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• Motor Sizing - Basic Help

Discussion in 'The main mechanical design forum' started by Ste_Mulv, Aug 1, 2013.

1. Ste_MulvMember

Joined:
Aug 2011
Posts:
9
0
Hi there,

I am trying to size the sprockets, wheels, chain and motor for this basic over view design.

You will see that the power train is made up for 5 wheels connected together using a sprocket and chain system. The mass of the material is 30kg. The power train shown will have another identical one directly above it, that the material will be passed to and from.

I am trying to calculate the forces to size these 4 components: wheels, chain, sprockets and motor.

The way that the material will be passed from 1 drive train to the next is shown here and I have hit a maths issue.

The average speed of the material between t0 and t1 is coming out as the same as the total average speed over tTot. This does not make sense as the material starts at 0m/s then is accelerated to Vpkm/s.

I will use the Vpk value to work out the acceleration and deceleration rate in order to get the total force required in the chain and thus on the motor.

Please ignore the Sprung loaded devices that will hold the material in place, as I have decided that I will achieve that in a different way. If anyone has any any ideas or examples about how to get the chain and sprocket system to work smoothly or even if moving to a belt system, I would love to hear about it.

Best regards,

Steve Mulvenna

2.
3. joninstjohnMember

Joined:
Aug 2012
Posts:
13
0
Hi Steve,

For sizing your motor I'd be looking to calculate the power required to do the job then consider losses for chain drive, bearings, gearbox etc. This will give you the motor size based on power.

The average speed between t0 and t1 is Vpk / 2 since your sketch assumes linear acceleration. Do you know what Vpk needs to be? The means by which the work piece is held may have a large effect on your power requirements depending on how you propose to do it.

A starting point for chains and sprockets might be machinery's handbook.

Looks an interesting problem you are solving!

Cheers

Jon