Hi Folks,
I have a project where an old pendulum clock needs an auto wind mechanism. I am using a electromagnetic clutch with a max of 20 ft lbs of force at the gearbox motor shaft. The clock shaft requires ~25 ft lbs to rewind the clock. If I use a 3:1 or 4:1 reduction gear w/ chain/sprockets what is the torque on the electromagnetic clutch?

T = rFSinθ
θ=90
Sinθ= 1
Tclock = 25 ft lbs
Rclock = 4(Rmotor) (or 3x) (36 tooth #40 chain)
Rmotor = ~2.7" (12-15 tooth # 40 chain)

Tmotor max = 20 ft lbs (The gearbox has ample torque, the clutch only puts out 20 ft lbs.)

Is this a linear relationship?
any help would greatly be appreciated.
regards,
dave62

 

25ftlb to wind a clock? Must be a pretty good size clock, unless it was always intended to be wound by a motor and thus requires only a few turns.

Ignoring friction losses, the torque times the shaft speed is a constant, so reducing the shaft speed by a factor of 3 would increase the torque by a factor of three. However, there WILL be those losses, but for what you are talking about they will probably be pretty minor.

What steps are you taking to avoid overwinding it?

 

25ftlb to wind a clock? Must be a pretty good size clock, ...
...

You're not kidding! 25 ft lb on even a large 2" clock key would be 150 lb of force your fingers need on the edges of the key. Maybe it's 25 in lbs?

 

Hi Folks,
Thanks for the replies. It's a pendulum clock in the tower of a church. The weights for the bell wind is about 200 lbs with cable/pullies. There is a second mechanism to wind the clock w/ about a 150 lb weight. The weights travel about 20-25 ' over a 6-8 day period after which they need to be rewound to the top again.

I am using a limit switch on the bottom to trigger a 555 timer monostable multivibrator configuration; set to wind for a given time and 2 safety limit switches on the top to deactivate the mechanism. It has an Ogura CT20 electromagnetic clutch which has a max torque rating of 20 ft lbs, therefore it needs to be stepped down to reduce the torque on the clutch.

I put a torque wrench on it and it's about 25 ft lbs.

 

super cool project man! did you get the answer you were after, or are you still needing help?

 

Let me confirm, the motor drives a 13 tooth #40 sprocket and the clock drum shaft is driven by a 36 tooth #40. 13/36= 2.7 1:2.7. If Torque is 25 ft lbs on drum (reduced sprocket) the torque on the motor/clutch will be 25 ft lbs/2.7 = ~9.2 ft lbs?

That should work since the clutch is rated for 20 ft lbs.
thanks again.

 

Let me confirm, the motor drives a 13 tooth #40 sprocket and the clock drum shaft is driven by a 36 tooth #40. 13/36= 2.7 1:2.7. If Torque is 25 ft lbs on drum (reduced sprocket) the torque on the motor/clutch will be 25 ft lbs/2.7 = ~9.2 ft lbs?

That should work since the clutch is rated for 20 ft lbs.
thanks again.

It'll probbaly be a bit higher than that owing to fiction in the gear-reduction system, but I would expect you to end up somewhere in the 10ftlb to 12ftlb range, perhaps even a bit lower.