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86
()
1
1
1
−−=
ω
−ω−
=
ω
ω
′
=
p
p
D
p
D
K
K
m
m
m
R (24)
The value of R is negative. This result means that the motion of gear
K relative to gear J in this planetary train is opposite in direction to the
absolute velocity of gear K. However, the direction of relative rotation
does not influence the magnitude of tooth mesh loss in the gear pair.
Therefore, the absolute value of R must be used in the calculation of train
losses and efficiency. Introducing this value into Equation 23 gives
−
η−+
=
−η−+
=
JAKB
JA
t
pt
t
dddd
dd
m
e
)1(1
1
)1)(1(1
1
(25)
Equation 25 shows that as the speed ratio of this train increases, the
absolute value of R and, consequently, the tooth mesh losses in the train
increase, while efficiency decreases. For trains of this type, having a
large speed ratio, the velocities of gear engagement are relatively high
and the efficiency relatively low. The power developed at the gears in
such trains is R times higher than in a conventional train having the
same size gears and the same speed and power at the output shaft.
For the train shown in Fig. 7a, assume that
21
<
<
K
BJA
dddd
.
Such a train wil1 reduce the speed and reverse the direction of rotation
of the output shaft, so that m
p
is now negative with an absolute value
greater than one (Equation 20). In the planetary train, Fig. 7a, the
angular velocity of the gear K is now negative, and the external resisting
torque acting on output shaft H is positive.
When the planet cage is stopped, the angular velocity of gear K
becomes
)()(
DKDKK
ω
+
ω
−
=
ω
−
+
ω−=ω
′
(26)
which is a negative value. The product of external torque acting on this
gear and angular velocity is negative. Therefore, gear K' is a driven
member in the equivalent train with a stopped planet cage. Gear A' is the
driver in this case.
The power developed by gear K is P
0
R. The power of engagement
of the driving gear A' is
tAB
RPP
η
=
′
0
. Tooth mesh loss in this train is
)1(
1
0 t
t
t
RPL η−
η
= (27)
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