Isochronous knife edge suspension?

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

dave8

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In which case, everything I have written so far is wasted effort

Please ignore me

MichaelG.

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

.…. and therefore, to a roller of zero radius [i.e. an hypothetical knife-edge]

MichaelG.

I have had another read through Woodward's analysis. He mainly looks at the case where the roller runs on the top of a flat plane (or knife edges in Dave's original proposal). The pendulum rod is fixed to a point on the roller so its axis passes through the roller centre, and the rod is significantly longer than the roller radius. He shows that for this case the circular deviation is increased by a factor (1+4r/L) compared to an ordinary suspension for small swing angles.

For the case where the roller is like a hoop suspended from the plane (by magnetism for example as Dave proposes) the same analysis applies, except that the circular deviation is reduced. Obviously you would want that to be reduced to zero, and he shows that for this to be true the bob mass has to be on the periphery of the hoop, when of course its motion is cycloidal. For longer rods I think it looks like the circular deviation will get larger the longer the rod and be possibly of opposite sign to its normal value.

I think the overall conclusion must be that roller suspension can only cancel CD for the case of a "suspended hoop" with the bob on its periphery, when the motion is cycloidal. This is only true for a simple pendulum anyway, all real pendulums are compound. But I guess Duncan's idea could be used.  Have a very light rod of carbon fibre with the bob at one end; and a steel "shoe" with a cylindrical "sole" at the other of radius equal to half the rod length and long enough to allow for the intended maximum angle.  Though the shoe would have significant weight it would not contribute much to the moment of inertia since the axis of rotation is on its upper surface.

Edited By John Haine on 04/09/2023 17:49:00

Apart from I should have said the curve at thr top of the pendulum should be radius L/2 I think the maths shows that your view is incorrect, whether you understand it or not.

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

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Agreed, Duncan … that’s the very essence of it.

There’s a difference between ‘a cycloid’ and ‘the appropriate cycloid’

MichaelG.

It would certainly be powerful if true but is demonstrably not! Michael's reductio ad absurdum is one demonstration.

I think this discussion got a bit sidetracked by the notion of "any point is cycloidal" - your original point about suspending a pendulum from a magnet is very valid and with the "shoe" would be a straightforward way to make a truly cycloidal pendulum. A bit of analysis needed to estimate the errors due to "compounding".

Even if the suspension gives cycloidal motion, the magnetic path through the curved shoe would be moving, and so magnetic hysteresis would cause damping. Whether this is more or less than the mechanical hysteresis in a spring suspension is beyond my pay grade.

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

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

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

dave8

I give up

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

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

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

dave8

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

Well done for posting that explanatory note, dave8

… it will be helpful to anyone thinking along similar lines in the future.

MichaelG.

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I should perhaps add that I had deliberately ignored the complexities around the use of magnets … basically because my interest was the geometry.