Isochronous knife edge suspension?

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

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

dave8

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

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

This discussion seems to have taken an unpleasant change in style

… I now regret that I unintentionally revived it at the end of page 4

MichaelG.

Michael, guilty. Me, not you. Its because of perceived rissole-like responses from one member. I would be happy to draw a line here.

dave8

Michael

My apologies if I have offended anyone - that was not my intent.

Regards,

Tony

o.k.

dave8

Michael

Duncan -

Thanks - that is very useful indeed - the evolute of a cycloid is a replica of that cycloid shifted by 180 degrees, which is what is needed for the cheeks to work. It also indicates that the generating circle for the cheeks must have the same radius as the desired cycloid's generating circle - as per previus discussions this must be half of the effective length of the pendulum. Pretty clear then that this was Huygens' intent.

Regards,

Tony

I think I now see where the confusion comes from. Much of this discussion has become about Huygens cycloidal cheeks and how they are constructed. The 1/2 pendulum length roller is needed for that.

The method I proposed is based on the rolling wheel principal to produce a cycloidal path. That is a completely different approach to Huygens. The question now is - does the diameter of the roller matter in this context?

dave8

There is only one cycloid which will give isochronous motion. This has been stated time and time again. Michael gave up on 05/09, now I give up.

I believe I answered that with the diagram I posted on 07/09/2023 13:41:13 and subsequent posts. The diameter of the roller has to be equal to the effective length of the pendulum, otherwise the bob *does not* follow a cycloid.

Thanks everybody for your help in this. I've finally got my head around the change of curve from extending the rod, and even the apparent paradox of a point in a circle being a cycloid one minute and not the next. Sorry for testing your patience so hard.

dave8

If this project wasn't already dead, here's another nail in the coffin. In preparing a cirular section to match the required circle ( 125mm radius now), placement of the magnet came up. The only place for the magnetic forces to even out would be in the middle of the bob! Not very practical.

dave8

I'm not sure I see that Dave. Why wouldn't the arrangement I sketched work?

John, I think you mean the sketch in the "general" forum? (If this works, it leads to the intrigueing idea that local magnetism can replace gravity in a clock)

Just from playing about with magnets and rollers they accelerate to seek the closest contact at the strongest pole point. This ruins any hope of a high Q and swamps the effect of gravity. I was trying to minimise this by putting the magnet in the centre of the rollers.It just felt wrong.

Indeed, that's the one. Maybe one can shape the corners of the polepiece to prevent this?

If I may, just to close this topic off, I did mention earlier that it is known that the cycloid does not make a compound pendulum isochronous (and in fact there is no curve that can do that). My friend Andrew Millington (who is a real mathematician) has looked at this and come up with an approximation for the circular deviation of a compound pendulum moving in a cycloid, and also checked this against Woodward's analysis. This would apply to David's magnetically suspended pendulum as well as a pendulum with cycloidal cheeks. There are two factors to consider. One is the distance from the suspension point to the CoG of the whole pendulum, which is "r". The other is the "radius of gyration" which is a measure of the pendulum's "compoundness", which is "k".If the amplitude of swing is "a" radians, then the fractional change in clock rate is:

k^2.a^2 / [16(r^2 + k^2)]

If the pendulum is "simple" then k = 0, so the rate change is zero.

Suppose we want a 1metre pendulum. The bob weight can't be too large because it has to be supported by the magnet. So let's suppose it was 1kg. The "shoe" which has the circular face and contains the magnet needs steel polepieces and the magnet has to be strong and probably quite heavy. For the sake of argument suppose the whole shoe weighs 100g. Assume the rod is say carbon fibre and effectively weightless. Given the length of the rod and the weights you can calculate the position of the CoG and the value of k. Putting them in the formula and calculating gives a fractional change in rate of:

(+a^2/16)/13

For a normal circular pendulum the standard expression for CD is:

-a^2/16.

So while the CD is significantly less than for a conventional pendulum it isn't zero and can't be made zero. I think your suggestion and the experiments you did were very ingenious, but the fact that one does lose the "ideal" anisochnonicity (is that a word?) combined with the practical difficulties you've highlighted make this approach of theoretical interest only.

Andrew pointed out a small error, my formula

(+a^2/16)/13 = a^2/208

should have been...

(+a^2/16)/11 = a^2/176

Clearly though this is for specific physical parameters and could be larger or smaller depending.

Edited By John Haine on 13/09/2023 10:53:43

John, could that error be "tuned out" with a rating nut?

dave8