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how to take picture of a virtual cube in matlab? i have generated the cube but do not know how to take picture in matlab

1 vue (au cours des 30 derniers jours)
i have generated a virtual 3D cube. i am asked to take a picture of this cube from 4 meters away and with a tilt angle of 30 degree. do i have to do this with rotation matrix R and translation matrix T? this is the code of the virtual 3D cube
%%Generate 3D calibration pattern:
%%Pw holds 32 points on two surfaces (Xw = 1 and Yw = 1) of a cube
%%Values are measured in meters.
%%There are 4x4 uniformly distributed points on each surface.
cnt = 1;
%%plane : Xw = 1
Pw = zeros(4,3);%preallocating Pw
for i=0.2:0.2:0.8,%it means it starts from 0.2 gives an interval 0.2 and goes till 0.8
for j=0.2:0.2:0.8,
Pw(cnt,:) = [1 i j];%is the cnt-th row of pw , pw is a matrix whos cnt-th row is a row matrix, cnt is a variable that starts from 1 and after forming
%the row matrix, it goes on till i and j conditions are fulfilled.
cnt = cnt + 1;
end
end
%%plane : Yw = 1
for i=0.2:0.2:0.8,
for j=0.2:0.2:0.8,
Pw(cnt,:) = [i 1 j];
cnt = cnt + 1;
end
end
N = cnt;
%%plot3(Pw(:,1), Pw(:,2), Pw(:,3), '+');
%%Virtual camera model
%%Extrinsic parameters : R = RaRbRr
gamma = 40.0*pi/180.0;
Rr = [ [cos(gamma) -sin(gamma) 0];
[sin(gamma) cos(gamma) 0];
[ 0 0 1]; ];
beta = 0.0*pi/180.0;
Rb = [ [cos(beta) 0 -sin(beta)];
[0 1 0];
[sin(beta) 0 cos(beta)]; ];
alpha = -120.0*pi/180.0;
Ra = [ [1 0 0];
[0 cos(alpha) -sin(alpha)];
[0 sin(alpha) cos(alpha)]; ];
R = Ra*Rb*Rr;
T = [0 0 4]';
%%Intrinsic parameters
f = 0.016;
Ox = 256;
Oy = 256;
Sx = 0.0088/512.0;
Sy = 0.0066/512.0;
Fx = f/Sx;
Fy = f/Sy;
%%asr is the aspect ratio
asr = Fx/Fy;
%%Generate Image coordinates
%%surface Xw = 1
cnt = 1;
Pc = zeros(16,3);
n=length([(Ox - Fx*Pc(cnt,1)/Pc(cnt,3)) (Oy - Fy*Pc(cnt,2)/Pc(cnt,3))]);
p=zeros(16,n);
for cnt = 1:1:16,
Pc(cnt,:) = (R*Pw(cnt,:)' + T)';
p(cnt,:) = [(Ox - Fx*Pc(cnt,1)/Pc(cnt,3)) (Oy - Fy*Pc(cnt,2)/Pc(cnt,3))];
end
plot(p(:,1), p(:,2), 'r+');
axis([0 512 0 512]);
grid;
hold;
%%surface Yw = 1
for cnt = 17:1:32,
Pc(cnt,:) = (R*Pw(cnt,:)' + T)';
p(cnt,:) = [(Ox - Fx*Pc(cnt,1)/Pc(cnt,3)) (Oy - Fy*Pc(cnt,2)/Pc(cnt,3))];
end
plot(p(17:32,1), p(17:32,2), 'g+');
%%plot3(Pc(:,1), Pc(:,2), Pc(:,3), '+');
grid;
please let me know. i can understand that i need to do a little modification in this code, but i don't understand how to take a picture of this virtual cube.

Réponses (1)

Walter Roberson
Walter Roberson le 17 Mar 2013
You can use getframe() to take an image. You can also use saveas() or print() if you want an image file (e.g., .jpg)
  2 commentaires
Sat m
Sat m le 17 Mar 2013
thank you. but can you please tell me how can i take picture of this virtual cube from 4 meters distance and from a tilt angle 30 degree? can i do this with getframe()? if you give me some code it will be good for me.

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