Isochronous knife edge suspension?

[John HaineParticipant @johnhaine32865]

Exactly.

[david bennett 8Participant @davidbennett8]

That's encouraging. Much less "lifting" needed for the pendulum. A circular curve based on a theoretical pendulum length should be easy to generate. Maybe a correction between physical and actual pendulum lenghts could be made with a computer. More reding to do though.

dave8

[John HaineParticipant @johnhaine32865]

Have you got copies of Philip Woodward's articles?

[david bennett 8Participant @davidbennett8]

Posted by John Haine on 31/08/2023 17:09:07:

Have you got copies of Philip Woodward's articles?

Hi John – no. They would be much appreciated

dave8

[david bennett 8Participant @davidbennett8]

The next problem I see is determining the radius needed for the roller. It doesn't seem possible to accurately measure the length of a compound pendulum i.e from suspension point to centre of mass. Therefore some way to adjust for this seems necessary. I am considering a rating nut on the bob, not for timekeeping, but for altering mass position to agree with isochronism. Does this sound reasonable?

dave8

[Mike EvermanParticipant @mikeeverman73009]

Hi Gents, first post. Just wanted to chime in with my experience in this area. I've trod this path fairly deeply and made some rigs for testing. I wrote a paper on the subject for HSN 2009-5, which I would attach if I could.

Since writing that paper, I made another version of it that was higher fidelity, but still needs some things, of course. I do love the concept of a purely mechanical inverted pendulum, and it was very satisfying that it worked.

My takeaway from the whole project is that I think a rolling circular suspension has promise, if extremely polished and clean, but doubtful it will ever be a precision clock. The bob can fairly easily be made to follow a cycloid path, but the support itself must be made simply harmonic with a balance spring properly placed, which I have not added yet, as I have moved on.

[david bennett 8Participant @davidbennett8]

This is still bugging me.Can anyone explain why, if any single point on a pendulum is isochronous, the rest of the pendulum is not?

dave8

Edited By david bennett 8 on 01/09/2023 05:40:11

[John HaineParticipant @johnhaine32865]

Obviously the whole thing has to be isochronous. You can compute the effective "length" of a compound pendulum by adding together all the moments of inertia and moments of mass of its components in a spreadsheet.

As for the roller radius, Woodward's second paper has the necessary analysis but that would need extending to derive the necessary radius (if indeed there is one).

I'll send you a PM.

[SillyOldDufferModerator @sillyoldduffer]

Posted by Mike Everman on 01/09/2023 00:25:15:

Hi Gents, first post. Just wanted to chime in with my experience in this area. I've trod this path fairly deeply and made some rigs for testing. I wrote a paper on the subject for HSN 2009-5, which I would attach if I could.

…

Welcome Mike, delighted to have you aboard. Model Engineering includes experimental engineering and much else.

There are four or five threads running on pendulums at the moment, and one on calculating Q-factor. Our projects mostly have electronics, microcontrollers, and magnetic impulsing and are focussed on high precision, but all contributions welcome.

Unfortunately the forum can't host anything other than jpg images. The workaround is to host it elsewhere on the web and link to it. Dropbox is OK for light interest sharing.

It is planned to upgrade the forum software soon. At the moment the new software hosts images, video and audio, but not documents. They're high on the wish list so fingers crossed we get them too.

Dave

[david bennett 8Participant @davidbennett8]

My point is that if any arbitrary point (on the rod) represented by my tuppenny roller on a flat plane is isochronous, what more do I need?

dave8

[John HaineParticipant @johnhaine32865]

Well I don't think isochronism is a concept applicable to a point! It's a property of the system as a whole. To put it another way, what do you mean by a point being isochronous?

[david bennett 8Participant @davidbennett8]

I mean that if a point on the roller (where the rod joins it) is following a cycloidal curve, then that point and all the rest must be isochronous.

dave8

[John HaineParticipant @johnhaine32865]

No, that doesn't follow. It's the CoG of the bob that has to follow a specific cycloid, with a generating circle diameter equal to the rod length. The generating circle for your configuration is the roller diameter. (This assumes that the mass is concentrated at the bob.)

[david bennett 8Participant @davidbennett8]

As I said before, if any point on a pendulum (in this case the roller) is isochronous,how can the other points on the pendulum not be?

dave8

[Michael GilliganParticipant @michaelgilligan61133]

At the risk of stating the obvious … It may be worth noting that for Huygens, a pendulum was a relatively heavy bob on the end of a very light and flexible string, making large angle swings … the modern style of clock pendulum evolved later.

The elegant geometrical truth that he describes depends fundamentally upon that … which [I think] goes a long way towards explaining the difficulty of using cycloidal cheeks on the modern style of pendulum with its long rod and short suspension spring [the dimensional tolerancing becomes very difficult]

How much of this reads-across to the roller idea, I am not sure.

MichaelG.

[david bennett 8Participant @davidbennett8]

Michael, I don't think the cycloidal cheeks are relevant here. The roller is quite a different idea. This is bugging me as it appears to be an unexplained anomaly.

dave8

[John HaineParticipant @johnhaine32865]

Think of a weight sliding down a symmetrical valley and up the other side with no friction. It will carry on going down and up again, oscillating to and fro with the same amplitude. What shape does the valley have to have such that the time taken for a single back and forth oscillation is the same irrespective of the amplitude? This shape is the "tautochrone" which is actually the evolute of a cycloid. But the evolute of a cycloid is itself the same cycloid but just shifted.

For a pendulum, the weight is the bob, and so it has to move in a cycloid. The effective pendulum length is the radius (sorry, not diameter) of the generating circle. So if the weight is suspended by a rod the suspension has to arrange that the CoG of the bob moves in a cycloid with a generating circle of the same radius as the (maximum) rod length, for example using curved cheeks. Or arrange that it does so at least over a range of amplitudes of interest. Woodward showed that a roller on top of a plane support can't achieve this, but it's an open question whether a roller underneath a plane would work.

For a roller arranged to roll on a plane, though a point on its circumference will roll in a cycloid, the generating circle has the same radius as the roller which is much smaller than the rod length. So the CG of the bob will not move in a cycloid (in fact it follows a trochoid) and the pendulum will not be isochronous (and nor in fact will the point on the roller where the rod is attached be).

If you could arrange a weightless roller with the bob attached to its circumference, then it would be isochronous.

There's an interesting mechanism that might be exploited that avoids magnets, called the Rolamite. I have seen an article discussing its application in a clock though not read in detail.

[Michael GilliganParticipant @michaelgilligan61133]

I must leave it for you to investigate, Dave … I will be very interested to see what you discover

MichaelG.

[david bennett 8Participant @davidbennett8]

Posted by John Haine on 01/09/2023 13:20:20:….. the pendulum will not be isochronous (and nor in fact will the point on the roller where the rod is attached be).

If you could arrange a weightless roller with the bob attached to its circumference, then it would be isochronous…….

John, thats the bit I don't get. If the point on the roller where the rod is attached rolls on a flat plane,surely it must trace a cycloid, and be isochronous.?

What does a roller weigh if it is suspended by a magnet?

I am just a simple amateur in all this, so bow to superior knowledge, but I'm losing faith in this project.

dave8

[david bennett 8Participant @davidbennett8]

John, have sent you a PM

dave8

[david bennett 8Participant @davidbennett8]

Please, can someone explain this paradox to me?

On the 2 penny pendulum I showed earlier, science tells us there is a point on the rod ( about 1/2" down from the suspension point) where  a bob on the shaft ifollows a cycloidal  path and is isochronous. If the rod is extended down to 39" and the bob placed there, why is that not following a cycloidal path? If any one point on the rod follows a cycloid , why aren't they all?

I know I am wrong, and I beleive all the theory that the path depends on pendulum length, but I would like to resolve this in my mind.

dave8

Edited By david bennett 8 on 03/09/2023 05:26:25

Edited By david bennett 8 on 03/09/2023 05:37:12

[Michael GilliganParticipant @michaelgilligan61133]

I will try, Dave … but it repeats things I have already mentioned, and l think your mind-set is currently different.

Please look carefully at Huygens’ description of how to construct the cycloid and [crucially] note that his style of pendulum does not use a rod … it is flexible.

Your 2p version works as a little self-contained mechanism … but extending it with a stiff rod will not.

MichaelG.

.

Edit: __ Here, in case you didn’t find it earlier, is a link to Ian Bruce’s translation:

https://www.17centurymaths.com/contents/huygenscontents.html

Edited By Michael Gilligan on 03/09/2023 06:59:31

[david bennett 8Participant @davidbennett8]

I was wondering if the science was based on a fixed suspension point, and needed an additional term to account for the moving suspension I presented which itself rolls in a cycloidal manner.

dave8

[John HaineParticipant @johnhaine32865]

Posted by david bennett 8 on 03/09/2023 05:10:27:

Please, can someone explain this paradox to me?

… If any one point on the rod follows a cycloid , why aren't they all?

…

dave8

Edited By david bennett 8 on 03/09/2023 05:26:25

Edited By david bennett 8 on 03/09/2023 05:37:12

Though the point where the rod is attached to the roller may describe a cycloid, it's the wrong size for the bob which is much further away. And the CG of the bob (supposing all the system mass is concentrated there) actually describes a prolate trochoid not a cycloid as Woodward shows. So the system cannot be isochronous. I'm not sure what more one can say.

Huygen's science does not assume a fixed suspension point, the bob is "suspended" from wherever the suspension cord instantaneously touched the cycloidal cheek.

[david bennett 8Participant @davidbennett8]

I am looking at this as a simple logic exercise. If a point anywhere on a moving pendulum is observed to move in a particular arc (no matter what it is called) wouldn't you expect all points on that pendulum to move in similar arcs?

dave8

Edited By david bennett 8 on 03/09/2023 20:02:42

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