arXiv:1301.4072v1  [math.AG]  17 Jan 2013
Classiﬁcation of Angle-Symmetric 6R Linkages
Zijia Li and Josef Schicho∗
Johann Radon Institute for Computational and Applied Mathematics Austrian Academy
of Sciences (RICAM), Altenbergerstrasse 69, 4040 Linz, Austria
Abstract
In this paper, we consider a special kind of overconstrained 6R closed linkages
which we call angle-symmetric 6R linkages.
These are linkages with the
property that the rotation angles are equal for each of the three pairs of
opposite joints. We give a classiﬁcation of these linkages. It turns that there
are three types. First, we have the linkages with line symmetry. The second
type is new. The third type is related to cubic motion polynomials.
Keywords:
Dual quaternion, overconstrained 6R linkages, classiﬁcation,
angle-symmetric
1. Introduction
Movable closed 6R linkages have been considered by many authors (see [1,
2, 3, 4, 5, 6]). In this paper, we give the complete classiﬁcation of a certain
class of such linkages, which we call angle-symmetric. This means that the
rotation angles at the three pairs of opposite joints are equal for all possible
conﬁgurations, or at least for inﬁnitely many conﬁgurations (it could be that
a certain linkage has two components, where only one of them is angle-
symmetric). It is well-known that the line symmetric linkage of Bricard [4]
is angle-symmetric. A second family is new; it can be characterized by the
presence of three pairs of parallel rotation axes. This ﬁlls a gap in [7, Section
3.8]. A third family was discovered in [8, 9] using factorizations of cubic
motion polynomials.
∗Corresponding author.
Email address: {zijia.li, josef.schicho}@oeaw.ac.at (Zijia Li and Josef
Schicho)
Preprint submitted to Mechanism and Machine Theory
September 25, 2018
Our main tool is the λ-matrix of a linkage, to be deﬁned in section 2,
and its rank r. Intuitively speaking, the conﬁguration set can be described
as the vanishing set of r equations in three variables, namely the cotangents
of the half of the rotation angles. We will show that r is either 2, 3, or 4.
If r = 2, then the linkage is line symmetric. If r = 3, then we get the new
linkage with three pairs of parallel axes. If r = 4, then we obtain the linkage
described in [8, 9] using motion polynomials.
We use Study’s description of Euclidean displacements by the algebra DH
of dual quaternions (see [8, 9]).
Structure of the paper. The remaining part of the paper is set up as follows.
In Section 2, we give the deﬁnition of the λ-matrix. We also show that the
rank of this matrix is 2, 3, or 4. Section 3 contains the main result and
examples
2. The λ-matrix
In this section we deﬁne, for a given linkage, a matrix Mλ whose rows
are related to an algebraic system deﬁning the conﬁguration space. In the
next section, we will see that the rank of this matrix is the basic criterion for
classifying angle-symmetric linkages.
The set of all possible motions of a closed 6R linkage is determined by
the position of the six rotation axes in some ﬁxed initial conﬁguration. (The
choice of the initial conﬁguration among all possible conﬁgurations is arbi-
trary. In some later steps in the classiﬁcation, we will occasionally change
the initial conﬁguration.)
The algebra DH of dual quaternions is the 8-dimensional real vector space
generated by 1, ǫ, i, j, k, ǫi, ǫj, ǫk (see [8, 9]). Following [8, 9], we can represent
a rotation by a dual quaternion of the form
 cot
 φ
2

−h

, where φ is the
rotation angle and h is a dual quaternion such that h2 = −1 depending only
on the rotation axis. We use projective representations, which means that
two dual quaternions represent the same Euclidean displacement if only if
one is a real scalar multiple of the other.
Let L be a 6R linkage given by 6 lines, represented by dual quaternions
h1, . . . , h6 such that h2
i = −1 for i = 1, . . . , 6. A conﬁguration (see [8, 9]) is
a 6-tuple (t1, . . . , t6), such that the closure condition
(t1 −h1)(t2 −h2)(t3 −h3)(t4 −h4)(t5 −h5)(t6 −h6) ∈R\{0}
2
holds. The conﬁguration parameters ti – the cotangents of the rotation angles
– may be real numbers or ∞, and in the second case we evaluate the expres-
sion (ti −hi) to 1, the rotation with angle 0. The set of all conﬁgurations of
L is denoted by KL.
There is a subset of KL, denoted by Ksym, deﬁned by the additional re-
strictions t1 = t4, t2 = t5, t3 = t6. We assume that Ksym is a one-dimensional
set, i.e.
the linkage has an angle-symmetric motion.
Mostly, we will as-
sume, slightly stronger, that there exists an irreducible one-dimensional set
for which none of the ti is ﬁxed. Such a component is called a non-degenerate
component. We also exclude the case dimC Ksym ≥2. Linkages with mobility
≥2 do exist, but they are well understood.
The closure condition is equivalent to
(t1 −h1)(t2 −h2)(t3 −h3) = λ(t3 + h6)(t2 + h5)(t1 + h4),
where λ is a nonzero real value depending on t1, t2, t3. By taking norm on
both sides, we get λ2 = 1, i.e. λ = ±1. By multiplying both sides with
(t1 + h1) from the left and with (t1 −h4) from the right, and afterwords
dividing by (t2
1 + 1), we obtain the equation
(t2 −h2)(t3 −h3)(t1 −h4) = λ(t1 + h1)(t3 + h6)(t2 + h5).
Similarly, we obtain
(t3 −h3)(t1 −h4)(t2 −h5) = λ(t2 + h2)(t1 + h1)(t3 + h6),
(t1 −h4)(t2 −h5)(t3 −h6) = λ(t3 + h3)(t2 + h2)(t1 + h1),
(t2 −h5)(t3 −h6)(t1 −h1) = λ(t1 + h4)(t3 + h3)(t2 + h2),
(t3 −h6)(t1 −h1)(t2 −h2) = λ(t2 + h5)(t1 + h4)(t3 + h3).
We may divide Ksym into two disjoint subsets K+
sym and K−
sym, according
to whether λ is equal to +1 or −1 in the equations above. Any irreducible
component of Ksym is either contained in K+
sym or in K−
sym. Note that ∞3 is
an element of K+
sym.
Remark 1. When we want to study some component K0 ⊂K−
sym, we may
proceed in the following way: we take a conﬁguration τ ∈K0, which deﬁnes a
set of rotations around the joint axes. Then we apply these rotations, obtain-
ing new positions for the 6 lines. In the transformed linkage, the component
corresponding to K0 contains ∞3. So we will always assume that λ = 1.
3
When λ = 1, after moving the right parts of the above equations to the
left, we get an equation
M†X = 0,
where X = [t1t2, t1t3, t2t3, t3, t2, t1, 1]T. If we denote h6 + h3, h5 + h2, h4 + h1
by g3, g2, g1 respectively, then the coeﬃcient matrix M† is


g3, g2, g1,
h5h4 −h1h2,
h6h4 −h1h3,
h6h5 −h2h3,
h6h5h4 + h1h2h3
g3, g2, g1,
h1h5 −h2h4,
h1h6 −h3h4,
h6h5 −h2h3,
h1h6h5 + h2h3h4
g3, g2, g1,
h2h1 −h4h5,
h1h6 −h3h4,
h2h6 −h3h5,
h2h1h6 + h3h4h5
g3, g2, g1,
h2h1 −h4h5,
h3h1 −h4h6,
h3h2 −h5h6,
h3h2h1 + h4h5h6
g3, g2, g1,
h4h2 −h5h1,
h4h3 −h6h1,
h3h2 −h5h6,
h4h3h2 + h5h6h1
g3, g2, g1,
h5h4 −h1h2,
h4h3 −h6h1,
h5h3 −h6h2,
h5h4h3 + h6h1h2


.
Note that M† is a 6 × 7 matrix with entries in dual quaternions. We also
consider M† to be a 48 × 7 matrix with real entries. It can be decomposed
into submatrices M†
1, · · · , M†
6, where M†
i is the real 8×7 matrix – or the row
vector with 7 dual quaternion entries – corresponding to the i−th equivalent
formulation of the closure condition above, for i = 1, . . . , 6.
Our classiﬁcation is based on the following theorem which gives the
bounds for the rank of M†.
Theorem 1. Assume that Ksym contains a non-degenerate component of
dimension 1. Then r := rank(M†) ∈{2, 3, 4}.
Before we prove Theorem 1, we give a lemma.
Lemma 1. Assume that Ksym contains a non-degenerate component K0 of
dimension 1 such that ∞3 ∈K0, and r ≥4. Then there exists a polynomial
of the form
bt1 + ct2 + d,
where b, c, d ∈R and bc ̸= 0, which vanishes on Ksym, maybe after some
permutation of the variables t1, t2, t3. Moreover, we can deﬁne a matrix N†
of rank ≥r −2 such that the projection of Ksym to (t1, t3) is deﬁned by
N†X′ = 0,
(1)
where X′ = [t2
1, t1t3, t1, t3, 1]T.
4
Proof. As r ≥4, we have at least four independent equations in three
variables (t1, t2, t3) of tridegree at most (1, 1, 1). We denote four of them by
F1, F2, F3, F4.
First, we assume that the F1 is irreducible. The resultants of F1 and Fi,
i = 2, 3, 4 with respect to the last variable t3 are denoted by F12, F13, F14.
The bidegrees of them are at most (2, 2). All these polynomials vanish on
Ksym.
If one of them is 0, such as F12 = 0, then F1 and F2 must have
a non-trivial common factor. This can only be F1, since F1 is irreducible.
Then the tridegree of F1 is less then (1, 1, 1). Because F1 vanishes on the
non-degenerate component K0, it must contain at least two variables, and
so F1 is a polynomial of degree (1, 1, 0), maybe after some permutation of
variables.
If none of the three resultants vanishes, then let G = gcd(F12, F13, F14).
The bidegree of G is in the set {(2, 2), (2, 1), (1, 1)}, up to permutation of
variables t1, t2. If it is (1, 1), then G can be considered as a polynomial of
tridegree (1, 1, 0) that vanishes on K0. If the bidegree of G is (2, 2) or (2, 1),
then we write F12 = GU2, F13 = GU3, F14 = GU4 with suitable polynomials
U2, U3, U4. The bidegrees of U2, U3, U4 are at most (0, 1), hence U2, U3, U4 are
linear dependent, which means that there are three real number λ2, λ3, λ4
such that
λ2F12 + λ3F13 + λ4F14 = 0.
As a consequence, we have
Res(F1, λ2F2 + λ3F3 + λ4F4) = 0,
where Res denotes the resultant. Then we can continue as in the case F12 = 0
above.
Again we get a polynomial of degree (1, 1, 0), maybe after some
permutation of variables.
Second, if F1 is reducible, then it has two factors with degree (1, 1, 0) and
(0, 0, 1), up to permutation of variables t1, t2, t3. Again, F1 vanishes on the
non-degenerate component K0, and so it must contain at least two variables,
and so it is a polynomial of degree (1, 1, 0), maybe after some permutation
of variables.
In all cases above, we have a polynomial of tridegree (1, 1, 0) vanishing on
K0. Since ∞3 is in Ksym, it is of the form bt1 + ct2 + d = 0, with b, c, d ∈R
and bc ̸= 0, as stated in the lemma. We can use it to eliminate t2: on K0,
we have t2 = −bt1+d
c
.
5
The equations for the projection of K0 to the (t1, t3)-plane can be obtained
by substituting. We get the equation N†X′ = 0, where N† := M†L, and
L =


−b
c
0
−d
c
0
0
0
1
0
0
0
0
−b
c
0
−d
c
0
0
0
0
1
0
0
0
−b
c
0
−d
c
0
0
1
0
0
0
0
0
0
1


.
This follows from the fact that on K0, we can replace X by LX′. Because
rank(L) = 5, we also get rank(N†) ≥rank(M†) −2.
□
Proof of Theorem 1. r ≥2: Assume, indirectly, that r ≤1. Then the
system M†X = 0 is equivalent to zero or only one single equation in three
variables, and it will have at least a two-dimensional complex conﬁguration
set, which contradicts our assumption.
r ≤4: Assume, indirectly, that r ≥5. Then from Lemma 1, the projec-
tion of Ksym to (t1, t3) is deﬁned by
N†X′ = 0,
(2)
where r1 := rank(N†) ≥r −2 ≥3. The equation (2) is equivalent to a
system of r1 polynomial equations of bidegree at most (2, 1). Because Ksym
is a curve and has non-degenerate components, the r1 polynomials have a
common factor with bidegree at least (1, 1). Then r1 ≤2 which contradicts
to r1 ≥3.
3. Classiﬁcation
This section contains three parts. First, we show that the existence of a
line symmetry implies r = 2. Second, we show that r = 2 or r = 3 implies a
line symmetry or another geometric consequence which we call the “parallel
property”. Third, we relate the case r = 4 to a linkage described in [8, 9].
3.1. Line Symmetric Linkages
We now describe line symmetric 6R linkages in terms of dual quaternions.
A 6R linkage L = [h1, h2, h3, h4, h5, h6] is line symmetric if and only if there
is a line represented by a dual quaternion l such that l2 = −1 and
h1 = lh4l−1,
h2 = lh5l−1,
h3 = lh6l−1,
(3)
6
where ll−1 = 1. Geometrically, the rotation around l by the angle π takes hi
to hi+3 for i = 1, 2, 3.
Proposition 1. If L is line symmetric, then r = 2.
Proof. As the norm of l is equal to 1, it follows l−1 = −l and we write (3)
as
h1 = −lh4l,
h2 = −lh5l,
h3 = −lh6l.
(4)
We deﬁne a map α from dual quaternion to itself as
α : DH −→DH,
h 7−→h + l¯hl,
where ¯h denotes the conjugate of h in dual quaternion. It is true that all
entries of M†
1 are in Im(α).
For instance, we have α(h1) = h1 −lh1l =
h1 + h4 = g1, α(h5h4) = h5h4 + lh4h5l = h5h4 −(lh4l)(−lh5l) = h5h4 −
h1h2, α(h6h5h4) = h6h5h4 −lh4h5h6l = h6h5h4 + (−lh4l)(−lh5l)(−lh6l) =
h6h5h4 + h1h2h3. It is not diﬃcult to prove that α is a R-linear map. If we
consider M†
1 to be an 8 × 7 matrix with real entries, then r2 := rank(M†
1)
is less or equal to the dimension of Im(α). W.l.o.g. we assume l = i. We
compute Im(α) as α(1) = 1+ii = 1−1 = 0, α(ǫ) = ǫ+ǫii = 0, α(i) = i−iii =
2i, α(j) = j −iji = 0, α(k) = k −iki = 0, α(ǫi) = ǫi −ǫiii = 2ǫi, α(ǫj) =
ǫj −ǫiji = 0, α(ǫk) = ǫk −ǫiki = 0. Therefore, the dimension of Im(α) is 2.
So we have r2 ≤2.
The next step is to prove that all M†
i for i = 1, ..., 6 are equal. It is true
that the ﬁrst three columns are equal in all M†
i for i = 1, ..., 6. As Im(α) is
equal to ⟨i, ǫi⟩R and g1, g2, g3, h6h5 −h2h3 ∈Im(α), we obtain
g1 × g2 = g1 × g3 = g2 × g3 = (h6h5 −h2h3) × g1 = 0,
(5)
where g × h denotes the cross product of purely vectorial dual quaternions
g, h. The equalities M†
1 = · · · = M†
6 can be shown from (5). For instance,
h5h4−h1h2−(h1h5−h2h4) = h5×h4−h1×h2−h1×h5+h2×h4 = g2×h4−h1×
g2 = g2×g1 = 0, h1h5−h2h4−(h4h2−h5h1) = h1h5−h2h4+(h1h5 −h2h4) = 0
or h6h5h4+h1h2h3−(h1h6h5+h2h3h4) = −⟨h6, h5⟩h4+⟨h2, h3⟩h4−⟨h2, h3⟩h1+
⟨h6, h5⟩h1 + (h6 ×h5) ×h4 + h1 ×(h2 ×h3) −h1 ×(h6 ×h5) −(h2 ×h3) ×h4 =
(h6 × h5 + h3 × h2) × g1 = (h6h5 −h2h3) × g1 = 0, where ⟨g, h⟩denotes the
inner product of purely vectorial dual quaternions g, h. As a consequence,
we have r = r2 ≤2. But we have r ≥2 by Theorem 1, so r = 2.
□
Remark 2. The well-known fact that line symmetric linkages are movable
can also be obtained as a corollary from Theorem 1. When r = 2, then the
conﬁguration set is deﬁned by 2 equations in 3 variables.
7
3.2. Linkages with Rank 2 and 3
In this subsection, we show that r = 2 or 3 implies either a line symmetry
or another property, deﬁned as follows. We say that L = [h1, . . . , h6] has the
parallel property if h1 ∥h4, h2 ∥h3, h5 ∥h6, maybe after some cyclic
permutation of indices. In this section, we always assume that the rank of
the λ-matrix of L is 2 or 3.
In the following, we use the technique of generic points of algebraic curves.
This simpliﬁes the analysis a lot. Let C be an irreducible algebraic curve.
Let F be a ﬁeld such C can be deﬁned by equations over F (for instance
F = Q). Following [10, Section 93], we say that some point p ∈C is generic
if it fulﬁlls no algebraic conditions deﬁned by polynomials with coeﬃcients in
F, excerpt those that are a consequence of the equations of C. The existence
of generic points is shown in [10, Section 93]; typically, the coordinates of a
generic point are transcendental numbers.
Let K0 ⊂K+
sym be an irreducible non-degenerate component of the linkage
L = [h1, . . . , h6], and let τ0 = (t′
1, t′
2, t′
3) be a generic point of K0.
The
conﬁguration τ0 corresponds to a set of rotations around the joint axes. When
we apply these rotations, we get new positions for the 6 lines, and we deﬁne
the transformed linkage by L′ = [h′
1, h′
2, h′
3, h′
4, h′
5, h′
6]. Note that L and L′
represent really the same linkage, just in diﬀerent initial positions.
Lemma 2. If primal(g′
1) = 0, then L has the parallel property. Here primal(h)
denotes the primal part of the dual quaternion h. More precisely, we will have
h1 ∥h4, h2 ∥h3, h5 ∥h6, in all conﬁgurations in K0.
Proof. Assume that primal(g′
1) = 0. The parallelity of the ﬁrst and fourth
axis can be expressed as a set of polynomial equations in the conﬁguration
parameters (t1, t2, t3). These equations are fulﬁlled for the generic point τ0.
By a well-known property of generic points it follows that they are fulﬁlled
for all points in K0. For this reason, the ﬁrst and fourth axis are parallel at
all position.
Let S = [p1, p2, p3, p4, p5, p6], where pi = primal(h′
i) for i = 1, . . . , 6.
Then S is a spherical linkage with the ﬁrst and fourth axis coinciding at
all positions. We can separate S into two 3R linkages S1 = [p1, p2, p3] and
S2 = [p4, p5, p6]. A 3R linkage is necessarily degenerate: either some angles
are constant or some axes coincide. Since t2 is not a constant in K0, we
obtain p2 = ±p3 or p1 = ±p2. Since t3 is not a constant in K0, we obtain
8
p2 = ±p3 or p1 = ±p3. If p2 ̸= ±p3, then we have p1 = ±p2 and p1 = ±p3, a
contradiction. So we obtain p2 = ±p3. Similarly, we also have p5 = ±p6.
Therefore, we get a linkage with h′
1 ∥h′
4, h′
2 ∥h′
3, h′
5 ∥h′
6. Since the
parallel property is fulﬁlled for the generic point of the conﬁguration curve,
it is fulﬁlled for all points in K0. In particular, the original linkage L has the
parallel property.
□
There is no i such that g′
i = 0 for i = 1, 2, 3, because if g′
i = 0 would be
true, then the lines h′
i and h′
i+3 would be equal; the initial conﬁguration was
chosen generically, so the lines hi and hi+3 would be equal for all conﬁgura-
tions in K0, and this is not possible. Moreover, it is not possible that two
of gi for i = 1, 2, 3 have 0 primal parts. In order to prove this, we assume
indirectly primal(g′
2) = 0 and primal(g′
3) = 0. By Lemma 2, we get h2 ∥h5,
h3 ∥h4, h1 ∥h6 and h3 ∥h6, h4 ∥h5, h1 ∥h2. It follows that L is a planar
6R Linkage which has mobility more than one.
Before the main theorem, we give several lemmas in the following.
Lemma 3. Let a, b be two purely vectorial dual quaternions. If a × b = 0,
then there is a dual number α such that b = αa or a = αb, or the primal
parts of a and b both vanish.
Proof. Straightforward.
□
In the next two proofs, we use the following argument from linear algebra.
Let 1 ≤i1 < · · · < ir < ir+1 < · · · < is ≤7 be integers.
Let A :=
a1M†
1 + · · · + a6M†
6 be some linear combination of the matrices M†
1, . . . , M†
6,
where a1, . . . , a6 ∈R. If the vector space generated by the columns (i1, . . . , is)
of M† is already generated by the columns (i1, . . . , ir) of M†, then the vector
space generated by the columns (i1, . . . , is) of A is also generated by the
columns (i1, . . . , ir) of A.
Lemma 4. If g′
3 × g′
1 = g′
2 × g′
1 = 0, then g′
2 × g′
3 = 0.
Proof. We distinguish two cases.
Case I: primal(g′
1) ̸= 0. By Lemma 3, there exist α2, α3 ∈D such that
g′
2 = α2g′
1 and g′
3 = α3g′
1, and it follows that g′
2 × g′
3 = 0.
Case II: primal(g′
1) = 0. Then primal(g′
2) ̸= 0 and primal(g′
3) ̸= 0. If
there exists α ∈D such that g′
3 = αg′
2, then g′
2 × g′
3 = 0. Otherwise, g′
1 is a
dual multiple of g′
2 but g′
3 is not, so g′
1, g′
2, g′
3 are linearly independent. Then
9
the ﬁrst three columns generate the column space of M†. By linear algebra,
the ﬁrst three columns of A := M†
1 +M†
4 −M†
3 −M†
6 also generate the column
space of A. But
A = [0, 0, 0, 0, 2g′
3 × g′
1, 2g′
3 × g′
2, ∗]
(6)
(we do not care about the last entry denoted by ∗), and it follows that
g′
2 × g′
3 = 0.
□
Lemma 5. We have g′
3 × g′
1 = g′
2 × g′
1 = g′
2 × g′
3 = 0.
Proof. Let r3 be the dimension of the vector space generated by g′
1, g′
2, g′
3.
If r3 = 1, then it follows that g′
3 × g′
1 = g′
2 × g′
1 = g′
2 × g′
3 = 0. If r3 = 2 or
r3 = 3, then the vector space V generated by the ﬁrst 6 columns of M† is
already generated by the ﬁrst three and one of the other three columns.
Assume, for instance, that V is generated by columns (1, 2, 3, 6).
By
linear algebra, the corresponding columns also generate the space of the ﬁrst
six columns of
M†
1 + M†
4 −M†
2 −M†
5 = [0, 0, 0, 2g′
2 × g′
1, 2g′
3 × g′
1, 0, ∗].
This implies g′
3 × g′
1 = g′
2 × g′
1 = 0, and by Lemma 4, we also get g′
2 × g′
3 = 0.
If V is generated by columns (1, 2, 3, 4), then the above linear algebra
argument shows g′
1 × g′
3 = g′
2 × g′
3 = 0. The equality g′
2 × g′
1 = 0 follows
again from by Lemma 4, applied to the linkage [h3, h4, h5, h6, h1, h2]. The
third case, when V is generated by columns (1, 2, 3, 5), is also similar.
□
Lemma 6. If primal(g′
i) ̸= 0 for i = 1, 2, 3, then L′ is line symmetric.
Proof. By Lemma 3, there exists a dual quaternion u and invertible dual
numbers α1, α2, α3 such that g′
i = αiu for i = 1, 2, 3. Let β := u¯u ∈D.
Because the primal part of u is nonzero, the primal part of β is positive, and
1
√β is deﬁned. We set l′ :=
1
√βu. Then l′2 = −1 and g′
ih′
i = h′
i
2 + h′
i+3h′
i =
h′
i+3
2 + h′
i+3h′
1 = h′
i+3g′
i, hence h′
i+3 = g′
ih′
ig′
i
−1 = l′h′
il′−1 for i = 1, 2, 3.
□
Theorem 2. If r = 2 or 3, then L has a line symmetry or the parallel
property.
Proof. Let K0 ⊂K+
sym be an irreducible non-degenerate component and
τ0 = (t1, t2, t3, t1, t2, t3) be a generic point of K0. We get L′ = [h′
1, h′
2, h′
3,
h′
4, h′
5, h′
6] by applying the rotations speciﬁed in τ. By Lemmas 4, 5, and 6,
10
we conclude that L′ has a line symmetry or the parallel property. If a line
symmetric linkage moves in an angle symmetric way, then the transformed
linkage is also angle symmetric. This implies that when L′ is line symmetric,
then L is also line symmetric.
On the other hand, if L′ has the parallel
property, then parallelity holds for all points in K0, in particular L has the
parallel property.
□
Theorem 3. If r = 2, then L is line symmetric.
Proof. By Theorem 1 and Theorem 2, we may assume that L has and
parallel property and r = 2. Let L′ = [h′
1, h′
2, h′
3, h′
4, h′
5, h′
6] be the linkage
transformed by a generic position. We may assume h′
1 ∥h′
4, h′
2 ∥h′
3, h′
5 ∥h′
6.
The primal part of g′
1 is 0 and the primal parts g′
2 and g′
3 are not. We deﬁne
l′ as
1
√
g′
2g′
2
g′
2. Then l′2 = −1. By Lemma 5, we also get h′
2 = −l′h′
5l′ and
h′
3 = −l′h′
6l′ (see also the proof of Lemma 6). Moreover, g′
1 is a real multiple
of ǫl′, and g′
1h′
1 = h′
4g′
1. By the last equation, the primal part of h′
1 + l′h′
4l′
is zero. The dual part of h′
1 + l′h′
4l′ is equal to u := g′
1 −h′
4 + l′h′
4l′. The
vectorial part of ul′ = g′
1l′ −h′
4l′ −lh′
4 vanishes, so u is a multiple of l′. On
the other hand, the scalar product of u with l′ also vanishes, hence u = 0
and h′
1 = −l′h′
4l′. It follows that L′ and L are same line symmetric.
□
In the end of this subsection, we give a construction of angle-symmetric
6R linkage with parallel property. The construction is based on the fact that
we have a partially line symmetry taking h2 to h5 and h3 to h6 (see Lemma
3 and Lemma 5 above).
Construction 1. (Angle-Symmetric 6R Linkage with Parallel Property)
I. Choose a rotation axis u such that u2 = −1.
II. Choose another rotation axis h1 such that h2
1 = −1 and it is perpen-
dicular to u.
III. Choose two parallel rotation axes h2 and h3 which are not perpendic-
ular to u such that h2
2 = h2
3 = −1.
IV. Set h4 = −uh1u + rǫu, where r is a random real number.
V. Set h5 = −uh2u and h6 = −uh3u.
VI. Our angle-symmetric 6R Linkage with parallel property is L = [h1, h2,
h3, h4, h5, h6].
□
11
Example 1. (Angle-Symmetric 6R Linkage with Parallel Property) We set
u = i,
h1 = −7
11ǫi + j,
h2 =

2ǫ −3
5

i −
3
2ǫ + 4
5

j −ǫk,
h3 =

−2ǫ + 3
5

i +
3
2ǫ + 4
5

j + 2ǫk,
r = 14
11,
h4 = 7
11ǫi −j,
h5 =

2ǫ −3
5

i +
3
2ǫ + 4
5

j + ǫk,
h6 =

−2ǫ + 3
5

i −
3
2ǫ + 4
5

j −2ǫk.
It can be seen that the axes of h1, h4 are parallel, and the axes of h2, h3
and h5, h6, respectively, are parallel. Furthermore, the conﬁguration curve
contains a non-degenerate component:
(t1, t2, t3, t4, t5, t6) =
5
4t, t, t, 5
4t, t, t

.
Thus, we have an example of angle-symmetric 6R linkage with parallel prop-
erty. The rank of M† is 3. In Figure 1, we present nine conﬁguration posi-
tions of this linkage produced by Maple.
□
Remark 3. A random instance of Construction 1 produces a linkage where
t1 is parametrized by a quadratic function in t = t2 = t3. This example is
special because t1 is linear in t. (There is a degenerate component of the
conﬁguration curve that is responsible for this drop of the degree.)
3.3. Linkages with Rank 4
In this subsection, we show that the angle-symmetric linkages with Rank 4
are exactly those that have been constructed in [9, Example 3] by factoriza-
tion of cubic motion polynomials.
12
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
Figure 1: These nine pictures which are produced by Maple are diﬀerent
positions of the linkage in Example 1. The six colored tetrahedra(gray, blue,
yellow, red, green, pink) represent six links in the linkage, and the joints are
common edges of connected tetrahedra.
13
Recall that a motion polynomial P is a polynomial in one variable t with
coeﬃcients in DH such that P ¯P is a real polynomial that does not vanish
identically. (Multiplication in DH[t] is deﬁned by requiring that t commutes
with the coeﬃcients in DH.) Motion polynomials parametrize motions: by
substituting a real number for t, we obtain an element in the Study quadric.
We give a brief sketch of the construction in [8, 9]. Linear motion polyno-
mials of the form (t −a −bh), a, b ∈R, b ̸= 0, h ∈DH, h2 = −1 parametrize
revolutions. When we multiply three such polynomials R1, R2, R3, we get a
cubic motion polynomial Q. Generically, there are 6 diﬀerent factorizations
into linear monic polynomials, and there is one of the form R6R5R4 such
that the equations R1 ¯
R1 = R4 ¯
R4, R2 ¯
R2 = R5 ¯
R5, R3 ¯
R3 = R6 ¯
R6 hold. The
three linear factors R4, R5, R6 are again motion polynomials parametrizing
revolutions. The six axes of R1, . . . , R6 deﬁne a closed 6R linkage; let us call
it a linkage of cubic polynomial type.
We set Ri(t) = t −ai −bihi for i = 1, . . . , 6, ai, bi ∈R, bi ̸= 0, hi ∈DH,
h2
i = −1. The equations above are equivalent to ai = ai+3 and b2
i = b2
i+3 for
i = 1, 2, 3. We may even assume bi = −bi+3; if not, we replace hi+3 and bi+3
by −hi+3 and −bi+3. We multiply R1R2R3 = R6R5R4 by ¯
R4 ¯
R5 ¯
R6 and get
that
(t−a1−b1h1)(t−a2−b2h2)(t−a3−b3h3)(t−a1−b1h4)(t−a2−b2h5)(t−a3−b3h6)
is a real polynomial. This shows that the conﬁguration curve is parametrized
by
(t1, t2, t3, t4, t5, t6) =
t −a1
b1
, t −a2
b2
, t −a3
b3
, t −a1
b1
, t −a2
b2
, t −a3
b3

.
In particular, the linkage of cubic polynomial type is angle symmetric.
Here is a converse of the above statement.
Theorem 4. If L is an angle-symmetric linkage such that the λ-matrix has
rank r = 4, then L is of cubic polynomial type.
Proof. By Lemma 1, there exist a polynomial of the form bt1 +ct2 +d that
vanishes on Ksym, b, c, d ∈R, bc ̸= 0, and the projection of Ksym to (t1, t3)
is in the common zero set of two linear independent polynomials of bidegree
(2, 1). The equation of the projection is therefore a common factor of these
two equations and must have bidegree smaller than (2, 1). Since Ksym has a
14
non-degenerate component, the common factor cannot be constant in t1 or t3,
hence it has bidegree (1, 1). Because (∞, ∞) is contained in the projection,
the common factor has the form b′t1 + c′t2 + d′ for b′, c′, d′ ∈R, b′c′ ̸= 0. This
allows to parametrize Ksym with linear functions
(t1, t2, t3) =
t −a1
b1
, t −a2
b2
, t −a3
b3

for a1, . . . , b3 ∈R, b1b2b3 ̸= 0. Now the linkage can be reconstructed from
the two factorizations of the cubic motion polynomial
(t−a1−b1h1)(t−a2−b2h2)(t−a3−b3h3) = (t−a3+b3h6)(t−a2+b2h5)(t−a1+b1h4),
so it is of cubic polynomial type.
□
4. Conclusion
In the analysis of the case r = 3, we obtained a new type of linkages (with
parallel property h1 ∥h4, h2 ∥h3, h5 ∥h6). It is not clear from the paper if
every linkage with parallel property is angle-symmetric. We know that this
is not the case. A complete analysis of linkages with parallel property will
be the topic of a future paper.
5. Acknowledgements
We would like to thank G´abor Heged¨us and Hans-Peter Schr¨ocker for
discussion and helpful remarks. The research was supported by the Austrian
Science Fund (FWF): W1214-N15, project DK9.
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16