How to mask a circle in perspective projection?

Question

Ma'ayan Gadot 2018-11-20

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https://ww2.mathworks.cn/matlabcentral/answers/430911-how-to-mask-a-circle-in-perspective-projection

评论： Ma'ayan Gadot 2018-11-27

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

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Bruno Luong 2018-11-20

编辑：Bruno Luong 2018-11-20

What is "world coordinates"? What are those red triangles in the second picture? Where are the "dots"?

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Answer 1

Bruno Luong 2018-11-20

1
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/430911-how-to-mask-a-circle-in-perspective-projection#answer_347994

编辑：Bruno Luong 2018-11-20

在 MATLAB Online 中打开

@Maayan please make an effort to give feedback to various answer/questions addressed to you.

Just update the persective code in this thread.

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

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Bruno Luong 2018-11-22

编辑：Bruno Luong 2018-11-22

在 MATLAB Online 中打开

For MATLAB version before R2016b, user needs to replace

XYZ-Camera.CP

with

bsxfun(@minus, XYZ, Camera.CP)

Ma'ayan Gadot 2018-11-27

Thank you very much!

Your response solvd the problem.

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Answer 2

Jan 2018-11-20

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/430911-how-to-mask-a-circle-in-perspective-projection#answer_347985

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: