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Projection matrices in OpenCV vs Multiple View Geometry

asked 2017-08-04 11:17:47 -0500

Milkboy gravatar image

I am trying to follow "Multiple View Geometry in Computer Vision" formula 13.2 for computing the homography between for a calibrated stereo rig. It should be simple math

H = K' (R - t. transpose(n) / d) inv(K)

Where H is the 3x3 homography. K and K' are 3x3 camera intrinsic matrices, R is a 3x3 rotation between the cameras. t is column vector translation between the cameras. n is a plane normal vector and d is the constant of the plane equation of a plane that both cameras are viewing. The idea here is that going right to left, a homogenous pixel coordinate is unprojected to a ray, the ray is intersected with a plane and the intersection point is projected to the other image.

When I plug numbers in and try to match up two captured images I can't get the math as shown to work. My main problem is that I don't understand how simply multiplying a 3D point by a camera matrix K can project the point. In the opencv documentation for the calibration module:

x' = x / z

y' = y / z

u = fx * x' + cx

v = fy * y' + cy

By the above math, x and y are divided by z before the focal length is multiplied. But I don't see how the math from the text accomplishes the same thing by merely multiplying a point by K. Where is the divide by z in the formula for H?

Can someone help with this problem that is probably just notation? Scott

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answered 2017-08-05 09:08:22 -0500

Eduardo gravatar image

See this course: Lecture 12: Camera Projection, Robert Collins, CSE486, Penn State.

Here the relevant slides uploaded:




Note: here the principal point is not considered (in this representation) but it is the same thing.

Another good reference but more oriented graphics rendering than computer vision.

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Asked: 2017-08-04 11:17:47 -0500

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Last updated: Aug 05 '17