Frictional Drive in Textile Frictional Drive in Textile Part 2:Gear DrivePart 2:Gear Drive

Frictional Drive in Textile Frictional Drive in Textile Part 2:Gear DrivePart 2:Gear Drive

Gear Drive1. What is gear drives2. Types of gear drives

1. Straight-tooth gear2. Helical gear3. Double helical gear4. Spiral gear5. Worm and worm wheel

Function of Gear wheel• A pair of toothed wheels carried on separate

shafts.• The teeth meshing together causes one

shaft to drive the other at an exact rate.• Normally, different size of gear (no. of teeth)

can be used to control the speed of the machine.

• Additional of gear can adjust the machine rotating direction.

Type of Gear wheel

Ratio of Gear Trains

Figure 1

Ratios of Gear Trains

• A series of gears in transmission system is called gear trains.

• The velocity ratio (first to last) of the complete train is:

(No. of rev. of the first driving gear)-----------------------------------------(No. of rev. of the final driven gear)

Question 1• Fig 1 (a), wheel A has 50 teeth and

wheel B has 30 teeth. If A rotates at 200 rev/min, at what speed does B rotate?

Answer 1• n1 x teeth1 = n2 x teeth2• 200 x 50 = n2 x 30• n2= 333 rev/min.• With two wheels, the driven gear

rotates in the opposite direction, velocity ratio is negative.

• Fig. 1 (b) has introduced a third wheel C, it will not affect the speed of rotation of B but would change its direction. Wheel C is called idler or carrier

Question 2• If wheel C is added as shown in Fig

1 (b), find out the rotational speed of wheel B if wheel A=50, wheel B=30 and wheel C=10, assume wheel A rotates at 200 rev/min.

Figure 2

Answer 2• Speed at wheel C is

– 200 x 50 = n2 x10; n2=1000

• Speed at wheel B is– 1000 x 10 = n3 x 30; n3= 333rev/min

• This is the same answer as in Q1.• The additional wheel will not change

the rotation speed, but only the direction.

Question 3Compound wheel carrier

• Fig 1 (d) shows a compound wheel carrier. Wheels F and G are keyed to the same shaft and hence rotate at the same speed.

• If E is the first driving wheel and have a speed of 120 rev/min. find out the speed of H?

Answer 3• Speed of wheel F:

– 120 x 60 =n2 x 40; n2=180

• which is the same speed of wheel G;

• Speed of wheel H:– 180 x 20 = n3 x 50; n3=72 rev/min.

General Equation

Rev/min of first driving wheel x (product of teeth of driving wheels) ---------------------------------------- (product of teeth of driven wheels)

Question 4

Figure 2

Question 4• Fig 2 shows a gear train. If the first

driving wheel, k, have a speed of 1,000 rev/min, find out the speed of rotation of final wheel, t.

• The numbers of teeth in gear wheels are: k=20, l=60, m=24, n=54, o=28, p=64, q=20, r=30 and t=40; s is a single-start worm, s=1

Answer 4• Driving gears are: k, m, o, q and s

1000 x (20 x 24 x 28 x 20 x 1) --------------------------- (60 x 54 x 64 x 30 x 40)

= 1.08 rev/min

This example shows the useful of compound carriers that reduce the speed from 1000 rev/min to 1 rev/min