Frictional Drive in Textile Frictional Drive in Textile Part 1:Drum drive, belt Part 1:Drum drive, belt drive and chain drivedrive and chain drive

Frictional Drive in Textile Frictional Drive in Textile Part 1:Drum drive, belt Part 1:Drum drive, belt drive and chain drivedrive and chain drive

Frictional Drive1. What is frictional drives2. Types of frictional drives

1. Drum drive (eg cheese winding)2. Belt drive (spinning, weaving)3. Vee-belt drive (motor power )4. Chain-and-sprocket drive

1) Frictional Drives• Many drives for textile machines rely on

the friction between the driving element and the driven element. Examples:– Belt and rope drives;– Drum drives to cone in yarn preparation– Drum drives to warping beams

• The common feature is one element has a hard surface and the other is a relatively soft surface

Drum Drive

Cone winding is a typical example of frictional drive

Driver

Driven

Drum drive• Figure 1 shows the cheese sits on top

of the drum (driven).• The surface speed of drum (driver) is

V=n1 x pi x d1• Assume no slippage between the drum

and cheese, the surface speed will be:• n1 x pi x d1 =n2 x pi x d2• n1 /n2 = d2 /d1

Drum drive• The last equation shows that the ratio

of the rotational speeds of the two elements is the inverse ratio of their diameter.

• It should be noted that the running speed (rev/min) for the empty cheese (smaller diameter) is higher than full cheese (bigger diameter).

ExampleCalculation for cheese

winding

Question 1• A drum has 12cm diameter and

rotates at 2650 rev/min., find out– A) the actual winding rate (m/min)– B) the rev/min of the cheese at start

(empty package)– C) the rev/min of the cheese at 18cm

diameter (full package)

Answer 1• 1) Actual winding rate:

– V= n1 x pi x d1– V= (2650 x 3.14 x 12)/100 =1000 m/min

• 2) Rev/min at start:– n1 x d1 x pi = n2 x d2 x pi– 2650 x 12 = n2 x 5; n2=6360 rev/min

• 3) Rev/min at 18 cm:– 2650 x 12 =n2 x 18; n2=1767 rev/min

• It should note that the rev/min for empty package is running faster than full package

2) Belt driveFlat belt drive

Driver

Driven

Figure 2

Flat belt drive• If the belt has no thickness, the

surface speed:• n1 x d1 x pi = n2 x d2 x pi• If the thickness is t, the surface

speed :• n1 x (d1 + t) = n2 x (d2 + t)

Question 2• In Fig. 2, pulley A rotates at 120

rev/min and has diameter of 30 cm. Pulley B has a diameter of 50 cm. If the belt is 0.6 cm thick, estimate the number of rev/min of pulley B, with slippage ignored.

• Suppose 4% slippage occurred, what is the rev/min of pulley B?

Answer 2• n1 x (d1 + t) = n2 x (d2 + t)• 120 x (30+0.6) = n2 x (50+0.6)• n2=72.6 rev/min• If 4% slippage, the pulley B will be • 72.6 x 0.96% =69.7 rev/min

Vee-belt drive• The flat belt drive as discussed has the

limitation to the power that it can transmit.• Owning to the “wedging” effect, the Vee-belt

drive can trasmit four times as much power as the flat belt for the same cross-sectional area.

• Vee-belt drive is a common transmitting power on textile machine, e.g from the electric motor to the main shaft of the machine.

Vee-belt drive

High transmitting power of Vee-belt drive

Question 3Vee-belt

• The linear velocity of a vee-belt is 1200 m/min. If the pitch diameters of the motor pulley and machine pulleys are 16 and 36 cm, calculate the motor speed and the machine speed

Answer 3• Surface speed V= n1 x pi x d1• Linear speed = speed of pulley

(rev/min) x pi x d1• 1200 m/min =(n1 x 3.14 x 16)/100;

n1=2385 rev/min. (motor speed)• 1200 m/min = (n2 x 3.14 x 36)/100;

n2=1060 rev/min (machine speed)

Chain and sprocket drive

• This is a positive drive with a driving sprocket connected to the driven sprocket by a roller chain.

• The driving chain from bicycle is an example of chain-and-sprocket drive.

• It is simple and versatile.

(Rev/min of driving / rev/min of driven) = (no. of teeth of driven / no of teeth of driver)

Chain-and-Sprocket drive

Question 4• A sprocket of 24 teeth rotates at

56 rev/min and, by roller-chain drives, causes a sprocket of 40 teeth to rotate. Calculate the number of rev/min of the driven sprocket.

Answer 4• 56 x (24/40) = 33.6 rev/min.• One advantage of a chain drive is

that several sprockets may be driven from one driving sprocket.

• One of these guide sprockets may be adjustable so that any stretch in the chain may be taken up.