Wenger P., Chablat D. et Baili M., “A DH-parameter based condition for 3R orthogonal manipulators to have 4 distinct 
inverse kinematic solutions”, Journal of Mechanical Design, Volume 127, pp. 150-155, Janvier 2005. 
A DH-parameter based condition for 3R orthogonal manipulators to have 4 
distinct inverse kinematic solutions 
P. Wenger, D. Chablat and M. Baili 
Institut de Recherche en Communications et Cybernétique de Nantes UMR CNRS 6597 
1, rue de la Noë, BP 92101, 44312 Nantes Cedex 03 France 
 
Positioning 3R manipulators may have two or four inverse kinematic solutions (IKS). This paper derives a necessary 
and sufficient condition for 3R positioning manipulators with orthogonal joint axes to have four distinct IKS. We show 
that the transition between manipulators with 2 and 4 IKS is defined by the set of manipulators with a quadruple root of 
their inverse kinematics. The resulting condition is explicit and states that the last link length of the manipulator must be 
greater than a quantity that depends on three of its remaining DH-parameters. This result is of interest for the design of 
new manipulators.  
1 Introduction 
This paper focuses on positioning 3R orthogonal manipulators i.e. positioning 3R manipulators with orthogonal joint 
axes. A positioning manipulator may be used as such for positioning tasks in the Cartesian space (x, y, z), or as a 
regional structure of a 6R manipulator with spherical wrist. Among the various kinematic criteria that can be used to 
assess the performances of a manipulator, the accessibility inside the workspace, i.e. the number of inverse kinematic 
solutions (IKS) in the workspace, is of primary interest. Positioning 3R manipulators are known to have at most four 
inverse kinematic solutions (IKS) in their workspace [1]. In general, the number of IKS varies from one point to another 
in the workspace [2-5], which may include regions with 0, 2 or 4 IKS [6-8]. Depending on its geometric parameters, a 
3R manipulator may be binary, i.e. may have at most two IKS in its workspace, or it may be quaternary, i.e. it may have 
up to four IKS [1]. We know from [9] that 3R manipulators with any two intersecting joint axes (i.e. a1=0 or a2=0 or 
a3=0) are quaternary; [13] showed that a 3R orthogonal manipulator with no offset at joint 3 (i.e. d3=0) is quaternary if 
the last link length is greater than the second one (i.e. a3>a2). But this condition is not necessary, that is, a manipulator 
such that d3=0  may be quaternary even if a3<a2. On the other hand, [10] stated a particular necessary and sufficient 
condition, namely, 3R orthogonal manipulators with no joint offsets (i.e. d2= d3=0) are quaternary if, and only if, a1≠a2 
and the link lengths do not satisfy a1>a2>a3. To the authors’ knowledge, no more general DH-parameter based necessary 
and sufficient condition has been derived for a manipulator to be quaternary.  
This paper derives an explicit DH-parameters based necessary and sufficient condition for a 3R manipulator with 
orthogonal joint axes to be quaternary. We show that the transition between binary and quaternary manipulators is 
defined by the set of manipulators with a quadruple root of their inverse kinematics. The set of such manipulators is 
shown to form a separating surface in the manipulator parameter space, which can be defined explicitly by an equation 
of the form a3=f(a1, a2, d2). This result is of interest for the design of new manipulators. 
2. Preliminaries 
2.1 Manipulators under study 
The length DH-parameters of an orthogonal manipulator are referred to as a1, a2, a3, d1, d2, d3 while the angle parameters 
α1 and α2 are set to –90° and 90°, respectively. From now on, a1 can be set to 1 without loss of generality. First joint 
Wenger P., Chablat D. et Baili M., “A DH-parameter based condition for 3R orthogonal manipulators to have 4 distinct 
inverse kinematic solutions”, Journal of Mechanical Design, Volume 127, pp. 150-155, Janvier 2005. 
offset d1 can be chosen equal to zero by an appropriate choice of the reference frame. Last joint offset d3 is set to 0 in 
this study. Thus, we need to handle only three design parameters, which will assume strictly positive values in this 
study. Fig. 1 shows the kinematic architecture of an orthogonal manipulator in its zero configuration. The three joint 
variables are referred to as θ1, θ2 and θ3, respectively. The position of the end-tip (or wrist center) is defined by the three 
Cartesian coordinates x, y and z of the operation point P with respect to a reference frame (O, x, y, z) attached to the 
manipulator base as shown in Fig.1.  
 
z 
y
O 
x
θ 1
θ2
a 1 
d2
a3
a 2 
P
θ3
 
Fig. 1 : Orthogonal manipulator in its zero configuration  
2.2 Singularities curves in the joint space and in the workspace 
The singularities of general 3R manipulators have been derived in [10,12]. They can be determined by calculating the 
determinant of the Jacobian matrix [10], or using a recursive method [12]. For the orthogonal manipulators under study, 
i.e. with α1, α2, a1 and d3 equal to –90°, 90°, 1 and 0, the determinant of the Jacobian matrix takes the following form 
[10]: 
 
2
3 3
2
3 2
3 2
3 1
det( )
(
)(
(
-
)
)
a
c a
c
s a
c d
s a
=
+
+
J
 
 
where ci=cos(θι) and si=sin(θι). A singularity occurs when det(J)=0. Since the singularities are independent of θ1, the 
contour plot of det(J)=0 can be displayed as curves in 
2
3
,
.
π
θ
π
π
θ
π
−
≤
<
−
≤
<
 The singularities can also be 
displayed in the Cartesian space by plotting the points where the inverse kinematics has double roots [3,7]. Thanks to 
their symmetry about the first joint axis, it is sufficient to draw a half cross-section of the workspace by plotting the 
points (
2
2
x
y
ρ =
+
, z).  
 
Two cases arise: 
• 
if a2>a3, the first factor of det(J) cannot vanish and the singularities form two distinct curves S1 and S2 in the 
joint space [10]. When the manipulator is in such a singularity, there is line that passes through the operation 
point and that cuts all joint axes [12 ]. The singularities form two disjoint sets of curves in the workspace. 
These two sets define the internal boundary WS1 and the external boundary WS2, respectively, with WS1=f(S1) 
and WS2=f(S2). Figure 2(a) shows the singularity curves when a2=2, a3=1.5, d2=1. For this manipulator, the 
internal boundary WS1 has four cusp points, where three IKS coincide [7]. It divides the workspace into one 
region with two IKS (the outer region) and one region with four IKS (the inner region), which means that this 
manipulator is quaternary. As shown in section 3, the left and right segments of the internal boundary may 
cross and define a void when d2 is decreased; if d2 is sufficiently small, the internal boundary has no cusp, the 
region with four IKS is replaced with a void and the manipulator is binary. 
• 
if a2≤a3, the operation point can meet the second joint axis whenever θ3=±arccos(-a2/a3) and two horizontal 
Wenger P., Chablat D. et Baili M., “A DH-parameter based condition for 3R orthogonal manipulators to have 4 distinct 
inverse kinematic solutions”, Journal of Mechanical Design, Volume 127, pp. 150-155, Janvier 2005. 
lines appear in the joint space. No additional curve appears in the workspace cross-section but only two points. 
This is because, since the operation point meets the second joint axis when θ3=±arccos(-a2/a3), the location of 
the operation point does not change when θ2 is rotated. Figure 2(b) shows the singularity curves of a 
manipulator such that a2=3, a3=4, d2=3. 
 
4 iks
2 iks
  
 
(a) a2=2, a3=1.5, d2=1 
 
 
 
 
 
(b) a2=3, a3=4, d2=3 
Fig. 2 : Singularity curves for a quaternary manipulator when a2>a3 (a) and when a2<a3 (b) 
3. Transition between binary and quaternary manipulators 
In this section, we show that the transition between binary and quaternary manipulators is the set of manipulators with a 
quadruple root of their inverse kinematics. This result is a consequence of a classification work conducted in [13,14]. 
Using Groebner Bases and Cylindrical Algebraic Decomposition, [14] derived the equations of several surfaces that 
divide the DH-parameters space into 105 domains of manipulators having the same number of cusps in their workspace. 
The systematic investigation of the 105 domains and their kinematic interpretation conducted in [13] showed that, (i) all 
manipulators satisfying a2≤a3 are quaternary, and (ii) the set of manipulators satisfying a2>a3 is composed of two 
adjacent domains, one of which being the set of all binary manipulators, the other one being composed of only 
quaternary manipulators with four cusps like the one shown in Fig. 2(a). In other words, binary manipulators exist only 
when a2>a3 and a boundary surface exists that divides the set of manipulators such that a2>a3 into two domains in the 
parameter space (a2, a3, d2). Now, we show in Fig. 3 how a quaternary manipulator with four cusps turns binary under 
the continuous deformation of the internal boundary of its workspace as a3 is progressively decreased (a2=1.5, d2=0.5). 
In Fig. 3(a), a3=1.1, the manipulator is quaternary and the internal boundary is like in Fig. 2(a) with four cusps and no 
void. When a3=0.9, the two lateral segments cross. Two nodes appear, which define a void and two separate regions 
with four IKS (Fig. 3(b)). The two cusps and the node of each such region get closer to each other as a3 is decreased 
(Figs. 3(c) and  3(d)). Then, they merge into one unique point with four coincident IKS (the region with four IKS is 
reduced to one point) and, finally, disappear and the manipulator turns binary (Fig 3(e)). Thus, the transition between a 
quaternary manipulator and a binary manipulator is characterized by the existence of a pair of four coincident IKS.     
Wenger P., Chablat D. et Baili M., “A DH-parameter based condition for 3R orthogonal manipulators to have 4 distinct 
inverse kinematic solutions”, Journal of Mechanical Design, Volume 127, pp. 150-155, Janvier 2005. 
4 IKS 
Cusp 
2 IKS
4 IKS
Void
Node
4 IKS 
Void 
 
(a) a3=1.1 
 
(b) a3=0.9 
 
 
(c) a3=0.7 
4 IKS
Void
Void 
2 IKS
 
(d) a3=0.5 
 
(e) a3=0.2 
Fig. 3 : Continuous deformation of the internal boundary as a3 is decreased (a2=1.5, d2=0.5). From 1.1 to 0.5, the 
manipulator is quaternary (a-c). From 0.5 to 0.2, two cusps and one node merge into one point with four equal 
IKS and then disappear : the manipulator turns binary (d-e).   
4. Existence condition of a point with four IKS 
To get the equation of the separating surface, we derive the existence condition of a point with four IKS. This can be 
done with the fourth-degree inverse kinematics univariate polynomial in t=tan(θ3/2). The fourth-degree inverse 
kinematics 
polynomial 
of 
3R 
manipulators 
was 
derived 
in 
[3]. 
It 
can 
be 
set 
in 
the 
form 
4
3
2
0
1
2
3
4
( )
4
6
4
0
P t
C t
C t
C t
C t
C
=
+
+
+
+
=
 where C0, C1, C2 and C4 are functions of 
2
2
x
y
ρ =
+
, z and the DH-
parameters [3]. For the orthogonal manipulators under study, i.e. α1, α2, a1 and d3 equal to –90°, 90°, 1 and 0, 
respectively, C0, C1, C2 and C4 can be written as: 
 
 
 =  
C0 
 − 
 −  +  + 
a2 
2 a3
2
a2 a3 V R
V
2
4
d2
2
 
 
= 
C1
a3d2 (
)
−
 +  + 
2a2 a3 V 2
2
 
 
 =  
C2 − 
 + 
 + 
 −  +  + 
a2 
2 a3
2
3
2a3
2 d2
2
3
2a3
2
3
R
3
V
2
12
d2
2
3  
= 
C3
a3 d2 (
)
 +  + 
2a2 a3 V 2
2
 
 
 = 
C4 
 + 
 −  +  + 
a2
2 a3
2
a2 a3 V R V
2
4
d2
2
 
Wenger P., Chablat D. et Baili M., “A DH-parameter based condition for 3R orthogonal manipulators to have 4 distinct 
inverse kinematic solutions”, Journal of Mechanical Design, Volume 127, pp. 150-155, Janvier 2005. 
where 
2
2
2
2
2
2
2
2
3
1
V
x
y
z
a
d
a
= −
−
−
−+
+
+
 and 
2
2.
R
x
y
=
+
  
[3] also derived the existence conditions of multiple IKS. For P(t) to have four equal roots, the following three equations 
must be simultaneously satisfied [3]:   
 
2
0
4
1
3
2
4
3
0
C C
C C
C =
−
+
 
(1) 
 
2
2
3
0
2
4
1
2
3
0
3
1
4
2
2
0
C C C
C C C
C C
C C
C
−
−
−
=
+
  
(2)  
 
2
0
2
1
0
C C
C =
−
 
(3) 
We need to eliminate the Cartesian coordinates x, y and z in order to write a condition on the DH-parameters only. Thus, 
V and R must be eliminated. This task is performed using computer algebra tools [11]. Such tools are available in 
symbolic commercial softwares. We have used the Maple function resultant. First, R is eliminated from (1) and (3). This 
yields a fourth-degree polynomial in V. Then, R is eliminated from (2) and (3). We get a third-degree polynomial in V. 
Finally, V is eliminated from the aforementioned two polynomials. The resulting polynomial is: 
 
a2
12a3
2d2
4Q1Q2Q3 
(4) 
where Q1, Q2 and Q3 are polynomials in a2, a3 and d2: 
 
6
2
4
2
4
4
2
4
2
2
2
2
4
1
2
3
2
3
2
3
2
2
3
2
3
2
3
2
2
4
4
2
2
4
2
2
2
4
4
2
4
4
6
2
2
2
2
2
3
2
2
3
3
2
3
2
3
2
3
2
3
2
3
2
2
2
3
2
2
3
2
2
Q
a a
a a
a a
d a a
a a
a a
d a a
d a a
a
d a
d a
d a
d a
d a
d a
=
−
−
+
+
+
−
+
−
+
−
+
−
+
−
 
(5) 
543 a3 
5 a2 
2 d2 
2 648 a3
5 a2 
4 81 a3 
5 a2 
2 32a3
4 d2
4 a2 8a3
4 d2
2 a2 1110a2
3 a3
4 d2
2
25d2 
2 a2 
5 
− 
 +  
 −  
 −  
 − 
 + 
 − 
47 a2 
2 a3 
3 d2 
2 141 a2 
2 d2 
4 a3 
3 47 a2
2 a3
3 d2
8
210a3d2
2 a2
6
486a3
3 a2
10
972a2
6 a3 
3 
 −  
 −  
 −  
 − 
 + 
 + 
1215 a3 
3 a2 
8 1458 a2 
5 a3 
4 8 a2 d2 
10 a3
4
48a2 a3
4 d2
6
32a2 a3
4 d2
8
141a2
2 d2
6 a3 
3 
 −  
 −  
 −  
 − 
 − 
 − 
243 a3 
4 a2 
3 81 a3
2 a2 
5 162 a3 
2 a2 
7 81a3
2 a2
9
243a3
3 a2
4
1224a3
4 d2
6 a2
3
300a3 
4 d2 
8 a2 
3 
 +  
 +  
 −  
 + 
 − 
 + 
 + 
1791 a3 
4 d2 
4 a2 
3 801 a2 
4 a3 
3 d2 
2 35a2
4 a3 d2
4
29d2
6 a2
3 a3
2
444a3
2 d2
2 a2
5
58a3 
2 d2 
4 a2 
3 
 +  
 − 
 +  
 + 
 + 
 + 
Q2= 
 
35 a3 d2 
2 a2 
4 16d2 
12 a3 
5 80 a3 
5 d2
10
160a3
5 d2
8
160d2
6 a3
5
80a3
5 d2
4
16a3
5 d2 
2 
 +  
 +  
 +  
 + 
 + 
 + 
 + 
177 a3 
2 d2 
4 a2 
5 813 a3 
2 d2 
2 a2 
7 2340a2
8 a3
5 d2
2
2052a3
5 d2
4 a2
4
2025a2
6 a3
5
3078 a2 
8 a3 
5 
 −  
 − 
 −  
 + 
 − 
 + 
1872 d2 
4 a2 
8 a3 
5 3096 d2 
4 a2 
7 a3 
4 1440d2
6 a2
5 a3
4
459a3
3 a2
6 d2
4
72a3
4 a2
9 d2
2
 +  
 − 
 + 
 − 
 + 
2268 a2 
10 a3 
5 d2 
2
648 a2 
12 a3 
5 972a2
11 a3
4
2268a2
10 a3
5
3159a2
7 a3
4
1188a2
5 a3 
4 d2 
4 
 +  
 +  
 +  
 − 
 + 
 − 
2601 a2 
6 a3 
3 d2 
2 72 d2 
6 a2 
6 a3 
5 180a3
5 d2
6 a2
4
2340a3
5 d2
4 a2
6
2352a3
5 d2
2a2
4
 +  
 − 
 +  
 + 
 + 
 
1773 a3 
3 d2
2 a2 
8 2682 a2 
5 a3 
4 d2
2
1557a3
5 d2
2 a2
6
1872a2
7 a3
4 d2
2
2916a2
9 a3 
4 
 −  
 −  
 − 
 + 
 − 
29 a2 
3 a3 
2 d2
2 168 a2 
4 a3 
5 d2 
8 552a2
2 d2
8 a3
5
1227a3
5 d2
4 a2
2
1233a2
2 a3
5 d2 
6 
 +  
 −  
 − 
 − 
 − 
84 a3 
5 d2 
10 a2 
2 1845 a3 
3 d2 
4 a2
4
1287d2
6 a2
4 a3
3
 −  
 −  
 − 
 
(6) 
 
2
3
2
3
2
3
Q
a
a d
a
= −
+
+
 
(7) 
Note : we would have obtained exactly the same equations for d3≠0. In effect, it turns out that, when d3≠0, coefficients 
C0, C1, C2 and C4 have the same expressions as function of V and R. This is because d3 appears only in V (more 
precisely, V(d3≠0)= V(d3=0)+ d3
2). Since V eliminated in Eqs. (1-3), the resulting condition does not change. Thus, the 
condition for a manipulator to have four equal IKS is independent of d3. 
5. Separating surface and the necessary and sufficient condition 
Since elimination may generate spurious solutions [11], solutions of (4) include, in addition to the surface that separates 
quaternary and binary manipulators in the parameter space (a2, a3, d2), other non-separating surfaces. We know that the 
surface that separates quaternary and binary manipulators is necessarily among the surfaces found in [14]. This is 
because [14] determined the surfaces that divide the parameter space into domains where the number of cusps is 
Wenger P., Chablat D. et Baili M., “A DH-parameter based condition for 3R orthogonal manipulators to have 4 distinct 
inverse kinematic solutions”, Journal of Mechanical Design, Volume 127, pp. 150-155, Janvier 2005. 
constant. The equations of these surfaces are [14]: 
 
2
2
2
2
3
2
0
a
a
d
−
+
=
 
(8) 
 
2
6
4
4
2
4
2
2
4
4
2
4
2
2
2
2
2
2
4
2
2
4
2
3
2
3
2
3
2
2
3
2
3
2
3
2
2
3
2
3
2
2
2
2
3
2
4
2
6
2
2
2
4
3
3
2
3
2
3
2
3
2
2
2
3
2
2
0
a a
a a
a a d
a a
a a
a a d
a a
a a d
a d
a d
a
a d
a d
a d
−
+
−
+
−
+
+
−
−
−
+
+
+
=
 
(9) 
 
2
2
2
3
4
2
2
2
2
2
2
2
2
2
3
2
3
2
3
2
2
0
a d
a
a
a
a
a a
a a
+
−
+
−
+
−
=
 
(10) 
 
2
2
2
3
4
2
2
2
2
2
2
2
2
2
3
2
3
2
3
2
2
0
a d
a
a
a
a
a a
a a
+
+
+
−
+
−
=
 
(11) 
 
Comparing Eqs. (8-11) with Eqs. (4-7) show that Eq. (5), i.e. Q1=0, is the same as Eq. (9). On the other hand, Eqs. (6) 
and (7) are different from Eqs. (8), (10) and (11). Thus, the only valid solution is Q1=0.  This equation can be put in an 
explicit form. In effect, this is a second-degree polynomial in a3
2. Solving this quadratics for a3
 yields the equations of 
two regular surfaces given by the following two explicit equations: 
 
2
2
2
2
2
2
2
2
2
2
2
3
2
2
1
2((
)
(
))
2
2
2
a
d
a
d
a
a
d
AB
+
−
−
=
+
−
 
(12) 
and 
 
 
2
2
2
2
2
2
2
2
2
2
2
3
2
2
1
2((
)
(
))
2
2
2
a
d
a
d
a
a
d
AB
+
−
−
=
+
+
 
(13)  
where  
 
2
2
2
2
2
2
2
2
(
1)
and
(
1)
A
a
d
B
a
d
=
+
+
=
−
+
 
 
Fig. 4 shows the graph of the aforementioned two surfaces. For more clarity, 2-dimensional sections of the surfaces are 
drawn in (a2, a3) for d2=0.5 and d2=1, respectively (Figs. 444444(a) and 4(b)), and in (d2, a3) for a2=0.5 and a2=1.5, 
respectively (Figs. 4(c) and 4(d)). 
a3 
a2 
(12)
(13) 
a3
a2
(12)
(13)
 
(a) d2=0.5 
 
 
 
(b) d2=1 
a3 
d2 
(12) 
(13) 
a3
d2
(12)
(13)
 
(c) a2=0.5 
 
 
 
(d) a2=1.5 
Wenger P., Chablat D. et Baili M., “A DH-parameter based condition for 3R orthogonal manipulators to have 4 distinct 
inverse kinematic solutions”, Journal of Mechanical Design, Volume 127, pp. 150-155, Janvier 2005. 
Fig. 4 : Graphs of Eqs. (12) and (13) shown in sections of the DH-parameter space. Graph of (12) is shown in bold 
lines 
In sections shown in Figs. 4(a), 4(b) and 4(d), the boundary between quaternary and binary manipulators is defined only 
by Eq. (12) since Eq. (13) has solutions only for a3> a2, i.e. for quaternary manipulators.  In section a2=0.5, on the other 
hand (Fig. 4(c)), the two graphs intersect and (13) has solutions in a3< a2 when d2 is small enough.  It turns out, 
however, that the surface defined by Eq. (13) does not play any role in the separation and the really separating surface is 
defined by Eq. (12) only. In effect, let choose three test manipulators (1), (2) and (3) in section a2=0.5, defined by 
(a3=0.15, d2=0.21), (a3=0.4, d2=0.1) and (a3=0.45, d2=0.4), respectively (Fig. 5a). These manipulators were chosen such 
that (1) and (2) are separated by (13), and (2) and (3) are separated by (12). Figure 5b shows the workspace of the three 
test manipulators. Manipulators (1) and (2) are binary whereas (3) is quaternary. Thus, the boundary surface that 
separates the binary from the quaternary manipulators is defined by (12) and the necessary and sufficient condition for 
an orthogonal  manipulator to be quaternary is 
2
2
2
2
2
2
2
2
2
2
2
3
2
2
1
2((
)
(
))
2
2
2
a
d
a
d
a
a
d
AB
+
−
−
>
+
−
. 
 
a3 
d2 
*
* *
Test manipulator (3)   
Test manipulator (2)   
Test manipulator (1)   
  
(a) Test manipulators in section a2=0.5 
2 IKS 
Void 
2 IKS
Void
4 IKS
Void
 
2 IKS 
Zoomed in 
view of 
internal 
boundary 
 
Test manipulator (1) 
 
Test manipulator (2) 
 
 
Test manipulator (3) 
(a2=0.5, a3=0.15, d2=0.21)  
(a2=0.5, a3=0.4, d2=0.1)  
 
(a2=0.5, a3=0.45, d2=0.4) 
(b) Workspaces of the test manipulators 
Fig. 5 : The three test manipulators (a) and their workspace (b). Manipulators (1) and (2) are binary whereas (3) 
is quaternary 
For verification purposes, we have written a procedure that plots all binary manipulators by scanning the parameter 
space. The procedure checks the existence of a cusp point in a cross section of the workspace: it scans the internal 
boundary and checks the existence of a triple root of the inverse kinematics polynomial [3]. If there is no cusp and if 
a3 < a2, the manipulator is binary and a mark is plotted. As soon as a cusp is found, no mark is plotted and next 
Wenger P., Chablat D. et Baili M., “A DH-parameter based condition for 3R orthogonal manipulators to have 4 distinct 
inverse kinematic solutions”, Journal of Mechanical Design, Volume 127, pp. 150-155, Janvier 2005. 
manipulator is checked. Figure 6 depicts the resulting plots in the same sections as in Fig. 4. Each parameter was 
scanned with a step of 0.03. Comparison of Figs. 6(a), 6(b), 6(c) and 6(d) with Figs. 4(a), 4(b), 4(c) and 4(d), 
respectively, confirms that the separating surface is defined by Eq. (12).  
 
For an orthogonal manipulator such that a1 ≠1, Eq. (12) is obtained by dividing the DH-parameters by a1. By doing so, 
we get: 
 
2
2
2
2
2
2
2
2
2
2
1
2
2
3
2
2
1
2((
)
(
))
2
2
2
a
d
a
a
d
a
a
d
AB
+
−
−
=
+
−
 
where 
2
2
2
2
2
1
2
2
1
2
(
)
and
(
)
A
a
a
d
B
a
a
d
=
+
+
=
−
+
. 
In summary, an orthogonal manipulator given by its four strictly positive DH-parameters
1
2
3
2
,
,
,
a a a d has four distinct 
IKS if and only if,  
 
2
2
2
2
2
2
2
2
2
2
1
2
2
3
2
2
2
2
2
2
2
1
2
2
1
2
1
2((
)
(
))
2
2
2
(
)
(
)
a
d
a
a
d
a
a
d
a
a
d
a
a
d
+
−
−
>
+
−
+
+
−
+
 
(14) 
If a1=0 or a2=0 or a3=0, the manipulator is quaternary [9]. If d2=d3=0, the manipulator is quaternary if and only if, a1≠a2 
and the link lengths do not satisfy a1>a2>a3 [10]. 
a3 
a2
a3
a2
 
(a) d2=0.5 
 
 
 
 
 
(b) d2=1 
a3 
d2
a3
d2
 
(c) a2=0.5 
 
 
 
 
 
(d) a2=1.5 
Fig. 6 : Numerical plots of binary manipulators in the same sections as in Fig. 4 
6. Conclusion and discussion 
A necessary and sufficient condition for an orthogonal manipulator to be quaternary, i.e., to have four distinct inverse 
kinematic solutions, was established as an explicit expression in the DH-parameters. An orthogonal manipulator given 
by its four strictly positive DH-parameters 
1
2
3
2
,
,
,
a a a d  is quaternary if and only if, 
2
2
2
2
2
2
2
2
2
2
1
2
2
3
2
2
2
2
2
2
2
1
2
2
1
2
1
2((
)
(
))
2
2
2
(
)
(
)
a
d
a
a
d
a
a
d
a
a
d
a
a
d
+
−
−
>
+
−
+
+
−
+
, 
and can be assessed easily when designing a new manipulator regional structure. This condition was confirmed 
numerically by scanning the parameter space. To the authors’ knowledge, this condition was never found before.  
Figure 7 shows the separating surface in the normalized parameter space (
2
3
2
,
,
a a d are divided by a1). The surface is flat 
Wenger P., Chablat D. et Baili M., “A DH-parameter based condition for 3R orthogonal manipulators to have 4 distinct 
inverse kinematic solutions”, Journal of Mechanical Design, Volume 127, pp. 150-155, Janvier 2005. 
and close to the plane a3=0, except in the vicinity of d2=0. Binary manipulators, which are below the surface, are much 
less numerous than their quaternary counterparts.  
a3 
a2 
d2
 
Fig. 7 : Plot of the separating surface 
This study assumed d3=0, i.e. no offset along the last joint, because our arguments referred to a previous classification of 
manipulators such that d3=0. But we have noticed in section 4 that the existence condition for a manipulator to have four 
IKS is independent of d3 and thus Eq. (12) is independent of d3 too. This shows that (12) still plays a role when d3≠0 but 
this does not prove that condition (14) is still necessary and sufficient. Writing the necessary and sufficient condition for 
d3≠0 requires to enlarge the classification of [13] and [14], which is under study. We have already found that condition 
(14) remains necessary and sufficient for any value of d3 provided that 
2
1 / 2
2,
d
a
≥
 or for 
3
2
2
.
d
d
≤
On the other 
hand, it turns out that condition (14) is always sufficient when d3≠0, namely, if (14) is true, then the manipulator will be 
quaternary for any value of d3.  
References 
[1] Pieper, B., 1968, “The kinematics of manipulators under computer control,” PhD thesis, Stanford University. 
[2] Gupta, K.C., and Roth, B., 1982, “Design Considerations for Manipulator Workspaces,” ASME Journal of 
Mechanical Design, Vol. 104, pp. 704-711. 
[3] Kholi D., Spanos J., 1985, “Workspace analysis of mechanical manipulators using polynomial discriminant,” ASME 
J. Mechanisms, Transmission and Automation in Design, Vol. 107, juin 1985, pp. 209-215. 
[4] Kumar A., Waldron K.J., 1981, “The workspace of a mechanical manipulator,” ASME Journal of Mechanical 
Design, Vol. 103., pp. 665-672. 
[5] Yang, D.C.H., and Lee, T.W., 1983, “On the workspace of mechanical manipulators,” ASME Journal of Mechanical 
Design, Vol. 105, pp. 62-69. 
[6] Rastegar, J., and Deravi, P., 1987, “Methods to determine workspace with different numbers of configurations and 
all the possible configurations of a manipulator”, Mechanisms and Machine Theory, Vol. 22, No 4, pp. 343-350. 
[7] Tsai, K.Y., and Kholi, D., 1993, “Trajectory planning in task space for general manipulators,” ASME Journal of 
Mechanical Design, Vol. 115, pp. 915-921. 
[8] Ceccarelli, M., 1995, “A Synthesis Algorithm for Three-Revolute Manipulators by Using an Algebraic Formulation 
of Workspace Boundary,” ASME Journal of Mechanical Design, Vol. 117, pp 298-302. 
[9] Burdick J. W., 1995, “A classification of 3R Positioning Manipulator Singularities and Geometries,” Mechanisms 
and Machine Theory, Vol. 30, pp 71-89. 
[10] El Omri J., 1996, “Kinematic analysis of robotic manipulators,” PhD Thesis, University of Nantes (in french). 
[11] Buchberger B., Collins G. E., and Loos R., 1982, Computer Algebra: Symbolic & Algebraic Computation, 
Wenger P., Chablat D. et Baili M., “A DH-parameter based condition for 3R orthogonal manipulators to have 4 distinct 
inverse kinematic solutions”, Journal of Mechanical Design, Volume 127, pp. 150-155, Janvier 2005. 
Springer-Verlag, Wien, pp. 115-138 
[12] Burdick J. W., 1995, “A recursive method for finding revolute-jointed manipulator singularities”, Mechanisms and 
Machine Theory, Vol. 117, pp 55-63. 
[13] Baili M., Wenger P. and Chablat D., 2003, “Classification of one family of 3R positioning manipulators”, in Proc. 
11th ICAR, pp 1849-1854. 
[14] Corvez S. and Rouillier F., 2002, “Using computer algebra tools to classify serial manipulators”, in Proc. Fourth 
International Workshop on Automated Deduction in Geometry, Linz, Austria.