Pendulum capacitor failing...

On Thu, 28 Aug 2008 09:12:30 -0400, PeterD put finger to keyboard and composed:

The OP stated that the "pendulum swing gradually increases then will eventually stop and get stuck to the magnet". I took this to mean that the capacitor fails as the circuit approaches resonance. If we assume a circuit resistance of 10 ohms, then, at resonance, this would result in a current of 11A which would probably challenge the fuse. Moreover, it would result in a capacitor voltage of around 3000V. I suspect that the actual DC resistance of the coil is much less than 10 ohms.

- Franc Zabkar

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On Thu, 28 Aug 2008 13:06:55 -0400, greenpjs put finger to keyboard and composed:

I'm having difficulty visualising the arrangement, too.

As for why the circuit is detuned, it may be to avoid the huge voltages and currents at resonance. Even so, I calculate that a 9.7uF capacitor would be subjected to a voltage of 3200V and a current of

12A. Surely that can't be right.

- Franc Zabkar

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On Thu, 28 Aug 2008 09:13:18 -0500, "David" put finger to keyboard and composed:

Oops, I see my mistake.

Z = jWL + 1/jWC = jWL - j/WC = j (WL - 1/WC)

So the magnitude of the impedance of L & C, assuming R=0, is given by ...

|Z| = WL - 1/WC = XL - XC

There should be no squared terms. Sorry.

PI = 3.14159265# C = 9.7 * .000001 L = .75 F = 60 W = 2 * PI * F XL = W * L XC = 1 / W / C Z = XL - XC I = 110 / Z VC = I * XC VL = I * XL E = .5 * L * I * I PRINT C, L, XL, XC, Z, I, VC, VL, E

The results are now ...

.0000097 .75 282.7433 273.4621 9.28125

11.85185 3241.032 3351.032 52.67489

So the voltage is 3200V, current is 12A, and the energy in the coil is

53J (I think). The current for a C of 10uF is 6A as Ross calculated. Sorry again for my mistake.

- Franc Zabkar

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Agreed... I don't like what he used for a core, myself.

Hey all, right now I'm getting ready for the start of school, so am up to ears in work (I'm still at work right now). I plan on changing the cap before classes start next week, when I do, I'll stick my meter in there get some readings for you. (probably this weekend).

On my choice of 1/16" welding rods for the core...The core had been a huge headache (long story)... I had ordered soft iron rods from a supplier who had them back ordered from Japan for several months (I was surprised you can't buy soft iron rods in the states - at least I could not find them). An engineer from a transformer company suggested using the welding rods (no flux). They work much better than the old steel core or anything else I tried to use.

Thanks to all of you, this has been very educational. I wish I had asked for your help 2 years ago when I had to rebuild this thing.

Thank you.

dersh.

As the OP said, the circuit is tuned "off resonance" when the pendulum is not close. Then when it approaches, its proximity alters the resonance in such a way that there is more attraction while it is approaching, and slightly less while it is receding. The net result is a transfer of momentum to the pendulum. Sort of like the way a "slingshot" orbit works.

Isaac

You might try using capacitors with a definite current rating. Perhaps several of them in parallel to distribute the current. And a sturdy varistor to clamp the peak voltage across the capacitor.

For example the Panasonic ECWF series of 1.0uF caps can take 5 amps. Put nine of them in parallel to handle the bulk of the current and you should be set up nicely, with a safety factor of five or so. Add a few 0.1uF caps of the same type to fine-tune the resonance.

As for clipping the voltage peaks, I'd put a few sturdy varistors across the coil and the capacitor.

Isaac, We're getting close. What I'm having trouble picturing is how the attraction can be slightly less while it is receding. I can understand the pendulum itself affecting the resonance of the circuit, but why wouldn't it do so symmetrically? In other words, I would think that whatever momentum is gained on the way in would be lost on the way out. Clearly I'm wrong, but would like to understand why. By the way, I have always wondered the same thing about slingshot orbits. So, if you can explain it, that will kill two birds with one stone.

Pat

On Thu, 28 Aug 2008 06:01:52 +1000, Franc Zabkar put finger to keyboard and composed:

FWIW, I found this pendulum article (not the same as the OP's):

Unfortunately it is pay-for-view.

===================================================================== Abstract. The pendulum described operates from a.c. mains. The bob consists of a heavy coil across which a capacity is connected, forming a circuit nearly resonant at mains frequency. This coil swings along a laminated iron bar between two magnetizing coils on the bar carrying alternating current; these are connected in series with opposite polarity. =====================================================================

Google's search summary states that ...

===================================================================== It was found that the capacitance had to be about. 5. %. above the value required for resonance (Le. the moving-. coil circuit had to be slightly inductive) ... =====================================================================

Therefore it seems to me that my suggestion probably won't work.

- Franc Zabkar

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That would probably spoil the high "Q" that makes the circuit work.

Isaac

On Tue, 26 Aug 2008 08:54:50 -0700 (PDT), "dersh.z" wrote: : :I work at a small college with a bottom electromagnet driven pendulum. :

During my Google searching I came across this interesting paper which details the Doubochinski Pendulum. It includes details for constructing an experimental system.

It relies on a permanent magnet attached to the pendulum and an ac coil to provide the magnetic field, but no capacitor. Perhaps it could be easily up-scaled to a size suitable to the OP.