Kinematic and Dynamic Analysis of Spatial Six Degree of Freedom ...

Using (2.7) and (2.12), one can find the magnitude of components of the moment
o
vector e ( P,
Q,
R)
and its time derivatives from known vectors ρ( x,
y,
z)
and e( L,
M , N)
and
their known time derivatives.
2.3 Kinematics of Two Unit Screws in Space
A rigid body in space can be described by two unit screws E1 and E2. Figure 2.3
shows two unit screws placed arbitrarily in space. The angle and the distance between these
~
two screws is defined with the dual angle, A = α + w ⋅ a whereα is the twist angle and a is the
shortest distance between the two screw axis.
Figure 2.3 – Two arbitrary unit screws in space
For these two unit screws, one can write the following equations:
2
~ 2 ~ 2 ~ 2
E = 1+
w ⋅ 0 ⎫ L
1
1 + M1
+ N1
= 1+
w⋅
0
2 ⎪ ~ 2 ~ 2 ~ 2
E2
= 1+
w ⋅ 0 ⎬ ⇒ L2
+ M 2 + N2
= 1+
w⋅
0
~ ⎪ ~ ~ ~ ~ ~ ~ ~
E1
⋅ E2
= cos A⎭
L1L2
+ M1M
2 + N1N
2 = cos A
From screw algebra we know that
a
α
~ ~ ~
L , M , N )
1(
1 1 1
(2.13)
~
cos( A) = cosα
− w ⋅ asinα
(2.14)
~ ~ ~ ~ ~ ~
From six dual coordinates L1, M1,
N1,
L2,
N2,
N2
that describe the position of a rigid
body in space, just three of them are independent as we have three constraint equations (2.13).
E
E
~ ~ ~
L , M , N )
2(
2 2 2
13