CS549-Homework 11 Solved

Inexact Match (7 pts): In stereo imaging, if the rays defined by 𝑋𝑋⃗𝐿𝐿 and 𝑋𝑋⃗𝑅𝑅 do not intersect, we can find 𝑋𝑋⃗𝑊𝑊 anyway by minimizing an error measure.  One way to do this is to project

𝑋𝑋⃗𝑊𝑊 into the left and right image planes to give 𝑋𝑋⃗𝐿𝐿′ and 𝑋𝑋⃗𝑅𝑅′. The error E is defined as the squared difference between the observed image points 𝑋𝑋⃗𝐿𝐿 and 𝑋𝑋⃗𝑅𝑅 and the projected image points 𝑋𝑋⃗𝐿𝐿′ and 𝑋𝑋⃗𝑅𝑅′.

a. Show that for the simplified parallel optical axis camera geometry used in class where

              𝑓𝑓⃗ ∙ 𝐵𝐵⃗ = 0, 𝑅𝑅 = 𝐼𝐼, 𝑇𝑇⃗      𝐵𝐵⃗

𝐸𝐸 ≡ 𝑋𝑋⃗𝐿𝐿′ − 𝑋𝑋⃗𝐿𝐿2 + 𝑋𝑋⃗𝑅𝑅′ − 𝑋𝑋⃗𝑅𝑅2

𝑓𝑓     𝑏𝑏        2                     𝑓𝑓        2                     𝑓𝑓        𝑏𝑏        2 =  𝑧𝑧𝑊𝑊 𝑥𝑥𝑊𝑊 + 2 − 𝑥𝑥𝐿𝐿 + 𝑧𝑧 𝑊𝑊 𝑦𝑦𝑊𝑊 − 𝑦𝑦𝐿𝐿 + 𝑧𝑧 𝑊𝑊 𝑥𝑥𝑊𝑊 − 2 − 𝑥𝑥𝑅𝑅

                                                                                 𝑓𝑓                  2

+ 𝑧𝑧 𝑊𝑊 𝑦𝑦𝑊𝑊 − 𝑦𝑦𝑅𝑅 
b.      By differentiating E with respect to 𝑥𝑥𝑊𝑊 and 𝑦𝑦𝑊𝑊, show that

𝑥𝑥𝑊𝑊 = 𝑥𝑥𝐿𝐿+𝑥𝑥𝑅𝑅 𝑧𝑧 𝑊𝑊 and 𝑦𝑦𝑊𝑊 = 𝑦𝑦𝐿𝐿+𝑦𝑦𝑅𝑅 𝑧𝑧 𝑊𝑊

                                                                                                 2          𝑓𝑓                                              2           𝑓𝑓

c.       By differentiating E with respect to 𝑧𝑧𝑊𝑊 and using the results of b., show that

 𝑊𝑊 = 𝑓𝑓𝑏𝑏

 𝑧𝑧

∆𝑥𝑥

and conclude that

 𝑊𝑊 = 𝑋𝑋⃗𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵⃗2

𝑋𝑋⃗
𝐵𝐵⃗ ∙ ∆⃗

2. Reference Image (4 pts): Sometimes we treat one image in a stereo pair as a “reference image” aligned with world coordinates.  The other image and disparity are taken with respect to the reference image.  Suppose that the right image is the reference image.  Then we have

𝑋𝑋⃗𝑅𝑅𝐶𝐶 = 𝑋𝑋⃗𝑊𝑊 and 𝑋𝑋⃗𝐿𝐿𝐶𝐶 = 𝑋𝑋⃗𝑊𝑊 + 𝐵𝐵⃗

Show that in this formulation, the world point is given by

 𝑊𝑊 = 𝑋𝑋⃗𝐴𝐴𝐴𝐴𝐴𝐴 𝐵𝐵⃗2 − 𝐵𝐵⃗

𝑋𝑋⃗
                                                                                                           𝐵𝐵⃗ ∙ ∆⃗   2

You may assume that the left and right image rays intersect, although as Problem 1. shows, they don’t have to.

 

3. Inexact Match, Alternative Version (10 pts): Another way to determine distance when 𝑋𝑋⃗𝐿𝐿 and 𝑋𝑋⃗𝑅𝑅 do not intersect is to find the point 𝑋𝑋⃗𝑊𝑊 closest to the rays defined by 𝑋𝑋⃗𝐿𝐿 and 𝑋𝑋⃗𝑅𝑅. The left image ray is defined by 

 𝑊𝑊 = − 𝐵𝐵⃗ + 𝑙𝑙𝑋𝑋⃗𝐿𝐿, 0 ≤ 𝑙𝑙

𝑋𝑋⃗𝐿𝐿
2

 

The goal is to find points 𝑋𝑋⃗𝐿𝐿𝑊𝑊 and 𝑋𝑋⃗𝑅𝑅𝑊𝑊 on the left and right image rays that are closest to each other; the point equidistant between them is the desired 𝑋𝑋⃗𝑊𝑊. We can find these points by minimizing the following:  

min𝑋𝑋⃗𝐿𝐿𝑊𝑊 − 𝑋𝑋⃗𝑅𝑅𝑊𝑊2

𝑙𝑙,𝑟𝑟

Show that the solution is given by

𝑋𝑋⃗𝑅𝑅2𝑋𝑋⃗𝐿𝐿 ∙ 𝐵𝐵⃗ − 𝑋𝑋⃗𝐿𝐿 ∙ 𝑋𝑋⃗𝑅𝑅𝑋𝑋⃗𝑅𝑅 ∙ 𝐵𝐵⃗𝑋𝑋⃗𝐿𝐿 + 𝑋𝑋⃗𝐿𝐿 ∙ 𝑋𝑋⃗𝑅𝑅𝑋𝑋⃗𝐿𝐿 ∙ 𝐵𝐵⃗ − 𝑋𝑋⃗𝐿𝐿2𝑋𝑋⃗𝑅𝑅 ∙ 𝐵𝐵⃗𝑋𝑋⃗𝑅𝑅

1

 

 𝑋𝑋⃗𝑊𝑊 = 2                                        𝑋𝑋⃗𝐿𝐿2𝑋𝑋⃗𝑅𝑅2 − 𝑋𝑋⃗𝐿𝐿 ∙ 𝑋𝑋⃗𝑅𝑅2