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11.4 Triangulation, Study notes of Geometry

(camera geometry). Measurements. Pose Estimation known estimate. 3D to 2D correspondences. Triangulation estimate known. 2D to 2D coorespondences.

Typology: Study notes

2022/2023

Uploaded on 05/11/2023

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Triangulation
16-385 Computer Vision (Kris Kitani)
Carnegie Mellon University
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Triangulation

16-385 Computer Vision (Kris Kitani) Carnegie Mellon University

Structure (scene geometry) Motion (camera geometry) Measurements Pose Estimation known^ estimate 3D to 2D correspondences Triangulation (^) estimate known^ 2D to 2D coorespondences Reconstruction (^) estimate estimate 2D to 2D coorespondences

Triangulation

x

x

0

C

C

0 image 1 image 2 Find 3D object point Will the lines intersect?

Triangulation

x

x

0

C

C

0 image 1 image 2 Find 3D object point (no single solution due to noise)

x = PX known known Can we compute X from a single correspondence x?

camera center

X

image plane x x y z z = f infinitely many 3D points along the ray no unique 3D point

x = PX known known Can we compute X from two correspondences x and x’? yes if perfect measurements

x = PX There will not be a point that satisfies both constraints because the measurements are usually noisy known known x 0 = P 0 X x^ =^ PX Need to find the best fit Can we compute X from two correspondences x and x’? yes if perfect measurements

x y z

p 1 p 2 p 3 p 4 p 5 p 6 p 7 p 8 p 9 p 10 p 11 p 12

X
Y
Z

How do we solve for unknowns in a similarity relation? Direct Linear Transform Remove scale factor, convert to linear system and solve with SVD. x = PX Also, this is a similarity relation because it involves homogeneous coordinates x = ↵PX Same ray direction but differs by a scale factor (inhomogeneous coordinate) (homogeneous coordinate)

x y z

p 1 p 2 p 3 p 4 p 5 p 6 p 7 p 8 p 9 p 10 p 11 p 12

X
Y
Z

How do we solve for unknowns in a similarity relation? Direct Linear Transform Remove scale factor, convert to linear system and solve with SVD. x = PX Also, this is a similarity relation because it involves homogeneous coordinates x = ↵PX Same ray direction but differs by a scale factor (inhomogeneous coordinate) (homogeneous coordinate)

Recall: Cross Product

a b c = a ⇥ b c · a = 0 c^ ·^ b^ = 0 Vector (cross) product takes two vectors and returns a vector perpendicular to both a ⇥ b =

a 2 b 3 a 3 b 2 a 3 b 1 a 1 b 3 a 1 b 2 a 2 b 1

cross product of two vectors in the same direction is zero a ⇥ a = 0 remember this!!!

x y z

p 1 p 2 p 3 p 4 p 5 p 6 p 7 p 8 p 9 p 10 p 11 p 12

X
Y
Z

x y z

—– p

1

—– p

2

—– p

3

X

x y z

p

1

X

p

2

X

p

3

X
x
y
p

1

X
p

2

X
p

3

X
yp

3

X p

2

X
p

1

X xp

3

X
xp

2

X yp

1

X

x ⇥ PX = 0 Using the fact that the cross product should be zero Third line is a linear combination of the first and second lines. (x times the first line plus y times the second line) One 2D to 3D point correspondence give you 2 equations

x
y
p

1

X
p

2

X
p

3

X
yp

3

X p

2

X
p

1

X xp

3

X
xp

2

X yp

1

X

x ⇥ PX = 0 Third line is a linear combination of the first and second lines. (x times the first line plus y times the second line) One 2D to 3D point correspondence give you 2 equations Using the fact that the cross product should be zero