Isochronous knife edge suspension?

[Michael GilliganParticipant @michaelgilligan61133]

[assuming that you are still considering a rodded pendulum]

Similar, Yes … but that alone is not sufficient to make the pendulum isochronous.

MichaelG.

.

Incidentally; just for clarity … can you explain why you describe your roller arrangement as a ‘knife-edge’ … it seems to be anything but.

Edited By Michael Gilligan on 03/09/2023 20:48:53

[david bennett 8Participant @davidbennett8]

Michael, sorry to be a nuisance about this, but if a point on the roller is travelling a cycloidal path, then all other points are too. That is the root of the paradox.

dave8

[John HaineParticipant @johnhaine32865]

But they aren't, as Philip's diagram shows.

[david bennett 8Participant @davidbennett8]

Knife edge – refers to the as yet uncut sharp V's on the edge of the coins which will be replaced by all steel replacements. I did say this was a quick and dirty first try.

dave8

[david bennett 8Participant @davidbennett8]

Posted by John Haine on 03/09/2023 21:05:35:

But they aren't, as Philip's diagram shows.

You don't find that paradoxical?

[Michael GilliganParticipant @michaelgilligan61133]

Posted by david bennett 8 on 03/09/2023 21:20:19:

Knife edge – refers to the as yet uncut sharp V's on the edge of the coins which will be replaced by all steel replacements. I did say this was a quick and dirty first try.

dave8

.

But that’s surely on an axis rotated 90° from what is normally considered a knife-edge suspension

… sorry, I’m too confused to think it through

MichaelG.

Edited By Michael Gilligan on 03/09/2023 21:32:40

[david bennett 8Participant @davidbennett8]

Michael,

— " I'm too confused to think it through "—

I know just how you feel.

dave8

[Michael GilliganParticipant @michaelgilligan61133]

Good morning, Dave

For the sake of your sanity, [and mine] … I will try once more, in bite-size steps:

1. Galileo told the world that the pendulum was isochronous … this [the story goes] being based on his personal observation of either a large thurible, or a lamp, in Pisa cathedral … the timing of which he checked with his pulse [!]
2. This was over-simplistic
3. Huygens put things right; by demonstrating that a simple pendulum was not isochronous, but that it could be made so by forcing it into cycloidal motion.
4. Please note that the hypothetical ‘simple pendulum’ comprises a massless and infinitely flexible ‘string’ hanging from a rigid support, and carrying a heavy bob of zero physical dimensions. [Reality is, of course, a little different]
5. Huygens’ clock used a verge escapement, which nicely suits the large angle of pendulum swing.
6. As horology developed, the anchor escapement and the dead-beat escapement superseded the verge, and the pendulum became a rod and bob, suspended on a ‘spring’ of shim.
7. This arrangement is not intrinsically isochronous, but can be considered so at the small angles of swing which suited the new escapements.
8. Further attempts to improve isochronism, by introducing cycloidal cheeks proved futile, because they could only act on the suspension spring … and that made appropriate dimensioning impractical … The problems far-outweighed any improvement.
9. Fedchenko saved the day with his astonishing spring arrangement, but that’s another story.
10. Your proposed roller arrangement, as the support for a rodded pendulum, cannot possibly mimic Huygens’ purist analysis …

I can only suggest that you work-through the clear instructions provided by Huygens, to get a ‘feel’ for the geometry.

MichaelG.

[SillyOldDufferModerator @sillyoldduffer]

My poor old bonehead can't accommodate this stuff so I started drawing it. Didn't get far! Before reaching the nature of the curve described by the bob's centre of mass, I think I spotted a fundamental practical problem. It's slip.

Surely the nature of the curve followed by the bob's centre of mass depends on the amount of slip between roller and flat? I think 100% slip would cause the roller to act as a simple hinge – so the pendulum won't be isochronous. To get isochronicity, I believe the roller mustn't slip at all? If that's right I can't think of a way of guaranteeing zero slip.

Other practical issues are the:

• amount of friction between roller and flat – I'd expect this to be high compared with a knife edge or spring suspension.
• need for the flat to be truly horizontal, otherwise not isochronous and the roller will wander downhill

Though I like David's idea and have merely failed to prove Messrs Huygens and Haines wrong myself, I think the cost of driving a high-friction suspension will outweigh the benefit of the pendulum being closer to isochronous. Nonetheless, I reckon David should build a better engineered version of his 2p proof-of-concept and measure it. My guess is friction will cause more trouble than the swing not being isochronous even if slip doesn't matter. I could be wrong; the experiment needs to tried!

Dave

[david bennett 8Participant @davidbennett8]

Dave, you may well be right that friction will be the problem – though it's not so much friction as the running through treacle feeling you get. I agree a better version is needed, though the theory nedds sorting first. I an coming to the conclusion that roller size doesn't matter. Imagine concentric circles ( each slightly distorted) drawn from the centre of a rolling cylinder. Eacg represents a pendulum length on a cycloidal path. The distorted circle determined the path , the length determines the period. I suggest you try the coins and magnet thing for yourself to get a feel for it. I have another problem going forward. I have no way to measure the period of a pendulum, neither the knowledge or equipment.

dave8

Edited By david bennett 8 on 04/09/2023 14:39:39

[duncan webster 1Participant @duncanwebster1]

Imagine David's set up, but continuously rotating anticlockwise at an angular velocity of w radians per second instead of oscillating. Define a point on the circumference of the cylinder (radius r) which, when the pendulum rod is vertical, is directly opposite the point of contact with the flat plane, call it P. The cylinder centre moves to the right at a velocity of w*R. Relative to the cylinder centre, point P has a horizontal velocity of w*r*cos(wt) where t is the time since the rod was vertical. The combined horizontal velocity relative to ground of point P is thus

w*r*(1+cos(wt)). This is =0 when wt=180 and a maximum when wt=0, and point P moves in a cycloid.

Now consider the motion of the bob, which is at a distance L from the centre of the cylinder. The velocity of the centre of the cylinder is as before, but the velocity of the bob relative to the cylinder centre is +w*L*cos(wt) so the combined motion is wr+wLcos(wt). This is clearly not =0 at wt=180, so no cycloid and no coconut.

However, if instead of the 2p pieces, the top of the pendulum was a curve (a sort of tee with a very long upright), this curve being equal to L. I think this then gives cycloidal motion to the bob.

Have I just repeated what has been said before? Probably, but it kept the grey cells working whilst waiting for SWMBO.

How do you stop it walking? Opposed tapes like the end of a Newcimen engine beam? Fine gear teeth? Neither appeals to me, but then I'm not a pendulista.

[david bennett 8Participant @davidbennett8]

Duncan,the maths behind that sent me to sleep, but I think you are saying the pendulum length decides rthe roller radius. That is the matter under discussion.

dave8

[david bennett 8Participant @davidbennett8]

Dave, just a PS to my last –I am not trying to disprove anyone.

dave8

[Michael GilliganParticipant @michaelgilligan61133]

Posted by david bennett 8 on 04/09/2023 14:38:52:

.

[…] I an coming to the conclusion that roller size doesn't matter. […]

.

In which case, everything I have written so far is wasted effort

Please ignore me

MichaelG.

[david bennett 8Participant @davidbennett8]

To clarify my view. The concept is really very simple.Geometry tells us that a point on the rim a roller rolling without slip on a flat plane describes a cycloid. Note that this applies to a roller of any radius. For various reasons, the roller has to be below the flat plane. My view is that a pendulum attached to that roller will be cycloidal  and therefore isochronous no matter what length it is.

dave8

Edited By david bennett 8 on 04/09/2023 17:16:07

[Michael GilliganParticipant @michaelgilligan61133]

Posted by david bennett 8 on 04/09/2023 17:14:24:

.

To clarify my view. The concept is really very simple.

[…] Note that this applies to a roller of any radius. […]

.…. and therefore, to a roller of zero radius [i.e. an hypothetical knife-edge]

MichaelG.

[John HaineParticipant @johnhaine32865]

I have had another read through Woodward's analysis. He mainly looks at the case where the roller runs on the top of a flat plane (or knife edges in Dave's original proposal). The pendulum rod is fixed to a point on the roller so its axis passes through the roller centre, and the rod is significantly longer than the roller radius. He shows that for this case the circular deviation is increased by a factor (1+4r/L) compared to an ordinary suspension for small swing angles.

For the case where the roller is like a hoop suspended from the plane (by magnetism for example as Dave proposes) the same analysis applies, except that the circular deviation is reduced. Obviously you would want that to be reduced to zero, and he shows that for this to be true the bob mass has to be on the periphery of the hoop, when of course its motion is cycloidal. For longer rods I think it looks like the circular deviation will get larger the longer the rod and be possibly of opposite sign to its normal value.

I think the overall conclusion must be that roller suspension can only cancel CD for the case of a "suspended hoop" with the bob on its periphery, when the motion is cycloidal. This is only true for a simple pendulum anyway, all real pendulums are compound. But I guess Duncan's idea could be used.  Have a very light rod of carbon fibre with the bob at one end; and a steel "shoe" with a cylindrical "sole" at the other of radius equal to half the rod length and long enough to allow for the intended maximum angle.  Though the shoe would have significant weight it would not contribute much to the moment of inertia since the axis of rotation is on its upper surface.

Edited By John Haine on 04/09/2023 17:49:00

[duncan webster 1Participant @duncanwebster1]

Posted by david bennett 8 on 04/09/2023 17:14:24:

To clarify my view. The concept is really very simple.Geometry tells us that a point on the rim a roller rolling without slip on a flat plane describes a cycloid. Note that this applies to a roller of any radius. For various reasons, the roller has to be below the flat plane. My view is that a pendulum attached to that roller will be cycloidal and therefore isochronous no matter what length it is.

dave8

Edited By david bennett 8 on 04/09/2023 17:16:07

Apart from I should have said the curve at thr top of the pendulum should be radius L/2 I think the maths shows that your view is incorrect, whether you understand it or not.

[david bennett 8Participant @davidbennett8]

John, yes, pretty much what I suggested on 31/8/23. I'm just guessing too, but the idea that if any point in any pendulum is cycloidal, then all points must be, is very powerful

dave8

[Michael GilliganParticipant @michaelgilligan61133]

Posted by duncan webster on 04/09/2023 17:55:35:

.

[…] the curve at thr top of the pendulum should be radius L/2 […]

.

Agreed, Duncan … that’s the very essence of it.

There’s a difference between ‘a cycloid’ and ‘the appropriate cycloid’

MichaelG.

[John HaineParticipant @johnhaine32865]

It would certainly be powerful if true but is demonstrably not! Michael's reductio ad absurdum is one demonstration.

I think this discussion got a bit sidetracked by the notion of "any point is cycloidal" – your original point about suspending a pendulum from a magnet is very valid and with the "shoe" would be a straightforward way to make a truly cycloidal pendulum. A bit of analysis needed to estimate the errors due to "compounding".

[duncan webster 1Participant @duncanwebster1]

Even if the suspension gives cycloidal motion, the magnetic path through the curved shoe would be moving, and so magnetic hysteresis would cause damping. Whether this is more or less than the mechanical hysteresis in a spring suspension is beyond my pay grade.

[david bennett 8Participant @davidbennett8]

Looks like we are getting towards the truth. I don't think Michaels zero radius roller would actually roll? maybe if not a sharp Vee, then maybe shaped to a small  cycloid? This is a side issuue (size of roller) and only considered to avoid extra work in making a first model. It is not fundamental. What we need is experimental work by someone who can measure if the periods of such a suspended pendulum is isochronous to check if the roller size matters. It only needs measurements between a tuppeny pendulum and a penny one.

dave8

Edited By david bennett 8 on 04/09/2023 21:09:50

Edited By david bennett 8 on 04/09/2023 21:10:46

[david bennett 8Participant @davidbennett8]

Further about a small cycloidal shaped Vee edge pendulum magnetically suspended. If that worked, and was isochronous it would prove that roller size doesn't matter.

dave8

please forget this idea – it might as well be a small circular shape.

Edited By david bennett 8 on 04/09/2023 21:37:14

[david bennett 8Participant @davidbennett8]

Posted by Michael Gilligan on 04/09/2023 18:07:44:

Posted by duncan webster on 04/09/2023 17:55:35:

.

[…] the curve at thr top of the pendulum should be radius L/2 […]

.

Agreed, Duncan … that’s the very essence of it.

There’s a difference between ‘a cycloid’ and ‘the appropriate cycloid’

MichaelG.

The cycloid is entirely appropriate to the roller -in this case 2p.

dave8

[Author]

Posts