A simple threespeed transmission is shown The power flow is

A simple three-speed transmission is shown. The power flow is as follows: (a) First gear: gear 4 is shifted to mesh with gear 7; power flows through gears 2, 5, 7, 4. (b) Second gear: gear 3 is shifted to mesh with gear 6; power flows through gears 2, 5, 6, 3. (c) Reverse gear: gear 4 is shifted to mesh with gear 9; power flows through gears 2, 5, 8, 9, 4. An automobile with this transmission has a differential ratio of 3:1 and a tire outside diameter of 24 in. If the automobile is traveling at 30 mph, determine the engine speed for the car under first gear, second gear and reverse gear respectively.

Solution

solution:

1)here vehicle is travelling at speed of 30 mile per hour

V=30 MpH or 13.4112 m/s

d=24 inch or .6096 m

2)angular velocity of wheel

V=(d/2)Ww

Ww=13.4112*2/.6096=44 rad/s

3)differential gear box operate at

G=3/1

G=Wd/Ww

Wd=3*44=132 rad/s

Wd=2*pi*Nd/60

Nd=Ngout=1260.5 rpm

4)here for first gear 2,5,7,4,gear ratio is

G1=N2/N4=N2*N7/N5*N4=T5*T4/T2*T7

G1=30*26/15*19=2.7368

G1=N2/N4

N4=Nd=1260.5

N2=2.7368*1260.5

N2=3449.73 rpm

N2 is speed of engine

5)for second gear

G2=N2/N3=N2*N6/N5*N3=T5*T3/T2*T6

G2=30*21/15*24

G2=1.75=N2/N3

N3=Nd

N2=1.75*1260.5=2205.87 rpm

6)for reverse gear

G3=N2/N4=N2*N8*N9/N5*N9*N4=T5*T9*T4/T2*T8*T9

G3=30*14*26/15*13*14=4

G3=4

G3=N2/N4

N4=Nd

N2=4*1260.5=5002.4 rpm

N2 is speed of engine for reverse gear speed transmission