IEEE Robotics & Automation Magazine - September 2010 - 91
examples in the literature where the linear coordinates precede
the angular ones. From a mathematical point of view, the difference is purely cosmetic: it is simply a consequence of the order in
which we have chosen to list the basis vectors in D and E. However, if you want to use spatial arithmetic software, then you will
have to comply with the order expected by the software.
The pattern of coordinate names appearing in (9) and (10) is
not the general case because we have used some special symbols. In
particular, we used the standard symbol x for angular velocity and
the (not quite so) standard symbol n for moment. In the general
case, the coordinates' names are derived from the name of the vector they describe. For a generic motion vector m,
^ the coordinates
would be called mx , my , mz , mOx , mOy , and mOz . Of course, you
could also number the coordinates (and basis vectors) if you prefer.
€ cker Transforms
Plu
Let A and B be two Cartesian frames defining two Pl€
ucker
coordinate systems, which we shall also call A and B. Let
A
m, B m, A f , B f 2 R 6 be coordinate vectors representing m 2 M 6
and f 2 F 6 in A and B coordinates, respectively. The coordinate
transformation rules for these vectors are as follows:
B
f ¼ BX ÃA A f ,
X ÃA ¼ (BX A )ÀT ,
[cf. (2)]. The formulae for BX A and BX ÃA depend only on the
location of B relative to A and are given by
E
XA ¼
0
0
E
1
Àr 3
0
,
1
(11)
and
X ÃA
E
¼
0
0
E
1 Àr 3
:
0
1
(12)
which maps any Euclidean vector v to the vector product
r 3 v . (It is the same idea as the matrix ~u that we encountered
in the 3-D example.)
Differentiation
Spatial vectors are differentiated in the same way as any other
vector, namely
2
2 3
0
rx
r 3 ¼ 4 ry 5 3 ¼ 4 rz
Àry
rz
Àrz
0
rx
3
ry
Àrx 5,
0
(13)
(14)
The derivative of a motion vector is a motion vector, and the
derivative of a force vector is a force vector. Unfortunately,
the situation gets a little more complicated when we come to
coordinate vectors because of the need to distinguish between
the derivative of a coordinate vector and the coordinate vector
representing a derivative.
To clarify this distinction, let m 2 M 6 be a motion vector,
and let A m 2 R 6 be a coordinate vector representing m in A
coordinates. We can now identify the following quantities:
u dm=dx: the derivative of m
A
u (dm=dx): the vector representing dm=dx in A coordinates
A
A
u d m=dx: the derivative of the coordinate vector m.
The derivative of a coordinate vector is always its componentwise derivative. If the basis vectors do not vary with x, then we
have A (dm=dx) ¼ d A m=dx (and similarly for forces); otherwise, these quantities will differ by a term depending on the
derivatives of the basis vectors.
For the special case of a time derivative in a moving Pl€
ucker
coordinate system, we have the following formulae:
A
In these equations, E is the coordinate transform from CA
to CB (the Cartesian coordinate systems defined by frames A
and B), and r locates the origin of frame B in CA coordinates
(see Figure 2).
The symbols 0 and 1 denote zero and identity matrices of
appropriate dimensions, and the expression r 3 is the skewsymmetric matrix
SEPTEMBER 2010
Figure 2. Location of frame B relative to A.
d
s(x þ dx) À s(x)
s(x) ¼ lim
:
dx!0
dx
dx
where BX A is the coordinate transformation matrix from A to
B coordinates for motion vectors, and BX ÃA is the corresponding matrix for force vectors. The two are related by
B
r
A
m ¼ X A m,
B
B
B
A
B
and
B
E
dm
d
¼ Am þ AvA 3 Am,
dt
dt
(15)
and
A
df
dt
¼
dA
f þ AvA 3Ã A f :
dt
(16)
In these equations, A is both the name of a Pl€
ucker coordinate
system and the name of the frame that defines it, while AvA is
the velocity of frame A expressed in A coordinates. These
equations introduce two new operators, 3 and 3Ã , which
are the spatial-vector equivalents of the cross-product operator
IEEE Robotics & Automation Magazine
91
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