use the 'patch' function with different results (using different matrices)

Hi! I have two matrices: 'V' and 'trace'.

Using patches with V generates a completely black surface on the figure and I'm fine with it.

When I use the 'trace' matrix the result is different. How come? Is there a way to display 'trace' in the image in the same way as 'V'?

load V

axes1 = axes('Parent',figure);
patch(V(:,1), V(:,2),V(:,3),'k');
view(axes1,[45.7625707544407 8.27114462456761]);
axis equal

%%
load trace

axes2 = axes('Parent',figure);
patch(trace(:,1), trace(:,2),trace(:,3),'r');
view(axes2,[31.7625655339418 23.9174311926605]);
axis equal

[Matt J]

When I use the 'trace' matrix the result is different. How come?

Why do we expect them to be the same? Is there some relationship between V and trace that you haven't mentioned? They certainly don't contain the same number of points.

load trace; load V

whos trace V

  Name         Size            Bytes  Class     Attributes

  V          360x3              8640  double              
  trace      962x3             23088  double              

[Alberto Acri]

The matrices are different. I expected that using the 'trace' matrix I would get a black geometry display similar to the one I got with the V matrix!

[Walter Roberson]

Why would you expect that?

Remember that for patch(), points are considered to be ordered

If you sort the original points by angle, and plot the sort order:

load trace

axes3 = axes('Parent', figure);
tc = mean(trace,1);
ang = atan2(trace(:,2) - tc(2), trace(:,1) - tc(1));
[~, order1] = sort(ang);
[~, order2] = sort(order1);
plot(order2)

adjacent points on input bounce back and forth in angle order -- and not even alternately.

figure();
plot(order2(1:50))

Answer 1

[Matt J]

If you are trying to recover the boundary from a surface triangulation, then,

load trace

[~,P]=freeBoundary(delaunayTriangulation(trace(:,1:2)));

Warning: Duplicate data points have been detected and removed.
The Triangulation indices are defined with respect to the unique set of points in delaunayTriangulation.

[~,loc]=ismember(P,trace(:,1:2),'rows');

V=num2cell([P,trace(loc,3)],1);

patch(V{:},'k'); 
view([31.7625655339418 23.9174311926605]);
axis equal

Answer 2

[Walter Roberson]

In your first patch() call you ask for 'k' -- black -- for the faces. The edge colors default to black as well.

In the second patch() call you ask for 'r' -- red -- for the faces. The edge colors default to black.

The V matrix has coordinates for a simple circle. The trace matrix has much more complex coordinates, going back and forth.

If you were to sort the coordinates then you could get a circle as well. But remember that the coordinates for patch() are ordered so those represent completely different graphing tasks.

load V

axes1 = axes('Parent',figure);
patch(axes1, V(:,1), V(:,2), V(:,3), 'facecolor', 'r', 'facealpha', 0.5, 'edgecolor', 'k');
view(axes1,[45.7625707544407 8.27114462456761]);
axis(axes1, 'equal');

%%
load trace

axes2 = axes('Parent',figure);
patch(axes2, trace(:,1), trace(:,2), trace(:,3), 'facecolor', 'r', 'facealpha', 0.5, 'edgecolor', 'k');
view(axes2,[31.7625655339418 23.9174311926605]);
axis(axes2, 'equal');

axes3 = axes('Parent', figure);
tc = mean(trace,1);
ang = atan2(trace(:,2) - tc(2), trace(:,1) - tc(1));
[~, order] = sort(ang);
patch(axes3, trace(order,1), trace(order,2), trace(order,3), 'facecolor', 'r', 'facealpha', 0.5, 'edgecolor', 'k');
view(axes3,[31.7625655339418 23.9174311926605]);
axis(axes3, 'equal')

[Walter Roberson]

Note that if you know you have scattered coordinates, then see also boundary instead of the sorting method I used.