# The  Kinematics Analysis of a Novel 3-DOF Cable-Driven Wind Tunnel

Authors: Yao Yu; Wu Hongtao
Journal: Journal of Robotics
 (2010)

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## Abstract

The kinematics analysis method of a novel 3-DOF wind tunnel mechanism based on cable-driven parallel mechanism is provided. Rodrigues' parameters are applied to express the transformation matrix of the wire-driven mechanism in the paper. The analytical forward kinematics model is described as three quadratic equations using three Rodridgues' parameters based on the fundamental theory of parallel mechanism. Elimination method is used to remove two of the variables, so that an eighth-order polynomial with one variable is derived. From the equation, the eight sets of Rodridgues' parameters and corresponding Euler angles for the forward kinematical problem can be obtained. In the end, numerical example of both forward and inverse kinematics is included to demonstrate the presented forward-kinematics solution method. The numerical results show that the method for the position analysis of this mechanism is effective.


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## Body

## 1. Introduction

Parallel manipulators have separate serial kinematic chains that are linked to the ground and the moving platform at the same time. They have some potential advantages over serial robot manipulators such as accuracy, greater load capacity, higher velocities, and accelerations. Parallel manipulators have been developed for applications in many fields [1–4].In the past few decades, parallel manipulators using cable transmission have been enthusiastically studied in a number of areas. In a cable suspended parallel robot, the moving platform is suspended and manipulated by the attached cables that are connected to the base; for example cable-suspended robots are Robocrane [5, 6], ultra-high-speed cable robot [7], dexterous hands [8, 9], parallel cable-suspended manipulators [10, 11], teleoperating robots [12], and robots for biological use [13]. The major advantage of using tendon transmission lies in that actuators can be installed on the remote base such that a lightweight and compact design can be realized.In recent years, many researchers paid great attention to the use of cable-driven mechanism in wind tunnel test due to its fewer interference on the streamline flow. Some successful achievements have been made in the Suspension ACtive pour Soufflerie (SACSO) project about the cable-driven parallel suspension system in low-speed wind tunnels with 8 cables [14], another application is WDPSS-8 developed by Huaqiao university for use in wind tunnel test [15].However, recent works about cable-driven wind tunnel mechanism (CDWTM) have focused on the design of 6 degree-of-freedom (DOF); further research is still under way to meet practical application. This article puts forward a new 3-DOF wind tunnel equipment based on cable-driven parallel mechanism, which can provide 3-DOF pure rotation for the scale model in wind tunnel. Because position analysis is one of the key complicated and important problems for the cable-driven parallel mechanism, the forward and inverse kinematics of the aforementioned mechanism is the main concern of the paper. In general, a numerical iterative scheme, such as Newton-Raphson method, can be applied to this problem. But such method not only demands an initial estimate that should be fairly close to the solution of the current configuration but also cannot guarantee the convergence to the actual solution. As is well known there are many methods that can be used to express transformation matrix in the closed-form solution for position analysis; in [16] the closed-form solution of a 3-DOF parallel manipulator is investigated by the Euler angles, resulting in a 16th degree polynomial expression in one single variable; screw theory [17, 18] is employed for the forward kinematics of the parallel mechanism; Husty [19] developed an algorithm for solving the direct kinematics of general Stewart-Gough platforms by using Euler parameters; an univariate polynomial of 40th degree is obtained; Lee and Shim et al. [20]. presented the closed-form forward kinematics of the 6-6 Stewart platform with Rodridgues’ parameters; 40 sets of solutions to describe the posture of the moving platform have been determined. Our goal is to develop an algebraic algorithm to provide all the solutions of forward kinematics for the 3-DOF mechanism by means of Rodridgues’ parameters, and a concise forward kinematics equation can be achieved.The reminder of the paper is the following. Section2 briefly outlines the system. Section 3 presents a method to obtain the analytical solution of forward kinematics. Sections 4 and 5 cover a numerical example, and conclusions are given in the last section.


## 2. Description of the 3-DOF CDWTM

Figure1 shows the kinematic model of the CDWTM with the scale model, spherical joint, five cables, and the base (the wind tunnel). For convenience sake, the scale model is substituted for moving platform a1a2a3. The spherical joint is fixed in the center of moving platform, one end of the cables Ob4, Ob5 is connected to the spherical joint, and the other ends b4 and b5 are attached to the upper and lower walls of the wind tunnel, respectively. The CDWTM can achieve 3-DOF rotational motion about point O when the other three cables a1b1,a2b2,anda3b3 are actuated cooperatively via the motors and pulleys fixed in wind tunnel, which is very suitable for changing the three angles of the scale model in wind tunnel test. The base coordinate frame {OXYZ} is located in spherical joint, with origin placed in the center of the spherical joint. The x-axis is against the streamline flow, the y-axis is along lift force, and the z-axis is coincident with side force. The frame {OX′Y′Z′} is attached to the moving platform, origin is the same with the base frame. Here, vector Li connects the couple vertices ai and bi (i=1,2,3). The forward kinematics of the CDWTM is to determine the orientation of the moving platform while the lengths of the three cables, L1,L2,andL3  are known. γ,β,andα are orientation of the scale model, which is formed by rotating the axis X, the axis Y, and the axis Z, respectively.Figure 1
Functional schematic of a 3-DOF CDWTM.


## 3. Analysis of Forward Kinematics

### 3.1. Rodridgues’ Parameter Description of the Rotational Matrix

Choose Rodridgues vectoru={x,y,z}T, and it’s opposite corresponding symmetry matrix is U=[0-zyz0-x-yx0]; here x,y,andz are called Rodridgues’ parameters. According to Cayley’s formula [21], the transformational matrix R can be written as follows:
(1)R=(I-U)-1(I+U).HereI is a unit 3 × 3 matrix.Furthermore, the inverse of[I-U] can be represented as a function of vector U, that is,
(2)(I-U)-1=I+U+uuT1+uTu.
Taking (2) into (1), then we have
(3)R=(1-uTu)I+2U+2uuT1+uTu.


### 3.2. Solution Procedure

From the geometric relationship in Figure1, we can have
(4)Li=Rai-bi(i=1,2,3),
where ai  and bi denote the position vectors ai and bi,respectively.So the following equation can be obtained:(5)Li2=ai2+bi2-2biTRai(i=1,2,3),
where, Li, ai, and bi are the norms for the corresponding vectors Li, ai, and bi,  respectively.For the convenience, letei  and fi be the unit vector for the vertices vectors ai and bi, respectively, that is,
(6)ei=aiai,fi=bibi(i=1,2,3).And also set(7)cosθi=-Li2+ai2+bi22aibi,cosθ0i=eiTfi(i=1,2,3).Here, the physical meaning ofθ0i is the initial angle of θi. So (5) can be rewritten in a concise form as
(8)fiTRei=cosθi(i=1,2,3).Substitute (3) into (8), and multiply both sides by the nonzero factor (1+uTu). (8) becomes
(9)uTWiu+2HiTu=cosθi-cosθ0i(i=1,2,3),
where,
(10)Wi=fieiT+eifiT-(cosθi+cosθ0i)I,Hi=ei×fi(i=1,2,3).There are three unknowns in (9), which are x,y, and z. By eliminating two of the variables, the eighth polynomial in one variable can be algebraically achieved.In order to obtain the polynomial, we letai=[ai1,ai2,ai3]T and bi=[bi1,bi2,bi3]T(i=1,2,3); taking the initial conditions into (10), we have
(11)Wi=[wi11wi12wi13wi21wi22wi23wi31wi32wi33](i=1,2,3),H1=(h1,h2,h3),H2=(h4,h5,h6),H3=(h7,h8,h9).
Here, Wi and Hi are known from the initial conditions and (10).According to (11), (9) can be rewritten in the following scalar equations:
(12)w111x2+w122y2+w133z2+(w112+w121)xy+(w113+w131)xz+(w123+w132)yz+2h1x+2h2y+2h3z=d1,w211x2+w222y2+w233z2+(w212+w221)xy+(w213+w231)xz+(w223+w232)yz+2h4x+2h5y+2h6z=d2,w311x2+w322y2+w333z2+(w312+w321)xy+(w313+w331)xz+(w323+w332)yz+2h7x+2h8y+2h9z=d3
with di=cosθi-cosθ0i,(i=1,2,3).To derive a univariate equation inx, simplify (12) as if x is a known constant; we have
(13)y2=p3+p2y+p1z,(14)z2=p6+p5y+p4z,(15)yz=p9+p8y+p7z,
where p1~p9are all the function of x; the details can be found in the appendix.From (15), we have
(16)y2z=p3z+p2yz+p1z2=p3z+p2(p9+p8y+p7z)+p1(p6-p5y-p4z)=y(yz)=p9y+p8y2+p7yz=p9y+p8(p3+p2y+p1z)+p7(p9+p8y+p7z)Amplify (14) by y; we can get
(17)yz2=y(p6-p5y-p4z)=p6y-p5(p3+p2y+p1z)-p4(p9+p8y+p7z)=z(yz)=p9z+p8yz+p7z2=p9z+p8(p9+p8y+p7z)+p7(p6-p5y-p4z).We also have (18) and (19) from (16) and (17)
(18)s11+s12y+s13z=0,(19)s21+s22y+s23z=0,
where
(20)s11=p1p6-p3p8+p2p9-p7p9,s12=p1p5-p7p8-p9,s13=p3+p1p4+p2p7-p72-p1p8,s22=p2p5+p6-p5p7+p4p8-p82,s23=p1p5-p7p8-p9.
Similarly, amplify (15) by z; we can get
(21)s11z+s12yz+s13z2=s11z+s12(p9+p8y+p7z)+s13(p6-p5y-p4z)=0,
that is,
(22)s31+s32y+s33z=0
with
(23)s31=p6(p3+p1p4+p2p7-p72-p1p8)+(p1p5-p7p8-p9)p9,s32=(p5(p3+p1p4+p2p7-p72-p1p8)+p8(p1p5-p7p8-p9)),s33=(p1p6-p3p8+p4(p3+p1p4+p2p7-p72-p1p8)+p7(p1p5-p7p8-p9)+p2p9-p7p9).
In matrix form, Equations (18)–(22) can be arranged as follows:
(24)[s11s12s13s21s22s23s31s32s33][1yz]=0.To get the nontrivial solution of (24),  the determinant of the coefficient must be zero, that is,
(25)|A|=((p1p5-p7p8-p9)2)×(p6(p3+p1p4+p2p7-p72-p1p8)-(p3+p1p4+p2p7-p72-p1p8)×(p2p5+p6-p5p7+p4p8-p82)+(p1p5-p7p8-p9)p9-(p5(p3+p1p4+p2p7-p72-p1p8)+p8(p1p5-p7p8-p9))×(p1p5-p7p8-p9)×(p1p6-p3p8+p2p9-p7p9)-(p3+p1p4+p2p7-p72-p1p8)×(p3p5-p6p7+p4p9-p8p9))+p1p6-p3p8+p4(p3+p1p4+p2p7-p72-p1p8)+p7(p1p5-p7p8-p9)+(p1p6-p3p8+p4(p3+p1p4+p2p7-p72-p1p8)+p7(p1p5-p7p8-p9)+p2p9-p7p9)×((p2p5+p6-p5p7+p4p8-p82)×(p1p6-p3p8+p2p9-p7p9)-(p1p5-p7p8-p9)×(p3p5-p6p7+p4p9-p8p9))=0.From the appendix and (18)–(22), s31 is a polynomial ofx4, while s11,s21,s32,ands33 are all polynomials ofx3, and s12,s13,s22,ands23 are only polynomials of x2, so (25) is an eighth polynomial in one variable.


## 3.1. Rodridgues’ Parameter Description of the Rotational Matrix

Choose Rodridgues vectoru={x,y,z}T, and it’s opposite corresponding symmetry matrix is U=[0-zyz0-x-yx0]; here x,y,andz are called Rodridgues’ parameters. According to Cayley’s formula [21], the transformational matrix R can be written as follows:
(1)R=(I-U)-1(I+U).HereI is a unit 3 × 3 matrix.Furthermore, the inverse of[I-U] can be represented as a function of vector U, that is,
(2)(I-U)-1=I+U+uuT1+uTu.
Taking (2) into (1), then we have
(3)R=(1-uTu)I+2U+2uuT1+uTu.


## 3.2. Solution Procedure

From the geometric relationship in Figure1, we can have
(4)Li=Rai-bi(i=1,2,3),
where ai  and bi denote the position vectors ai and bi,respectively.So the following equation can be obtained:(5)Li2=ai2+bi2-2biTRai(i=1,2,3),
where, Li, ai, and bi are the norms for the corresponding vectors Li, ai, and bi,  respectively.For the convenience, letei  and fi be the unit vector for the vertices vectors ai and bi, respectively, that is,
(6)ei=aiai,fi=bibi(i=1,2,3).And also set(7)cosθi=-Li2+ai2+bi22aibi,cosθ0i=eiTfi(i=1,2,3).Here, the physical meaning ofθ0i is the initial angle of θi. So (5) can be rewritten in a concise form as
(8)fiTRei=cosθi(i=1,2,3).Substitute (3) into (8), and multiply both sides by the nonzero factor (1+uTu). (8) becomes
(9)uTWiu+2HiTu=cosθi-cosθ0i(i=1,2,3),
where,
(10)Wi=fieiT+eifiT-(cosθi+cosθ0i)I,Hi=ei×fi(i=1,2,3).There are three unknowns in (9), which are x,y, and z. By eliminating two of the variables, the eighth polynomial in one variable can be algebraically achieved.In order to obtain the polynomial, we letai=[ai1,ai2,ai3]T and bi=[bi1,bi2,bi3]T(i=1,2,3); taking the initial conditions into (10), we have
(11)Wi=[wi11wi12wi13wi21wi22wi23wi31wi32wi33](i=1,2,3),H1=(h1,h2,h3),H2=(h4,h5,h6),H3=(h7,h8,h9).
Here, Wi and Hi are known from the initial conditions and (10).According to (11), (9) can be rewritten in the following scalar equations:
(12)w111x2+w122y2+w133z2+(w112+w121)xy+(w113+w131)xz+(w123+w132)yz+2h1x+2h2y+2h3z=d1,w211x2+w222y2+w233z2+(w212+w221)xy+(w213+w231)xz+(w223+w232)yz+2h4x+2h5y+2h6z=d2,w311x2+w322y2+w333z2+(w312+w321)xy+(w313+w331)xz+(w323+w332)yz+2h7x+2h8y+2h9z=d3
with di=cosθi-cosθ0i,(i=1,2,3).To derive a univariate equation inx, simplify (12) as if x is a known constant; we have
(13)y2=p3+p2y+p1z,(14)z2=p6+p5y+p4z,(15)yz=p9+p8y+p7z,
where p1~p9are all the function of x; the details can be found in the appendix.From (15), we have
(16)y2z=p3z+p2yz+p1z2=p3z+p2(p9+p8y+p7z)+p1(p6-p5y-p4z)=y(yz)=p9y+p8y2+p7yz=p9y+p8(p3+p2y+p1z)+p7(p9+p8y+p7z)Amplify (14) by y; we can get
(17)yz2=y(p6-p5y-p4z)=p6y-p5(p3+p2y+p1z)-p4(p9+p8y+p7z)=z(yz)=p9z+p8yz+p7z2=p9z+p8(p9+p8y+p7z)+p7(p6-p5y-p4z).We also have (18) and (19) from (16) and (17)
(18)s11+s12y+s13z=0,(19)s21+s22y+s23z=0,
where
(20)s11=p1p6-p3p8+p2p9-p7p9,s12=p1p5-p7p8-p9,s13=p3+p1p4+p2p7-p72-p1p8,s22=p2p5+p6-p5p7+p4p8-p82,s23=p1p5-p7p8-p9.
Similarly, amplify (15) by z; we can get
(21)s11z+s12yz+s13z2=s11z+s12(p9+p8y+p7z)+s13(p6-p5y-p4z)=0,
that is,
(22)s31+s32y+s33z=0
with
(23)s31=p6(p3+p1p4+p2p7-p72-p1p8)+(p1p5-p7p8-p9)p9,s32=(p5(p3+p1p4+p2p7-p72-p1p8)+p8(p1p5-p7p8-p9)),s33=(p1p6-p3p8+p4(p3+p1p4+p2p7-p72-p1p8)+p7(p1p5-p7p8-p9)+p2p9-p7p9).
In matrix form, Equations (18)–(22) can be arranged as follows:
(24)[s11s12s13s21s22s23s31s32s33][1yz]=0.To get the nontrivial solution of (24),  the determinant of the coefficient must be zero, that is,
(25)|A|=((p1p5-p7p8-p9)2)×(p6(p3+p1p4+p2p7-p72-p1p8)-(p3+p1p4+p2p7-p72-p1p8)×(p2p5+p6-p5p7+p4p8-p82)+(p1p5-p7p8-p9)p9-(p5(p3+p1p4+p2p7-p72-p1p8)+p8(p1p5-p7p8-p9))×(p1p5-p7p8-p9)×(p1p6-p3p8+p2p9-p7p9)-(p3+p1p4+p2p7-p72-p1p8)×(p3p5-p6p7+p4p9-p8p9))+p1p6-p3p8+p4(p3+p1p4+p2p7-p72-p1p8)+p7(p1p5-p7p8-p9)+(p1p6-p3p8+p4(p3+p1p4+p2p7-p72-p1p8)+p7(p1p5-p7p8-p9)+p2p9-p7p9)×((p2p5+p6-p5p7+p4p8-p82)×(p1p6-p3p8+p2p9-p7p9)-(p1p5-p7p8-p9)×(p3p5-p6p7+p4p9-p8p9))=0.From the appendix and (18)–(22), s31 is a polynomial ofx4, while s11,s21,s32,ands33 are all polynomials ofx3, and s12,s13,s22,ands23 are only polynomials of x2, so (25) is an eighth polynomial in one variable.


## 4. Example

As an example, we consider a 3-DOF CDWTM with the following initial structural parameters, the vertices vectors of the base areb1=[1.6,1.25,1.3]T, b2=[1.6,1.25,-1.3]T, and b3=[-2,1.25,0]T, the vertices vectors of the moving platform are a1=[0.571581674,0.386554,0.11777264]T,a2=[0.497384168,0.408871083,-0.27465146]T, anda3=[-0.791685337,-0.038439224,0.14750171]T. The lengths of the three cables are L1=1.789090488,L2=1.724702626 and L3=1.77252834. The final eighth equation about x in the example is
(26)1.7746×109x8+1.59847×106x7+5.61258×108x6+1.4869×107x5+6.75153×107x4+2.93121×106x3+2.20763×106x2+145723x-8357.62=0.Solving the equation, the eight sets of Rodridgues’ parameters can be achieved in Table1.Table 1
Eight sets of Rodridgues’ parameters.
xyz1-0.0926620.01434440.099612520.0001063-0.32055i-0.282708-0.268418i0.228011-0.218445i30.0001063+0.32055i-0.282708+0.268418i0.228011+0.218445i40.011217-0.256745i0.209228-0.145614i0.130454+0.129199i50.011217+0.256745i0.209228+0.145614i0.130454-0.129199i60.0163316-0.559761i0.029079+0.313876i-0.536719+0.0142575i70.0163316+0.559761i0.029079-0.313876i-0.536719-0.0142575i80.03599460.0912780.0836409According to (3), their eight corresponding orientations are showed in Table 2.Table 2
The eight corresponding orientations (Unit : Degree).
αβγ111.13742.65279-10.3294219.7905-14.0228i-26.5992-24.9282i-7.64571-39.2758i319.7905+14.0228i-26.5992+24.9282i-7.64571+39.2758i412.3115 + 8.6638i21.2364-13.6688i4.71965-29.9883i512.3115-8.6638i21.2364+13.6688i4.71965+29.9883i6-55.4203-0.91328i3.87121+1.79706i0.70938-73.4251i7-55.4203+0.91328i3.87121-1.79706i0.70938+73.4251i810105Next we will demonstrate the validity of the analytic forward kinematics by the numerical example of inverse kinematics. Let the initial orientations of the scale model be zero, the corresponding vertices vectors of the moving platform area10=[0.6,0.3,0.2]T, a20=[0.6,0.3,-0.2]T, and a30=[-0.8,0.1,0]T. When the moving platform has a posture of α=100,β=100, andγ=50, the transformation matrix expressed by means of Euler angles can be showed as following
(27)R=[cαcβcαsβsγ-sαcγcαsβcγ+sαsγsαcβsαsβsγ+cαcγsαsβcγ-cαsγ-sβcβsγcβcγ]=[0.969846-0.1580830.1854940.171010.983688-0.0557927-0.1736480.08583170.98106],
where cα=cosα, sα=sinα, cβ=cosβ, sβ=sinβ, cγ=cosγ, sγ=sinγWith the help ofR, we can calculate the corresponding vertices vectors of the moving platform in the base frame
(28)ai=Rai0(i=1,2,3).From the above equation, we havea1=[0.571581674,0.386554,0.11777264]T,a2=[0.497384168,0.408871083,-0.27465146]T, and a3=[-0.791685337,-0.038439224,0.14750171]T.So the lengths of the three cables are the norms of vectorsaibi, a2b2, a2b3, that is,  L1=1.789090488, L2=1.724702626, and L3=1.77252834.The inverse kinematic results are the same with the initial structural parameters of the forward kinematics, which verifies that the analysis of forward kinematics is correct.


## 5. Conclusion

This paper presents the kinematics analysis method of a novel 3-DOF cable-driven parallel mechanism used in wind tunnel test, and then we employ elimination method to solve the analytic forward kinematics of the mechanism by using of Rodridgues’ parameters, so an eighth polynomial in one variable is derived finally. A numerical example is included to verify the effectiveness and accuracy of the developed algorithm for real-time computation and control.



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