Pendulum - How it affects gear train torque?

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Lengthening the pendulum in a weight-driven clock does not change the torque applied by the gear train or escapement to the pendulum. The pendulum acts as a driven oscillator, relying on the gear train to provide torque to maintain its motion against air resistance. The gear train must balance the torque to prevent excessive arc increases while overcoming losses. The discussion also raises a question about the impact of pendulum length on downstream gears when the escapement is repositioned. Overall, the torque dynamics remain consistent regardless of pendulum length.
jryer
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Does the gear train in a classic weight driven clock see a torque decrease when we lengthen the pendulum?

thanks guys for your help - (inventor question)
 
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Are you asking this as a physics question or as a design question?

The physics version would be: "How would lengthening the part of the pendulum below the pivot point change the torque delivered by the escapement"

The design question would be: "How would lengthening the part of the pendulum below the pivot point change the torque required from the escapement".
 
jbriggs444 said:
Are you asking this as a physics question or as a design question?

The physics version would be: "How would lengthening the part of the pendulum below the pivot point change the torque delivered by the escapement"

The design question would be: "How would lengthening the part of the pendulum below the pivot point change the torque required from the escapement".

The first one - "How would lengthening the part of the pendulum below the pivot point change the torque delivered by the escapement; would it increase or decrease?"

Also, I'm assuming the change in the torque delivered is transferred (seen/realized) to the gear train, is this a correct assumption?
 
Possibly we are still speaking at cross-purposes.

To my way of thinking, the pendulum is a driven oscillator. It does not apply torque to the gear train. The gear train applies torque to it. In the absence of any applied torque, it would continue move back and forth in an arc that slowly decreases over time due primarily to air resistance.

The gear train has to provide enough torque to overcome the losses due to air resistance. But it must not provide so much so that the pendulum's arc increases too greatly. [The escapement assures that this torque is applied in proper synchronization with the pendulum's motion].

So my answer is that the torque applied by the gear train/escapement on the pendulum is unchanged by the length of the pendulum. From Newton's third law, the torque applied by the pendulum to the escapement/gear train is also unchanged.
 
jbriggs444 said:
Possibly we are still speaking at cross-purposes.

To my way of thinking, the pendulum is a driven oscillator. It does not apply torque to the gear train. The gear train applies torque to it. In the absence of any applied torque, it would continue move back and forth in an arc that slowly decreases over time due primarily to air resistance.

The gear train has to provide enough torque to overcome the losses due to air resistance. But it must not provide so much so that the pendulum's arc increases too greatly. [The escapement assures that this torque is applied in proper synchronization with the pendulum's motion].

So my answer is that the torque applied by the gear train/escapement on the pendulum is unchanged by the length of the pendulum. From Newton's third law, the torque applied by the pendulum to the escapement/gear train is also unchanged.

JBriggs, take a look at this link

http://electronics.howstuffworks.com/gadgets/clocks-watches/clock4.htm

What happens if we put the escapement/pendulum assembly between the blue (1st) and purple (2nd) gears? Now does lengthening the pendulum affect the torque on the downstream gears (purple, green, and plum gears)?

thanks again
 
Bump. Anyone care to contribute on the 7:50pm post? thanks again
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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