Gear Geometry and Applied Theory Episode 3 Part 5 ppt

P1: JXR CB672-23 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2:6 23.3 Conditions of Assembly 703 It is easy to prove that m (c) 31 =−1. This result is obtained from the following con- siderations [Fig. 23.2.5(b)]. Suppose that the carrier is fixed and gears 1 and 2, and 3 and 2 are in contact at points A and B, respectively. Vectors V A and V B represent linear velocities of corresponding gears at points A and B. Taking into account that N 1 = N 3 and V A =−V B , we get that m (c) 31 =−1. The negative sign of m (c) 31 means that gears 1 and 3 of the inverted mechanism are rotated in opposite directions. Equation (23.2.13) with m (c) 31 =−1 yields that ω c = ω 1 + ω 3 2 . (23.2.14) Let us consider the following cases of transformation of motion: (1) Assume that one of the sun gears (of gears 1 and 3), for instance gear 1, is fixed. Equation (23.2.14) with ω 1 = 0 yields ω c = ω 3 2 . (23.2.15) The discussed mechanism works as a planetary gear train. (2) Consider now that gears 1 and 3 are rotated with equal angular velocities in the same direction. Equation (23.2.14) with ω 1 = ω 3 yields that ω c = ω 1 = ω 3 . (23.2.16) Consequently, gear 1, 3, and the carrier c are rotated with the same angular velocity. The gear train is like a clutch: all movable links are rotated as one rigid body. (3) Considering that gears 1 and 3 are rotated with equal angular velocities in oppo- site directions (ω 1 =−ω 3 ), we get that ω c = 0 [see Eq. (23.2.14)]. The discussed mechanism operates as a gear train with fixed axes of rotation. 23.3 CONDITIONS OF ASSEMBLY Observation of Assigned Backlash Between Planet Gears [Litvin et al., 2002e] We consider the condition of assembly for the planetary mechanism shown in Fig. 23.2.4. The obtained results may be extended for other planetary gear trains. Fig- ure 23.3.1 shows two neighboring planet gears with the backlash k b m, where m is the module of the gears and k b is the unitless coefficient. Our goal is to derive an equation that relates N 1 , k b , and the gear ratio m (3) c1 = ω c /ω 1 of a planetary gear train wherein gear 3 is fixed. The derivation is based on application of the following equation: r 2a = E 12 sin  π n  − k b m 2 . (23.3.1) Here, r 2a is the radius of the addendum circle of gear 2; E 12 is the shortest distance; n is the number of planet gears. It is easy to verify that E 12 = N 1 + N 2 2 m (23.3.2) r 2a =  N 2 2 + 1  m. (23.3.3) P1: JXR CB672-23 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2:6 704 Planetary Gear Trains Figure 23.3.1: For derivation of distance between two neighboring planet gears. In addition to Eqs. (23.3.1) to (23.3.3), we use equation N 2 = N 3 − N 1 2 (23.3.4) obtained from Fig. 23.2.4, and the equation [see Eq. (23.2.12)] ω c ω 1 = N 1 N 1 + N 3 = m (3) c1 . (23.3.5) Using the system of equations (23.3.1) to (23.3.5), we obtain the following relations between N 1 , m (3) c1 , and k b : N 1 = 2m (3) c1 (2 + k b ) 2m (3) c1 + sin  π n  − 1 . (23.3.6) Because N 1 > 0, we obtain that m (3) c1 > 1 − sin  π n  2 . (23.3.7) Inequality (23.3.7) represents the restriction for the minimum value of m (3) c1 considering as given the number n of planet gears. Relation Between Tooth Numbers of Planetary Train of Fig. 23.2.4 The conditions of assembly of the planetary gear train shown in Fig. 23.2.4 yield, as shown below, a relation between tooth numbers N 1 and N 3 and the number n of planet gears. The number of teeth N 2 of planet gears does not affect the conditions of assembly. The derivations are based on the following considerations [Litvin et al., 2002e]: Step 1: Consider initially the assembly of a train that is formed by gears 1, 3, and planet gear 2 (1) [Fig. 23.3.2(a)]. Carrier c is in the position shown in the figure and the axes of tooth symmetry of gear 2 (1) coincide with reference line O 3 O (1) 2 and the axes of spaces of gears 1 and 3. P1: JXR CB672-23 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2:6 23.3 Conditions of Assembly 705 Figure 23.3.2: Installment of planet gears 2 (1) and 2 (2) . NOTE: The drawings correspond to the case wherein the tooth number of 2 (1) (i = 1, ,n) is even, but the following derivations are true for gear 2 (i ) with an odd number of teeth. Step 2: Consider now that the neighboring planet gear 2 (2) has to be installed in the gear train wherein gears 1, 3, and 2 (1) have the positions shown in Fig. 23.3.2(a). Gear 2 (2) is mounted on carrier c; the axis of symmetry of gear 2 (2) teeth coincides with O 3 O (2) 2 that forms with O 3 O (1) 2 angle φ c = 2π/N. The axis of space symmetry of gear 3 forms (i) angle m (2) 3 (2π/N 3 ) with line O 3 O (1) 2 (m (2) 3 is an integer number), and (ii) angle δ (2) 3 with the line O 3 O (2) 2 . Similarly, the axis of space symmetry of gear 1 forms (i) angle m (2) 1 (2π/N 1 ) with line O 3 O (1) 2 (m (2) 1 is an integer number), and (ii) angle δ (2) 1 with the line O 3 O (2) 2 . The superscript “(2)” in the designations m (2) 3 and m (2) 1 , δ (2) 3 and δ (2) 1 indicates that planet gear 2 (2) is considered. Angles m (2) k (2π/N j ), δ (2) k (k = 1, 3) and φ c are measured counterclockwise from line O 3 O (1) 2 of center distance. It is evident that P1: JXR CB672-23 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2:6 706 Planetary Gear Trains the planet gear 2 (2) cannot be put into mesh with gears 1 and 3 [Fig. 23.3.2(b)] because δ (2) 3 and δ (2) 1 differ from zero. Step 3: To put gear 2 (2) into mesh with gears 1 and 3, it is necessary to turn gears 1, 3, and 2 (1) holding at rest carrier c. Gears 1 and 3 are turned in opposite directions and therefore angles δ (2) 1 , δ (3) 3 indicate deviations from O 3 O (2) 2 of axes of symmetry of spaces of gears 1 and 3 in opposite directions. The ratio δ (2) 3 /δ (2) 1 is determined as N 1 /N 3 that is the gear ratio of the inverted gear drive formed by movable gears 1, 2 (1) , 3, and the fixed carrier c. The magnitude δ (2) k (k = 1, 3) must be less than the angular distance between neighboring teeth. The magnitude m (2) k (k = 1, 3) represents the integer number of spaces of gear 1 and 3 located in the area formed by lines O 3 O (1) 2 and O 3 O (2) 2 [Fig. 23.3.2(b)]. Step 4: Figure 23.3.2(b) enables us to determine the magnitude of related turns δ (2) 3 and δ (2) 1 required for the assembly of planet gear 2 (2) with gears 1 and 3. The generalized conditions of assembly of a planet gear 2 (k) (k = 2, ,n) in a gear drive with n planet gears are represented by the following equations: (i) Figure 23.3.2(b) extended for an assembly of planet gear 2 (k) yields m (k) 1 2π N 1 − δ (k) 1 = (k − 1)2π n  δ (k) 1 < 2π N 1  (23.3.8) m (k) 3 2π N 3 − δ (k) 3 = (k − 1)2π n  δ (k) 3 < 2π N 3  (23.3.9) δ (k) 3 δ (k) 1 = N 1 N 3 . (23.3.10) (ii) Equations (23.3.8) to (23.3.10) yield the following relation: (k − 1)(N 1 + N 3 ) n = m (k) 1 + m (k) 3 (k = 2, ,n) (23.3.11) (iii) Taking into account that (m (k) 1 + m (k) 3 ) is an integer number, we obtain that (N 1 + N 3 )/n has to be an integer number as well. This condition is observed, for instance, in the case where N 1 = 62, N 3 = 228, and n = 5. Determination of m (k) 1 ,m (k) 3 , δ (k) 1 , and δ (k) 3 (k = 1, ,n) Equations (23.3.8) and (23.3.9) and inequalities for δ (i ) 1 and δ (i ) 3 yield the following inequalities for determination of m (k) 1 and m (k) 3 : m (k) 1 − (k − 1)N 1 n < 1(k = 2, ,n) (23.3.12) (k − 1)N 3 n − m (k) 3 < 1(k = 2, ,n) (23.3.13) where m (k) 1 and m (k) 3 are integer numbers. We recall that m (i ) 1 and m (i ) 3 are the integer number of spaces of gears 1 and 3 that neighbor to the line of center distance O 3 O (i ) 2 . Figure 23.3.2(b) shows m (2) 3 and m (2) 1 of such spaces that neighbor to the line O 3 O (2) 2 . P1: JXR CB672-23 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2:6 23.4 Phase Angle of Planet Gears 707 Table 23.3.1: Parameters m (k) 1 , m (k) 3 , δ (k) 1 , and δ (k) 3 i m (k) 1 m (k) 3 δ (k) 1 δ (k) 3 10 0 0 0 213 45 3 5·62 2π 3 5·228 2π 325 91 1 5·62 2π 1 5·228 2π 4 38 136 4 5·62 2π 4 5·228 2π 5 50 182 2 5·62 2π 2 5·228 2π Determination of δ (k) 1 and δ (k) 3 that represent the turning angles of gears 1 and 3 for the assembly of planet gear 2 (i ) with gears 1 and 3 is based on the following equations: δ (k) 1 = m (k) 1 2π N 1 − (k − 1)2π n (k = 2, ,n) (23.3.14) δ (k) 3 = (k − 1)2π n − m (k) 3 2π N 3 (k = 2, ,n). (23.3.15) Numerical Example 23.3.1 A planetary gear drive with N 1 = 62, N 3 = 228, and n = 5 is considered. It is easy to verify that (N 1 + N 3 )/n is an integer number and the requirement (23.3.11) is ob- served indeed. The results of computations of m (k) 1 , m (k) 3 , δ (k) 1 , and δ (k) 3 are presented in Table 23.3.1. 23.4 PHASE ANGLE OF PLANET GEARS The concept of the phase angle is used in this chapter for computation of transmission errors (see Sections 23.7 and 23.8). A phase angle determines the angle that is formed by the axis of symmetry of the tooth (space) with the respective line of the center distance. The phase angle is zero wherein the axis of tooth (space) symmetry coincides with the respective line of center distance as shown in Fig. 23.3.2(a). Figure 23.3.2(b) shows that axes of tooth (space) symmetry of gears 1, 2, and 3 will coincide with O 3 O (2) 2 after the related turns through angles δ (2) 1 and δ (2) 3 of gear 1 and 3 are accomplished (the gear drive formed by gears 1, 2 (1) , and 3 is considered). However, such a turn will cause the respective axes of tooth (space) symmetry of gears 1, 2 (1) , and 3 to be deviated from the line of center distance O 3 O (2) 2 . To restore the orientation of tooth (space) axis of symmetry of gears 1, 2 (1) , and 3 as shown in Fig. 23.3.2(a), we provide turns of gear 1 and 3 of the gear drive formed by 1, 2 (1) ,2 (2) , and 3 whereas carrier c is held at rest. The turn of gears 1 and 3 of the gear drive (1, 2 (1) ,2 (2) ,3)is performed in the direction that is opposite to the direction of the turns δ (2) 1 and δ (2) 2 . The P1: JXR CB672-23 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2:6 708 Planetary Gear Trains Figure 23.4.1: Illustration of orientation of tooth (spaces) axes of symmetry of gears 1, 2 (2) , and 3 with respect to center distance O 3 O (2) 2 . turn mentioned above is accomplished for gears 1 and 3 of the gear drive formed by (1, 2 (1) , 3). Figure 23.4.1 shows the obtained orientation of respective axes of tooth (space) sym- metry after the two sets of related turns are accomplished. Axes of tooth (space) symme- try of gears 1, 2 (1) , and 3 are located on line O 3 O (1) 2 . Angles µ (2) 1 , µ (2) 2 , and µ (2) 3 indicate the deviation of the respective axes of tooth (space) symmetry of gears 1, 2 (2) , and 3 from O 3 O (2) 2 . Figure 23.4.2 represents in enlarged scale the orientation of spaces of gear 1 in the area determined by spaces of numbers 1 and m (k) 1 (k = 2, 3, 4, 5). The phase angle  (k) 1 is formed by line O 3 O (k) 2 and the space number (m (k) 1 − 1) that neighbors to O 3 O (k) 2 and is measured clockwise, in the direction of rotation of gear 1 in the gear drive with the carrier held at rest (Fig. 23.2.4). Figure 23.4.2 yields the following equation for determination of the phase angle:  (k) 1 = (k − 1)2π n − (m (k) 1 − 1) 2π N 1 (k = 2, ,5) (23.4.1) Using the input data for numerical example 23.3.1, we obtain the following results:  (2) 1 = 2 5 · 62 2π;  (3) 1 = 4 5 · 62 2π;  (4) 1 = 1 5 · 62 2π;  (2) 1 = 3 5 · 62 2π. P1: JXR CB672-23 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2:6 23.5 Efficiency of a Planetary Gear Train 709 Figure 23.4.2: For derivation of phase angle  (k) 1 . 23.5 EFFICIENCY OF A PLANETARY GEAR TRAIN Let us compare a planetary gear train with a conventional one, with fixed gear axes, designed for the same gear ratio of angular velocities of the input and output mechanism links. The comparison shows the following: (i) The planetary gear train has smaller dimensions than the conventional one. A conventional design requires application of a set of conventional gear drives but not a single train is applied for the assigned reduction of speed. (ii) However, a planetary gear train usually has a much lower efficiency in comparison with a conventional gear train. An exception is the planetary gear train shown in Fig. 23.2.4 (see Example 23.5.2 below). The determination of the efficiency of a planetary gear train is a complex problem. A simple solution to this problem is proposed by Kudrjavtzev et al. [1993] and is based on the following considerations: (i) The efficiency of a planetary gear train is related to the efficiency of an inverted train when the relative velocities of the planetary train and the inverted one are observed to be the same. The inverted train is obtained from the planetary train wherein the carrier is held at rest and gear j , which has been fixed in the planetary train, is released. (ii) Two cases of efficiency of a planetary train designated by η ( j) ic and η ( j) ci may be considered. Here, designations in η ( j) ic indicate that gear i and carrier c are the driving and driven links of the planetary train. Respectively, designations in η ( j) ci P1: JXR CB672-23 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2:6 710 Planetary Gear Trains indicate that carrier c and gear i are the driving and driven links of the planetary train. In both cases, the superscript ( j ) indicates that gear j is the fixed one. (iii) We consider gear i or carrier c as the driving link of the planetary gear train if M k ω k > 0, (k = 1, c) (23.5.1) where M k is the torque applied to link k; ω k is the angular velocity of link i in absolute motion, in rotation about the frame of the planetary gear train. (iv) It is assumed that a torque M i of the same magnitude is applied to link i of the planetary and inverted gear trains. Torque M i is considered as positive if i (but not c) is the driving link. In the case in which the driving link is the carrier, torque M i is the resisting moment and M i < 0. The ratio (ω c /ω i ) may be obtained from the kinematics of the planetary gear train using the following equation for the train with fixed gear j : ω 1 − ω c −ω c = m (c) ij . (23.5.2) The determination of the efficiency of planetary gear trains is considered for two typical examples. The discussed approach may be extended and applied to various examples of planetary gear trains. Example 23.5.1 The planetary gear train shown in Fig. 23.2.4 is considered for the following conditions: (i) gear 3 is fixed ( j = 3); and (ii) gear 1 and carrier c are the driving and driven links, respectively. Equation (23.5.2) yields ω c ω 1 = N 1 N 3 + N 1 . (23.5.3) Consider now the inverted gear train where rotation is provided from gear 1 to gear 3 wherein the carrier is fixed. Torque M 1 is applied to gear 1 and M 1 is positive because gear 1 is the driving gear in the planetary gear train. The angular velocity ω 1 of gear 1 of the planetary gear train is of the same direction as M 1 , and M 1 ω 1 > 0. The efficiency η (3) 1c of the planetary train is determined as η (3) 1c = M 1 ω 1 − P l M 1 ω 1 . (23.5.4) Here, (M 1 ω 1 − P l ) is the output power, and M 1 ω 1 is the input power (M 1 ω 1 > 0). The key for determination of η (3) 1c is that the power P l lost in the planetary gear train is determined as the power lost in the inverted train. The input power of the inverted train is M 1 (ω 1 − ω c ) > 0, because M 1 > 0 and (ω 1 − ω c ) > 0. We consider as known the coefficient  (c) = 1 − η (c) of the inverted train. Then, we may determine the power lost in the inverted gear train as P l =  (c) M 1 (ω 1 − ω c ). (23.5.5) Equations (23.5.3), (23.5.4), and (23.5.5) yield η (3) 1c = 1 − ψ (c)  N 3 N 1 + N 3  . (23.5.6) P1: JXR CB672-23 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2:6 23.6 Modifications of Gear Tooth Geometry 711 Example 23.5.2 The same planetary gear train is considered given the conditions that the driving and driven links of the planetary gear train are the carrier and link 1, respectively. Gear 1 is now the driven link of the planetary gear train; M 1 < 0 because M 1 is the resisting moment. We consider now the inverted train taking into account that M 1 (ω 1 − ω c ) < 0, because M 1 < 0 and (ω 1 − ω c ) > 0. The power P l lost in the inverted train is determined as follows: P l = 1 η (c) (−M 1 )(ω 1 − ω c ) − (−M 1 )(ω 1 − ω c ) = 1 − η (c) η (c) (−M 1 )(ω 1 − ω c ). (23.5.7) Equation (23.5.7) provides that the lost power P l is positive. (Recall that M 1 < 0 and (ω 1 − ω c ) > 0.) Then we obtain η (3) c1 = P driven P driven + P l = −M 1 ω 1 −M 1 ω 1 + 1 − η (c) η (c) (−M 1 )(ω 1 − ω c ) = 1 1 + 1 − η (c) η (c)  1 − ω c ω 1  = 1 1 + 1 − η (c) η (c)  N 3 N 1 + N 3  . (23.5.8) It follows from Eq. (23.5.8) that the efficiency η (3) c1 >η (c) . This means that the planetary gear train shown in Fig. 23.2.4 has higher efficiency than the inverted gear train if the rotation in the planetary gear train is transformed from the carrier c to the sun gear 1. 23.6 MODIFIC ATIONS OF GEAR TOOTH GEOMETRY We limit the discussion to the modification of gear tooth geometry for the planetary gear train shown in Fig. 23.2.4. Spur gears of involute profile are applied in the existing design. The purposes of modification of tooth geometry are as follows: (i) Improvement of bearing contact and reduction of transmission errors. This goal is achieved by application of double-crowned planet gears. (ii) Reduction of backlash between the planet gears and sun gear 1 and ring gear 3. The reduction of backlash enables us to obtain a more uniform distribution of load between the planet gears (see below). Modification of Geometry of Planet Gears The developed modification is based on double crowning of planet gears accomplished as a combination of profile crowning and longitudinal crowning (see Litvin et al. [2001c] and Chapter 15). Longitudinal crowning enables us to substitute instantaneous line contact of tooth surfaces by point contact and avoid an edge contact. Longitudinal crowning is achieved by tool plunging. Profile crowning of a planet gear enables us to substitute the involute profile by a profile that is conjugated to a parabolic profile of a rack-cutter (see Chapter 15). Then, a parabolic function of transmission errors can P1: JXR CB672-23 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2:6 712 Planetary Gear Trains be predesigned. Such a function of transmission errors is able to absorb almost linear functions of transmission errors caused by errors of alignment (see Section 9.2). Modification of Tooth Geometry of Gears 1 and 3 The purpose of modification is to regulate backlash caused by angular errors of install- ment of the planet gears on the carrier [Litvin et al., 2002e]. The goal mentioned above is achieved as follows: (i) The tooth surface of  1 is designed as an external screw involute one of a small helix angle. Respectively, the tooth surface of  3 is designed as an internal screw involute one of the same helix angle and direction as  1 . (ii) The regulation of backlash between a planet gear 2 (i ) and gears 1 and 3 is achieved by axial translation of gear 2 (i ) that is accomplished during the assembly. The regulation has to be performed for the whole set of planet gears. Figure 23.6.1 illustrates schematically how the backlash between planet gear 2 (i ) and gears 1 and 3 is regulated. Figure 23.6.1(a) shows the backlash x (i ) existing before regulation. Figure 23.6.1(b) shows that the backlash is eliminated by axial displacement z (i ) of planet gear 2 (i ) . The regulation described above has to be accomplished for all planet gears of the set 2 (i ) (i = 1, ,n). 23.7 TOOTH CONTACT ANALYSIS (TCA) The TCA computer program enables simulation of misaligned gear drives for determi- nation of transmission errors and conditions of contact. Conventional Gear Drive In the case of a conventional gear drive formed by two gears, there are two contacting gear tooth surfaces and the simulation of meshing is based on the following procedure (see Section 9.4). (i) The gear tooth surfaces are represented in a mutual coordinate system S f rigidly connected to the housing of the gear drive. (ii) The instantaneous tangency of gear tooth surfaces  1 and  2 is represented by the following vector equations: r (1) f (u 1 ,θ 1 ,φ 1 ) − r (2) f (u 2 ,θ 2 ,φ 2 ) = 0 (23.7.1) n (1) f (u 1 ,θ 1 ,φ 1 ) − n (2) f (u 2 ,θ 2 ,φ 2 ) = 0. (23.7.2) Here, designations (u i ,θ i )(i = 1, 2) indicate surface parameters; φ i (i = 1, 2) are generalized parameters of motion of gears. Vector equation (23.7.1) means that at a point M of tangency, surfaces  1 and  2 have a common position vector. Vector equation (23.7.2) confirms that the surfaces have a common unit vector of the normals to the surfaces. [...]... rotation of planetary (i ) gear 2 with respect to the carrier c, and φc is the angle of rotation of the carrier Step 3: The conditions of tangency of gears 1 and 2(i ) , and gears 2(i ) and 3 provide ten independent scalar equations These equations contain eight surface parameters of (i ) gears 1, 2(i ) , and 3 and three motion parameters φ1 , φ2c , and φc Considering φ1 as the input parameter, we may obtain...P1: JXR CB672- 23 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2:6 23. 7 Tooth Contact Analysis (TCA) 7 13 Figure 23. 6.1: Schematic illustration of regulation of backlash: (a) backlash between gears 1, 3, and 3( i ) before regulation; (b) elimination of backlash by axial displacement of z (i ) of planet gear 2(i ) Vector equations ( 23. 7.1) and ( 23. 7.2) yield a system of only five... errors and the backlash as well Then, it becomes possible to minimize and equalize the backlash of five planetary gears by regulation The resulting function of transmission errors of the sub -gear drive (1, 2(i ) , 3) is determined as 3 (φ1 ) = 3 (φ2 (φ1 )) − N1 φ1 N3 ( 23. 7.4) where 3 (φ2 (φ1 )) is obtained from the TCA computer program An approximate solution for 3 (φ1 ) is represented as 3 (φ1... crowning and errors of P1: JXR CB672- 23 CB672/Litvin 716 CB672/Litvin-v2.cls February 27, 2004 2:6 Planetary Gear Trains 23. 8 ILLUSTRATION OF THE EFFECT OF REGULATION OF BACKLASH A planetary gear drive with n = 5 planet gears is considered (Fig 23. 2.4) Considering n = 5 sets of sub -gear drives, we may obtain functions of transmission errors for all sub -gear drives of the set of five planet gears as...  Yp         (24 .5. 4)  Z  = M pc  Z  =  0 sin γ cos γc 0   Zc  c    c   p 1 1 1 0 0 0 1 P1: JXT CB672-24 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2 :30 24 .5 Generation by Disk-Shaped Tool: Workpiece Surface is Given 731 Equations (24 .5 .3) and (24 .5. 4) yield −Zc sin γc − y p Zc cos γc − z p Ec − x p = = Nxp N yp Nzp (24 .5. 5) Using Eq (24 .5. 5), we can eliminate Zc Then,... derivations of matrices Mc f and M f p the drawings of Fig 24.1 .3) Step 3: Determine ρ using the equation 2 2 ρ = xc + yc 0 .5 = ρ(u p , θ p ) (24 .5. 10) P1: JXT CB672-24 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2 :30 732 Generation of Helicoids Figure 24 .5. 1: For derivation of diskshaped tool profile Step 4: Considering θ p as the input parameter and using Eqs (24 .5. 9) and (24 .5. 10), we determine the... the input one Surfaces 1 and 2 are in point contact if the respective Jacobian for the system of Eqs ( 23. 7.1) and ( 23. 7.2) differs from zero (see Section 9.4) Then, at the point of tangency of 1 and 2 , the system of equations ( 23. 7.1) and ( 23. 7.2) can be solved by functions (see Section 9.4) {u 1 (φ1 ), θ1 (φ1 ), u 2 (φ1 ), θ2 (φ1 ), φ2 (φ1 )} ∈ C 1 ( 23. 7 .3) P1: JXR CB672- 23 CB672/Litvin CB672/Litvin-v2.cls... the set of five is in mesh with gear 3, whereas the remaining planet gears are not in mesh with gear 3 The graphs of Fig 23. 8.1 show that the planet gear that is in mesh is Figure 23. 8.2: Illustration of functions of transmission errors for sub-drives and integrated function of transmission error P1: JXR CB672- 23 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2:6 23. 8 Illustration of the Effect... 2004 2:6 23. 8 Illustration of the Effect of Regulation of Backlash 717 ∗ gear 2(2) The backlash at the position φ1 = φ1 between gears 2(k) (k = 1, 3, 4, 5) is determined by 3 = M2 Mk (k = 1, 3, 4, 5) The backlash in sub -gear drives can be reduced by regulation of installment of planet gears on the carrier (see Fig 23. 6.1) Figure 23. 8.2 shows the integrated function φ (φ1 ) of transmission errors obtained... February 27, 2004 2:6 714 Planetary Gear Trains Using functions ( 23. 7 .3) and surface equations, we can determine the path of contact on surfaces 1 and 2 , and the function of transmission errors (see Section 9.4) Application of TCA for a Planetary Gear Drive (Fig 23. 2.4) A planetary gear drive with several planet gears is a multi-body system Considering the misaligned gear drive as a system of rigid bodies, . ω c ). ( 23. 5. 5) Equations ( 23. 5 .3) , ( 23. 5. 4), and ( 23. 5. 5) yield η (3) 1c = 1 − ψ (c)  N 3 N 1 + N 3  . ( 23. 5. 6) P1: JXR CB672- 23 CB672/Litvin CB672/Litvin-v2.cls February 27, 2004 2:6 23. 6 Modifications. 2:6 23. 4 Phase Angle of Planet Gears 707 Table 23. 3.1: Parameters m (k) 1 , m (k) 3 , δ (k) 1 , and δ (k) 3 i m (k) 1 m (k) 3 δ (k) 1 δ (k) 3 10 0 0 0 2 13 45 3 5 62 2π 3 5 228 2π 32 5 91 1 5 62 2π 1 5 228 2π 4. to Eqs. ( 23. 3.1) to ( 23. 3 .3) , we use equation N 2 = N 3 − N 1 2 ( 23. 3.4) obtained from Fig. 23. 2.4, and the equation [see Eq. ( 23. 2.12)] ω c ω 1 = N 1 N 1 + N 3 = m (3) c1 . ( 23. 3 .5) Using the