Arduino Pendulum Clock Design – Comments Welcome

[SillyOldDufferModerator @sillyoldduffer]

Everyone is doing better than me!

Today I modified the tripod to stiffen it and allow space underneath for wires to leave the vacuum. Not too bad apart going out into the storm to buy bits, and forgot the important one – super-glue. Also, no studding.

Wouldn't have mattered except I broke the pendulum rod reassembling the clock…

Spent this evening blowing my mind trying to solve the following problem. Given the period of a pendulum, and it's amplitude, how long does it take the bob to travel to BDC from a beam breaker offset by 'n' millimetres. (Time side to side, 833mS, amplitude 5°, offset 5mm, rod about 240mm.)

Dave

[Michael GilliganParticipant @michaelgilligan61133]

Posted by SillyOldDuffer on 02/10/2020 21:25:34:

[…]]

Spent this evening blowing my mind trying to solve the following problem. Given the period of a pendulum, and it's amplitude, how long does it take the bob to travel to BDC from a beam breaker offset by 'n' millimetres. (Time side to side, 833mS, amplitude 5°, offset 5mm, rod about 240mm.)

Dave

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It is, or at least should be near enough, simple harmonic motion … so there’s a sine wave lurking in there.

… I will leave you to do the geometry for your particular installation

MichaelG.

[duncan webster 1Participant @duncanwebster1]

first define some variables

x = distance from centre

Xmax = maximum distance from centre, half the amplitude

w = frequency in radians per second

For SOD case period = 2*833ms, so w = 2*pi/(2*0.833) = 3.77 rads/sec

then x = Xmax * sin (w*t) where t is the time since passing the centre position, and w*t is radians.

If the swing is 5 degrees total, 2.5 degrees each side, and the rod length is 244mm, then Xmax = 244*sin(2.5 degrees) = 10.64 mm, so when the bob gets to the beam breaker

5 = 10.64 * sin(3.77*t), from which we get sin(wt) = 5/10.64 = 0.46, wt = asin(0.46) = 0.489 (remember the 0.46 is radians) and so t = 0.489/3.77 = 0.13 sec = 130 ms.

It causes me all sorts of confusion when clockies talk about a 1 second pendulum when they mean it takes 1 second to go from extreme left to extreme right. Engineers would talk about the period being from extreme left over to extreme right and back again, hence a 2 second pendulum, but I don't suppose clockies are going to change so I'd better get used to it.

[duncan webster 1Participant @duncanwebster1]

Just realised there is something amiss. If the pendulum is 240 mm long it should have a period of about 1 second, not 1.66 secs as described.

[Michael GilliganParticipant @michaelgilligan61133]

Forgive the digression but I just found a vintage [2004] paper, which might be of background interest: **LINK**

https://www.researchgate.net/publication/2171311_Simple_Pendulum_Revisited

MichaelG.

[SillyOldDufferModerator @sillyoldduffer]

Posted by duncan webster on 02/10/2020 23:25:18:

first define some variables

x = distance from centre

Xmax = maximum distance from centre, half the amplitude

w = frequency in radians per second

For SOD case period = 2*833ms, so w = 2*pi/(2*0.833) = 3.77 rads/sec

then x = Xmax * sin (w*t) where t is the time since passing the centre position, and w*t is radians.

If the swing is 5 degrees total, 2.5 degrees each side, and the rod length is 244mm, then Xmax = 244*sin(2.5 degrees) = 10.64 mm, so when the bob gets to the beam breaker

5 = 10.64 * sin(3.77*t), from which we get sin(wt) = 5/10.64 = 0.46, wt = asin(0.46) = 0.489 (remember the 0.46 is radians) and so t = 0.489/3.77 = 0.13 sec = 130 ms.

It causes me all sorts of confusion when clockies talk about a 1 second pendulum when they mean it takes 1 second to go from extreme left to extreme right. Engineers would talk about the period being from extreme left over to extreme right and back again, hence a 2 second pendulum, but I don't suppose clockies are going to change so I'd better get used to it.

Rushing to do something else! First many thanks!

My pendulum is to this design:

Me too confused by clockies halving the period. My ticks are detected on one side of the bob only, so the time is for a full swing there and back. It's about 833mS. So Duncan's assumption it's a 1.6 second pendulum in his calculation was unnecessary – we were on the same page at the beginning.

Another issue is my pendulum is part spring. A conventional pendulum rod is rigid and typically hangs from a short spring at the top. My carbon fibre rod is flexible, and there is no separate spring. I expect the usual formula to be approximately right, but not spot on.

More on this later, but I want to repair the broken pendulum before lunch.

Ta,

Dave

[John HaineParticipant @johnhaine32865]

Though the usual period formula may not apply, as long as the swings are small the motion will be near enough sinusoidal "for all practical purposes" (FAPP). As you know the period and amplitude Duncan's method should work.

[Roger HartParticipant @rogerhart88496]

I know very little but FWIW I think this is a single ended tuning fork. Had a poke around and no convenient formula came up on the web. Lord Raleigh did one but can't find it. But Wiki has a formula for a conventional tuning fork.

Taking E as 100GPa, rho as 1.5g/cm^3 and the radius of the rod(s) as 3mm and taking the Wiki round rod option for a 240mm long unloaded rod I get 3.78Hz resonance, other options are available.

Now I don't think losing one rod makes a big difference (!!) but the loading lump on the end will drop the resonant frequency, but not smart enough to figure out how much. What bothers me about this design is the round rod – it has no favoured direction of vibration – could go anywhere. Takes us into Sturm-Liouville territory where chaos reigns.

So, a rectangular rod and much heavier stiffer mountings is my feeling – or buy a rubidium oscillator.

[Roger HartParticipant @rogerhart88496]

Whoops 0.387Hz.

[Roger HartParticipant @rogerhart88496]

Whoops again stick to 3.78Hz. Blame scruffy notes.

[duncan webster 1Participant @duncanwebster1]

I told you that clockies confuse it! No doubt SOD will be able to redo the sums with the correct value for w

[Roger HartParticipant @rogerhart88496]

Apologies for useless calculation. With a very thin rod this setup looks more like a Foucault pendulum and those who set up that type go to a great deal of trouble to control wobbles and uncertainties. Even then Foucault pendulums are liable to start following a figure 8 pattern unless controlled.

I do feel that some of the trouble lies in the lack of a conventional suspension constraining the degree of freedom. I also worry about interactions between an iron electromagnet coil and the iron pendulum – symmetry and residual magnetism. I would be thinking to keep the electromagnet very symmetrical and possibly air cored.

Interesting project, good luck with it.

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Roger Hart on 04/10/2020 10:32:11:

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I do feel that some of the trouble lies in the lack of a conventional suspension constraining the degree of freedom.

 

+1

MichaelG.

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_________
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As I wrote on 15-September:

Bold assertion … Feel free to contradict me:

It doesn’t matter a jot what path the pendulum describes, or the extent of its displacement … provided that it is consistent … The real problem is finding ways to achieve that consistent behaviour.

That was the logic behind my first post, on p1 of this thread.

MichaelG.

Edited By Michael Gilligan on 04/10/2020 11:16:47

[SillyOldDufferModerator @sillyoldduffer]

Posted by Roger Hart on 04/10/2020 10:32:11:

…

I do feel that some of the trouble lies in the lack of a conventional suspension constraining the degree of freedom. I also worry about interactions between an iron electromagnet coil and the iron pendulum – symmetry and residual magnetism. I would be thinking to keep the electromagnet very symmetrical and possibly air cored.

Interesting project, good luck with it.

You may well be right Roger! All very experimental. As of this morning it doesn't work at all!

One proposition is that energising the bob at the bottom with an electromagnetic should tend to keep the path straight. As the impulse pulls the bob it should (?) take the minimum energy route. If true, it means no energy is wasted mechanically forcing the path straight, resulting in higher Q. Unproven!

Does it matter if the bob becomes magnetised? I thought not, but hadn't thought of symmetry. Another question to be answered.

The design has interesting possibilities, which may allow the build to be optimised such that I get closer to a free-pendulum clock without mechanical complications. As well as timing the impulse accurately I can alter pulse strength by:

• Increasing or decreasing pulse width
• Putting a resistor (or transistor) in series with the coil
• Moving the bob closer, or not, to the electromagnet.

There are more prospects involving manipulation of the data at run-time. For example I can use the microcontroller to average the tick time, making the conversion to hh:mm:ss more accurate as experience is gained, or to detect rate changes. Not difficult to add temperature and barometric compensation in the same way. But this is more software than pendulum!

I don't know if it will out-perform other pendulum clocks. If it does I shall be very smug indeed, but failure is always an option…

Lot's to do and I'm a slow worker.

Dave

[Martin KyteParticipant @martinkyte99762]

Even if it eventually not a success as a timekeeper it will have served a very useful purpose as an experimental test bed to explore the limitations and effects of things like degrees of freedom, electro-magnetic impulsing etc. In some ways if you build a thing and it just works it doesn't really teach you anything. Harrison learned enormous amounts about mechanical timekeeping from his original three sea clocks even though as marine timekeepers they were failures. Failures are good, they teach us the 'why'.

It may turn out to be a success in the end, we shall have to see, but I suspect that the greatest value will not be the clock at the end of it all but all the stuff discovered on the way. There is lots of usefull insight just in theis thread to date so well done for generating all the interest.

best regards Martin

[John HaineParticipant @johnhaine32865]

Central impulsing at BDC is definitely preferred since it causes minimum change in the pendulum's phase – ideally zero. It might also help with symmetry but that isn't the main reason to do it.

With a coil below the bob you can either attract the bob just before BDC, repel it just after, or both though that requires reversing the coil current at BDC (assuming you have a small magnet on the end of the rod).

In that configuration, ideally the coil should be air cored, so there is no interaction between bob and coil except when the latter is energised. It does mean you need more current. A core could be OK if you impulse every swing since the force "blip" will just merge with the impulse.

Just as a reference point my pendulum above is impulsed with a current of about 15 mA applied to an air-cored coil for 15 ms – you shouldn't need much energy. The impulse is applied roughly every 3rd swing on average I reckon.

[SillyOldDufferModerator @sillyoldduffer]

Update: clock feeling better – fixed a bug in the software cause by the last bug fix, blush, and fitted the correct pendulum mount, rather than the next size up.

Duncan's formula for calculating timing is working except I assumed the bob would break the beam halfway across, in practice it triggers sooner, so I'm still guessing about the exact timing point. 40mS isn't quite right, but the clock runs.

The 0.5mm diameter rod I'm using came from Easy Composites Ltd, and is described as pultruded matrix carbon fibre rod.

It's deflection characteristic is:

This data enables the springiness of the rod to be calculated.

Dave

[Michael GilliganParticipant @michaelgilligan61133]

Posted by SillyOldDuffer on 04/10/2020 19:14:35:

Update: clock feeling better – […]

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Good News

For the benefit of those descended from ‘rivet counters’, Dave … do you happen to know what matrix resin is used, and what proportion of carbon is in there ?

MichaelG.

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Edit: Just found the supplier:

https://www.easycomposites.co.uk/1mm-carbon-fibre-rod

… all looks very impressive.

Edited By Michael Gilligan on 04/10/2020 20:27:05

[John HaineParticipant @johnhaine32865]

Best part of 2 days spent chasing gremlin in the rate correction system, eventually re-wrote the code and it sprang into life first try. Having measured the average rate over 15000 seconds as 0.988701 s (but only measured using the micros() function on an Arduino Nano so questionable precision/accuracy) I have set the correction counter to 88, so every 88 "clock seconds" it will omit one dial pulse. Now leaving clock until tomorrow to see how the timekeeping fares.

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Michael Gilligan on 04/10/2020 20:20:59:

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Edit: Just found the supplier:

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Sorry, I linked the 1mm product … Here, for correctness, is the 0.5mm:

**LINK** https://www.easycomposites.co.uk/0-5mm-carbon-fibre-rod

and a similar product from another UK source:

https://www.hyperflight.co.uk/products.asp?code=CARBON-ROD-05MM&name=0-5mm-carbon-rod

MichaelG.

Edited By Michael Gilligan on 04/10/2020 22:32:44

[Michael GilliganParticipant @michaelgilligan61133]

@Duncan

If you have a moment, could you please do a comparison of those two specifications ^^^

… My little Imperial brain is struggling with the units and the mixed terminology

MichaelG.

[duncan webster 1Participant @duncanwebster1]

I strongly suspect that this is into angels dancing on pins territory, but according to my ruminations the frequency of a pendulum with springiness is

f = 1/(2*pi) * sqrt ( g/L + k/M)

where g = 9.81 m/sec^2, L is length in metres, k is stiffness in Newtons/metre, M is mass in kg.

In the unlikely event of anyone wanting a copy of the sum. send me a pm with your email address.

[John HaineParticipant @johnhaine32865]

I'd be interested please Duncan – I think you have my email address?

But for this pendulum where the rod stiffness is that of a bending beam, k isn't the YM of the material but has to be computed from the YM and the dimensions?

[duncan webster 1Participant @duncanwebster1]

k is the stiffness of the actual rod, which SOD has measured, but as you say if the dimensions and Young's Modulus are known you could work it out.

I'll make a legible copy and send it to you

[SillyOldDufferModerator @sillyoldduffer]

Centre of bob to suspension point about 223mm giving 877mS per complete swing if anyone has time to check Duncan's formula against reality.

Spent most of today clock-watching and tweaking software in hope of making the thing reliable. Plenty of problems!

• Clock and computer are on my dining table. Bumping the table does the clock no good!
• Clock base needs to be level and my table isn't.
• Pendulum takes about a minute to settle down after starting. (I guess all clocks suffer this.)
• I expected increasing pulse width would strengthen the impulse. Yeah but! I now see longer impulses braking the pendulum if the magnet is still on when the bob passes BDC. (My bob is attracted by the electromagnet, rather than carrying a permanent magnet to be repulsed.)
• Sensor upset by sunlight more subtly than first thought. Sunlight at an angle to the bob causes early or late triggering, and I now believe reflections off the bob cause jitter. Maybe Snopake wiil help.
• Timing's far too critical. Need to redesign the holder so the magnet is positioned accurately under BDC and stays put. To allow experimentation the electromagnet is wedged into the holder with matchstick wedges and the holder is Blue Tacked to the clock base. Not solid enough!

Two steps forward, three back…

Dave

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