Designing a Gearing System for an Analog Clock: Need Help!

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SUMMARY

The forum discussion centers on designing a gearing system for an analog clock using a stepper motor and a 1/64 turn servo. The user proposes that 1/4 of a turn of the stepper motor should equal one second, but struggles with the gearing calculations necessary to drive the minute and hour hands. The suggested approach involves using four gears and four pinions to achieve the required reductions, with a recommendation to consider a step frequency of every 0.9375 seconds to minimize motor wear. The discussion concludes that a reduction ratio of 60 for the minute hand and 12 for the hour hand is standard in clockmaking.

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Addohm
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This task seems to me like it should be rather menial but I can't seem to wrap my head around it. I've already designed a circuit to run an analog clock with components I have already got sitting around my work bench. When it comes to designing the gearing though, I am struggling.

My basic idea is this...
1/4 of a turn of my stepper motor will be equal to one second

My servo is a 1/64 turn servo so the simple way of doing this is out the window. The math for the programming is simple. The math for the gearing eludes me. I'm not sure really what makes sense. I think I need to have 4 gears to reduce the gear size. So, 4 gears and 4 pinions. I expect that my servo will drive my minutes hand, and my hours hand will be geared off of that.

So, 1 turn for 4 seconds.
15 turns for 1 minute
900 turns for 1 hour
10,800 turns for 12 hours

Would someone explain to my mechanically simpleton mind a simplified version of how to properly gear this?

Edit: It doesn't need to remain a 1/4 turn drive. It would probably be easier for each step to somehow represent a bit of time, or I could pass a single step every .9375 seconds (60/.9375=64 "Steps"). It would be less wear on the motor as well if I use it less often, or more often with smaller steps.
 
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Addohm said:
This task seems to me like it should be rather menial but I can't seem to wrap my head around it. I've already designed a circuit to run an analog clock with components I have already got sitting around my work bench. When it comes to designing the gearing though, I am struggling.

My basic idea is this...
1/4 of a turn of my stepper motor will be equal to one second

My servo is a 1/64 turn servo so the simple way of doing this is out the window. The math for the programming is simple. The math for the gearing eludes me. I'm not sure really what makes sense. I think I need to have 4 gears to reduce the gear size. So, 4 gears and 4 pinions. I expect that my servo will drive my minutes hand, and my hours hand will be geared off of that.

So, 1 turn for 4 seconds.
15 turns for 1 minute
900 turns for 1 hour
10,800 turns for 12 hours

Would someone explain to my mechanically simpleton mind a simplified version of how to properly gear this?

Edit: It doesn't need to remain a 1/4 turn drive. It would probably be easier for each step to somehow represent a bit of time, or I could pass a single step every .9375 seconds (60/.9375=64 "Steps"). It would be less wear on the motor as well if I use it less often, or more often with smaller steps.
If you choose the frequency (as you suggest) so the motor turns once a minute, you would need a reduction by 60 to drive a minute hand followed by a reduction of 12 to drive the hour hand. These ratios must surely be standard in clock making.
 

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