John Haine | 10/08/2023 06:47:59 |
5563 forum posts 322 photos | I think S K's points about measurement error are good. If your amplitude measurement is based on a pulse length which is the small difference between two numbers, then every measurement has an error. If you then difference successive amplitude measurements then again you get errors stacking up. When I has estimating the Q of my Arduinome IIRC I got reasonably consistent results looking at decay over the 60s impulse cycle where one could at least average 30 numbers - Q ~12,000 I think. |
Joseph Noci 1 | 10/08/2023 11:44:45 |
1323 forum posts 1431 photos | I am very green on this so please forgive questions that may seem very simple to those skilled in the art... looking at decay over the 60s impulse cycle What does that mean, and how do you do it? My pendulum has a 3.5kg bob, inside of which is an XYZ accelerometer , and it has a slotted opto detector at BDC. Ignoring the magnetic impulse mechanism fitted, for the moment, how do I practically measure Q with this setup? I hope to have a pendulum swing of 2 degrees ( 1deg either side) . Can I measure Q by setting the pendulum swinging manually, and monitor the pulse and pendulum period say on a 'scope? I have been playing and using SK's method I get Q's of 18000 to 20000 and I do not believe it.... A gentle help please! |
duncan webster | 10/08/2023 11:58:49 |
5307 forum posts 83 photos | You could use the slotted opto, as amplitude decays the ratio of interrupted to open increases, then a bit of dimple maths, but I'll have to fire up the PC, trying to type it on a phone will drive me potty. |
John Haine | 10/08/2023 12:19:18 |
5563 forum posts 322 photos | Posted by Joseph Noci 1 on 10/08/2023 11:44:45:
I am very green on this so please forgive questions that may seem very simple to those skilled in the art... "looking at decay over the 60s impulse cycle" What does that mean, and how do you do it? The pendulum is impulsed once every minute by a gravity arm and the amplitude slowly decays until the next impulse (though not by very much). In equilibrium the max and min amplitude over the impulse cycle are constant from (impulse) cycle to cycle. The amplitude is estimated for each pendulum cycle by an opto and a picPET, deriving the amplitude from the pulse length. The decay of the maximum over 30 cycles closely follows a reverse-exponential law: An/Ao = exp(-n*pi/Q) where An is the amplitude of the n'th cycle and Ao that of the first. Just invert the expression to get Q. My pendulum has a 3.5kg bob, inside of which is an XYZ accelerometer , and it has a slotted opto detector at BDC. Excellent that you have an accelerometer! It will be very interesting to see the results. How wide is the slot (or vane)? A simple and reasonably accurate way to calculate amplitude as long as it is fairly large compared to the width and on-centre is just to compute the velocity and derive the amplitude from that and the period. Ignoring the magnetic impulse mechanism fitted, for the moment, how do I practically measure Q with this setup? What I do is to use the impulsing to get the amplitude up to a decent value a little larger than your target, and switch it off, then just observe the amplitude as it decays and apply the formula above. I hope to have a pendulum swing of 2 degrees ( 1deg either side) . Can I measure Q by setting the pendulum swinging manually, and monitor the pulse and pendulum period say on a 'scope? You could use a 'scope but I know you're a time-nut too so something like a picPET would be better! I have been playing and using SK's method I get Q's of 18000 to 20000 and I do not believe it.... Re-calculate using the above formula, I am pretty certain it is right (and a mathematician friend agrees). A gentle help please!
|
Joseph Noci 1 | 10/08/2023 13:49:45 |
1323 forum posts 1431 photos | 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: 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 Haine | 10/08/2023 14:16:22 |
5563 forum posts 322 photos | Yes, the length of the opto pulse between transitions. The time between successive say positive edges would be the period. |
Joseph Noci 1 | 10/08/2023 15:21:21 |
1323 forum posts 1431 photos | 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 Haine | 10/08/2023 15:43:26 |
5563 forum posts 322 photos | 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 1 | 10/08/2023 16:31:35 |
1323 forum posts 1431 photos | 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..
|
SillyOldDuffer | 10/08/2023 17:11:44 |
10668 forum posts 2415 photos | 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 |
SillyOldDuffer | 10/08/2023 18:04:05 |
10668 forum posts 2415 photos | 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 K | 10/08/2023 18:51:03 |
288 forum posts 42 photos | 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 Gilligan | 10/08/2023 20:32:03 |
![]() 23121 forum posts 1360 photos | 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 K | 11/08/2023 00:50:00 |
288 forum posts 42 photos | Yeah, no, let's not! Thank you! |
Michael Gilligan | 11/08/2023 03:04:45 |
![]() 23121 forum posts 1360 photos | |
John Haine | 11/08/2023 10:26:34 |
5563 forum posts 322 photos | 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 Haine | 11/08/2023 10:37:50 |
5563 forum posts 322 photos | 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.. |
SillyOldDuffer | 11/08/2023 11:44:10 |
10668 forum posts 2415 photos | 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:
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 Gilligan | 11/08/2023 13:31:47 |
![]() 23121 forum posts 1360 photos | 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: 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 |
SillyOldDuffer | 11/08/2023 16:18:15 |
10668 forum posts 2415 photos | 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 |
Please login to post a reply.
Want the latest issue of Model Engineer or Model Engineers' Workshop? Use our magazine locator links to find your nearest stockist!
Sign up to our newsletter and get a free digital issue.
You can unsubscribe at anytime. View our privacy policy at www.mortons.co.uk/privacy
You can contact us by phone, mail or email about the magazines including becoming a contributor, submitting reader's letters or making queries about articles. You can also get in touch about this website, advertising or other general issues.
Click THIS LINK for full contact details.
For subscription issues please see THIS LINK.