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
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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
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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
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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
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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)
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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
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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
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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
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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
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36
Fig. 5 The trajectory tracking simulation results. a Result of YC, b result of ZC, c result of c
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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
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