How to mask a circle in perspective projection?
3 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Ma'ayan Gadot
am 20 Nov. 2018
Kommentiert: Ma'ayan Gadot
am 27 Nov. 2018
Hi,
I simulate a 3-D scene using matlab camproj perspective.
This procedure creates a perspective projection of the shapes (triangles), but the stimuli coordinates remain in orthographic representation (world coordinates).
I want to mark all the triangles inside a circle. Using inpolygon, marke the triangles as if this was done in orthographic projection and not perspective.
See attached files for demostration.
Any ideas on how to get the triangles screen coordinates?
Thank you
1 Kommentar
Bruno Luong
am 20 Nov. 2018
Bearbeitet: Bruno Luong
am 20 Nov. 2018
What is "world coordinates"? What are those red triangles in the second picture? Where are the "dots"?
Akzeptierte Antwort
Bruno Luong
am 20 Nov. 2018
Bearbeitet: Bruno Luong
am 20 Nov. 2018
@Maayan please make an effort to give feedback to various answer/questions addressed to you.
Z = peaks;
x = 1:size(Z,2);
y = 1:size(Z,1);
[X,Y] = meshgrid(x,y);
figure(1);
clf
ax=subplot(1,2,1);
surface(X,Y,Z)
view(rand*360,(rand-0.5)*180)
camproj(ax,'perspective');
axis(ax,'equal');
drawnow;
CT = get(ax,'CameraTarget');
CP = get(ax,'CameraPosition');
% Get the camera closer in order to amphasize the 3D perspective effect
NewCP = CT + 0.2*(CP-CT);
set(ax,'CameraPosition',NewCP);
CP = get(ax,'CameraPosition');
CU = get(ax,'CameraUpVector');
cadeg = get(ax,'CameraViewAngle');
CT = CT(:);
CP = CP(:);
CU = CU(:);
ca = cadeg*pi/180;
%
c = CT-CP;
d = norm(c);
c = c / d;
u = CU - (c'*CU)*c;
u = u/norm(u);
p = cross(c,u);
R = [c,p,u];
Camera = struct('R', R, 'CP', CP);
[xcam, ycam] = CamProj(X, Y, Z, Camera);
[xc, yc] = CamProj(CT, Camera);
radius = 0.2;
incircle = (xcam-xc).^2+(ycam-yc).^2 < radius^2;
subplot(1,2,2);
plot(xcam(~incircle),ycam(~incircle),'.b',...
xcam(incircle),ycam(incircle),'.r');
hold on
rectangle('Position',[xc-radius yc-radius 2*radius 2*radius],'Curvature',[1 1],'EdgeColor','r')
axis equal
% This doesn't seem to crop the projection correctly
camyangle = sin(ca/2)*[-1 1];
camylim = [min(ycam),max(ycam)];
camxlim = [min(xcam),max(xcam)];
%%
function [xcam, ycam] = CamProj(varargin)
% Do perspective projection
% [xcam, ycam] = CamProj(X, Y, Z, Camera)
% [xcam, ycam] = CamProj(XYZ, Camera), XYZ is (3 x n) array
if nargin >= 4
[X,Y,Z,Camera] = deal(varargin{:});
sz = size(X);
XYZ = [X(:),Y(:),Z(:)]';
else
[XYZ,Camera] = deal(varargin{:});
[m,n] = size(XYZ);
if m~=3 && n==3
XYZ = XYZ.';
n = m;
end
sz = [1,n];
end
CPU = Camera.R'*(XYZ-Camera.CP);
XYcam = CPU(2:3,:)./CPU(1,:);
xcam = reshape(XYcam(1,:),sz);
ycam = reshape(XYcam(2,:),sz);
end % CamProj
3 Kommentare
Bruno Luong
am 22 Nov. 2018
Bearbeitet: Bruno Luong
am 22 Nov. 2018
For MATLAB version before R2016b, user needs to replace
XYZ-Camera.CP
with
bsxfun(@minus, XYZ, Camera.CP)
Weitere Antworten (1)
Jan
am 20 Nov. 2018
I assume the triangles are the "dots" and they are read, because they are inside a certain distance to the line through the center of the circle, which is normal in the plane of the circle.
The solution will be to calculate a 2D projection of the coordinates at first. This is done by the graphics renderer already, but not complicated by maths also. Afterwards the selection of the points inside the circle is trivial again.
This might help:
1 Kommentar
Bruno Luong
am 20 Nov. 2018
Read through the first link, it seems the projection is only valid in 'orthographic' projection mode, and not 'perspective' mode.
For persective the correct projection is answered yesterday in this thread
Siehe auch
Kategorien
Mehr zu Camera Views finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!