Spatially Isotropic Configuration of Stewart Platform-based Force Sensor

7/28/2019 Spatially Isotropic Configuration of Stewart Platform-based Force Sensor

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relative error in the force mapping [12]. Isotropic design of multi-axis force sensor is one of the important issues in research.

Uchiyama et al. [1315] proposed an index for the evaluation of structural isotropy of the force sensor and studied a systematic

design procedure to minimize the performance index for the force sensor. Xiong [16] presented the concept of isotropy on the

basis of Fisher's information matrix. Wang et al. [17] established a model of the solution space for the Stewart platform-based force

sensor and conducted an optimal design by using the isotropy performance atlases. Liu and Tzo [18] presented a six-axis force

sensor with its force-sensing element in the form of four identical T-shaped bars, which has obtained good results in terms of the

measurement isotropy and measurement sensitivity. Hou et al. [19] studied a modified Stewart platform-based force sensor and

discussed the performance analysis and comprehensive index optimization of the sensor.

The isotropic performance of the Stewart platform-based force sensor substantially depends on the structure isotropy of the

Stewart platform. Pittens and Podhorodeski [20] studied the Stewart Platform's isotropy using the index of condition number.

Stoughton and Arai [21] presented a modification of the Stewart platform and optimized the structural design with respect to a

weighted sum of dexterity and workspace volume. Zanganeh and Angeles [22] have worked on the Jacobian matrix of Stewart

Platform and provided conditions for attaining isotropic parallel manipulators. Fattah and Ghasemi [23] studied the isotropic

conditions using a numeric method. Tsai and Huang [24] studied the isotropy conditions in order to get certain isotropy

generators. Bandyopadhyay and Ghosal [25] presented an algebraic formulation method to design isotropic spatial parallel

manipulators using the concept of combined isotropy.

Few results, however, have been published on the study of spatially isotropic configuration of Stewart platform-based force

sensor using the mathematical analytic approach. As far as we know, there is no Stewart platform-based force sensor reported

with a spatially isotropic structure except for the force sensor with orthogonal parallel architecture reported in [8]. Studying the

isotropic performance of the Stewart platform-based force sensor and finding classes of spatially isotropic configurations of the

Stewart platform-based force sensor are the main goals of the paper.

The organization of this paper is as follows. Following the introduction, in Section 2, the mathematical expression of the force

mapping matrix for the generalized Stewart platform-based force sensor is built by using the screw theory, and the conditions

leading to spatially isotropic configuration are introduced. In Section 3, the isotropic performance of the classical Stewart platform-

based force sensor is studied in detail. In Section 4, the isotropy analysis of a modified Stewart platform-based force sensor is

presented, and the analytic relations of spatially isotropic configuration are obtained. Finally, the conclusion is presented in

Section 5, summarizing the present works.

2. Basic theory

2.1. Mathematical model of the generalized Stewart platform-based force sensor

Thedrawing of the generalized Stewart platform-based force sensor is shown in Fig. 1, which is composed of an upper platform,

a lower platform and six elastic legs connecting the two platforms with spherical joints. The Cartesian coordinates Og-XYZ, called

frame{g}, is set up with its origin located at the geometrical center of the upper platform. pi(i =1, 2, 6) is the center of the i-th

spherical joint of the upper platform, Pi(i =1, 2, 6) is the center of the i-th spherical joint of the lower platform. F and M

represent the force vector and torque vector acted on the center of the upper platform respectively.

The force and torque applied on the upper platform are distributed to the six legs, for the equilibrium of the upper platform, the

following equation can be obtained.

Fw = F + M= 6

i = 1fi$i 1

p3

p6

p5

p2

p1p

4

P3

P6

P5

P2

P1

P4

f3

f6

f5

f2

f1

f4

X

Y

Z

Og

F M

Fig. 1. The generalized Stewart platform-based force sensor.

143J. Yao et al. / Mechanism and Machine Theory 46 (2011) 142155


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where Fw= (F,M)T= {Fx FYFZ MX MYMZ}

T is the vector of six-dimension external force applied on the upper platform. is the dual

sign;fi represents the reacting force produced on the i-th elastic leg; $i= (Si, S0i) represents the unit line vectoralong the axis ofthe

i-th leg, Si= (piPi)/|piPi|, S0i=(Pi pi)/|piPi|; and pi and Pi are the position vectors of the center of the i-th spherical joints piand Pi with respect to frame {g} respectively. Then Eq. (1) can be expressed as

Fw = G f 2

where f={f1

f2

f3

f4

f5

f6

}T is the vector composed of the forces of the six legs; G is the force Jacobian matrix [26] which can be

expressed in the form as

G =S1 S2 S3 S4 S5 S6

S01 S02 S03 S04 S05 S06

!3

where G is a 66 matrix, the first three rows of the matrix G are corresponding to force vector, while the last three rows are

corresponding to torque vector. These two parts of matrix Ghave different dimensional units, then the matrix Gcan be rewritten

as

G =GFGM

!4

where GF is the force mapping matrix, and GM is the torque mapping matrix.

The mapping relation between the six-dimension external force and the reacting forces of the six legs can be given as

f = CF 5

where C = G1, similarly, the mapping matrix Cis partitioned into the matrix CFand CM, which are corresponding to force vector

and torque vector respectively. The matrix Ccan be rewritten as

C = CF CM : 6

2.2. The conditions leading to spatially isotropic configuration

Generally speaking, a good six-axis force sensor is designed for measuring all applying force components in the isotropic mode,

so it is expected that the same measurement sensitivities are achieved in order to give approximately uniform measurement

accuracies for all force components. While if the mapping matrix is singular or seriously ill-conditioned, it would result in high

divergence among leg forces in response to loads in certain directions. Ideally, the isotropic sensors would be considered as

maximally informative [16] and most slightly affected by the interferential noise, the machining error, and other error sources

[12].

As we know, if we use the condition number of mapping matrix Gor Cto evaluate the isotropic performance, the measurement

index is dimensional-related, and it is non-sense in physics to discuss isotropic performance between the force and the torque. So

it is usually to discuss the force isotropy and torque isotropy separately [16].

A sensor configuration is considered to be force isotropic, if the following condition is satisfied.

GFGTF = m1I33 7

A sensor configuration is considered to be torque isotropic, if the following condition is satisfied.

GMGTM = m2I33 8

Especially, the sensor configuration is considered to be spatially isotropic if the mapping matrix G satisfies the following

condition.

GGT

=GFG

TF GFG

TM

GMGTF GMG

TM

" #=

m1I33 00 m2I33

!9

where m1, and m2 are scalars and I33 is a 33 identity matrix.

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It can be seen that, from Eq. (7), in order to obtain spatially isotropic configuration, the matrix GGT must be a diagonal matrix and the

conditions of force isotropy and torque isotropy are satisfied simultaneously. Then the following equivalent relations can be

obtained.

GGT

=m1I33 0

0 m2I33

" #

(GFG

TF = m1I33; GMG

TM = m2I33

GFGTM = 0; GMG

TF = 0

GGT =m

1I

330

0 m2I33

" #CTC =

I33

= m1

0

0 I33 = m2

" #10

3. Isotropy analysis of the classical Stewart platform-based force sensor

3.1. Static mathematical model of the classical Stewart platform-based force sensor

The drawing of the classical Stewart platform-based force sensor is shown in Fig. 2. ui(i =1,2,6) is the center of the i-th

spherical joint of the upper platform, Ui(i =1,2,6) is the center of the i-th spherical joint of the lower platform; the Cartesian

coordinate Oc-XYZ, called frame{c}, is set up with its origin located at the geometrical center of the upper platform, the X axis is

perpendicular to the line connecting the twojoints U1 and U6; thecentral angles between u1 and u3 and between u3 and u5 are 2/3,

and the central angles between U1 and U3 and between U3 and U5 are2/3 too; 1 denotes the central angle between U6 and U1,the

central angles between U2

and U3

,U4

and U5

areequal to1

respectively;2

denotes the central anglebetween u6

and u1

, thecentral

angles between u2 and u3, u4 and u5 are equal to2 respectively; R1 denotes the radius of the lower platform; R2 denotes the radius

of the upper platform; and Hc denotes the distance between the upper and lower platform.

For the classical Stewart platform-based force sensor, the force Jacobian matrix (3) can be expressed as

Gc =

u1U1u1U1j j

u2U2u2U2j j

u3U3u3U3j j

u4U4u4U4j j

u5U5u5U5j j

u6U6u6U6j j

U1 u1u1U1j j

U2 u2u2U2j j

U3 u3u3U3j j

U4 u4u4U4j j

U5 u5u5U5j j

U6 u6u6U6j j

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37775 11

where

uiUi = R2 cosiR1cos i R2 sin iR1 sin i Hc T; i1; 2;; 6

Ui ui = R2Hc siniR2Hccosi R1R2sin ii T; i1; 2;; 6uiUij j =

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR2 cosiR1 cosi 2 + R2 siniR1 sini 2 + H2c

q; i1; 2;; 6

= 1 2 3 4 5 6 =12

2

3

12

2

3+

12

2

3+

12

2

3

12

12

!

= 1 2 3 4 5 6 =22

2

3

22

2

3+

22

2

3+

22

2

3

22

22

!

Z

Y

X

Hc

Oc

u1

u2

u3

u4

u5 u6

U1

U2U3

U4

U5 U6

R2

R1

2

1

M

F

Fig. 2. The classical Stewart platform-based force sensor.



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where ui and Ui (i=1,2,,6) are the position vectors of the center of the i-th spherical joints ui and Ui with respect to the frame {c},

respectively.

3.2. Isotropic performance of the classical Stewart platform-based force sensor

For the classical Stewart platform-based force sensor, the matrix GcGcT can be expressed as

GcGTc =

F1 0 0 0 R 0

0 F1 0 R 0 00 0 F2 0 0 00 R 0 M1 0 0R 0 0 0 M1 00 0 0 0 0 M2

2666666437777775

1

L2c

!12

where

F1 = 3 L2c H

2c

; F2 = 6H

2c ; M1 = 3H

2c R

22; M2 = 3R

21R

22 1cosd ;

R = 3Hc R1R2cos d = 2 R22

; Lc =ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

R21 + R22 + H

2c 2R1R2cos d = 2

q;d12

where Lc denotes the length of the leg, d is the angle difference between 1 and 2.

By comparing Eqs. (12) and (9), it can be seen easily that the isotropic performance of the classical Stewart platform-basedforce sensor is determined by four parameters R1, R2, Hc and d.

To obtain spatially isotropic configuration, the matrix GcGcT must be a diagonal matrix, namely the expression of R in Eq. (12)

must be equal to zero, then we have

R1R2cos d = 2 R22 = 0: 13

Solving for R2 from Eq. (13), we obtain

R2 = R1cos d = 2 : 14

Furthermore, to obtain spatially isotropic configuration, the following condition must be satisfied.

F1 = F2M

1

= M2

R21 + R

222R1R2cos d = 2 = 2H2c

H2

c = R2

1 1cosd ((15

Solving for R2 from Eq. (15), we obtain

R2 = 2R1sin

6 d

2

: 16

It can be seen that both Eqs. (14) and (16) must be satisfied simultaneously, then we have

cos d = 2 = 2sin

6 d

2

: 17

Eq. (17) can be simplified as

ffiffiffi3p

sin d= 2

= 0:

18

The equation above implies d=0. Under this condition, the Stewart platform becomes architectural singular [27].

Therefore, we can conclude that there are no spatially isotropic configurations for the classical Stewart platform-based force

sensor.

4. Spatially isotropic modified Stewart platform-based force sensor

4.1. The modified Stewart platform-based force sensor

In order to improve the dexterity of the Stewart platform, Stoughton and Arai [21] proposed a modified Stewart platform

manipulator with two different rings in the base. Based on the idea, the modified Stewart platform-based force sensor is

introduced to improve the isotropy performance (see Fig. 3). The Cartesian coordinate Om-XYZ, called frame{m},is set up with its

origin located at the geometrical center of the upper platform; Ra1 and Ra2 denote the radii of the two concentric circles on thelower platform, on which the centers of spherical joint located; Rb1 and Rb2 denote the radii of the radii of the twoconcentric circles



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on the upper platform, on which the centers of spherical joint located; Ai(i =1,2,6) is the center of the i-th spherical joint of the

lower platform, A1, A2 and A3 are placed symmetrically on the circle of radius Ra1, A4,A5 and A6 are placed symmetrically on the

circle of radius Ra2; Bi(i =1,2,6) is the center of the i-th spherical joint of the upper platform, B1, B2 and B3 are placed

symmetrically on the circle of radius Rb1, B4, B5 and B6 are placed symmetrically on the circle of Rb2; Hm denotes the distance

between the upper and lower platform; 1 and 1 denote the directional angles of the first spherical joint of the lower platform

and the upper platform with respect to X axis of the frame{m} respectively; and 2 and 2 denote the directional angles of the

fourth spherical joints of the lower platform and the upper platform with respect to X axis of the frame{m} respectively.

For the modified Stewart platform-based force sensor, the force Jacobian matrix (3) can be expressed as

Gm =

B1A1B1A1j j

B2A2B2A2j j

B3A3B3A3j j

B4A4B4A4j j

B5A5B5A5j j

B6A6B6A6j j

A1 B1

B1A1j jA2 B2

B2A2j jA3 B3

B3A3j jA4 B4

B4A4j jA5 B5

B5A5j jA6 B6

B6A6j j

26664

37775

19

where

BiAi = Rb1cosiRa1cosi Rb1siniRa1sini Hm T; i = 1; 2; 3 ;BjAj = Rb2cosjRa2cosj Rb2sinj Ra2sinj Hm

T; j = 4; 5; 6 ;

BiAij j =ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

R2a1 + R2b12Ra1Rb1cos ii + H2m

q; i = 1; 2; 3 ;

BjAj

= ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR2a2 + R2b22Ra2Rb2cos jj + H2mr

; j = 4; 5; 6 ;

Ai Bi = Rb1Hmsini Rb1Hmcosi Ra1Rb1sin ii T; i = 1; 2; 3 ;

Aj Bj = Rb2Hmsinj Rb2Hmcosj Ra2Rb2sin jj h i

T; j = 4; 5; 6 ;

= 1 2 3 4 5 6 = 1 1 +2

31 +

4

32 2 +

2

32 +

4

3

!;

= 1 2 3 4 5 6 = 1 1 +2

31 +

4

32 2 +

2

32 +

4

3

!:

where Ai and Bi (i =1,2,3) are the position vectors of the center of the i-th spherical joints Ai and Bi with respect to the frame{m}

respectively, and Aj and Bj (j =4,5,6) are the position vectors of the center of the j-th spherical joints Aj and Bj with respect to the

frame{m} respectively.

The leg length ofLi (i =1,2,3) and Lj (j = 4,5,6) can be expressed as

Lm1 = Li = BiAij j =ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

R2a1 + R2b12Ra1Rb1cos 11 + H2mq

i = 1; 2; 3 20

X

Y

Z

1

1

B1B2

B3

B4

B5 B6

A1A2

A3

A4

A5

A6

2

2

Rb1

Rb2

Ra1

Ra2

Hm

Om

Fig. 3. The modified Stewart platform-based force sensor.



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Lm2 = Lj = BjAj

= ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR2a2 + R2b22Ra2Rb2cos 22 + H2mq

j = 4; 5; 6 21

From Eqs. (20) and (21), it can be seen that not all legs are of the same length, in general, the first three legs have identical

length Lm1, the last three legs have identical length Lm2.

Then Eq. (19) can be expressed in the form as

Gm =G

F1G

F2GM1 GM2 ! 1 = L

m10

0 1 = Lm2 ! 22

where

GF1 = B1A1 B2A2 B3A3 ; GF2 = B4A4 B5A5 B6A6 ;GM1 = A1 B1 A2 B2 A3 B3 ; GM2 = A4 B4 A5 B5 A6 B6 :

The matrix [Gm][Gm]T can be expressed in the form as

Gm Gm T =

GF1GTF1

L2m1+

GF2GTF2

L2m2

GF1GTM1

L2m1+

GF2GTM2

L2m2

GM1GTF1

L2m1+

GM2GTF2

L2m2

GM1GTM1

L2m1+

GM2GTM2

L2m2

266664

377775: 23

Wetry tofind the analytic relations of the key structural parameters using Eq. (23), but the complex expressions ofLm1 and Lm2make the conditions leading to spatially isotropic configuration much complicated.

However, if the following constraint condition is satisfied,

Lm1 = Lm2 = Lm 24

Eq. (23) can be simplified as follow.

Gm Gm T =1

L2m

GF1GTF1 + GF2G

TF2 GF1G

TM1 + GF2G

TM2

GM1GTF1 + GM2G

TF2 GM1G

TM1 + GM2G

TM2

" #T: 25

By comparing Eqs. (25) and (23), it can be seen that the introduction of the constraint condition (24) makes the expression of

matrix [Gm][Gm]T much simpler, and makes the isotropy analysis of the sensor much easier.

4.2. The analytic solution of spatially isotropic configuration

By introducing the constraint condition Lm1= Lm2= Lm, the matrix [Gm][Gm]Tcan be expressed as

Gm Gm T =

F1 0 0 R1 R2 00 F1 0 R2 R1 00 0 F2 0 0 2R1

R1 R2 0 M1 0 0R2 R1 0 0 M1 00 0 2R1 0 0 M2

26666664

37777775

1

L2m26

where

F1 =3

2R

2a1 + R

2b12Ra1Rb1cos1 + R

2a2 + R

2b22Ra2Rb2cos2

; F2 = 6H

2m;

M1 =3

2H

2m R

2b1 + R

2b2

; M2 = 3 R

2a1R

2b1sin

21 + R

2a2R

2b2sin

22

;

R1 = 3

2Hm Ra1Rb1sin1 + Ra2Rb2sin2 ;

R2 =3

2Hm Ra1Rb1cos1 + Ra2Rb2cos2R

2b1R

2b2

;

1 = 11; 2 = 22

where, 1 is the angle difference between 1 and 1, and 2 is the angle difference between 2 and 2.



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To obtain spatially isotropic configuration, the matrix (26) must be a diagonal matrix, then we have

R10 = 0

R20 = 0

Ra1Rb1sin1 + Ra2Rb2sin1 = 0

Ra1Rb1cos1 + Ra2Rb2cos2R2b1R

2b2 = 0:

((27

Solving for Ra1 and Ra2 from Eq. (27), we obtain

Ra1 = R2b1 + R2b2

sin2

Rb1sin 12 28

Ra2 =R

2b1 + R

2b2

sin1

Rb2sin 12 29

Substituting Eqs. (20) and (21) into the constraint condition Lm1= Lm2, we obtain

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR2a1 + R

2b12Ra1Rb1cos 11 + H2m

q

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiR2a2 + R

2b22Ra2Rb2cos 22 + H2m

q= 0: 30

The Eq. (30) above can be simplified as

R2a1 + R2b12Ra1Rb1cos1 R

2a2 + R

2b22Ra2Rb2cos2

= 0: 31

Substituting Eqs. (28) and (29) into (31), we obtain

R2b1R

2b2

2sin1 R2b1 + R

2b2

cos2sin2

sin 12 +

R2b1 + R2b2

2

R2b2sin22R

2b1sin

21

R2b1R

2b2sin

2 12 = 0: 32

Solving for Rb1 from Eq. (32), six solutions of Rb1 can be obtained as follows.

Rb1 = Rb2 33

Rb1 = Rb2

34

Rb1 =

ffiffiffiffiffi1

p+

ffiffiffiffiffi2

p2

Rb2 35

Rb1 =

ffiffiffiffiffi1

p

ffiffiffiffiffi2

p2

Rb2 36

Rb1 =

ffiffiffiffiffi1

p+

ffiffiffiffiffi2

p2

Rb2 37

Rb1 =

ffiffiffiffiffi1

p

ffiffiffiffiffi2

p2

Rb2 38

Where

1 =2sin2cos12 + 1

sin1;2 =

2sin2cos121sin1

:

By observing Eqs. (28) and (29), it may be seen easily that sin(12) can't be equal to 0, then we can obtain the following

inequation based on the value range of cosine function.

cos 12 + 1 N 0cos 12 1b0

(39

The inequality (39) implies that E1 and E2 have different mathematical signs and the last four solutions (35)

(38) ofRb1 arecomplex numbers, hence the last four solutions are not practical.



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Eq. (34) is also not practical, so only the solution Rb1= Rb2 has practical meaning, which implies that the six joints on the upper

platform must be placed on the same circle. Then we can set

Rb1 = Rb2 = Rb: 40

Under the conditions (40) and (24), the modified Stewart platform-based force sensor can be simplified as shown in Fig. 4. Rbdenotes the radius of the circle on the upper platform, on which the centers of six spherical joints located. The other parameters

Ra1, Ra2, Hm, 1, 1, 2, and 2 have the same meanings as shown in Fig. 3, and all the six legs have the identical length Lm.

Substituting Rb1=Rb2= Rb into the matrix (26), we obtain

GmGTm =

F1 0 0 R1 R2 00 F1 0 R2 R1 00 0 F2 0 0 2R1

R1 R2 0 M1 0 0R2 R1 0 0 M1 00 0 2R1 0 0 M2

26666664

37777775

1

L2m41

where

F1 = 3Rb Ra1cos1 + Ra2cos2 + 3R2b + 3 = 2 R2a1 + R2a2

; F2 = 6H2m;

M1 = 3H2mR

2b; M2 = 3R

2b R

2a1sin

21 + R

2a2sin

22

;

R1 = 32

HmRbRa1sin132

HmRbRa2sin2;

R2 =3

2HmRbRa1cos1 +

3

2HmRbRa2cos23HmR

2b; R3 = 2R1:

By comparing Eqs. (41) and (9), it can be seen that the isotropy performance is determined by the six parameters Ra1, Ra2, Rb,

Hm, 1 and 2.

To obtain spatially isotropic configuration, the matrix (41) must be a diagonal matrix, then we obtain

R1 = 0R2 = 0

3

2HmRbRa1sin1 +

3

2HmRbRa2sin2 = 0

3

2HmRbRa1cos1 +

3

2HmRbRa2cos23HmR

2b = 0:

8>>>:

8>>>:42

Solving for Ra1 and Ra2 from Eq. (42), we obtain

Ra1 = 2Rbsin2

sin 12 43

Ra2 =2Rbsin1

sin 12 44

1

1

Rb

Ra1

Ra2

X

Y

Z

Hm

2

2A1A2

A3

A4

b1

b2

b3

b4

b5b6

Om

A5 A6

Fig. 4. The modified Stewart platform-based force sensor with all the six joints on the upper platform placed on the same circle.



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In addition, to obtain spatially isotropic configuration, the following condition must be satisfied.

F1F2 = 06H2m + 3Rb Ra1cos1 + Ra2cos2 3R2b3= 2 R2a1 + R2a2

= 0 45

Substituting Eqs. (43) and (44) into Eq. (45) and solving for Hm, we have

Hm =Rb

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi6 + 2cos 2122 cos 42 2cos 22 + 2cos 41 + 22 6cos 21 + 22 2cos 21 + 2cos 21 + 42 cos 41 s

2cos22 2cos21

46

where | | represents the absolute value of the corresponding expression.

Furthermore, to obtain spatially isotropic configuration, the following condition must be satisfied.

M1M2 = 03H2mR

2b3R

2b R

2a1sin

21 + R

2a2sin

22

= 0: 47

Substituting Eqs. (43), (44) and (46) into Eq. (47) and solving for 1, we obtain four solutions for 1 as follows.

1 = atan2 sin2ffiffiffiffiffiffiffi

Q4p

; cos222Q2 ffiffiffiffiffiffiffi

Q4p

cos2

24

35 = atan2 sin2; cos222Q2cos2

" #48


Q4p

;cos

222Q2

ffiffiffiffiffiffiffiQ4

pcos2

24

35 = atan2 sin2; cos222Q2cos2

" #49


Q3p

; cos

22 + 2Q2

ffiffiffiffiffiffiffiQ3

pcos2

24

35 = atan2 sin2; cos22 + 2Q2cos2

" #50


Q3p

;

cos22 + 2Q2 ffiffiffiffiffiffiffiQ3p

cos224 35

= atan2 sin2;cos

22 + 2Q2cos2

" # 51where

Q1 = 169360cos22 + 192cos

42; Q2 =

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi3 1cos22

cos22

q;

Q3 = 13 + 12cos

22 + 4Q2

Q1; Q4 =

1312cos22 + 4Q2

Q1

where atan2(y,x) is the four quadrant inverse tangent of the real parts of the elements of x and y, atan2(y,x).

It can be seen that 1 can't be obtained from Eqs. (48)(51) when 2=/2, then substituting (43), (44), (46) and 2=/2

into Eq. (47) and solving for 1, we obtain four solutions for 1 as follows.

1 = arccos 2

ffiffiffiffiffiffi3

13

r 52

1 = arccos 2

ffiffiffiffiffiffi3

13

r 53

1 = arccos 2

ffiffiffiffiffiffi3

13

r 54

1

= arccos 2 ffiffiffiffiffiffi313

r 55



11/14

4.3. Determination of the valid isotropic solution

From the analysis above, it can be seen that there are four solutions for 1 with respect to any value of2, but we can't makesure

all the four solutions have practical meaning. Then an algorithm is presented to select the valid solutions and determine the valid

range of the solutions:

1. Choose arbitrary Rb, which can be specified by the design requirement.

2. Choose 2 in the given range, here we set 2 in the range from to .

3. Calculate 1 by using Eqs. (48)(51) and (52)(55).

4. Calculate Ra1, Ra2, and Hm by Eqs. (43), (44) and (46) respectively.

5. Describe the variation graphs of variables 1, Hm, Ra1 and Ra2 versus 2 in graph respectively.

6. Select the valid solutions and determine the valid range of1 by observing whether the values of variables Hm, Ra1 and Ra2 are

practical or not.

For example, the valid range of the solution (51) can be determined by the algorithm above. Set Rb=1 and 2 varies in the

range (, ), and describe the variation graphs of variables 1, Hm, Ra1 and Ra2 versus 2 which are shown in Fig. 5a

drespectively.

As shown in Fig. 5ad, it can be seen that all the values of variables Hm, Ra1 and Ra2 are positive when 2 is in the range from /

2 to 0 or from 0 to /2, which implies that the solution (47) has practical meaning in the range 2 ; 0

0; 2

.

The valid range of other solutions (48), (49) and (50) can be determined by using a similar method.

Furthermore, the validity of solutions (52)(55) can be determined by substituting them into (43), (44) and (46) respectively

and observing whether the values of Hm, Ra1 and Ra2 are all positive or not.

2/3

2/3

0

0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Hm

Ra1 0

1

2

3

1

2

3

b

c

a

d

0

0.5

1.0

1.5

0.5

1.0

1.5

Ra2

/3

/3

2/3 0 /3 2/3/3 2/3 0 /3 2/3/3

2/3 0 /3 2/3/3 2/3 0 /3 2/3/3

1

2 2

2 2

Fig. 5. The variations of1, Hm, Ra2 and Ra1 versus 2 when the spatially isotropic conditions are satisfied: (a) the variation of1 versus 2; (b) the variation ofHmversus 2; (c) the variation of Ra2 versus 2; and (d) the variation of Ra2 versus 2.

Table 1

The values of structural parameters which can lead to spatially isotropic configuration.

Solution No. Rb 1 2 1 2 Hm Ra1 Ra2

Solution 1 1 /6 0 0.4254 +/6 /18 0.3590 0.6151 1.4619

Solution 2 1 /6 0 0.1090 +/6 /18 0.1910 1.2413 0.7779



12/14

By using the algorithm above, we know that there are only two solutions having practical meaning, the valid solutions and the

correspondingly valid ranges are listed as follows.

1 = atan2 sin2;cos

222Q2

cos2

" #; 2

5

6;

2

6; 0

0;

6

2;

5

6

1 = atan2 sin2;

cos22 + 2Q2

cos2" #

; 2

2 ; 0

0;

2

1 = arccos 2

ffiffiffiffiffiffi3

13

r ; 2 =

2

1 = arccos 2

ffiffiffiffiffiffi3

13

r ; 2 =

2

:

8>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>:

56

In conclusion, spatially isotropic configuration is determined by the six parameters 2, 1, Rb, Hm, Ra1 and Ra2. The parameters Rband 2 can be specified according to the design requirement, 1 can be determined by Eq. (56), and parameters Ra1, Ra2 and Hm can

be determined by Eqs. (43), (44) and (46) respectively. Classes of spatially isotropic configuration can be obtained easily by these

analytic expressions, and these results are very important for the structural design of the Stewart platform-based force sensor.

4.4. Examples of spatially isotropic configuration

The configuration of the force sensor is determined by eight parameters Ra1, Ra2, Rb, Hm, 1,1, 2 and2, the values of2 and 2can be chosen arbitrarily only if the angle difference between 2 and 2 equals the value of2, and the values 1 and 1 can be

chosen arbitrarily only if the angle difference between 1 and 1 equals the value of1.

Given Rb=1 and 2=/18, the values of structural parameters 1, Hm, Ra1, and Ra2 can be obtained by corresponding

expressions, and we set 1=/6, 2=0, then the values of all the structural parameters are determined, which are given in

Table 1.

The corresponding spatially isotropic configurations for solutions 1 and 2 are shown in Fig. 6a and b respectively.

The force Jacobian matrix for the configuration in Fig. 6a is

G=

0:8164966 0:4082483 0:4082483 0:7071068 0:0000000 0:70710680:0000000 0:7071068 0:7071068 0:4082483 0:8164966 0:4082483

0:5773503 0:5773503 0:5773503 0:5773503 0:5773503 0:57735030:2886751 0:2886751 0:5773503 0 0:5000000 0:5000000

0:5000000 0:5000000 0:0000000 0:5773503 0:2886751 0:28867510:4082483 0:4082483 0:4082483 0:4082483 0:4082483 0:4082483

2666666437777775

57

and the matrix GGT is

GGT

=

2:0000000 0:0000000 0:0000000 0:0000000 0:0000000 0:00000000:0000000 2:0000000 0:0000000 0:0000000 0:0000000 0:00000000:0000000 0:0000000 1:9999999 0 0:0000000 0

0:0000000 0:0000000 0 1:0000000 0:0000000 00:0000000 0:0000000 0:0000000 0:0000000 1:0000000 0:00000000:0000000 0:0000000 0 0 0:0000000 1:0000001

26666664

37777775: 58

Fig. 6. Spatially isotropic configurations of the Stewart platform-based force sensor.



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The force Jacobian matrix for the configuration in Fig. 6b is

G =

0:4082483 0:8164966 0:4082483 0:7071068 0:7071068 0:00000000:7071068 0:0000000 0:7071068 0:4082483 0:4082483 0:8164966

0:5773503 0:5773503 0:5773503 0:5773503 0:5773503 0:57735030:2886751 0:2886751 0:5773503 0:0000000 0:5000000 0:5000000

0:5000000 0:5000000 0:0000000 0:5773503 0:2886751 0:28867510:4082483 0:4082483 0:4082483 0:4082483 0:4082483 0:4082483

26666664

37777775

59

and the matrix GGT is

GGT

=

2:0000000 0:0000000 0:0000000 0:0000000 0:0000000 0:00000000:0000000 2:0000000 0:0000000 0:0000000 0:0000000 0:0000000

0:0000000 0:0000000 2:0000002 0:0000001 0:0000001 00:0000000 0:0000000 0:0000001 1:0000000 0:0000000 0:0000000

0:0000000 0:0000000 0:0000001 0:0000000 1:0000000 0:00000000:0000000 0:0000000 0 0:0000000 0:0000000 1:0000000

26666664

37777775: 60

It can be seen that, from Eqs. (58) and (60), the conditions leading to spatially isotropic configuration are satisfied, and the

numerical example shows the correctness of the analytic results.

5. Conclusion

In order to obtain spatially isotropic configuration of the Stewart platform-based force sensor, a modified Stewart platform-

based force sensor which can achieve spatial isotropy is proposed. The isotropic performances of the classical and modified

Stewart platform-based force sensor are studied using an analytic approach. The results indicate that, for the classical Stewart

platform-based force sensor, it is impossible to realize the spatially isotropic configuration in theory; for the modified Stewart

platform-based force sensor, there does exist the spatially isotropic configuration, and the analytic relations of the key structural

parameters leading to spatially isotropic configuration are obtained. Classes of spatially isotropic configuration can be obtained by

adopting the analytic results. The contents of this paper should be useful for the further research and practical application of the

six-axis force sensor.

Acknowledgements

This paper is supported by the National Science Foundation of China under Grant No. 50975245, and the research and

development project of science and technology of Qinhuangdao city under Grant No. 201001A11.

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Spatially Isotropic Configuration of Stewart Platform-based Force Sensor