RSSR Mechanism with Constant Transmission Angle

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1 華岡工程學報第 1 7 期中國文化大學工學院民國 9 2 年 6 月第頁 HUA KANG JO URNAL OF ENGINEERING COLLE GE OF ENGINEERING CHIN ESE CULTURE UNIVERSITY Jun e,2003,vo1.17,p.69-p.78 固定傳力角之 RS SR 機構 RSSR Mechanism with Constant Transmission Angle 鐘文遠 Wen-Yeuan Chung 中國文化大學機械工程學系 Department of Mechanical Engineering, Chinese Culture Dni. Abstract The existence of the RSSR linkage with constant transmission angle is investigated. The transmission angle and other parameters of the linkage are related by an equation of second degree. The curve describ ed by this equation is then compared to a unit circle to find the existence of the RSSR linkage with constant transmission angle. The synthesis of such a linkage is also discuss ed. Keywords: Transmission Angle, SYnthesis, Linkage 摘要本文對具有固定傳力角之 RS S R 連桿組的存在作探討首先, 傳力角和連桿組之幾何尺寸以一個二次方程式加以表示再將該二次方程式與一單位圓加以比較, 以驗證固定傳力角之 R S SR 連桿組之存在隨後探討該固定傳力角連桿組的合成關鍵語 : 傳力角, 合成, 連桿組 69

2 I. Introduction The transmission angle is defined as the angle between the vector along the coupler link and the velocity vector of the end point of the output link. The cosine value ofthis angle also represents the ratio ofthe force component that acts on the output link to rotate the link to the force exerted by the coupler link. In designing a linkage, we usually keep the transmission angle small and almost constant to get a good force transmission performance, The transmission angle is a significant index in synthesizing a linkage. When an RSSR mechanism is used as a function generator, the objective function to be optimized may include both transmission ratio and the error [1][ 2]. For a skew crank-and-rocker linkage, design charts showing the variation of the transmission angle with respect to the specified swing angles were dev eloped [3]. On the other hand, the transmission angle may also be used to analyze the mobility of the linkage [4]. The extreme values of the transmission angle of RSSR linkage were investigated [5]. Rec ently, the existence of the RSSR linkage with constant or partially constant transmission angle received attention [6][7]. Both previous works began with differentiating the transmission angle with resp ect to the input angle. In this work, fully rotatable RSSR linkage with constant transmission angle is investigated by simply comparing a planar curve of two degree to a unit circle. The synthesis of a nontrivial case is also discussed. II. General RSSR Linkage A general RSSR linkage spherical joints. Both A o and X-Y-Z and x-y-z frames. The input link AoABB o is shown in Figure 1. A and Bare B o are revolute joints and also the origin for AoA, with length a, rotates along Z axis. The output link BoB, with length c, rotates along z axis. The line PJP Z is the common perpendicular to Z axis and z axis. The skewed angle and distance between both Z and z axes are a and d. The length of the coupler link AB is defined as b. The distances of offsets are AoP J = g and BoPz = h. Both axes Y and yare parallel to the line P 1P2. The input angl e 8 is measured counterclockwise from the positive X-axis to the input link. The angl e ~ is the output angle measured from the positive x-axis to the output link. 70

3 z x Figure 1. A general RSSR linkage. The coordinates of points A o and A shown in Figure 1 can be listed as A o (0,0,0) and A(a cos 8, asin8, 0) with respect to the X-Y-Z frame. The coordinate of points Bo and Bare Bo(O,O,O) and B(ccos~,csin~,O) with respect to the x-y-z frame. For the purpose of analysis, the coordinate of point A can be represented with respect to the x-y-z frame as A: (a cos o cos O- gsina, asin 8 - d, a sin acos8 + g cosce - h). The input-output equation can then be derived from the length of the coupler link, that is -AB 1 2 = b 2,and this leads to ' E cos ~ + F siu] + G = 0 (1) where E = 2c(gsina - acos8cosa) F=2c(d-asin8) G = a 2 - b 2 + c 2 + d 2 + g2 + h 2-2ah cos8 sin a - 2adsin8-2ghcosa (2) III. Transmission Angle The transmission angle 8 is defined as the angle between the vector Fe ' along the coupler link, and VB' the velocity vector of the point B. Both vectors Fe and VB are shown in Figure 2. The vector of VB is along the direction of (+ sin ~, ±coso, 0), where the sign of "-" or "+" is dependent on the rotating direction of the output link. The direction of the vector Fe can be found from vector AB. As a result, the angle 8 can be derived by taking inner product of 71

4 both vectors Fe and VB and is represented as bc cos S = ±accosssin~cosa + cg sin o.sin ~ + acsin Scos~± dc cos o (3) By making use ofequations (2), Equation (3) can also be rewritten as bccoso=±f cos~+esin~ (4) 2 2 To avoid the sign of ± or +, this equation can be squared and modified as 4b 2c2 cos 2 0 = F 2 cos 2 ~ + E 2 sin 2 ~ - 2EFcos~sin ~ (5) By utilizing Equation (1), we may also derive that E 2 cos 2 ~ + F 2 sin 2 ~ + 2EFcos~sin~ = G 2 (6) Therefore, from both Equations (5) and (6), the angle 0 can then be expressed as 4b 2c2 cos 2 0 = E 2 + F 2 - G 2 (7) This equation means that the transmission angle is a function of the input angle and the dimension of the linkage. On the other hand, the RSSR linkage usually has two branches when an input angle S is specified. If both branches are marked as AoABB o and AoAB'B o' dyads ABB o and AB'B o are symmetrical to a plane that includes point A and the rotating axis of the output link. Hence, the values oftransmission angle for both branches are the same. A Figure 2. Illustration ofthe transmission angle o. IV. Existence oflinkage with Constant Transmission Angle The investigation of the existence of constant transmission angle is based on. Equation (7). By referring to Equation (2), Equation (7) can be expanded as where Csssin2S+CscsinScosS+Ccccos2S+CssinS+CccosS+Co =0 (8) C = 4a 2 d 2-4a 2 c 2 ss Csc = 8a 2dhsina 72

5 C cc = 4a 2 h 2 sin 2 a - 4a 2c2 cos 2 a C, = - 4adH + 8ac 2d C, = -4ahH sina + 8ac 2 gsinacosa Co = H 2 + 4b 2 c 2 cos 2 8-4c 2g2 sin 2 a - 4c 2 d 2 H=a2-b2 +c2+d2+ g 2 +h2-2ghcosa (9) By setting ~ = cos e and y = sin e, Equation (8) can be treated as an equation of the second degree and may describe a hyperbola, parabola, ellipse or even circle on the ~ - y plane. When the dimensions of the linkage are given and a transmission angle 8* has been specified, the curv e describ ed by Equation (8) can be plotted. If this curve intersects a unit circle with the origin as the center, that is ~ 2 + / = 1, at some points, this means that the linkage can reach the specified transmission angle 8*. The corresponding input angles e of the linkage can be found in two ways. One is to find graphically from the intersection points of both unit circle and the curve. The other analytical method is to solve Equation (8) directly. A special case worth mentioned is when the curve describes exactly the same as the unit circle, ~ 2 + / = 1. This special case represents that the transmission angles of the linkage equal to the speci fied value 8* for all input angles e. In other words, the valu es of the transmission angle of this linkage are a constant. The existence of the linkage with constant transmission angle can be investigated by anal yzing the coefficients described in Equation (9). When Equation (8) describ es a unit circle, the necessary conditions can be listed as Csc = 0 (10) C, = C c =0 (11) C ss = C cc = -Co (12) To satisfy Equation (10), this leads to three different categories: d = 0, a = 0, and h = O. Further, both C, and C, equal to zero if a 2 _ b 2 _ c 2 +d 2 + g2 + h 2-2gh cosa = 0 h - gcosa = 0 In order to satisfy Equation (12), two equations can be derived as (13-1) (13-2) (c 2+h2)sin 2a=d2 (14-1) a 2h2 sin 2 a - a 2c2 cos 2 a + c 4 + b 2 c 2 cos 2 8 _c 2 g 2 sin 2 a _c 2 d 2 = 0 (14-2) Therefore, if the parameters of an RSSR linkage can satisfy all Equations (10), (13), and (14), the transmission angle of this linkage is a constant. Several cases and possibilities will be discussed and analyz ed as follows. Category 1: When d = O. From Equation (14-1), we got two possibilities. One is when d = c = h = 0 and it is evidently a trivial case. The other one is when d = 0 and a = O. By looking at Equations (13), both a 2 - b 2 _ c 2 = 0 and 73

6 h = g can be derived. As a result, both rotating axes of input and output links are coincident and both points A o and B, are the same. This linkage is just a right triangle i1a oab or i1b oab, and is also a trivial case. Category 2: When a = O. The length d has to be zero in order to satisfy Equation (14-1). Thus, this case is the same as just discussed and is a trivial one. Category 3: When h = O. Either a = 90 or g = 0 can be derived from Equation (13-2). For h = 0 and a = 90, the constraint equations are c = d and a 2 - b 2 + g 2 = 0. The configuration of this RSSR linkage is shown in Figure 3. It is obvious that i1a oab always keeps as a right triangle and the point B is stationary and lies on the Z-axis when the linkage moves. Thus, this case can be treated as another trivial one. z (~ ) x g y z Figure 3. A trivial case ofrssr linkage. The last case to be discussed is when h = 0 and g = O. From Equations (13) and (14), the other constraint equations are (15) (16) This case is a nontrivial one. The linkage is a simply skewed four-bar and will be analyzed in the next section. Although equations similar to Equations (15) and (16) were also partly derived in previous works [6][7], the methods of approach are different and the derivation of the present paper is more explicit than previous 74 (17)

7 works. Furthermore, both previou s works didn' t mention the relationship between the value of the constant transmission angle and other parameters of the linkage, such as Equ ation (17). V. Analysis of Linkage with Constant Transmission Angle When the parameters of a simply skewed linkage can satisfy both Equations (15) and (16), the transmi ssion angle is a constant and can be calculated by using Equation (17). In order to synth esize this linkage efficiently, furth er analysis would be conducted. By adding Equations (15) and (16), this leads to This Equation tells that a, the length of the input link, is always longer than b, the length ofthe coupler link. Ifwe add both Equations (15) and (17), we may get (18) a 2 sin 2 a=b 2sin 2 0 (19) By looking at Equ ations (18) and (19), the value of the transmission angle 101 should be always larger than lal, that is 101 ~ lal. In synthesizing this special linkage, one may prefer to assign the values of the angle a and the transmission angle 0. Then, the relationship between the lengths of the frame link and other links can be found from Equations (15) - (17) as a d b Isin cos a l = '--,=== ==== Isinal~sin 2 -sin 2 a Icos al =-;============ d ~sin 2 -sin 2 a c 1 d - Isin al From the ratio of link lengths listed in Equation (20), the longest link is the input link and the frame link is the shortest one. Further, the Grasho f's rule can be prov ed to be satisfi ed. Therefore, this special linkage is certainly a drag link [8]. The ratio of link lengths a / d and b / d are also plotted as shown in Figures 4 and 5, respectively. The abscissa and ordinate are the skewed angle a and the transmission angle 0. The ratio of c / d depends on the angle a only as describ ed in Equation (16). The lengths of both input and coupler links approach infinite if the transmission angle is close to the angle a. Besides, the length ofthe input link is infinite if a is zero or for a planar four-bar. As a result, to get a linkage with acceptabl e constant transmission angle and reasonable link lengths, it is recommended that the angles a and are chos en to be within 75 (20)

8 oo ) Ang'e o.(deg) Figure 4. Ratio of link lengths a / d so ,, '.....-:'. ",/ ~ : ~.l.'.< ",:'Y 1 - " c " b:'.?<.> :/ :.,,,i "~o ld.»-:'/ ' " ' 0 o 0...>";':., / /. 1 ~ : 1d : /1> ~ I",,,./ s" ~ W 70 W 90 Ang!9ct {dag J Figure 5. Ratio of link lengths b / d A numerical example is given when a = 30 and () = 40, the lengths of the links are found by using Equation (20) as a = d, b = d, and c = Zd, The input angle versus the output angle is also plotted as in Figure 6. This linkage is a drag link as expected. The values of the transmission angle can be checked to be a constant 40 interval and for both branches. A=2.76 8=2.14 C=2 0 =1 G=O H=O a=30deg, for the whole rotation 350 0> '" ẉ.j (!) :!e. 250 ~ 200 f ::> go i/ ' INTPlIT ANGLE (deg) / ~ OO Figure 6. Output angle versus input angle of example corresponding to constant transmission angle 8=40. 76

9 VI. Conclusion A linkage with constant transmission angle has been found. This linkage is a simply skewed-four bar with both offsets being zero. There are five parameters subjecting to three equalities in synthesizing this linkage. Three of the parameters are ratios of link lengths, and the other two are the skewed angle as well as the desired value of the constant transmission angle. As a result, two degrees of freedom or parameters are left and can be chosen by the designers. Furthermore, the value of the transmission angle is always larger than that of the skewed angle for this special linkage. 77

10 Reference 1. Gupta, K. and Kazerounian, S., "Synthesis of Fully Rotatable R-S-S-R Linkages, " Mechanism and Machine Theory, Vol. 18, No.3, pp , Sutherland, G. and Roth, B., "A Transmission Index for Spatial Mechanisms," Transactions of the ASME, Journal of Engineering for Industry, Vol. 98, pp , Soylemez, E. and Freudenstein, F, "Transmission Optimization of Spatial 4-Link," Mechanism and Machine Theory, Vol. 17, No. 4, pp , Kazerounian, K. and Solecki, R., "Mobility Ana lysis of General Bi-Modal Four-Bar Linkages Based on Their Transmission Angle," Mechanism and Machine Theory, Vol. 28, No.3, pp , Zhang, W., "The Limit Values of Pressure Angle and Positions of Maximal Pressure Angle ofthe Spatial Linkage," Mechanism and Machine Theory, Vol.. 28, No.4, pp , Zhan g, W., "On the Finding of the Spatial Linkage with Constant Pressure Angle," Mechanism and Machine Theory, Vol. 32, No.8, pp , Saka, Z, "The RSSR Mechanisms with Partially Constant Transmission Angle," Mechanism and Machine Theory, Vol. 31, No.6, pp , Chung, W., "Mobility Analysis of RSSR Mechanisms by Working Volume," DETC 2001/DAC-21045, ASME 2001 Design Engineering Technical Conferences,

0 0 = 1 0 = 0 1 = = 1 1 = 0 0 = 1

0 0 = 1 0 = 0 1 = = 1 1 = 0 0 = 1 0 0 = 1 0 = 0 1 = 0 1 1 = 1 1 = 0 0 = 1 : = {0, 1} : 3 (,, ) = + (,, ) = + + (, ) = + (,,, ) = ( + )( + ) + ( + )( + ) + = + = = + + = + = ( + ) + = + ( + ) () = () ( + ) = + + = ( + )( + ) + = = + 0