Okay this has has been driving me crazy.
I would like to know how a conveyor speed can be calculated on the basis of chain pitch, sprocket PCD & no. of teeth and output RPM of gearmotor.
So here is a practical example I'm having to solve at the moment:
I have to drive a conveyor at a speed of 760mm/min. To drive this, I am using a drive system which consists of a small sprocket mounted on the output shaft of the gearbox. This sprocket is connected to a larger sprocket by means of a chain.
A shaft from the larger driven sprocket is connected to the main conveyor sprocket on which the conveyor chain is mounted. This chain needs to be driven at 760mm/min or in other words, the part at one station needs to travel to the next station 760mm away in one minute.
I have enclosed a sketch of this arrangement for reference.
I am going to use a VFD to achieve the required speed. Available specifications:-
Motor Speed - 915 RPM (6-pole)
Gearbox Reduction - 73.6 (this I am yet to finalize, Can go for a different reduction ratio if required)
Output RPM - 12.43
PCD of Small drive sprocket - 106.14mm
No. of teeth on small drive sprocket - 13
PCD of large driven sprocket - 461mm
No. of teeth on large driven sprocket - 57
PCD of conveyor sprocket (mounted on same shaft as driven sprocket) - 598mm
No. of teeth on conveyor sprocket - 74
Chain Pitch - 1"
Any help for this would be highly appreciated!