Frictional Drive in Textile Part 2:Gear Drive

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Presentation transcript:

Frictional Drive in Textile Part 2:Gear Drive

Gear Drive What is gear drives Types of gear drives Straight-tooth gear Helical gear Double helical gear Spiral gear 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 Speed at wheel B 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 3 Compound 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: which is the same speed of wheel G; 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