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CS549-Homework 12 Solved

Stereo Motion (7 pts): In stereo imaging we can compute a point's world coordinates from left and right images as
๐ต๐ตโƒ—2

๐‘‹๐‘‹โƒ—๐‘Š๐‘Š = ๐‘‹๐‘‹โƒ—๐ด๐ด๐ด๐ด๐ด๐ด

๐ต๐ตโƒ— โˆ™ โˆ†โƒ—

Show that if the world point is in motion, we can compute its velocity as
๐ต๐ตโƒ—2                 ๐ต๐ตโƒ—

๐‘‰๐‘‰โƒ—๐‘Š๐‘Š = ๐‘‰๐‘‰โƒ—๐ด๐ด๐ด๐ด๐ด๐ด  − ๐‘‹๐‘‹โƒ—๐‘Š๐‘Š  

๐ต๐ตโƒ— โˆ™ โˆ†โƒ—       ๐ต๐ตโƒ— โˆ™ โˆ†โƒ—

 

Suppose that a moving world point is imaged as
๐‘๐‘               −๐‘๐‘

๐‘‹๐‘‹โƒ—๐ฟ๐ฟ๐ผ๐ผ = 20๐‘‘๐‘‘ , ๐‘‹๐‘‹โƒ—๐‘…๐‘…๐ผ๐ผ = 20๐‘‘๐‘‘

๐‘“๐‘“               ๐‘“๐‘“

With the usual imaging geometry of ๐ต๐ตโƒ— = [๐‘๐‘  0  0]๐‘‡๐‘‡, ๐น๐นโƒ— = [0  0 ๐‘“๐‘“]๐‘‡๐‘‡, what is ๐‘‰๐‘‰โƒ—๐‘Š๐‘Š?

Express ๐‘‰๐‘‰โƒ—๐‘Š๐‘Š in the simplest terms.

 

 

Shading (6 pts): Consider the two surfaces
๐‘ง๐‘ง1 = (๐‘ฅ๐‘ฅ2 + ๐‘ฆ๐‘ฆ2) and ๐‘ง๐‘ง2 = 2๐‘ฅ๐‘ฅ๐‘ฆ๐‘ฆ

Find ๐‘๐‘(๐‘ฅ๐‘ฅ, ๐‘ฆ๐‘ฆ) and ๐‘ž๐‘ž(๐‘ฅ๐‘ฅ, ๐‘ฆ๐‘ฆ) for both surfaces.
Show that ๐‘ง๐‘ง1 and ๐‘ง๐‘ง2 give rise to the same shading when a rotationally symmetric reflectance map applies, that is, when ๐‘…๐‘…(๐‘๐‘, ๐‘ž๐‘ž) = ๐‘…๐‘…(๐‘๐‘2 + ๐‘ž๐‘ž2).
 

SLAM: Following Bailey & Durrant-Whyte Part II, at time k the vehicle pose ๐ฑ๐ฑ๐‘ฃ๐‘ฃ๐‘˜๐‘˜and landmark locations m can be combined into a single state ๐ฑ๐ฑ๐‘˜๐‘˜.
๐ฑ๐ฑ๐‘ฃ๐‘ฃ๐‘˜๐‘˜

๐ฑ๐ฑ๐‘˜๐‘˜ =  ๐ฆ๐ฆ

The covariance matrix of ๐ฑ๐ฑ๐‘˜๐‘˜, denoted by ๐๐๐‘˜๐‘˜|๐‘˜๐‘˜, is partitioned into

๐๐๐‘˜๐‘˜|๐‘˜๐‘˜ = ๐๐๐๐๐‘ฃ๐‘ฃ๐‘ฃ๐‘ฃ๐‘‡๐‘‡๐‘ฃ๐‘ฃ๐‘ฃ๐‘ฃ ๐๐๐๐๐‘ฃ๐‘ฃ๐‘ฃ๐‘ฃ๐‘ฃ๐‘ฃ๐‘ฃ๐‘ฃ ๐‘˜๐‘˜|๐‘˜๐‘˜

Control input ๐ฎ๐ฎ๐‘˜๐‘˜ only affects the vehicle pose, not the landmark locations, via ๐ฑ๐ฑ๐‘ฃ๐‘ฃ๐‘˜๐‘˜ =

๐Ÿ๐Ÿ๐’—๐’—(๐ฑ๐ฑ๐‘ฃ๐‘ฃ๐‘˜๐‘˜−1, ๐ฎ๐ฎ๐‘˜๐‘˜). Thus, the update equation for state ๐ฑ๐ฑ๐‘˜๐‘˜ is

 

๐ฑ๐ฑ๐‘˜๐‘˜ = ๐Ÿ๐Ÿ(๐ฑ๐ฑ๐‘˜๐‘˜−1, ๐ฎ๐ฎ๐‘˜๐‘˜) = ๐Ÿ๐Ÿ๐’—๐’—(๐ฑ๐ฑ๐‘ฃ๐‘ฃ๐‘˜๐‘˜−1, ๐ฎ๐ฎ๐‘˜๐‘˜)

๐ฆ๐ฆ

When control ๐ฎ๐ฎ๐‘˜๐‘˜ is applied, but before any landmark measurement updates are made, the covariance prediction ๐๐๐‘˜๐‘˜|๐‘˜๐‘˜−1 is given by

๐๐๐‘˜๐‘˜|๐‘˜๐‘˜−1 = ∇๐Ÿ๐Ÿ๐ฑ๐ฑ๐๐๐‘˜๐‘˜−1|๐‘˜๐‘˜−1∇๐Ÿ๐Ÿ๐ฑ๐ฑ๐‘‡๐‘‡ + ∇๐Ÿ๐Ÿ๐ฎ๐ฎ๐”๐”๐‘˜๐‘˜∇๐Ÿ๐Ÿ๐ฎ๐ฎ๐‘‡๐‘‡

๐๐๐Ÿ๐Ÿ                       ๐๐๐Ÿ๐Ÿ where ∇๐Ÿ๐Ÿ๐ฑ๐ฑ = , ∇๐Ÿ๐Ÿ๐ฎ๐ฎ = , and ๐”๐”๐‘˜๐‘˜ is the control convariance. Show that the covariance

 

๐๐๐ฑ๐ฑ๐‘˜๐‘˜−1               ๐๐๐ฎ๐ฎ๐‘˜๐‘˜

∇๐Ÿ๐Ÿ๐’—๐’—๐ฑ๐ฑ๐๐๐‘ฃ๐‘ฃ๐‘ฃ๐‘ฃ,๐‘˜๐‘˜−1|๐‘˜๐‘˜−1∇๐Ÿ๐Ÿ๐’—๐’—๐‘‡๐‘‡๐ฑ๐ฑ + ∇๐Ÿ๐Ÿ๐’—๐’—๐ฎ๐ฎ๐”๐”๐‘˜๐‘˜∇๐Ÿ๐Ÿ๐’—๐’—๐‘‡๐‘‡๐ฎ๐ฎ

๐๐

๐๐๐‘˜๐‘˜|๐‘˜๐‘˜−1 =        ๐‘ฃ๐‘ฃ๐‘ฃ๐‘ฃ๐‘‡๐‘‡ ,๐‘˜๐‘˜−1|๐‘˜๐‘˜−1∇๐Ÿ๐Ÿ๐’—๐’—๐‘‡๐‘‡๐ฑ๐ฑ

๐๐๐Ÿ๐Ÿ

where ∇๐Ÿ๐Ÿ๐’—๐’—๐ฑ๐ฑ = ๐๐๐ฑ๐ฑ๐‘ฃ๐‘ฃ๐’—๐’— and ∇๐Ÿ๐Ÿ๐’—๐’—๐ฎ๐ฎ = ๐๐๐ฎ๐ฎ๐๐๐Ÿ๐Ÿ๐’—๐’—๐‘˜๐‘˜.
∇๐Ÿ๐Ÿ๐’—๐’—๐ฑ๐ฑ๐๐๐‘ฃ๐‘ฃ๐‘ฃ๐‘ฃ,๐‘˜๐‘˜−1|๐‘˜๐‘˜−1

 

๐๐๐‘ฃ๐‘ฃ๐‘ฃ๐‘ฃ,๐‘˜๐‘˜−1|๐‘˜๐‘˜−1
prediction simplifies (!) to

๐‘˜๐‘˜−1

 

Hint: Expand  ∇๐Ÿ๐Ÿ๐ฑ๐ฑ and ∇๐Ÿ๐Ÿ๐ฎ๐ฎ. This problem is not as hard as it looks once you understand what is being asked.

 

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