Int. J. Mach. Learn. & Cyber. (2012) 3:27–37 DOI 10.1007/s13042-011-0040-1

ORIGINAL ARTICLE

The design and dynamic analysis of a novel 6-DOF parallel mechanism Xiuling Liu • Qingquan Wang • Alexander Malikov Hongrui Wang

•

Received: 5 January 2011 / Accepted: 13 July 2011 / Published online: 23 August 2011  Springer-Verlag 2011

Abstract In this paper, we present the kinematical characteristics of a novel 6-DOF orthogonal parallel mechanism. The dynamic equation of the parallel mechanism is first established using Lagrangian approach. Then, the complex dynamics model of the mechanism is simplified for the ease of analysis and implementation. The dynamic property of the simplified model has been studied through simulation. Keywords Parallel robot  Inverse dynamic  Trajectory tracking

1 Introduction The 6-DOF parallel mechanism (also known as Stewart platform) was proposed by Stewart and Hunt. Parallel mechanisms have some advantages compared with their serial counterparts, such as higher stiffness coefficient, payload capacity, more accurate positioning, higher speed X. Liu (&)  H. Wang School of Electronic and Informational Engineering, Hebei University, Baoding 071002, China e-mail: [email protected] H. Wang e-mail: [email protected] Q. Wang Key Lab of Industrial Computer Control Engineering of Hebei Province, Yanshan University, Qin Huang Dao 066004, China e-mail: [email protected] A. Malikov Kazan State Technical University, Kazan, Tatarstan 420111, Russia e-mail: [email protected]

and acceleration operation. In recent years, parallel mechanisms have been applied in many areas, such as 2-DOF three branches parallel mechanism, singularities parallel mechanism ect [1–3]. Existing research on 6-DOF parallel mechanisms mainly focuses on variations of conventional Stewart platform [4]. Conventional 6-DOF Stewart platforms have some weakness in practical use, such as highly coupling between six chains and complex nonlinearity, which make the analysis difficulty and also complicates the design of control algorithms [5]. In this paper, we propose a novel orthogonal 6-DOF parallel mechanism [6]. The mechanism is a parallel PSS (Prismatic Spherical Spherical) platform of 3-2-1 type which is made of motion platform, 6 fork chains and pedestal. Each fork chain is connected with the platform through spherical hinge which is perpendicular to the motion platform. As a parallel mechanism, this new mechanism inherits the advantages of conventional Stewart platform, such as stronger payload capacity, higher speed and acceleration operation. In addition, the unique orthogonal construction of 6 fork chains reduces the coupling among them and the nonlinearity of the system, which makes the analysis and control algorithm easier [7]. Proper dynamic analysis is important for parallel mechanisms to achieve accurate control [8, 9]. Due to the structural characteristics of the parallel mechanism, the dynamic analysis is very complex. In general, existing methods include Newton–Euler, Lagrange [10], virtual work principle, Arsenault et al. [11]. Among them, Lagrange method applies the Lagrangian formulation to describe the behavior of a dynamic system in terms of energy, which results in relatively simple derivation of the dynamic equations. In addition, the closed-form dynamical equations can be derived systematically in any coordinate system [12, 13]. In this paper, we use Lagrange method to

123

28

Int. J. Mach. Learn. & Cyber. (2012) 3:27–37

get the dynamic model of the new parallel mechanism [14]. This model can also be used in other related experiments using parallel mechanism, such as motion planning, control and simulation [15]. The paper is organized as follows. First, the structure of the parallel mechanism is described. Then, the dynamic model of mechanism is established in Sect. 3. In Sect. 4, we simplify the dynamic model and conduct simulation studies using the simplified model. Conclusions are summarized in Sect. 5.

2 Structure of the 6-DOF parallel mechanism Fig. 1 The structure of the parallel platform

The 6-DOF parallel mechanism is a PSS 3-2-1 type parallel platform [16]. The platform has six kinematic links (Fig. 1). Each kinematic link is a branch and the six links are divided into three groups as 3, 2, and 1 which are connected with the three verticals. The spherical hinges of the first group set kinematic links (B1, B2, B3) coincide with three vertexes of an isosceles triangle whose center is the center of the plane surface. The spherical hinges of the second group set kinematic links (B4, B5), coincide the two vertexes of the section of line which through the center of the plane surface and are parallel with the edges of the platform. The third group spherical hinge (B6) is set in the center of plane. In each branch, linear movement in three orthogonal related directions is produced by six AC servo motors. Dli ði ¼ 1; . . .; 6Þ is used to present the movement of each drive. It is defined as the original state of the orthogonal parallel mechanism that the position of platform is orthogonal and Dli ¼ 0ði ¼ 1; . . .; 6Þ. The length of platform is 76 cm, width is 70 cm, height is 4 cm and weight is 240 kg.

3 Dynamic model of the parallel mechanism In this section, we derive the dynamic model of the parallel mechanism, where the energy of the system is calculated based on the inverse kinematics analysis of the mechanism. From [17], we have the result of inverse kinematics: Dl1 ¼ XC  aa11 þ a þ c qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  c2  ðYC  aa12 Þ2  ðZC  aa13 Þ2

ð1Þ

Dl2 ¼ YC  ea21  ba22 þ b þ c qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  c2  ðXC  ea11  ba12 þ eÞ2  ðZC  ea31  ba32 Þ2 ð2Þ Dl3 ¼ YC þ ea21  ba22 þ b þ c qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  c2  ðXC þ ea11  ba12  eÞ2  ðZC þ ea31  ba32 Þ2 ð3Þ

Dl4 ¼ Zc þ da31 þ ma32  ha33 þ h þ c qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  c2  ðXC þ da11 þ ma12  ha13  dÞ2  ðYc þ da21 þ ma22  ha23  mÞ2

ð4Þ

Dl5 ¼ ZC  da31 þ ma32  ha33 þ h þ c qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  c2  ðXC  da11 þ ma12  ha13 þ dÞ2  ðYC  da21 þ ma22  ha23  mÞ2

ð5Þ

Dl6 ¼ ZC  na32  ha33 þ h þ c qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  c2  ðXC  na12  ha13 Þ2  ðYC  na22  ha23 þ nÞ2

ð6Þ

123

Int. J. Mach. Learn. & Cyber. (2012) 3:27–37

29

0

3.1 Kinematics relations The velocity equal of the platform mass center can be represented as ~_ VC ¼ e0 R C ~ where R_ C ¼ ð X_ C

ð7Þ

Z_C Þ: The platform angular veloc ¼ ex ~ is defined by x  ¼ c_ þ b_ þ a: _ ity x The speeds of drives equal to speeds of points Ci ði ¼ 1; . . .; 6Þ, and angular speeds of rods Bi Ci ði ¼ 1; . . .; 6Þ [18]. For this purpose, we shall first find speeds of points Bi ði ¼ 1; . . .; 6Þ of the platform:

1 X_ B2 B C V~B2 ¼ @ Y_B2 A _ 0 ZB2 1 X_ C  xx ba13 þ xy ea13 þ xz ðba11  ea12 Þ B C ¼ @ Y_C  xx ba23 þ xy ea23 þ xz ðba21  ea22 Þ A Z_C  xx ba33 þ xy ea33 þ xz ðba31  ea32 Þ

Y_C

_ _ ~_ þ AR ~_ þ eR  rBi  ¼ e0 R ~ ¼ e0 ðR ~ VBi ¼ VC þ ½x; C Bi x C Bi xÞ ¼ e0 V~Bi

ð10Þ 0

1

X_ B3 B C V~B3 ¼ @ Y_B3 A Z_ 0 B3 1 X_ C  xx ba13  xy ea13 þ xz ðba11 þ ea12 Þ B C ¼ @ Y_C  xx ba23  xy ea23 þ xz ðba21 þ ea22 Þ A Z_C  xx ba33  xy ea33 þ xz ðba31 þ ea32 Þ

ð8Þ where V~Bi ¼ ð X_ Bi Y_Bi 0 1 0 0 0 _ RB1 ¼ @ 0 0 a A; 0 a 0

T ~ ~ Z_Bi Þ ¼ R_ C þ ARBi x; 0 1 0 0 b _ RB2 ¼ @ 0 0 e A; b e 0 _

ð11Þ Similarly: Based on pair B1, let’s consider kinematical pair C1 (Fig. 2). Choosing point B1 as a pole, the speed of point C1 can be expressed to:

0

1 X_ C þ xx ðha12 þ ma13 Þ  xy ðha11 þ da13 Þ  xz ðma11  da12 Þ V~B4 ¼ @ Y_C þ xx ðha22 þ ma23 Þ  xy ðha21 þ da23 Þ  xz ðma21  da22 Þ A Z_C þ xx ðha32 þ ma33 Þ  xy ðha31 þ da33 Þ  xz ðma31  da32 Þ 0

1 X_ C þ xx ðha12 þ ma13 Þ  xy ðha11  da13 Þ  xz ðma11 þ da12 Þ V~B5 ¼ @ Y_C þ xx ðha22 þ ma23 Þ  xy ðha21  da23 Þ  xz ðma21 þ da22 Þ A Z_C þ xx ðha32 þ ma33 Þ  xy ðha31  da33 Þ  xz ðma31 þ da32 Þ 0

1 X_ C þ xx ðha12  na13 Þ  xy ha11 þ xz na11 V~B6 ¼ @ Y_C þ xx ðha22  na23 Þ  xy ha21 þ xz na21 A Z_C þ xx ðha32  na33 Þ  xy ha31 þ xz na31

0

1

0

0

b

_ B RB 3 ¼ @ 0

0

C e A;

b 0

_

RB 5

0 ¼@h m

e h 0 d

0

h

_ B RB 4 ¼ @ h

0 1 m d A; 0

0

m 0

_

RB6

m

 1 ; B1 C1  VC1 ¼ VB1 þ ½x

C d A;

0 d

0 ¼@ h n

1

h 0 0

where VC1 ¼ VC1 i0 ;

0 1 n 0 A: 0

Z X 1

From Eq. 8, we have: 1 0 1 0 X_ C  aða12 xz  a13 xy Þ X_ B1 V~B1 ¼ @ Y_B1 A ¼ @ Y_C  aða22 xz  a23 xy Þ A Z_C  aða32 xz  a33 xy Þ Z_B1

ð12Þ

B1

ϕ1

A1 VC1

Y X

A1

Z

ϕ1

1

ϕ1

ϑ1

ϑ1 Y

B1

ð9Þ Fig. 2 Kinematical pair C1

123

30

Int. J. Mach. Learn. & Cyber. (2012) 3:27–37

 1 ¼ /_ 1 þ #_ 1 ¼ /_ 1 k0 þ #_ 1 ð sin /1 i0 þ cos /1 j0 Þ x ~ 1; ¼ e0 x ~ 1 ¼ ð #_ 1 sin /1 x

#_ 1 cos /1

T /_ 1 Þ ;

B1 C1 ¼ RC1  RB1 ¼ e0 ðR~C1  R~B1 Þ Let us write down Eq. 12 in projections to axes OXYZ: VC1 ¼ X_ B1

þ #_ 1 cos /1 ðZC1  ZB1 Þ  /_ 1 ðYC1  YB1 Þ;

0 ¼ Y_B1 þ /_ 1 ðXC1  XB1 Þ þ #_ 1 sin /1 ðZC1  ZB1 Þ; 0 ¼ Z_B1  #_ 1 sin /1 ðYC1  YB1 Þ  #_ 1 cos /1 ðXC1  XB1 Þ: From here we find: #_ 1 ¼

Z_B1 ;  YB1 Þ þ cos /1 ðXC1  XB1 Þ

sin /1 ðYC1 1 /_ 1 ¼  ðXC1  XB1 Þ   Z_B1 sin /1 ðZC1  ZB1 Þ _  YB 1 þ sin /1 ðYC1  YB1 Þ þ cos /1 ðXC1  XB1 Þ ðYC1  YB1 Þ ðZC1  ZB1 Þ þ Z_B1 : VC1 ¼ X_ B1 þ Y_B1 ðXC1  XB1 Þ ðXC1  XB1 Þ And after substitutions of values V~B1 and ðXC1  XB1 Þ ¼ Dl1  ða þ cÞ þ aa11  XC ; ðYC1  YB1 Þ ¼ aa21  YC ; ðZC1  ZB1 Þ ¼ aa31  ZC : We have Z_C aða32 xz a33 xy Þ sin/1 ðaa21 YC Þþcos/1 ðDl1 ðaþcÞþaa11 XC Þ ð13Þ  1 Y_C aða22 xz a23 xy Þ /_ 1 ¼  Dl1 ðaþcÞþaa11 XC  ðZ_C aða32 xz a33 xy ÞÞsin/1 ðaa31 ZC Þ þ sin/1 ðaa21 YC Þþcos/1 ðDl1 ðaþcÞþaa11 XC Þ

#_ 1 ¼

ð14Þ VC1 ¼ X_ C  aða12 xz  a13 xy Þ K þ Dl1  ða þ cÞ þ aa11  XC where

Fig. 3 Kinematical pair Ci (i = 2, 3)

K ¼ Y_C ðaa21  YC Þ þ Z_C ðaa31  ZC Þ þ aðxz ðaa11 a12 þ YC a22 þ Zc a32 Þ  xy ðaa11 a13 þ YC a23 þ Zc a33 ÞÞ Similarly with pair C1, for kinematical pair Ci(i = 2, 3) (Fig. 3) choosing point Bi remove it.Let us write down the speed of point Ci(i = 2, 3) in projections to axes OXYZ directly: 0 ¼ X_ Bi þ #_ i sin /i ðZCi  ZBi Þ  /_ i ðYCi  YBi Þ; VCi ¼ Y_Bi þ /_ i ðXCi  XBi Þ  #_ i cos /i ðZCi  ZBi Þ; 0 ¼ Z_Bi þ #_ i cos /i ðYCi  YBi Þ  #_ i sin /i ðXCi  XBi Þ: From here we have: Z_Bi #_ i ¼  cos /i ðYCi  YBi Þ þ sin /i ðXCi  XBi Þ 1 /_ i ¼ ðYCi  YBi Þ   Z_Bi sin /i ðZCi  ZBi Þ _  XB i þ  cos /i ðYCi  YBi Þ þ sin /i ðXCi  XBi Þ ðX ðZCi  ZBi Þ Ci  XBi Þ þ Z_Bi ; VCi ¼ Y_Bi þ X_ Bi ðYCi  YBi Þ ðYCi  YBi Þ and after substitution of values V~B2 ; V~B3 and ðXC2  XB2 Þ ¼ e þ ea11 þ ba12  XC ; ðYC2  YB2 Þ ¼ Dl2  ðb þ cÞ þ ea21 þ ba22  YC ; ðZC2  ZB2 Þ ¼ ea31 þ ba32  ZC ; ðXC3  XB3 Þ ¼ e  ea11 þ ba12  XC ; ðYC3  YB3 Þ ¼ Dl3  ðb þ cÞ  ea21 þ ba22  YC ; ðZC3  ZB3 Þ ¼ ea31 þ ba32  ZC :

ð15Þ

We have #_ 2 ¼

Z_C  xx ba33 þ xy ea33 þ xz ðba31  ea32 Þ sin /2 ðe þ ea11 þ ba12  XC Þ  H1  cos /2

 _/ ¼ 1  X_ C  xx ba13 þ xy ea13 þ xz ðba11  ea12 Þ 2 H  ðZ_C  xx ba33 þ xy ea33 þ xz ðba31  ea32 ÞÞ sin /2 ðea31 þ ba32  ZC Þ þ sin /2 ðe þ ea11 þ ba12  XC Þ  H1  cos /2

123

Int. J. Mach. Learn. & Cyber. (2012) 3:27–37

VC2 ¼ Y_C xx ba23 þxy ea23 þxz ðba21 ea22 Þþ

31

1 H1

fX_ C ðeþea11 þba12 XC Þþ Z_C ðea31 þba32 ZC Þ þðbxx exy Þða13 ðeþXC Þþa23 ðea21 þba22 Þþa33 ZC Þ xz ððba11 ea12 ÞðeþXC Þþðba21 ea22 Þðea21 þba22 Þ þðba31 ea32 ÞZC Þg Z_C  xx ba33  xy ea33 þ xz ðba31 þ ea32 Þ #_ 3 ¼ sin /3 ðe  ea11 þ ba12  XC Þ  H2  cos /3

#_ 4 ¼ X_ C þxx ðha12 þma13 Þxy ðha11 þda13 Þxz ðma11 da12 Þ sin/4 ðmda21 ma22 þha23 YC ÞH3 cos/4 1 /_ 4 ¼ ðDl ðhþcÞda 4 31 ma32 þha33 ZC Þ   Y_C þxx ðha22 þma23 Þxy ðha21 þda23 Þ xz ðma21 da22 Þ þsin/4 ðd da11 ma12 þha13 XC Þ 

ðX_ C þxx ðha12 þma13 Þxy ðha11 þda13 Þxz ðma11 da12 ÞÞ sin/4 ðmda21 ma22 þha23 YC ÞH3 cos/4



 _/ ¼ 1 X_ C  xx ba13  xy ea13 þ xz ðba11 þ ea12 Þ 3 H2  ðZ_C  xx ba33  xy ea33 þ xz ðba31 þ ea32 ÞÞ sin /3 ðea31 þ ba32  ZC Þ þ sin /3 ðe  ea11 þ ba12  XC Þ  H2  cos /3

VC3 ¼ Y_C xx ba23 xy ea23 þxz ðba21 þea22 Þþ

1 H2

VC4 ¼ Z_C þ xx ðha32 þ ma33 Þ  xy ðha31 þ da33 Þ 1 _ fYC ðm  da21  ma22 H3

fX_ C ðeea11 þba12 XC Þþ Z_C ðea31 þba32 ZC Þ

 xz ðma31  da32 Þ þ

þðbxx þexy Þða13 ðeþXC Þ

þ ha23  YC Þ þ X_ C ðd  da11  ma12 þ ha13  XC Þ

þa23 ðea21 þba22 Þþa33 ZC Þ

þ xx ððha22 þ ma23 Þðm  YC Þ þ ðha12 þ ma13 Þ

xz ððba11 þea12 ÞðeþXC Þþðba21

 ðd  XC Þ  ðha32 þ ma33 Þðda31  ma32 þ ha33 ÞÞ

þea22 Þðea21 þba22 Þþðba31 þea32 ÞZC Þg

 xy ððha21 þ da23 Þðm  YC Þ þ ðha11 þ da13 Þðd  XC Þ

where H1 ¼ Dl2 ðbþcÞþea21 þba22 YC

 ðha31 þ da33 Þðda31  ma32 þ ha33 ÞÞ

H2 ¼ Dl3  ðb þ cÞ  ea21 þ ba22  YC For kinematical pair Ci(i = 4, 5, 6) (Fig. 4) choose point Bi(i = 4, 5, 6) as pole. Similarly with the expressions of Ci(i = 1, 2, 3), we have:

 xz ððma21  da22 Þðm  YC Þ þ ðma11  da12 Þ  ðd  XC Þ  ðma31  da32 Þðda31  ma32 þ ha33 ÞÞg; #_ 5 ¼ X_ C þ xx ðha12 þ ma13 Þ  xy ðha11  da13 Þ  xz ðma11 þ da12 Þ sin/5 ðm þ da21  ma22 þ ha23  YC Þ  H4 cos/5  1 _ /_ 5 ¼ YC þxx ðha22 þma23 Þxy ðha21 da23 Þ H4 xz ðma21 þda22 Þþsin/5 ðd þda11 ma12 þha13 XC Þ 

X_ C þxx ðha12 þma13 Þxy ðha11 da13 Þxz ðma11 þda12 Þ sin/5 ðmþda21 ma22 þha23 YC ÞH5 cos/5



Fig. 4 Kinematical pair Ci (i = 4, 5, 6)

123

32

Int. J. Mach. Learn. & Cyber. (2012) 3:27–37

VC5 ¼ Z_C þ xx ðha32 þ ma33 Þ  xy ðha31  da33 Þ 1  xz ðma31 þ da32 Þ þ fY_C ðm þ da21  ma22 H4 þ ha23  YC Þ þ X_ C ðd þ da11  ma12 þ ha13  XC Þ þ xx ððha22 þ ma23 Þðm  YC Þ þ ðha12 þ ma13 Þ  ðd  XC Þ  ðha32 þ ma33 Þðda31  ma32 þ ha33 ÞÞ  xy ðha21  da23 Þðm  YC Þ þ ðha11  da13 Þ  ðd  XC Þ  ðha31  da33 Þðda31  ma32 þ ha33 Þ  xz ðma21 þ da22 Þðm  YC Þ þ ðma11 þ da12 Þ  ðd  XC Þ  ðma31 þ da32 Þðda31  ma32 þ ha33 Þg; X_ C þ xx ðha12  na13 Þ  xy ha11 þ xz na11 sin /6 ðn þ na22 þ h a23  YC Þ  H5  cos /6  1 Y_C þ xx ðha22  na23 Þ  xy ha21 /_ 6 ¼ ðZC6  ZB6 Þ

to the relational angular speed (Excluding viscous friction), with proportionality factor pB. For mathematical model construction, we use the Lagrangian equations:   d oT oT ¼ Qi ði ¼ 1; . . .; sÞ ð16Þ  dt oq_ i oqi As the generalized coordinates, we choose the platform’s center of mass and rotational angle as the generalized coordinates: q1 ¼ XC ; q6 ¼ a

q2 ¼ YC ;

q3 ¼ Z C ;

q4 ¼ c;

q5 ¼ b;

#_ 6 ¼

þxz na21 þ sin /6 ðna12 þ h a13  XC Þ  X_ C þ xx ðha12  na13 Þ  xy ha11 þ xz na11  ; sin /6 ðn þ na22 þ h a23  YC Þ  H5  cos /6 VC6 ¼ Z_C þ xx ðha32  na33 Þ  xy ha31 þ xz na31 1 _ þ fYC ðn þ na22 þ ha23  YC Þ H5 þ X_ C ðna12 þ ha13  XC Þ þ xx ððha22  na23 Þ  ðn  YC Þ  ðha12  na13 ÞXC  ðha32  na33 Þ  ðna32 þ ha33 ÞÞ  xy ðha21 ðn  YC Þ  ha11 XC  ha31 ðna32 þ h a33 ÞÞ þ xz ðna21 ðn  YC Þ  na11 XC  na31 ðna32 þ h a33 ÞÞg: Here H3 ¼ Dl4  ðh þ cÞ  da31  ma32 þ ha33  ZC H4 ¼ Dl5  ðh þ cÞ þ da31  ma32 þ ha33  ZC H5 ¼ Dl6  ðh þ cÞ þ na32 þ ha33  ZC 3.2 Derivation of spatial movement differential equations 1.

Now we make the following assumptions: Coordinate system OXYZ is inertial.

Mass of rods and actuator are ignored. The mass of parallel mechanism platform is M. Axes Cx, Cy, Cz are the main central inertia axes. Jx, Jy, and Jz are the rotational moments inertia of the platform to these axes. 2.

In points Ai(i - 1, …, 6) the forces are applied F1 ¼ F1 i0 ; F2 ¼ F2 j0 ; F3 ¼ F3 j0 ; F4 ¼ F4 k0 ; F5 ¼ F5 k0 ; F6 ¼ F6 k0 : The rotation angle for any hinge Ci is proportional to the angular velocity (Excluding elastic and viscous friction), Calculating according to factors of proportionality kc and pc. Any hinge Bi(i - 1, …, 6) is proportional

123

ð17Þ Kinetic energy is 1 T ¼ ðM X_ C2 þ M Y_C2 þ M Z_C2 þ Jx x2x þ Jy x2y þ Jz x2z Þ 2 ð18Þ _ _  ¼ c_ þ b þa where in according to x xx ¼ b_ sin a þ c_ cos b cos a; xy ¼ b_ cos a  c_ cos b sin a;

ð19Þ

xz ¼ a_ þ c_ sin b: Components of the generalized forces, caused by active force action, will be calculated according to Lagrangian second identity: o rk oVk Qi ¼ Fk  ¼ Fk  ð20Þ oqi oq_ i  FÞ;  FA ¼ And caused by pair forces actions ðF;  FB ¼ F with moment M  ¼ ½AB; FB  ¼ ½BA; FA  in the F; following way, o rA  o rB o rA  oð rA þ ABÞ þ FB  ¼ FA  þ FB  oqi oqi oqi oqi o rA  o rA  oAB ¼ F  F þ FB  oqi oqi oqi   dAB o dt  AB  o½x; ox    ¼ FB  ¼ FB  ¼ FB  ; AB oq_ i oq_ i oq_ i   ox   ox ¼  ½AB; FB  ¼ M : oq_ i oq_ i ð21Þ

Qi ¼ FA 

In Eq. 21, it follows those components of the generalized forces, which caused by resistance moments are defined as Qi ¼ Mconp 

ox oq_ i

ð22Þ

Int. J. Mach. Learn. & Cyber. (2012) 3:27–37

33

From relations Eqs. 20 and 22 we have: 6 oVAj oVC X o#_ j Fj  þ  ðkC #j þ pC #_ j Þ  oq_ i oq_ i oq_ i j¼1 !  6  X o/_ j  x  jÞ oðx  x  jÞ  ðkC /j þ pC /_ j Þ  pB ð x :  oq_ i oq_ i j¼1

Q X C ¼ F1 þ

Qi ¼ M g 

ð23Þ The vector of a platform angular speed is expressed in basis e0 : ~  ¼ ex ~ ¼ e0 Ax ~ ¼ e0 X; x 0 1 0 1 c_ þ a_ sin b XX C B C ~ ¼B X @ XY A ¼ @ b_ cos c  a_ cos b sin c A:

r _ 3 ðk / þp / X C j C j pB ðXBj ÞZ Þ



6  X

sin/j ðXCj XBj Þ þðkC /j þpC /_ j pB ðXrBj ÞX Þ ZCj ZBj ! 1  sin/j ðYCj YBj Þcos/j ðZCj ZBj Þ

1 ðXrBj ÞX r C ~ r ¼ e0 B  x  i ¼ e0 X x @ ðXBj ÞY A Bi r ðXBj ÞZ 0 1 c_ þ a_ sin b þ #_ 1 sin /1 B C  x  1 ¼ e0 @ b_ cos c  a_ cos b sin c  #_ 1 cos /1 A x b_ sin c þ a_ cos b cos c  /_ 1 0 1 c_ þ a_ sin b  #_ j cos /j B C  x  j ¼ e0 @ b_ cos c  a_ cos b sin c  #_ j sin /j A; x b_ sin c þ a_ cos b cos c  /_

þ 

1 c_ þ a_ sin b  /_ j B C  x  j ¼ e0 @ b_ cos c  a_ cos b sin c  #_ j cos /j A x b_ sin c þ a_ cos b cos c  #_ j sin /j ðj ¼ 4; 5; 6Þ

j¼1

þðXrBj ÞZ 

oðXrBj ÞZ



oq_ i

þ



1 ZCj  ZBj

ð26Þ

3 ZC  ZBj ZC1  ZB1 X þ Fj j XC1  XB1 j¼2 YCj  YBj

The expressions of Qc, Qb, Qa is too complicated, so leave them out here. From Eqs. 16, 18, and 19, we have:

Qi ¼ M g 

oðXrBj ÞY

ðkC /j þ pC /_ j  pB ðXrBj ÞX Þ

ð27Þ

Since VAi ¼ VCi , expression 23 takes the form of:

ðXrBj ÞY

6 X

þF4 þ F5 þ F6  n  kC #1 þ pC #_ 1 þ pB ððXrB1 ÞX sin /1  ðXrB1 ÞY cos /j Þ  sin /1 ðZC1  ZB1 Þ r _  ðkC /1 þ pC /1  pB ðXB1 ÞZ Þ X C1  X B 1   1  sin /1 ðYC1  YB1 Þ þ cos /1 ðXC1  XB1 Þ

0

oðXrBj ÞX

ðkC /1 þ pC /_ 1  pB ðXrB1 ÞZ Þ X C1  X B 1

QZC ¼ Mg þ F1

j

6 oVCj oVC X þ  ðkC #j þ pC #_ j Þ Fj  _ oq_ i o q i j¼1 ! _ _ o / o# j j   ðkC /j þ pC /_ j Þ  oq_ i oq_ i

6 X YC  YBj YC1  YB1 þ F2 þ F3 þ Fj j XC1  XB1 ZCj  ZBj j¼4

j¼4

ðj ¼ 2; 3Þ

ðXrBj ÞX



ð25Þ

0

6 X

ðkC #j þpC #_ j pB ððXrBj ÞY cos/j þðXrBj ÞZ sin/j ÞÞ

j¼4

Q YC ¼ F1

Then

YCj YBj

j¼2

b_ sin c þ a_ cos b cos c

XZ

 pB



3 6 X XC XBj X XC XBj Fj j þ Fj j YCj YBj j¼4 ZCj ZBj j¼2

ð24Þ

oq_ i

!

oq_ i

According to Eq. 24 the generalized forces are calculated as:

oT oT oT ¼ 0; ¼ 0; ¼ 0; oXC oYC oZC oT oT oT ¼ M X_ C ; ¼ M Y_C ; ¼ M Z_C ; _ _ o XC oY C oZ_C       d oT d oT d oT € € ¼ M XC ; ¼ M YC ; ¼ M Z€C dt oX_ C dt oY_C dt oZ_C oT ¼ 0; oc oT ¼ Jz xz sin b  Jy xy cos b sin a þ Jx xx cos b cos a; o_c

123

34

Int. J. Mach. Learn. & Cyber. (2012) 3:27–37

  d oT ¼ ðJx x_ x  ðJy  Jz Þxy xz Þ cos b cos a dt o_c  ðJy x_ y  ðJz  Jx Þxx xz Þ cos b sin a

ðJx x_ x  ðJy  Jz Þxy xz Þ sin a þ ðJy x_ y  ðJz  Jx Þxx xz Þ cos a ¼ Qb Jz x_ z  ðJx  Jy Þxx xy ¼ Qa :

þ ðJz x_ z  ðJx  Jy Þxx xy Þ sin b oT ¼ ððJy xy sin a  Jx xx cos aÞtgb þ Jz xz Þðxx cos a ob  xy sin aÞ oT ¼ Jx xx sin a þ Jy xy cos a; ob_   d oT ¼ Jx x_ x sin a þ Jy x_ y cos a dt ob_

Based on kinematics equations, we have X_ C ¼ VX ;

Z_C ¼ VZ ;

Let us lead last equations to Cauchy normal form: M V_X ¼ QXC ;

ð31Þ

M V_Y ¼ QYC ;

ð32Þ

M V_Z þ Mg ¼ QZC ;

ð33Þ

xx cos a  xy sin a ; cos b

ð34Þ

 ðJx xx cos a  Jy xy sin aÞ

b_ ¼ xx sin a þ xy cos a;

ð35Þ

 ðxx cos a  xy sin aÞtgb

a_ ¼ xz  ðxx cos a  xy sin aÞtgb;  1 ðJy  Jz Þxy xz þ Qb sin a x_ x ¼ Jx  cos a þ ðQc  Qa sin bÞ cos b

ð36Þ

þ ðJx xx cos a  Jy xy sin aÞxz

oT ¼ Jx xx ðb_ cos a  c_ cos b sin aÞ oa  Jy xy ðb_ sin a þ c_ cos b cos aÞ ¼ ðJx  Jy Þxx xy ;   oT d oT ¼ Jz x z ; ¼ Jz x_ z : oa_ dt oa_

c_ ¼

P ¼ MgZC ¼ ½ 0

0

Mg

0

ð37Þ

1 ðJz  Jx Þxx xz þ Qb cos a Jy  sin a ðQc  Qa sin bÞ  cos b

ð38Þ

1 ððJx  Jy Þxx xy þ Qa Þ: Jz

ð39Þ

x_ y ¼

Potential energy of the platform is: 0

0 T q

x_ z ¼

The derivatives are: oP oP oP ¼ 0; ¼ 0; ¼ Mg; oXC oYC oZC oP oP oP ¼ 0; ¼ 0; ¼ 0; _ _ o XC oYC oZ_C       d oP d oP d oP ¼ ¼ ¼ 0; dt oX_ C dt oY_C dt oZ_C oP oP oP oP oP oP ¼ ¼ 0; ¼ ¼0 ¼ oc ob oa o_c ob_ oa_       d oP d oP d oP ¼ 0; ¼ 0; ¼0 dt o_c dt ob_ dt oa_ Substituting the derivatives in the Eq. 16, we have: M X€C ¼ QXC ; M Y€C ¼ QYC ; M Z€C þ Mg ¼ QZC ; ðJx x_ x  ðJy  Jz Þxy xz Þcos bcos a  ðJy x_ y  ðJz  Jx Þxx xz Þ  cos bsin a þ ðJz x_ z  ðJx  Jy Þxx xy Þ sin b ¼ Qc

123

Y_C ¼ VY ;

Equations 31–39, together with the kinematical relation Eqs. 13–15 describe the dynamics of spatial movements of the parallel mechanism platform. 4 Simplified dynamics model and simulation study The dynamics model of the parallel mechanism in the previous section is very complex, which makes it very hard to design a controller. Thus, in this section, we simplify the dynamics model of the parallel mechanism. To this end, consider the dynamic model of the platform movement in a symmetry plane discarding influence of one of the drives. We also make the following assumptions: 1. 2. 3. 4.

a = 0, b = 0, XC = 0 Elasticity and viscosity in hinges are not considered; Influence of a drive 1 is not considered; Actuators 2 and 3, 4 and 5 create equal forces, F2 = F3, F4 = F5. Let us introduce some notions:

MYðcÞ ¼ m  m cos c  h sin c  YC ;

Int. J. Mach. Learn. & Cyber. (2012) 3:27–37

35

NYðcÞ ¼ n  n cos c þ h sin c þ YC ; BZðcÞ ¼ ZC  b sin c; In view of the specified assumptions and designations from the general equations, the equations for given plane movement are: M Y€C ¼ QYC

ð40Þ

M Z€C ¼ QZC

ð41Þ

JX €c ¼ Qc

ð42Þ

MYðcÞ NYðcÞ QYC ¼ Q2 ¼ F2 þF4 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þF6 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi c2 ðMYðcÞÞ c2 ðNYðcÞÞ2 kC /4 þpC /_ 4 kC /6 þpC /_ 6 ffi 2 þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi c2 ðMYðcÞÞ c2 ðNYðcÞÞ2 pB ð_c  /_ 4 Þ pB ð_c  /_ 6 Þ ffi 2  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 c ðMYðcÞÞ c2 ðNYðcÞÞ2 BZðcÞ QZC ¼ Q3 ¼ Mg þ F2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ F4 þ F6 2 c  ðBZðcÞÞ _ k C h 2 þ pC h2 pB ð_c  h_ 2 Þ  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 c2  ðBZðcÞÞ c2  ðBZðcÞÞ " # cos cBZðcÞ Qc ¼ Q4 ¼ F2 b sin c  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 c2  ðBZðcÞÞ 2 3 m cos c þ h sin cþ 6 7 sin c  h cos cÞ 7 þ F4 6 4 þ MYðcÞðm 5 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi c2  ðMYðcÞÞ2 2 3 NYðcÞðn sin c þ h cos cÞ7 6 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 þ F6 4h sin c  n cos c  c2  ðNYðcÞÞ2 ðkC h2 þ pC h_ 2 Þb cos c pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 c2  ðBZðcÞÞ ðkC /4 þ pC /_ 4 Þðm sin c  h cos cÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2  c2  ðMYðcÞÞ ðkC /6 þ pC /_ 6 Þðn sin c þ h cos cÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ  pB ð_c  h_ 2 Þ c2  ðNYðcÞÞ ! b cos c  1 þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2  pB ð_c  /_ 4 Þ c2  ðBZðcÞÞ ! m sin c  h cos c  1  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 c2  ðMYðcÞÞ ! n sin c þ h cos c _  pB ð_c  /6 Þ 1 þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ; c2  ðNYðcÞÞ þ

Variables h2, /4, /6 are defined in the general model of spatial movement.

Without elasticity and viscosity forces in hinges, we obtain the simplified model of movements in a considered plane in the form of the Eqs. 66–68 where the generalized forces become: QYC ¼ Q2 MYðcÞ NYðcÞ ¼ F2 þ F4 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ F6 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; c2  ðMYðcÞÞ c2  ðNYðcÞÞ2 BZðcÞ QZC ¼ Q3 ¼ Mg þ F2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 þ F4 þ F6 ; c2  ðBZðcÞÞ 2 3 cos cBZðcÞ 7 6 Qc ¼ Q4 ¼ F2 b4sin c  qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 c2  ðBZðcÞÞ2 2

3

MYðcÞðm sin c  h cos cÞ7 6 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ F4 4m cos c þ h sin c þ 5 c2  ðMYðcÞÞ2 2 3 NYðcÞðn sin c þ h cos cÞ7 6 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5: þ F6 4h sin c  n cos c þ c2  ðNYðcÞÞ2 According to the corresponding parameters which have been tested, we verify the results of the model by simulation of the kinematics and dynamics. Parameters as follows: a = 0.38 m, b = 0.35 m, c = 0.3 m, h = 0.02 m, a = 0.38 m, m = 0.129 m, n = 0.241 m, M = 4 kg. Make the parallel platform to move in an elliptic orbit whose center is the origin of coordinates and equation of locus is 2 YCd 4

2 þ ZCd ¼ 1. So in Y direction, the platform does a sinusoidal trace as YCd ¼ 2 sinðt þ pi =6Þ and does a cosinusoid as ZCd ¼ cosðt þ pi =6Þ in the Z direction, c = 0. So the simplified dynamics model is: 2 32 3 2 3 M 0 0 0 Y€C 6 76 7 4 0 M 0 54 Q€C 5 þ 4 Mg 5 0 €c 0 20 M 3 Y YC ffi C pffiffiffiffiffiffiffiffiffi ffi pffiffiffiffiffiffiffiffiffi 1 2 3 c2 YC2 c2 YC2 6 7 F2 6 pffiffiffiffiffiffiffiffiffi 7 ZC ffi 1 1 74 F4 5 ¼6 6 c2 ZC2 7 4 5 F hY hZ 6 C ffi C ffi C ffi pbZ ffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffi p ffiffiffiffiffiffiffiffiffi mþ n þ c2 ZC2

c2 YC2

c2 YC2

Controlled value: q ¼ ½YC ; ZC ; cT ; input variable: s ¼ ½F2 ; F4 ; F6 T . In this way, we deduce the standard form equation of the dynamics model. Let the initial position of mass center of the platform is (-1, -1, -1) since the actual initial position is different from theoretic position. Designing a nonlinear controller for the system, and then the trajectory tracking control of the system is studied. The simulation result shows the validity of the dynamical model. Figure 5 shows the

123

36

Fig. 5 The trajectory tracking simulation results. a Result of YC, b result of ZC, c result of c

123

Int. J. Mach. Learn. & Cyber. (2012) 3:27–37

Fig. 6 The trajectory tracking error curve. a Result of YC, b result of ZC, c result of c

Int. J. Mach. Learn. & Cyber. (2012) 3:27–37

37 Program of Science and Technology Agency of Hebei Province (08243531D).

References

Fig. 7 6-DOF bionic robot horse

trajectory tracking results. Figure 6 is the trajectory tracking error curve of YC, ZC, c. It can be seen that trajectory tracking can be achieved quickly, although the error curves fluctuate slightly around zero. This simulation results illustrate the effectiveness of the derived dynamic model of the parallel mechanism.

5 Conclusion and application In this paper, we first analyze the kinematic characteristics of the novel PSS 6-DOF robot in rectangular. Cartesian coordinate system and establish the dynamic equation of parallel mechanism using the Lagrange method. Then, based on the requirement of motion control, the dynamics model is simplified for the ease of analysis and controller design. Simulation studies have been conducted using the simplified dynamic model which illustrates the effectiveness of the model. As an experimental application of the proposed parallel mechanism, we have designed a bionic robot horse, as shown in Fig. 7. The robot horse aims to simulate the movement of a real horse, and can be used in various applications such as entertainment, game, health, etc. The dynamics model established in this paper provides the basis for the design the control system for this robot horse. It also allows us to evaluate and optimize the performance of the system. Acknowledgments This work is supported by the project supported by National Natural Science Foundation of China (61074175), International Science and Technology Cooperation Project of ministry of Science (2008DFR10530), Science and Technology Support

1. Schuster A, Yamaguchi Y (2011) From foundational issues in artificial intelligence to intelligent memristive nano-devices. Int J Mach Learn Cybern 2(2):75–87 2. Huang G-B, Wang DH, Lan Y (2011) Extreme learning machines: a survey. Int J Mach Learn Cybern 2(2):107–122 3. Tielong S (2000) Robust control foundation of robot. Qinghua University Press, Beijing 4. Zhenlin J (2002) Design philosophy and applied technology of a novel orthogonal 6-DOF parallel robot. Yanshan University, Qinhuangdao 5. Wei H (2005) Robot dynamics and control. Higher Education Press, Beijing 6. Frolov KV, Popov SA, Musaitov AK (2001) The mechanisms theory and machines mechanic. Vishaya shkola 7. Yang T (2009) An evolutionary simulation–optimization approach in solving parallel-machine scheduling problems—a case study. Comput Ind Eng 56(3):1126–1136 8. Staicua S, Zhang Dan (2006) A novel dynamic modelling approach for parallel mechanisms analysis. Politehnica University, Bucharest 9. Renaud P, Vivas A (2006) Kinematic and dynamic identification of parallel mechanisms. Control Eng Pract 14(9):1099–1109 10. Alavandar S, Nigam MJ (2008) Inverse kinematics solution of 3DOF planar robot using ANFIS. In: Proceedings of ICCCC, vol 3, pp 150–155 11. Arsenault M, Gosselin CM (2006) Kinematic static and dynamic analysis of a planar 2-DOF tensegrity mechanism. Mech Mach Theory 41:1072–1089 12. Geikea T, McPhee J (2003) Inverse dynamic analysis of parallel manipulators with full mobility. Mech Mach Theory 38:549–562 13. Yen P-L, Lai C-C (2009) Dynamic modeling and control of a 3-DOF Cartesian parallel manipulator. Mechatronics 19:390–398 14. Yen P-L (2008) A two-loop robust controller for compensation of the variant friction force in an over-constrained parallel kinematic machine. Int J Mach Tools Manuf 48:1354–1365 15. Yu-bin L, Yong-sheng G, Jie Z (2006) Singularity analysis of a 6-PRRS parallel robot based on geometrical method. In: The sixth world congress on intelligent control and automation, Dalian, pp 9217–9211 16. Guo HB, Li HR (2006) Dynamic analysis and simulation of a six degree of freedom Stewart platform manipulator. In: Proc Inst Mech Eng C J Mech Eng Sci 220(1):61–72 17. Wang H, Wang Q, Liu X, Dong Bin (2010) The inverse kinematics analysis of the novel 6-DOF parallel mechanism. In: Proceedings of the ninth international conference on machine learning and cybernetics, Qingdao, vol 11, issue no 14, pp 2015–2010 18. Wang Y, de Silva CW (2008) A machine-learning approach to multi-robot coordination. Eng Appl Artif Intell 21:470–484

123