The document proposes a new method for characterizing singular configurations of the 6/6 Stewart platform (SP) using the 3D Kennedy theorem. It defines the singular configuration as when the intersection line m of two planes defined by parallel leg lines is perpendicular to the common normal of two other leg lines. This characterization is proved using reciprocal screw theory and properties of lines intersecting screws. The method is also applicable for analyzing the mobility of the SP and is shown to relate to other reported methods of singularity analysis for parallel robots.

1. Proceedings of the ASME 2016 International Design Engineering Technical Conferences &
Computers and Information in Engineering Conference
IDETC/CIE 2016
August 21-24, 2016, Charlotte Convention Center, Charlotte, North Carolina, USA
DETC2016-59613
A GEOMETRIC SINGULAR CHARACTERIZATION OF THE 6/6 STEWART
PLATFORM
Michael Slavutin
School of Mechanical
Engineering
Tel Aviv University,
Ramat Aviv 69978, Israel
Avshalom Sheffer
School of Mechanical
Engineering
Tel Aviv University,
Ramat Aviv 69978, Israel
Offer Shai
School of Mechanical
Engineering,
Tel Aviv University,
Ramat Aviv 69978, Israel
ABSTRACT
The paper introduces the 3D Kennedy theorem and applies
it for characterizing of the singular configuration of 6/6 Stewart
Platform (SP). The main idea underlying the proposed singular
characterization is as follows: we search for two lines, which
cross four of the six leg lines of the robot. For these two lines
we find two parallel lines that cross the remaining leg lines 5
and 6. Each pair of parallel lines defines a plane. Let 𝑚 be the
intersection line of these two planes. The proposed singular
characterization is: the 6/6 SP is in a singular configuration if
and only if the line 𝑚 is perpendicular to the common normal
of leg lines 5 and 6.
In addition, the method developed for the singular
characterization is also used for the analysis of the mobility of
SP. Finally, the proposed method is compared to other
singularity analysis methods, such as of Hunt’s and Fichter’s
singular configuration and the 3/6 Stewart Platform singularity.
The relation between the reported characterizations of the
6/6 SP and other reported works is highlighted. Moreover, it is
shown that the known 3/6 singular characterization is a special
case of the work reported in the paper.
KEYWORDS: Singular characterization, 3D Kennedy theorem,
Instantaneous screw axis (ISA), screw theory, 3/6 and 6/6 SP
NOMENCLATURE
ISA Instantaneous screw axes
SP Stewart Platform
∨ Join
∧ Meet
∘ Reciprocal product
2D Two-dimensional
3D Three-dimensional
3/6 SP 3/6 Stewart Platform
1. INTRODUCTION
This paper introduces the characterization of the singular
configuration of 6/6 Stewart Platform. Stewart Platform
consists of two bodies connected by six legs, which can vary
their lengths. One of the bodies is called the base and the other
is called the platform. One of the important problems in parallel
robots is characterization of the singular positions or special
configurationsI
. It is one of the main concerns in the analysis
and design of manipulators [1]. One of the known singular
configurations of SP is when all the six leg lines of the
mechanism cross one line [2]. An additional configuration is
when the moving platform rotates by ±90° around the vertical
axis [3]. Merlet [4] classified the singular configurations
especially for 3/6 SP by using Grassmann algebra. This analysis
results are not available for the most general case of Gough-
Stewart platform, especially for general complex singularity
(5A) and special complex singularity (5B).
The main contribution of this paper is in the analysis of the
singularity of SP in general. The work introduced in the paper
can be used (as appears in the examples) as necessary condition
for singularity, such as architectural singularity that came up by
Ma and Angeles [5]. Some of the papers dealt with kinematics
to find the singular conditions of the mechanism [6] while
others dealt with statics [7, 8]. The main topic of this paper is
focusing on the statics (forces) in the SP from which the
singular characterization is derived.
In section two we provide a brief explanation of the 3D
Kennedy theorem and how the 2D Kennedy theorem is derived
from it. The proposed singular characterization of SP 6/6 is
introduced in section three by using reciprocal product (screw
theory) and 3D Kennedy theorem. Section four introduces the
I
The term “singularity” originates from mathematics and the term
“special configuration” originates from mechanical engineering. In this
paper we chose to adopt and use the term “singularity”.

2. 2 Copyright © 2016 by ASME
analysis of the mobility of SP 6/6. The mobility of the ISA is
shown to be the combination mobility of two lines. Section five
introduces a comparison to other singularity analysis methods
that appear in the literature.
2. THE 3D AND 2D KENNEDY THEOREMS
The 3D Kennedy theorem is defined in this paper through
screw theory, and for this we have to use the following two
lemmas that will also be used in the singular characterization in
section ‎3
Lemma 1: The reciprocal product of two lines, 𝐿1 and 𝐿2, which
cross each other is zero [9].
Lemma 2: Let $ be a screw, 𝐿 a line, and three conditions:
a. Reciprocal of a screw and line 𝐿 is zero.
b. The screw $ and line 𝐿 cross each other.
c. The directions of the screw $ and line 𝐿 are perpendicular.
Given two of the above three conditions results in the third
condition.
Proof. Let us start from the definition of a reciprocal product of
a screw and a line:
$ ∘ 𝐿 = ((
𝑆̅$
𝑆̅0$
) + ℎ (
0
𝑆̅$
)) ∘ (
𝑆̅𝐿
𝑆̅0 𝐿
) = (
𝑆̅$
𝑆̅0$
) ∘ (
𝑆̅𝐿
𝑆̅0 𝐿
) + ℎ𝑆̅$ ∙ 𝑆̅𝐿
The symbol ∘ denotes the reciprocal product.
The equation consists of three parts:
1. $ ∘ 𝐿, the reciprocal product of a screw $ with a line 𝐿
2. (
𝑆̅$
𝑆̅0$
) ∘ (
𝑆̅𝐿
𝑆̅0 𝐿
) is equal to zero if the screw $ and the
line 𝐿 cross each other.
3. ℎ𝑆̅$ ∙ 𝑆̅𝐿 is equal to zero if the direction of the screw $
and the direction of the line 𝐿 are perpendicular
It can be verified that given two of the above conditions
results in the third condition. Having these two lemmas it is
now possible to prove 3D Kennedy theorem.
Kennedy theorem in 3D [10, 2]: for any two screws, $1 and $2,
the relative screw $1,2 crosses and is perpendicular to the
common normal 𝑙 between the two screws $1 and $2.
Proof: Let 𝑙 be the common perpendicular of $1 and $2 thus
$1 ∘ 𝑙 = 0 and $2 ∘ 𝑙 = 0. The reciprocal product of the relative
screw $1,2 with the common normal 𝑙 is also zero since:
$1 ∘ 𝑙 = $2 ∘ 𝑙 = 0 → ($2 − $1) ∘ 𝑙 = $1,2 ∘ 𝑙 = 0
The common normal 𝑙 is perpendicular to $1 and $2 thus it
is also perpendicular to the relative screw:
𝑆̅$1
⋅ 𝑆̅𝑙 = 𝑆̅$2
⋅ 𝑆̅𝑙 = 0 → (𝑆̅$2
− 𝑆̅$1
) ⋅ 𝑆̅𝑙 = 𝑆̅$1,2
⋅ 𝑆̅𝑙 = 0
Since the reciprocal product of $1,2 and line 𝑙 is zero and
they are perpendicular it follows from Lemma 2 that they cross
each other.
The Aronhold-Kennedy theorem in 3D applies both for
kinematics and statics. Most of the works reported in the
literature about the use of the 3D Kennedy theorem are in the
context of kinematics [10, 11, 12], while in this paper we use it
in statics.
It is interesting to note that the 2D Aronhold-Kennedy
theorems in kinematics [10] and statics [13] are the projection
of the 3D theorem to a plane.
The 2D Aronhold-Kennedy theorem in kinematics states
that the relative instant centers of any three bodies 𝑖, 𝑗 and 𝑘 of
the mechanism, 𝐼(𝑖𝑗), 𝐼(𝑖𝑘) and 𝐼(𝑗𝑘), must lie on a straight line,
as shown in Figure 1.
This theorem follows from the 3D theorem by setting
ℎ = 0 and the screw lines parallel.
Applying the duality relation between kinematics and
statics, the 2D Aronhold-Kennedy theorem in statics is derived
as follows:
Theorem 1: The dual Kennedy Theorem in statics in 2D [13]:
given any two forces in the plane 𝐹̅1, 𝐹̅2 and their subtraction
𝐹̅1,2 the three forces intersect at a single point.
This theorem is derived from the 3D theorem by setting
ℎ = 0, thus we are talking about two lines instead of screws
and their subtraction. The intersection point is the projection of
the common normal to the two screws and their relative, as
shown in Figure 1.

3. 3 Copyright © 2016 by ASME
Figure 1. The Kennedy Theorem and its projection in 2D: statics and kinematics
3. THE SINGULAR CHARACTERIZATION OF THE 6/6
STEWART PLATFORM (SP)
This section introduces the method of the singular
characterization of the 6/6 SP and its proof. The theorem
assumes that in the singular configuration of 6/6 SP there are at
least two lines which cross four of the six leg lines of the SP.
During the research it was found that all the singular
configurations of the 6/6 SP had these two lines. The authors
are working to prove that the assumption about the two lines is
correct in any singular position.
Figure 2. The Stewart Platform 6/6 in a generic configuration.
Theorem 2: Let the lines 𝐿1 and 𝐿2, be the two lines that cross
the four of the six leg lines 1–4 of the SP.
Let the lines 𝐿1
′
and 𝐿2
′
be parallel to 𝐿1 and 𝐿2,
respectively, and cross the leg lines 5, 6.
Let 𝑚 be the line intersection of the two planes
𝜋 𝐿1,𝐿1
′ and 𝜋 𝐿2,𝐿2
′ .
The 6/6 Stewart Platform is in a singular
configuration if and only if line 𝑚 is perpendicular
to the common normal of leg lines 5 and 6, i.e.
𝑚 ⊥ 𝐶𝑁(5,6) as shown in Figure 3c.
The proof of theorem 2 is based on the force and moments
equations around the lines 𝐿1 and 𝐿2, when those two lines
cross four leg lines 1-4 of the six leg lines of the SP, as shown
in Figure 3a.
(𝑓1 𝑙1 + 𝑓2 𝑙2 + 𝑓3 𝑙3 + 𝑓4 𝑙4 + 𝑓5 𝑙5 + 𝑓6 𝑙6) ∘ (
𝑆̅𝐿1
𝑆̅01
) = 0 (1)
(𝑓1 𝑙1 + 𝑓2 𝑙2 + 𝑓3 𝑙3 + 𝑓4 𝑙4 + 𝑓5 𝑙5 + 𝑓6 𝑙6) ∘ (
𝑆̅𝐿2
𝑆̅02
) = 0 (2)
Let us focus on equation (1).
(𝑓1 𝑙1 + 𝑓2 𝑙2 + 𝑓3 𝑙3 + 𝑓4 𝑙4) ∘ (
𝑆̅ 𝐿1
𝑆̅01
) + (𝑓5 𝑙5 + 𝑓6 𝑙6) ∘ (
𝑆̅ 𝐿1
𝑆̅01
) = 0 (3)

4. 4 Copyright © 2016 by ASME
Since 𝐿1 crosses lines 𝑙1, … , 𝑙4, thus after applying Lemma
1 on equation (3) we derive:
(𝑓5 𝑙5 + 𝑓6 𝑙6) ∘ (
𝑆̅𝐿1
𝑆̅01
) = 0 (4)
The line 𝐿1
′
is parallel to the line 𝐿1 and crosses the line
forces 𝐹5 = 𝑓5 𝑙5 and 𝐹6 = 𝑓6 𝑙6, as shown in Figure 3a. Since 𝐿1
′
crosses the line forces 𝑙5 and 𝑙6 we conclude due to Lemma 1
that:
(𝑓5 𝑙5 + 𝑓6 𝑙6) ∘ (
𝑆̅𝐿1
𝑆̅01
′ ) = 0 (5)
Subtraction of equations (4) and (5) results in the following
equation:
(𝑓5 𝑙5 + 𝑓6 𝑙6) ∘ (
𝑆̅𝐿1
𝑆̅01
) − (𝑓5 𝑙5 + 𝑓6 𝑙6) ∘ (
𝑆̅𝐿1
𝑆̅01
′ ) = 0 (6)
The wrench screw $5,6
𝑤
is the sum of the two line forces,
𝐹5 = 𝑓5 𝑙5 and 𝐹6 = 𝑓6 𝑙6, as follows:
$5,6
𝑤
= 𝑓5 𝑙5 + 𝑓6 𝑙6
$5,6
𝑤
∘ [(
𝑆̅𝐿1
𝑆̅01
) − (
𝑆̅𝐿1
𝑆̅01
′ )] = $5,6
𝑤
∘ (
0
𝑆̅01
− 𝑆̅01
′ ) = 0
The subtraction of the two moments of the lines 𝐿1, 𝐿1
′
results in a vector normal to the plane 𝜋 𝐿1,𝐿1
′ since the two
forces are parallel, as follows:
$5,6
𝑤
∘ (
0
(𝑅̅ 𝐿1
− 𝑅̅ 𝐿1
′ ) × 𝑆̅𝐿1
) = 0 (7)
𝑅̅ 𝐿1
, 𝑅̅ 𝐿1
′ and 𝑆̅𝐿1
belongs or parallel to the plane 𝜋 𝐿1,𝐿1
′ , so
(𝑅̅ 𝐿1
− 𝑅̅ 𝐿1
′ ) × 𝑆̅𝐿1
= 𝑁̅ 𝐿1,𝐿1
′ , where 𝑁̅ 𝐿1,𝐿1
′ is normal to the
plane 𝜋 𝐿1,𝐿1
′ . Substitution of the normal to the plane in equation
(7) and applying the reciprocal product results in:
(
𝑆̅5,6
𝑤
𝑆̅5,6
𝑤 0) ∘ (
0
𝑁̅ 𝐿1,𝐿1
′
) = 𝑆̅5,6
𝑤
∙ 𝑁̅ 𝐿1,𝐿1
′ = 0 (8)
The latter multiplication is equal to zero iff:
𝑆̅5,6
𝑤
⊥ 𝑁̅ 𝐿1,𝐿1
′ ∥ 𝜋 𝐿1,𝐿1
′ (9)
The same process is applied to equation (2) resulting in:
$5,6
𝑤
∥ 𝜋 𝐿1,𝐿1
′ ∥ 𝜋 𝐿2,𝐿2
′
Since the screw is parallel to the two planes it should be
parallel to the intersection line, designated by 𝑚, of these two
planes, as follows:
$5,6
𝑤
∥ 𝑚 (𝑚 = 𝜋 𝐿1,𝐿1
′ ∧ 𝜋 𝐿2,𝐿2
′ )
As was mentioned in the introduction, this method relies
on the existence of inner forces in the singular configuration.
equations (1) and (2) ensure the existence of inner forces in all
of the six legs of the SP indicating that the SP is in the singular
configuration. Based on 3D Kennedy theorem the two forces 𝐹5
and 𝐹6 and their relative screw $5,6
𝑤
, share a common normal.
Since $5,6
𝑤
is parallel to 𝑚 and perpendicular to the common
normal, the theorem is proved.
Note, there is no need that the lines 𝐶𝑁(5,6) and 𝑚 cross
each other, but just to be perpendicular.
a. b. c.
Figure 3. A schematic explanation of the singular characterization of the 6/6 SP
a. The lines 𝐿1 and 𝐿2 cross the four leg lines 1–4. The lines 𝐿1
′
and 𝐿2
′
are parallel to 𝐿1 and 𝐿2, respectively, and cross the
leg lines 5, 6.
b. The line 𝑚 is the intersection of the two planes 𝜋 𝐿1,𝐿1
′ and 𝜋 𝐿2,𝐿2
′ .
c. The common normal to the leg lines 5, 6 is perpendicular to line 𝑚.

5. 5 Copyright © 2016 by ASME
4. THE MOBILITY ANALYSIS OF THE 6/6 SP
Relying on the proposed method for characterizing the
singular configuration of 6/6 SP introduced in previous section,
in this section we introduce the mobility analysis of the
platform of 6/6 SP. The location of the ISA, $ 𝑡
, of the
mechanism is determined through the following steps:
Theorem 3: Let the lines 𝐿1 and 𝐿2, be the two lines that cross
the four leg lines 1–4 of the six leg lines of the SP.
Let the lines 𝐿5
′
and 𝐿6
′
be parallel to 𝑙5 and 𝑙6, and
cross the lines 𝐿1 and 𝐿2, respectively.
Let 𝑚′
= 𝜋𝑙5,𝐿5
′ ∧ 𝜋𝑙6,𝐿6
′ be the line intersection of
the two planes 𝜋𝑙5,𝐿5
′ and 𝜋𝑙6,𝐿6
′ .
Let 𝑀′
be any point on line 𝑚′
.
Let lines 𝐿5
′′
and 𝐿6
′′
lie on the planes 𝜋𝑙5,𝐿5
′ , 𝜋𝑙6,𝐿6
′
and parallel to 𝐿5
′
, 𝐿6
′
, respectively, and both pass
through point 𝑀′
.
Let 𝜋 𝑀′ be the plane defined by the two lines 𝐿5
′′
and 𝐿6
′′
.
Let line 𝑚′⊥
lie on the plane 𝜋 𝑀′ and perpendicular
to the line 𝑚′
.
The ISA, $ 𝑡
, is the common perpendicular to the
common normal of the lines 𝐿1 and 𝐿2, 𝐶𝑁𝐿1,𝐿2
,
and to 𝑚′⊥
, as shown in Figure 4.
Any screw can be written as the sum of two lines [9]. In
this section, we will prove that the mobility of the ISA is a sum
of the mobility of two lines with angular velocities, 𝜔1 and 𝜔2,
as follows:
$ 𝑡
= 𝜔1 𝐿1 + 𝜔2 𝐿2 (10)
The projection of the platform velocity along each leg is
equal to zero since the legs are rigid bodies. This can be
formulated as follows:
$ 𝑡
∘ 𝑙𝑖 = 0 , 𝑖 = 1, … ,6 (11)
In this section we will see how to calculate all the
quantities: $ 𝑡
, 𝜔1 and 𝜔2. We use the same two lines 𝐿1 and 𝐿2
introduced in previous section that cross the four leg lines 1–4
of the SP, as shown in Figure 4. Equation (10) is proved to be
correct for the four leg lines since they intersect lines 𝐿1 and 𝐿2,
thus according to Lemma 1
(𝜔1 𝐿1 + 𝜔2 𝐿2) ∘ 𝑙𝑖 = 0 , 𝑖 = 1, … ,4
Now, we have to check that equation (11) is valid also for
leg lines 𝑙5 and 𝑙6 . Let 𝐿5
′
, 𝐿6
′
be two lines parallel to the leg
lines 𝑙5, 𝑙6 that cross both lines 𝐿1 and 𝐿2, respectively, as
shown in Figure 4b. According to equation (11) $ 𝑡
∘ 𝑙5 = $ 𝑡
∘
𝑙6 = 0 and due to Lemma 1
$ 𝑡
∘ 𝐿5
′
= $ 𝑡
∘ 𝐿6
′
= 0 (12)
Thus we derive the following equations:
{
$ 𝑡
∘ (𝑙5 − 𝐿5
′ ) = 0
$ 𝑡
∘ (𝑙6 − 𝐿6
′ ) = 0
(13)
$ 𝑡
∘ [(
𝑆̅𝑙5
𝑆̅05
) − (
𝑆̅𝑙5
𝑆̅05
′ )] = $ 𝑡
∘ (
0
𝑆̅05
− 𝑆̅05
′ ) = 0 (14)
$ 𝑡
∘ [(
𝑆̅𝑙6
𝑆̅06
) − (
𝑆̅𝑙6
𝑆̅06
′ )] = $ 𝑡
∘ (
0
𝑆̅06
− 𝑆̅06
′ ) = 0 (15)
Let us focus on equation (14) and the same is applied to
equation (15).
$ 𝑡
∘ (
0
(𝑅̅𝑙5
− 𝑅̅ 𝐿5
′ ) × 𝑆̅𝑙5
) = 0 (16)
𝑅̅ 𝑙5
, 𝑅̅ 𝐿5
′ and 𝑆̅𝑙5
belong or parallel to the plane 𝜋𝑙5,𝐿5
′ , thus
(𝑅̅𝑙5
− 𝑅̅ 𝐿5
′ ) × 𝑆̅𝑙5
= 𝑁̅𝑙5,𝐿5
′ , where 𝑁̅𝑙5,𝐿5
′ is normal to the plane
𝜋𝑙5,𝐿5
′ .
After substituting the normal to equation (16) we obtain:
(
𝑆̅ 𝑡
𝑆̅ 𝑡0) ∘ (
0
𝑁̅𝑙5,𝐿5
′
) = 𝑆̅ 𝑡
∙ 𝑁̅𝑙5,𝐿5
′ = 0 (17)
The latter multiplication is equal to zero iff:
$ 𝑡
⊥ 𝑁̅𝑙5,𝐿5
′ ∥ 𝜋𝑙5,𝐿5
′ (18)
Applying the same process to equation (15) results in:
$ 𝑡
∥ 𝜋𝑙5,𝐿5
′ ∥ 𝜋𝑙6,𝐿6
′ (19)
Let 𝑚′ be the intersection line of the two planes
𝜋𝑙5,𝐿5
′ and 𝜋𝑙6,𝐿6
′
Since the twist screw is parallel to both planes it should
also be parallel to their intersection line, 𝑚′
$ 𝑡
∥ 𝑚′ (20)
In order to find the exact position of the ISA, we should
follow the next steps:
Let 𝑀′
be a point on the line 𝑚′
. The lines 𝐿5
′′
and 𝐿6
′′
lie
on the planes 𝜋𝑙5,𝐿5
′ and 𝜋𝑙6,𝐿6
′ and parallel to 𝐿5
′
and 𝐿6
′
,
respectively. The two lines 𝐿5
′′
and 𝐿6
′′
should pass through point
𝑀′
thus together with the point 𝑀′ they define the plane 𝜋 𝑀′.
Since 𝐿5
′′
is parallel to the lines 𝑙5 and 𝐿5
′
and all the three lines
lie in the plane 𝜋𝑙5,𝐿5
′ , 𝐿5
′′
is a linear combination of 𝑙5 and 𝐿5
′
as
follows:
𝐿5
′′
= 𝛼𝑙5 + 𝛽𝐿5
′
(21)
Where 𝛼 and 𝛽, are scalars.
Following equation (11) and equation (12) we derive:
$ 𝑡
∘ 𝐿5
′′
= $ 𝑡
∘ 𝛼𝑙5 + $ 𝑡
∘ 𝛽𝐿5
′
= 0 (22)
Meaning that the reciprocal product of $ 𝑡
and 𝐿5
′′
is equal
to zero.
The same process is applied to 𝐿6
′′
, resulting in:
$ 𝑡
∘ 𝐿6
′′
= $ 𝑡
∘ 𝛼𝑙6 + $ 𝑡
∘ 𝛽𝐿6
′
= 0 (23)
Let any line 𝐿 lie on the plane 𝜋 𝑀′ that pass through the
point 𝑀′
. Notice that since 𝐿 belongs to flat pencil defined by
𝐿5
′′
and 𝐿6
′′
and 𝑀′
it is a linear combination of the lines 𝐿5
′′
, 𝐿6
′′
,
so, $ 𝑡
∘ 𝐿 = 0, as follows:
$ 𝑡
∘ 𝐿 = $ 𝑡
∘ (𝜆𝐿5
′′
+ 𝜇𝐿6
′′) = 0 (24)
Where 𝜆, 𝜇 are scalars.

6. 6 Copyright © 2016 by ASME
We will prove now the existence of a special line,
designated 𝑚′⊥
, which is perpendicular both to 𝑚′ and the
screw $ 𝑡
and crosses the screw.
The plane 𝜋 𝑀′ crosses any line that is not parallel to it thus
it crosses $ 𝑡
, at point 𝑀′′. We draw a line between 𝑀′
and 𝑀′′
and call it 𝑚′⊥
with the above special properties. Due to the
construction, 𝑚′⊥
lies in the plane 𝜋 𝑀′ and belongs to the flat
pencil of the point 𝑀′
thus according to equation (24) the
reciprocal product of 𝑚′⊥
with the screw is equal to zero. Line
𝑚′⊥
crosses the screw (it was so constructed), so two
conditions of Lemma 2 are satisfied thus it follows that 𝑚′⊥
is
perpendicular to $ 𝑡
. Since the $ 𝑡
is parallel to 𝑚′
due to
equation (20), 𝑚′⊥
is also perpendicular to 𝑚′
.
Based on the Kennedy theorem the two lines 𝐿1 and 𝐿2 and
their relative screw $1,2
𝑡
share a common normal. Thus, the ISA,
$ 𝑡
=$1,2
𝑡
, should be perpendicular to the common normal of the
lines 𝐿1 and 𝐿2, 𝐶𝑁𝐿1,𝐿2
, and to 𝑚′⊥
, as shown in Figure 4c.
As mentioned in equation (10), the ISA is the sum of two
lines, i.e the mobility of the ISA consists of the mobility of the
two lines, 𝐿1 and 𝐿2.
The method introduced in this section allows us to
calculate the angular velocities 𝜔1 and 𝜔2 and the pitch ℎ of
the ISA.
When we define the origin as shown in Figure 5, the ISA
can be written as follows:
$ 𝑡
=
(𝐼)
(𝐼𝐼)
(
𝜔 𝑡 𝑆̅𝑡
𝜔 𝑡ℎ𝑆̅𝑡
) = (
𝜔1 𝑆̅𝐿1
+ 𝜔2 𝑆̅𝐿2
𝑎𝑘̂ × 𝜔1 𝑆̅𝐿1
+ (−𝑏𝑘̂) × 𝜔2 𝑆̅𝐿2
) (25)
According to Figure 5a and Figure 5b, the lines can be
presented as follows:
𝑆̅t = (
1
0
0
) , 𝑆̅𝐿1
= (
cos 𝛼
sin 𝛼
0
) , 𝑆̅𝐿2
= (
cos 𝛽
sin 𝛽
0
)
By simplifying the equation (25)-I we get the angular
velocities 𝜔1 and 𝜔2
𝜔1 =
𝜔 𝑡 sin 𝛽
sin(𝛽 − 𝛼)
(26)
𝜔2 =
𝜔 𝑡 sin 𝛼
sin(𝛼 − 𝛽)
(27)
By simplifying the equation (25)-II we get the pitch ℎ of
the ISA and the ratio between a and b
ℎ =
−𝑎𝜔1 sin 𝛼 + 𝑏𝜔2 sin 𝛼
𝜔 𝑡
= (𝑎 + 𝑏)
sin 𝛼 sin 𝛽
sin(𝛼 − 𝛽)
(28)
𝑎
𝑏
=
𝜔2 cos 𝛽
𝜔1 cos 𝛼
= −
tan 𝛼
tan 𝛽
(29)
In the example in Figure 5 𝛼 = 37.33°, 𝛽 = −71.43°,
𝑎 = 30.34 [𝑚𝑚], 𝑏 = 118.44 [𝑚𝑚].
Given that 𝜔 𝑡 = 1 [
𝑟𝑎𝑑
𝑠𝑒𝑐
]
According to equations (26), (27), (28) and Figure 5
𝜔1 = 1 [
𝑟𝑎𝑑
𝑠𝑒𝑐
] , 𝜔2 = 0.64 [
𝑟𝑎𝑑
𝑠𝑒𝑐
] , ℎ = −90.32 [
𝑚𝑚
𝑠𝑒𝑐
]

7. 7 Copyright © 2016 by ASME
a. b. c.
d. e. f.
Figure 4. The analysis of the mobility of SP 6/6
a. The lines 𝐿1 and 𝐿2 cross the four leg lines 1–4 and their common normal, 𝐶𝑁𝐿1,𝐿2
.
b. The lines 𝐿5
′
and 𝐿6
′
parallel to 𝑙5 and 𝑙6, respectively, and cross the lines 𝐿1 and 𝐿2.
c. 𝑚′
= 𝜋𝑙5,𝐿5
′ ∧ 𝜋𝑙6,𝐿6
′ is the intersection planes line of 𝜋𝑙5,𝐿5
′ and 𝜋𝑙6,𝐿6
′ .
d. The lines 𝐿5
′′
and 𝐿6
′′
lie on the plane 𝜋𝑙5,𝐿5
′ , 𝜋𝑙6,𝐿6
′ and parallel to 𝐿5
′
and 𝐿6
′
, respectively. The point 𝑀′
lie on the line 𝑚′
and lines 𝐿5
′′
and 𝐿6
′′
pass through this point and construct the plane 𝜋 𝑀′.
e. The line 𝑚′⊥
lie on the plane 𝜋 𝑀′ and perpendicular to the line 𝑚′
.
f. The ISA $ 𝑡
is perpendicular to the common normal of 𝐿1 and 𝐿2, i.e. 𝐶𝑁𝐿1,𝐿2
, and to 𝑚′⊥
.
a. b.
Figure 5. Determining the angular velocities 𝝎 𝟏 and 𝝎 𝟐 and the pitch 𝒉 of the ISA $𝒕
a. The angle of the lines, 𝐿1 and 𝐿2, relative to the ISA.
b. The distance of the lines, 𝐿1 and 𝐿2, from the ISA along the 𝑘̂ axis.

8. 8 Copyright © 2016 by ASME
5. COMPARISON TO OTHER KNOWN SINGULARITY
ANALYSIS METHODS
One of the known singular configurations of Stewart
Platform is when all the leg lines meet one line [2]. Through the
proposed method, this singular configuration can be derived
since the reciprocal product of all the leg lines with the
common line is zero. In Table 1 it is shown to be a special case
of the proposed characterization when the four lines, 𝐿1, 𝐿2 and
the corresponding parallel lines: 𝐿1
′
, 𝐿2
′
become one line.
Another known singular configuration is when the moving
platform rotates by ±90° around the vertical axis [3]. As
proven by [14], this singular condition is applicable to any
manipulator with a coplanar base and platform (irregular
hexagons) and the angle of rotation is not necessarily ±90° but
is a function of the geometric parameters. After examining this
case, no special characterization was found, but the known
singular configuration proved to satisfy the condition in
theorem 2, thus confirmed its singularity. Merlet [4] studied the
singularity of the six-DOF 3/6 Stewart Platform, called for
short 3/6 SP, based on the Grassmann line geometry. He
discovered many new singularities, including 3C, 4B, 4D, 5A,
and 5B. Hunt et al. [15] followed by [16] proved relying this
time on Grassmann–Cayley algebra that all the combinations of
3/6 SP in which their singularity condition is delineated to be
four planes intersecting at a single point. They showed that all
the singular configurations of Merlet are special cases of this
condition. The geometric condition consists of four planes,
defined by the actuator lines and the position of the spherical
joints, which intersect at a single point.
Table 1. Comparison to other singularity analysis methods
Analysis of the mobility
The singular characterization
introduced in this paper
Singular configuration
The common line is the platform
twist, $ 𝑡
The analysis of the singularity of
SP 6/6 is based on the following
equation:
(𝑓1 𝑙1 + 𝑓2 𝑙2 + 𝑓3 𝑙3 + 𝑓4 𝑙4
+𝑓5 𝑙5 + 𝑓6 𝑙6) ∘ 𝐿 = 0 (30)
In this Hunt’s singularity,
equation (30) is satisfied due to
Lemma 1, resulting in a special case:
the two lines 𝐿1, 𝐿2 and their parallel
lines 𝐿1
′
, 𝐿2
′
become one line, i.e.
𝐿1 = 𝐿2 = 𝐿1
′
= 𝐿2
′
= 𝐿
The common line crosses all
the leg lines of the SP
Hunt’s
Singular
Configuration
The SP is in a singular configuration if and only if line 𝑚 is normal to the
common normal of the leg lines 5 and 6, 𝐶𝑁(5,6)
The moving platform rotates
by ±90° around the vertical axis
Fichter’s
Singular
Configuration

9. 9 Copyright © 2016 by ASME
In this topology the two lines 𝐿1 and 𝐿2 are found easily. The two lines
cross the four leg lines of the robot — 1, 2, 3, 4. Line 𝐿1 (𝐿2) passes through the
four joints connecting the four legs to the base (moveable platform) through the
joints: 𝐴1,... 𝐴4 (𝐵1,... 𝐵4), respectively. Both 𝐿1
′
and 𝐿2
′
pass through joint 𝐶
connecting the legs 5 and 6 to the movable platform so that both cross leg lines
5 and 6. Plane 𝜋 𝐿2,𝐿2
′ coincides with the movable platform. In this topology the
intersection line 𝑚 has a special geometry meaning. The 𝐶𝑁(5,6) is
perpendicular to the plane 𝜋5,6, created by the legs 5 and 6. In the proposed
characterization 𝑚 is perpedicular to the 𝐶𝑁(5,6) thus 𝑚 lies on the plane 𝜋5,6.
Thus, for this topology 𝑚 can be defined as the intersection line of the planes
𝜋5,6 and the movable platform.
Moreover, since 𝑚 ∈ 𝜋 𝐿1,𝐿1
′ it assures that line m intersects 𝐿1 as shown in
the figure.
Singular
Configuration
of SP 6/6
introduced in
[17, 18]
The ISA, $ 𝑡
, should be
perpendicular to the common normal
of the lines 𝐿1 and 𝐿2, 𝐶𝑁𝐿1,𝐿2
, and to
𝑚′⊥For the topology of the SP 3/6
the two lines 𝐿1 and 𝐿2 are well
defined. For example, line 𝐿1 is the
intersection of planes 𝜋1,2 and 𝜋3,4.
Line 𝐿2 is defined to pass through
joints 𝐴 and 𝐵. Both 𝐿1
′
and 𝐿2
′
pass
through joint 𝐶. Therefore the plane
𝜋 𝐿2,𝐿2
′ coincides with the movable
platform and line 𝑚 lies on the
platform and passes through joint 𝐶.
Four planes intersect at a
single point
The four planes: three pairs
of leg lines (𝜋1,2, 𝜋3,4, 𝜋5,6) and
the platform plane intersect at a
single point.
3/6 Stewart
Platform

10. 10 Copyright © 2016 by ASME
In addition, because it is perpendicular
to the 𝐶𝑁(5,6), it also lies on the plane
𝜋5,6. Thus, line 𝑚 lies both on the
moveable platform and on 𝜋5,6. The
intersection of lines 𝐿1 and 𝑚 is the
known point where all the four planes:
(𝜋1,2, 𝜋3,4, 𝜋5,6) and the platform
intersect.
6. CONCLUSIONS
The paper introduces the characterization of the singular
configuration of Stewart platform 6/6. This technique uses
among others the dual form of the well-known Aronhold-
Kennedy theorem in kinematics and is believed to be a
significant contribution to the literature. The main advantage of
the proposed method is that it is also applicable to other types
of parallel mechanisms and not limited to SPs. The method
presented in the paper is consistent with other approaches that
appear in the literature. It is important to note that both 6/6 and
3/6 SP are Assur Graphs. In 2D it was proved [19] that the
singularity of AG has special properties and only in AGs the
singular configuration possess the following properties: the
system is both mobile (has an infinitesimal motion) in all the
joints with 1DOF and has a self-stress, i.e. inner forces in all
the elements. A conjecture says that the special singularity also
exist in 3D Assur Graphs. In 3/6 SP (3D Triad) and in 6/6 SP
(3D Body-bar atom) the conjecture is proved to be correct.
The authors believe that the method introduced in this paper
has a significant potential in characterizing singularity of
spatial parallel mechanisms. The authors continue to explore
this topic and in the forthcoming paper they will present new
singular conditions of others complex mechanisms in three-
dimensions.
ACKNOWLEDGMENT
This research was supported by a grant from the Ministry
of Science & Technology, Israel & the Russian Foundation for
Basic Research, the Russian Federation.
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