Member postings for david bennett 8

John, I was aware of problems with using "reluctance",but I was using in the non-scientific way. I did search for another word to convey my meaning, but still cannot think of one.

dave8

Edited By david bennett 8 on 07/09/2023 17:51:06

Visualise a series of concentric circles, each distorted at the base by their appropriate cycloid . draw that on a card and pin it's centre to the axis of the 2p . Compare

pendulum paths of a large circle and a small one (2p size )

dave8

Edited By david bennett 8 on 07/09/2023 16:28:54

If I may clarify my decision without intending to upset anybody - None of your formulae take account of the influence of magnetising the roller or the shoe. Compare a non magnetised set-up with a magnetised one. At each end of a swing, one arm of a magnetised curved shoe is in close contact with a steel plate.the other (balancing) arm is some distance away.The "reluctance" of the magnet to let go will greatly affect the natural motion of the pendulum. This will happen at the other extreme too.I think that this is another example of a close look at what is actually happening serves better than blindly following the accepted formulae.

dave8

I am abandoning this build. Any imbalance in the magnets will interfere with the gravitational effect on the pendulum. It is therefore impractical.

dave8

Gentlemen, I originated this post, and the confusion. First , I should have said 39" diameter. Sorry. I neededd the best way to do this in principle.That has now been answered. Many of the quoted sizes are adjustable .Thanks.

dave8

I've just had a very dispiriting thought -

If tthese "coins" do not give a cycloidal movement to the pendulum, then the whole idea will not work. I can see no way to centralise the magnets on a shoe that is not circular.

dave8

Just a practical observation, for if anyone else is investigating this idea -

It seems that the magnets need to be circlar in shape, and probably set into shallow "pockets" cut into the rollers. If the magnets are not held central they can accelerate as they near steel plate and upset the timing.

dave8

Thanks for the correction - I think I was suffering from a thick brain.

dave8

Jason, thanks for that.

dave8

Thanks for all the ideas. Perhaps you now understand the lack of dimensions. This is for a clock pendulum suspension, and needs to be accurate.

I am leaning towards making this of two identical pieces of, say, 2mm thick steel plate. The only critical dimension is the radius and the need for the rolling edge to be at 90 degrees to the faces. I am hoping to avoid hand working the rolling faces to preserve accuracy. If I had the choice of machines, Ithink I would be looking at a pantograph engaving machine.

dave8

Edited By david bennett 8 on 05/09/2023 13:29:58

Edited By david bennett 8 on 05/09/2023 13:44:12

O.K, I amend my question to "any sensible suggestions"

dave8

Following a current discussion on the "clocks" forum here, I would welcome .any suggestions on the best way to form a piece of steel, about 2" long by 1/2" wide into a section of a circle with a 39" outside radius. The finish has to be good, at least ground.I have an old Myford lathe.

dave8

Edited By david bennett 8 on 04/09/2023 23:57:52

The cycloid is entirely appropriate to the roller -in this case 2p.

dave8

Further about a small cycloidal shaped Vee edge pendulum magnetically suspended. If that worked, and was isochronous it would prove that roller size doesn't matter.

dave8

please forget this idea - it might as well be a small circular shape.

Edited By david bennett 8 on 04/09/2023 21:37:14

Looks like we are getting towards the truth. I don't think Michaels zero radius roller would actually roll? maybe if not a sharp Vee, then maybe shaped to a small  cycloid? This is a side issuue (size of roller) and only considered to avoid extra work in making a first model. It is not fundamental. What we need is experimental work by someone who can measure if the periods of such a suspended pendulum is isochronous to check if the roller size matters. It only needs measurements between a tuppeny pendulum and a penny one.

dave8

Edited By david bennett 8 on 04/09/2023 21:09:50

Edited By david bennett 8 on 04/09/2023 21:10:46

John, yes, pretty much what I suggested on 31/8/23. I'm just guessing too, but the idea that if any point in any pendulum is cycloidal, then all points must be, is very powerful

dave8

To clarify my view. The concept is really very simple.Geometry tells us that a point on the rim a roller rolling without slip on a flat plane describes a cycloid. Note that this applies to a roller of any radius. For various reasons, the roller has to be below the flat plane. My view is that a pendulum attached to that roller will be cycloidal  and therefore isochronous no matter what length it is.

dave8

Edited By david bennett 8 on 04/09/2023 17:16:07

Dave, just a PS to my last --I am not trying to disprove anyone.

dave8

Duncan,the maths behind that sent me to sleep, but I think you are saying the pendulum length decides rthe roller radius. That is the matter under discussion.

dave8

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