Transcript Document

A differential gear train
General solution for differential gears
Gears 2 and 3 rotate relative to
the frame. Therefore
Gear 3 is fixed to the carrier of the
planetary drive. Considering gears
5 and 6,
Relative to the carrier,
Solution
1
5  3
1
6  3
1
1

N6
N5
 N 5 5  N 5 3   N 6 6  N 6 3
1
1
1
1
 N 5 5  N 6 6   N 6 3  N 5 3
1
1
1
1

1
ω3 =
1
1
N 5 ω5 + N 6 ω6
N5 + N6
Solution
1
1
ω3 =
1
N 5 ω5 + N 6 ω6
N5 + N6
This defines the relation between the ring gear speed and
the wheel speeds. The ring gear speed is in turn related to
the driveshaft speed which comes from the engine.
Normally
N5 = N6
1
 ω3 =
1
Ring gear speed is the
mean of the two wheel
speeds
2N 5
1
 ω3 =
1
1
N 5 ω5 + N 5 ω6
1
ω5 + ω6
2
Solution
1
1
ω3 =
1
ω5 + ω6
2
For straight line
motion both wheels
must have the same
speed
 ω5 = - ω6 + 2 ω 3
1
1
1
1
1
i .e . ω 5 = ω 6
 ω6 = - ω6 + 2 ω 3
1
1
1
 2 ω6 = 2 ω 3
1
1
 ω5 = ω6 = ω 3
1
1
1
For straight line motion ring gear must rotate at the same
speed as the wheels
Solution
1
4  3  4
1
3
3
3
1
4  3
1
6  3
1
1

4
6
N6
N4
 N 4 4  N 4 3   N 6 6  N 6 3
1
1
1
1
 N 4 4   N 6  N 4  3  N 6 6
1
1
 N 6  N 4   3  N 61 6
1
 4 
1
1
N4

N6
N4
Solution
For straight line
motion
1
1
1
ω5 = ω6 = ω 3
 N 6  N 4   3  N 61 6
1
 4 
1
N4
N 4 3
1
 4 
1
N4
 4  3
1
1
For straight line motion all gears must rotate at the same
absolute speed.
Hence for straight line motion the differential moves like
a rigid shaft
Sample Problem involving differential gear train
The differential for a rear wheel-driven vehicle
is shown schematically. If the drive shaft turns
at 900 rpm, what is the speed of the vehicle if
neither wheel slips and the outside diameter of
the wheels is 24 in?
Solution to sample Problem involving differential gear train
The differential for a rear wheel-driven vehicle is shown schematically. If
the drive shaft turns at 900 rpm, what is the speed of the vehicle if neither
wheel slips and the outside diameter of the wheels is 24 in?
1
 2  900 rpm
1
3  2
1
N2
 900 
N3
28
 273.91  (ignoring sign)
92
For straight line m otion
1
ω 3 = ω5  ω6
1
1
  3   5   6  273.91 rpm
1
1
1
Sample Problem involving differential gear train
Assume that the vehicle is stopped so that the right wheel
sits on a small icy patch and can spin freely while the left
wheel does not spin. Determine the angular velocity of the
right wheel if the angular speed of the drive shaft is 500
rpm.
Solution to sample Problem involving differential gear train
Assume that the vehicle is stopped so that the right wheel sits on a small icy patch
and can spin freely while the left wheel does not spin. Determine the angular
velocity of the right wheel if the angular speed of the drive shaft is 500 rpm.
1
 2  5 0 0 rp m
1
3  2
1
1
N2
28
 500 
N3
 1 5 2 .7 4 (ig n o rin g sig n )
92
5  0
1
1
F o r g en eral m o tio n
0  6
1
 3 
1
1

2
  3  3 0 4 .3 5 rp m
1
ω3 =
6
2
1
ω5 + ω6
2
  6  2  3  2  1 5 2 .7 4
1
1
Sample Problem involving differential gear train
Assume that the vehicle is traveling at 35 mph and turns
around a curve with a radius of 50 ft from the centerline of
the vehicle. The center-to-center distance between the
treads of the right and left wheels is 60 in. Compute the
rotational speed of each rear wheel, the rotational speed of
the ring gear, and the rotational speed of the drive shaft.
Solution to sample Problem involving differential gear train
Assume that the vehicle is traveling at 35 mph and turns around a curve with a radius of 50
ft from the centerline of the vehicle. The center-to-center distance between the treads of
the right and left wheels is 60 in. Compute the rotational speed of each rear wheel, the
rotational speed of the ring gear, and the rotational speed of the drive shaft.
S peed of center of vehicle= 35
m ile
 35
hr
m ile
 5280
hr
ft
/ 3600
m ile
s
hr

1 5 4 ft
3
 51.33
s
ft
s
A ssum ing a left turn,

1
in 
R adius of circle traversed by left w heel =  50  60 in / 12
  47.5 ft
2
ft 


1
in 
R adius of circle traversed by right w heel =  50  60 in / 12
  52.5 ft
2
ft 

For no slip for both w heels
L inear velocity of left w heel =
154 ft
3
L inear velocity of right w heel =

47.5
s
50
154 ft
3
 48.77

s
52.5
s
ft
 53.9
50
ft
s
w heel radius = 1 f t   5  48.77 rad / s ,  6  53.9 rad / s ,
1
1
1
For general m otion
1
2  3
1
N3
N2
1
ω3 =
1
ω5 + ω6
2
 3 
1
48.77  53.9
 51.335 rad / s
2
 (ignoring sign)   2  168.67 rad / s  10120 rad/m in = 1610.65 rpm
1