The Kinematic Image of 2R Dyads
and Exact Synthesis of 5R Linkages
Tudor-Dan Rad and Hans-Peter Schröcker
Unit Geometry and CAD
University of Innsbruck, Austria
September 14, 2018
We characterise the kinematic image of the constraint variety of a 2R dyad
as a regular ruled quadric in a three-space that contains a “null quadrilateral”.
Three prescribed poses determine, in general, two such quadrics. This allows
us to modify a recent algorithm for the synthesis of 6R linkages in such a
way that two consecutive revolute axes coincide, thus producing a 5R linkage.
Using the classical geometry of twisted cubics on a quadric, we explain some
of the peculiar properties of the the resulting synthesis procedure for 5R
linkages.
Keywords: 2R dyad, kinematic map, 5R linkage, Goldberg linkage, linkage synthesis,
dual quaternions
MSC 2010: 70B15 (primary); 70B10, 12D05, 51N15
1. Introduction
Given three poses of a rigid body, there exists, in general, a unique closed kinematic loop
of four revolute joints with one degree of freedom, a so-called Bennett or 4R linkage,
one of whose links visits the three poses.
Suh (1969) solved this synthesis problem
by constructing two 2R dyads to the three given poses and proving that they can be
combined to form a Bennett linkage. The three pose synthesis problem for 4R linkages
was re-visited by Brunnthaler et al. (2005) in a dual quaternion setting. A generalisation
of their approach led to a factorisation theory for “motion polynomials” (Hegedüs et al.,
2013a). It was used by Hegedüs et al. (2014) to synthesise spatial 6R linkages – closed
kinematic chains with six revolute joints – to four prescribed poses in general position.
More precisely, the four poses determine two families of (not necessary real) rational
coupler motions, each giving rise to nine families of overconstrained 6R linkages.
It
should also be mentioned that the construction is only capable of producing a special
type of 6R linkages.
1
arXiv:1503.04566v2  [math.MG]  27 Jul 2015
In this paper we intend to close the gap in this sequence of exact factorisation based
synthesis procedures for closed loop linkages and present a construction of 5R linkages
(Goldberg linkages) to “more than three but less than four poses”. Our synthesis al-
gorithm is based on a characterisation of the image of open 2R chains under Study’s
kinematic mapping and the observation that the coupler motion of a 5R linkage is also
parameterised by a cubic motion polynomial (see Hegedüs et al., 2013b). This allows to
treat 5R synthesis as a special case of above mentioned synthesis of 6R linkages.
We continue this text with a quick introduction to dual quaternions and kinematics
in Section 2. In Section 3 we provide a geometric characterisation for the kinematic
image of a 2R dyad. While necessary properties can be derived easily, the proof of their
sufficiency requires more work and comprises the main part of this text. The synthesis
of 5R linkage in Section 4 is then merely a corollary to previous results. Note however,
that it is not obvious how to use the additional degrees of freedom in exact 5R synthesis.
2. Preliminaries
This article’s scene is the projectivised dual quaternion model of spatial kinematics.
Here, we give a very brief introduction to this model for the purpose of settling our
notation. More details will be introduced in the text as needed. For more thorough in-
troductions to dual quaternions and there relations to kinematics we refer to (Klawitter,
2015, Section 3) or (Selig, 2005, Section 11). Familiarity with Hegedüs et al. (2013a) is
recommended too.
The dual quaternions, denoted by DH, form an associative algebra in R8 where mul-
tiplication of the base elements 1, i, j, k, ε, εi, εk, and εk is defined by the rules
i2 = j2 = k2 = ijk = −1,
ε2 = 0,
iε = εi,
jε = εj,
kε = εk.
An element q ∈DH may be written as q = p + εd with quaternions p, d ∈H := ⟨1, i, j, k⟩
(angled brackets denote linear span). In this case the quaternions p and d are referred to
as primal and dual part of q, respectively. The conjugate dual quaternion is q = p + εd
and conjugation of quaternions is done by multiplying the coefficients of i, j, and k with
−1. It satisfies the rule qr = r q for any q, r ∈DH. The dual quaternion norm is defined
as ∥q∥= qq. We readily verify that it is a dual number, that is, an element of D := ⟨1, ε⟩.
We identify linearly dependent non-zero dual quaternions and thus arrive at the pro-
jective space P 7 = P(R8). Writing [q] for the point in P 7 that is represented by q ∈DH,
the Study quadric is defined as S := {[q] ∈P 7 : ∥q∥∈R}. With q = p+εd, the algebraic
condition for [q] ∈S is pd + dp = 0.
Identifying P 3 with the projective subspace generated by ⟨1, εi, εj, εk⟩, a point [q] =
[p + εd] ∈S with non-zero primal part acts on [x] ∈P 3 via
[x] 7→[(p + εd)x(p −εd)].
(1)
The map (1) is the projective extension of a rigid body displacement in R3. Composi-
tion of displacements corresponds to dual quaternion multiplication. This isomorphism
between SE(3), the group of rigid body displacements, and the projectivisation of the
2
group of dual quaternions of real norm and non-zero primal part provides a rich and
solid algebraic and geometric basis for investigations in kinematics.
3. A characterisation of 2R spaces
We complement the Study quadric by the quadric N = {[q]: ∥q∥∈εR} of points repre-
sented by dual quaternions whose norm has vanishing primal part. It is a quadric of rank
four. The set E = {[q]: q ∈εH} of its singular points is a projective space of dimension
three and it is contained in the Study quadric. Only the points of E are real points of
N whence we are led to consider the complex extension P(C8) of P 7. We call N the
null cone and refer to the set E of its singular points as the exceptional three space.
Given two non coplanar lines ℓ1, ℓ2 in Euclidean three space, we consider the set of all
displacements obtained as composition of a rotation around ℓ2, followed by a rotation
about ℓ1. Its kinematic image is known to lie in a three space (Selig, 2005, Section 11.4)
which we call a 2R space.
By construction, a 2R space contains the point [1] (the
identity displacement) and intersects S in a doubly ruled regular quadric. The rulings are
obtained by varying one revolute angle and fixing the other. In order to characterise 2R
spaces among all three-dimensional subspaces of P 7 with these properties, we introduce
the notion of null quadrilaterals. These are spatial quadrilaterals contained in both, S
and N. We call their edges null lines.
Theorem 1. A three space P ⊂P 7 with [1] ∈P is a 2R space if and only if it
• intersects the Study quadric in a regular ruled quadric Q,
• does not intersect the exceptional three space E, and
• contains a null quadrilateral.
The first and second item in Theorem 1 exclude exceptional cases with coplanar,
complex, or “infinite” revolute axes, the latter corresponding to prismatic joints. The
important point is existence of a null quadrilateral. We split the proof of Theorem 1
into a series of lemmas. It will be finished by the end of this section.
Lemma 1. The straight line [x] ∨[y] is contained in S ∩N if and only if xx = yy =
xy + yx = 0.
We omit the straightforward computational proof but note that the left-hand sides of
each of the three conditions in Lemma 1 are dual numbers. Hence, they give just six
independent conditions on the real coefficients of x and y.
Lemma 2. The conditions of Theorem 1 are necessary for 2R spaces.
Proof. The constraint variety of a 2R chain can be parameterised as
R(t1, t2) = (t1 −h1)(t2 −h2)
(2)
3
with two dual quaternions h1, h2 that satisfy h1h1 = h2h2 = 1, h1 + h1 = h2 + h2 = 0
and h1h2 ̸= h2h1. The first condition ensures that the dual quaternions are suitably
normalised and [h1], [h2] ∈S. The second condition means that h1 and h2 describe
half-turns (rotations through an angle of π). The third condition ensures that the axes
of these half turns are different. Equation (2) describes a composition of two rotations
for all values of t1, t2 in R ∪{∞} with ∞corresponding to zero rotation angle. Even if
their kinematic meaning is unclear, we also allow complex parameter values. Expanding
(2) yields
R(t1, t2) = t1t2 −t1h2 −t2h1 + h1h2.
We see that the kinematic image of the 2R dyad lies in the three space spanned by [1],
[h1], [h2], [h1h2]. In a suitable projective reference with these points as base points,
the surface parameterised by (2) is the quadric with equation x0x3 −x1x2 = 0 which is
indeed regular and ruled.
The intersection of P with the exceptional three space E is non empty if and only
if the primal part of R(t1, t2) vanishes for certain parameter values t1, t2. This can
only happen if the primal parts of h1 and h2 are linearly dependent over R but then
the revolute axes are parallel and P, contrary to our assumption, is contained in S.
The other possibility for P ⊂S, intersecting revolute axes, has been excluded as well.
Clearly, P is not contained in the null cone N either. We claim that the intersection of
P and N consists of the four lines given by t1 = ±i, t2 = ±i. Indeed, they are null lines.
Take for example x = (i −h1)(t2 −h2) and, in view of Lemma 1, compute
xx = (i −h1)(t2 −h2)(t2 −h2)(i −h1) = (t2 −h2)(t2 −h2)(i −h1)(i −h1)
= (t2 −h2)(t2 −h2)(−i2 −i(h1 + h1) + h1h1) = (t2 −h2)(t2 −h2)(−1 −0 + 1) = 0.
This derivation uses the fact that (t2 −h2)(t2 −h2) is a real number and thus commutes
with all other factors. The cases t1 = −i, t2 = ±i are similar so that we have verified all
conditions of Theorem 1.
The proof of sufficiency is more involved. We need two additional lemmas from pro-
jective geometry which are formulated and proved in Appendix A.
Lemma 3. A three space P that satisfies all conditions of Theorem 1 is a 2R space.
Proof. Denote the vertices of the null quadrilateral (in that order) by [u1], [v1], [u2], and
[v2]. Intersecting the Study null quadrilateral with the tangent hyperplane of S at [1]
yields points
[m1] ∈[u1] ∨[v1],
[n1] ∈[v1] ∨[u2],
[m2] ∈[u2] ∨[v2],
[n2] ∈[v2] ∨[u1]
of a planar and non-degenerate quadrilateral (Figure 1).
The joins [m1] ∨[m2] and
[n1] ∨[n2] are real lines and their real points correspond to rotations about fixed axes.
Pick real rotation quaternions [h1] ∈[m1] ∨[m2] and [h2] ∈[n1] ∨[n2]. The axes of
these rotations are the only candidates for our 2R dyad. Hence, we have to show that
either [h1h2] ∈P or [h2h1] ∈P. It is easy to see that this is the case if and only if
4
[1]
[m1]
[m2]
[n1]
[n2]
[h1]
[h2]
[u1]
[u2]
[v1]
[v2]
[n2m1]
[n1m1]
[n1m2]
[n2m2]
[m2n1]
[m1n1]
[m1n2]
[m2n2]
Figure 1: Null lines and null quadrilaterals in the proof of Lemma 3.
either [m1n1] ∈P or [n1m1] ∈P. In fact, we will even show that one of these products
equals v1.
We claim that both M1 := [m1] ∨[m1n1] and M2 := [m1] ∨[n1m1] are null lines. In
order to show this, we have to verify the conditions of Lemma 1:
(m1n1)(m1n1) = m1(n1n1
| {z }
=0
)m1 = 0,
(n1m1)(n1m1) = n1(m1m1
| {z }
=0
)n1 = 0,
m1(m1n1) + (m1n1)m1 = m1(n1 + n1
|
{z
}
∈C
)m1 = (n1 + n1)(m1m1
| {z }
=0
) = 0,
m1(n1m1) + (n1m1)m1 = (m1m1
| {z }
=0
)n1 + n1(m1m1
| {z }
=0
) = 0.
Similarly, we see that also N1 := [n1] ∨[m1n1] and N2 := [n1] ∨[n1m1] are null lines.
Thus, we are in the situation depicted in Figure 1 where we have three null quadrilaterals
with respective vertices
[u1], [v1], [u2], [v2];
[m1n1], [m2n1], [m2n2], [m1n2];
[n1m1], [n1m2], [n2m2], [n2m1].
The second and third quadrilateral are different because m1 and n1 do not commute
(otherwise they would lie on the same line through [1] which contradicts the regularity
of the quadric Q := P ∩S). Our proof will be finished as soon as we have shown that
the first and the second or the first and the third quadrilateral are equal. For this, it is
sufficient to show that [v1] = [m1n1] or [v1] = [n1m1].
At first, we argue that the primal part of [v1] equals the primal part of [m1n1] or of
[n1m1]. Because P does not intersect E, the projection on the primal part is a regular
projectivity P →[H] with centre E. We denote projected objects by a prime, that is,
we write u′
1, v′
1, m′
1 etc., for the primal parts of u1, v1, m1 etc. The quadric Q is regular
and ruled and so is its primal projection Q′. Hence, the point [m′
1] ∈Q′ is incident with
precisely two lines contained in Q′, say M′
1 and M′
2. But [m′
1] ∨[v′
1] is contained in Q′.
Hence [v′
1] ∈M′
1 or [v′
1] ∈M′
2. Similarly, [v′
1] is also incident with one of the two lines
contained in Q′ and incident with [n′
1]. Thus, [v′
1] = [m′
1n′
1] or [v′
1] = [n′
1m′
1].
Now we have to lift this result to the dual part and show that [v1] = [m1n1] or
[v1] = [n1m1]. The alternative being similar, we assume [v′
1] = [m′
1n′
1]. This means
5
that we have two null quadrilaterals, one with vertices [u1], [v1], [u2], [v2] and one with
vertices [m1n1], [m2n1], [m2n2], [m1n2] such that their primal projections are equal and
corresponding sides intersect in the vertices [m1], [n1], [m2], [n2] of a planar quadrilateral.
But then Lemma 5 in Appendix A asserts their equality and [v1] = [n1m1] follows.
4. Synthesis procedures
In this section we discuss the exact synthesis of 5R linkages, based on our geometric
characterisation of 2R spaces. The attentive reader will note that we neither provide a
synthesis algorithm nor a concrete example. The reason for this (and one of the main
points of this paper) is that the synthesis of 5R linkages can be treated as special case
of a recently introduced synthesis procedure for 6R linkages (Hegedüs et al., 2014). In
this context, 2R spaces play a crucial role.
At first, we re-prove a classical result of Suh (1969) on the synthesis of 2R dyads to
three poses. Its proof is just a variant of our computations in the proof of Theorem 1.
Theorem 2. Three poses [p0] = [1], [p1], [p2] ∈S that span a plane that is not tangent
to S and does not intersect E are incident with precisely two 2R spaces.
Proof. By Theorem 1, the sought 2R space P contains a null quadrilateral. We compute
its vertices [u1], [v1], [u2], [v2] as follows. The plane [p0] ∨[p1] ∨[p2] and the Study
quadric S have a regular conic C in common. This conic intersects the null cone in two
pairs m1, m2, and n1, n2 of conjugate complex points that can be computed from the
roots of a univariate quartic polynomial. Again, we denote projection on the primal
part by a prime. We can determine the primal part u′
1 of u1 by solving the system
u′
1u′
1 = u′
1m′
1 + m′
1u′
1 = u′
1n′
2 + u′
1n′
2 = 0. It consists of two linear and one quadratic
equation. Hence, there are precisely two solutions in projective sense. Picking one of
them, the primal parts of the remaining vertices can be computed unambiguously: Write,
for example, v′
1 = u′
1 + λm′
1 and solve v′
1n′
1 + n′
1v′
1 = 0 for λ. Finally, from [u′
1], [v′
1],
[u′
2], [v′
2] the points [u1], [v1], [u2], [v2] can be uniquely recovered as in the proof of
Lemma 5.
Remark 1. The two solutions in Theorem 2 both contain the conic C. This conic is
the kinematic image of the coupler motion of a Bennett linkage (Hamann, 2011). Thus,
as already mentioned by Suh (1969), the two 2R dyads of Theorem 2 can be combined
to form a Bennett linkage. Moreover, the computation of Lemma 3 shows that the two
2R spaces are [1] ∨[h1] ∨[h2] ∨[h1h2] and [1] ∨[h1] ∨[h2] ∨[h2h1] with suitable rotation
quaternions h1, h2. In other words, they differ only by the ordering of their axes. This
shows that the figures formed by opposite axes in a Bennett linkage are congruent.
In a 5R linkage there are two possibilities to assign a coupler to a fixed base. The two
choices can be distinguished by the degree of their relative motion with respect to the
base. As has been shown recently by (Hegedüs et al., 2013b, Theorem 6), the degrees
of relative motions between non-adjacent links in the dual quaternion model of rigid
body kinematics are two and three, respectively. In order to gain additional degrees of
6
freedoms over the synthesis of Bennett linkages, it is mandatory to use the link to the
degree three motion as coupler. In this sense, we can say that the coupler motion of a
5R linkage admits a rational cubic parameterisation (a cubic motion polynomial) in the
dual quaternion model of SE(3). However, not every twisted cubic in the Study quadric
gives rise to a 5R linkage. The necessary and sufficient condition is that the coupler
motion can also be generated by a 2R dyad, that is, the twisted cubic lies in a 2R space.
Using the factorisation theory for rational motions (Hegedüs et al., 2013a), a generic
cubic motion polynomial C can be written, in six different ways, as C = (t −h1)(t −
h2)(t−h3) with linear motion polynomials t−hi, i ∈{1, 2, 3}, that parameterise rotations
about fixed axes. Suitable combinations of such factorisations give an overconstrained
6R linkage whose coupler follows the prescribed cubic motion. For details on how to
compute cubic interpolants on quadrics, how to factor the resulting motions and how to
pick factorisations suitable for linkage synthesis we refer to Hegedüs et al. (2014). We
just highlight the major differences to the case of 5R linkage synthesis.
By a fundamental property of the underlying theory, the factorisations of a cubic
motion polynomial are, at least in generic situations, in bijection with the permutations
of the real quadratic factors of CC (Hegedüs et al., 2013a). Pairings for “admissible”
kinematic chains correspond to permutations with different factors at the beginning or at
the end. In the case of 5R synthesis, the situation is special. At least one factorisation
is of the shape C = (t −h1)(t −h2)(t −h3) where either the axes of h1 and h2 or
of h2 and h3 are identical. Lets assume the first case. Then h1 and h2 commute so
that C = (t −h2)(t −h1)(t −h3) is another factorisation, different in algebraic sense
but identical in kinematic sense. It can be paired with just two of the remaining four
factorisations to form a closed 5R linkage.
If h2 and h3 have identical axis, above
discussion can be repeated with C = (t −h3)(t −h2)(t −h1) instead of C.
For a more detailed discussion of some properties of 5R linkages in this context, we
need some results of twisted cubics that can be found in classical texts (for example
Salmon, 1882; Cayley, 1885). The intersection of a 2R space with the Study quadric
is a regular ruled quadric Q. It carries two families of rulings. One family contains a
ruling corresponding, via Study’s kinematic mapping, to all rotations about the axis at
the base of the 2R dyad, the other family contains a ruling corresponding to the moving
revolute axes. We may accordingly speak of a “first” and a “second” family of rulings.
The twisted cubics on Q can be partitioned into two classes: Members of the first class
intersect every ruling of the first family in two and every ruling of the second family
in just one point. For cubics of second class, the situation is just the other way round.
Finally, four points on Q determine two one-parametric families of twisted cubics, one
of first class and one of second. A member of each family is uniquely determined by the
choice of one further point on a ruling through one of the four given points.
These considerations clearly show that four poses in general position cannot be reached
by a 5R linkage unless their kinematic images span a 2R space. On the other hand, three
poses in general position already determine, by Theorem 2, two 2R spaces but also a
Bennett linkage. By above discussion, we have three degrees of freedom to find a cubic
curve through three quadric points. In other words, the number of free degrees in 5R
linkage synthesis exceed that of 4R linkages by three. Unfortunately, only restricted use
7
can be made of these. It is, for example, not possible to arbitrarily prescribe one further
orientation: While there is a dual quaternion with prescribed primal part (orientation) in
each 2R space, it will, in general, not lie on the Study quadric. An additional prescribed
position is prevented for similar reasons.
Finally, also the appearance of two types of factorisations with commuting factors on
the left (h1, h2) or on the right (h2, h3) can be explained by the two families of twisted
cubics on Q. If the commuting factors are on the left, C0(t) = (t −h1)(t −h2) is a
quadratic parameterisation of the line [1]∨[h1] and every point on that line corresponds
to two parameter values. In particular, there exist a second parameter value t0 (besides
∞) such that [C0(t0)] = [1]. But then C(t0) yields a second curve point (besides [1]) on
[1] ∨[h3], that is, the cubic belongs to the second family. In similar manner we see that
the cubic belongs to the first family, if h2 and h3 commute.
5. Conclusion
We characterised the kinematic image of 2R dyads and used it to specialise a recently
developed synthesis procedure for 6R linkages in such a way that it yields 5R linkages.
Even if sufficient degrees of freedom are available, it is not possible in general to find a 5R
linkage that visits three and a half poses (three poses plus one position or one orientation)
because of geometric obstructions. Thus, a natural selection of interpolation data for the
envisaged coupler motion is not entirely clear. Hopefully, future research will provide us
with more ideas in this direction.
A. Auxiliary results
Here, we prove two technical results. The formulation of Lemma 4 could be simplified
but in its present form its application in the proof of Lemma 5 is apparent.
Lemma 4. Given a three-space E ⊂P 7 and four points [u′
1], [v′
1], [u′
2], [v′
2] that span a
three-space F ⊂P 7 with E ∩F = ∅, set U1 := [u′
1] ∨E, V1 := [v′
1] ∨E, U2 := [u′
2] ∨E,
V2 := [v′
2] ∨E and consider four projections
µ1 : U1 →V1,
ν1 : V1 →U2,
µ2 : U2 →V2,
ν2 : V2 →U1
with respective centres [m1], [n1], [m2], and [n2]. If the centres span a plane L, the
composition κ := ν2 ◦µ2 ◦ν1 ◦µ1 of the four projections is the identity.
Proof. Note that dim U1 ∨V1 = 5 so that the projection µ1 (and also ν1, µ2, ν2) from a
suitable point is well defined. Take an arbitrary point [u1] ∈U1 but not in E and not in
L and set [v1] := µ1([u1]), [u2] := ν1([v2]), [v2] := µ2([u2]). Because these points are not
coplanar, we may take them, in above order, as base points of a projective coordinate
system in G := [u1]∨[v1]∨[u2]∨[v2] and complement them by four further base points in
E and a suitable unit point to a projective coordinate system of P 7. We select the unit
point such that its projection into G from E gives ([m1]∨[m2])∩([n1]∨[n2]) and denote
induced projective coordinates in U1, V1, U2, V2, and G (obtained by dropping three
8
or four zero coordinates) by writing the subspace as subscript. With this convention,
the projection centres are [m1] = [1, 1, 0, 0]G, [n1] = [0, 1, 1, 0]G, [m1] = [0, 0, 1, 1]G,
[n2] = [1, 0, 0, 1]G.
Moreover, we denote the coordinate vector of an arbitrary point
x ∈U1 by [x0, ⋆]U1 where the star abbreviates four fixed coordinates. We then have
[x0, ⋆]U1
µ1
7→[−x0, ⋆]V1
ν1
7→[x0, ⋆]U2
µ2
7→[−x0, ⋆]V2
ν2
7→[x0, ⋆]U1
and the claim follows.
Lemma 5. Let E, F ⊂P 7 be non-intersecting three spaces, the latter spanned by
points [u′
1], [v′
1], [u′
2], [v′
2]. Consider further a regular quadric Q containing E and four
vertices [m1], [n1], [m2], [n2] ∈Q of a planar quadrilateral. Then there exists a unique
spatial quadrilateral with vertices [u1], [v1], [u2], [v2] that is contained in Q, whose sides
[u1] ∨[v1], [v1] ∨[u2], [u2] ∨[v2], [v2] ∨[u1] are, in that order, incident with [m1], [n1],
[m2], [n2] and whose projections from E into F are [u′
1], [v′
1], [u′
2], and [v′
2].
Proof. Denote a quadratic form associated to Q by ω. Given [u′
1], [v′
1], [u′
2], [v′
2], we
have to reconstruct [u1], [v1], [u2], [v2] subject to the constraints
ω(ui, ui) = ω(vi, vi) = ω(ui, vj) = 0,
i, j ∈{1, 2}
(3)
and
[m1] ∈[u1] ∨[v1],
[n1] ∈[v1] ∨[u2],
[m2] ∈[u2] ∨[v2],
[n2] ∈[v2] ∨[u1].
(4)
If [u1] is given, we find [v1] by projecting [u1] from centre [m1] onto E ∨[v′
1]. Lemma 4
tells us that we can find [u2] and [v2] in similar manner such that (4) is satisfied. Then,
some of the conditions in (3) become redundant and it is sufficient to consider only
ω(u1, u1) = ω(v1, v1) = ω(u2, u2) = ω(v2, v2) = 0.
(5)
In a projective coordinate system with base points [u′
1], [v′
1], [u′
2], [v′
2] ∈F and further
base points in E the quadratic form ω (and the quadric Q) is described by a matrix of
the shape
"
A
B
B
O
#
where A, B and O are matrices of dimension 4×4, O is the zero matrix and B is regular.
Now (5) gives rise to a linear system for the unknown coordinates of [u1], [v1], [u2], and
[v2]. The matrix of the linear system is equivalent to the matrix B by elementary row
transformations, whence existence and uniqueness of a solution hinges solely on the
regularity of B.
Acknowledgement
This work was supported by the Austrian Science Fund (FWF): P 26607 (Algebraic
Methods in Kinematics: Motion Factorisation and Bond Theory).
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