1. Course Motion

2. Course 7 Motion 3D motion expression:
Motion of a rigid object in 3D is usually expressed
as a rotation around system origin followed by a
translation.
Let be a 3D point of an object at time
Let be the point after motion at

3. be rotation , a 3x3 orthonormal matrix.
be translation vector.
Then, from time t1 to t2 :

4. r112+r122+r132=1
r212+r222+r232=1
r312+r322+r332=1
r11r21+r12r22+r13r23=0
r21r31+r22r32+r21r33=0
r31r11+r32r12+r33r13=0

5. Note:
There are only 3 independent parameters in R.
Properties of Rotation matrix:
(1)
(2)
(3)
(4)

6. 2) R is expressed by rotation axis and rotation angle.
Let rotation axis be
where: n12+n22+n32=1.
Rotation is made by a rotating with an angle  around a rotation axis. 

7. Then the elements of R are:

8. 3) Express R by 3 rotation angles.

9. 4) Quaternion form of rotation:
Quaternion is a four-element vector,
which can be used to express a rotation:
Let a rotation around axis by angle 

10. Then:
Where
-----scalar part
-----vector part
Quaternion conjugation

11. Quaternion product:
A 3D vector can be expressed as a quaternion with scalar part being zero:
Pure rotation in 3D:
Express the rotation by quaternion:

12. 5) Homogenous coordinate systems:
Then
in 3D coordinate system is written as
in homogenous coordinate system.

13. 2. Motion from 3D PCs: (1) method 1:
3 or more point on a rigid object (at least 3 point not collinear) can be seen over two-time instances t1 and t2:
(1)
(2)
(3)
Subtract (1) from (2), (3) respectively:
(4)
(5)

14. (5):
(6)
From (4),(5) and (6), R can be solved (never forget rotation matrix constraints).
After finding R,
(2) method 2:
From eq (4), vector to
subjects to a pure rotation

15. The vector
is perpendicular to rotation axis
The same is true for

16. Thus,
And rotation angle can be determined by the angle of two planes
At least 3 not collinear 3D points on rigid object is needed to determine 3D motion!

17. 3. Motion from 2D PC: ------- From 2D images to determine 3D motion.
At least 5-point correspondences over two-image view are required.
D translation can only be determined over a scale factor.
Degeneration case: 3D points are on a quadratic surface.

18. Assume: ------ Single stationary camera.
Central projection model.
Rigid moving object.
Focus length f = 1, thus
3D point:
2D image:

19. Let 3D motion from of time to of
(1)
Where
From equ (1)
(2)
Apply to both sides of equ (2)
(3)

20. Apply to both sides of equ (3)
(4)
Let we define
(5)
The eq(4) can be rewritten as
(6)
Note: eq.(6) is a homogeneous scalar equation.
is a matrix containing only motion parameters, or more PCs can uniquely determine E, subject to:

21. After matrix E is found ,translation can be solved:Or
(7)
can be determined from eq(7) subject to

22. Once is obtained, rotation R can be obtained by least-square method:
(8)
Or let

23. Note, 180o reflection of motion is still a solution of equ (7) (homogeneous equation). In this case, object is moving behind the camera. To check for a real solution, we apply to both sides of equ (2).
Therefore if z > 0, it must hold that
Thus if,
let

24. Remark: at least 8 non-degenerated PC’s over two image frames are needed to solve for a 3D motion using a linear method!
4. Motion from LC’s:
from two image frames of a single camera,
3D motion can never be solved.
over 3 frames, at least 6 LC’s are required.
motion models

25. Model A

26. Model B
Relation between model A and B:

27. Now, let we consider model the case B:
At time t1: (10)
At time t2 : (11)

28. At time t3:
(12)
From equ. (11)
(13)
Applying to both sides of equ. (13) and notice that

29. We get
(14)
In the same way
(15)
Eliminate from eqs (14) and (15), we obtain:
(16)
If we define (17)

30. F, G, H are 3x3 matrices.
Then equ. (16) can be written in compact form:
(18)
Where
Note: equ. (18) is a vector equation containing 3 linear homogeneous equations. And only two of them are linear independent (prove it).

31. Therefore, 13 LC’s over 3 frames are needed to linearly solve for F, G, and H. 
Let we define:
We have
After E is found, translation can be solved by:
Subject to , let be

32. Similarly, we define
Then
And
Subject to , let the solution be

33. In solving for R12 and R13, we rather reconstruct E and E’ for consistence (remember E and E’ were column by column).
are chosen such that
Rotations can then be solved by:

34. Remark: check revered rotations:if
Next, we determine relative amplitudes of translations
Let
substitute them into eqs (17):

35. when m, n are solved, translations are:
Build structures of 3D lines:
Direction of 3D line :
Position of :
Choose sign to make

36. Check translation reflection. Evidently,
3D Line
Check translation reflection. Evidently,
So for and
For and

37. 5. Motion from other image clues:If
Else if
5. Motion from other image clues:
Optical flow
Texture
Other clues
Will be discussed later.