Pendulum ‘Q’ value and measurement methods

[Joseph Noci 1Participant @josephnoci1]

Thanks John – that is a good start!

I don't have a picpet – I do have a lot of other circuitry around the pendulum , some not quite working yet, nor relevant to the basic principles I still hope to discover. There is a differential, rotary vane variable capacitor , a sort of rotary encoder, that attaches to the pivot – this method because only the vane attaches to the pivot point – so there is no added friction from encoder bearings, etc – I hope to achieve 0.1deg resolution from this device over 4deg angle. A variation of the concept in this pdf:

Seismometer

This to get a sinusoid per cycle and hope to integrate to rate to get energy loss/cycle – but that is for MUCH later.

I also have a time-to-digital converter ( Texas TDC7200) that gives delta between two edges, down to 10 or so ns resolution. I intend to feed the start pulse from the 1Hz output of my GPSDO (note, NOT GPS 1PPS, but the 1Hz divided down from the 10MHz disciplined osc) , which is quite good quality, and the stop pulse from the opto on the pendulum. A log of that variation of delta data should give the decay rate?

That is my version of the picPET…

The accelerometer is complex- basic mems devices won't work – DC accelerometers measure gravity and that swamps the small accelerations in a pendulum. Also, the sensitivity of these devices seems to be around +-0.5G, with 0.02mg/bit – still not good enough, even in Tilt measurement mode using two axes. 'True' vibration type sensors, piezo with charge amps are the better option, with bandwidth reduced to 5Hz or so, but the devices are two big to fit in the pendulum and there is a non-piezo coax cable to contend with, and a large price! I have fitted the 0.5G device – it gives usable data, but not good enough for deep analysis..

The Pendulum pivot is knife edge type.

If this thing gets of the ground and is even half as good as the rest of the pendulums on the forum, I will post photo's etc – else I will just go quiet..

 

Thanks John

 

edit:

This is not clear to me-

The amplitude is estimated for each pendulum cycle by an opto and a picPET, deriving the amplitude from the pulse length.

What is the pulse length?  Is it the length of the pulse out of the opto-interrupter? ie, low when the vane blocks the aperture, and high when it opens again, so the time between the hi-to-lo and lo-to-hi edges ( ie, the width of the light-blocking vane), or the time between two rising (for example) edges. If the latter, is that edge always in the one direction of swing, ie, the edge when the vane passes from left to right through the aperture, or both, ie, from left to right and then back from right to left

I have difficulty making myself clear here-

 

Edited By Joseph Noci 1 on 10/08/2023 13:58:25

[John HaineParticipant @johnhaine32865]

Yes, the length of the opto pulse between transitions. The time between successive say positive edges would be the period.

[Joseph Noci 1Participant @josephnoci1]

Sorry John…

I become confused with the terms used by different people – sometimes the same term or word does not mean the same thing…

I understand period to be one sinusoid of the pendulum. That is a complete swing, left to right to left.

A 2 second pendulum, approx 1meter long, has a period of 2 sec. A left to right only is a half period = 1sec.

The opto generates 2 positive edges per 2 second PERIOD. Are these the two positive edges you refer to?

one on the left to right swing ( HALF PERIOD) and one on the right to left swing?

Those two edges are then one half period ( 1sec) apart …but…there is a jitter, or rather a delta, equal to the width of the light interrupting vane, since one direction triggers on the left of the vane edge, the other swing direction on the right vane edge. Is this delta not significant in the small numbers you are chasing? It's not like the jitter on the GPS 1PPS, which does vary positive and negative around a middle value ( and suffers from hanging bridges) – it is a constant delta between swing direction periods, the value changing only with pendulum decreasing velocity. or do you compensate for this 'vane width' period somehow? ( How?).

[John HaineParticipant @johnhaine32865]

Sorry for the confusion, horologists tend to use words slightly differently.

CYCLE is a whole sinusoid, left-right-back to start. This is two SWINGS or BEATS.

My usage of an "impulse cycle" is because the Arduinome (Synchronome derivative) applies one impulse for every 30 pendulum cycles, i.e. 60s.

The interrupter generates a short pulse, ideally symmetric around the central position of the bob, or bottom dead centre, "BDC". This has two edges of opposite polarity (e.g. -+ / +-). The pulse length depends on the velocity of the bob and width of the flag (or slot, depending).

The time between successive edges of the same sign (e.g. -+) is the semi-period. You can calculate the semi-period from either or both. Double it to get the period.

Horologists confusingly call a 2s period pendulum a SECONDS pendulum as most mechanical escapements release on both directions

Though a simple velocity-based calculation can give a reasonable estimate of amplitude when the pulse is short compared with the period, there are more complex algorithms which will correct for the small "trigonometric" error which can become significant as the amplitude falls; and also for errors caused by the sensor not being precisely on the centreline of the pendulum's swing.

There has been a lot of discussion of this over on the HSN forum recently.

[Joseph Noci 1Participant @josephnoci1]

Thanks John.

also for errors caused by the sensor not being precisely on the centreline of the pendulum's swing.

What do you mean by centerline – in line with the pivotal center of swing ( in line with the arc swept by the pendulum shaft?) or is it BDC you mean?

If inline with the swing is meant, what happens when not inline – if the sensor and vane are not directly below the through axis of the pendulum shaft say, but offset (1cm) away from that line, away from the pendulum swing direction – that should not have any effect surely? The swing and interrupt position is still the same, with only a lateral shift in sensor position, on the side of the swing direction. ( its really difficult to explain placements and positions in a theoretical space on the page without much waving of hands…!)

If not at BDC the period of the two swings are not seen the same, ie, successive rising edges are not 1 second apart ( correct?).

SO the BDC position of the sensor and vane is critical, also correct?

You did not comment –

If successive +ve edges are used, once on left to right swing, then the next on right to left, the '1sec' period is reduced by the vane width time period – is this not significant?

Again, forgive my many questions please..

[SillyOldDufferModerator @sillyoldduffer]

Posted by S K on 09/08/2023 22:34:22:

SOD, I know you believe you are measuring Q, and you are getting "numbers," but I don't believe you are actually measuring Q. The old "garbage in, garbage out" problem, to my eye. If your data was Gaussian and well behaved, and if the obtained value of Q was stable, maybe, but at this point none of that seems evident.

Also, I am not sure you are using thousands of samples. Sure, you are collecting thousands, but per Q measurement you are throwing nearly all of them out after selecting only a few (I don't know the details, however, so I could be wrong on this point).

In addition, you are subtracting two numbers that are very close together from each other, getting a very small number (i.e., 6 or so orders of magnitude smaller), and then dividing that very small number into a comparatively large one again, tempting the gods of mathematical fate. In this scenario, minute deviations can cause huge impacts on the end result, as it seems you are seeing.

Try using the decay method to check. It can't be that hard, others have done it, and the decay method is much more intuitively related to the loss of energy per swing anyway. And also, the value of Q obtained this way should be quite stable from trial to trial (as I believe one would expect from a macroscopic pendulum of this sort).

I plead not guilty to all charges!

My data has a normal distribution, or at least close to it – admittedly there's a slight bias to the right:

As the distribution is close to normal I am confident that the Q calculation is valid.

I most certainly am using thousands of samples. The distribution above, was derived from the clock's log at 15:22 today when it contained 2,386,218 samples after a run of over 618 hours. Everey beat of the pendulum writes a record to the log. The Q varying by Hour graph slices the same data into hourly segments, each of which contains about 3,800 samples, from which Q in that hour is calculated. May I remind the jury that 3,800 samples is far more than the decay method collects: one of John Haine's posts mentions 30 samples, which is line with my experience.

I don't believe the calculation is ill-formed, not when I have 64 bit floating point anyway. The method uses 3 percentiles taken from a large data set. Percentiles are related to averages – one of them is the mean. True Q tends to infinity as the distribution sharpens but my distribution isn't particularly sharp. Worth investigating further though, because I haven't checked how numpy's percentile function works – maybe it loses precision. I'm not seeing huge variatons – up to about 30%, usually much less,

I have used the decay method in the past and it produces similar values. Decay method accuracy is limited by observational error and the small number of samples. As far as I know no-one has measured Q repeatedly and confirmed that Q doesn't vary over time. Although one might assume the Q of a macroscopic pendulum would remain stable, my data log suggest it doesn't. Given that temperature, pressure, humidity and other factors vary over time, I'm not surprised Q changes too.

Dave

Edited By SillyOldDuffer on 10/08/2023 17:14:14

[SillyOldDufferModerator @sillyoldduffer]

Posted by Michael Gilligan on 09/08/2023 21:45:50:

This is not a criticism, Dave … far from it !

But something is nagging me: By [ quite understandably] scaling your pressure graph from 970, are you perhaps exaggerating its significance ?

… what would it look like if you scaled it from zero ?

MichaelG.

The graph plotting software scales Q and pressure separately so both fit on screen, which I agree is artificial. Gets worse: not comparing like-with-like. Q-factor is a dimension-less quantity, which in my dataset happens to between about 15,000 and 32,000, whilst pressure in is millibar, the weather providing a low of 970mb and a high of 1008mb.

Both data sets occurred in the same time, but as their dimensions and Y-scales are unrelated, the eye is required to look for a pattern. The pattern doesn't depend on scale and is usually more obvious if both lines use all the space available.

In the Q and Pressure graph the two plots lines don't appear to have a relationship, so the variation of Q I've apparently detected is unlikely to be caused by air pressure.

Dave

[S KParticipant @sk20060]

Although one might assume the Q of a macroscopic pendulum would remain stable, my data log suggest it doesn't. Given that temperature, pressure, humidity and other factors vary over time, I'm not surprised Q changes too.

Dave

No one is suggesting that Q should be a constant. But it changing by factors of as high as 2 over relatively short time periods does not sound physically realistic to me, and it rather sounds like noise. But maybe I'm wrong?

 

Edited By S K on 10/08/2023 18:58:28

[Michael GilliganParticipant @michaelgilligan61133]

Posted by SillyOldDuffer on 10/08/2023 18:04:05:

Posted by Michael Gilligan on 09/08/2023 21:45:50:

[…{

… what would it look like if you scaled it from zero ?

MichaelG.

[…]

Both data sets occurred in the same time, but as their dimensions and Y-scales are unrelated, the eye is required to look for a pattern. The pattern doesn't depend on scale and is usually more obvious if both lines use all the space available.

 

.

An admirable reply, Dave … but please let me be the first to bring Politics into this discussion

… I am sure I have seen the Liberal Democrats using similar techniques to manipulate the way that viewers perceive presented data.

MichaelG.

.

Edit: ___ ferinstance:

https://www.onlondon.co.uk/fixed-that-for-you-dodgy-liberal-democrat-election-graphs-2017/

Edited By Michael Gilligan on 10/08/2023 20:42:23

[S KParticipant @sk20060]

Yeah, no, let's not! Thank you!

[Michael GilliganParticipant @michaelgilligan61133]

[John HaineParticipant @johnhaine32865]

Posted by SillyOldDuffer on 08/08/2023 20:25:21:

…

Bandwidth is calculated from the 29.3, 50,0 and 70.7 percentiles.

….

Dave

Why do you choose those Dave? The bandwidth would normally be calculated from the "3dB down" points but it isn't obvious to me how "3dB down" = 0.7071 maps on to a percentile of a period distribution.

Edited By John Haine on 11/08/2023 10:26:47

[John HaineParticipant @johnhaine32865]

Thanks John.

also for errors caused by the sensor not being precisely on the centreline of the pendulum's swing.

What do you mean by centerline – in line with the pivotal center of swing ( in line with the arc swept by the pendulum shaft?) or is it BDC you mean?

Correct – centre of the sensor lies on the vertical axis through the suspension or pivot point.

If inline with the swing is meant, what happens when not inline – if the sensor and vane are not directly below the through axis of the pendulum shaft say, but offset (1cm) away from that line, away from the pendulum swing direction – that should not have any effect surely? The swing and interrupt position is still the same, with only a lateral shift in sensor position, on the side of the swing direction. ( its really difficult to explain placements and positions in a theoretical space on the page without much waving of hands…!)

When you analyse the edge times that result they change as a result of such a shift, but if you know how large it is this can be allowed for in the calculation. Actually an offset can be helpful since it allows you to "know" which side the pendulum is in the calculation

If not at BDC the period of the two swings are not seen the same, ie, successive rising edges are not 1 second apart ( correct?).

SO the BDC position of the sensor and vane is critical, also correct?

Yes – on my current build the sensor is on a carriage on rails with micrometer adjustment.

You did not comment –

If successive +ve edges are used, once on left to right swing, then the next on right to left, the '1sec' period is reduced by the vane width time period – is this not significant?

Yes, sorry of course it is. So if the edge times over a complete cycle were say T1, T2, T3, T4, (-+, +-, -+, +-) you can calculate the period from T3-T1 and T4-T2 and average to take out the effect of any small offset.

Again, forgive my many questions please..

[SillyOldDufferModerator @sillyoldduffer]

Posted by Michael Gilligan on 10/08/2023 20:32:03:

Posted by SillyOldDuffer on 10/08/2023 18:04:05:

Posted by Michael Gilligan on 09/08/2023 21:45:50:

[…{

… what would it look like if you scaled it from zero ?

MichaelG.

[…]

Both data sets occurred in the same time, but as their dimensions and Y-scales are unrelated, the eye is required to look for a pattern. The pattern doesn't depend on scale and is usually more obvious if both lines use all the space available.

.

An admirable reply, Dave … but please let me be the first to bring Politics into this discussion

…

The essential difference is I have no motive for deceiving anyone! Statistics are often used to mislead, but I'm open and honest about:

• the equipment and environment used to collect the data
• what the data is (it's available to anyone else who wants to play with it)
• the statistical methods I used, and why
• the program used to analyse the data (also available)
• faults and anomalies I've spotted
• how I interpreted the results, and if I came to a conclusion, the reasoning is explained. (With many examples of me unable to conclude anything due to lack of evidence or my failure to understand it).

Important to be open because all stages include the possibility of mistakes and others are much better at spotting blunders than the author!

In many cases I'm confident the statistics tell the absolute truth. For example, the effect of temperature on period. Others, like 'Variation of Q By Hour' are controversial, and a bit surprising. There's doubt: are the sums right, has the data been sampled appropriately, has SOD used the right statistical techniques, is his code correct? The statistics suggest something is going on, but what is unknown. In these cases statistics mean further investigation is needed, not that a truth has been revealed. SK suggested a number of reasons why 'Variation of Q By Hour' is wrong. One of them was that the Q calculation I using only works if the data distribution is 'normal'. It's one of several possible has 'SOD used the right statistical techniques' questions. Actually another statistical test shows that my data distribution is normal, so that particular suggestion can be dismissed. A step forward, but not proof that 'Variation of Q By Hour' is trustworthy.

Statistics are a bit like drawing a map with inadequate data. Early versions can be misleading, and it's necessary to say "Here Be Dragons". The quality of the map improves as more data arrives, and different ways of applying the data are applied. If new data arrives that contradicts existing data, and both are sound, discard the fist proposition, and start searching for a relationship that satisfies both. When more than one alternative is available, the simplest is usually right etc. Another analogy is statistics is rather like a homing missile. The missile is fired vaguely in the right direction and after a bit it's sensors collect enough data to start aligning with the target. More data, and more accurate data arrives, each time the alignment improves, until the missile gets a lock.

So I run the statistics, have a think about what they might mean, and use them to progressively improve understanding of what's going on. It's a process in which it's expected that earlier statistics might have been wrong or misinterpreted.

Very different from what politicians do. This morning I heard a Minister say that £2bn had been allocated to one aspect of a big problem he's managing. Jolly helpful if it were true, but it's very unlikely to be new money. More likely he's transferred £2Bn from one part of the system to another which doesn't help when the real problem is underfunding across the board. Sounds as if the politician has done something big and decisive, and maybe he has, but more likely he is fobbing us off by re-arranging the deck-chairs on the Titanic. His £2bn statistic isn't a lie, but it probably misrepresents the truth. He wasn't asked where the money is coming from, and that's what we need to know.

Dave

[Michael GilliganParticipant @michaelgilligan61133]

Posted by SillyOldDuffer on 11/08/2023 11:44:10:

.

[…]

.

Oh dear

That’s a surprisingly lengthy reply, Dave … I hope I have not offended you.

Not for one second have I doubted your integrity, and I apologise if that was inadvertently implied.

I was merely elaborating upon my previous suggestion that you might try showing zero on your pressure scale.

… Whether done by accident or design, it is possible to mislead the viewer by emphasising small variations [in the way that your automatic scaling does].

MichaelG.

.

Edit:__ one good thing has come out of this … I have just found a downloadable [and therefore legible] copy of Mr Huff’s little masterpiece:

https://www.researchgate.net/profile/Grigori-Evreinov/post/Data-visualization-which-is-best-for-within-and-cross-source-data/attachment/5b0690b6b53d2f63c3cdcad5/AS%3A629755908993031%401527156917765/download/How-to-Lie-with-Statistics.pdf

Edit: __ Chapter 5 is the relevant one …

Edited By Michael Gilligan on 11/08/2023 13:46:12

Edited By Michael Gilligan on 11/08/2023 14:15:07

[SillyOldDufferModerator @sillyoldduffer]

Posted by Michael Gilligan on 11/08/2023 13:31:47:

Posted by SillyOldDuffer on 11/08/2023 11:44:10:

.

[…]

.

Oh dear

That’s a surprisingly lengthy reply, Dave … I hope I have not offended you.

…

Not offended at all Michael. Statistics have bad name, as in 'lies, damn lies, and statistics', so I took the opportunity to explain why I'm keen on applying them to pendula.

Dave

[Michael GilliganParticipant @michaelgilligan61133]

< phew >

[SillyOldDufferModerator @sillyoldduffer]

Posted by John Haine on 11/08/2023 10:26:34:

Posted by SillyOldDuffer on 08/08/2023 20:25:21:

…

Bandwidth is calculated from the 29.3, 50,0 and 70.7 percentiles.

….

Dave

Why do you choose those Dave? The bandwidth would normally be calculated from the "3dB down" points but it isn't obvious to me how "3dB down" = 0.7071 maps on to a percentile of a period distribution.

…

All my clock books have disappeared! Luckily the gremlins forgot to hide my Radio library, and the same idea is in Dennison, Lorek (editors), 2005, Radio Communication Handbook 8th Edition, Radio Society of Great Britain

Snip from page 1.16:

And also discussed on this website.

Might seem surprising that a pendulum and tuned circuits follow the same Physical Laws, but both involve oscillations.

I have more background in communications than clocks, so tend to think of pendula as oscillators, emitting signals with a bandwidth, allowing me to use the maths and electronic techniques I'm familiar with to model them.

Dave

Dave

[SillyOldDufferModerator @sillyoldduffer]

I also came across Estill I Green, (1955) "The Story of Q", a not too mathematical history of Q and its wider adoption.

This comparative table is useful:

and it mentions the relationship between Q and Power Factor, which I hadn't twigged must exist.

Dave

[Michael GilliganParticipant @michaelgilligan61133]

An excellent find, Dave … I will read it properly after some essential gardening

Weather is suitably mild at last !

MichaelG.

[Michael GilliganParticipant @michaelgilligan61133]

For the sake of completeness [and probably nothing more], here is Johnson’s patent: **LINK**

https://worldwide.espacenet.com/patent/search?q=pn%3DUS1628983A

MichaelG.

[S KParticipant @sk20060]

Dave:

Something continues to bother me about your method and your obtained value of Q. I am not doubting the fundamentals of the method (much), but the numbers do not seem to make sense in relative terms.

You appear to be getting about 1ms standard deviation in your period measurements, whereas in my last test I'm getting about 4us; a factor of 250 difference. Not all is equal, but assuming that the bandwidth is strongly tied to those numbers (i.e., 250 times lower S.D. means 250 times narrower bandwidth), then using your method I should get a Q that is very approximately 250 times yours. That would not remotely make sense.

What am I getting wrong or not seeing?

Edited By S K on 16/08/2023 14:37:23

[S KParticipant @sk20060]

Well, I used Dave's method on my pendulum data, as far as I understand it (and I may not), and got a "Q" of 190,000. That's not possible.

[John HaineParticipant @johnhaine32865]

If I may I'd like to comment on Dave's method. It seems to me that the resemblance of the normal distribution curve to a resonance curve is tempting but unfortunate. Assuming that Dave's period distribution is normal, its width is characterised by just one parameter, the standard deviation. The "-3dB" points are related to this but superfluous.

In practice the width of the period distribution must depend on the amplitude and noise level (i.e. the "signal to noise ratio" as well as on Q. So buried in Dave's computation is this SNR, which itself is also determined by Q (as this determines the bandwidth of the noise which is affecting the period). So one can't take the width of the distribution compared to the mean as the Q since actually if the noise was zero Q would apparently be infinite! Some time back I looked at this and I think this equation for the rms period variation is accurate for a pendulum with small swing, in equilibrium amplitude, at frequency Fo, Q, amplitude a, and mass m, where N is the spectral density of the noise force component of the impulse.

To compute Q from this you would need to know N. Pendulum Q can be measured most directly by a run-down test. Electronics Q is generally best measured as bandwidth or sometimes magnification factor.

[SillyOldDufferModerator @sillyoldduffer]

After reading SK and John's posts I decided to pour myself a couple of large sherries and go to bed early!

Dave

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