How can I wrap data points around an object in Octave

Question

I'm trying to wrap a signal around a hyperbolic cone (the cone looks like this http://www.sectioaurea.com/sectioaurea/the_golden_angle.htm) I've found away to wrap data around a cylinder.

See cylinder with wrapped data around it below.

%% // Generate sample data
x = linspace(0,10*pi) ;
y2 = cos(x) ;
y1 = 10*cos(x/10) ;
y = y1+y2 ; y = y-min(y) ;
figure, 
plot(x,y,'-o') ;

%% // Basic cylinder (just for background)
[Xc,Yc,Zc] = cylinder(1*ones(1,50),50);
Zc = Zc * max(y) ;
surf(Xc,Yc,Zc) ;
hold on 

%% // Fold the points around the cylinder
Number_of_turn = 2 ;
xrange = [min(x),max(x)] ;
xspan = xrange(2)-xrange(1) ;
xc = x / xspan * 2*pi * Number_of_turn ;

Xp = cos(xc) ;
Zp = y ;
Yp = sin(xc) ;

plot3(Xp,Yp,Zp,'-ok') ;

But I'm not sure how to get it to wrap data around a hyperbolic cone (the cone looks like this http://www.sectioaurea.com/sectioaurea/the_golden_angle.htm)

I'm using Octave 4.2.2 in Ubuntu 18.04 64bit

There doesn't seem to be anything magical happening here with regard to 'wrapping'. It's just a `surf` object and a `plot3` object, plotted on the same `axes` object (via `hold on`). As long as you have the correct equations, you can use the same approach to plot a 'wrapping function' `plot3` object on top of a such a 'hyperbolic cone' `surf`object in the same way. — Tasos Papastylianou, Jan 01 '20 at 16:48
@TasosPapastylianou that's the issue. The equations http://www.sectioaurea.com/sectioaurea/the_golden_angle.htm . The issue starts with the `cylinder` command how to make a hyperbolic cone using it. — Rick T, Jan 02 '20 at 08:14

Tasos Papastylianou · Accepted Answer · 2020-01-04T15:38:18.103

Here's an example

% plot surface
[X,Y] = ndgrid( -0.1:0.001:0.1, -0.1:0.001:0.1 );
Z = 1 ./ sqrt(X.^2 + Y.^2);   % The height at each X,Y grid point is
                              % the inverse of its norm

surf(X,Y,Z, 'edgecolor', 'none');
colormap cool
camlight

%plot spiral
Theta = [1:0.1:500];
Z = Theta;    % height increases same as angle
Norm = 1 ./ Z;   % the definition of the hyperbolic cone
X = cos(Theta) .* Norm;
Y = sin(Theta) .* Norm;

hold on
plot3(X,Y,Z, 'r', 'linewidth', 1.5 )
hold off

axis([-0.1, 0.1, -0.1, 0.1, 0, 500])

UPDATE

Here is another example, wrapping a specific function; hopefully it demonstrates what you had in mind. Note I had to offset the height a little bit, otherwise a hyperbolic cone at close-to-zero heights is very close to infinity, which is difficult to graph.

%% Generate a hyperbolic cone surface, and 'wrap' an arbitrary 1D function to it, by
%% treating the function's x-coordinate as an angle, and for each such
%% angle, projecting the corresponding height onto the surface of the cone at the same
%% height.

%%%%%%%%%%%%%%%%%%%%%%
% Generate sample data
%%%%%%%%%%%%%%%%%%%%%%

   x = linspace( 0, 10 * pi );   % Generate 100 points (default) from 0 to 10π
   h = cos(x) + 10 * cos( x / 10 );   h = h - min(h);
   h = 30 * h + 30;   % make it a bit taller, and small offset so that low values
                      % don't end up near infinity when mapped onto the hyperbolic
                      % cone surface

%%%%%%%%%%%%%%%%%%
% Generate Surface
%%%%%%%%%%%%%%%%%%

 % Generate a suitable XY grid
   [GridX, GridY] = ndgrid( -0.1:0.0005:0.1, -0.1:0.0005:0.1 );

 % For each point on the grid, calculate a distance Δ from the (0,0) origin.
   Delta = sqrt( GridX .^ 2  +  GridY .^ 2 );

 % According to the definition of the hyperbolic cone, the height Z at each gridpoint
 % is equal to 1 over Δ
   Z = 1 ./ Delta;

 % Plot the resulting surface with height Z at each gridpoint X,Y
   surf( GridX, GridY, Z, 'edgecolor', 'none' );

 % Give the surface a nicer appearance
   colormap( 'ocean' ); caxis([0, 25]); camlight;


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Wrap data around generated surface
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

   NumberOfTurns = 2;
   XRange  = [ min( x ), max( x ) ];
   XSpan   = XRange(2) - XRange(1);
   XOffset = XRange(1)

 % Convert x inputs to angles Theta ('xc' in previous example code)
   Theta  = (NumberOfTurns * 2 * pi / XSpan) * (x - XOffset);

  % Since, according to the hyperbolic cone equation, 'H = 1 / Δ' (where H denotes
  % Height, and Δ denotes distance from the [0,0] origin), it follows that for a
  % given H, we can find the corresponding Δ as Δ = 1 / H. Therefore, for each
  % [θ, h] pair, we can find a distance Δ, such that the cartesian point
  % [ Δcos(θ), Δsin(θ), h ] sits on the cone. We use this fact to 'wrap' the
  % trajectory function around the cone.

  Delta = 1 ./ h;   % overwriting old Delta definition for notational convenience.

  WrappedX = Delta .* cos( Theta );
  WrappedY = Delta .* sin( Theta );
  WrappedZ = h;

 % Plot the wrapped datapoints over the previous surface graph
   hold on
   plot3( WrappedX, WrappedY, WrappedZ, '-or', 'linewidth', 1.5, 'markersize', 4, 'markerfacecolor', [0.5,0,0], 'markeredgecolor', [0.25,0,0]);
   hold off

 % Set appropriate axis limits to visualise relevant range
   axis([-0.1, 0.1, -0.1, 0.1, 0, 1000]);
   set(gca, 'color', [0.5,0.5,0.5]);

Papastlianou thanks but when I try and map the points around the curve by changing the sections in `%% // Fold the points around the cylinder` they don't "fold" around the object like in the example with the cylinder. Any ideas why? — Rick T, Jan 03 '20 at 16:40
@RickT Presumably it has something to do with the way you're changing the sections? — Tasos Papastylianou, Jan 03 '20 at 17:52
A cylinder is governed by one set of equations, and a hyperbolic cone by another. A cylinder is no more relevant here than a pina colada. Why are you so keen to use a `cylinder` function as part of this problem? If you have an equation for the surface, and one for the trajectory function, your problem is solved. Do you understand how the basic hyperbolic function gives rise to the cone surface and this spiral trajectory in the first place? — Tasos Papastylianou, Jan 03 '20 at 17:56
sorry if there was a miss communication. The code you so kindly posted got me past that I was referring to the `section` titled `%% // Fold the points around cylinder` in my code which from the original question `doesn't` use the cylinder command in there (at-least that section). At the moment I've narrowed it down to the `xc` variable at the end. — Rick T, Jan 03 '20 at 19:02
Ok I think I see where the misunderstranding lies. You don't just want to recreate the spiral trajectory shown in the link. You want the sinusoidal trajectory instead. The principle is the same, but now that I understand what you're after, I'll rewrite the example with relevant comments. — Tasos Papastylianou, Jan 04 '20 at 12:39

How can I wrap data points around an object in Octave

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