recovery plane from 4 point imaging using calibration camera - matlab

Question

I have a camera and its K matrix (calibration matrix) also I have image of plane, I know the real points of the 4 corners and thier correspondence pixel. I know how to compute the H matrix if z=0 (H is homography matrix between Image and the real plane).
And Now I try to get the real point of the plane (3D point) with the rotation matrix and the transltion vector
I follow this paper :Calibrating an Overhead Video Camera by Raul Rojas in section 3 - 3.3.
My code is:
ImagePointsScreen=[16,8,1;505,55,1;505,248,1;44,301,1;];
screenImage=imread( 'screen.jpg');
RealPointsMirror=[0,0,1;9,0,1;9,6,1;0,6,1]; %Mirror
RealPointsScreen=[0,0,1;47.5,0,1;47.5,20,1;0,20,1];%Screen
imagesc(screenImage);
hold on
for i=1:4
drawBubble(ImagePointsScreen(i,1),ImagePointsScreen(i,2),1,'g',int2str(i),'r')
end
Points3DScreen=Get3DpointSurface(RealPointsScreen,ImagePointsScreen,'Screen');
figure
hold on
plot3(Points3DScreen(:,1),Points3DScreen(:,2),Points3DScreen(:,3));
for i=1:4
drawBubble(Points3DScreen(i,1),Points3DScreen(i,2),1,'g',int2str(i),'r')
end
function [ Points3D ] = Get3DpointSurface( RealPoints,ImagePoints,name)
M=zeros(8,9);
for i=1:4
M((i*2)-1,1:3)=-RealPoints(i,:);
M((i*2)-1,7:9)=RealPoints(i,:)*ImagePoints(i,1);
M(i*2,4:6)=-RealPoints(i,:);
M(i*2,7:9)=RealPoints(i,:)*ImagePoints(i,2);
end
[U S V] = svd(M);
X = V(:,end);
H(1,:)=X(1:3,1)';
H(2,:)=X(4:6,1)';
H(3,:)=X(7:9,1)';
K=[680.561906875074,0,360.536967117290;0,682.250270165388,249.568615725655;0,0,1;];
newRO=pinv(K)*H;
h1=newRO(1:3,1);
h2=newRO(1:3,2);
scaleFactor=(norm(h1)+norm(h2))/2;
newRO=newRO./scaleFactor;
r1=newRO(1:3,1);
r2=newRO(1:3,2);
r3=cross(r1,r2);
r3=r3/norm(r3);
R=[r1,r2,r3];
RInv=pinv(R);
O=-RInv*newRO(1:3,3);
M=K*[R,-R*O];
for i=1:4
res=pinv(M)* [ImagePoints(i,1),ImagePoints(i,2),1]';
res=res';
res=res*(1/res(1,4));
Points3D(i,:)=res';
end
Points3D(i+1,:)=Points3D(1,:); %just add the first point to the end of the array for draw square
end
My result is :
Now I have two problem :
1.The point 1 is at (0,0,0) and this is not the real location
2.the points are upside down
What I am doing worng?

Answer 1

A homography is normally the transform of a plane in two positions/rotations.
The position in camera coordinates of a plane is normally called pose or extrinsic parameters
opencv has a solvePnP() function which uses Ransac to estimate the position of a known plane.
ps. Sorry don't know the equivalent matlab but Bouguet has a matlab version of the openCV 3D functions on his site

Answer 2

I found the answer in the paper: Calibrating an Overhead Video Camera by Raul Rojas in section 3 - 3.3.
for the start: H=K^-1*H
Given four points in the image and their known coordinates in the world, the
matrix H can be recovered, up to a scaling factor . We know that the first
two columns of the rotation matrix R must be the first two columns of the
transformation matrix. Let us denote by h1, h2, and h3 the three columns of
the matrix H.Due to the scaling factor we then have that
xr1 = h1
and
xr2 = h2
Since |r1| = 1, then x= |h1|/|r1| = |h1| and x = |h2|/|r2| = |h2|. We can thus
compute the factor and eliminate it from the recovered matrix H. We just set
H'= H/x
In this way we recover the first two columns of the rotation matrix R.
The third column of R can be found remembering that any column in a rotation
matrix is the cross product of the other two columns (times the appropriate
plus or minus sign). In particular
r3 = r1 × r2
Therefore, we can recover from H the rotation matrix R. We can also recover
the translation vector (the position of the camera in field coordinates). Just
remember that
h'3 = −R^t
Therefore the position vector of the camera pin-hole t is given by
t = −R^-1 h3