Isochronous knife edge suspension?

[david bennett 8Participant @davidbennett8]

I don't think Huygens suggested that the cylinder had to be half pendulum size in diameter. It was just convenient. He was just using that to present his proof for his particular pendulum. Others since him have mis-interpreted his intentions.He wasn't trying to establish a rule for all cycloids, as we are.

dave8

[Michael GilliganParticipant @michaelgilligan61133]

Tony, Duncan, Mike, Dave

Thanks for your comments/observations

Just to add a personal ‘position statement’ … I am well aware that other avenues have already led to more accurate pendulums, but I have long been interested in Huygens’ approach.

The mathematical proofs, I regret to say, cause me to glaze-over … so the simple geometrical exercise is much more to my taste.

The other difficulty, of course, is my trivially small command of Latin … There is no way that I could translate Huygens’ text, so I am reliant upon Ian Bruce for the English and Huygens for the pictures !

… The best I can manage is a “sanity check” on a few individual words from Bruce’s translation, but the grammar of ‘Scientific Latin’ can be full of subtleties beyond my comprehension.

I look forward to any further insights.

MichaelG.

[John HaineParticipant @johnhaine32865]

I think the killer to cycloidal approaches is the fact that all real pendulums are compound to some extent, as the bobs aren't point masses, rods have mass, whatever the pendulum uses to support the top of the rod has mass, and there could be things like weight trays and so on fitted on the rod. As Mike alludes to, there is no path the CG of a compound pendulum can follow which will make it isochronous. What the error might be for the magnetic configuration being discussed here I'm trying to ascertain.

I first came across the term "isochronous" probably in 1972 when I was working on digital data transmission at the PO Research labs in my first job, so it has been around a long time.

[Tony JeffreeParticipant @tonyjeffree56510]

Posted by duncan webster on 07/09/2023 22:58:08:

It's covered in this article from Wikipedia. If it has stood the test of 300 years, and mathematicians such as Lagrange and Euler have been involved in providing proofs I think we can take it as read.

To save flogging through it the significant sentence in my view is

there were much more significant sources of timing errors that overwhelmed any theoretical improvements that traveling on the tautochrone curve helps. Finally, the "circular error" of a pendulum decreases as length of the swing decreases, so better clock escapements could greatly reduce this source of inaccuracy.

In other words, keep the amplitude low and constant and forget about isochroism (if there is such a word)

Correct me if I am wrong, but the proofs seem to be focused on proving that if the particle (pendulum bob) follows a cycloidal path, then it will be isochronous. I couldn't see (maybe I didn't look hard enough) any discussion of how the use of cycloidal cheeks corrected the bob's path to be cycloidal, and what parameters the cheels needed to have in order for this to happen.

[duncan webster 1Participant @duncanwebster1]

I'm getting a lot out of my depth here, but my reading of it is that the cheeks have to be cycloidal.

[Tony JeffreeParticipant @tonyjeffree56510]

Posted by duncan webster on 08/09/2023 12:04:40:

I'm getting a lot out of my depth here, but my reading of it is that the cheeks have to be cycloidal.

I don't dispute that, but I haven't seen that proved, or what radius of generating circle for the cycloid is required in order for it to work.

[david bennett 8Participant @davidbennett8]

Posted by david bennett 8 on 08/09/2023 01:11:46:

I don't think Huygens suggested that the cylinder had to be half pendulum size in diameter. It was just convenient. He was just using that to present his proof for his particular pendulum. Others since him have mis-interpreted his intentions.He wasn't trying to establish a rule for all cycloids, as we are.

dave8

Further to this – We had an example on this site where totally unnecessary dimensions can be specified , which could ne misinterpretd. when I enquired on the "general" forum for the best way to produce a 39" curve, I was required to give 3-dimensions for the part. They wheren't needed, but arbitrary sizes where given just so the problem could be visualised. Perhaps that is why Huygens gave his 1/2 pendulum size for producing a cycloid.

dave8

[duncan webster 1Participant @duncanwebster1]

Posted by Tony Jeffree on 08/09/2023 12:34:16:

……….or what radius of generating circle for the cycloid is required in order for it to work.

Same as the pendulum cycloid.

[Michael GilliganParticipant @michaelgilligan61133]

I don’t think this does anything to resolve the issue raised by dave8

[regarding a small generating circle with a long rod appended radially]
.

But it’s a neat interactive demonstration, with a simple explanation of the mathematics

… so I commend it to all readers

**LINK**

https://www.geogebra.org/m/QeQ9aA5e

MichaelG.

.

P.S. __ @dave8 … Please do correct me if I have misrepresented you !!

[Tony JeffreeParticipant @tonyjeffree56510]

Posted by duncan webster on 08/09/2023 13:51:38:

Posted by Tony Jeffree on 08/09/2023 12:34:16:

……….or what radius of generating circle for the cycloid is required in order for it to work.

Same as the pendulum cycloid.

Plausible answer, but where's the proof?

[david bennett 8Participant @davidbennett8]

Posted by Tony Jeffree on 07/09/2023 14:04:58:

Of course, a moment's thought about the wheels on a car would have brought you to the same conclusion. A point on the tread follows a near-enough cycloidal path (give or take the flexibility of the tyre), but you'd better hope that the wheel axle doesn't follow a cycloid, or it is going to be a rather bumpy ride!

Edited By Tony Jeffree on 07/09/2023 14:05:35

No. A point on the tread follows a circular path.

dave8

[duncan webster 1Participant @duncanwebster1]

Only if the wheels are skidding and the vehicle isn't moving.

[duncan webster 1Participant @duncanwebster1]

Posted by Tony Jeffree on 08/09/2023 18:38:03:

Posted by duncan webster on 08/09/2023 13:51:38:

Posted by Tony Jeffree on 08/09/2023 12:34:16:

……….or what radius of generating circle for the cycloid is required in order for it to work.

Same as the pendulum cycloid.

Plausible answer, but where's the proof?

I think it's covered in that link I posted yesterday to Wikepedia. If it's good enough for Lagrange, it's good enough for me

[Tony JeffreeParticipant @tonyjeffree56510]

Posted by david bennett 8 on 08/09/2023 18:51:16:

Posted by Tony Jeffree on 07/09/2023 14:04:58:

Of course, a moment's thought about the wheels on a car would have brought you to the same conclusion. A point on the tread follows a near-enough cycloidal path (give or take the flexibility of the tyre), but you'd better hope that the wheel axle doesn't follow a cycloid, or it is going to be a rather bumpy ride!

Edited By Tony Jeffree on 07/09/2023 14:05:35

No. A point on the tread follows a circular path.

dave8

Only if your car is up on bricks. Mind you, if it is, the wheels have probably been nicked…

Edited By Tony Jeffree on 08/09/2023 19:17:15

[duncan webster 1Participant @duncanwebster1]

Only if your car is up on bricks. Mind you, if it is, the wheels have probably been nicked…

Edited By Tony Jeffree on 08/09/2023 19:17:15

Is that a Liverpool cycloid? When Liverpool was European Capital of Culture you used to come back and find your car propped up on books. Goes away and hides now from irate Scousers (including SWMBO and family)

[Tony JeffreeParticipant @tonyjeffree56510]

Posted by duncan webster on 08/09/2023 19:14:46:

Posted by Tony Jeffree on 08/09/2023 18:38:03:

Posted by duncan webster on 08/09/2023 13:51:38:

Posted by Tony Jeffree on 08/09/2023 12:34:16:

……….or what radius of generating circle for the cycloid is required in order for it to work.

Same as the pendulum cycloid.

Plausible answer, but where's the proof?

I think it's covered in that link I posted yesterday to Wikepedia. If it's good enough for Lagrange, it's good enough for me

I don't think it is. To repeat – I may be wrong, but it seems to me that the Wiki discussion is all about demonstrating that if the pendulum bob follows a cycloidal path, then the pendulum will be isochronous, and not about how the cycloidal path is created. Where I get off the bus is the apparent assumption that the cycloidal cheeks get you there (modifying the bob's path such that it travels along acycloid). If I've missed it in the Wiki explanation, well and good, but right now I'm not seeing it.

[Tony JeffreeParticipant @tonyjeffree56510]

Posted by duncan webster on 08/09/2023 19:22:00:

Only if your car is up on bricks. Mind you, if it is, the wheels have probably been nicked…

Edited By Tony Jeffree on 08/09/2023 19:17:15

Is that a Liverpool cycloid? When Liverpool was European Capital of Culture you used to come back and find your car propped up on books. Goes away and hides now from irate Scousers (including SWMBO and family)

That's the one

[david bennett 8Participant @davidbennett8]

If you are inside a moving car, observing the wheel will show a circular path (you may need a mirror) If you are outside the car, and stationary, a point on the wheel  of a moving car will be seen to have followed a cycloidal path.

dave8

Edited By david bennett 8 on 08/09/2023 19:41:01

[duncan webster 1Participant @duncanwebster1]

OK, try this, which is so far above my head it might as well be in orbit

[Michael GilliganParticipant @michaelgilligan61133]

Posted by Tony Jeffree on 08/09/2023 19:25:22:
.

[…]Where I get off the bus is the apparent assumption that the cycloidal cheeks get you there (modifying the bob's path such that it travels along acycloid). If I've missed it in the Wiki explanation, well and good, but right now I'm not seeing it.

.

I hesitate to write this … but here goes:

I think it is ‘intuitively obvious’ from Huygens

But, of course, that intuition only applies to a simple pendulum with flexible string.

MichaelG.

[david bennett 8Participant @davidbennett8]

Posted by Tony Jeffree on 08/09/2023 18:38:03:

Posted by duncan webster on 08/09/2023 13:51:38:

Posted by Tony Jeffree on 08/09/2023 12:34:16:

……….or what radius of generating circle for the cycloid is required in order for it to work.

Same as the pendulum cycloid.

Plausible answer, but where's the proof?

As I tried to show in my post yesterday,at 16:28 the generating circle doesn't matter. A cycloid is a cycloid. Pick any convenient size.

dave8

[Tony JeffreeParticipant @tonyjeffree56510]

Posted by david bennett 8 on 08/09/2023 20:39:40:

Posted by Tony Jeffree on 08/09/2023 18:38:03:

Posted by duncan webster on 08/09/2023 13:51:38:

Posted by Tony Jeffree on 08/09/2023 12:34:16:

……….or what radius of generating circle for the cycloid is required in order for it to work.

Same as the pendulum cycloid.

Plausible answer, but where's the proof?

As I tried to show in my post yesterday,at 16:28 the generating circle doesn't matter. A cycloid is a cycloid. Pick any convenient size.

dave8

I'm sorry, that's absurd. What if I choose a convenient size of 0.000001mm as the radius of the generating circle?

[david bennett 8Participant @davidbennett8]

Then it wouldn't be a convenient size – would it?

dave8

[david bennett 8Participant @davidbennett8]

On second thoughts, if you find that is a convenient size, please go ahead. Be sure to do a write-up so we can follow your procedures, especially the measuring. It's sure to attract a lot of attention. You'd like that, wouldn't you?

dave8

[david bennett 8Participant @davidbennett8]

Posted by Tony Jeffree on 07/09/2023 14:04:58:

Of course, a moment's thought about the wheels on a car would have brought you to the same conclusion. A point on the tread follows a near-enough cycloidal path (give or take the flexibility of the tyre), but you'd better hope that the wheel axle doesn't follow a cycloid, or it is going to be a rather bumpy ride!

Edited By Tony Jeffree on 07/09/2023 14:05:35

If a point on your tread follows a near-enough cycloidal path, you probably need new bearings.

dave8

Edited By david bennett 8 on 09/09/2023 04:45:55

Edited By david bennett 8 on 09/09/2023 04:58:52

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