Gear Geometry and Applied Theory Episode 3 Part 5 ppt

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 21[Fig.

23.3 Conditions of Assembly 703

It is easy to prove that m (c)31= −1 This result is obtained from the following siderations [Fig 23.2.5(b)] Suppose that the carrier is fixed and gears 1 and 2, and 3

con-and 2 are in contact at points A con-and B, respectively Vectors V Aand VBrepresent linear

velocities of corresponding gears at points A and B Taking into account that N1= N3

and VA= −VB , we get that m (c)31= −1 The negative sign of m (c)

31means 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= ω1+ ω3

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) withω1= 0 yields

ω c =ω3

The discussed mechanism works as a planetary gear train

(2) Consider now that gears 1 and 3 are rotated with equal angular velocities in thesame direction Equation (23.2.14) withω1= ω3yields that

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 (ω1= −ω3), we get thatω c= 0 [see Eq (23.2.14)] The discussedmechanism 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 inFig 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 bis the unitless coefficient Our goal is to derive an equation

that relates N1, k b , and the gear ratio m(3)c1 = ω c /ω1 of a planetary gear train whereingear 3 is fixed The derivation is based on application of the following equation:

Here, r 2a is the radius of the addendum circle of gear 2; E12is the shortest distance;

n is the number of planet gears It is easy to verify that

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

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 N1and N3and the number n of planet gears The number of teeth N2of 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 O3O2(1)and the axes ofspaces of gears 1 and 3

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 thegear 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

O3O2(2) that forms with O3O2(1) angle φ c = 2π/N The axis of space symmetry of gear 3 forms (i) angle m(2)3 (2π/N3) with line O3O2(1)(m(2)3 is an integer number), and(ii) angleδ(2)

3 with the line O3O2(2) Similarly, the axis of space symmetry of gear 1 forms

(i) angle m(2)1 (2π/N1) with line O3O2(1) (m(2)1 is an integer number), and (ii) angleδ(2)

the planet gear 2(2)cannot be put into mesh with gears 1 and 3 [Fig 23.3.2(b)] because

δ(2)

3 andδ(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 O3O2(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

N1/N3that 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 O3O2(1)and O3O2(2)

[Fig 23.3.2(b)]

Step 4: Figure 23.3.2(b) enables us to determine the magnitude of related turns δ(2)

3andδ(2)

1 required for the assembly of planet gear 2(2)with gears 1 and 3 The generalizedconditions 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

(iii) Taking into account that (m (k)1 + m (k)

3 ) is an integer number, we obtain that (N1+

N3)/n has to be an integer number as well This condition is observed, for instance,

in the case where N1= 62, N3= 228, and n = 5.

3 yield the following

inequalities for determination of m (k)1 and m (k)3 :

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 O3O2(i )

Figure 23.3.2(b) shows m(2)and m(2)of such spaces that neighbor to the line O3O(2)

23.4 Phase Angle of Planet Gears 707

A planetary gear drive with N1= 62, N3= 228, and n = 5 is considered It is easy

to verify that (N1+ N3)/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 , andδ (k)

3 are presented inTable 23.3.1

23.4 PHASE ANGLE OF PLANET GEARS

The concept of the phase angle is used in this chapter for computation of transmissionerrors (see Sections 23.7 and 23.8) A phase angle determines the angle that is formed bythe 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 therespective 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 O3O2(2)after the related turns through anglesδ(2)

1 andδ(2)

3 of gear 1 and 3are 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 O3O2(2) To restore the orientation oftooth (space) axis of symmetry of gears 1, 2(1), and 3 as shown in Fig 23.3.2(a), weprovide 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) isperformed in the direction that is opposite to the direction of the turnsδ(2)

andδ(2) The

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 O3O2(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) 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 O3O2(1) Anglesµ(2)

sym-1 ,µ(2)

2 , andµ(2)

3 indicatethe deviation of the respective axes of tooth (space) symmetry of gears 1, 2(2), and 3

from O3O2(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 O3O2(k) and the space number (m (k)1 − 1) that neighbors to O3O2(k)

and is measured clockwise, in the direction of rotation of gear 1 in the gear drive withthe carrier held at rest (Fig 23.2.4)

Figure 23.4.2 yields the following equation for determination of the phase angle:

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 mechanismlinks 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 comparisonwith a conventional gear train An exception is the planetary gear train shown inFig 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 invertedtrain when the relative velocities of the planetary train and the inverted one areobserved 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 )

i c and η ( j )

ci may beconsidered Here, designations in η ( j )

i c indicate that gear i and carrier c are the

driving and driven links of the planetary train Respectively, designations inη ( j )

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 thekinematics of the planetary gear train using the following equation for the train

with fixed gear j :

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

ω1 = N1

Consider now the inverted gear train where rotation is provided from gear 1 to gear 3

wherein the carrier is fixed Torque M1is applied to gear 1 and M1is positive becausegear 1 is the driving gear in the planetary gear train The angular velocityω1of gear 1

of the planetary gear train is of the same direction as M1, and M1ω1> 0.

Here, (M1ω1− P l ) is the output power, and M1ω1is the input power (M1ω1> 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 M1(ω1− ω c)> 0, because M1> 0 and (ω1− ω c)> 0 We consider as known

the coefficient (c) = 1 − η (c) of the inverted train Then, we may determine the powerlost in the inverted gear train as

P l =  (c) M1(ω1− ω c). (23.5.5)Equations (23.5.3), (23.5.4), and (23.5.5) yield

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 anddriven 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; M1< 0 because M1is the resisting

moment We consider now the inverted train taking into account that M1(ω1− ω c)< 0,

It follows from Eq (23.5.8) that the efficiencyη(3)

c1 > η (c) This means that the planetarygear 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 MODIFICATIONS OF GEAR TOOTH GEOMETRY

We limit the discussion to the modification of gear tooth geometry for the planetarygear train shown in Fig 23.2.4 Spur gears of involute profile are applied in the existingdesign The purposes of modification of tooth geometry are as follows:

(i) Improvement of bearing contact and reduction of transmission errors This goal isachieved 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 loadbetween 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 linecontact of tooth surfaces by point contact and avoid an edge contact Longitudinalcrowning is achieved by tool plunging Profile crowning of a planet gear enables us tosubstitute 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

be predesigned Such a function of transmission errors is able to absorb almost linearfunctions 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 install-mentioned above

is achieved as follows:

(i) The tooth surface of1is designed as an external screw involute one of a smallhelix angle Respectively, the tooth surface of3is designed as an internal screwinvolute one of the same helix angle and direction as1

(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 Theregulation 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 )andgears 1 and 3 is regulated Figure 23.6.1(a) shows the backlash x (i ) existing beforeregulation 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 allplanet 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 nation of transmission errors and conditions of contact

determi-Conventional Gear Drive

In the case of a conventional gear drive formed by two gears, there are two contactinggear 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 rigidlyconnected to the housing of the gear drive

(ii) The instantaneous tangency of gear tooth surfaces1and2is represented by thefollowing vector equations:

r(1)f (u1, θ1, φ1)− r(2)

f (u2, θ2, φ2)= 0 (23.7.1)

n(1)f (u1, θ1, φ1)− n(2)

f (u2, θ2, φ2)= 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 1and2have a common position vector Vector equation (23.7.2) confirms that the surfaceshave a common unit vector of the normals to the surfaces

23.7 Tooth Contact Analysis (TCA) 713

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 ofz (i )of planet gear 2(i ).Vector equations (23.7.1) and (23.7.2) yield a system of only five independent scalarequations because|n(1)

f | = |n(2)

f | = 1 One of the parameters, let us say φ1, may be chosen

as the input one Surfaces1and2are in point contact if the respective Jacobian forthe system of Eqs (23.7.1) and (23.7.2) differs from zero (see Section 9.4) Then, at thepoint of tangency of1 and2, the system of equations (23.7.1) and (23.7.2) can besolved by functions (see Section 9.4)

{u1(φ1), θ1(φ1), u2(φ1), θ2(φ1), φ2(φ1)} ∈ C1. (23.7.3)

Using functions (23.7.3) and surface equations, we can determine the path of contact

on surfaces1and2, 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 themisaligned gear drive as a system of rigid bodies, we may find out that only one planetgear is in mesh at every instant Conditions of tangency are determined as follows:

Step 1: The tooth surfaces of gears 1, 2(i ), and 3 are represented in fixed coordinate

system S3

Step 2: The rotation of gears 1 and 2(i )is determined by three parametersφ1,φ (i )

2c, and

φ c Here,φ1is the angle of rotation of gear 1,φ (i )

2c is the angle of rotation of planetarygear 2(i ) with respect to the carrier c, and φ cis 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 provideten independent scalar equations These equations contain eight surface parameters ofgears 1, 2(i ), and 3 and three motion parametersφ1, φ (i )

2c, and φ c Consideringφ1 asthe input parameter, we may obtain from the TCA computer program the transmissionfunctionφ c(φ1) and then determine the function of transmission errors of the misalignedplanetary gear train

The solution of ten non-linear equations can be simplified by representing them astwo subsystems of five equations each and then applying an iterative process of solution.Applying tooth contact analysis (TCA) for various planet gears 2(i ), we may determinewhich of the planet gears is in mesh at the considered positionφ1

Function of Transmission Errors of Sub-Gear Drives

For the purpose of regulation of backlash, we may determine the backlash consideringsub-gear drives but not the planetary gear drive The sub-gear drives used for such anapproach are formed by gears (1, 2 (i )), (2(i ) , 3), and (1, 2 (i ) ,3) (i = 1, n).

Transformation of rotation of the sub-gear drives is performed whereas the carrier isheld at rest Applying TCA for the sub-gear drives, it becomes possible to determine thefunctions of transmission errors and the backlash as well Then, it becomes possible tominimize 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 mined as

23.7 Tooth Contact Analysis (TCA) 715

Figure 23.8.1: Functions of transmission errorsφ (i )

3 (φ1 ) caused by double crowning and errors of location of planetary gears.

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 allsub-gear drives of the set of five planet gears as shown in Fig 23.8.1 The transmissionerrors are caused as the result of double crowning of planet gears and errors of location

of planet gears on the carrier The designations xindicates an error of location of theplanet gear measured in a direction that is perpendicular to the shortest distance betweenthe planet gear and gear 1 (Fig 23.2.4) Functions of transmission errors are represented

in Fig 23.8.1 taking into account the phase angles of planet gears (see Section 23.4) It

is important to recognize that errors of functionsφ (i )

3 (φ1) might be positive or negative(Fig 23.8.1)

Analysis of the shape of the functions of transmission errors represented in Fig 23.8.1shows the following:

(a) Only one planetary gear of the set of five planets is in tangency with gear 3

(b) It can happen, as shown in Fig 23.8.1, that the same planet gear of the set of

five is in mesh with gear 3, whereas the remaining planet gears are not in meshwith 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.

23.8 Illustration of the Effect of Regulation of Backlash 717

gear 2(2) The backlash at the positionφ1= φ∗