Martin Kyte | 03/02/2017 15:27:36 |
![]() 3445 forum posts 62 photos | OK chaps how about this for something to muse on. Consider a pendulum constrained to swing with the rotation of the Earth (as normal clock arrangement). How does the period vary from that of a Foucault pendulum of the same length (which is not so constrained). Also what is the effect or not of latitude. I would be interested on anyones take on energy/momentum considerations. Intuitively it must slow down as a result of earth rotation and I would say by a factor of one complete period per year. Before anyone tells me I know this is totally irrelevant regarding clock making as it has to be a constant effect (subject to the rotation staying practically constant). It's just something I happened to be musing on. Have a good weekend Martin |
Martin Kyte | 03/02/2017 16:14:35 |
![]() 3445 forum posts 62 photos | I obviously meant one period per day. Martin |
John Haine | 03/02/2017 16:17:43 |
5563 forum posts 322 photos | When you say "with the rotation of the Earth", do you mean, say, east-west at the equator? |
David Jupp | 03/02/2017 17:05:46 |
978 forum posts 26 photos | What exactly do you mean by 'slow down' ? The period of a pendulum is essentially constant (as long as the angle of swing is modest), as the energy in the system decays, the height of swing and speed at mid swing both reduce but period remains constant. I can imagine additional energy losses at the pivot for the fixed (not Foucault) pendulum - but the period should not change. That's if I'm remembering the derivation of the equation for period of a pendulum correctly from school Physics lessons many years ago. |
SillyOldDuffer | 03/02/2017 17:34:20 |
10668 forum posts 2415 photos | This sort of thing hurts my head. The Earth is spinning on it's axis and orbiting the sun. The sun is in a Galaxy, and the whole Universe is expanding. Everything is moving. The forces acting on a unconstrained pendulum (ie one only partly effected by the earth's movement) must be different from those acting on a constrained pendulum where the whole assembly is physically linked to the earth and there is only one degree of freedom. For many years I was happy with the pendulum formula taught at school until I discovered by accident that it's only an approximation. A very good one for small displacements, but nonetheless slightly inaccurate. The same book said that plumb lines are only approximately vertical, another illusion shattered. Latitude might make a difference too. Gravity weakens as you move towards the poles because the earth is an oblate spheroid. I've no idea how big a difference any of this would make to the periods of pendulums in practice. Any clever people out there? Dave |
john carruthers | 03/02/2017 17:56:42 |
![]() 617 forum posts 180 photos | I've often wondered about buying a precious material (say, gold) at the equator, and selling it at the poles? |
duncan webster | 03/02/2017 18:01:37 |
5307 forum posts 83 photos | Posted by john carruthers on 03/02/2017 17:56:42:
I've often wondered about buying a precious material (say, gold) at the equator, and selling it at the poles? If the bullion dealer has any sense he will measure the MASS using standard weights, and so different gravity won't matter |
pgk pgk | 03/02/2017 18:21:20 |
2661 forum posts 294 photos | It makes my head hurt to think about such things but presumably there's also a need to account for altitude and any physical effects acting on the pendulum material since persumably even non-magnetic materials can have eddy currents induced in them...and I suppose the shape of the pendulum might effect how ions flow. |
Gary Wooding | 04/02/2017 07:58:52 |
1074 forum posts 290 photos | Does the swing of a Foucault pendulum at the Equator appear to rotate? |
martin perman | 04/02/2017 10:34:31 |
![]() 2095 forum posts 75 photos | Martin, Are you describing the very large pendulum that hangs in the Science Museum in London, that swings with relation to the earth rotation. Martin P |
Geoff Theasby | 04/02/2017 12:28:55 |
615 forum posts 21 photos | Yes. But rather the earth rotates underneath it... Anyway, surely, a clock pendulum is constrained in that it can only swing in one plane, unlike the Foucault. Any tendency to twist from that plane is resisted by its construction and suspension. Yes? Geoff |
John Haine | 04/02/2017 12:33:21 |
5563 forum posts 322 photos | Posted by Martin Kyte on 03/02/2017 15:27:36:
Intuitively it must slow down as a result of earth rotation and I would say by a factor of one complete period per Martin Martin, please say why it is intuitive? A Foucault pendulum swinging in the equatorial plane at the equator would be completely "unaware" that it isn't constrained in the north-south direction. And why would it be intuitive that any slowing effect is just one period irrespective of the pendulum period? |
martin perman | 04/02/2017 12:56:13 |
![]() 2095 forum posts 75 photos | Gentlemen, There is a good explanation of how a Foucault Pendulum works here https://en.wikipedia.org/wiki/Foucault_pendulum Martin P |
Martin Kyte | 06/02/2017 09:24:17 |
![]() 3445 forum posts 62 photos | Hi Chaps. What I was really interested in was a way of thinking about the problem. OK. so I've thought about it some more and my conclusion is that there is no variation between the two systems as far as pendulum period is concerned. Here is my thought process.Assume that the period of the pendulum is much smaller than the period of earth rotation. Put the pendulum at the pole for simplicity. Rotation axis vertical through the mount period one day. Take a plane of reference the initial plane of the pendulum swing. An unconstrained pendulum swings for P periods per day and remains in the plane of reference. The constrained pendulum also swings for P periods but effectively completes one extra period in the day due to the rotation forced on it by the mount. (That's why I initially thought the pendulum must be slowed as it has to travel further) The moves in a curve close to an ellipse but precessing through 360 degrees in 24hours, think spirograph patterns. Each period the bob has travelled further than one complete swing of an unconstrained pendulum but has been accelerated a small amount by the torque in the suspension mount (usually a flat spring strip for non clock makers) so the period has remained the same. The suspension torque in this thought model compensates for the additional distance travelled by the bob. So there is no difference intrinsically in the period between the two systems (constrained and unconstrained). For the clock, however the escapement rotates with the mount so it must count one less swing per day. I am happy to listen to better explanations to my musings and as I said I'm mainly interested in a way of thinking about it. regards Martin
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Geoff Theasby | 06/02/2017 09:58:02 |
615 forum posts 21 photos | I think that the lack of accuracy in a clockwork assembly would be much greater than the difference of one extra swing of the pendulum. If it swings once a second, that is 60 x 60 x 24 times per day, or 86,400, but he clock mechanism is accurate only to within several seconds a day, even in the best clocks. One extra swing of the pendulum is 0.001% of the day. Geoff |
Martin Kyte | 06/02/2017 10:03:47 |
![]() 3445 forum posts 62 photos | I did say at the start that this has no bearing on practical clockmaking Geoff. It's just an exercise in thinking about the dynamics of pendulums. I still haven't sorted the mechanism by which the precession of Foucalt systems change with latitude so I've got a long way to go yet. :0) Martin |
Russell Eberhardt | 06/02/2017 10:49:46 |
![]() 2785 forum posts 87 photos | Posted by Geoff Theasby on 06/02/2017 09:58:02:
I think that the lack of accuracy in a clockwork assembly would be much greater than the difference of one extra swing of the pendulum. If it swings once a second, that is 60 x 60 x 24 times per day, or 86,400, but he clock mechanism is accurate only to within several seconds a day, even in the best clocks. One extra swing of the pendulum is 0.001% of the day. Really? A relatively recent reconstruction of Harrison's 250 year old design of pendulum clock lost less than one second in a 100 day trial at the Royal Observatory. Conventional English Regulator clocks can easily achieve 1/2 s a week. Russell. |
Martin Kyte | 06/02/2017 10:57:47 |
![]() 3445 forum posts 62 photos | It's irrelevant to clockmaking because any change in period would be constant. Martin |
roy entwistle | 06/02/2017 11:04:19 |
1716 forum posts | Removed because covered in original post Edited By roy entwistle on 06/02/2017 11:08:14 |
Russell Eberhardt | 06/02/2017 11:05:42 |
![]() 2785 forum posts 87 photos | Posted by Martin Kyte on 06/02/2017 09:24:17:
For the clock, however the escapement rotates with the mount so it must count one less swing per day. Yourbargument seems right to me. However that effect is much less than that due to the change of gravity with altitude and with latitude. The latter being a result of the rotation of the earth flattening the earth to an oblate spheroid. Russell. P.S. Your effect will vary with the sine of the latitude. Edited By Russell Eberhardt on 06/02/2017 11:55:16 |
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