Assignment 2: Single-view Reconstruction

16822: Geometry-Based Methods in Vision

Andrew id: shubhikg

Q1. (a) Stanford Bunny

The projection matrix obtained from the correspondences is:

\[ P = \begin{bmatrix} 6.36922229e^3 & -2.86047769e^3 & 1.23277412e^3 & 2.22744066e^3\\ -1.00866303e^3 & -6.72507991e^3 & 2.13180464e^3 & 1.82406274e^3\\ 5.36375510e^1 & -1.37865245e^0 & -7.01487421e^{-1} & 1.00000000e^0\\ \end{bmatrix} \]

Annotated Image	Surface Point Image	Boudning Box Image

Q1. (b) Cuboid

Input Image	Annotated Image	Edge Image

Q2. (a) Camera calibration from vanishing points

Annotated Image	Vanishing points and principal point

The K obtained is:

\[ K = \begin{bmatrix} 1.15417802e^3 & 0.0 & 5.75066005e^2\\ 0.0 & 1.15417802e^3 & 4.31939090e^2\\ 0.0 & 0.0 & 1.0\\ \end{bmatrix} \]

The principant point obtained is (575, 431).

Algorithm

1. We start with the assumption that K is sqaure matrix and has zero skew. Hence \(w = K^{-T} K^{-1}\) will be of the form:

\[ w = \begin{bmatrix} a & 0 & b \\ 0 & a & c \\ b & c & d \\ \end{bmatrix} \]

2. For each pair of parallel lines we then find the vanishing point by computing their intersection.

3. We have 2 sets of perpendicular vanishing points and each pair gives us one contrainst that is \(v_i^T w v_j = 0\)

4. With 3 unknows(-1 for scaling factor) of w and 3 equations we can expand and write it in the form AX=0 and then solve for w using SVD.

5. We can then obtain K by doing Cholesky decomposition as \(w = K^{-T} K^{-1}\)

Q2. (b) Camera calibration from metric planess

Input Image	Annotated Square 1	Annotated Square 2	Annotated Square 3

The K obtained is:

\[ K = \begin{bmatrix} 1.08054358e^3 & -9.20405422e^0 & 5.16504953e^2 \\ 0.00000000e^0 & 1.07783437e^3 & 4.00688382e^2\\ 0.0 & 0.0 & 1.0\\ \end{bmatrix} \]

The angle between each plane is:

Angle between plane 1 and 2 is 67.44565183.

Angle between plane 3 and 1 is 92.22894955.

Angle between plane 2 and 3 is 94.73850609.

Algorithm

1. In this we start with finding the honography that maps the four corners of a square (0, 1), (1, 1), (0, 1) and (0, 0) to the annotatons provided in the image.

2. We have 3 such annotations and hence will obtain 3 homographies \(H_1, H_2, H_3\).

3. For each homography we have the contraint \(h_1^T w h_2=0\) and \(h_1^T w h_1=h_2^T w h_2\) where \(h_i\) is the ith column of H.

4. Hence every h gives us 2 contraints and we have 6 unknows.

5. Solve for w using these 6 contraints forming AX=0 and then doing SVD.

6. We can then obtain K by doing Cholesky decomposition as \(w = K^{-T} K^{-1}\).

Q3: Single View Reconstruction

Annotated Image

Reconstructed Image view 1	Reconstructed Image view 2	Reconstructed Image view 3

The K obtained is:

\[ K = \begin{bmatrix} 808.52021208 & 0. & 510.71184892\\ 0. & 808.52021208 & 363.63610862\\ 0.0 & 0.0 & 1.0\\ \end{bmatrix} \]

Algorithm

1. We are given annotations of 6 planes. Using this we find 3 pair of perpendicular lines and use q2a to find the K matrix.

2. For each plane we then extract the normal using \(n = K^T v\) where v is the vanishing line for each plane.

3. We then unproject the first corner point using \(d_{direction} = K^{-1}x\) and fix its depth to 1.

4. Obtain the eqaution of every plane as \(n^TX + a = 0\) and find the value of a using points of known depth.

5. For every plane then we then get all the point lying on that plane in the image, unproject them to get their direction vectors.

6. We then fimd the true 3d points of every plane by intersection of all the rays and the true 3d eaution of the plane obtained in step 4.

7. Update the corners of this plane so that they can be used as reference for the next plane.

8. Do this for all the planes.