'EdgeColor','k','FaceAlpha',0.7);
end
view(2)
axis([14,17,-2,2]*1e-4)

●

The four element Color "property" is undocumented AFAIK. line doesn't have a FaceAlpha or Alpha property.

[Robert Demyanovich]

If N = 2000, which is what I've changed it to for the purposes of creating a figure, then I think it is certainly smooth enough. The color property for the code is for plot3.