Self-motions of pentapods with linear platform
Georg Nawratila, Josef Schichob
aInstitute of Discrete Mathematics and Geometry, Vienna University of Technology, Wiedner Hauptstrasse 8-10/104, 1040 Vienna, Austria
bJohann Radon Institute for Computational and Applied Mathematics, Austrian Academy of Sciences, Altenberger Strasse 69, 4040 Linz, Austria
Abstract
We give a full classification of all pentapods with linear platform possessing a self-motion beside the trivial rotation
about the platform. Recent research necessitates a contemporary and accurate re-examination of old results on this
topic given by Darboux, Mannheim, Duporcq and Bricard, which also takes the coincidence of platform anchor points
into account. For our study we use bond theory with respect to a novel kinematic mapping for pentapods with linear
platform, beside the method of singular-invariant leg-rearrangements. Based on our results we design pentapods with
linear platform, which have a simplified direct kinematics concerning their number of (real) solutions.
Keywords: Pentapod, Borel Bricard Problem, Bond Theory, Self-Motion, Direct Kinematic, Kinematic Mapping
1. Introduction
The geometry of a pentapod (see Fig. 2a) is given by the five base anchor points Mi with coordinates Mi :=
(Ai, Bi,Ci)T with respect to the fixed system Σ0 and by the five collinear platform anchor points mi with coordinates
mi := (ai, 0, 0)T with respect to the moving system Σ (for i = 1, . . . , 5). All pairs (Mi, mi) of corresponding anchor
points are connected by SPS-legs (or alternatively SPU-legs with aligned universal joints; cf. [1, Fig. 1]), where only
the prismatic joints are active.
If the geometry of the manipulator is given, as well as the lengths of the five pairwise distinct legs, a pentapod has
generically mobility 1 according to the formula of Gr¨ubler. In the discussed case of pentapods with linear platform
the degree of freedom corresponds to the rotation about the carrier line p of the five platform anchor points. This
rotational motion is irrelevant for applications with axial symmetry as e.g. 5-axis milling, spot-welding, laser or
water-jet engraving/cutting, spray-based painting, etc. (cf. [2, 3]). Therefore these mechanisms are of great practical
interest. In this context configurations should be avoided, where the manipulator gains an additional uncontrollable
mobility (beside the rotational motion around p), which is referred as self-motion within this article. Before we give
a review on pentapods possessing these special motions in Section 1.2, we repeat a few basics in geometry, which are
essential for the understanding of the paper.
1.1. Geometric basics
We consider the projective closure of the Euclidean 3-space, which means that we add a point at infinity to each
line, which is a so-called ideal point. Moreover two lines are parallel or coincide if and only if they have the same
ideal point. The set of ideal points of a pencil of lines (cf. Fig. 1a) form a so-called ideal line of the carrier plane
of this pencil. Again two planes are parallel or coincide if and only if they have the ideal line in common. The set
of ideal points of a bundle of lines (cf. Fig. 1b) constitute the so-called ideal plane. Summed up we can say that we
obtain the projective closure by addition of the ideal plane. Points, lines and planes, which are no ideal elements are
called finite.
We can also introduce projective point coordinates by homogenizing the coordinates (x, y, z) of a finite point by
(1 : x : y : z). The ideal point of a finite line in direction (u, v, w) has the coordinates (0 : u : v : w). Now we can
define a regular/singular projectivity (projective mapping) by multiplication of the projective point coordinates with a
regular/singular 4 × 4 matrix. If the set of ideal points is mapped onto itself, then the projectivity is called an affinity
(affine transformation).
arXiv version
2015
arXiv:1407.6126v4  [cs.RO]  16 Oct 2015
(a)
(b)
gκ
Giκ G jκ
Gkκ
Glκ
g
Gi
G j
Gk
Gl
(c)
Figure 1: (a) A pencil of lines is the 1-parametric set of all coplanar lines through a common point, which is the co-called vertex of the pencil. (b)
A bundle of lines is the 2-parametric set of all lines through a common point, which is the co-called vertex of the bundle. (c) A regulus of lines.
Under a regular projectivity a line g is mapped onto a line gκ due to the linearity of the mapping. The set of
lines [Gi, Giκ] with Gi ∈g is called a regulus of lines (cf. Fig. 1c) if g and gκ are skew. In this context it should be
noted that the cross-ratio (CR) of four collinear points (cf. [4, page 20]) is invariant under a regular projectivity; i.e.
CR(Gi, G j, Gk, Gl) = CR(Giκ, G jκ, Gkκ, Glκ).
Conic sections are well-known geometric objects. If the ideal line of the carrier plane of the conic section touches
the conic, then it is a parabola. If there are conjugate complex (resp. two real) intersection points, we get an ellipse
(resp. a hyperbola). The conjugate complex intersection points of a circle with the ideal line are the co-called cyclic
points of the circle’s carrier plane.
A further geometric object used within the article at hand is a so-called cubic ellipse. According to [5] a cubic
ellipse is a space curve of degree 3, which intersects the ideal plane in one real and two conjugate complex ideal
points1. If the latter ones are the cyclic points of a plane (not) orthogonal to the direction of the real ideal point, then
the cubic ellipse is called straight cubic circle (skew cubic circle). For more details we refer to [6].
Finally we need the notation of a so-called M¨obius transformation τ of the plane. If we combine the planar
Cartesian coordinates (u, v) to a complex number w := u + iv, then τ(w) can be defined as a rational function of the
form
τ :
w 7→z1w + z2
z3w + z4
,
(1)
with complex numbers z1, . . . , z4 satisfying z1z4 −z2z3 , 0. Moreover it should be noted that τ maps straight lines
onto straight lines or circles and that a M¨obius transformation is uniquely defined by three pairwise distinct points
w1, w2, w3 and their pairwise distinct images.
1.2. Review, motivation and outline
The self-motions of pentapods with linear platform represent interesting solutions to the still unsolved problem
posed by the French Academy of Science for the Prix Vaillant of the year 1904, which is also known as Borel-Bricard
problem (cf. [7, 8, 9]) and reads as follows: ”Determine and study all displacements of a rigid body in which distinct
points of the body move on spherical paths.”
For the special case of five collinear points the Borel-Bricard problem seemed to be solved since more than 100
years, due to the following results (cf. [10, page 415]): If five points of a line have spherical trajectories then this
property holds for all points of the line. The centers are located on a straight line (cf. Darboux [11, page 222]), a conic
section (cf. Mannheim [12, pages 180ff]) or a straight cubic circle (cf. Duporcq [13]; see also Bricard [8, Chapter
III]).
In a recent publication [14] the authors determined all pentapods with mobility 2, where neither all platform anchor
points nor all base anchor points are collinear. As a side result of this study we obtained the following three designs
of pentapods with linear platform possessing a self-motion of p:
1Therefore the projection in direction of the real ideal point yields an ellipse.
2
M3
M1
M2
M5
M4
m5
m4
p
m3m2
m1
(a)
M4
M1
M5
M3
M2
m5
= m3
m4
m1 = m2
h
(b)
M4
M1
M5 M3
M2
m5
m3
m4
m1 = m2
h
(c)
M4
M1
M5 M3
M2
m5
m3
m2
m4
m1
h
(d)
Figure 2: (a) Pentapod with linear platform p (and planar base). (b,c,d) The designs (α, β, γ) are illustrated in a pose of a self-motion.
(α) m1 = m2 = m3: The self-motion is obtained if m1 = m2 = m3 is located on the line h spanned by M4 and M5.
(β) m1 = m2 and M3, M4, M5 collinear: The self-motion is obtained if m1 = m2 is located on the line h spanned by
M3, M4, M5.
(γ) M2, M3, M4, M5 collinear: The self-motion is obtained if m1 is located on the line h spanned by M2, M3, M4, M5.
In all three cases, which are illustrated in Fig. 2, the following legs can be added without restricting the spherical
self-motion: Every point of p can be connected with any point of the line h with exception of the point m1 (= center
of spherical motion), which can be linked with any point of the fixed 3-space.
This already shows that the cases (α, β, γ) are not covered by the above cited results of Darboux, Mannheim,
Duporcq and Bricard, respectively. The reason for this is partially hidden in the formulation of the problem, as 100
years ago they studied the conditions for five points of a line to run on spherical trajectories, which already implies
that the collinear points are pairwise distinct. Therefore they missed the cases (α, β), but this still does not explain the
absence of case (γ). All in all this shows the need of a contemporary and accurate re-examination of the old results,
which also takes the coincidence of platform anchor points into account. This is done within the paper at hand, which
is structured as follows:
In Section 1.3 we give a short review on architecturally singular pentapods as they imply further solutions to our
problem. Based on the method of singular-invariant leg-replacements we determine in Section 2 all non-architecturally
singular pentapods with linear platform and planar base possessing self-motions. The same method is used in Section
3 to distinguish five different types of pentapods with linear platform and non-planar base. In Section 4 we introduce
the theory of bonds with respect to a novel kinematic mapping for pentapods with linear platform. This theory is used
in Section 5 for the determination of all non-architecturally singular pentapods with linear platform and non-planar
base possessing self-motions. Finally we use the presented results/methods to design pentapods with a linear platform,
which have a simplified direct kinematics with respect to the number of (real) solutions (cf. Section 6).
1.3. Architecturally singular pentapods
A pentapod is called architecturally singular if in any pose of the platform the rank of its Jacobian matrix is less
than five. This condition also has a line-geometric interpretation as the Jacobian is composed of the Pl¨ucker coordi-
nates of the five carrier lines of the legs (cf. [4]). All architecturally singular pentapods are known (cf. [15, Theorem
3] under consideration of [16]) as well as their properties of being redundant and invariant under projectivities of the
platform and the base. If we take additionally the collinearity of the platform into account (i.e. m1, . . . , m5 collinear)
we end up with the following list:
Corollary 1. If a pentapod with linear platform is architecturally singular, then it has one of the following designs 2:
2After a possible necessary renumbering of anchor points.
3
1. M1 = M2 = M3: The first three legs belong to a pencil of lines.
2. m1 = m2 = m3 and M1, M2, M3 are collinear: The first three legs belong to a pencil of lines.
3. M1, . . . , M4 are collinear and the following cross-ratio condition holds:
CR(m1, m2, m3, m4) = CR(M1, M2, M3, M4).
(2)
The first four lines belong to a regulus of lines. In the special case m1 = m2 and M3 = M4 the regulus splits up
into two pencils of lines.
4. m1 = m2 = m3 = m4: The first four legs belong to a bundle of lines.
5. M1, . . . , M5 are collinear.
6. m1 = m2 = m3 and M4 = M5.
7. m1 = m2 and m4 = m5. Moreover M1, M2, M5 are collinear and M3, M4, M5 are collinear.
8. m1, . . . , m5 are pairwise distinct, M1, . . . , M5 are coplanar and no three of them are collinear. Moreover there
is a projective correspondence between the points mi and Mi for i = 1, . . . , 5. For more details on this most
complicated case we refer to [15, Item 10 of Theorem 3] and [17].
9. m4 = m5 holds and M1, . . . , M5 are coplanar, where M1, M2, M3 are collinear. Moreover the following cross-
ratio condition holds:
CR(m1, m2, m3, m4 = m5) = CR(M1, M2, M3, M),
(3)
with M denoting the intersection point of [M4, M5] and the carrier line of M1, M2, M3.
In the cases (5-9) the five legs belong to a so-called congruence of lines (cf. [4, Section 3.2.1]).
Due to the above mentioned redundancy all these nine cases imply solutions to our problem, as they have a self-
motion in each pose of p (over C). Moreover, the architecturally singular cases 2,4,6,7,9 are also not covered by the
old results due to the coincidence of platform anchor points. The remaining cases are discussed in more detail:
Ad 3: All lines of the regulus can be added without restricting the self-motion. Therefore this case corresponds to the
result of Darboux [11, page 222].
Ad 8: The projective correspondence can be extended to all points of the linear platform p and therefore they are
mapped onto a conic determined by M1, . . . , M5. Now all legs connecting corresponding anchor points can be
attached without changing the self-motion. This equals the solution given by Mannheim [12, pages 180ff].
Ad 1: This case can be interpreted as a special case of Mannheim’s solution, as following legs can be added without
restricting the self-motion: Every point of p can be connected with M1 = M2 = M3 with exception of the point
mi which can be linked with any point of the line [M1, Mi] for i = 4, 5. Therefore the conic splits up into the
two lines [M1, M4] and [M1, M5].
Ad 5: This trivial case can also be seen as a special case of Mannheim’s solution, as the conic degenerates into the
double counted carrier line of M1, . . . , M5.
As all architecturally singular pentapods with linear platform are already known, we can restrict our study done
in the remainder of the article to non-architecturally singular manipulators. Moreover, as the designs (α, β, γ) are not
architecturally singular, we can make the following three additional assumptions in order to exclude these already
known cases:
(i) No three platform anchor points coincide.
(ii) If two platform anchor points coincide, the remaining three base anchor points are not collinear.
4
(iii) No four base anchor points are collinear.
Definition 1. We define by P the set of all non-architecturally singular pentapods with a linear platform, which fulfill
the assumptions (i,ii,iii).
We split the determination of all elements of P with self-motions in two parts with respect to the criterion if the
base anchor points are coplanar (= planar pentapod; e.g. Fig. 2a,d) or not (= non-planar pentapod; e.g. Fig. 2b,c).
2. Planar pentapods of P with self-motions
Within this section we prove the following theorem:
Theorem 1. A planar pentapod of P has a self-motion only in the following case: There exists an orthogonal-
projection πε of the base plane ε and an orthogonal-projection πp of p in a way that the projected point sets are
congruent. In this case p can perform a circular translation.
For the proof of this theorem the following preparatory work has to be done:
Lemma 1. The anchor points of a planar pentapod with a linear platform, which fulfills the assumptions (i,ii,iii), can
always be relabeled in a way that the following conditions hold:
M1 , M2,
M1, M2, M3 not collinear,
M1, M2, M4 not collinear,
m3 , m4.
Proof: If two platform anchor points coincide we denote them with m4 = m5. Then due to assumption (ii) M1, M2, M3
are not collinear. Due two assumption (iii) one of the remaining two base points is not on the line spanned by M1 and
M2. We denote this point by M4 and we are done.
Now we discuss the case where all five platform anchor points are pairwise distinct:
⋆If two base points coincide3 then we denote them with M3 = M4. Due to assumption (iii) there are at least two
further base points which span together with M3 = M4 a plane. We denote these points by M1 and M2, respectively.
⋆If no base points coincide, but three of them are collinear, then we denote them by M1, M2 and M5. Due to
assumption (iii) we are done.
⋆If no three base points are collinear, we can label the points arbitrarily.
□
Moreover we can choose the moving frame Σ in a way that m1 equals its origin. The fixed frame Σ0 is selected in a
way that M1 equals the origin, M2 is located on the x-axis and the remaining points belong to the xy-plane. Due to
Lemma 1 we can assume w.l.o.g. that A2B3B4(a3 −a4) , 0 holds.
The proof of Theorem 1 is based on the following result obtained by Borras et al. [17]: A leg of a given planar
pentapod with linear platform is replaced by a leg with platform anchor point (a, 0, 0) and base anchor point (A, B, 0)
fulfilling Eq. (6) of [17], which reads as follows under consideration of our special choice of coordinate systems Σ
and Σ0:
[(D2, D3, 0) + a(D4, D5, D1)]

A
B
1
= 0
(4)
with
D1 := det(A, B, aA, aB),
D2 := −det(a, B, aA, aB),
D3 := det(a, A, aA, aB),
D4 := −det(a, A, B, aB),
D5 := det(a, A, B, aA),
(5)
3No three base points can coincide as it yields a contradiction to assumption (iii).
5
and
a :=

a2
a3
a4
a5

,
A :=

A2
A3
A4
A5

,
B :=

0
B3
B4
B5

,
aA :=

a2A2
a3A3
a4A4
a5A5

,
aB :=

0
a3B3
a4B4
a5B5

,
(6)
then the resulting planar pentapod with linear platform has the same singularity set (and direct kinematics solution)
if it is not architecturally singular.
In the following we study Eq. (4) in more detail:
Lemma 2. A planar pentapod with linear platform fulfilling the assumptions (i,ii,iii) is architecturally singular if and
only if D1 = D4 = 0 holds (with respect to Σ and Σ0).
Proof: Due to Lemma 1 we can assume w.l.o.g. that the following determinant

A2
0
0
A3
B3
a3B3
A4
B4
a4B4

= A2B3B4(a3 −a4)
(7)
is different from zero. Therefore D1 = D4 = 0 implies rk(a, A, B, aA, aB) < 5 which characterizes architecturally
singularity (cf. [16, 17, 18]).
□
Based on this lemma we can prove the next one, which reads as follows:
Lemma 3. For a planar pentapod of P the condition D2 = D3 = 0 cannot hold (with respect to Σ and Σ0).
Proof: The proof is done by contradiction as follows: We show that for a planar pentapod with linear platform the
condition D2 = D3 = 0 (with respect to Σ and Σ0 ⇒A2B3B4(a3 −a4) , 0) implies either an architecturally singular
design (⇔D1 = D4 = 0; cf. Lemma 2) or a contradiction to the assumptions (i,ii,iii).
It can easily be seen that D2 = 0 is fulfilled for a2 = 0 (⇔m1 = m2). Now D3 = 0 simplifies to:
−a3a4a5A2(A4B5 −A5B4 −A3B5 + A5B3 + A3B4 −A4B3).
(8)
a3a4a5 = 0 contradicts assumption (i), A2 cannot vanish, and the last factor implies the collinearity of M3, M4, M5,
which contradicts assumption (ii). Therefore we can assume for the remaining discussion that a2 , 0 holds. We
distinguish the following cases:
1. a5 , 0: Under this assumption we can solve D2 = 0 for A5. Then the numerator of D3 factors into
a3a4(A2B3 −A2B4 + A3B4 −A4B3)F[20],
(9)
where the number in the brackets gives the number of terms. As F is also a factor of D1 and D4, the condition F = 0
implies an architecturally singular design. Moreover ai = 0 (⇔m1 = mi) implies the collinearity of M2, Mj, M5 for
pairwise distinct i, j ∈{3, 4}. Therefore a3a4 = 0 contradicts assumption (ii). The vanishing of the third factor of
Eq. (9) implies the collinearity of M2, . . . , M5, which contradicts assumption (iii).
2. a5 = 0 (⇔m1 = m5): Now D2 and D3 factors into:
−a2a3a4B5(A2B3 −A2B4 + A3B4 −A4B3),
a2a3a4A5(A2B3 −A2B4 + A3B4 −A4B3).
(10)
a2a3a4 = 0 contradicts assumption (i) and the last factor implies the collinearity of M2, M3, M4, which contradicts
assumption (ii). Therefore A5 = B5 = 0 has to hold, but in this case the first and the fifth leg coincide. This closes
the proof of Lemma 3.
□
6
M3
M1
M2
V
M5
M4
m4 = m5
p
m2 = m3
m1
(a)
M◦
1
M◦
1
M◦
1
M◦
1
M◦
1
M◦
1
M◦
1
M◦
1
M◦
1
M◦
1
M◦
1
M◦
1
M◦
1
M◦
1
M◦
1
M◦
1
M◦
1
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
2 = M◦
3
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
M◦
4 = M◦
5
m◦
1
m◦
1
m◦
1
m◦
1
m◦
1
m◦
1
m◦
1
m◦
1
m◦
1
m◦
1
m◦
1
m◦
1
m◦
1
m◦
1
m◦
1
m◦
1
m◦
1
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
2 = m◦
3
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
m◦
4 = m◦
5
S 2
S 2
S 2
S 2
S 2
S 2
S 2
S 2
S 2
S 2
S 2
S 2
S 2
S 2
S 2
S 2
S 2
(b)
M3
M1
M2
M5
M4
m4 = m5
p
m2 = m3
m1
(c)
M′
4 = M′
5
M′
2 = M′
3
M′
1
m′
4 = m′
5
m′
2 = m′
3
m′
1
(d)
Figure 3: (a) V is a finite point. (b) Spherical image of the pentapod given in (a) with respect to the unit sphere S 2 centered in V. (c) V is an ideal
point. (d) Orthogonal projection of the pentapod given in (c) onto a plane orthogonal to V.
Due to this lemma Eq. (4) determines for all planar pentapods of P a bijection between points of p and lines in the
base plane.4 Due to the linear relation the lines generate a pencil with vertex V, which can also be an ideal point (⇒
parallel line pencil). According to [19] we are now able to perform a series of leg-replacements in a way that we end
up with a non-architecturally singular pentapod of the following type (see Fig. 3a,c):
m2 = m3,
m4 = m5,
M2, M3, V are collinear,
M4, M5, V are collinear.
(11)
Based on this preparatory work we can prove Theorem 1 as follows:
Proof of Theorem 1: We distinguish the following two cases:
1. V is a finite point: In this case the motion of p can only be spherical with center V. Therefore we can also consider
the spherical 3-legged manipulator, which we obtain by projecting the pentapod onto the unit sphere S 2 centered
in V. Therefore the spherical 3-legged manipulator has to have a self-motion.
Note that the projected base points M◦
i as well as the projected platform points m◦
i are located on great circles of S 2.
It is well known (cf. [20, Lemma 2] and [21, Theorem 5]) that the 3-legged spherical manipulators illustrated in
(see Fig. 3b) can only have self-motions if two platform or base anchor points coincide. Now this is only possible
if the line p contains V. In this case the platform of the spherical manipulator collapse into a point and we only get
the trivial rotation about the line p as uncontrollable motion while p itself remains fix. Therefore this case does not
yield a solution.
2. V is an ideal point: Now the motion of p can only be a planar one orthogonal to the direction of V. Therefore the
corresponding planar 3-legged manipulator, which is obtained by an orthogonal projection of the pentapod onto a
plane orthogonal to V, also has to have a self-motion. Note that the projected base points M′
i as well as the projected
platform points m′
i are collinear (see Fig. 3d). According to [20, 21] this planar 3-legged manipulator can only
have a self-motion in one of the following two cases:
(a) Two platform or base anchor points coincide: This is only possible if the line p contains V. Analogous
considerations as in the spherical case show that we do not get a solution.
(b) The platform and the base are congruent and all legs have equal lengths: In this case the planar 3-legged
manipulator has a circular translation. This already implies the solution given in Theorem 1.
□
Remark 1. Note that the case given in Theorem 1 is also not covered by the more than 100 year old results of
Darboux, Mannheim, Duporcq and Bricard, respectively, even though no platform anchor points have to coincide.
Therefore our study reveals a further lost case beside design (γ).
4For D2 = D3 = 0 Eq. (4) would factor into a(D4A + D5B + D1) = 0, which does not imply such a bijection.
7
But this case is not novel, as it is already contained within the more general characterization given in Theo-
rem 25 of [21], which reads as follows: A pentapod has a 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 condition
rk(−−−−→
M2m2, . . . , −−−−→
M5m5) ≤1. Therefore a pentapod with a linear platform and a translational self-motion has to have a
planar base.
⋄
3. Types of non-planar pentapods
Based on the idea of singular-invariant leg-replacements, which can also be extended to the non-planar case (cf.
[2]), one can distinguish different types introduced in this section.
Lemma 4. The anchor points of a non-planar pentapod with a linear platform can always be relabeled in a way that
the following conditions hold:
1. M1, . . . , M4 span a tetrahedron.
2. If M1, . . . , M5 are pairwise distinct, then M1, Mi, M5 are not collinear for all i ∈{2, 3, 4}.
If two base anchor points coincide, then M1 is none of them.
As the proof is trivial it is left to the reader and we proceed with the coordinatization used within this section:
Again we choose the origin of the moving system Σ in m1. The fixed frame Σ0 is selected in a way that M1 equals the
origin, M2 belongs to the x-axis and M3 is located in the xy-plane. All in all this yields A2B3C4 , 0 and M5 , o (zero
vector).
Moreover we introduce the following notation: Di jk denotes the determinant of the 4 × 7 matrix
(a, A, B, C, aA, aB, aC)
with
C = (0, 0,C4,C5)T,
aC = (0, 0, a4C4, a5C5)T
(12)
after removing the i-th, j-th and k-th column. Now we can state the following lemma:
Lemma 5. For a non-planar pentapod of P the condition D167 = D157 = D156 = 0 cannot hold (with respect to Σ and
Σ0).
Proof: As M2, M3, M4 are linearly independent there exist a unique triple (λ2, λ3, λ4) with λ2M2+λ3M3+λ4M4 = M5.
Due to M5 , o we have (λ2, λ3, λ4) , (0, 0, 0). Now it can easily be seen that the following equivalences hold:
D167 = 0 ⇔
4
X
i=2
λiaiAi = a5A5,
D157 = 0 ⇔
4
X
i=2
λiaiBi = a5B5,
D156 = 0 ⇔
4
X
i=2
λiaiCi = a5C5.
(13)
Therefore D167 = D157 = D156 = 0 implies λ2a2M2 + λ3a3M3 + λ4a4M4 = a5M5. As a5 cannot equal zero6 we can
divide both sides by a5, which shows that the following relation has to hold:
(λ2, λ3, λ4) =
 a2
a5
λ2, a3
a5
λ3, a4
a5
λ4
!
.
(14)
In order to get no contradiction with assumption (i) and M5 , o the implied three equations only have the following
solution: ai = a5 and λj = λk = 0 with pairwise distinct i, j, k ∈{2, 3, 4}. For λi , 1 the points M1, Mi, M5 are collinear,
which contradicts Lemma 4 and for λi = 1 the i-th leg and the fifth leg coincide; a contradiction.
□
In the following we distinguish two cases with respect to the criterion whether D567 vanishes or not. This subdivision
was also used by Bricard in [8, Items 12 and 13 of Chapter III] as D567 = 0 is equivalent to the following geometric
condition, which we call the affine relation (AR):
(AR) There exists a singular affinity κ with Mi 7→mi for i = 1, . . . , 5.
5This theorem is originally stated for hexapods but it also holds for pentapods, as its proof is also valid for 5-legged manipulators.
6For a5 = 0 we get m1 = m5 and therefore a2a3a4 , 0 has to hold, as otherwise we get a contradiction to assumption (i).
8
3.1. D567 , 0
Under this assumption we can use the following result of Borras and Thomas [2]: A leg of a given non-planar
pentapod with linear platform is replaced by a leg with platform anchor point (a, 0, 0) and base anchor point (A, B,C)
fulfilling Eq. (7) of [2], which reads as follows within our notation:

D267 −aD567
−D367
D467
D257
−D357 −aD567
D457
D256
−D356
D456 −aD567


A
B
C
= a

D167
D157
D156

(15)
then the resulting pentapod has the same singularity set (and direct kinematics solution) if it is not architecturally
singular.
Solving Eq. (15) with Crammer’s rule yields:
A = d1(a)
d0(a),
B = d2(a)
d0(a),
C = d3(a)
d0(a).
(16)
Due to the assumption D567 , 0, the polynomial d0 is cubic in the unknown a. The other polynomials d1, d2, d3 are of
degree 3 or less in a, but due to Lemma 5 one of them has to be cubic, which shows the following result:
Corollary 2. The locus of base anchor points of singular-invariant leg-replacements of a non-planar pentapod of P,
which does not fulfill the affine relation (AR), is a cubic space curve.
According to Borras and Thomas [2] we can distinguish different types with respect to the number of roots of
d0 = 0, for which the system Eq. (15) is consistent. This yields the following classification:
Theorem 2. A non-planar pentapod of P, which does not fulfill the affine relation (AR), belongs to one of the following
four types: The cubic of Corollary 2:
Type 1 is irreducible: There is a bijection σ between p and this space curve s (see Fig. 4a).
Type 2 splits up into an irreducible conic q, located in the finite plane ε and a finite line g1 < ε, which intersects q in
the point Q: There is a bijection σ between p \ {P1} and q \ {Q}. Moreover the finite point P1 is mapped to g1
(see Fig. 4b).
Type 3 splits up into the finite lines l and the finite skew lines g1, g2, which intersects l in the point L1 and L2, respec-
tively: There is a bijection σ between p \ {P1, P2} and l \ {L1, L2}. Moreover the finite point Pi is mapped to gi
for i = 1, 2 (see Fig. 4c).
Type 4 splits up into the finite lines g1, g2, g3, which are not coplanar but intersect each other in the finite point V:
All points of p \ {P1, P2, P3} are mapped to V. Moreover the finite point Pi is mapped to gi for i = 1, 2, 3 (see
Fig. 4d).
Note that in Type i we have 4 −i points W1σ−1, . . . , W4−iσ−1 on p (counted with algebraic multiplicity) for
i = 1, . . . , 4, which are mapped by σ to ideal points W1, . . . , W4−i of the base. These points of p have the special
property that their trajectory is in a plane orthogonal to the respective ideal point. Therefore each of these point pairs
determines a so-called Darboux condition (cf. [8, Item 6 of Chapter II] and [22, Section 4.1]). Any other finite point
of p determines a so-called sphere condition; i.e. it is located on a sphere centered in the corresponding finite base
anchor point.
The points Pi have the special property of possessing circular trajectories, i.e. their path is planar and spherical at
the same time.
Note that the ideal point U of p is in all four cases mapped by σ onto a finite point Uσ. Therefore this point pair
determines a so-called Mannheim condition, which is the inverse of the Darboux condition; i.e. a plane of the moving
system orthogonal to p slides through a finite point of the base.
Remark 2. The above given correspondence between points on p and points on the base can also be seen as the
correspondence of point paths of p and the centers of their osculating spheres. It is an old result of Sch¨onflies [23]
that this correspondence is cubic.
⋄
9
W3σ−1
W2σ−1
W1σ−1
p
U
W1
W2
W3
Uσ
s
(a)
W2σ−1
P1
W1σ−1
p
U
g1
q
Uσ
W1
W2
Q
ε
(b)
P2
W1σ−1
P1
p
U
L1
g1
L2
W1
Uσ
g2
l
(c)
P3
P2
P1
g3
g2
p
U
V = Uσ
g1
(d)
Figure 4: Sketch of the types of non-planar pentapods of P, which do not fulfill the affine relation: (a) Type 1. (b) Type 2. (c) Type 3. (d) Type 4.
Pentapods fulfilling the affine relation belong to Type 5.
3.2. D567 = 0
Theorem 3. For a non-planar pentapod of P, which fulfills the affine relation (AR), the ideal point U of p is related
to an ideal element of the base within the correspondence implied by singular-invariant leg-replacements. Pentapods
with this property belong to Type 5.
Proof: Due to Lemma 5 not all three determinants D156, D157, D167 can be equal to zero. Therefore we can assume
e.g. that D167 , 07 holds. Then similar considerations to those given in [2] yield the following correspondence:

D267
−D367
D467
D126 −aD156
−D136
D146 + aD167
D127 −aD157
aD167 −D137
D147


A
B
C
= a

D167
0
0
.
(17)
Introducing homogeneous coordinates to the first of these three equations show that the ideal point U of p is mapped
to an ideal element (ideal point, ideal line or the complete ideal plane) of the base.
□
Lemma 6. If a pentapod of Type 5 has a self-motion, then the ideal element has to be a point W. Moreover the
self-motion is a Sch¨onflies motion with axis direction W.
Proof: We prove by contradiction that U cannot be mapped to more than one ideal point W of the base: Assume that
U can be connected with two distinct ideal points W1 and W2 of the base. Then these special two ”legs” correspond to
two angle conditions (cf. [8, Item 7 of Chapter II] and [22, Section 4.1]); i.e. the angle enclosed by U and Wi (i = 1, 2)
has to be constant during the self-motion. As this already fixes the orientation of p, the pentapod can only have a
translational self-motion, which implies the coplanarity of the base (cf. Remark 1); a contradiction.
As the angle enclosed by U and W has to be constant, the self-motion of p can only be a Sch¨onflies motion with
axis direction W.
□
In the following we analyze the Types 1-5 separately with respect to the existence of self-motions. Type 4 can be
discussed similar to the planar case, as singular-invariant leg-replacements can be used to get the same pentapod
7For D156 , 0 or D157 , 0 the argumentation can be done analogously.
10
illustrated in Fig. 3a with the sole difference that the three lines through the finite point V are not coplanar. Under
consideration of [21, Theorem 6] an analogous argumentation as in item 1 of the proof of Theorem 1 can be done for
Type 4, which shows the following result:
Theorem 4. A pentapod of Type 4 cannot possess a self-motion.
For the discussion of the remaining four types we apply the theory of bonds for pentapods with linear platform,
which is the content of the next section.
4. Bond theory for pentapods with linear platform
Readers, who are not familiar with the following terms of algebraic geometry (ideal, variety, Gr¨obner base, radi-
cal), are recommended to look up these basics in the articles [24] and [25], before studying this section.
It was shown by Husty [26] that the sphere condition equals a homogeneous quadratic equations in the Study
parameters (e0 : e1 : e2 : e3 : f0 : f1 : f2 : f3). For our choice of the moving frame Σ the sphere condition Λi simplifies
to:
Λi :
(a2
i + A2
i + B2
i + C2
i −R2
i )(e2
0 + e2
1 + e2
2 + e2
3) −2aiAi(e2
0 + e2
1 −e2
2 −e2
3) −4aiBi(e0e3 + e1e2)
+ 4aiCi(e0e2 −e1e3) −4ai(e0 f1 −e1 f0 −e2 f3 + e3 f2) + 4Ai(e0 f1 −e1 f0 + e2 f3 −e3 f2)
+ 4Bi(e0 f2 −e1 f3 −e2 f0 + e3 f1) + 4Ci(e0 f3 + e1 f2 −e2 f1 −e3 f0) + 4( f 2
0 + f 2
1 + f 2
2 + f 2
3 ) = 0,
(18)
where Ri denotes the radius of the sphere centered in Mi on which mi is located.
Now, all real points of the Study parameter space P7 (7-dimensional projective space), which are located on the so-
called Study quadric Ψ : P3
i=0 ei fi = 0, correspond to an Euclidean displacement, with exception of the 3-dimensional
subspace e0 = e1 = e2 = e3 = 0, as its points cannot fulfill the condition e2
0 + e2
1 + e2
2 + e2
3 , 0. The translation vector
t := (t1, t2, t3)T and the rotation matrix R of the corresponding Euclidean displacement mi 7→Rmi + t are given by:
t1 = −2(e0 f1 −e1 f0 + e2 f3 −e3 f2),
t2 = −2(e0 f2 −e2 f0 + e3 f1 −e1 f3),
t3 = −2(e0 f3 −e3 f0 + e1 f2 −e2 f1),
and
R =

e2
0 + e2
1 −e2
2 −e2
3
2(e1e2 −e0e3)
2(e1e3 + e0e2)
2(e1e2 + e0e3)
e2
0 −e2
1 + e2
2 −e2
3
2(e2e3 −e0e1)
2(e1e3 −e0e2)
2(e2e3 + e0e1)
e2
0 −e2
1 −e2
2 + e2
3
,
(19)
if the normalizing condition e2
0 + e2
1 + e2
2 + e2
3 = 1 is fulfilled.
Now the solution for the direct kinematics of a pentapod with linear platform can be written as the algebraic
variety of the ideal spanned by Ψ, Λ1, . . . , Λ5, e2
0 + e2
1 + e2
2 + e2
3 = 1. In the general case the dimension of this variety
equals 1, as we also obtain the 1-dimensional rotation about the line p. In the following we present another kinematic
mapping, where we get rid of this redundancy.
4.1. Kinematic mapping for pentapods with linear platform
With respect to the following nine homogeneous motion parameters (n0 : x0 : . . . : x3 : y0 : . . . : y3) with
n0 := f 2
0 + f 2
1 + f 2
2 + f 2
3 and
x0 := 2(e2
0 + e2
1 + e2
2 + e2
3),
y0 := 4(−e0 f1 + e1 f0 + e2 f3 −e3 f2),
x1 := 2(−e2
0 −e2
1 + e2
2 + e2
3),
y1 := 4(e0 f1 −e1 f0 + e2 f3 −e3 f2),
x2 := −4(e0e3 + e1e2),
y2 := 4(e0 f2 −e1 f3 −e2 f0 + e3 f1),
x3 := 4(e0e2 −e1e3)
y3 := 4(e0 f3 + e1 f2 −e2 f1 −e3 f0),
the sphere condition of Eq. (18) is linear; i.e.
Λi :
1
2(a2
i + A2
i + B2
i + C2
i −R2
i )x0 + aiAix1 + aiBix2 + aiCix3 + aiy0 + Aiy1 + Biy2 + Ciy3 + 4n0 = 0.
(20)
11
For x0 = 1 the Euclidean displacement of mi is given by:

ai
0
0
7→

−aix1 −y1
−aix2 −y2
−aix3 −y3
.
(21)
Note that (n0 : x0 : . . . : x3 : y0 : . . . : y3) can be interpreted as a point in the 8-dimensional projective space
P8. In order to determine the image of the kinematic mapping, we compute a Gr¨obner bases of the ideal generated
by n0 −( f 2
0 + f 2
1 + f 2
2 + f 2
3 ), . . . , y3 −4(e0 f3 + e1 f2 −e2 f1 −e3 f0) and e0 f0 + e1 f1 + e2 f2 + e3 f3 eliminating the Study
parameters e0, . . . , f3. It contains three quadric elements in the remaining variables, namely:
Φ1 :
x2
1 + x2
2 + x2
3 −x2
0 = 0,
Φ2 :
y2
1 + y2
2 + y2
3 −8x0n0 = 0,
Φ3 :
x1y1 + x2y2 + x3y3 −x0y0 = 0.
(22)
Because the degree of the image variety (which is computed from the Gr¨obner bases) is equal to 8, the three quadrics
generate the ideal of the image variety I. The image itself consists of all real points in the zero set with x0 , 0. Note
that I is of dimension 5 instead of 6, as we removed the rotations around the line p, which do not change the spherical
condition.
Definition 2. The intersection of I with the five hyperplanes of Eq. (20) is the complex configuration set C of the
pentapod. Its real points are called real configurations.
Lemma 7. A generic pentapod with linear platform has eight solutions for the direct kinematics over C.
Proof: For the solution of the direct kinematics we have to intersect the image variety I with the five hyperplanes of
P8 given by Eq. (20). In the generic case the five hyperplanes have a linear 3-space L in common, which intersects I
in a finite number of points (= complex configuration set C) whose cardinality equals the degree of I.
□
Remark 3. Lemma 7 fits with the results obtained in [27], where the number of eight solutions for the direct kine-
matics problem over C was given for planar pentapods with linear platform. Due to Lemma 7 this number also holds
for the non-planar case.
⋄
4.2. Bonds
Let us assume that the pentapod under consideration has a 1-dimensional self-motion. In this case the complex
configuration set C of Definition 2 is a curve. The points on this configuration curve, which do not correspond to
Euclidean displacements as they fulfill x0 = 0, are the so-called bonds. This gives a rough idea of bonds, which is
developed in detail within this section.
The intersection of the 5-fold I with the hyperplane x0 = 0 yields the so-called boundary of I. A Gr¨obner bases
computation of the ideal generated by the generators of the ideal of I and by x0 reveals that some perfect squares,
for instance (x2y3 −x3y2)2, are contained in this ideal. Therefore the radical contains the elements Γ4, Γ5, Γ6 below.
Its zero set is a 4-fold of degree 4, hence a variety of minimal degree (degree = codimension + 1). It is given by the
following set of equations:
Γ1 :
x2
1 + x2
2 + x2
3 = 0,
Γ2 :
y2
1 + y2
2 + y2
3 = 0,
Γ3 :
x1y1 + x2y2 + x3y3 = 0,
(23)
Γ4 :
x1y2 −x2y1 = 0,
Γ5 :
x1y3 −x3y1 = 0,
Γ6 :
x2y3 −x3y2 = 0,
(24)
and therefore it is independent of y0 and n0.
Definition 3. The intersection of C with the boundary of I is the set B of bonds.
Due to x0 = 0 the set B of bonds is independent of the leg lengths Ri (cf. Eq. (20)), and therefore B only depends
on the geometry of the pentapod with linear platform.
Remark 4. Bonds of pentapods where already introduced in [21] and [28]. In [21] they were defined with respect
to the Study parameters and in [28] with respect to a special compactification of SE(3), which can be seen as a
generalization of the method presented above. But both approaches are not suited for the study of pentapods with
linear platform, due to the rotational redundancy about p.
⋄
12
Type
Necessary Condition 1
Necessary Condition 2
Sufficiency & Leg Parameters
1
Theorem 6
Theorem 9
Theorem 13
2
Theorem 7
Theorem 10
Theorem 14
3
Theorem 8
5
Theorem 5
Theorem 11
Theorem 12
Table 1: Overview of the Theorems, where the first and second necessary condition for mobility arising from bond theory are applied to the
remaining types (for Type 4 see Theorem 4) of non-planar pentapods of P. Moreover the theorems are listed, where the sufficiency of the obtained
necessary conditions is proven. Within these theorems also the leg parameters for a self-motion are given.
In the following we state two necessary conditions for the existence of self-motions in terms of bonds:
Assume that a pentapod with linear platform has a 1-dimensional configuration set C; i.e. there exists a configura-
tion curve c on the 5-fold I. Therefore the corresponding bond set B contains at least one bond β (up to conjugation).
Therefore the existence of a bond is the first necessary condition.
As we have mobility 1, the pentapod with linear platform fulfills the necessary condition of being infinitesimal
movable in each pose of the 1-dimensional motion of p. This is equivalent with the existence of a 1-dimensional
tangent space in each point of the configuration curve c. As β ∈c holds, this implies a second necessary condition for
mobility 1.
Remark 5. Therefore the 3-space L has to intersect the 5-fold I in the bond β at least of multiplicity 2. As this also
holds for the conjugate of β, only one pair of conjugate complex bonds can exist (due to Bezout’s theorem). Moreover
if β (and its conjugate) is a singular point of I then the second necessary condition is trivially fulfilled.
⋄
5. Non-planar pentapods of P with self-motions
The following determination of all non-planar pentapods of P with self-motions is based on the two necessary
conditions given in Section 4.2. For an overview of the workflow/results structured by types please see Table 1.
5.1. First necessary condition
In this section we only apply the first necessary condition, namely the existence of a bond.
Theorem 5. The base anchor points of possible leg-replacements of a pentapod of Type 5 with self-motions have to
be located on an irreducible cubic ellipse s∗on a cylinder of revolution.
Proof: Given is a pentapod of Type 5. W.l.o.g. we can assume that M1, . . . , M4 span a tetrahedron. Due to the
properties of Type 5 (cf. Lemma 6) we can replace the sphere condition implied by the fifth leg by an angle condition;
i.e. the angle ϕ enclosed by the ideal point m5 of p and the ideal point M5 of the base is constant.
Now we can choose the fixed frame Σ0 that M1 is the origin, M5 is the ideal point of the x-axis, and M2 is located
in the xy-plane. The moving frame Σ is chosen in a way that m1 is its origin. With respect to these coordinate systems
the angle condition reads as follows:
∢5 :
x1 + wx0 = 0
(25)
where w denotes arccos (ϕ). The equations of Λi for i = 1, 2, 3, 4 are given in Eq. (20) under consideration of
A1 = B1 = C1 = C2 = a1 = 0.
We set x0 = 0 and start the computation of the bonds: As M1, . . . , M4 are non-planar (⇔K , 0 with
K = A2(B3C4 −B4C3) + B2(A4C3 −A3C4)
(26)
holds) we can solve Λ1, . . . , Λ4, ∢5 for n0, x1, y1, y2, y3 w.l.o.g.. As we get x1 = 0, the condition Γ1 implies x2 = ±ix3.
W.l.o.g. we only discuss the upper branch as the lower one can be done analogously.
Now the numerator of Γ3 factors into x3F3[26] and the numerator of Γ4 splits up into x3F4[10]. Therefore we have
to distinguish two cases:
13
1. x3 = 0: Now the numerator of Γ2 factors into y2
0F2[41], where F2 is quadratic with respect to a4. The discriminant
of F2 with respect to a4 equals:
−K2 h
(a2A3 −a3A2)2 + (a2B3 −a3B2)2 + a2
2C2
3
i
.
(27)
This expression can never be greater than zero. It is equal to zero if the following three conditions are fulfilled:
a2A3 −a3A2 = 0,
a2B3 −a3B2 = 0,
a2C3 = 0.
(28)
As a2 = a3 = 0 contradicts assumption (i) we have to discuss the following cases:
(a) a2 = 0: This implies A2 = B2 = 0 and therefore the first and second leg coincide; a contradiction.
(b) a3 = 0, a2 , 0: We get A3 = B3 = C3 = 0 and therefore the first and third leg coincide; a contradiction.
(c) a2a3 , 0: We get C3 = 0, A3 = a3A2/a2 and B3 = a3B2/a2 but this implies K = 0; a contradiction.
As a consequence the case x3 = 0 does not imply any solution to our problem.
2. x3 , 0: In this case F3[26] = 0 and F4[10] = 0 have to hold. Computing the resultant of these two expressions
with respect to y0 yields −x3K(Gr + iGc) with
Gr =
[a4B3(a2 −a3) + a3B4(a4 −a2)] B2 + a2(a3 −a4)(B3B4 −C3C4),
Gc = −[a4C3(a2 −a3) + a3C4(a4 −a2)] B2 −a2(a3 −a4)(B3C4 + B4C3).
(29)
These two expressions do not depend on the Ai coordinates. We denote the orthogonal projection of the base
anchor point Mi onto the yz-plane of the fixed frame Σ0 by M′
i for i = 1, . . . , 4. In the following we want to show
that M′
1, . . . , M′
4 has to be pairwise distinct as well as m1, . . . , m4. The proof is done by contradiction, where the
following cases have to be discussed:
(a) Two platform anchor points coincide: W.l.o.g. we can set m1 = m2; i.e. a2 = 0. Then Gr and Gc simplify to
Gr = −a3a4B2(B3 −B4),
Gc = a3a4B2(C3 −C4).
(30)
As a3a4 = 0 contradicts assumption (i), and M′
3 = M′
4 (⇒M3, M4, M5 collinear) implies a contradiction to
assumption (ii) we remain with the case M′
1 = M′
2:
In this case m1 = m2 is located on a circle in a plane orthogonal to the axis of the Sch¨onflies motion. As the
point m1 = m2 (and therefore the complete line p) cannot be translated in direction of M5, the problem reduces
to a planar one (projection to the yz-plane of Σ0). As a circular translation of the resulting planar manipulator
is not possible (as otherwise the base has to be planar; cf. Remark 1), anchor points have to coincide (cf. item
2 in the proof of Theorem 1).
i. If further base anchor points coincide beside M′
1 = M′
2 we get again a contradiction to the assumption
M1, . . . , M4 are non-planar.
ii. Further platform anchor points (beside m′
1 = m′
2) can only coincide without contradicting assumption (i)
or (ii) if p is parallel to the axis of the Sch¨onflies motion. In this case the platform of the planar manipulator
collapse into a point and we only get the trivial rotation about the line p as uncontrollable motion while p
itself remains fix.
(b) No platform anchor points coincide and two projected base anchor points coincide: W.l.o.g. we can assume
that M′
1 = M′
2 holds; i.e. B2 = 0. Then Gr and Gc simplify to
Gr = a2(a3 −a4)(B3B4 −C3C4),
Gc = −a2(a3 −a4)(B3C4 + B4C3),
(31)
which cannot vanish without contradiction.
14
As a consequence M′
1, . . . , M′
4 have to be pairwise distinct, as well as m1, . . . , m4. Therefore there exists a uniquely
defined M¨obius transformation τ with mi 7→M′
i for i = 1, 2, 3 (cf. Section 1.1); i.e. Bi + iCi = τ(ai) holds with τ of
Eq. (1) and
z1 =
a2 −a3
B2 −B3 −iC3
,
z2 = 0,
z3 =
a2B3 −B2a3 + ia2C3
(B2 −B3 −iC3)(B3 + iC3)B2
,
z4 = a2a3B2
B3 + iC3
.
(32)
Now it can easily be verified that the conditions Gr = 0 and Gc = 0 of Eq. (29) imply that τ also maps m4 onto
M′
4. As M1, . . . , M4 span a tetrahedron, M′
1, . . . , M′
4 has to be located on a circle. Therefore the base anchor points
of possible leg-replacements have to belong to the cylinder of revolution Θ through M1, . . . , M4 with generators in
direction of M5.
In the following we study the possible leg-replacements for this case in more detail: By homogenizing the matrix of
Eq. (12) it is not difficult to see that the corresponding matrix reads as follows:

a2
A2
B2
0
a2A2
a2B2
0
a3
A3
B3
C3
a3A3
a3B3
a3C3
a4
A4
B4
C4
a4A4
a4B4
a4C4
0
0
0
0
1
0
0

.
(33)
As M1, . . . , M4 are not coplanar, we have D167 , 0. But due to D156 = 0 and D157 = 0 Eq. (17) simplifies to

D267
−D367
D467
D126
−D136
D146 + aD167
D127
aD167 −D137
D147


A
B
C
= a

D167
0
0
.
(34)
Solving this system of linear equations yields a solution of the form given in Eq. (16), but now d0(a) is a quadratic
expression in a and d1(a) a cubic one. Therefore the base anchor points belong to a cubic curve s∗, which has to be
located on Θ.
Moreover s∗cannot split up into three generators as M′
1, . . . , M′
4 are pairwise distinct. The cubic s∗can also not
split up into a conic q and a generator g1 for the following reason: This case equals Type 2 where P1 is not a finite
point but the ideal point U of p. Therefore Mi has to be located on q \ {Q}, as otherwise mi equals U, which does not
yield a sphere condition. As this has to hold for i = 1, . . . , 4 we get a contradiction to the non-planarity assumption.
Therefore the cubic curve has to be an irreducible cubic curve s∗on a cylinder of revolution. From the following
theorem of projective geometry it is clear that the ideal plane cannot be an osculating plane of s∗or be tangent to it:
The osculating plane in a point X of a cubic equals the tangent plane to the cone of chords with respect X along the
tangent of X (which is a generator of the cone of chords). Therefore the remaining two intersection points of s∗with
the plane at infinity are conjugate complex.
□
Theorem 6. The irreducible cubic s of a pentapod of Type 1 with self-motions has to be a cubic ellipse located on a
cylinder of revolution.
Proof: s has at least one real intersection point with the ideal plane, which is denoted by M4. As the ideal point m5
of p is mapped to a finite point M5 of the base, the point M4σ−1 has to be a finite point of p, which is denoted by
m4. Therefore this point pair (M4, m4) determines a Darboux condition Ω4 and the point pair (M5, m5) a Mannheim
condition Π5. The remaining three point pairs (Mi, mi) imply sphere conditions Λi for i = 1, 2, 3.
Moreover we choose the fixed frame Σ0 that M1 is the origin, M4 the ideal point of the x-axis and M2 is located
in the xy-plane. Moreover we can define the moving frame Σ in a way that m1 is its origin. With respect to these
coordinate systems our conditions can be written as:
Ω4 :
p4x0 + a4x1 + y1 = 0,
Π5 :
p5x0 + A5x1 + B5x2 + C5x3 + y0 = 0,
(35)
where (p4, 0, 0)T are the coordinates of the intersection point of the Darboux plane and the x-axis of Σ0, and (p5, 0, 0)T
are the coordinates of the intersection point of the Mannheim plane and the x-axis of Σ. The equations of Λi for
i = 1, 2, 3 are given in Eq. (20) under consideration of A1 = B1 = C1 = C2 = a1 = 0.
15
We set x0 = 0 and start the computation of the bonds: Due to the properties of Type 1 no four base points can be
coplanar (⇒B2C3C5 , 0) and no two platform anchor points can coincide. Under these assumptions we can solve
the equations Λ1, Λ2, Λ3, Ω4, Π5 for y0, y1, y2, y3, n0. Now the numerator of Γ4 factors into x1F with
F := (a2A2 −a4A2 −a2A5)x1 + (a2B2 −a4B2 −a2B5)x2 −a2C5x3.
(36)
Therefore we have to distinguish two cases:
1. x1 , 0: In this case F = 0 has to hold, which can be solve w.l.o.g. for x3. Then the numerator of Γ6 factors into
x2G with
G := (A2x1 + B2x2)(a2 −a4)(a3C3 −a3C5 −a4C3)+(A3x1 + B3x2)(a3 −a4)a2C5 −(A5x1 + B5x2)(a3 −a4)a2C3. (37)
We distinguish two cases:
(a) x2 , 0: In this case G = 0 has to hold. We define H := A2(a2 −a4)(a3C3 −a3C5 −a4C3) + A3(a3 −a4)a2C5 −
A5(a3 −a4)a2C3 and discuss the following two cases:
i. H , 0: Under this assumption we can solve G = 0 for x1. Then the remaining equations only imply
one condition which is quadratic with respect to B3. The discriminant of this condition with respect to B3
equals:
−H2 h
(a2A2 −a4A2 −a2A5)2 + (a2B2 −a4B2 −a2B5)2 + a2
2C2
5
i
.
(38)
Therefore B3 cannot be real; a contradiction.
ii. H = 0: We can solve H = 0 for A3 w.l.o.g.. Then we can solve the G = 0 for B3 w.l.o.g., which already
yields the contradiction, as now the points M1, M2, M3, M5 are coplanar.
(b) x2 = 0: Now the numerator of Γ1 factors into
x2
1
h
A2
2(a2 −a4)2 −2a2A2A5(a2 −a4) + (A2
5 + C2
5)a2
2
i
.
(39)
The discriminant with respect to A2 equals −C2
5 and therefore we get a contradiction.
2. x1 = 0: From Γ1 we get x2 = ±x3i. In the following we only discuss the case x2 = x3i, as the other one can be done
analogously. Now the numerator of Γ6 factors into x2
3(Gr + iGc) with
Gr := (B2B3 −B3B5 + C3C5)a2 −B2(B3 −B5)a3,
Gc := B2(C3 −C5)a3 −(B2C3 −B3C5 −B5C3)a2.
(40)
This are the corresponding expressions to Eq. (29). Therefore the two point sets M′
1, M′
2, M′
3, M′
5 and m1, m2, m3, m5
are again M¨obius equivalent. As a consequence s has to be located on the cylinder of revolution Θ through
M′
1, M′
2, M′
3, M′
5 with generators in direction of M4. By using again the theorem of projective geometry given at the
end of the last proof we are done.
□
Theorem 7. The conic q of a pentapod of Type 2 with self-motions has to be located on a cylinder of revolution,
where one generator is the line g1.
Proof: The proof can be done analogously to the one of Theorem 6. In order to streamline the presentation it is given
in Appendix A. It should only be noted that q can be an ellipse or a circle, respectively, as q is the planar section of a
cylinder of revolution.
□
Theorem 8. A pentapod of Type 3 cannot possess a self-motion.
Proof: In order to improve the readability of the paper the proof of the non-existence of pentapods of Type 3 with
self-motions is given in Appendix B.
□
16
5.2. Second necessary condition
Due to the obtained results only pentapods of Type 1,2,5 remain as candidates for self-motions. In this section we
check them with respect to the second necessary condition.
Theorem 9. The irreducible cubic s of a pentapod of Type 1 with self-motions has to be a straight cubic circle.
Proof: Due to Theorem 6 the cubic s has three pairwise distinct points at infinity, which are denoted by M2, M3, M4.
Note that M4 is real and that M2, M3 are conjugate complex; i.e. M2 = M3. The corresponding platform anchor points
are denoted by m2, m3, m4 where m2 = m3 holds. Therefore we get three Darboux conditions Ωi implied by the point
pairs (Mi, mi) for i = 2, 3, 4. The ideal point of the line p is again denoted by m5 and its corresponding base anchor
point with M5. Therefore this point pair implies one Mannheim condition Π5. The pentapod is completed by a sphere
condition Λ1 determined by the two finite points M1 and m1.
The fixed frame Σ0 is chosen that M1 is the origin and that M2 and M3 are located in the xy-plane in direction
(1, B2, 0) and (1, B2, 0), respectively. As M2, M3, M4 cannot be collinear, M4 is the ideal point in direction of (A4, B4, 1).
Moreover we locate the origin of the moving frame Σ in m1. With respect to these coordinate systems Σ and Σ0 our
conditions can be written as:
Ωj :
pjx0 + ajx1 + a jBjx2 + y1 + Bjy2 = 0,
Ω4 :
p4x0 + a4A4x1 + a4B4x2 + a4x3 + A4y1 + B4y2 + y3 = 0,
Π5 :
p5x0 + A5x1 + B5x2 + C5x3 + y0 = 0,
(41)
where (pj, 0, 0)T for j = 2, 3 are the coordinates of the intersection point of the Darboux plane and the x-axis of Σ0,
(0, 0, p4)T are the coordinates of the intersection point of the Darboux plane and the z-axis of Σ0, and (p5, 0, 0)T are
the coordinates of the intersection point of the Mannheim plane and the x-axis of Σ. The equation of Λ1 is given in
Eq. (20) under consideration of A1 = B1 = C1 = a1 = 0.
With respect to the chosen frames Σ and Σ0 the first necessary condition is fulfilled if
Br = A4B4
A2
4 + 1,
Bc = ±
q
A2
4 + B2
4 + 1
A2
4 + 1
,
(42)
holds with B2 = Br + iBc and Bc , 0. Then the bond β reads as follows:
n0 = 0,
x0 = 0,
x1 = −B2,
x2 = 1,
x3 = A4B2 −B4,
(43)
y0 = B2(A5 −A4C5) −B5 + B4C5,
y1 = B2a2,
y2 = −a2,
y3 = −a2(A4B2 −B4).
(44)
Now we apply the second necessary condition; i.e. the eight tangent-hyperplanes to Φ1, Φ2, Φ3, Λ1, Ω2, Ω3, Ω4, Π5
in the bond β have to have a line in common. Therefore we compute the gradients of these eight hypersurfaces with
respect to the unknown n0, x0, . . . , x3, y0, . . . , y3 in the bond β. The resulting 8×9 matrix J has rank 8 (⇒β is a regular
point of the 5-fold I). For the necessary condition rk(J) < 8 the determinants of all 8 × 8 submatrices of J have to
vanish. The numerator of the determinant of the 8 × 8 submatrix of J, obtained by removing the column steaming
from the partial derivative with respect to y2, factors into (iar −ac −ia4)(iar + ac)(A2
4 + B2
4 + 1)L1L2 with
L1 = A4
q
A2
4 + B2
4 + 1 ∓iB4,
L2 = B5 + B5A2
4 −B4C5 −A4A5B4 ± i(A5 −A4C5)
q
A2
4 + B2
4 + 1,
(45)
and a2 = ar + iac, where ac , 0 holds. It can easily be seen that L2 can only vanish if M1, M4, M5 are collinear, which
yields a contradiction. Therefore L1 = 0 has to hold, which can only be the case for A4 = B4 = 0 (⇒rk(J) = 7). This
implies Br = 0 and Bc = ±i, which shows that s is a straight cubic circle.
□
Theorem 10. The conic q of a pentapod of Type 2 with self-motions has to be a circle and the line g1 is orthogonal
to its carrier plane (= degenerated case of a straight cubic circle).
17
Proof: The proof can exactly be done as for Theorem 9 under consideration of a4 = 0, C5 = 0 and that no two
platform anchor points can coincide beside m1 and m4.
It should only be noted that in this case L2 vanishes for M1 = M5, which yields a contradiction to the properties of
Type 2.8 Therefore we remain again with the solution A4 = B4 = 0, which implies Br = 0 and Bc = ±i. This proves
the theorem.
□
Theorem 11. The cubic s∗of a pentapod of Type 5 with self-motions has to be an irreducible straight cubic circle.
Proof: The proof of this theorem can be done in a similar fashion as those of Theorems 9 and 10. In order to stream-
line the presentation the proof of Theorem 11 is given in Appendix C.
□
Due to the Theorems 4,8,9,10,11 the condition of Duprocq [13], that the centers of the spheres have to be located on a
straight cubic circle, is valid for non-planar pentapods of P. Note that Duporcq also mentioned explicitly the special
cases of Type 2 and Type 5 beside the general case of Type 1. Therefore it remains to show if this so-called Duporcq
condition is already sufficient for the existence of a self-motion. This is done in the next section.
5.3. Sufficiency of the Duporcq condition
The sufficiency is proven separately for the Types 1,2,5. Moreover in each of the three proofs also the leg param-
eters for a self-motion are given.
Theorem 12. A pentapod of Type 5 fulfilling the Duporcq condition has a 1-parametric set of self-motions (over C).
With respect to the coordinatization used in the proof of Theorem 11 (under consideration of A4 = B4 = 0 and B2 = i)
the leg parameters are given by:
w = C5
a5
,
p2 = −(a3 −a5)(A5 −iB5)
a5
,
p3 = −(a2 −a5)(A5 + iB5)
a5
,
(46)
and the following condition remains in R1 and R5:
(a2
5 + B2
5 + C2
5)(a2 + a3 −a5) + (R2
1 −R2
5 −a2a5 −a3a5 + a2
5)a5 = 0.
(47)
Proof: We use the same coordinatization as given in the first two paragraphs of the proof of Theorem 11 under
consideration of A4 = B4 = 0 and B2 = i. We distinguish two cases:
1. C5 −a5w , 0: Under this assumption we can solve Λ1, Ω2, Ω3, ∢4, Λ5, Φ3 for x3, y0, y1, y2, y3, n0. Plugging the
obtained expressions into Φi yields the equations Φ∗
i in x0, x1, x2 for i = 1, 2. Now we compute the resultant Ξ of
the numerator of Φ∗
1 (which is quadratic in x0, x1, x2) and the numerator of Φ∗
2 (which is quartic in x0, x1, x2) with
respect to x2. Ξ factors into x4
0N[11464], where N is quartic in x0, x1. As x0 , 0 has to hold (cf. Section 4.1), N has
to be fulfilled identically. Therefore we denote the coefficient of xi
0x j
1 of N by Ni j. Then N04 factors into 16E2
1E2
2
with
E1 = (a3 −a5)(A5 −iB5) + a5p2,
E2 = (a2 −a5)(A5 + iB5) + a5p3.
(48)
W.l.o.g. we can set E1 equal to zero and solve it for p2. Then the numerator of N13 factors into 32a3(a3 −a5)(A5 −
iB5)(C5 −a5w)2E2
2. Therefore E2 = 0 has to hold, which can be solved for p3 w.l.o.g.. Now the numerator of N22
splits up into
64(a2
5 + B2
5)a2a3(a2 −a5)(a3 −a5)(C5 −a5w)4,
(49)
which cannot vanish without contradiction.
8In this case the cubic of base points splits up into three lines; one real and two conjugate complex ones, which intersect each other in M1 = M5
(complex version of Type 4).
18
2. w = C5/a5: We can solve the equations Λ1, Ω2, Ω3, ∢4, Λ5 for x3, y1, y2, y3, n0 and plug the obtained expressions
into Φi, which yields quadratic equation Φ∗
i in x0, x1, x2, y0 for i = 1, 2, 3. Now then numerator of Φ∗
1 and Φ∗
3 do not
depend on y0 in contrast to the numerator of Φ∗
2 (coefficient of y2
0 equals 4a2
5). Therefore we compute the resultant
Ξ of the numerators of Φ∗
1 and the numerator of Φ∗
3 with respect to x2. Ξ factors into x2
0a2
5N[208], where N is
quadratic in x0, x1. We denote the coefficient of xi
0x j
1 of N again by Ni j. In the following we show that E1 = 0 and
E2 = 0 has to hold:
N02 factors into 4a2
5E1E2, which can be solve w.l.o.g. for p2. Then N11 splits up into
−2a5E2
h
a3(A2
5 + B2
5) + (a2 + a3 −a5)C2
5 + (1 −a2 −a3)a2
5 + (R2
1 −R2
5)a5 −(A5 −iB5)a5p3
i
.
(50)
Either E2 = 0 holds and we are done or the last factor vanishes. In the latter case we can compute p3 w.l.o.g..
Then N20 factors into (C2
5 −a2
5)E2
2. For C5 = ±a5 the condition Φ1 equals x2
1 + x2
2 = 0, which cannot yield a real
self-motion. Therefore E2 = 0 has to hold.
Summed up we have proven that E1 = 0 and E2 = 0 have to hold. These equations can be solved for p2, p3 w.l.o.g..
Plugging the obtained expressions into N shows that only the condition given in Eq. (47) remains. This condition
can always be solved for R2
5 and the self-motion is again obtained by back-substitution, which finishes the proof of
the sufficiency.
□
Theorem 13. A pentapod of Type 1 fulfilling the Duporcq condition has a 1-parametric set of self-motions (over C).
With respect to the coordinatization used in the proof of Theorem 9 (under consideration of A4 = B4 = 0 and B2 = i)
the leg parameters are given by:
p2 = −A5(a2a3 −a2
4) −i(a2a3 −a2
4)B5
(a3 −a4)2
,
p3 = −A5(a2a3 −a2
4) + i(a2a3 −a2
4)B5
(a2 −a4)2
,
p4 =
C5(a2a3 −a2
4)
(a2 −a4)(a3 −a4),
(51)
and the following condition remains in R1 and p5:
(a2 −a4)2(a3 −a4)2 h
2(a2a3 −a2
4)p5 −(a2 + a3 −2a4)R2
1 −(2a2a3 −a2a4 −a3a4)a4
i
+
(a2a3 −a2
4)2(a2 + a3 −2a4)(A2
5 + B2
5 + C2
5) = 0.
(52)
Theorem 14. A pentapod of Type 2 fulfilling the Duporcq condition has a 1-parametric set of self-motions (over C).
With respect to the coordinatization used in the proof of Theorem 13 (under consideration of a4 = C5 = 0) the leg
parameters are given by:
p2 = −a2(A5 −iB5)
a3
,
p3 = −a3(A5 + iB5)
a2
,
p4 = 0,
(53)
and the following condition remains in R1 and p5:
(A2
5 + B2
5)(a2 + a3) + 2a2a3p5 −R2
1(a2 + a3) = 0.
(54)
Proof of Theorem 13 and Theorem 14: The proofs of these theorems can be done in a fashion similar to the one
of Theorem 12. For this reason and in order to streamline the presentation the corresponding proofs are given in
Appendix D and Appendix E, respectively.
□
Note that the proof of the sufficiency in the Theorems 12, 13 and 14 was only done over C; i.e. the self-motion has
not to be real.9 Arguments of reality were only used to exclude some special cases.
This can best be seen for the self-motions obtained for Type 5, as they belong to the class of Borel-Bricard motions.
These are the only non-trivial motions where all points of the moving space have spherical trajectories (cf. [8, Chapter
VI]; see also [29]). Note that this special case was also discussed in detail by Krames [10, Section 5]. In this case Φ∗
1
equals x2
0(C2
5 −a2
5) + a2
5(x2
1 + x2
2) = 0. This already shows the following result:
9In contrast the conditions given in Theorem 1 and those for the cases (α, β, γ) of Section 1 are even sufficient for the existence of real self-
motions.
19
Corollary 3. The 1-parametric set of self-motions given in Theorem 12 is real if |C5| < |a5| holds; complex otherwise.
Until now we are not able to give a corresponding easy characterization for designs of Type 1 and 2 with real/complex
self-motions. But the following two examples prove that real self-motions exist:
Example 1. This example of a pentapod of Type 1 with a real self-motion is based on the formulas of Λ1, Ω2, Ω3, Ω4, Π5
given in the proof of Theorem 9. The geometry of the pentapod is determined by:
a2 = B2 = i,
a3 = B3 = −i,
a4 = 2,
A4 = B4 = 0,
A5 = B5 = C5 = 1.
(55)
For the leg-parameters:
R1 =
√
3,
p2 = −3
25 −21
25i,
p3 = −3
25 + 21
25i,
p4 = −3
5,
p5 = 46
75,
(56)
which are in accordance with Theorem 13, the pentapod has the following self-motion (under consideration of x0 = 1;
cf. Eq. (21)):
x1 = 7
4t2 −7
5t −161
300 ∓T
300,
x2 = 1
4t2 −1
5t −23
300 ± 7T
300,
x3 = t,
(57)
y1 = −1
4t2 + 1
5t + 59
300 ∓7T
300,
y2 = 7
4t2 −7
5t −413
300 ∓T
300,
y3 = −2t + 3
5,
(58)
with T =
p
−(75t2 −30t −41)(75t2 −90t + 31). Both branches (upper and lower one) are real for t ∈t−, t+ with
t−= 1
5 −2
15
√
33,
t+ = 1
5 + 2
15
√
33.
(59)
For this example we also show how to compute the finite base anchor point M (with coordinates (A, B,C)T), the finite
platform anchor point m (with coordinates (a, 0, 0)T) and the leg length R of a further leg. Its corresponding sphere
condition Λ has to be a linear combination of the given equations Λ1, Ω2, Ω3, Ω4, Π5; i.e.
µ1Λ1 + µ2Ω2 + µ3Ω3 + µ4Ω4 + µ5Π5 −Λ = 0
(60)
for any choice of n0, x0, . . . , x3, y0, . . . , y3. Therefore their nine coefficients imply nine equations in the ten unknowns
µ1, . . . , µ5, R, A, B,C, a. This system has the following solution (in dependence of a):
µ1 = 1,
µ2 = A + Bi,
µ3 = A −Bi,
µ4 = 2C,
µ5 = 2a,
R2 = (3a2 −8a + 9)(25a4 −64a3 + 146a2 −136a + 100)
75(a2 + 1)(a −2)2
,
(61)
with
A = a(a −1)
a2 + 1 ,
B = a(a + 1)
a2 + 1 ,
C =
a
a −2.
(62)
The last equation gives the bijection σ between points m of p and points M of the irreducible straight cubic circle s
(cf. Eq. (16)).
In Fig. 5a the trajectories of the platform anchor points m1 (a = 0), m6 (a = 1), m7 (a = 3), m8 (a = −1) and m9
(a = −2) are displayed for the upper branch of the self-motion.
⋄
Example 2. This example of a pentapod of Type 2 with a real self-motion is based on the formulas of Λ1, Ω2, Ω3, Ω4, Π5
given in the proof of Theorem 9 under consideration of a4 = 0 and C5 = 0. The geometry of the pentapod is determined
by:
B2 = i,
B3 = −i,
a2 = 1 + i,
a3 = 1 −i,
A4 = B4 = 0,
A5 = B5 = 1.
(63)
For the leg-parameters:
R1 = 2,
p2 = −1 −i,
p3 = −1 + i,
p4 = 0,
p5 = 1,
(64)
20
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
1
m−
1
m−
1
m−
1
m−
1
m−
1
m−
1
m−
1
m−
1
m−
1
m−
1
m−
1
m−
1
m−
1
m−
1
m−
1
m−
1
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
9
m−
9
m−
9
m−
9
m−
9
m−
9
m−
9
m−
9
m−
9
m−
9
m−
9
m−
9
m−
9
m−
9
m−
9
m−
9
m−
9
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
1
m+
1
m+
1
m+
1
m+
1
m+
1
m+
1
m+
1
m+
1
m+
1
m+
1
m+
1
m+
1
m+
1
m+
1
m+
1
m+
1
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
9
m+
9
m+
9
m+
9
m+
9
m+
9
m+
9
m+
9
m+
9
m+
9
m+
9
m+
9
m+
9
m+
9
m+
9
m+
9
m+
9
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
sssssssssssssssss
(a)
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
6
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
8
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m−
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
7
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m+
6
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m−
1 = m+
1
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
m+
8
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M7
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M8
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M1
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M9
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
M6
qqqqqqqqqqqqqqqqq
(b)
Figure 5: Trajectories of the upper branch of the self-motion, where the starting pose at time t−and the end pose at time t+ are illus-
trated: (a) Example 1 and (b) Example 2. Animations of these two self-motions can be downloaded from the homepage of the first author
(http://www.geometrie.tuwien.ac.at/nawratil).
which are in accordance with Theorem 14, the pentapod has the following self-motion (under consideration of x0 = 1;
cf. Eq. (21)):
x1 = −1
2t2,
x2 = ±T
2 ,
x3 = t,
y1 = 1
2t2 + 1 ∓T
2 ,
y2 = −1
2t2 −1 ∓T
2 ,
y3 = 0,
(65)
with T =
√
−t4 −4t2 + 4. Both branches (upper and lower one) are real for t ∈t−, t+ with
t−= −
q
2
√
2 −2,
t+ =
q
2
√
2 −2.
(66)
Analogous considerations as in Example 1 show the following bijection σ between points m of p \ {m1} and points M
of the circle q \ {M1}:
A =
a(a −2)
a2 −2a + 2,
B =
a2
a2 −2a + 2,
C = 0.
(67)
The point m1 is mapped to the line [M1, M9], which equals the z-axis of Σ0.
In Fig. 5b the trajectories of the platform anchor points m1 = m9 (a = 0), m6 (a = 1), m7 (a = 2) and m8 (a = −1)
are displayed for the upper branch of the self-motion.
⋄
Remark 6. Further examples of pentapods of Type 1 and Type 2 are implied by a remarkable motion, where all
points of a hyperboloid, which carries two reguli of lines, have spherical trajectories. This well-studied motion is
also known as BBM-II motion in the literature (e.g. [30]). It is known (cf. [30, page 24] and [31, page 188]) that the
21
corresponding sphere centers of lines, belonging to one regulus10, are located on irreducible straight cubic circles,
which imply examples of self-motions of Type 1. Note that there also exist degenerated cases where the hyperboloid
splits up into two orthogonal planes, which imply examples of self-motions of Type 2.
⋄
6. Conclusions for practical applications
We introduced a novel kinematic mapping for pentapods with linear platform (cf. Section 4.1), which can be used
for an efficient solution of the direct kinematics problem. Beside this achievement we listed all pentapods with linear
platform which
⋆are architecturally singular (cf. Corollary 1),
⋆possess self-motion (over C) without having an architecture singularity.
The latter are either the designs (α, β, γ) given in Section 1, pentapods of Types 1,2,5 fulfilling the Duporcq condition
(cf. Section 5) or the manipulator given in Theorem 1.
Clearly, architecturally singular pentapods are not suited for practical application and therefore engineers should
be aware of these designs. The usage of pentapods with self-motions within the design process is a double-edged
sword; on the one side they should be avoided for reasons of safety11 and on the other side they have a simplified
direct kinematics for the following reason:
As the bonds are independent of the set of leg lengths (cf. Section 4.2), they always appear as solution of the direct
kinematics problem even though the given set R1, . . . , R5 does not cause a self-motion of the manipulator. Recall
that a bond of a self-motion corresponds to a complex configuration on the boundary (cf. Remark 5). Therefore four
solutions (= two conjugate complex bonds of multiplicity 2; cf. Remark 5) of the direct kinematics of a pentapod with
linear platform possessing a self-motion are always located on the boundary of I. This yields the following corollary:
Corollary 4. A pentapod with linear platform possessing a self-motion (over C) can have a maximum of four real
configurations (instead of generically 8, cf. Theorem 7), if the given set of leg lengths does not imply a self-motion.
The direct kinematic problem of these pentapods reduces to the solution of a polynomial of degree 4.
As this quartic equation can be solved explicitly, these pentapods seem to be of special interest for practical
application. We demonstrate this result on the basis of the following example:
Example 3. Continuation of Example 1: We consider the pentapod with platform anchor points m1, m6, . . . , m9 and
base anchor points M1, M6, . . . , M9 of Example 1 (see Fig. 5a). But now we want to solve the direct kinematics problem
of this pentapod of Type 1 for the following given set of leg lengths:
R1 = 2,
R6 = 1,
R7 = 5,
R8 = 3,
R9 = 4,
(68)
which does not cause a self-motion. We can solve the corresponding system of equations Λ1, Λ6, . . . , Λ9 for n0, y0, y1, y2
and y3. Moreover we can set x0 = 1. Now Φ3 of Eq. (22) is only linear in x1 and x2 and we can solve it for x1. Then
Φ1 and Φ2 are only quadratic in x2, and therefore the resultant of these two expressions with respect to x2 yields:
4316636297 + 69486876480x3 + 241133479200x2
3 −291209472000x3
3 + 76425120000x4
3 = 0,
(69)
which is only of degree 4 in x3.
⋄
But it is even possible to use this advantage of self-motions without any risk (cf. footnote 9), by designing pen-
tapods with linear platform, which only have complex self-motions. In the following we list two sets of such designs:
(I) Pentapods of Type 5 with |C5| ≥|a5| (cf. Corollary 3).
10The corresponding sphere centers of lines belonging to the other regulus are again located on a line (cf. [30, page 24]), which imply architec-
turally singular pentapod designs.
11A self-motion is dangerous because it is uncontrollable and thus a hazard to man and machine.
22
(II) We can also solve the planar case (cf. Section 2) with the bond based approach used for the non-planar one.
This study shows, that a planar pentapod of P (cf. Definition 1) has a bond if and only if the vertex V (cf. proof
of Theorem 1) is an ideal point. Moreover the second necessary condition implied by the theory of bonds is
only fulfilled if the affine relation (AR) holds12.
Remark 7. Note that the condition (AR) equals the linear constraint given in [1], where the observation was
reported that these planar pentapods with linear platform only possess a maximum of four real solutions of the
direct kinematics problem (without giving an explanation for this behavior). This problem can even be solved
quadratically as the solutions are symmetric with respect to the base plane (cf. [1]). Therefore this also holds
for the design (γ) and for the designs (α, β) under the extra condition of a planar base.
⋄
The constraint (AR) is even sufficient for the existence of a self-motion (over C). Now we design the pentapods
in a way that the distance between the parallel lines [Mi, V] and [Mj, V], which are fibers of the affinity κ, is
• equal or less than the distance dist(mi, mj) between their images: This yields the pentapods characterized
in Theorem 1, which all have real self-motions (cf. footnote 8).
• greater than the distance dist(mi, mj) between their images: Then the line p cannot be oriented that
dist(M′
i, M′
j)=dist(m′
i, m′
j) holds, which already shows that this design-set (II) is free of real self-motion.
Therefore the authors recommend engineers to design pentapods with linear platform within the set (I) in the non-
planar case and within the set (II) in the planar one, respectively. For reasons of completeness we also give the
following corollary:
Corollary 5. The following pentapods with linear platform, which do not possess a self-motion, have a maximum of
six real configurations (instead of generically 8, cf. Theorem 7):
1. The irreducible cubic s of a pentapod of Type 1 is a cubic ellipse located on a cylinder of revolution.
2. The conic q of a pentapod of Type 2 belongs to a cylinder of revolution, where one generator is the line g1.
3. A pentapod of Type 5 with one of the following two properties:
(a) Either the base anchor points of possible leg-replacements are located on an irreducible cubic ellipse s∗
on a cylinder of revolution.
(b) or it has the following design (under consideration of footnote 1): m2 = m3, m4 = m5 and [M2, M3] is
parallel to [M4, M5].
4. Planar pentapod of P, where the associated point V is an ideal point (cf. Fig. 3c).
The direct kinematic problem of the listed pentapods reduces to the solution of a polynomial of degree 6.
Proof: Items 1, 2 and 3(a) are a direct consequence of the Theorems 6, 7 and 5, respectively. Item 4 follows from the
above given discussion of design-set (II) and item 3(b) can be seen as its corresponding non-planar case (cf. Fig. 3c).
In the latter case the cubic splits up into 3 parallel (but non-planar) lines. This case is hidden in item 2(a) of the proof
of Theorem 5.
Finally, it should be noted that the direct kinematics problem is only cubic for item 4 as the pentapod is planar (cf.
Remark 7).
□
We want to close the paper by referring to [32] where the remaining problem of characterizing designs of Type 1 and
Type 2 without real self-motions is discussed, as they imply further ”save” designs with a closed form solution.
12In this case the bond (and its conjugate) are singular points of the 5-fold I (cf. Remark 5).
23
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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[26] Husty, M.L.: An algorithm for solving the direct kinematics of general Stewart-Gough platforms. Mechanism and Machine Theory 31(4)
365–380 (1996)
[27] Zhang, C.-de, Song, S.-M.: Forward kinematics of a class of parallel (Stewart) platforms with closed-form solution. Proceedings of IEEE
International Conference on Robotics and Automation, Sacramento, California, April 9-11, 1991, pages 2676–2681
[28] Gallet, M., Nawratil, G., Schicho, J.: Bond Theory for Pentapods and Hexapods. Journal of Geometry 106(2) 211–228 (2015)
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[31] Krames, J.: Die Borel-Bricard-Bewegung mit punktweise gekoppelten orthogonalen Hyperboloiden ( ¨Uber symmetrische Schrotungen VI).
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[32] Nawratil, G.: On the line-symmetry of self-motions of linear pentapods. [arXiv:1510.03567]
24
Appendix A: Proof of Theorem 7
We can start with the same set of equations as in the proof of Theorem 6, but we have the extra conditions m1 = m4
and that M1, M2, M3, M5 are located in a plane, which does not contain M4. Due to the properties of Type 2 not all
base points are coplanar (⇒B2C3C5 , 0) and no two platform anchor points can coincide beside m1 and m4 (⇒
a1 = a4 = 0). Therefore we can express the coplanarity of M1, M2, M3, M5 as
A3 = A2B3C5 −A2B5C3 + B2C3A5
B2C5
.
(70)
We set x0 = 0 and start the computation of the bonds: We can solve the equations Λ1, Λ2, Λ3, Ω4, Π5 for
y0, y1, y2, y3, n0 w.l.o.g.. Now the numerator of Γ4 factors into x1F with F of Eq. (36). Therefore we have to dis-
tinguish two cases:
1. x1 , 0: In this case F = 0 has to hold, which can be solve w.l.o.g. for x3. Then the numerator of Γ6 factors into:
x2a3(x2B2 + x1A2)(B3C5 −B5C3 + B2C3 −B2C5).
(71)
The last factor cannot vanish, as otherwise the base anchor points M2, M3, M5 are collinear, a contradiction. There-
fore we remain with two cases:
a. x2 = −x1A2/B2. Now the numerator of Γ1 factors into:
x2
1
h
A2
2(B2
5 + C2
5) −2A2A5B2B5 + B2
2(A2
5 + C2
5)
i
.
(72)
The discriminant with respect to A2 equals −C2
5(A2
5 + B2
5 + C2
5) and therefore we get a contradiction.
b. x2 = 0: In this case the numerator of Γ1 factors into x2
1[C2
5 + (A2 −A5)2] which can also not vanish without
contradiction.
2. x1 = 0: This case is exactly the same as the one discussed in item 2 of the proof of Theorem 6, which already
yields the result.
□
Appendix B: Proof of Theorem 8
We can make leg-replacements such that m1 = m2 = P1, m3 = m4 = P2, M1, M2 ∈L1 and M3, M4 ∈L2. Moreover
we can choose M2 and M3 as ideal points of L1 and L2, respectively. Therefore the point pairs (Mi, mi) determine
sphere conditions Λi for i = 1, 4 and Darboux condition Ωi for i = 2, 3. Moreover we can assume that M1 is located
in the Darboux plane of (M2, m2) and that M4 is located in the Darboux plane of (M3, m3). Finally we can assume that
m5 is the ideal point of the line p. Therefore the point pairs (M5, m5) determines a Mannheim condition Π5.
W.l.o.g. we choose the fixed frame Σ0 that M1 equals its origin and that M2 and M3 are located in the xy-plane
symmetric with respect to the x-axis. Therefore the directions of M2 and M3 are given by (1, B2, 0) and (1, −B2, 0),
respectively. Moreover we can define the moving frame Σ in a way that m1 is its origin. With respect to these
coordinate systems our conditions can be written as:
Ω2 :
y1 + B2y2 = 0,
Ω3 :
(A4 −B2B4)x0 + a3x1 −a3B2x2 + y1 −B2y2 = 0,
Π5 :
p5x0 + A5x1 + B5x2 + C5x3 + y0 = 0,
(73)
where (p5, 0, 0)T are the coordinates of the intersection point of the Mannheim plane and the x-axis of Σ. The equations
of Λi for i = 1, 4 are given in Eq. (20) under consideration of A1 = B1 = C1 = a1 = 0 and a4 = a3.
Now we set x0 = 0 and prove that no bonds can exist. W.l.o.g. we can solve Λ1, Ω2, Ω3, Π5 for n0, y0, y1, y2. Then
the numerator of Γ4 can only vanish in the following two cases:
25
1. x1 = B2x2: The numerator of Γ2 implies y3 = 0. Now the numerator of Λ4 equals:
a4 [x2(B2A4 −B2A5 + B4 −B5) + (C4 −C5)x3] .
(74)
We distinguish two cases:
(a) C4 , C5: Under this assumption we can solve the last factor for x3. Then x2
2 factors out from the numerator
of Γ1 and we remain with only one condition, which is quadratic with respect to A5. The corresponding
discriminant equals −(C4 −C5)2(B2
2 + 1) and therefore no solution exists.
(b) C4 = C5: In this case Λ4 can only vanish for x2 = 0 as B2A4 −B2A5 + B4 −B5 = 0 implies the collinearity of
M3, M4, M5, a contradiction. Then Γ1 cannot vanish without contradiction.
2. x1 = −B2x2: Now the numerator of Γ6 equals −x2(x3a4 + y3). Therefore we have to distinguish two cases:
(a) x2 = 0: Now Γ1 and Γ2 imply x3 = y3 = 0, a contradiction.
(b) y3 = −x3a4: Then Λ4 factors into
a4 [x2(B2A5 −B5) −C5x3] .
(75)
We distinguish two cases:
i. C5 , 0: Under this assumption we can solve the last factor for x3. Then x2
2 factors out from the numerator
of Γ1 and we remain with only one condition, which is quadratic with respect to A5. The corresponding
discriminant equals −C2
5(B2
2 + 1) and therefore no solution exists.
ii. C5 = 0: In this case Λ4 can only vanish for x2 = 0 as B2A5 −B5 = 0 implies the collinearity of M1, M2, M5,
a contradiction. Then Γ1 cannot vanish without contradiction.
□
Appendix C: Proof of Theorem 11
Due to Theorem 5 the irreducible cubic s∗has three pairwise distinct points at infinity, which are denoted by
M2, M3, M4. Note that M4 is real and that M2, M3 are conjugate complex; i.e. M2 = M3. The corresponding platform
anchor points are denoted by m2, m3, m4 where m2 = m3 holds. Therefore we get two Darboux conditions Ωi implied
by the point pairs (Mi, mi) for i = 2, 3. Moreover the point pair (M4, m4) implies the angle condition ∢4. The pentapod
is completed by two sphere conditions Λj, which are determined by the two finite points Mj and mj for j = 1, 5.
We choose the fixed frame Σ0 that M1 is the origin and that M2 and M3 are located in the xy-plane in direction
(1, B2, 0) and (1, B2, 0), respectively. As M2, M3, M4 cannot be collinear, M4 is the ideal point in direction of (A4, B4, 1).
Moreover we can define the moving frame Σ in a way that m1 is its origin. With respect to these coordinate systems
our conditions can be written as:
Ωi :
pix0 + aix1 + aiBix2 + y1 + Biy2 = 0,
∢4 :
wx0 + A4x1 + B4x2 + x3 = 0
(76)
where (pi, 0, 0)T for i = 2, 3 are the coordinates of the intersection point of the Darboux plane and the x-axis of Σ0
and w denotes arccos (ϕ). The equations of Λj is given in Eq. (20) under consideration of A1 = B1 = C1 = a1 = 0 for
j = 1, 5.
In the following we substitute B2 = Br + iBc and a2 = ar + iac with Br, Bc, ar, ac ∈R and Bcac , 0. Then we set
x0 = 0 and start the computation of bonds: We can solve the equations Λ1, Ω2, Ω3, ∢4, Λ5 for x3, y1, y2, y3, n0 w.l.o.g..
Now the numerator of Γ4 can only vanish without contradiction for x1 = (−Br ± iBc)x2. We only discuss the upper
sign as the conjugate solution can be done analogously. Then the numerator of Γ6 can be solved for y0 w.l.o.g.. From
the numerator of Γ1 we can factor out x2
2 and we remain with only one condition. From its imaginary part and real
part we can compute Br and Bc given in Eq. (42).
Then we compute again the 8 × 9 matrix J with respect to the obtained bond. For the necessary condition rk(J) <
8 the determinants of all 8 × 8 submatrices of J have to vanish. The numerator of the determinant of the 8 × 8
submatrix of J, obtained by removing the column steaming from the partial derivative with respect to x2, factors into
(iar +ac −ia5)(iar +ac)(A2
4 + B2
4 +1)L1L2 with L1 and L2 of Eq. (45). Analogous arguments as in the proof of Theorem
9 shows that s∗has to be a straight cubic circle.
□
26
Appendix D: Proof of Theorem 13
We use the same coordinatization as given in the first three paragraphs of the proof of Theorem 9 under consider-
ation of A4 = B4 = 0 and B2 = i.
W.l.o.g. we can solve the equations Λ1, Ω2, Ω3, Ω4, Π5 for y0, y1, y2, y3, n0. Plugging the obtained expressions
into Φi yields quadratic equation Φ∗
i in x0, . . . , x3 for i = 1, 2, 3. Therefore Φ∗
i corresponds with a quadric in the
homogeneous 3-space (spanned by x0, . . . , x3 ). In the following we want to determine the parameters R1, p2, p3, p4, p5
in a way that these three quadrics have a curve in common. This can be done as follows:
We compute the resultant Ξk of Φ∗
i and Φ∗
j with respect to x1 for pairwise distinct i, j, k ∈{1, 2, 3}. Ξ1, Ξ2 and Ξ3 are
homogeneous quartic expressions in x0, x2, x3, but they are only quadratic with respect to x2. Therefore we eliminate
this unknown by computing the resultant Υk of Ξi and Ξj for pairwise distinct i, j, k ∈{1, 2, 3}. Then the greatest
common divisor N of Υ1, Υ2 and Υ3 has to vanish. It turns out that N has 411 terms and that it is homogeneous of
degree 4 in x0, x3. For a self-motion of the line p the expression N has to be fulfilled independently of x0, x3. Therefore
we denote the coefficient of xi
0x j
3 of N by Nij.
In the following we show that the three conditions E3 = 0, E4 = 0, E5 = 0 have to be fulfilled with:
E3 = A5(a2a3 −a2
4) + p2(a3 −a4)2 −i(a2a3 −a2
4)B5,
E4 = A5(a2a3 −a2
4) + p3(a2 −a4)2 + i(a2a3 −a2
4)B5,
E5 = C5(a2a3 −a2
4) −p4(a2 −a4)(a3 −a4).
(77)
This can be seen as follows: N04 = 0 splits up into E3E4. W.l.o.g. we can set E3 equal to zero and solve it for p2.
Then the numerator of N13 factors into −2a3(A5 −iB5)E4E5. Therefore we have to distinguish two cases:
1. E4 = 0: W.l.o.g. we can solve this equation for p3. Then the numerator of N22 splits up into 4a2a3(A2
5 + B2
5)E2
5.
Therefore E5 = 0 has to hold and we are done.
2. E5 = 0: W.l.o.g. we can solve this equation for p4. Then the numerator of N22 and N31 splits up into:
a3(A5 −iB5)E4F22[83],
a3a4C5(A5 −iB5)E4F31[73].
(78)
Either E4 = 0 holds and we are done or F22 = 0 and F31 = 0 have to hold. In the latter case we compute
F22 + F31, which factors into a3(a2 −a4)2(A5 −iB5)E4. Therefore again E4 = 0 has to be fulfilled.
Summed up we have proven that E3 = 0, E4 = 0, E5 = 0 have to hold. These equations can be solved for p2, p3, p4
w.l.o.g.. Plugging the obtained expressions into N shows that only the condition given in Eq. (52) remains. We
distinguish two cases:
• a2a3 −a2
4 , 0: In this case Eq. (52) can always be solved for p5. Then the self-motion can be computed by
back-substitution; i.e. we compute the common factor of Ξ1, Ξ2, Ξ3 and solve it for x2 (which only appears
quadratic) and finally the common factor of Φ∗
1, Φ∗
2 and Φ∗
3 (which is linear in x1) gives the self-motion.
• For the special case a2a3 −a2
4 = 0 we get R2
1 = a2
4 from Eq. (52). Moreover if we set a2 = ar +iac with ar, ac ∈R
and ac , 0 the condition a2a3 −a2
4 = 0 is equivalent with a2
r + a2
c −a2
4 = 0, which can be solved for a4 w.l.o.g..
Now it can easily be seen that Φ∗
1(a2
r + a2
c) = Φ∗
2 holds. Therefore we only remain with the following condition
beside Φ∗
1 (which equals Φ1 given in Eq. (22)):
Φ∗
3 :
x2
0p5 −(x2
1 + x2
2)ar −x2
3
q
a2r + a2c + x0(x1A5 + x2B5 + x3C5) = 0.
(79)
This finishes the proof of the sufficiency.
□
27
Appendix E: Proof of Theorem 14
We start with the same set of equations as in the proof of Theorem 13 under consideration of a4 = 0, C5 = 0 and
that no two platform anchor points can coincide beside m1 and m4.
Moreover we can compute analogously the corresponding expression N, which has in this case only 204 terms.
We denote the coefficient of xi
0x j
3 of N again by Ni j. In the following we show that the three conditions E6 = 0, E7 = 0
and p4 = 0 have to be fulfilled with:
E6 = a2(A5 −iB5) + a3p2,
E7 = a3(A5 + iB5) + a2p3.
(80)
This can be seen as follows: N04 = 0 splits up into a2a3E6E7. W.l.o.g. we can set E6 equal to zero and solve it for p2.
Then the numerator of N13 factors into 2a2
2(A5 −iB5)p4E7. Therefore we have to distinguish two cases:
1. E7 = 0: W.l.o.g. we can solve this equation for p3. Then the numerator of N22 splits up into 4a2a3(A2
5 + B2
5)p2
4.
Therefore p4 = 0 has to hold and we are done.
2. p4 = 0: Now the numerator of N22 splits up into:
a2(A5 −iB5)E7
h
a2p3(A5 −iB5) + R2
1(a2 + a3) −a2(A2
5 + B2
5 + 2a3p5)
i
.
(81)
If E7 = 0 holds we are done. Therefore we express p3 from the last factor, which can be done w.l.o.g.. Then
N31 is fulfilled identically and we only remain with N40. Its numerator factors into:
a2R2
1
h
2a2a3p5 + (a2 + a3)(A2
5 + B2
5 −R2
1)
i
.
(82)
Now we have to distinguish two cases:
(a) We can solve the last factor for p5 w.l.o.g.. Then it can easily be checked that E7 is fulfilled and we are
done.
(b) R1 = 0: This condition implies a spherical self-motion of the line p with center M1 = m1. Now a one-
parametric set of legs with base anchor points on the circle q (the corresponding platform anchor points
on p are given by σ−1) can be attached without restricting the spherical self-motion. It can easily be seen
that any two legs of this set already fix the pose of p; hence no real self-motion of p exists.
Summed up we have proven that E6 = 0, E7 = 0 and p4 = 0 have to hold. E6 = 0, E7 = 0 can be solved for p2, p3
w.l.o.g.. Plugging the obtained expressions into N shows that only the condition given in Eq. (54) remains. This
condition can always be solved for p5 and the self-motion is again obtained by back-substitution, which finishes the
proof of the sufficiency.
□
28