pendulum timing conundrum

Discussion in 'General Electronics Chat' started by cornishlad, Mar 23, 2015.

Jul 31, 2013
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Please bear with me on this question which has stretched my ageing analogue mind.
I'm building a pendulum clock. One swing - say, left to right -(t) takes 1 sec and the "period" (2t) - start position back to the same position takes 2 seconds.
It's in an early stage and the pendulum is not yet driven - but will swing freely for a while.
For mechanical reasons I need to get the pendulum adjusted to be be as nearly timed as possible - ie before it can be rated as a clock over several hours.
The pendulum is sensed by opto-interupters. At present I have just one which happens to be at about the centre of the swing (but I don't think it matters where it is). The timing width of the pulse from it is about 200 milliseconds but the width varies a bit depending on the amplitude of the swing (and hence its speed )
My only method of measuring the measuring (t), or (2t) is with a Racal-Dana 9904 counter/timer in "stopwatch" mode. This can start and stop a count by selecting edge polarities. It will easily measure the opto pulse width but that's no good to me.
My attempts at a timing diagram are confounded by the fact that a +ve edge is a -ve edge on the next pass as the pendulum is traveling the other way.
If you are still reading this - thank you
So, the question is how can I accurately measure (t) or (2t) As far as I can work it out I will need a dual "D" type flip flop (4013) set to divide by 4 and measure one positive or negative going output pulse.
Confirmation, or other suggestions please !
I'm asking because, after an accident, the opto has failed and a replacement will take 2 weeks from China..

2. JDT Well-Known Member

Feb 12, 2009
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OK, so if you have an opt-interrupter in the middle, that say, gives an output when it is interrupted, you will get a pulse every "zero crossing". Or, to put it another way, 2 pulses per cycle.

So, if you have a frequency counter, just divide the reading by two. A more accurate way might be to count the pulses over a fixed time. Hopefully 3600 x 2 = 7200 per hour.

Or you could construct a simple divide-by-two circuit (a flip-flop) to give you a square wave at the actual frequency of the pendulum.

Jan 15, 2015
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The length of the pulse should be a function of the width of the pendulum passing the optical pickup and thus a function of the speed of the pendulum. To accurately measure pendulum travel end to end you really need two optical pickups and measure time interval A/B. Using a single pickup at one end of the travel you could measure again time interval A/B then divide by 2 to approximate the one way transition time. I don't understand why the opto is an issue? I would think dozens of optical arrangements would work that should be inexpensive and readily available in the UK.

Yes, you can also use a /2 flip flop. I see JDT posted as I slowly typed along.

Ron

4. JDT Well-Known Member

Feb 12, 2009
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You had an optical sensor that is interrupted by the pendulum arm as it passes through the vertical position and you get two pulses per cycle. Also, the time that the opto is blocked depends on the amplitude of the swing. If the swing is small, the opto will be blocked for a greater proportion of the cycle. If is it is stopped it will be blocked all the time.

If you fitted a magnet to the pendulum bob (and have a stationary coil) you could use this fact to decide whether to give a "push" or not to keep it going. I can see that a simple micro-processor circuit could do this. And the micro-processor could have counters and a LED driver to display the time. Timed by your pendulum. A kind of modern version of a grandfather clock!

Jul 31, 2013
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I missed that simply dividing by 2 should be enough. The Racal counter will measure periods between edges so setting +ve edge to start and -ve edge to stop should indeed do it.
Freqency counting is not an option as the lowest frequency it will count is 10Hz.
Counting pulses over a period of time does not seem possible with this instrument unless there is tricky intervention with the manual "hold" function - and the pulses would have to be counted manually.
Thank you for that response...

6. JDT Well-Known Member

Feb 12, 2009
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Measuring the period in this way (once you have connected a flip-flop) will give the period of a single half-cycle. This is probably the most accurate method.

7. Alec_t AAC Fanatic!

Sep 17, 2013
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If you always trigger the timer on the same edge of the opto pulses, then won't the sum of two succesive interval measurements be the required period, independent of pulse width (hence swing amplitude) and sensor position?

Jul 31, 2013
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Thank you for thinking about the problem...now I think solved by JDT and yourself..
One practical thought though is it isn't a good idea to place an opto sensor at the end of the swing at this stage of construction (free swinging, non-motored). Later, in the design, there be one at one end of the swing but that will effectively set the amplitude of the swing by the "Hipp Toggle" principle. You may need to Google that if you're not familiar. An opto positioned at/near an extemity will, as the swing decreases, give 2 quick pulses, then 1 pulse, then no pulses.

Jul 31, 2013
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I mentioned that a digital implemtation of the 1842 Hipp Toggle will decide when it needs a push . I will be doing it with hardware (4013 and some gates) I have just bought an Arduino but have no idea how to program it although I can see that that would be a tidy solution . But it may take me a year to learn how.. lol

Jul 31, 2013
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Yes indeed...I've understood that now..R

11. MrChips Moderator

Oct 2, 2009
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I presume that you will use the mechanical pendulum to trigger a photo sensor which will then be used to clock a digital circuit at 1Hz.
If the opto-interrupter is placed in the middle then the pendulum has to have a period of 2 seconds.
The width of the trigger has no effect if the sensor is placed exactly at the mid-point of the swing. The issue about leading/trailing edge is a non-issue.
If the sensor is not exactly at the mid-point, the time recorded for a 2-second period will still be consistent.

Jul 31, 2013
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Line 1: yes indeed..actually there will be 3 opto sensors. The central one will do the clock pulses (which will be divided down to 1 pulse per 30 seconds to drive traditional slave clock(s) The other two are to implement the pendulum drive. I've got a scheme for that.
line 2: yes
line 3: yes..I got that figured ! The issue of leading and trailing edges was only an issue in triggering the Racal counter/timer, not in the proposed clock circuit.
line 4:got that as well..

incidentally the optos I'm using are tinyi pcb's with onboard comparators. Not so readily obtainable as the bare bones optos listed on ebay....Thanks again for your input. R

13. JDT Well-Known Member

Feb 12, 2009
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Actually, if you are going to keep the pendulum swinging using a magnet and coil, then the induced EMF can also be used to trigger the energising of the coil and extract a timing "tick". No optos required!

This has already been done to construct a cardboard clock. See here. Actually, extracting a timing pulse is not shown but would be a simple addition.

Bernard likes this.

Jul 31, 2013
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My clock is based on the master clocks that did sterling service in factories, schools etc from 1900 until around 1970. the purely electrical ones are excellent timekeepers due the the very low interference with the pendulum (by wheelwork etc)
Many run for 30 seconds without a pendulum impulse (Synchronome, Gents) and the PO N036 uses the "hipp Toggle" I mentioned already, which pulses the pendulum only when it's needed (sufficient diminution of swing) My clock will work like that.
Novelty electrical mantel clocks which pulsed the pendulum every swing (like the Bulle clock) were usually pretty poor timekeepers (relatively) due the constant interference (every swing) with the pendulums natural rhythm. (which is how the back emf clocks work)
Don't want to rattle on too much as it's not a clock forum R

Impulse is by electro magnet beneath and armature on the pendulum btw..

JDT likes this.
15. MrChips Moderator

Oct 2, 2009
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Using a mechanical pendulum for accurate time keeping is not a bad idea. But you still will not be able to compete with a simple quartz crystal oscillator which can be as accurate as 1ppm.
A thermally stable crystal oscillator will do even better.