How does CAM Mechanism work

how to draw a cam mechanism and how does a drop cam mechanism work and how to build a cam mechanism and how does a cam and follower mechanism work
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Dr.NaveenBansal,India,Teacher
Published Date:25-10-2017
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Ch4Frame Page 103 Friday, June 2, 2000 6:41 PM Cams 4 4.1 INTRODUCTION A cam mechanism is a two-link system in which the cam is always a driving link. This mechanism transforms a rotational or translational motion of the cam into a prescribed translational or angular motion of the follower. An example of a cam mechanism is shown in Figure 4.1 where 1 is the cam, 2 is the follower, 3 is the spring, and 4 is the camshaft. In Figure 4.1 the cam is a plate and the follower is a pin. The transfer of motion is achieved through the contact between the cam and the follower. As is clear, this contact exists only if there is a compressive force between the cam and the follower. The function of the prestressed spring in this respect is to ensure that this contact is maintained during the cycle of cam rotation. The motion of the follower reflects the shape of the cam profile, and thus the main objective of cam design is to find a cam profile needed to obtain a desired follower motion. The follower in Figure 4.1 is called the knife-edge follower. The specific feature of this mechanism is that at the cam–follower interface a relative motion (sliding) takes place. Since the transfer of motion involves transfer of forces through the cam–follower interface, the contact stresses at this interface in the presence of sliding may be unacceptable from the point of view of wear. Given that the contact stresses vary during the rotation, the wear is not uniform along the profile, thus leading to a deviation from the designed follower motion. Such potential for the deterioration of motion transformation in systems with knife-edge followers has led to cam mechanisms with flat-faced (Figure 4.2) or roller (Figure 4.3) followers. With the flat-faced follower, the contact stresses are lower while the sliding takes place. With the roller follower the sliding is eliminated at the expense of a more complex design. Note that the axis of the follower may pass through the camshaft center (Figure 4.1) or be at some distance from this center (Figure 4.3). In all cases shown in Figures 4.1 through 4.3 the rotational motion of the cam is transformed into the reciprocating (oscillating) follower motion. In Figure 4.4 another transformation of motion, namely, from rotational to angular oscillation, is shown. Instead of the roller follower shown in Figure 4.4, a flat-faced follower can be used. Another type of cam mechanism is shown in Figures 4.5 and 4.6, where a reciprocating motion of the cam is transformed into either an angular oscillation (Figure 4.5) or translational oscillation (Figure 4.6). The functional role of the spring in the above cam designs is to ensure a constant contact between the cam and the follower. The presence of the spring complicates the design and results in increased contact stresses. An alternative design solution is to insert the roller inside a groove, Figure 4.7. The outside and inside profiles of this groove are in fact the profiles of cams, one moving the follower up and another down. Since clearance between the roller and the groove is needed to allow for free roller rotation, the transition from up to down motion of the follower may be associated with discontinuity in motion, additional parasitic forces, and noise. 103 Ch4Frame Page 104 Friday, June 2, 2000 6:41 PM 104 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms FIGURE 4.1 Cam with a knife-edge follower. FIGURE 4.2 Cam with a flat-faced follower. 4.2 CIRCULAR CAM PROFILE A circular cam is made by mounting a circular plate on a camshaft at some distance d away from the circle center (Figure 4.8). This gives the simplest cam profile. The eccentric attachment of the circular plate produces a reciprocating motion of the follower. The problem of direct analysis is to find the follower displacement given the rotation of a cam with known radius R and eccentricity d. Place the origin of the global coordinate system in the camshaft center. Then, the distance D from this center (Figure 4.9) characterizes the follower position. The tip of the follower can also be reached following the vectors d and R. The three Ch4Frame Page 105 Friday, June 2, 2000 6:41 PM Cams 105 FIGURE 4.3 Cam with a roller follower. FIGURE 4.4 Cam with a rocking follower. vectors, d, R, and D, form a loop at any cam position. This loop is identical to a loop for a slider-crank mechanism, in which d plays the role of a crank, R of a connecting rod, and D characterizes the slider position. This analogy means that the analysis of this cam mechanism is identical to that of the slider-crank mechanism. Indeed, the loop-closure equation in this case is T 3p 3p T T - - - - d cosgg , sin++ R cosqq , sin D cos , sin =0 (4.1) Łł Łł 2 2 where the angles q and g are shown in Figure 4.9. In the above equation the unknowns are the distance D and the angle g . Thus, this equation falls into the second case category according to the analysis of various cases in Chapter 2. From the equiva- lency of Equations 2.28 and 4.1, it follows that the corresponding solutions for the former, namely, Equations 2.31 and 2.32, can be used in this case. It is necessary Ch4Frame Page 106 Friday, June 2, 2000 6:41 PM 106 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms FIGURE 4.5 Reciprocating cam with a flat follower. FIGURE 4.6 Reciprocating cam with a knife follower. FIGURE 4.7 Cam with a roller inside a groove. to substitute in these solutions r by D, q by q , q by 3p /2, a by g , r by R, and i j i j b by –d. As a result, the solution for the follower displacement is 2 2 2 Dd = sing – R – d cos g (4.2) and the angle q is q if –sinq 0 and cosq 0 p – q if –sinq 0 and cosq 0 q = (4.3) p + q if –sinq 0 and cosq 0 Ł 2p – q if –sinq 0 and cosq 0 Ch4Frame Page 107 Friday, June 2, 2000 6:41 PM Cams 107 FIGURE 4.8 Circular cam. FIGURE 4.9 Loop-closure equation for a circular cam. where 3p d cosg q = - - + a r csin (4.4) - 2 R d cosq = – - cosg (4.5) R and d D sinq = – - sing + - - (4.6) R R Ch4Frame Page 108 Friday, June 2, 2000 6:41 PM 108 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms D 5 4 3 2 1 g 1 2 3 4 5 6 FIGURE 4.10 Position of the follower during one cycle of circular cam rotation. u 5 4 3 2 1 g 1 2 3 4 5 6 FIGURE 4.11 Displacement diagram for a circular cam. If, at the extreme, d = R, then Equation 4.2 gives DR = sinq – Rsinq . It follows from the above that the correct sign in Equation 4.2 must be plus for this case. At the other extreme, if d R, then the second term under the square root can be neglected and the result is Dd = sing – R . Again, the correct sign must be plus for D to be positive. Thus, one can assume that the sign in Equation 4.2 must be plus for any value of d. 2 2 2 Dd = sing + R – d cos g (4.7) In Figure 4.10 the position of the follower, D, is shown for one cycle of cam rotation for the case of d = 1 cm and R = 4 cm. The difference between the maximum follower displacement and its minimum is called the lift. If the minimum follower coordinate (position) is subtracted from its current position, the resulting diagramCh4Frame Page 109 Friday, June 2, 2000 6:41 PM Cams 109 FIGURE 4.12 Typical displacement diagram. is called the displacement diagram. Such a diagram for a circular cam is shown in Figure 4.11. A minimum follower position constitutes a reference level for the follower displacement. This reference level is a circle with a constant radius r = D – D , b min which is called the base radius. Thus, a cam profile can be viewed as a displacement diagram wrapped around the base circle. 4.3 DISPLACEMENT DIAGRAM The displacement diagram serves as an input into the cam mechanism design. Consider the cam mechanism in Figure 1.1. The function of the cam might be, as is the case in the internal combustion engine, to open the valve, to keep it open during some part of the cycle (this is called dwell), and then to close it and to keep it closed for some time (to dwell again). A generic displacement diagram may look as shown in Figure 4.12. The requirements of how long it should take to rise, to dwell, to return, and to dwell again, and also of what the lift should be define the size and the shape of the cam. The function depicted in Figure 4.12 is a piecewise function, which means that special attention should be paid to transition from one continuous function to another, for example, from rise to dwell. This represents another objective of cam design, to ensure a smooth transition of the follower from one part of the displacement diagram to another. Consider, for example, a transition from rise to dwell in Figure 4.12. The rise curve is described by some function u() q , while the dwell is described by another 1 function u() q = const. For a smooth transition from the rise to dwell, it is needed 2 that at qq = , A 2 du() q d u() q 1 1 u() q== u() q , 0, and -= 0 (4.8) 1 A 2 A 2 dq dq Since q = w t (where w is the angular velocity of the cam, and t is the time), the requirements for the equality of first and second derivatives is equivalent to the requirements that the velocity and acceleration of the follower does not experience jumps at the point of transition. The same requirement should be met at the other transitional points in Figure 4.12: q = 0 , qq = , and qq = . These latter B C requirements put limitations on what type of functions can be used to generate theCh4Frame Page 110 Friday, June 2, 2000 6:41 PM 110 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms FIGURE 4.13 Inflection points on the displacement diagram. cam profile for a given displacement diagram. The requirements (Equation 4.8) are called smoothness requirements. 4.4 CYCLOID, HARMONIC, AND FOUR-SPLINE CAMS Any function that meets the type of requirements given in Equation 4.8 is a suitable cam profile function. For example, the suitable function for the rising part of the diagram must be tangential to the q -axis at point 0 and at point A (Figure 4.13). It means that it must change the curvature from concave to convex, and thus it must have an inflection point. A few analytical functions are used to describe the rise and return parts of the displacement diagram while meeting the above smoothness con- ditions. These are cycloid, harmonic, and polynomial functions. In the following an application of all these functions to the displacement diagram design is considered. A cycloid is a curve traced by a point on a circle rolling along a straight line. In Figure 4.14 an example of a cycloid function is shown, where point A is embedded into the circle, and angle a corresponds to arc a r, which is the distance traveled by the circle. The x- and y-coordinates of the cycloid are parametric functions of angle a xr =() aa – sin (4.9) and yr =() 1 – cosa (4.10) Both functions can be used in cam profile design. A cam in which the first function, Equation 4.9, is used is called the cycloid cam, while a cam in which the second function, Equation 4.10, is used is called the harmonic cam. 4.4.1 CYCLOID CAMS The Rise Part of the Displacement Diagram One can utilize the function given by Equation 4.9 to describe the rise of the follower from 0 to point A in Figure 4.13. Note that Equation 4.9 comprises two components: ra and –r sina . Thus, it is a superposition of a straight line and a sinusoidal function.Ch4Frame Page 111 Friday, June 2, 2000 6:41 PM Cams 111 x — r A 6 5 4 3 2 1 α 1 2 3 4 5 6 FIGURE 4.14 Normalized functions a (dashed line) and (a - sin a) ( solid line). In normalized coordinates, x/r, these functions are shown in Figure 4.14 in the interval 0 to 2p . The straight line must go from 0 to point A, and then the sinusoidal function is drawn with respect to this line (see Figure 4.14). It can now be shown that this function meets the above smoothness requirements. But first it is necessary to transform Equation 4.9 from the (x,a )- to the (u,q )-coordinates. The needed trans- formation is achieved by mapping the a -range (0–2p ) onto the q -range (0–q ). Such A mapping follows from the relationship a q rise - - = - (4.11) 2p q A where a is the transformed angle a . rise Now substituting a from Equation 4.11 into Equation 4.9 gives the transformed rise cycloid equation: q q - - ur = 2p – sin2p (4.12) Łł q q A A The radius of the circle r in Equation 4.12 remains undetermined. It should be treated as a parameter to be determined from the requirement that at q = q u = L A (see Figure 4.12). Then it follows that L r = - - (4.13) 2p Thus, the function describing the rise of the follower from 0 to L when the cam rotates from q = 0 to q = q is ACh4Frame Page 112 Friday, June 2, 2000 6:41 PM 112 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms L q q u = - - 2p - – sin2p - (4.14) 1 Łł 2p q q A A Now, check that the requirements given by Equation 4.8 are met. The first and second derivatives of the function Equation 4.14 are du L q 1 - - - - = 1 – cos 2p (4.15) Łł q q dq A A and 2 d u 2p L q 1 - - - - - = sin2p (4.16) 2 2 q A dq q A It is easy to see that both derivatives are equal to zero at q = 0 and q = q . A The Return Part of the Displacement Diagram Now, one can utilize the same function given by Equation 4.9 to describe the return of the follower from q to q in Figure 4.13. The procedure is the same. One maps B C the a -range (0 to 2p ) onto the q -range (q to q ). The coordinate transformation in B C this case is a qq– return C - - = - (4.17) 2p q – q B C where a is the transformed angle a . return Since the expression for r is known (see Equation 4.13), the return part of the displacement diagram is described by Equation 4.12, where angle a is, according to Equation 4.17, qq– qq– L C C u = - - 2p - – sin2p - (4.18) 3 Łł 2p q – q q – q B C B C The first and second derivatives in this case are as follows: du qq– L 3 C - - = - 1 – cos 2p - (4.19) Łł dq q – q q – q B C B C and 2 d u qq– 2p L C 3 - - = - - sin2p - (4.20) 2 2 q – q B C dq () q – q B CCh4Frame Page 113 Friday, June 2, 2000 6:41 PM Cams 113 u — L 1 0.8 0.6 0.4 0.2 θ 1 2 3 4 5 6 FIGURE 4.15 Normalized displacement diagram for a cycloid cam. It is easy to check that at q = q and q = q both derivatives are equal to zero. C B Thus, the cycloid function can be used to describe the cam profile and satisfy the smoothness requirements. An example of the normalized displacement diagram for q = p/2 , q = 5p/4, and q = 7p/4 is shown in Figure 4.15. In Figures 4.16 and 4.17 A B C the corresponding normalized velocity and acceleration diagrams are shown. The normalized displacements, velocities, and accelerations are, respectively, as follows: On the rise part: u u = - (4.21) 1 L u ˙q ˙ A u = - (4.22) 1 L 2 u ˙˙q ˙˙ A u = - (4.23) 1 L On the return part: u u = - (4.24) 3 L u ˙() q – q ˙ B C u = - (4.25) 3 L 2 u ˙˙() q – q ˙˙ B C u = - (4.26) 3 LCh4Frame Page 114 Friday, June 2, 2000 6:41 PM 114 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms Velocity 2 1.5 1 0.5 θ 1 2 3 4 5 6 FIGURE 4.16 Normalized velocity diagram for a cycloid cam. Acceleration 6 4 2 θ 1 2 3 4 5 6 -2 -4 -6 FIGURE 4.17 Normalized acceleration diagram for a cycloid cam. It is denoted above: 2 du() q d u() q u ˙ = - - and u ˙˙ = 2 dq dq The displacement, velocity, and acceleration of the follower, shown in Figures 4.15 through 4.17, are functions of the angle of rotation of the cam, and thus they do not depend on the angular speed of rotation. The displacement as a function of the angle of rotation allowed one to find a proper cam profile. It does not, however, answer the question of what is the real acceleration of the follower, which one has to know to choose the proper spring stiffness (see Figure 4.1). If one substitutes the angle of rotation qw = t into Equations 4.14 and 4.18, one will have displacements as functions of time and angular velocity w . L w t w t - - - - - - u = 2p – sin2p (4.27) 1 Łł 2p q q A ACh4Frame Page 115 Friday, June 2, 2000 6:41 PM Cams 115 and w t – q w t – q L C C u = - - 2p - - – sin2p - - (4.28) 3 Łł 2p q – q q – q B C B C where for a single cycle the time t changes from 0 to T, and T = 2p/w is the period of the cycle. It is clear from Equations 4.17 through 4.28 that the displacement of the follower does not depend on the angular velocity of the cam. However, the velocity and acceleration of the follower do depend on it. Indeed, the velocity of the follower as a function of time is proportional to w , since du q () t dq () t du q du q () t - - - - -== w (4.29) dq dt dq dt and du q ⁄() dq is the velocity as the function of the angle of rotation. Similarly, assuming that the angular velocity is constant, the acceleration of the follower is proportional to the square of angular velocity, since 2 2 d u q () t d du q () t 2d u q -== w - - w (4.30) 2 2 Łł dt dq dt dq Because the angular velocity of the cam is a scaling factor in both velocity and acceleration of the follower, the solutions given by Equations 4.12 through 4.20 are general solutions of the kinematics of the cycloid cam. In this respect, the plots of the follower displacement (Figure 4.15), velocity (Figure 4.16), and acceleration (Figure 4.17), except for the specific values of the angles q , q , and q mentioned A B C above, are generic plots characterizing the cycloid cam. 4.4.2 HARMONIC CAMS Now it will be shown that Equation 4.10 can also be used to describe the cam profile and it meets the requirement of smoothness. The plot of the function given by Equation 4.10 in normalized, y/r, coordinates over the interval 0 to 2p is shown in Figure 4.18. As can be seen, a part of this function within the interval 0 to p can be used for the rise part of the displacement diagram, whereas the second part, within the interval p to 2p for the return part of the diagram. Again, points q , q , and q (see Figure 4.12) will be used as transition points A B C from one continuous function to another on a cam displacement diagram. However, in this case the 0 to q interval will be mapped on 0 to p and q to q on p to 2p A B C of the harmonic function. The corresponding mapping relationships are similar to Equations 4.11 and 4.17 except, instead of 2p , a p is used in both. Thus, the displacements of the follower during the rise and return parts of the cycle are obtained from Equation 4.10 by substituting corresponding expressions for aCh4Frame Page 116 Friday, June 2, 2000 6:41 PM 116 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms y — r 2 1.5 1 0.5 α 1 2 3 4 5 6 FIGURE 4.18 Function 1 – cos a. q a = p - (4.31) rise q A and qq– C - a = p (4.32) return q – q B C Now the corresponding displacement formulas are as follows: L q - - u = 1 – cos p (4.33) 1 Łł 2 q A and qq– L C u = - 1 – cos p - (4.34) 3 Łł 2 q – q B C It is easy to check that at q = q and q = q , u = L, whereas at q = q , u = 0. The A B C normalized displacement diagram is shown in Figure 4.19. Now, one can check whether the smoothness requirements are satisfied. The first and the second derivatives of the function Equation 4.10 are r sin a and r cos a, respectively. For the rise part, a from Equation 4.31 is substituted, and for the return one, a from Equation 4.32 is substituted for both derivatives. One can see that at both points A and B (see Figure 4.12) the velocities are equal to zero, while accel- erations are not. Thus, the harmonic cam does not satisfy all the smoothness require- ments. The jump in acceleration while passing through these points means a jump in inertial forces. For high-speed cams when there are design constraints on forces or noise, this cam may not be acceptable.Ch4Frame Page 117 Friday, June 2, 2000 6:41 PM Cams 117 u — L 1 0.8 0.6 0.4 0.2 θ 1 2 3 4 5 6 FIGURE 4.19 Normalized displacement diagram for a harmonic cam. Velocity 1.5 1.25 1 0.75 0.5 0.25 θ 1 2 3 4 5 6 FIGURE 4.20 Normalized velocity plot for a harmonic cam. In Figure 4.20 the normalized velocity and in Figure 4.21 the normalized accel- eration plots are shown. These plots confirm that there is a jump in accelerations at the end of the rise and the beginning of the return. One can also see the amount of this jump in normalized coordinates. The normalization of displacements, velocities, and accelerations in this case is as for a cycloid cam (Equations 4.21 through 4.26). 4.4.3 COMPARISON OF TWO CAMS: CYCLOID VS. HARMONIC Here the kinematic properties of two cams are compared: displacements, velocities, and accelerations in the same normalized coordinates. The displacements are shown in Figure 4.22. One can see that the displacement curves in Figure 4.22 look sufficiently close. However, the differences at specific angles may be significant. One can check the displacements, for example, at q = q /4. They are 0.0908451 and A 0.146447 for the cycloid and harmonic cams, respectively; i.e., the difference is 38%.Ch4Frame Page 118 Friday, June 2, 2000 6:41 PM 118 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms Acceleration 4 2 θ 1 2 3 4 5 6 -2 -4 FIGURE 4.21 Normalized acceleration plot for a harmonic cam. u — L 1 0.8 0.6 0.4 0.2 θ 1 2 3 4 5 6 FIGURE 4.22 Comparison of normalized displacement diagrams for two cams: cycloid (solid line) and harmonic (dashed line). It is also of interest to compare the accelerations of two cams at this point. They are 6.28319 and 3.48943 for the cycloid and harmonic cams, respectively; i.e., the difference is 45%. The velocities differ most significantly at the inflection point q = q /2. They are 2 and 1.57 for the cycloid and harmonic cams, respectively. A The comparison of normalized velocities and accelerations for the cycloid and harmonic cams are shown in Figures 4.23 and 4.24. 4.4.4 CUBIC SPLINE CAMS The cubic spline method of designing cams is based on using cubic polynomials to fit a given displacement diagram at a predetermined number of points. This method is used for designing nonstandard cams. In general, the design of cams based on this approach requires a numerical solution of a system of linear algebraic equations. This book will limit itself to a simplified version of the method, which retains all the conceptual elements of it but is more manageable from the analytical point of view.Ch4Frame Page 119 Friday, June 2, 2000 6:41 PM Cams 119 Velocity 2 1.5 1 0.5 θ 1 2 3 4 5 6 FIGURE 4.23 Comparison of normalized velocity diagrams for two cams: cycloid (solid line) and harmonic (dashed line). Acceleration 6 4 2 θ 1 2 3 4 5 6 -2 -4 -6 FIGURE 4.24 Comparison of normalized acceleration diagrams for two cams: cycloid (solid line) and harmonic (dashed line). One can design a displacement diagram comprising six piecewise continuous functions, which are identified in Figure 4.25 by numbers. The first two, 1 and 2, are cubic splines describing the rise, the constant function 3 describes the dwell, the next two, 4 and 5, describe the return, and 6 is again a constant describing the dwell. It is necessary to find such cubic polynomials that meet the smoothness criteria for the cam. The general form of the cubic polynomial is 3 2 u = a q++ b q c q+ h i =1, 2, 4, 5 (4.35) i i i i i i i i where a , b , c, and h are constants to be determined for each spline. In total, for i i i i four splines there are 16 unknown constants. For each spline, one may request that it meet the displacement, velocity, and acceleration requirements at its boundaries (in this case at points: 0, 0.5q , q , q , and 0.5 (q + q )). For each boundary (interface A A B B C of two piecewise functions), one will have three equations defining the smoothnessCh4Frame Page 120 Friday, June 2, 2000 6:41 PM 120 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms FIGURE 4.25 Piecewise continuous functions describing a displacement diagram. requirements. In total, for six boundaries, there will be 18 requirements. For 16 unknown constants there is need for only 16 equations, which means that not all smoothness requirements will be satisfied. Thus, there is a freedom to choose which smoothness requirements to satisfy when using the cubic polynomial splines. First, write down the first and second derivatives for the polynomial given by Equation 4.35. As before, they represent the velocity and acceleration of the follower in normalized (with respect to the angular velocity) coordinates. 2 u ˙ = 3a q++ 2b q c i =1, 2, 4, 5 (4.36) i i i i i i and u ˙˙ = 6a q + 2b i =1, 2, 4, 5 (4.37) i i i i Now, consider spline 1 in Figure 4.25, and assume that at q = 0, the displacement, velocity, and acceleration are zeros. 2 du() q d u() q 1 1 u() 0 = 0, - - = 0, and - - = 0 (4.38) 1 2 dq dq and that at q = 0.5q the displacement is equal to 0.5L: A L u() q = - (4.39) 1 2 This gives four equations, which is sufficient to find four constants for the first spline. Similarly, for the second spline, if one declares that at q = q the displacement, A velocity, and acceleration are, respectively, 2 du() q d u() q 2 2 - - - - u() q = L, = 0, and = 0 (4.40) 2 2 dq dqCh4Frame Page 121 Friday, June 2, 2000 6:41 PM Cams 121 and that at q = 0.5q the displacement is equal to 0.5L: A L u() q = - (4.41) 2 2 there will be four equations needed to find four constants for the second spline. Before finding the spline constants, it is worth making two comments. First, at the inflection point q = 0.5q the requirement of equal displacements only is satisfied; A the equality of velocities and accelerations is not guaranteed. The second point is more technical. Namely, if the rise part of the displacement diagram is approximated with two splines only, the system of equations defining constants can be decoupled. In other words, one can find these constants for each spline independently. The decoupling is achieved by specifying the boundary conditions for the splines as above. If, however, one requested that at q = 0, u = 0 and at q = q , u = L, and at 1 A 2 q = 0.5q A du() q du() q 1 2 - - = - - (4.42) dq dq and 2 2 d u() q d u() q 1 2 - - = - - (4.43) 2 2 dq dq then the equations for two splines would be coupled. Moreover, now the smoothness requirements would be met at the inflection point, and not guaranteed at q = 0 and at q = q . A Only the first case, when the smoothness is satisfied at q = 0, q = q , q = q , A B and q = q , will be discussed. C From Equations 4.38 it follows that b = c = h = 0, and from Equation 4.39, 1 1 1 the last unknown is found: 4L a = - - (4.44) 1 3 q A Thus, the first spline is described by the formula: 3 q u = 4L - (4.45) 1 3 q A For the second spline from Equations 4.40, it follows that 2 3 b==–3 3a q , c a q , and h=La – q (4.46) 2 2 A 2 2 A 2 2 ACh4Frame Page 122 Friday, June 2, 2000 6:41 PM 122 Fundamentals of Kinematics and Dynamics of Machines and Mechanisms and from Equation 4.41, 4L a = - - (4.47) 2 3 q A Thus, the second spline is described by the formula: 3 2 q q q - - - u = 4L–3 3 + 3 – L (4.48) 2 Łł Łł Łł q q q A A A The boundary conditions for the fourth spline at q = q are B 2 du() q d u() q 4 4 u() q = L , - - = 0, and - - = 0 (4.49) 4 2 dq dq and at qq =() + q ⁄ 2 is B C L u() q = - (4.50) 4 2 Satisfying the first set of boundary conditions, Equations 4.49 gives the expressions for the three constants: 2 3 b==–3 3a q , c a q , and h=La – q (4.51) 4 4 B 4 4 B 4 4 B Satisfying the fourth boundary condition, Equation 4.50 gives the expression for a : 4 4L a = - - (4.52) 4 3 () q – q B C Thus the formula for the fourth spline is 2 2 3 3 12Lq q 12Lq q 4Lq 4Lq B B B u = L + - - – - - + - - – - - (4.53) 4 3 3 3 3 () q – q () q – q () q – q () q – q B C B C B C B C The boundary conditions for the fifth spline at q = q are C 2 du() q d u() q 5 5 u() q = 0, - -== 0 , and - - 0 (4.54) 5 2 dq dq and at qq =() + q ⁄ 2 is B C L u() q = - (4.55) 5 2