2. Chain Dynamics
Tooth Force
Pressure Angle
Chain Tension
Link Tension
Frictional Tooth Force
Figure 2.7 The Balance of Forces Around the Roller
But actually, sprocket teeth need some inclination so that the teeth can engage and slip off of the roller. The balance of forces that exist around the roller are shown in Figure 2.7, and it is easy to calculate the required back tension. For example, assume a coefficient of friction µ = 0, and you can calculate the back tension (T k ) that is needed at sprocket tooth number k with this for- mula: T k = T 0 3 sin ø k-1 sin( ø + 2 b ) Where: T k = back tension at tooth k T 0 = chain tension ø = sprocket minimum pressure angle 17 – 64/N(˚) N = number of teeth 2 b = sprocket tooth angle (360/N) k = the number of engaged teeth (angle of wrap 3 N/360); round down to the nearest whole number to be safe By this formula, if the chain is wrapped halfway around the sprocket, the back tension at sprocket tooth number six is only 0.96 N. This is 1 percent of the amount of a flat belt. Using chains and sprockets, the required back tension is much lower than a flat belt. Now let’s compare chains and sprockets with a toothed-belt back tension. Although in toothed belts the allowable tension can differ with the number of pulley teeth and the revolutions per minute (rpm), the general recommen- dation is to use 1/3.5 of the allowable tension as the back tension (F). This is shown in Figure 2.8. Therefore, our 257 N force will require 257/3.5 = 73 N of back tension. Both toothed belts and chains engage by means of teeth, but chain’s back tension is only 1/75 that of toothed belts. { }
13
Made with FlippingBook Digital Proposal Creator