Gear Geometry and Applied Theory Episode 3 Part 5 ppt

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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 VA and VB represent 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)

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

ωc = ω1 + ω3

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 ω1 = 0 yields

ωc = ω3

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 ω1 = ω3 yields that

ωc = ω1 = ω3. (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 (ω1 = −ω3), we get that ωc = 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 kbm, where m is the

module of the gears and kb is the unitless coefficient. Our goal is to derive an equation

that relates N1, kb, and the gear ratio m(3)

c1 = ωc/ω1 of a planetary gear train wherein

gear 3 is fixed. The derivation is based on application of the following equation:

r2a = E12 sin $π

n

%

− kbm

2 . (23.3.1)

Here, r2a is the radius of the addendum circle of gear 2; E12 is the shortest distance;

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

E12 = N1 + N2

2

m (23.3.2)

r2a =

 N2

2 + 1



m. (23.3.3)

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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

N2 = N3 − N1

2 (23.3.4)

obtained from Fig. 23.2.4, and the equation [see Eq. (23.2.12)]

ωc

ω1

= N1

N1 + N3

= m(3)

c1 . (23.3.5)

Using the system of equations (23.3.1) to (23.3.5), we obtain the following relations

between N1, m(3)

c1 , and kb:

N1 = 2m(3)

c1 (2 + kb)

2m(3)

c1 + sin $π

n

%

− 1

. (23.3.6)

Because N1 > 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 N1 and N3 and the number n of planet

gears. The number of teeth N2 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 O3O(1)

2 and the axes of

spaces of gears 1 and 3.

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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

O3O (2)

2 that forms with O3O(1)

2 angle φc = 2π/N. The axis of space symmetry of

gear 3 forms (i) angle m(2)

3 (2π/N3) with line O3O(1)

2 (m(2)

3 is an integer number), and

(ii) angle δ

(2)

3 with the line O3O (2)

2 . Similarly, the axis of space symmetry of gear 1 forms

(i) angle m(2)

1 (2π/N1) with line O3O (1)

2 (m(2)

1 is an integer number), and (ii) angle δ

(2)

1

with the line O3O (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/Nj), δ

(2)

k (k = 1, 3) and

φc are measured counterclockwise from line O3O(1)

2 of center distance. It is evident that