I've got a weight-driven pendulum clock, I increased the driving weight by 10% and the period decreased by 1%. Why?

What's the equation?

The pendulum is length = 32cm; bob weight = 7g; rod weight = 7g. So it's a compound pendulum

The amplitude of the swing is 0.05rad.

The Q factor is

Q = 2pi(energy at start of swing)/(energy lost during swing) = 200

Apparently a clock's pendulum mostly loses energy through viscous drag; i.e. assume drag is proportional to velocity.

The energy is replenished by the escapement. It's a deadbeat escapement so (I think) it can be modelled by adding a constant torque to the pendulum. The torque is in the direction of the swing and only operates when the angle of the pendulum is in the range

-0.03 < angle < 0.03

Outside that range, the escapement applies no torque to the pendulum.

On the web, I've found equations for a compound pendulum with drag, and a compound pendulum driven by an external sine wave but none of those describe a real pendulum clock.

You can assume that angle=sin(angle). Horologists call the error that introduces "circular error" and it's tiny for the sort of angles I've got. And it would increase the period by, say, 50ppm whereas I see a 1% decrease in the period.

1% is a big number. I want to understand it but the maths is beyond me.



I would simply accept that your initial estimates are off by a small percentage, and recalibrate by moving the bob-weight. –

I'm wondering if I can build electronics to alter the force in the driving chain to regulate the clock. So I quite like the 1% change in period that I observed - it's big enough to be useful. My original question was "what's the equation". I want to know whether the effect I've seen is common to all pendulum clocks with a deadbeat escapement. I think the math involved is the sort of thing engineers do every day.

you can easily calibrate the clock after building it.

I already have the clock. I'm wanting to regulate it.

How would you alter the driving force using electronics?

I have built an auto-winder, https://www.instructables.com/An-Auto-winder-for-a-Weight-driven-Clock/

You'll see that the weight moves slightly side-to-side. In other words, the tension due to the weight shifts from the left-hand chain (the taut chain) to the right-hand chain (the loose chain). So with a little cleverness inside the winder I can change the "weight" or, more precisely, the torque on the chain wheel inside the clock. Will the clock mind? No. The weight is 550g and the chain is 50g so originally as the weight descended, the tension increased by 20%. The clock is happy with a varying weight. How will the clock react? I think it will change it's speed.

The winder will know whether the clock is running slow or fast by measuring how far the chain-wheel on the winder has turned. It will adjust the speed to compensate by moving the weight to the left or right (on average).

I know how to do the electronics and I should end up with a clock that's as maintenance-free and accurate as a quartz clock I can buy for £1.

... if you want your result to be so precise, you would need to know all of the parameters in the equations very precisely as well.

The result of the calculation does not need to be precise. The winder/regulator measures what the clock is doing and adjusts its speed up/down. It's a feedback system. The mechanics and electronics will be extremely simple and the software is straightforward.

I want to know what the math is given the information in the original question. I want to know because if I come up with a successful design, will it apply to a wide range of clocks or is my clock idiosyncratic in some way? If the math explains what I measured then I'll know how generally my solution can be applied.

  • $\begingroup$ Looks like you have the formula wrong, see omnicalculator.com/physics/simple-pendulum $\endgroup$
    – Solar Mike
    Mar 6 at 15:07
  • $\begingroup$ I'm not familiar with all of the terms you used. But based on "it's a compound pendulum" and "I increased the driving weight by 10% and the period decreased by 1%": Did you increase only the weight of the bob-weight? In that case, the effective length of your pendulum increased, since it's center-of-gravity moves towards the bob. Increasing the (effective) length of your pendulum can alter your period. $\endgroup$
    – Chris_abc
    Mar 6 at 17:28
  • $\begingroup$ @PeterBalch Clear. I misinterpreted, sorry. I don't have any experience designing clocks myself, but I think I would simply accept the 1% change in period, and re-calibrate my clock by moving the bob-weight up or down. Moving the bob-weight will alter your pendulum period, giving you a simple method for calibration. $\endgroup$
    – Chris_abc
    Mar 7 at 9:13
  • $\begingroup$ @PeterBalch 1% may result in a large deviation of time during the day, but in mechanical engineering terms, being within 1% of your calculated value is very precise. As you say: the escapement introduces all sorts of extra effects. I don't think you can expect more precise results, unless you have the resources of a master clock/watch-maker. So I would simply accept that your initial estimates are off by a small percentage, and recalibrate by moving the bob-weight. $\endgroup$
    – Chris_abc
    Mar 7 at 9:15
  • $\begingroup$ @PeterBalch How would you alter the driving force using electronics? It seems like a complicated workaround to a simple problem to me. The math involved may be reasonable for any engineer, but if you want your result to be so precise, you would need to know all of the parameters in the equations very precisely as well. It's unrealistic to know such parameters that precisely. (And unnecessary, since you can easily calibrate the clock after building it.) $\endgroup$
    – Chris_abc
    Mar 7 at 12:42

1 Answer 1


I've read quite a few papers and I think I'm starting to understand why and how the driving weight affects the escapement error: why my clock runs faster when I increase the weight. Whether what I think matches reality is a different matter!

This paper is tough going but in it he says "any forces applied to the pendulum and acting toward the position of equilibrium reduce the period of the pendulum, whereas forces acting away from the position of equilibrium increase the period".

His paper seems to be entirely about applying an "impulse" at different times during the pendulums cycle and what effect that has. An impulse is an instantaneous increase in the pendulum's velocity - like flicking it with your fingernail. I think he applies his ("positive") impulse in the direction of movement of the (theoretical) pendulum - he's always speeding it up.

Presumably a "negative" impulse does the opposite. He doesn't discuss how the size of the impulse affects the period - only when it is applied.

A real escapement isn't an impulse. It applies a torque over a certain section of the pendulum's cycle. That torque may vary during the cycle but let's assume it's constant and is always "on" or "off".

Consider a verge escapement. The torque is always "on" but changes direction as one pallet then the other is pushed by the crown wheel. The direction of the torque varies like this:

verge escapement

The dotted lines show when the escapement advances by one tooth. The red and blue arrows show the direction of the torque. The torque in the red and blue parts between the dotted lines are equal and opposite so cancel and do not speed-up or slow-down the clock. In the parts outside the dotted lines (the overswing), the torque is towards the centre and so decreases the period of the pendulum. Increasing the torque will (presumably) speed-up the clock.

This paper discusses when torques are applied by a verge or a dead-beat escapement. They say a dead-beat escapement is like this

enter image description here

Note that they think the torque is not applied symmetrically about the rest position of the pendulum. (I don't know why they think so but it's clearly there in Fig 3.) The same assertion is made in this paper (Fig 5). On average, more torque is applied while the pendulum is moving away from the centre so increasing the torque will (presumably) slow down the clock.

My clock has a strip-pallet recoil escapement. According to Fig 3 of the same paper the torque is applied very much like a verge. I suppose that's true of all recoil escapements.

So the result is that if you increase the driving-weight and if the clock has a recoil escapement then it will speed-up. If it has a deadbeat escapement it will slow-down.

The amount of recoil speed-up depends on how big the overswing is. The amount of deadbeat slow-down depends on how asymmetric the deadbeat pallets are.

Is that what happens in practice with a real clock?

I don't have lots of clocks to test but I've written a simulation. It behaves as I've described. With a simulated recoil escapement, increasing the torque provided by the escapement speeds-up the clock. The amount of gain depends on the size of the overswing and how much extra torque is applied. A typical result is that 10% extra on the driving-weight makes the clock gain by 1%. Which is what I've measured in real-life.

With a simulated symmetrical deadbeat escapement, increasing the torque makes a tiny difference to the speed of the clock. That tiny difference is due to the swing angle increasing which leads to increased circular error (a loss of a few dozen parts per million).

With a simulated asymmetrical deadbeat escapement like the one above, increasing the torque makes a small difference to the speed of the clock. The size of that small difference depends on how big the asymmetry is but a typical value is that the clock loses by 0.02%.

Of course, I'm not an expert and may be talking complete nonsense.



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