Rigid-Body Displacements
1
Rigid-Body Displacement
Consider Frame A and vector PQ.
P Q = q1 a1 + q2 a2 + q3 a3
3
a3
2.5
2
1.5
Q
a2
1
0.5
0
0
P
3
0.5
2.5
1
2
1.5
1.5
2
2.5
3
a1
1
0.5
0
2
Rigid-Body Displacement
Let Frame B be Frame A, after rigid-body displacement g.
b2
3
b3
2.5
a3
Q′
2
1.5
P Q = q 1 a1 + q 2 a2 + q 3 a3
Q
1
g
P′
0.5
a2
P
P 0 Q 0 = q1 b 1 + q2 b 2 + q3 b 3
0
b1
0
1
a1
2
3
4
-5
-4
-3
-2
-1
0
1
2
3
5
Rigid-Body Displacement
Let Frame B be Frame A, after rigid-body displacement g.
b2
3
b3
2.5
P Q = q1 a1 + q2 a2 + q3 a3
a3
P Q0 = q10 a1 + q20 a2 + q30 a3
Q′
2
1.5
Q
1
P′
a2
P
0.5
0
b1
0
1
a1
2
3
4
-5
-4
-3
-2
-1
0
1
2
3
6
Rigid-Body Displacement
Let Frame B be Frame A, after rigid-body displacement g.
b2
3
b3
2.5
P Q = q1 a1 + q2 a2 + q3 a3
a3
P Q0 = q10 a1 + q20 a2 + q30 a3
Q′
2
1.5
Q
1
P′
a2
P
0.5
0
b1
0
1
a1
2
3
4
-5
-4
-3
-2
-1
0
1
2
3
2 03
2 3
q1
q1
4q20 5 = R 4q2 5 + d
q30
q3
7
Rotation
Translate frame B so reference frames share an origin.
P Q = q1 a1 + q2 a2 + q3 a3
a3
3
b2
P Q00 = q1 b1 + q2 b2 + q3 b3
2.5
2
Q′′
b3
= q100 a1 + q200 a2 + q300 a3
1.5
Q
1
a2
P
0.5
0
0
3
0.5
1
b1
a
1.5
0
2
2.5
3
-1
2
1
2 00 3 2
q1
R11
4q200 5 = 4R21
q300
R31
R12
R22
R32
32 3
R13
q1
R23 5 4q2 5
R33
q3
-2
8
Rotation
Translate Frame B so reference frames share an origin.
a3
3
b2
2.5
2
Q′′
b3
1.5
Q
1
a2
P
0.5
0
0
3
0.5
1
b1
a
1.5
0
2
2.5
3
-1
-2
2
1
b 1 = a1
9
Rotation
Translate Frame B so reference frames share an origin.
b2 = |b2 | cos(✓)a2 + |b2 | sin(✓)a3
a3
3
= cos(✓)a2 + sin(✓)a3
2.5
2
b3
b2
1.5
1
0.5
b1
a
0
0123 -2
-1.5
-1
-0.5
0
|b2 | sin(✓)
✓
0.5
a2
1
1.5
|b2 | cos(✓)
2
2.5
3
10
Rotation
Frame B so the two frames share an origin.
2.5
2
b3 =
a3
3
b3
=
|b3 | sin(✓)
|b3 | cos(✓)
|b3 | sin(✓)a2 + |b3 | cos(✓)a3
sin(✓)a2 + cos(✓)a3
b2
1.5
✓
1
0.5
b1
a
0
0123 -2
-1.5
-1
-0.5
0
a2
0.5
1
1.5
2
2.5
3
11
Rotation
Translate frame B so reference frames share an origin.
P Q00 = q1 b1 + q2 b2 + q3 b3
a3
3
= q100 a1 + q200 a2 + q300 a3
b2
2.5
2
Q′′
b3
P Q00 = q1 (a1 ) + q2 (cos(✓)a2 + sin(✓)a3 )
1.5
Q
1
+ q3 ( sin(✓)a2 + cos(✓)a3 )
a2
P
0.5
= q1 a1 + (q2 cos(✓)
0
0
q3 sin(✓))a2
3
0.5
1
b1
a
1.5
0
2
2.5
3
2
1
+ (q2 sin(✓) + q3 cos(✓))a3
-1
-2
12
Rotation
Translate frame B so reference frames share an origin.
P Q00 = q100 a1 + q200 a2 + q300 a3
a3
3
P Q00 = q1 a1 + (q2 cos(✓)
b2
2.5
2
+ (q2 sin(✓) + q3 cos(✓))a3
Q′′
b3
1.5
Q
1
q100 = q1
a2
P
0.5
q3 sin(✓))a2
q200 = q2 cos(✓)
0
0
q3 sin(✓)
3
0.5
1
b1
a
1.5
0
2
2.5
3
-1
2
1
q300 = q2 sin(✓) + q3 cos(✓)
-2
13
Rotation
Translate frame B so reference frames share an origin.
P Q00 = q1 (a1 ) + q2 (cos(✓)a2 + sin(✓)a3 )
+ q3 ( sin(✓)a2 + cos(✓)a3 )
a3
3
b2
P Q00 = q100 a1 + q200 a2 + q300 a3
2.5
2
Q′′
b3
1.5
Q
1
a2
P
0.5
0
0
3
0.5
1
b1
a
1.5
0
2
2.5
3
2
1
2 00 3 2
q1
1
4q200 5 = 40
q300
0
0
cos(✓)
sin(✓)
32 3
0
q1
sin(✓)5 4q2 5
cos(✓)
q3
-1
-2
14
Rotation
Translate frame B so reference frames share an origin.
2
1
0
R = 40 cos(✓)
0 sin(✓)
a3
3
b2
2.5
2
Q′′
b3
1.5
⇡
✓= !
4
Q
1
a2
P
0.5
0
0
3
0.5
1
b1
a
1.5
0
2
2.5
3
-1
-2
2
1
2
1
6
R = 40
0
3
0
sin(✓)5 = Rot(x, ✓)
cos(✓)
0
p
2
p2
2
2
0p
3
27
p2 5
2
2
15
Translation
Let d be the vector from the origin of Frame A to the origin
of Frame B, expressed in terms of Frame A.
a
b2 3
3
d = 1a1
b3
2.5
3a2 + 1a3
2
1.5
a2
1
0.5
b1
0
0
d
a1
1
2
3
4
-5
-4
-3
-2
-1
0
1
2
3
We can characterize a rigid-body displacement with a rotation
matrix and translation vector.
16
Rigid-Body Displacement
Let Frame B be Frame A, after rigid-body displacement g.
b2
3
b3
2.5
P Q = q1 a1 + q2 a2 + q3 a3
a3
P Q0 = q10 a1 + q20 a2 + q30 a3
Q′
2
1.5
Q
1
P′
a2
P
0.5
0
b1
0
1
a1
2
3
4
-5
-4
-3
-2
-1
0
1
2
3
2 03
2 3
q1
q1
4q20 5 = R 4q2 5 + d
q30
q3
17
Rigid-Body Displacement
Let Frame B be Frame A, after rigid-body displacement g.
b2
3
b3
2.5
P Q = q1 a1 + q2 a2 + q3 a3
a3
P Q0 = q10 a1 + q20 a2 + q30 a3
Q′
2
1.5
Q
1
P′
a2
P
0.5
0
b1
0
1
a1
2
3
4
-5
-4
-3
-2
-1
0
1
2 03 2
1
q1
4q20 5 = 6
40
q30
0
2
3
2
0
p
2
p2
2
2
3
2
= 4 2.295
3.12
32 3 2 3
0p
1
1
27 4 5
4 35
2
+
5
2
p
2
1
1
2
18
Rigid-Body Displacement
Let Frame B be Frame A, after rigid-body displacement g.
b2
3
b3
2.5
P Q = q1 a1 + q2 a2 + q3 a3
a3
P Q0 = q10 a1 + q20 a2 + q30 a3
Q′
2
1.5
Q
1
P′
a2
P
0.5
0
b1
0
1
a1
2
3
4
-5
-4
-3
-2
-1
0
1
2
3
2 3
02 0 3
q1
q1
4q2 5 = RT @4q20 5
q3
q30
1
dA
19