Transcript Revisit Quaternion

1
Motivation (Horn87)
Assume
correspondence
has been
determined…
• Registration of 3D Shapes
n
min  Rpi  b  qi
R ,b
2
i 1
Any other better solutions for this least-square problem?
2
Exercise
• Configuration A
• Configuration B
– p1 = (3,0,0)
– p2 = (0,3,0)
– p3 = (0,0,3)
– q1 = (3,3,-3)
– q2 = (3,0,0)
– q3 = (6,0,-3)
Find the rotation and translation that takes
configuration A to configuration B
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Solution
 3 3 6
3 0 0 
R 0 3 0  d d d    3 0 0 
  3 0  3
0 0 3 
 4  1  3 
d   1   1   0 
 2 1  3
This is a badly chosen example,
where the three points are
linearly independent.
If this is not so (such as three
coplanar points), this simple
approach cannot be used.
3  0 0 3
3
3 0 0  3 3 6   3
0   3 0 0
0
R 0 3 0   3 0 0    0
0 0 3  3 0  3  3  3  3 0 3 0
0 0 1 
R  1 0 0
0 1 0
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Recap in CG Course (link)
•
•
•
•
•
New algebra defined (addition, multiplication)
Length; unit quaternion as rotation
Perform rotations
Converting to/from rotation matrices
slerp
Rot(nˆ, )  q  cos 2  sin 2 nˆ
x  qxq*  ...
 

 
 (q02  q  q ) x  2q0 q  x  2q (q  x)
x  pxp*


x  qxq*  q pxp* q*  (qp) x(qp)*
sin  (1  t )
sin t
s(t ) 
q
r
sin 
sin 
cos  q0 r0  q1r1  q2 r2  q3r3
q
r
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History
• Quaternions were introduced by Irish
mathematician Sir William Rowan Hamilton in
1843. Hamilton was looking for ways of extending
complex numbers to higher spatial dimensions. He
could not do so for 3 dimensions, but 4 dimensions
produce quaternions.
• According to the story Hamilton told, on October
16, he was out walking along the Royal Canal in
Dublin with his wife when the solution in the form
of the equation
suddenly occurred to him; Hamilton then promptly
carved this equation into the side of the nearby
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Brougham Bridge
Definitions1
Addition
Multiplication
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Definitions2
Complex conjugate
Norm (of quaternion)
 pq  p q
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Alternative Formulation
A quaternion can be conveniently thought of
as either:
• A vector with four components;
• A scalar plus a vector with three
components; or
• A complex number with three different
“imaginary” parts
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Notation: we use
to denote quaternion
Treat quaternion as a vector in R4
Product in Matrix Form (Horn)
R : Expand first quaternion
 r0  rx  ry
r
r0  rz
x

R
ry rz
r0

 rz  ry rx
RRT  r  rI  RT R
R R
*
 rz 
 r0
ry  T   rx
,R 
 ry
 rx 


r0 
  rz
“R is orthogonal”!
Expand second quaternion : R
Note the matrices are different!
T
rx
ry
r0
rz
 rz
r0
ry
 rx
rz 
 ry 
rx 

r0 
This is the “matrix
way” to compute
quaternion product
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Details
R*  RT
r  r0 , rx , ry , rz 
 r0
r
x
R 
ry

 rz
 rx
r0
 rz
ry
 ry
rz
r0
 rx
r *  r0 , rx , ry , rz 
 rz 
 r0
 ry  T   rx
R 

 ry
rx


r0 
  rz
rx
r0
rz
 ry
ry
 rz
r0
rx
 (rx )  (ry )  ( rz )   r0
 r0
(r )
  r
r
(

r
)

(

r
)
x
0
z
y
 x
R*  
(ry )  ( rz )
r0
(rx )   ry

 
(

r
)
(

r
)

(

r
)
r
y
x
0
 z
   rz
rz 
ry 
 rx 

r0 
rx
r0
rz
 ry
ry
 rz
r0
rx
rz 
ry 
 rx 

r0 
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This is new!
Dot Product of Quaternions
Def:
4D vector viewpoint
p.9
 pq  p q
 pq r  Q p  r  Q p T r  pT Q T r
 
 
 pT Q *r  p T rq *  p  rq *
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Unit Quaternion as Rotation1
Observations: (why Lq(v) is a rotation…)
1. Length preserved after operation
2. If v is along q, it is left unchanged.
[Horn] dot product and triple product preserved …
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Unit Quaternion as Rotation2
Lq is linear over R3
Lq(v) = Lq(n+a) = Lq(n) + Lq(a) = Lq(n) + a
v
u
n
a
q
v=Lp(u) , w=Lq(v)
w= Lq(v)=qvq*=q(pup*)q*=(qp)u(qp)*=Lqp(u)
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Application: Data Registration
Find the most appropriate
rotation R and translation b
Iterative (least square) solution
or Closed form solution
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Convert global coordinates to local coordinate (from centroid)
1 n
1 n
centroidp   pi , q   qi
n i 1
n i 1
pi  pi  p ; qi  qi  q
1 n


pi    pi  p    pi  np   pi  n  pi   0

i 1
i 1
i 1
i 1
 n i 1 
Similarly,
n
n
n
n
n
 q  0
i 1
i
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 v  w   v  wv  w
  v  v   2v  w  w  w
  v  2 v  w   w
2
Objective function (minimization)
2
2
First, determine the rotation R, then the optimal translation by:
Optimal rotation R:
2
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Convert to quaternion:
This
correspond
to R and R
in p.8
is symmetric, with real eigenvalues l1, l2, l3, l4
and corresponding orthogonal unit eigenvectors v1,
v2, v3, v4 in R4
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Simplified to:
maxqT Mq
q
Represent q in eigen basis
q  q  1v1   2 v2   3v3   4 v4   1v1   2 v2   3v3   4 v4   12   22   32   42
q is rotation q  1  12   22   32   42  1
Therefore,
qT Mq  l112  l2 22  l3 32  l4 42  l112  l1 22  l1 32  l1 42  l1
maximumoccurs at : 1  1,  2   3   4  0
q  v1
l1: largest eigenvalue
The optimum rotation is the eigenvector with
largest eigenvalue. (It is already normalize).
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Details
T
Pi Qi is symmetric
p1
p2
p3 
 0
 0  q1  q2  q3 
 p

q

0

p
p
0

q
q
1
3
2
1
3
2
T
 Q

Pi  
i
  p2
q2 q3
p3
0
 p1 
0
 q1 





p

p
p
0
q

q
q
0
2
1
2
1
 3

 3

p2 q3  p3q2
p3q1  p1q3
 p1q1  p2 q2  p3q3
 pq pq
p1q1  p2 q2  p3q3
p1q2  p2 q1
3 2
2 3
T

Pi Qi 

p3q1  p1q3
p1q2  p2 q1
 p1q1  p2 q2  p3q3

p1q3  p3q1
p3q2  p2 q3
  p2 q1  p1q2
 p2 q1  p1q2


p1q3  p3q1


p3q2  p2 q3

 p1q1  p2 q2  p3q3 
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The Real Exercise
• Configuration A
• Configuration B
– p1 = (3,0,0)
– p2 = (0,3,0)
– p3 = (3,3,0)
– q1 = (1,4,1)
– q2 = (1,1,4)
– q3 = (1,4,4)
Find the rotation and translation that takes
configuration A to configuration B
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Quaternion in SVL
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