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A new class of rocker-belt mechanisms
Mechanism and Machine Theory 40 (2005) 963–976
Mechanism and Machine Theory www.elsevier.com/locate/mechmt
A new class of rocker-belt mechanisms Paolo Gallina
*
Department of Energetics, University of Trieste, Via A. Valerio 10-34127 Trieste, Italy Accepted 16 December 2004 Available online 11 March 2005
Abstract This paper describes a new rocker-belt mechanism. The mechanism is made up of a rocker, a timing belt that operates the rocker, a series of sprockets and a curved element. The latter, which is referred to as profile, represents the heart of the mechanism. One point of the belt is attached to a point on the rocker. The belt is wound around the sprockets and the profile. If the profile is properly shaped, for a given range of rotation angles of the rocker, the belt does not become slack. This work shows how to design such a profile; it turns out that the profile shape, which is related to several geometric parameters, is expressed by means of a non-linear differential equation. Eventually, the proposed approach is validated by an example. 2005 Elsevier Ltd. All rights reserved. Keywords: Rocker mechanism; Belt transmission
1. Introduction The motion of a four-bar linkage is often characterized by the term crank rocker to indicate that the crank rotates completely and the other link oscillates. In a similar manner, the term double rocker indicates that both links oscillate [1]. A rocker can be present in many other planar mechanisms or it can be itself a simple mechanism. In the following of the paper, the term rocker will indicate a rigid body which can rotate around a fixed point; the rotation angle is limited.
*
Tel.: +39 040 5582540; fax: +39 040 568469. E-mail address: [email protected]
0094-114X/$ - see front matter 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.mechmachtheory.2004.12.018
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In many applications, the rocker is the driving link of the linkage mechanism [2]. In this case it is operated by means of an actuator trough a mechanical transmission system. There exist different combinations of actuators and transmissions. For sake of simplicity, in order to stress the innovation and the advantages claimed by the rocker-belt mechanism proposed in this paper, three typical different solutions of actuated rockers are discussed (see Fig. 1). In configuration (a), the rocker is fixed to a spur gear. Therefore, the rocker can rotate around the a–a axis which is fixed with respect to the reference frame. The rocker is driven by an electric motor. The transmission-reducer module is made up of a pinion (r1 = pinion radius) driving the external gear (r2 = gear radius). As the matter of fact, more sophisticated reducers could be employed, but the substance of physical considerations would not change. It is assumed that the rocker is loaded by a force F acting at its free end, perpendicular to the rocker axis. In first analysis, the force the pinion exerts on the gear tooth is F t ¼ ðlF Þ=ðr2 cosð/ÞÞ
ð1Þ
where l is the rocker length and / is the pressure angle of the two gears in mesh.
ROCKER GEAR
A
REDUCER
F
F
a
a SPROCKET 2
ELETRIC MOTOR
SPROCKET 1
(a)
(b)
A
D
B
σ
F
σ
F
C E (c)
(d)
Fig. 1. Three different solutions to drive the rocker.
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In configuration (b), the rocker is also driven by an electric motor. The transmission-reducer module is made up of a timing belt looped around two sprockets. (r1 = sprocket 1 radius, r2 = sprocket 2 radius). As in case (a), it is assumed that the rocker is loaded by a force F at its free end, perpendicular to the rocker axis. In first analysis, the belt tension is T t ¼ ðl=r2 ÞF
ð2Þ
Both these two typical solutions present the following drawbacks: • the force between the teeth in contact (case (a)) and the belt tension (case (b)) are functions of the rocker length and of the driven sprocket radius: the higher is l and the lower is r2, the higher is the force/tension; • configuration (a) may exhibits backlash; • configuration (b) is cumbersome; • the rocker tends to bend because of the force F; in fact it behaves as a clamped-free beam These disadvantages could be overcome by employing the solution (c). In this case, two belts (or equivalently, two ropes or two chains) and two motors are required. Both belts have one end fixed to the free end of the rocker, while the other end is wound around a drum. As the matter of fact, ropes or chains are more suited to be wound around a drum, but from a kinematic point of view, ropes, belts or chains are equivalent. In fact, what is relevant to our analysis here, are the tensions and the mechanism geometry. For sake of simplicity, let us focus on a solution that employs belts. It is reminded that results we obtain are valid also for rope or chain transmissions. The belt tension (of the lower belt) is T ¼ ð1= sinðrÞÞF
ð3Þ
According to the load direction (force F direction), for the static equilibrium of the mechanism, one belt is tensioned, while the tension on the other one is null. The tensioned belt is referred to as loaded belt. r is the angle between the rocker axis and the loaded belt. r is a function of the rocker angular position. Comparing Eqs. (1) and (2) with (3), it can be inferred that, unlike solutions (a) and (b), in solution (c), the belt tension is not a function of the rocker length. As a consequence, even if the rocker length is long, the belt tension remains low for a given range of the rockerÕs angular position. This statement has to be considered very carefully in order to avoid misunderstandings. Let us assume to employ a longer rocker in order to produce the same force F at its free end. Besides assume that the two motors are located with respect to the rocker axis in order to maintain original proportions. In other words, let as focus on a scale enlarged mechanism. Although such a mechanism will have a longer rocker, the tension on the belt necessary to balance the same force F does not change, since it depends on the geometry of the whole mechanism and not on its dimensions. This is the reason why the rocker length r does not appear into Eq. (3). However, it is remarked that the whole torque the rocker can exert with respect to the rocker axis is a increasing function of the rocker axis: C = rF. Note that no backslash is present. Moreover for some applications the rocker bending is mitigated by the belt. Also this last statement needs to be clarified with an example. Let as assume
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that the rocker is part of a complete mechanism such as a four-bar linkage. Cases (a) and (c) are considered. If the four-bar linkage is employed to operate a general machine, it is subjected to a set of load forces which generate a force on the free end of the rocker where the revolute joint is located. Because of this force, in case (a) the rocker could bend according to its compliance. On the contrary, in case (c) this force is balanced by the belt tension and by the rocker which produces a reaction force oriented along its longitudinal axis. As a consequence, no bending moment acts on the rocker. As the matter of fact, the bending phenomenon depends on the particular application the rocker is employed for. For example, if the rocker itself constitutes the mechanism and it is used to balance a load torque applied to the rocker axis, solutions (a) and (c) show the same behaviour, as far as bending phenomenon is concerned. Of course, although this last solution presents many advantages, it should be avoided because it requires two motors for just 1 d.o.f. A simple way to eliminate one motor, without abandoning the benefits introduced by solution (c) is given by scheme (d). In fact, in (d) only one motor is employed. In this case, there is only one timing belt looped around two sprockets and the rocker. One point of the belt is fixed to the free end of the rocker. The second motor has been replaced with the sprocket A. Belt tensioners B and C play the role of keeping the belt always tight. They consist in a spring-like mechanism to push against the belt by means of a tightening pulley. The main drawback of solution (d) is that it shows backlash. In fact, when the force F is oriented as in figure, the belt segment ED tends to straighten while segment AD is bent because of the tensioner; conversely, when F assumes the opposite direction, the belt segment AD tends to straighten. As a consequence, the rocker rotates according to the load direction F even if the motor is stopped. In order to improve solution (d), in the next section a new mechanism is presented. Whit respect to the previous solution, it does not show backlash. In conclusion, the proposed solution has all the benefits given by solution (c), but it requires just one motor to operate the mechanism. Such a mechanism can represent an effective solution in all that applications where a rocker is required to operate a mechanism such as a four-bar linkage or the rocker itself has a mechanical function. For example it could be employed to position launching pad, orient an antenna or modify the mechanical component tilt. After the description of the mathematical model, an example to validate the theory is given.
2. Profile shape The proposed mechanism is sketched in Fig. 2. It is made up of the rocker, a belt that operates the rocker, two sprockets and a curved element attached to the free end of the rocker (point pE), which is referred to as profile. The rocker and the profile constitute a rigid body. One side of the belt is directly attached to the point pE of the rocker. The other side of the belt is first wound \ around the profile (along the arc pT pE ) and then fixed to the point pE. pT is the tangent point of the belt to the profile. Sprockets can rotate around the point pi and pj which are fixed with respect to the absolute reference frame (XY). The rocker can rotate around the point O, which is assumed to be the frame origin. a is the rocker angular position.
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p iT
pi
Tu
vT F
pT
PROFILE
y
CT
x
pE
Y
α
O
X Tl
pj p jT
Fig. 2. The proposed rocker-belt mechanism.
If the profile is properly shaped, for a limited range of rocker rotation angles, the belt does not become slack. In other words, the profile plays the role of the tensioners B and C in the solution (d) of Fig. 1. A xy local reference frame is introduced, where the unit vector x is oriented as in figure and the origin of the local frame is located at the free end of the rocker pE. pT is the tangent point of the upper side of the belt to the profile. piT and pjT are the tangent points of the two sides of the belt to the sprockets. Sprocket radii are assumed to be low enough to approximate sprocket tangent points with sprocket centres: piT ffi pi ¼ f xi y i gT and pjT ffi pj ¼ f xj y j gT . Let us assume that the tangent point pT is a function of the rocker angular position a fpT gl ¼ f xðaÞ
T
yðaÞ g
ð4Þ
where the symbol {*}l indicates that the co-ordinates of the point pT are expressed with respect to the local reference frame. x(a) and y(a) are two scalar functions. The aim of this mathematical treatment consists in finding the two function x(a), y(a) so that to prevent the belt from going slack for every angular position a. The point pT expressed with respect to the absolute frame is r xðaÞ cosðaÞ sinðaÞ r xðaÞ ¼ ð5Þ pT ¼ RðaÞ yðaÞ sinðaÞ cosðaÞ yðaÞ where R(a) is a rotation matrix and r is the rocker length. The derivatives of functions x(a) y(a), represent the components of a tangent vector to the profile with respect to the local frame fvT gl ¼ f dxðaÞ=da dyðaÞ=da gT . The same vector, expressed with respect to the absolute frame is
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8 9 dxðaÞ > > < = da vT ¼ RðaÞ ð6Þ > : dyðaÞ > ; da Note that the vector vT is not a unit vector. The goal functions x(a) and y(a), which define the profile shape, will be obtained by numerically solving a system made up of two equations. The two equations come from two constraints the profile geometry has to fulfil: 1. First of all, the belt has to be always tangent to the profile; 2. As the rocker angle a changes, the belt has to remain tight. 2.1. First constrain: belt tangent to the profile The first mathematical constrain comes from the following consideration: the belt segment pipT has to be always tangent to the profile at the point pT. Mathematically, this consideration can be expressed by the relationship pT pi ¼ kvT
ð7Þ
where k is a scalar number. Note that the vector vT is oriented in such a way thatk is a negative number. In other words, it is required that the tangent vector vT be aligned to the upper segment of the belt, where the segment is represented by the vector pTpi. By replacing Eqs. (5) and (6) into (7) and eliminating the parameter k by simple mathematical simplifications, one gets the relationship between the derivative of y with respect to a and the derivative of x with respect to a dy dx ¼ Cðx; y; aÞ ð8Þ da da where Cðx; y; aÞ ¼
y y i cosðaÞ þ xi sinðaÞ xi cosðaÞ þ y i sinðaÞ þ x r
ð9Þ
2.2. Second constrain: the belt remains tight The second mathematical constrain comes form the belt. It is assumed that the belt length is \ constant; in order to remain always tight, the sum of the segments pTpi, pEpj and the arc pT pE has to remain constant for every value of a sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Z a 2 2 dx dy da ¼ lk þ ð10Þ kpT pi k þ kpE pj k þ d a da 0 The symbol kvk = vTv indicates the length of the vector v. The integral term represents the length \ of the arc pT pE , while lk is a constant.
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The free end of the rocker pE expressed with respect to the absolute frame is 0 gT
ð11Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi T T Considering that kpT pi k ¼ ðpT pi Þ ðpT pi Þ and kpE pj k ¼ ðpE pj Þ ðpE pj Þ the derivative of Eq. (10) with respects to a leads to ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi s 2 2 T ðpE pj ÞT dpdaE ðpT pi Þ dpdaT dx dy qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ ¼0 ð12Þ T da da ðpE pj Þ ðpE pj Þ ðpT pi ÞT ðpT pi Þ pE ¼ RðaÞf r
where
( ) dx da dpT dRðaÞ r x þ RðaÞ dy ¼ da da y da sinðaÞ dpE ¼r da cosðaÞ
ð13Þ
dy dx ; y_ ¼ da and R_ ¼ dRðaÞ , it yields Replacing Eq. (13) into (12) and using the notation x_ ¼ da da qffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðpE pj ÞT dpdaE _x ðpT pi ÞT _ r x þ x_ 2 þ y_ 2 ¼ 0 R þR þ kpT pi k kpE pj k y y
It is shown now that some terms of Eq. (14) are null. In fact, let us consider the term qffiffiffiffiffiffiffiffiffiffiffiffiffiffi T _x ðpT pi Þ n¼ þ x_ 2 þ y_ 2 R kpT pi k y
ð14Þ
ð15Þ
By replacing Eqs. (6) and (7) into the term (15), it yields T
n¼
_x ðpT pi Þ R kpT pi k y
qffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi _x ðkvT ÞT þ x_ 2 þ y_ 2 ¼ R þ x_ 2 þ y_ 2 ¼ kpT pi k y
T _x qffiffiffiffiffiffiffiffiffiffiffiffiffiffi kR _x y_ R þ x_ 2 þ y_ 2 _ x y kR y_
ð16Þ
Exploiting the fact that k is a negative number, the term (15) becomes
n¼
k kkk
f _x
_x y_ gRT R qffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi k x_ 2 þ y_ 2 k y 2 2 2 2 2 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ x_ þ y_ ¼ x_ þ y_ þ1 ¼0 þ x_ þ y_ ¼ kkk x_ 2 þ y_ 2 kkk R _x y_ ð17Þ
As a consequence, Eq. (14) can be simplified into the following ðpE pj ÞT dpdaE ðpT pi ÞT _ r x R þ ¼0 kpT pi k kpE pj k y
ð18Þ
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In the Appendix A, it will be shown that Eq. (18) is equivalent to an algebric nonlinear elliptical equation of the form Uðx; y; aÞ ¼ 0
ð19Þ
3. Numeric solution of the profile shape The system to solve is made up of the Eqs. (8) and (19) dy dx ¼ Cðx; y; aÞ da da Uðx; y; aÞ ¼ 0
ð20Þ
A numeric solution can be found. In order to find a solution, the initial values of x and y are given: x0, y0. They represent the initial conditions of system (20). Note that the first equation of the system is a nonlinear first order differential equation, while the second one is an algebraic non linear equation in the tree variables x,y and a. This equation can be represented in an explicit form a ¼ /ðx; yÞ
ð21Þ
In other words, /(x,y) represent a root of Eq. (19), when x and y are given. Such a root can be carried out by an iterative numeric solution. Eventually, the iterative method to solve system (20) is given by the following iteration formula xnþ1 ¼ xn þ dx anþ1 ¼ /ðxn ; y n Þ
ð22Þ
y nþ1 ¼ y n þ Cðxn ; y n ; anþ1 Þdx This is not the only way to solve system (20), but it has been employed for its simplicity and effectiveness. In the Section 5, this iterative method will be employed to solve a real case. 4. Belt tensions This section deals with the force analysis of the mechanism. The belt transmission has the function to move the rocker against a resistive load. The resistive load can be given by a torque CT acting on the rocker. If the torque CT is assumed to be positive when counterclockwise oriented. Note that, if CT > 0, the lower side of the belt has to be tensioned in order to balance the torque, while the tension in the upper side of the belt is null. Conversely, if CT < 0, it is the upper side of the belt to be tensioned, while the tension on the lower side of the belt is null. In the former case, the equilibrium is given by the relationship * +! pE pj pE ; ð23Þ C T ¼ T l r sin kpE k kpE pj k p p
where Tl is the lower side belt tension. kppE k and kpE pj k are unit vectors that represent respectively E E j p p the rocker orientation and the lower side belt orientation, while the symbol hkppE k ; kpE pj ki represents E
E
j
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the angle between the two unit vectors, or in other words, the angle between the lower side belt and the rocker. Instead, the tension of the upper side belt Tu is null. In case CT < 0, the equilibrium is given by the relationship
pT pT pi C T ¼ T u kpT k sin ; ð24Þ kpT k kpT pi k where Tu is the upper side belt tension. In conclusion the belt tensions are Tl ¼
CT 1 D E pE pj pE r sin ; kp k kp p k E
E
8C T > 0 ð25Þ
j
Tu ¼ 0 and Tl ¼ 0 Tu ¼
CT 1 D E pT pi pT kpT k sin ; kp k kp p k T
T
8C T < 0
ð26Þ
i
Note that the tension changes with respect to the rocker angular position. Therefore, it is necessary to check the belt tensions versus the rocker angular position a in order to prevent the tension from reaching dangerous values.
5. Example In the proposed example, the centres of the two sprockets are set to pi ¼ f 0:1 0:1 gm and pj ¼ f 0 0:2 gm, and the rocker length to r = 0.3m. The increment employed in the relationships (22) has been set to dx = 0.001m. The initial conditions are x0 = 0 and y0 = 0.The integration of system (20) by means to the iterative scheme (22), leads to the solution represented in Fig. 3. The plot represents the y co-ordinate of a profile point versus its x co-ordinate, where x and y are the point co-ordinates with respect to the local frame. The curve goes trough the frame origin since the initial conditions were x0 = 0 and y0 = 0. By employing different initial conditions, one could get other profile shapes. From the iterative formula (22), it can be inferred that, at each iterative step, the co-ordinate x is incremented by a constant value dx. Then the new value of a is calculated followed by the new value of y. As a consequence the x co-ordinate can be thought of as the independent variable. Therefore, it is natural to show the plot of y versus x (Fig. 3) as well as the plot of a versus x (Fig. 4). Note that, a decreases as x increases. There exists an upper limit xmax for the co-ordinate x. In fact, if the value x during the iteration process overcomes xmax, Eq. (19) does not admit a solution. As a consequence, the profile has a limited length and the rocker has a limited rotation angle Da. The upper limit xmax depends on the
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P. Gallina / Mechanism and Machine Theory 40 (2005) 963–976 0.05 0.045 0.04 0.035
y [m]
0.03 0.025 0.02 0.015 0.01 0.005 0 0
0.01
0.02
0.03
0.04
0.05 x [m]
0.06
0.07
0.08
0.09
0.1
Fig. 3. Profile contour described in terms of y versus x.
-0.4
-0.5
-0.6
α [rad]
-0.7
-0.8
-0.9
-1
-1.1
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
x [m]
Fig. 4. Rocker position a versus x.
points pi, pj, the rocker length r and on the initial conditions x0 and y0. In the proposed example, xmax has been calculated by means of a trial and error technique.
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In particular it has been set xmax = xj, where xj fulfils the following requirement: Eq. (19) admits a solution for xj and it does not admit a solution for xj + dx. As it can be read from Fig. 3, xmax = 0.098 while, from Fig. 4, the maximum rocker rotation angle is Da = k1.15 (0.54)k = 0.61rad. Fig. 5 shows 5 different kinematic configurations of the rocker mechanism. Thick solid line represents the rocker and the profile. Thin solid lines represent the upper and lower side of the belt. The belt piece that connects the two sprockets is not represented for sake of clarity. In the initial configuration (configuration 1), the tangent point of the upper side of the belt to the profile pT coincides with the free end of the rocker pE. In fact, this kinematic configuration corresponds to the case x = 0 and y = 0. As the co-ordinate x increases, the tangent point pT moves along the profile, and at the same time, the rocker clockwise rotates. In configuration (5) the point pT reaches the free end of the profile. Finally, Fig. 6 represents the belt tensions versus the rocker angular position. In particular, solid line represents the normalized upper side belt tension Tu/jCT/rj versus a when CT < 0; dashed line represents the normalized lower side belt tension Tl/jCT/rj versus a when CT > 0. It can be inferred that the lower side belt tension is poorly affected by the rocker angular position. Conversely, the upper side belt tension decreases as the rocker rotates clockwise. Moreover, for the considered range of angular positions, the normalized upper side belt tension is always greater of the lower side belt tension.
Fig. 5. Five different rocker angular positions; thick solid line represents the rocker and the profile in 5 different configurations; thin solid lines represent the upper and lower side of the belt. The belt piece that connects the two pulley is not represented for sake of clarity.
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P. Gallina / Mechanism and Machine Theory 40 (2005) 963–976 2.8
2.6
2.4
2.2
2
1.8
1.6
1.4 -1 .1
-1
-0 .9
-0 .8
-0 .7
-0 .6
-0 .5
-0 .4
α [rad]
Fig. 6. Solid line represents the normalized upper side belt tension jjCTTu=rjj with versus a when CT < 0; dashed line represents the normalized lower side belt tension jjCTT l=rjj with versus a when CT > 0.
6. Conclusion This paper has introduced, described and validated a new mechanism to operate a crank. To the best of our knowledge, such a mechanism has never described before in literature. This solution show many advantages with respect to similar ones. Among others, the belt tension is not a function of the rocker length, no backslash, only one motor is required. Even if the mechanism is made up by few mechanical components, the mathematical description of the profile requires the solution of a first order non-linear differential equation. For a given example, such an equation has been solved numerically.
Appendix A Eq. (18) can be expressed as a function of x and y. By expanding the terms pT,pE and gets ( ) ( ) ) !T !T ( rx cosðaÞ sinðaÞ ( ) pi pj r R r rx y sinðaÞ cosðaÞ _R þ ¼0 kpE pj k kpT pi k y
dpE , da
one
ðA:1Þ
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The second factor can be simplified ( ) ( ) !T rx sinðaÞ ( ) rpTj pi R ðA:2Þ y cosðaÞ rx R_ ¼0 kpE pj k kpT pi k y 0 rx 0 1 T _ T _ Since R R ¼ is null. As a consequence, Eq. (A.2) , the term f r x y gR R y 1 0 becomes !T ( ) ( ) sinðaÞ rx ( ) rpTj R pi rx cosðaÞ y R_ ¼ kpT pi k kpE pj k y ( ) ( ) ( ) sinðaÞ r x r x f r x y gRT R_ rpTj pTi R_ ðA:3Þ cosðaÞ y y ¼ kpE pj k kpT pi k ( ) ( ) sinðaÞ rx pTi R_ rpTj cosðaÞ y ¼ kpE pj k kpT pi k Since the term kpTpikcontains the functions x and y, it has to be expanded. The previous equation can be rearranged in a better form by calculating its power fr x ( R
rx y
y gR_
T
( pi pTi R_
!T
) pi
rx
)
(
0 rpTj
sinðaÞ
) 12
C B C B cosðaÞ B C ( ) ! ¼ B ( ) C @ cosðaÞ rx A r p R pi j sinðaÞ y y
Finally, Eq. (A.4) can be rearranged in a nonlinear elliptical equation rx rx f r x y gA þ 2B C ¼0 y y
ðA:4Þ
ðA:5Þ
Coefficients A, B and C are functions of a T
A ¼ AðaÞ ¼ R_ pi pTi R_ aI B ¼ BðaÞ ¼ apTi R C ¼ CðaÞ ¼ apTi pi
ðA:6Þ
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P. Gallina / Mechanism and Machine Theory 40 (2005) 963–976
and 0 rpTj
sinðaÞ
12
C B cosðaÞ B C a ¼ B C @ r cosðaÞ p A j sinðaÞ and I is the eye matrix. For sake of shortness, Eq. (A.5) is expressed by the symbol U(x, y, a) = 0.
References [1] B. Shrinivas, C. Satish, Transmission angle in mechanisms, Mechanism and Machine Theory 37 (2) (2002) 175–195. [2] W.-J. Zhang, On the finding of the spatial linkage with constant pressure angle, Mechanism and Machine Theory 32 (8) (1997) 933–940.
Mechanism and Machine Theory www.elsevier.com/locate/mechmt
A new class of rocker-belt mechanisms Paolo Gallina
*
Department of Energetics, University of Trieste, Via A. Valerio 10-34127 Trieste, Italy Accepted 16 December 2004 Available online 11 March 2005
Abstract This paper describes a new rocker-belt mechanism. The mechanism is made up of a rocker, a timing belt that operates the rocker, a series of sprockets and a curved element. The latter, which is referred to as profile, represents the heart of the mechanism. One point of the belt is attached to a point on the rocker. The belt is wound around the sprockets and the profile. If the profile is properly shaped, for a given range of rotation angles of the rocker, the belt does not become slack. This work shows how to design such a profile; it turns out that the profile shape, which is related to several geometric parameters, is expressed by means of a non-linear differential equation. Eventually, the proposed approach is validated by an example. 2005 Elsevier Ltd. All rights reserved. Keywords: Rocker mechanism; Belt transmission
1. Introduction The motion of a four-bar linkage is often characterized by the term crank rocker to indicate that the crank rotates completely and the other link oscillates. In a similar manner, the term double rocker indicates that both links oscillate [1]. A rocker can be present in many other planar mechanisms or it can be itself a simple mechanism. In the following of the paper, the term rocker will indicate a rigid body which can rotate around a fixed point; the rotation angle is limited.
*
Tel.: +39 040 5582540; fax: +39 040 568469. E-mail address: [email protected]
0094-114X/$ - see front matter 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.mechmachtheory.2004.12.018
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P. Gallina / Mechanism and Machine Theory 40 (2005) 963–976
In many applications, the rocker is the driving link of the linkage mechanism [2]. In this case it is operated by means of an actuator trough a mechanical transmission system. There exist different combinations of actuators and transmissions. For sake of simplicity, in order to stress the innovation and the advantages claimed by the rocker-belt mechanism proposed in this paper, three typical different solutions of actuated rockers are discussed (see Fig. 1). In configuration (a), the rocker is fixed to a spur gear. Therefore, the rocker can rotate around the a–a axis which is fixed with respect to the reference frame. The rocker is driven by an electric motor. The transmission-reducer module is made up of a pinion (r1 = pinion radius) driving the external gear (r2 = gear radius). As the matter of fact, more sophisticated reducers could be employed, but the substance of physical considerations would not change. It is assumed that the rocker is loaded by a force F acting at its free end, perpendicular to the rocker axis. In first analysis, the force the pinion exerts on the gear tooth is F t ¼ ðlF Þ=ðr2 cosð/ÞÞ
ð1Þ
where l is the rocker length and / is the pressure angle of the two gears in mesh.
ROCKER GEAR
A
REDUCER
F
F
a
a SPROCKET 2
ELETRIC MOTOR
SPROCKET 1
(a)
(b)
A
D
B
σ
F
σ
F
C E (c)
(d)
Fig. 1. Three different solutions to drive the rocker.
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In configuration (b), the rocker is also driven by an electric motor. The transmission-reducer module is made up of a timing belt looped around two sprockets. (r1 = sprocket 1 radius, r2 = sprocket 2 radius). As in case (a), it is assumed that the rocker is loaded by a force F at its free end, perpendicular to the rocker axis. In first analysis, the belt tension is T t ¼ ðl=r2 ÞF
ð2Þ
Both these two typical solutions present the following drawbacks: • the force between the teeth in contact (case (a)) and the belt tension (case (b)) are functions of the rocker length and of the driven sprocket radius: the higher is l and the lower is r2, the higher is the force/tension; • configuration (a) may exhibits backlash; • configuration (b) is cumbersome; • the rocker tends to bend because of the force F; in fact it behaves as a clamped-free beam These disadvantages could be overcome by employing the solution (c). In this case, two belts (or equivalently, two ropes or two chains) and two motors are required. Both belts have one end fixed to the free end of the rocker, while the other end is wound around a drum. As the matter of fact, ropes or chains are more suited to be wound around a drum, but from a kinematic point of view, ropes, belts or chains are equivalent. In fact, what is relevant to our analysis here, are the tensions and the mechanism geometry. For sake of simplicity, let us focus on a solution that employs belts. It is reminded that results we obtain are valid also for rope or chain transmissions. The belt tension (of the lower belt) is T ¼ ð1= sinðrÞÞF
ð3Þ
According to the load direction (force F direction), for the static equilibrium of the mechanism, one belt is tensioned, while the tension on the other one is null. The tensioned belt is referred to as loaded belt. r is the angle between the rocker axis and the loaded belt. r is a function of the rocker angular position. Comparing Eqs. (1) and (2) with (3), it can be inferred that, unlike solutions (a) and (b), in solution (c), the belt tension is not a function of the rocker length. As a consequence, even if the rocker length is long, the belt tension remains low for a given range of the rockerÕs angular position. This statement has to be considered very carefully in order to avoid misunderstandings. Let us assume to employ a longer rocker in order to produce the same force F at its free end. Besides assume that the two motors are located with respect to the rocker axis in order to maintain original proportions. In other words, let as focus on a scale enlarged mechanism. Although such a mechanism will have a longer rocker, the tension on the belt necessary to balance the same force F does not change, since it depends on the geometry of the whole mechanism and not on its dimensions. This is the reason why the rocker length r does not appear into Eq. (3). However, it is remarked that the whole torque the rocker can exert with respect to the rocker axis is a increasing function of the rocker axis: C = rF. Note that no backslash is present. Moreover for some applications the rocker bending is mitigated by the belt. Also this last statement needs to be clarified with an example. Let as assume
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that the rocker is part of a complete mechanism such as a four-bar linkage. Cases (a) and (c) are considered. If the four-bar linkage is employed to operate a general machine, it is subjected to a set of load forces which generate a force on the free end of the rocker where the revolute joint is located. Because of this force, in case (a) the rocker could bend according to its compliance. On the contrary, in case (c) this force is balanced by the belt tension and by the rocker which produces a reaction force oriented along its longitudinal axis. As a consequence, no bending moment acts on the rocker. As the matter of fact, the bending phenomenon depends on the particular application the rocker is employed for. For example, if the rocker itself constitutes the mechanism and it is used to balance a load torque applied to the rocker axis, solutions (a) and (c) show the same behaviour, as far as bending phenomenon is concerned. Of course, although this last solution presents many advantages, it should be avoided because it requires two motors for just 1 d.o.f. A simple way to eliminate one motor, without abandoning the benefits introduced by solution (c) is given by scheme (d). In fact, in (d) only one motor is employed. In this case, there is only one timing belt looped around two sprockets and the rocker. One point of the belt is fixed to the free end of the rocker. The second motor has been replaced with the sprocket A. Belt tensioners B and C play the role of keeping the belt always tight. They consist in a spring-like mechanism to push against the belt by means of a tightening pulley. The main drawback of solution (d) is that it shows backlash. In fact, when the force F is oriented as in figure, the belt segment ED tends to straighten while segment AD is bent because of the tensioner; conversely, when F assumes the opposite direction, the belt segment AD tends to straighten. As a consequence, the rocker rotates according to the load direction F even if the motor is stopped. In order to improve solution (d), in the next section a new mechanism is presented. Whit respect to the previous solution, it does not show backlash. In conclusion, the proposed solution has all the benefits given by solution (c), but it requires just one motor to operate the mechanism. Such a mechanism can represent an effective solution in all that applications where a rocker is required to operate a mechanism such as a four-bar linkage or the rocker itself has a mechanical function. For example it could be employed to position launching pad, orient an antenna or modify the mechanical component tilt. After the description of the mathematical model, an example to validate the theory is given.
2. Profile shape The proposed mechanism is sketched in Fig. 2. It is made up of the rocker, a belt that operates the rocker, two sprockets and a curved element attached to the free end of the rocker (point pE), which is referred to as profile. The rocker and the profile constitute a rigid body. One side of the belt is directly attached to the point pE of the rocker. The other side of the belt is first wound \ around the profile (along the arc pT pE ) and then fixed to the point pE. pT is the tangent point of the belt to the profile. Sprockets can rotate around the point pi and pj which are fixed with respect to the absolute reference frame (XY). The rocker can rotate around the point O, which is assumed to be the frame origin. a is the rocker angular position.
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p iT
pi
Tu
vT F
pT
PROFILE
y
CT
x
pE
Y
α
O
X Tl
pj p jT
Fig. 2. The proposed rocker-belt mechanism.
If the profile is properly shaped, for a limited range of rocker rotation angles, the belt does not become slack. In other words, the profile plays the role of the tensioners B and C in the solution (d) of Fig. 1. A xy local reference frame is introduced, where the unit vector x is oriented as in figure and the origin of the local frame is located at the free end of the rocker pE. pT is the tangent point of the upper side of the belt to the profile. piT and pjT are the tangent points of the two sides of the belt to the sprockets. Sprocket radii are assumed to be low enough to approximate sprocket tangent points with sprocket centres: piT ffi pi ¼ f xi y i gT and pjT ffi pj ¼ f xj y j gT . Let us assume that the tangent point pT is a function of the rocker angular position a fpT gl ¼ f xðaÞ
T
yðaÞ g
ð4Þ
where the symbol {*}l indicates that the co-ordinates of the point pT are expressed with respect to the local reference frame. x(a) and y(a) are two scalar functions. The aim of this mathematical treatment consists in finding the two function x(a), y(a) so that to prevent the belt from going slack for every angular position a. The point pT expressed with respect to the absolute frame is r xðaÞ cosðaÞ sinðaÞ r xðaÞ ¼ ð5Þ pT ¼ RðaÞ yðaÞ sinðaÞ cosðaÞ yðaÞ where R(a) is a rotation matrix and r is the rocker length. The derivatives of functions x(a) y(a), represent the components of a tangent vector to the profile with respect to the local frame fvT gl ¼ f dxðaÞ=da dyðaÞ=da gT . The same vector, expressed with respect to the absolute frame is
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8 9 dxðaÞ > > < = da vT ¼ RðaÞ ð6Þ > : dyðaÞ > ; da Note that the vector vT is not a unit vector. The goal functions x(a) and y(a), which define the profile shape, will be obtained by numerically solving a system made up of two equations. The two equations come from two constraints the profile geometry has to fulfil: 1. First of all, the belt has to be always tangent to the profile; 2. As the rocker angle a changes, the belt has to remain tight. 2.1. First constrain: belt tangent to the profile The first mathematical constrain comes from the following consideration: the belt segment pipT has to be always tangent to the profile at the point pT. Mathematically, this consideration can be expressed by the relationship pT pi ¼ kvT
ð7Þ
where k is a scalar number. Note that the vector vT is oriented in such a way thatk is a negative number. In other words, it is required that the tangent vector vT be aligned to the upper segment of the belt, where the segment is represented by the vector pTpi. By replacing Eqs. (5) and (6) into (7) and eliminating the parameter k by simple mathematical simplifications, one gets the relationship between the derivative of y with respect to a and the derivative of x with respect to a dy dx ¼ Cðx; y; aÞ ð8Þ da da where Cðx; y; aÞ ¼
y y i cosðaÞ þ xi sinðaÞ xi cosðaÞ þ y i sinðaÞ þ x r
ð9Þ
2.2. Second constrain: the belt remains tight The second mathematical constrain comes form the belt. It is assumed that the belt length is \ constant; in order to remain always tight, the sum of the segments pTpi, pEpj and the arc pT pE has to remain constant for every value of a sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Z a 2 2 dx dy da ¼ lk þ ð10Þ kpT pi k þ kpE pj k þ d a da 0 The symbol kvk = vTv indicates the length of the vector v. The integral term represents the length \ of the arc pT pE , while lk is a constant.
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The free end of the rocker pE expressed with respect to the absolute frame is 0 gT
ð11Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi T T Considering that kpT pi k ¼ ðpT pi Þ ðpT pi Þ and kpE pj k ¼ ðpE pj Þ ðpE pj Þ the derivative of Eq. (10) with respects to a leads to ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi s 2 2 T ðpE pj ÞT dpdaE ðpT pi Þ dpdaT dx dy qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ þ ¼0 ð12Þ T da da ðpE pj Þ ðpE pj Þ ðpT pi ÞT ðpT pi Þ pE ¼ RðaÞf r
where
( ) dx da dpT dRðaÞ r x þ RðaÞ dy ¼ da da y da sinðaÞ dpE ¼r da cosðaÞ
ð13Þ
dy dx ; y_ ¼ da and R_ ¼ dRðaÞ , it yields Replacing Eq. (13) into (12) and using the notation x_ ¼ da da qffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðpE pj ÞT dpdaE _x ðpT pi ÞT _ r x þ x_ 2 þ y_ 2 ¼ 0 R þR þ kpT pi k kpE pj k y y
It is shown now that some terms of Eq. (14) are null. In fact, let us consider the term qffiffiffiffiffiffiffiffiffiffiffiffiffiffi T _x ðpT pi Þ n¼ þ x_ 2 þ y_ 2 R kpT pi k y
ð14Þ
ð15Þ
By replacing Eqs. (6) and (7) into the term (15), it yields T
n¼
_x ðpT pi Þ R kpT pi k y
qffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi _x ðkvT ÞT þ x_ 2 þ y_ 2 ¼ R þ x_ 2 þ y_ 2 ¼ kpT pi k y
T _x qffiffiffiffiffiffiffiffiffiffiffiffiffiffi kR _x y_ R þ x_ 2 þ y_ 2 _ x y kR y_
ð16Þ
Exploiting the fact that k is a negative number, the term (15) becomes
n¼
k kkk
f _x
_x y_ gRT R qffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffi k x_ 2 þ y_ 2 k y 2 2 2 2 2 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ x_ þ y_ ¼ x_ þ y_ þ1 ¼0 þ x_ þ y_ ¼ kkk x_ 2 þ y_ 2 kkk R _x y_ ð17Þ
As a consequence, Eq. (14) can be simplified into the following ðpE pj ÞT dpdaE ðpT pi ÞT _ r x R þ ¼0 kpT pi k kpE pj k y
ð18Þ
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P. Gallina / Mechanism and Machine Theory 40 (2005) 963–976
In the Appendix A, it will be shown that Eq. (18) is equivalent to an algebric nonlinear elliptical equation of the form Uðx; y; aÞ ¼ 0
ð19Þ
3. Numeric solution of the profile shape The system to solve is made up of the Eqs. (8) and (19) dy dx ¼ Cðx; y; aÞ da da Uðx; y; aÞ ¼ 0
ð20Þ
A numeric solution can be found. In order to find a solution, the initial values of x and y are given: x0, y0. They represent the initial conditions of system (20). Note that the first equation of the system is a nonlinear first order differential equation, while the second one is an algebraic non linear equation in the tree variables x,y and a. This equation can be represented in an explicit form a ¼ /ðx; yÞ
ð21Þ
In other words, /(x,y) represent a root of Eq. (19), when x and y are given. Such a root can be carried out by an iterative numeric solution. Eventually, the iterative method to solve system (20) is given by the following iteration formula xnþ1 ¼ xn þ dx anþ1 ¼ /ðxn ; y n Þ
ð22Þ
y nþ1 ¼ y n þ Cðxn ; y n ; anþ1 Þdx This is not the only way to solve system (20), but it has been employed for its simplicity and effectiveness. In the Section 5, this iterative method will be employed to solve a real case. 4. Belt tensions This section deals with the force analysis of the mechanism. The belt transmission has the function to move the rocker against a resistive load. The resistive load can be given by a torque CT acting on the rocker. If the torque CT is assumed to be positive when counterclockwise oriented. Note that, if CT > 0, the lower side of the belt has to be tensioned in order to balance the torque, while the tension in the upper side of the belt is null. Conversely, if CT < 0, it is the upper side of the belt to be tensioned, while the tension on the lower side of the belt is null. In the former case, the equilibrium is given by the relationship * +! pE pj pE ; ð23Þ C T ¼ T l r sin kpE k kpE pj k p p
where Tl is the lower side belt tension. kppE k and kpE pj k are unit vectors that represent respectively E E j p p the rocker orientation and the lower side belt orientation, while the symbol hkppE k ; kpE pj ki represents E
E
j
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the angle between the two unit vectors, or in other words, the angle between the lower side belt and the rocker. Instead, the tension of the upper side belt Tu is null. In case CT < 0, the equilibrium is given by the relationship
pT pT pi C T ¼ T u kpT k sin ; ð24Þ kpT k kpT pi k where Tu is the upper side belt tension. In conclusion the belt tensions are Tl ¼
CT 1 D E pE pj pE r sin ; kp k kp p k E
E
8C T > 0 ð25Þ
j
Tu ¼ 0 and Tl ¼ 0 Tu ¼
CT 1 D E pT pi pT kpT k sin ; kp k kp p k T
T
8C T < 0
ð26Þ
i
Note that the tension changes with respect to the rocker angular position. Therefore, it is necessary to check the belt tensions versus the rocker angular position a in order to prevent the tension from reaching dangerous values.
5. Example In the proposed example, the centres of the two sprockets are set to pi ¼ f 0:1 0:1 gm and pj ¼ f 0 0:2 gm, and the rocker length to r = 0.3m. The increment employed in the relationships (22) has been set to dx = 0.001m. The initial conditions are x0 = 0 and y0 = 0.The integration of system (20) by means to the iterative scheme (22), leads to the solution represented in Fig. 3. The plot represents the y co-ordinate of a profile point versus its x co-ordinate, where x and y are the point co-ordinates with respect to the local frame. The curve goes trough the frame origin since the initial conditions were x0 = 0 and y0 = 0. By employing different initial conditions, one could get other profile shapes. From the iterative formula (22), it can be inferred that, at each iterative step, the co-ordinate x is incremented by a constant value dx. Then the new value of a is calculated followed by the new value of y. As a consequence the x co-ordinate can be thought of as the independent variable. Therefore, it is natural to show the plot of y versus x (Fig. 3) as well as the plot of a versus x (Fig. 4). Note that, a decreases as x increases. There exists an upper limit xmax for the co-ordinate x. In fact, if the value x during the iteration process overcomes xmax, Eq. (19) does not admit a solution. As a consequence, the profile has a limited length and the rocker has a limited rotation angle Da. The upper limit xmax depends on the
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P. Gallina / Mechanism and Machine Theory 40 (2005) 963–976 0.05 0.045 0.04 0.035
y [m]
0.03 0.025 0.02 0.015 0.01 0.005 0 0
0.01
0.02
0.03
0.04
0.05 x [m]
0.06
0.07
0.08
0.09
0.1
Fig. 3. Profile contour described in terms of y versus x.
-0.4
-0.5
-0.6
α [rad]
-0.7
-0.8
-0.9
-1
-1.1
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
x [m]
Fig. 4. Rocker position a versus x.
points pi, pj, the rocker length r and on the initial conditions x0 and y0. In the proposed example, xmax has been calculated by means of a trial and error technique.
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In particular it has been set xmax = xj, where xj fulfils the following requirement: Eq. (19) admits a solution for xj and it does not admit a solution for xj + dx. As it can be read from Fig. 3, xmax = 0.098 while, from Fig. 4, the maximum rocker rotation angle is Da = k1.15 (0.54)k = 0.61rad. Fig. 5 shows 5 different kinematic configurations of the rocker mechanism. Thick solid line represents the rocker and the profile. Thin solid lines represent the upper and lower side of the belt. The belt piece that connects the two sprockets is not represented for sake of clarity. In the initial configuration (configuration 1), the tangent point of the upper side of the belt to the profile pT coincides with the free end of the rocker pE. In fact, this kinematic configuration corresponds to the case x = 0 and y = 0. As the co-ordinate x increases, the tangent point pT moves along the profile, and at the same time, the rocker clockwise rotates. In configuration (5) the point pT reaches the free end of the profile. Finally, Fig. 6 represents the belt tensions versus the rocker angular position. In particular, solid line represents the normalized upper side belt tension Tu/jCT/rj versus a when CT < 0; dashed line represents the normalized lower side belt tension Tl/jCT/rj versus a when CT > 0. It can be inferred that the lower side belt tension is poorly affected by the rocker angular position. Conversely, the upper side belt tension decreases as the rocker rotates clockwise. Moreover, for the considered range of angular positions, the normalized upper side belt tension is always greater of the lower side belt tension.
Fig. 5. Five different rocker angular positions; thick solid line represents the rocker and the profile in 5 different configurations; thin solid lines represent the upper and lower side of the belt. The belt piece that connects the two pulley is not represented for sake of clarity.
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P. Gallina / Mechanism and Machine Theory 40 (2005) 963–976 2.8
2.6
2.4
2.2
2
1.8
1.6
1.4 -1 .1
-1
-0 .9
-0 .8
-0 .7
-0 .6
-0 .5
-0 .4
α [rad]
Fig. 6. Solid line represents the normalized upper side belt tension jjCTTu=rjj with versus a when CT < 0; dashed line represents the normalized lower side belt tension jjCTT l=rjj with versus a when CT > 0.
6. Conclusion This paper has introduced, described and validated a new mechanism to operate a crank. To the best of our knowledge, such a mechanism has never described before in literature. This solution show many advantages with respect to similar ones. Among others, the belt tension is not a function of the rocker length, no backslash, only one motor is required. Even if the mechanism is made up by few mechanical components, the mathematical description of the profile requires the solution of a first order non-linear differential equation. For a given example, such an equation has been solved numerically.
Appendix A Eq. (18) can be expressed as a function of x and y. By expanding the terms pT,pE and gets ( ) ( ) ) !T !T ( rx cosðaÞ sinðaÞ ( ) pi pj r R r rx y sinðaÞ cosðaÞ _R þ ¼0 kpE pj k kpT pi k y
dpE , da
one
ðA:1Þ
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The second factor can be simplified ( ) ( ) !T rx sinðaÞ ( ) rpTj pi R ðA:2Þ y cosðaÞ rx R_ ¼0 kpE pj k kpT pi k y 0 rx 0 1 T _ T _ Since R R ¼ is null. As a consequence, Eq. (A.2) , the term f r x y gR R y 1 0 becomes !T ( ) ( ) sinðaÞ rx ( ) rpTj R pi rx cosðaÞ y R_ ¼ kpT pi k kpE pj k y ( ) ( ) ( ) sinðaÞ r x r x f r x y gRT R_ rpTj pTi R_ ðA:3Þ cosðaÞ y y ¼ kpE pj k kpT pi k ( ) ( ) sinðaÞ rx pTi R_ rpTj cosðaÞ y ¼ kpE pj k kpT pi k Since the term kpTpikcontains the functions x and y, it has to be expanded. The previous equation can be rearranged in a better form by calculating its power fr x ( R
rx y
y gR_
T
( pi pTi R_
!T
) pi
rx
)
(
0 rpTj
sinðaÞ
) 12
C B C B cosðaÞ B C ( ) ! ¼ B ( ) C @ cosðaÞ rx A r p R pi j sinðaÞ y y
Finally, Eq. (A.4) can be rearranged in a nonlinear elliptical equation rx rx f r x y gA þ 2B C ¼0 y y
ðA:4Þ
ðA:5Þ
Coefficients A, B and C are functions of a T
A ¼ AðaÞ ¼ R_ pi pTi R_ aI B ¼ BðaÞ ¼ apTi R C ¼ CðaÞ ¼ apTi pi
ðA:6Þ
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and 0 rpTj
sinðaÞ
12
C B cosðaÞ B C a ¼ B C @ r cosðaÞ p A j sinðaÞ and I is the eye matrix. For sake of shortness, Eq. (A.5) is expressed by the symbol U(x, y, a) = 0.
References [1] B. Shrinivas, C. Satish, Transmission angle in mechanisms, Mechanism and Machine Theory 37 (2) (2002) 175–195. [2] W.-J. Zhang, On the finding of the spatial linkage with constant pressure angle, Mechanism and Machine Theory 32 (8) (1997) 933–940.