Isochronous knife edge suspension?

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.

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.

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.

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.

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

Same as the pendulum cycloid.

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 !!

Plausible answer, but where's the proof?

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

dave8

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

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

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)

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.

That's the one

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

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

.

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.

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