Uncalibrated 1D projective camera and 3D affine reconstruction of lines
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Citations
Multiple View Geometry in Computer Vision.
Self-calibration of a 1D projective camera and its application to the self-calibration of a 2D projective camera
Image Analysis and Computer Vision
Self-Calibration of a 1D Projective Camera and Its Application to the Self-Calibration of a 2D Projective Camera
Linear 2D localization and mapping for single and multiple robot scenarios
References
Shape and motion from image streams under orthography: a factorization method
A computer algorithm for reconstructing a scene from two projections
Model-based object pose in 25 lines of code
What can be seen in three dimensions with an uncalibrated stereo rig
Geometric invariance in computer vision
Related Papers (5)
Frequently Asked Questions (9)
Q2. What is the main idea of the thirteen-line linear algorithm?
Using line segments instead of points as features has attracted the attention of many researchers [11, 2, 29, 28, 27, 1] for various tasks such as pose estimation, stereo and structure from motion.
Q3. How many complicated algebraic constraints are there?
the 27 tensor components that are introduced as intermediate parameters are still subject to 9 complicated algebraic constraints.
Q4. What is the determinant of the 2 2 matrix?
Its 2 2 determinant must vanish, i.e. jT jke2j = 0:As each entry of the 2 2 matrix is homogeneous linear in e2 = (u1; u2)T , the expansion of jT jke2j gives a homogeneous quadratic u21 + u1u2 + u22 = 0; (11) where ; ; are known in terms of Tijk .
Q5. How can the authors convert the affine structures obtained to Euclidean ones?
The affine structures obtained can be converted to Euclidean ones (up to a global scaling factor) as soon as the authors know the aspect ratio of the camera [17].
Q6. What is the tensor component of the projective camera?
The authors have just proven that recovering the directions of affine lines in 3D space is equivalent to 2D projective reconstruction from one-dimensional projective images.
Q7. How can the authors recover the projection matrices without loss of generality?
Without loss of generality, the authors can always take the following normal forms for the 3 projection matricesM = I2 2 0 ; M0 = A2 2 c ; M00 = D2 2 f : (12)It is straightforward to verify that the projection center of the first view is Ker(M1) = (0; 0; 1)T , so that e01 = c and e001 = f.
Q8. What is the main advantage of using simplified camera models?
In such cases, it is not only easier to use these simplified models but also advisable to do so, as by explicitly eliminating the ambiguities from the algorithm, one avoids computing parameters that are inherently ill-conditioned.
Q9. How many minors of the joint projection matrix are there?
One way to explicitly recover the scale factors( ; 0; 00)T is to notice that the rescaled image coordinates( u; 0u0; 00u00)T should lie in the joint image, or alternatively to observe the following matrix identity:0@M uM0 0u0M00 00u001A = 0@MM0M001A I3 3 x : The rank of the left matrix is therefore at most 3.