Addendum to Pentapods with Mobility 2
Georg Nawratil
Institute of Discrete Mathematics and Geometry
Vienna University of Technology
Wiedner Hauptstrasse 8-10/104, Vienna 1040, Austria
Email: nawratil@geometrie.tuwien.ac.at
Josef Schicho
Johann Radon Institute for Computational and Applied Mathematics
Austrian Academy of Sciences
Altenberger Strasse 69, Linz 4040, Austria
Email: josef.schicho@ricam.oeaw.ac.at
In a foregoing publication the authors studied pentapods
with mobility 2, where neither all platform anchor points nor
all base anchor points are located on a line. It turned out
that the given classiﬁcation is incomplete. This addendum
is devoted to the discussion of the missing cases resulting in
additional solutions already known to Duporcq.
1
Introduction
The geometry of a pentapod is given by the ﬁve base
anchor points Mi with coordinates Mi := (Ai,Bi,Ci)T with
respect to the ﬁxed system and by the ﬁve platform anchor
points mi with coordinates mi := (ai,bi,ci)T with respect
to the moving system (for i = 1,...,5). Each pair (Mi,mi)
of corresponding anchor points is connected by a SPS-leg,
where only the prismatic joint (P) is active and the spherical
joints (S) are passive.
If the geometry of the manipulator is given, as well as
the lengths of the ﬁve pairwise distinct legs, the pentapod has
generically mobility 1 according to the formula of Gr¨ubler.
The corresponding motion is called a 1-dimensional self-
motion of the pentapod. But, under particular conditions, the
manipulator can gain additional mobility. We can focus on
pentapods with mobility 2, as those with higher-dimensional
self-motions are already known (cf. [1, Corollary 1]).
1.1
Reason for the Addendum
The classiﬁcation of pentapods with mobility 2 given in
[1] was based on the following theorem of [2]:
Theorem 1. If the mobility of a pentapod is 2 or higher,
then one of the following conditions holds 1:
1After a possible necessary renumbering of anchor points and exchange
of the platform and the base.
(a) The platform and the base are similar. This is a so-called
equiform pentapod.
(b) The platform and the base are planar and afﬁne equiva-
lent. This is a so-called planar afﬁne pentapod.
(c) There exists p ≤5 such that m1,...,mp are collinear and
Mp+1,...,M5 are equal; i.e., Mp+1 = ... = M5.
(d) M1,M2,M3 are located on the line g which is parallel to
the line h spanned by M4 and M5. Moreover m1,m2,m3
are located on the line g′ which is parallel to the line h′
spanned by m4 and m5.
During the literature research for the article [3, Section 1],
we came across the work [4] of Duporcq, which describes
the following remarkable motion (see Fig. 1):
Let M1,...,M6 and m1,...,m6 be the vertices of two
complete quadrilaterals, which are congruent. Moreover the
vertices are labeled in a way that mi is the opposite vertex
of Mi for i ∈{1,...,6}. Then there exist a two-parametric
line-symmetric motion where each mi is running on spheres
centered in Mi.
M1
M2
M3
M5
M4
M6
m4
m5
m6
m2
m1
m3
Fig. 1.
Illustration of Duporcq’s complete quadrilaterals.
It can easily be checked that this conﬁguration of anchor
points corresponds to an architecturally singular hexapod
(e.g. [5] or [6]). As architecturally singular manipulators are
redundant we can remove any leg — without loss of general-
ity (w.l.o.g.) we suppose that this is the sixth leg — without
changing the direct kinematics of the mechanism. Therefore
arXiv:1602.00932v1  [cs.RO]  2 Feb 2016
the resulting pentapod M1,...,M5 and m1,...,m5, which we
call a Duporcq pentapod for short, has also a two-parametric
line-symmetric self-motion. This yields a counter-example
to Theorem 1, but the ﬂaw can be ﬁxed by adding the fol-
lowing case to Theorem 1 (cf. [7]):
(e) The following triples of points are collinear:
M1,M2,M3,
M3,M4,M5,
m3,m1,mi,
m3,m j,mk,
with pairwise distinct i, j,k ∈{2,4,5}.
Moreover the
points M1,...,M5 are pairwise distinct as well as the
points m1,...,m5.
As Theorem 1 only gives necessary conditions, the ad-
dendum is devoted to the determination of sufﬁcient ones for
the 2-dimensional mobility of pentapods belonging to item
(e). In detail the paper is structured as follows:
In Section 2 further necessary conditions are obtained by
means of M¨obius photogrammetry, which restrict the penta-
pod designs of (e) to three possible cases, up to afﬁnities of
the planar platform and the planar base. In Section 3 we re-
peat the theory of bonds based on two different embeddings
of SE(3) and prove Lemmata 1 and 2 as well as Corollaries
1 and 2. Based on these results we show in Section 4 that
only the Duporcq pentapods were missed by our classiﬁca-
tion given in [1]. The consequence of this result for article [1]
are summed up in the conclusions (Section 5).
2
M¨obius photogrammetric considerations
In Subsection 2.1 we recall some basics of M¨obius pho-
togrammetry, which are needed for the construction of the
three possible pentapod designs (up to afﬁnities of the planar
platform and the planar base) given in Subsection 2.2.
2.1
Basics
First of all we need the notation of a so-called M¨obius
transformation γ of the plane.
If we combine the planar
Cartesian coordinates (x,y) to a complex number z := x+iy,
then γ(z) can be deﬁned as a rational function of the form
γ :
z 7→z1z+z2
z3z+z4
,
(1)
with complex numbers z1,...,z4 satisfying z1z4 −z2z3 ̸= 0.
Therefore M¨obius transformations can be seen as the projec-
tive transformations of the complex projective line P1
C.
We identify by the mapping ι the unit-sphere S2 of
the Euclidean 3-space R3 with an algebraic curve C :=

x2 +y2 +z2 = 0
	
in P2
C. In detail this identiﬁcation works
as follows: Let u ∈S2 with u = (u1,u2,u3). Then determine
v,w ∈S2 with v = (v1,v2,v3) and w = (w1,w2,w3) in a way
that u,v,w determine a right-handed basis of R3. Then the
map ι : S2 →P2
C is given by:
ι : (u1,u2,u3) 7→(v1 +iw1 : v2 +iw2 : v3 +iw3)
(2)
as a different choice of v,w ∈S2 yields to the same point in
P2
C.
By denoting the vector (M1,...,M5) of ﬁve points in
R3 by M, the orthogonal parallel projection π of M along
the direction associated with c ∈C is given by πc(M). By
writing the planar Cartesian coordinates of each projected
point as a complex number we get πc(M) ∈(P1
C)5.
Remark 1. Assume that M1,...,M5 is known to be copla-
nar, the 5-tuple can be reconstructed from πc(M) only up to
afﬁnity, as the orientation of the carrier plane of the 5 points
with respect to ι−1(c) is not known. This also corrects [7],
where ”similarity” is written instead of ”afﬁnity”.
⋄
The equivalence class under the action of the M¨obius
group Γ on πc(M) is the so-called M¨obius picture [πc(M)]Γ
of M along the direction associated with c ∈C.
The set of all these equivalence classes [(P1
C)5]Γ can be
viewed as a quintic surface P5 ∈P5
C known as Del Pezzo sur-
face. For πc(M) with coordinates (x1 + iy1,...,x5 + iy5) ∈
(P1
C)5 the corresponding point of the Del Pezzo surface is
deﬁned as (ϕ0 : ϕ1 : ϕ2 : ϕ3 : ϕ4 : ϕ5) with:
ϕ0 := D12D23D34D45D15,
ϕ1 := D12D25D15D34D34,
ϕ2 := D12D23D13D45D45,
ϕ3 := D23D34D24D15D15,
ϕ4 := D34D45D35D12D12,
ϕ5 := D14D45D15D23D23,
(3)
and Dij := xiyj −xjyi. For details of this construction of P5
we refer to [2, Section 3.1], but it is important to note that P5
carries 10 lines Lij corresponding to equivalence classes for
which the projection of the ith and the jth point coincide (⇔
Dij = 0) for pairwise distinct i, j ∈{1,...,5}.
We are interested in the set of M¨obius pictures of M
under all c ∈C. By applying the so-called photographic map
fM of M given by
fM : C →P5
with
c 7→[πc(M)]Γ
(4)
we can compute the so-called proﬁle pM of M as the Zariski
closure of fM(C); i.e. ZarClo( fM(C)). Note that the proﬁle
is a curve on P5.
According to [2, Remark 3.5] the M¨obius picture can-
not be deﬁned for those values c ∈C, for which the associ-
ated directions are parallel to three collinear points of M, as
in this case all ﬁve ϕi’s are equal to zero. In our case two
such directions exist, which are parallel to the carrier line of
M1,M2,M3 (i.e. the metallic direction m) and M3,M4,M5
(i.e. the blue direction b), respectively (cf. Fig. 2). How-
ever, we can extend fM also to these directions by cancel-
ing out the common vanishing factor. For our given base
(cf. Fig. 2) this common factor is D12 = D13 = D23 (resp.
D34 = D35 = D45) for the metallic (resp. blue) direction,
thus Eq. (3) yields (0 : D25D34 : 0 : D24D15 : 0 : 0) (resp.
(0 : 0 : 0 : D24D15 : 0 : D14D23)). Therefore the m-direction
(resp. b-direction) is mapped on a point of L45 (resp. L12);
i.e.
[πm(M)]Γ ∈L45,
[πb(M)]Γ ∈L12.
(5)
green (g)
orange (o)
metallic (m)
yellow (y)
pink (p)
blue (b)
M3
M1
M2
M4
M5
Fig. 2.
The photographic map sends any direction vector parallel to
a line through 2 (but not 3) of the points Mi, M j to the unique point
in the M¨obius picture on the line Lij of the quintic surface P5. In the
base conﬁguration above, green (g) is sent to L25, orange (o) is sent
to L24, yellow (y) is sent to L15 and pink (p) is sent to L14. It is not
clear whether the directions blue (b) and metallic (m) are being sent;
later, we will show that b is sent to L12 and m is sent to L45.
2.2
Three possible designs
It is known (see [2, Section 4]) that for a pentapod with
mobility 2, which belongs to item (e) of Theorem 1, the pro-
ﬁles pM and pm have to coincide, where m denotes the vec-
tor of ﬁve points (m1,...,m5). As a consequence there has to
be a one-to-one correspondence between pM and pm, which
is used to reconstruct in three ways m (up to afﬁnity; cf. Re-
mark 1), under the assumption that M is given (cf. Fig. 2).
Assumption 1. W.l.o.g. we can assume that the recon-
struction m is afﬁnely transformed in a way that the M¨obius
pictures (and their extension in the case of three collinear
points) of m and M with respect to any direction c are iden-
tical.
First of all we have to distinguish the following three
cases, which are implied by the three possible collinearity
conﬁgurations stated in (e):
1. i = 2: W.l.o.g. we can set j = 4 and k = 5.
One can select m1 arbitrarily. As L14 ∩pm has to coin-
cide with L14 ∩pM the line m1m4 has to be parallel to
M1M4. Now we can select any point (̸= m1) on the par-
allel line to M1M4 through m1 as m4. The direction of
m1m2 is not uniquely determined as the line M1M2 also
contains the point M3. Due to the one-to-one correspon-
dence between the two proﬁles, L12 ∩pm has to corre-
spond with one of the two points on pM, which do not
admit a M¨obius picture. Therefore there are the follow-
ing two possibilities:
(a) m1m2 is parallel to M1M2: As a consequence m4m5
has to be parallel to M4M5.
Moreover as L24 ∩pm has to coincide with L24 ∩pM
the line m2m4 has to be parallel to M2M4. There-
fore we get m2 as the intersection point of a parallel
line to M1M2 through m1 and a parallel line to M2M4
through m4.
In the same way L15 ∩pm has to coincide with L15 ∩
pM and therefore m5 can be obtained as the intersec-
tion point of a parallel line to M4M5 through m4 and
a parallel line to M1M5 through m1.
Although all points are reconstructed, we have to
check if the last remaining condition is fulﬁlled,
namely if m2m5 is parallel to M2M5. As this can eas-
ily be veriﬁed, we get reconstruction 1 illustrated in
Fig. 3(left), which is in fact identical with M (cf. Fig.
2).
m3
m1
m2
m4
m5
m1
m2
m3
m4
m5
Fig. 3.
The two possible reconstructions of the platform conﬁgu-
ration from the M¨obius picture, under the additional assumption that
m3 is the intersection of lines m1m2 and m3m4. Note that the line
m2m5 must have direction g, the line m2m4 must have direction o,
the line m1m5 must have direction y and the line m1m4 must have
direction p. The left conﬁguration coincides with the base conﬁgura-
tion. We will see later that the right conﬁguration is not compatible be-
cause the lines through m1,m2,m3 and m3,m4,m5, respectively,
do not have the correct directions.
(b) m1m2 is parallel to M4M5: As a consequence m4m5
has to be parallel to M1M2.
Analogous arguments as in the above case with re-
spect to the swapped directions yield a further can-
didate platform m illustrated in Fig. 3(right). Calcu-
lation of the M¨obius picture of m with respect to the
directions m and b according to Eq. (3) shows
[πm(m)]Γ ∈L12,
[πb(m)]Γ ∈L45.
(6)
Due to Eq. (5), m and M do not have the same
M¨obius picture with respect to the directions m and
b; a contradiction.
2. i = 5: W.l.o.g. we can set j = 2 and k = 4.
For the same reasons as in item 1 we can select m1 ar-
bitrarily and can choose any point (̸= m1) on the parallel
line to M1M4 through m1 as m4. Moreover the following
two subcases can also be reasoned analogously to item
1:
(a) m1m2 is parallel to M1M2: As a consequence m4m5
has to be parallel to M4M5.
In this case we also get Eq. (6), which implies the
same contradiction as in case 1(b).
(b) m1m2 is parallel to M4M5: As a consequence m4m5
has to be parallel to M1M2.
As now the line m2m4 also contains the point m3
the corresponding direction does not admit a M¨obius
picture. Due to the one-to-one correspondence be-
tween pM and pm again two cases have to be distin-
guished:
i. m2m4 is parallel to M1,M5: As a consequence
m1m5 has to be parallel to M2M4.
Therefore m2 can be obtained as the intersection
point of the parallel line to M1M2 through m1
and the parallel line to M1M5 through m4. More-
over m5 equals the intersection point of the par-
allel line to M4M5 through m4 and the parallel
line to M2M4 through m1.
Although all points are reconstructed, we have to
check again if the last remaining condition is ful-
ﬁlled, namely if m2m5 is parallel to M2M5. As
this can easily be veriﬁed, we get reconstruction
2 illustrated in Fig. 4(right).
m3
m1
m2
m5
m4
m2
m1
m3
m5
m4
Fig. 4.
The two possible reconstructions of the platform conﬁgu-
ration from the M¨obius picture, under the additional assumption that
m3 is the intersection of lines m1m5 and m2m4. Here the direc-
tions of lines m1m2, m4m5, m1m4 and m2m5 are ﬁxed to b, m, p
and g, respectively. We will later see that the left conﬁguration is not
compatible. The right conﬁguration leads to a Duporcq pentapod.
ii. m2m4 is parallel to M2M4: As a consequence
m1m5 has to be parallel to M1M5.
Analogous considerations as in item 2(b)i yields
the candidate platform illustrated in Fig. 4(left).
Now the calculation of the M¨obius picture of this
candidate with respect to the orange direction o
yields (0 : 0 : D13D45 : 0 : D35D12 : 0). There-
fore we have [πo(m)]Γ ∈L15, which contradicts
[πo(M)]Γ ∈L24, thus we have no valid recon-
struction.
3. i = 4: W.l.o.g. we can set j = 2 and k = 5.
The discussion of cases is exactly the same as in item 2 if
one exchanges the indices 4 and 5. The resulting recon-
struction 3 as well as the corresponding non-valid candi-
date platform are illustrated in Fig. 5.
m4
m1
m2
m5
m3
m3
m1
m2
m4
m5
Fig. 5.
The two possible reconstructions of the platform conﬁgu-
ration from the M¨obius picture, under the additional assumption that
m3 is the intersection of lines m1m4 and m2m5. Here the directions
of lines m1m2, m4m5, m1m5 and m2m4 are ﬁxed to b, m, y, and
o, respectively. The left conﬁguration is not compatible. The right
conﬁguration leads to a Duporcq pentapod.
Moreover for the ith reconstruction (i = 1,2,3) there exists
an afﬁne relation κi between the set {M1,M2,M4,M5} and
the set {m1,m2,m4,m5}. In detail these afﬁne mappings κi
are given by:
κ1 :
M1 7→m1
M2 7→m2
M4 7→m4
M5 7→m5
κ2 :
M1 7→m4
M2 7→m5
M4 7→m1
M5 7→m2
κ3 :
M1 7→m5
M2 7→m4
M4 7→m2
M5 7→m1
For all three cases the validity of these afﬁne mappings can
be proven by direct computation.
Moreover it should be
noted that κ1 maps M3 7→m3 in addition. As a consequence
the pentapod design resulting from reconstruction 1 is a pla-
nar afﬁne pentapod belonging to item (b) of Theorem 1,
which was already discussed in [1]. Therefore we remain
only with reconstruction 2 and 3.
Remark 2. Note that the Duporcq pentapods ﬁt with recon-
struction 2 and 3 for the following reason: Assumed the base
(cf. Fig. 2) is given, then two lines of the complete quadri-
lateral through the points M1,M2,M4,M5 are already de-
termined by the collinearity of the triples M1,M2,M3 and
M3,M4,M5, respectively.
Therefore the quadrilateral is
completed either by the lines M1M5 and M2M4, which cor-
responds with reconstruction 2, or by the lines M1M4 and
M2M5, which corresponds with reconstruction 3.
⋄
3
Bond Theory
In this section we shortly repeat two different ap-
proaches for deﬁning so-called bonds.
The ﬁrst one dis-
cussed in Subsection 3.1 is based on the Study parametriza-
tion of SE(3) in contrast to the one presented in Subsection
3.2, which uses the so-called conformal embedding of SE(3).
In Section 3.3 a relation between the bonds based on these
different embeddings is given.
3.1
Bonds based on the Study Embedding of SE(3)
We denote the eight homogenous Study parameters by
(e0 : e1 : e2 : e3 : f0 : f1 : f2 : f3), where the ﬁrst four ho-
mogeneous coordinates (e0 : e1 : e2 : e3) are the so-called
Euler parameters.
Now, all real points of the Study pa-
rameter space P7, which are located on the so-called Study
quadric S : ∑3
i=0 ei fi = 0, correspond to an Euclidean dis-
placement, with exception of the 3-dimensional subspace
e0 = e1 = e2 = e3 = 0, as its points cannot fulﬁll the con-
dition N ̸= 0 with N = e2
0 + e2
1 + e2
2 + e2
3. All points of the
complex extension P7
C of P7, which cannot fulﬁll this nor-
malizing condition, are located on the so-called exceptional
quadric N = 0.
By using the Study parametrization of Euclidean dis-
placements, the condition that the point mi is located on a
sphere centered in Mi with radius Ri is a quadratic homoge-
neous equation according to Husty [8]. For the explicit for-
mula of this so-called sphere condition Qi we refer to [1, Eq.
(2)]. Now the solution for the direct kinematics over C of
a pentapod can be written as the algebraic variety V of the
ideal spanned by S,Q1,...,Q5,N = 1. In the case of pen-
tapods with mobility 2 the variety V is 2-dimensional.
We consider the algebraic motion of the pentapod,
which is deﬁned as the set of points on the Study quadric
determined by the constraints; i.e., the common points of
the six quadrics S,Q1,...,Q5. Now the points of the alge-
braic motion with N ̸= 0 equal the kinematic image of the
algebraic variety V. But we can also consider the set B of
points of the algebraic motion, which belong to the excep-
tional quadric N = 0. For an exact mathematical deﬁnition
of these so-called bonds we refer to [1, Deﬁnition 1]. In the
case of pentapods with mobility 2 the set B is of dimension
1; i.e., a bonding curve.
We use the following approach for the computation of
bonds: In a ﬁrst step we project the algebraic motion of the
pentapod into the Euler parameter space P3
C by the elimina-
tion of f0,..., f3. This projection is denoted by ς. In a sec-
ond step we determine the set Bς of projected bonds as those
points of the projected point set ς(V), which are located on
the quadric N = 0; i.e.,
Bς := ZarClo(ς (V))∩

(e0 : ... : e3) ∈P3
C | N = 0
	
. (7)
3.2
Bonds based on the Conformal Embedding of SE(3)
As shown in [9, Section 2.1], it is possible to construct a
projective compactiﬁcation X in P16
C for the complexiﬁcation
SE(3)C of the group SE(3) in a way that the sphere condition
is linear in the coordinates of P16
C . The map SE(3) ,→P16
C
is the so-called conformal embedding of SE(3) and X is a
projective variety of dimension 6 and degree 40.
Now the ﬁve linear sphere conditions determine a linear
subspace F ⊆P16
C of codimension 5. The intersection K =
X ∩F is deﬁned to be the complex conﬁguration set of the
pentapod.
It is also known that X can be written as the disjoint
union SE(3)C ∪BX, where the so-called boundary BX is ob-
tained as the intersection of X and a hyperplane H. More-
over the boundary can be decomposed into the following 5
subsets:
Vertex: This is the only real point in BX, a singular point
with multiplicity 20; it is never contained in K.
Collinearity points: If K contains such a point, then either
the platform points or the base points are collinear.
Similarity points: If K contains such a point, then there are
normal projections of platform and base to a plane such
that the images are similar.
Inversion points: If K contains such a point, then there are
normal projections of platform and base to a plane such
that the images are related by an inversion.
Butterﬂy points: If K contains such a point, then there are
two lines, one in the base and one in the platform, such
that any leg has either its base point on the base line or
its platform point on the platform line.
Now the set of bonds BK is obtained as the intersec-
tion of K and the boundary BX. Moreover it should be men-
tioned that the intersection multiplicity of K and H is at least
2 in each bond. Note that for pentapods with mobility 2, the
bondset BK is 1-dimensional.
3.3
Relation between Bonds based on different Embed-
dings
If ρ : SE(3) −→SO(3) is the map sending a direct isom-
etry to its rotational part, then there exists a linear projection
ξ : P16
C 99K P9
C such that the following diagram is commuta-
tive (cf. [10, Section 1]):
SE(3)
#
/
ρ

X ⊆P16
C
ξ

S ⊆P7
C
ς

SO(3)
/ P3
C
v3,2
/ V3,2 ⊆P9
C
(8)
where v3,2 is the Veronese embedding of P3
C and V3,2 is its
image in P9
C. The center of ξ is the linear space spanned by
similarity points, which contains also the collinearity points
and the vertex.
Lemma 1. For reconstruction 2 and 3 the complex conﬁg-
uration set K does not contain collinearity bonds, but four
butterﬂy bonds and one similarity bond.
Proof. The numbers of collinearity and butterﬂy bonds are
trivial. The reasoning for the existence of exactly one simi-
larity bond is as follows (cf. Fig. 6):
As M1,M2,M3 are collinear the ratio TV(M1,M2,M3)
remains constant under parallel projections (with projection
directions not parallel to the carrier line of the collinear
points). Therefore one can construct the point m′
3 on the line
m1m2 such that TV(m1,m2,m′
3) = TV(M1,M2,M3) holds.
In the same way one can construct the point m′′
3 on the line
m4m5 such that TV(m′′
3,m4,m5) = TV(M3,M4,M5) holds.
It can be checked by direct computations that m3,m′
3,m′′
3 are
located on a line g, which gives the direction of the projec-
tion direction of the platform.
The reverse construction from the platform to the base
yields the points M3,M′
3,M′′
3 located on a line G, which gives
the direction of the projection of the base.
Moreover g and G have to be parallel due to Assumption
1, which can also be checked by straightforward computa-
tions.
□
M3
M1
M2
M4
M5
M′
3
M′′
3
G
m2
m1
m3
m′
3
m′′
3
m5
m4
g
Fig. 6.
Projection along the black direction leads to one-
dimensional conﬁgurations that are similar. This shows that the pen-
tapod with such base/platform conﬁguration has a similarity bond.
Corollary 1. If reconstruction 2 or 3 has mobility 2, then
Ke has to be a surface.
Proof. First of all we want to recall the known characteri-
zation for a pure translational self-motion (according to [11,
Theorem 2] under consideration of [1, Footnote 4]): A pen-
tapod possesses a pure translational self-motion, if and only
if the platform can be rotated about the center m1 = M1 into
a pose, where the vectors −−−→
Mimi for i = 2,...,5 fulﬁll the con-
dition rk(−−−→
M2m2,...,−−−→
M5m5) ≤1. Note that this implies the
existence of a similarity bond. Moreover all 1-dimensional
self-motions are circular translations in planes orthogonal to
the parallel vectors −−−→
Mimi for i = 2,...,5.
If Ke is a point, then the orientation during the self-
motion is ﬁxed and we can only obtain a 2-dimensional trans-
lational self-motion. It is well-known [11] that in this case
the platform and the base have to be directly congruent.
If Ke is a curve, then for each corresponding platform
orientation a 1-dimensional translational sub-self-motion has
to exist. This implies a 1-dimensional set of similarity bonds,
which contradicts Lemma 1.
□
Lemma 2. Assume that K is a surface. Assume that the
projection v−1
3,2 ◦ξ: K 7→Ke is birational. Assume that the
pod has inﬁnitely many inversion bonds. Then the intersec-
tion of Ke and the exceptional quadric N = 0 has a curve of
multiplicity bigger than 1.
Proof. The mobility surface K has a tangential intersection
with the boundary hyperplane H at all inversion bonds. Since
the projection K →Ke is birational, the projection ξ : X ⊂
P16 99K P9 is locally an isomorphism for almost all inver-
sion points. Hence the image of K intersects the image of
the hyperplane H tangentially at almost all images of inver-
sion points. This is equivalent to saying that Ke intersects the
exceptional quadric N = 0 at almost all images of inversion
points. The closure of these points would be the curve with
intersection multiplicity bigger than 1.
□
Corollary 2. If in addition to the above assumptions Ke is
a quadric surface, then the intersection is totally tangential
along an irreducible quadric.
Proof. By degree, the part D of the intersection which has
multiplicity bigger than one can only be a line or a conic.
Since D is deﬁned over R, and any line contains real points,
and the exceptional quadric does not have real points, D is
not a line. If D were reducible, then it is the union of two
lines. If the two lines intersect, then the intersection point
would be real, but the exceptional quadric N = 0 contains no
real points; a contradiction. If the two lines do not intersect,
then we have a complete intersection of dimension 1 which
is disconnected, and this is also in contradiction to a well-
known theorem in algebraic geometry (see [12]). Therefore
we remain with the case stated in Corollary 2.
□
4
Computations in Study and Euler parameter space
Within this section we prove computationally that recon-
struction 2 and 3 can only have a 2-dimensional self-motion
in the case already known to Duporcq.
4.1
Computation of Ke
We parametrize the base as follows:
M1 := (0,0,0)T,
M2 := (1,0,0)T,
M4 := (A4,B4,0)T,
M5 := (A5,B5,0)T.
(9)
Then M3 is already determined as the intersection point of
M1M2 and M4M5, thus we get:
M3 :=
B4A5 −A4B5
B4 −B5
,0,0
T
.
(10)
As M3 has to be a ﬁnite point which is not allowed to collapse
with one of the other four given points we have B4B5U1 ̸= 0
with:
U1 := (B4 −B5)(B4A5 −A4B5)(B4A5 −A4B5 −B4 +B5).
(11)
Now we compute the platform with the help of the mapping
κi. In the remainder of this section we restrict to the case
i = 2 as the case i = 3 can be done in a total analogous way.
W.l.o.g. we can assume that the matrix A of κ2 has the form:
A :=

µ1 µ2
0 µ3

with
µ1µ3 ̸= 0
and
µ1 > 0.
(12)
Therefore we get m j = AM j for j = 1,2,4,5 and obtain m3
as the intersection point of m2m4 and m1m5, which yields:
m3 :=
 B4(A5µ1 +B5µ2)
B4A5 +B5 −A4B5
,
B4B5µ3
B4A5 +B5 −A4B5
,0
T
. (13)
As this point also has to be a ﬁnite point we get additionally
the assumption U2 ̸= 0 with
U2 := B4A5 +B5 −A4B5.
(14)
Moreover M3 cannot be located on M4M5 which yields U3 ̸=
0 with
U3 := B4A5 −B4 −A4B5.
(15)
Let us denote the numerator of the difference Q1 −Qi (i =
2,...,5) of sphere conditions by ∆i. Then we can compute
the linear combination
B4B5U1∆2 +U3∆3 +B5U1U2∆4 −B4U1U2∆5
(16)
which yields a quadratic expression Ke[1356] in the Euler
parameters (free of Study parameters f0,..., f3), where the
number in the bracket gives the number of terms.
Remark 3. Ke cannot be fulﬁlled identically for the follow-
ing reason: In this case the sphere conditions Q1,...,Q5 are
linearly dependent and therefore we would end up with an
degenerated architectural singular manipulator [13], which
has to have 4 collinear points in the platform or the base (see
also [14]) contradicting the design under consideration.
⋄
4.2
Determining the pentapod’s geometry
In the following we show that the projection K →Ke
cannot be birational. This is done by contradiction; i.e. we
assume that K →Ke is birational, and show that Ke cannot
intersect N = 0 totally tangential along an irreducible quadric
(cf. Corollary 2).
If Ke touches N = 0 along a quadric then there has to
exist a double-counted plane ε: ν0e0 +ν1e1 +ν2e2 +ν3e3 =
0 within the pencil of quadrics spanned by Ke = 0 and N = 0.
Thus we can make the following ansatz W = 0 with:
W := Ke +νN +(ν0e0 +ν1e1 +ν2e2 +ν3e3)2.
(17)
In the following we denote the coefﬁcient of ei
0ej
1ek
2el
3 of W
by Wijkl. We consider:
W1100 = 2ν0ν1,
W1010 = 2ν0ν2,
W0101 = 2ν1ν3,
W0011 = 2ν2ν3.
(18)
which implies the following two cases:
• ν0 = ν3 = 0: We compute
W0200 −W2000 = ν2
1,
W0020 −W0002 = ν2
2,
(19)
which implies that all ν0,...,ν3 are equal to zero, a con-
tradiction.
• ν1 = ν2 = 0: Now we compute
W2000 −W0200 = ν2
0,
W0002 −W0020 = ν2
3,
(20)
yielding the same contradiction.
This shows that the projection K →Ke cannot be birational,
which is equivalent with the condition that the system of
equations S = ∆2 = ∆3 = ∆4 = ∆5 = 0 linear in f0,..., f3
is linear dependent (rank of the coefﬁcient matrix is less than
4). By means of linear algebra it can easily be veriﬁed that
this can only be the case if T = 0 holds with
T := ε01e0e1 +ε02e0e2 +ε13e1e3 +ε23e2e3
(21)
and
ε01 := µ3(1+ µ1)B,
ε02 := µ1A(µ3 +1)−µ2B,
(22)
ε23 := µ3(1−µ1)B,
ε13 := µ1A(µ3 −1)+ µ2B,
(23)
by using the following abbreviations:
A := A5 −A4 +1,
B := B4 −B5.
(24)
Note that T = 0 is also a quadric in the Euler parameter
space. In the general case T = 0 and Ke = 0 intersect along
a curve, but due to Corollary 1 they have to possess at least a
common 2-dimensional component (i.e. a plane) or they are
even identical.
Necessary conditions for this circumstance are obtained
by determining the intersection of Ke = T = N = 0 by resul-
tant method; in detail this works as follows: We compute the
resultant RKe of T and N with respect to e0. In the same way
we compute the resultants RT and RN. Then we calculate
the resultant RKeT of RKe and RT with respect to e3. Anal-
ogously we obtain RKeN and RTN. Finally we compute the
greatest common divisor of RKeT, RKeN and RTN, which can
only vanish for B4B5U1U2µ1µ3F1F2 = 0 with
F1 :=[B2µ2 −AB(µ1 + µ3)](e2
2 −e2
1)+
[2A2µ1 −2B2µ3 −2ABµ2]e1e2
(25)
and
F2 := (1+ µ1)(µ3 −1)e2
1 +(1+ µ3)(µ1 −1)e2
2 −2µ2e1e2.
(26)
In order that the three quadrics N = T = Ke = 0 have a curve
(projected bonding curve) in common either F1 or F2 has to
be fulﬁlled identically.
ad F1: We solve the coefﬁcient of e2
2 of F1 for µ2 and plug the
obtained expression in the coefﬁcient of e1e2 of F1. The
resulting expression has only the real solution A = B = 0,
a contradiction.
ad F2: It can easily be seen that F2 is fulﬁlled identically if
and only if µ1 = µ3 = 1 and µ2 = 0 holds (due to our
assumptions with respect to A of Eq. (12)).
Therefore κ2 has to be the identity, which already im-
plies the geometric properties of the Duporcq pentapods.
We only remain to show that these pentapods possess 2-
dimensional self-motions, which are line-symmetric in ad-
dition.
4.3
Determining the pentapod’s self-motion
Plugging µ1 = µ3 = 1 and µ2 = 0 in T = 0 yields:
−2e0(Be1 +Ae2) = 0. Therefore we have to distinguish two
cases:
• e1 = −Ae2/B: Now Ke possesses 1397 terms and has to
vanish identically as dim(Ke) = 2 has to hold according
to Corollary 1.
We consider the coefﬁcient of e0e3 of
Ke which equals −8B4B5U1U2 and cannot vanish without
contradiction.
• e0 = 0: Now Ke factors into (e2
1 + e2
2 + e2
3)G[185] where
G is of the form g0 + g1R2
1 + g2R2
2 + g3R2
3 + g4R2
4 + g5R2
5,
where all gi are functions in the geometry of the platform
and the base. This equation can be solved for R2
3 w.l.o.g.
as g3 equals B2U2
2U3.
Now we compute f0, f1, f3 from S = ∆2 = ∆4 = 0 w.l.o.g..
Plugging the obtained expressions into ∆3 and ∆5 imply in
the numerators the following expressions:
−B4U1U2(e2
1 +e2
2 +e2
3)H
and
(e2
1 +e2
2 +e2
3)H, (27)
respectively, with H := h1e1 +h2e2 and
h1 := (R2
2 −R2
5)A4 +(R2
4 −R2
1)(A5 −1),
h2 := (R2
2 −R2
5)B4 +(R2
4 −R2
1)B5.
(28)
As dim(Ke) = 2 has to hold according to Corollary 1 the
expression H has to be fulﬁlled identically.
Under the
assumption R2
1 ̸= R2
4 we can compute A5 and B5 from
h1 = h2 = 0 but then we get U3 = 0; a contradiction. As
a consequence R2
1 = R2
4 has to hold, which implies to-
gether with h1 = h2 = 0 the condition R2
2 = R2
5. This al-
ready yields a 2-dimensional self-motion. Moreover, due
to R2
1 = R2
4 we get f0 = 0, which proves the line-symmetry
of this self-motion (cf. [3, Section 1]).
Remark 4. It should be noted that due to the existence
of the similarity bond and κ2 = id the Duporcq pentapods
also have pure translational 1-dimensional self-motions (cf.
proof of Corollary 1). Moreover each 2-dimensional self-
motion of a Duporcq pentapod contains a pure translational
1-dimensional sub-self-motion (obtained by e1 = e2 = 0). ⋄
5
Conclusions
In light of the results of this paper, Theorem 4 of [1] is
not correct, but the ﬂaw can be ﬁxed by rewriting the phrase
”which is not listed in Theorems 2 and 3” by ”which is nei-
ther a Duporcq pentapod nor listed in Theorems 2 and 3”.
Furthermore we have to check if the Duporcq pentapods
do not imply a further case in the list of non-architecturally
singular hexapods with 2-dimensional self-motions given in
[1, Theorem 5]. Starting with a Duprocq pentapod, the two
complete quadrilaterals are already determined and there is
only one further point which has the same geometric prop-
erties with respect to this quadrilateral as the third anchor
point. This is exactly the sixth anchor point illustrated in Fig.
1. But the resulting hexapod is architecturally singular as al-
ready mentioned in Subsection 1.1, thus Theorem 5 of [1] is
correct.
Acknowledgments
The ﬁrst author’s research is funded by the Austrian
Science Fund (FWF): P24927-N25 - “Stewart Gough plat-
forms with self-motions”. The second author’s research is
supported by the Austrian Science Fund (FWF): W1214-
N15/DK9 and P26607 - “Algebraic Methods in Kinematics:
Motion Factorisation and Bond Theory”.
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