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88
KB
JA
p
dd
dd
m −=1 (30)
The value of m
p
is positive or negative depending upon the value of
KBJA
dddd
. Assume input shaft H, connected to gear K, revolves
clockwise. The external torque applied to this shaft acts in a positive
direction. Assume also, that
1
<
KBJA
dddd so that 1>m
p
>0. This
arrangement will provide a speed increase without reversing the
direction of rotation. A positive speed ratio greater than one is
theoretically impossible for this train, because it requires the value of
KBJA
dddd to be less than zero.
To "stop" planet cage G, it will be necessary to revolve the entire
system at an angular velocity, — ω
G
. Then, the angular velocity of gear
K becomes
−ω−=
ω
−ω=ω−+ω=ω
′
1
1
)(
p
K
p
K
KGKK
mm
(31)
For the assumed conditions, the value of ω
K
' is negative. Therefore, the
product of positive torque and negative angular velocity is negative, and
gear K' will be a driven member in the equivalent train with a stopped
planet cage. Gear A' will thus be the driver.
If gear A' is the driver,
tiAB
RPP
η
=
′
/
and the tooth mesh loss in the
planetary train is
)1(
1
ti
t
t
RPL η−
η
= (32)
Fig. 8–Differential planetary gear system showing, a, arrangement of system
elements and, b, relationship between efficiency and speed ratio. This train is
basically the same as the one shown in Fig.7a except that direction of power
flow has been reversed
(a)
Output shaft, D
Planet gear, B
Planet gear, J
Input shaft, H
Sun gear, A
(fixed)
Planet cage, G
Sun gear, K
Speed ratio
Efficiency, per cent
4
0
2
0
6
0
8
0
10
0
- 0 1-
2
-
2
(b)
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