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Asked: 2026-06-10T18:02:31+00:00

I have a camera and its K matrix (calibration matrix) also I have image

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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 :
enter image description here

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?

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1 Answer

  1. Editorial Team

    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

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