Isochronous knife edge suspension?

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.

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

MichaelG.

John, have sent you a PM

dave8

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

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

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

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.

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

[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

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

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

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

Michael,

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

I know just how you feel.

dave8

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.

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

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

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.