Isochronous knife edge suspension?

Afraid not. The whole concept is to mimic a cycloidal path, which is defined by a circle rolling on a straight (flat) plane.

dave8

Edited By david bennett 8 on 30/08/2023 23:31:33

A small improvement - by using 1p coins and only 2 magnets half amplitude is about 3 minutes. At least it suggests a way forwards.

dave8

To be clear, there is only one cycloidal path that makes the pendulum isochronous and that's one where the roller radius (or diameter?) equals the pendulum length. Anything else just approximates the cycloid for small angles in a way that can correct the isochronism. Anyway the path of the bob for this setup is epi-trochoidal. As far as I know everyone who has looked at this (mainly Woodward) just assumed that the magnetic version would be impracticable and the calculations of the required roller diameter haven't been done, you're pioneering here Dave! Also one would have to factor in that real pendulums are compound just to make it more interesting.

.

As described by Huygens in the page that I posted 28/08/2023 16:15:31

”Upon a flat table … “

MichaelG.

Exactly.

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

Have you got copies of Philip Woodward's articles?

Hi John - no. They would be much appreciated

dave8

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

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.

https://www.youtube.com/watch?v=RZyIPPZ6GXU

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.

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

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.

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

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

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?

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

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.)

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

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.

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