Pendulum capacitor failing...

[Franc Zabkar]

On Thu, 28 Aug 2008 13:06:55 -0400, greenpjs <greenpjs@neo.rr.com> put
finger to keyboard and composed:

I am afraid I can't help the original poster, but I find this thread
very interesting. There is a pendulum at a local museum and I always
wondered how it worked. I assume that an electromagnet is used to
replace the small amount of energy lost during each swing of the
pendulum, but how is it actually connected and where is the
electromagnet? What does having an resonant circuit tuned to
slightly below the power line frequency have to do it? I guess I'm
asking for someone to explain the theory. I don't suppose there's a
"how stuff works" article on the subject, but I'll go check that out
now.

Thanks,
Pat

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
--
Please remove one 'i' from my address when replying by email.

 

[Franc Zabkar]

On Thu, 28 Aug 2008 09:13:18 -0500, "David" <someone@some-where.com>
put finger to keyboard and composed:

When I input the OP's numbers into the following QBASIC
formulae, I
get only 21.3mA for the current. I'm wondering whether
I've
misunderstood the OP's circuit.

XL = W * L
XC = 1 / W / C
Z = XL * XL - XC * XC

Frank,
You made a major mistake calculating things with your
program. XL is i*W*L (or j*W*L if you are an EE) which is an
imaginary number where i or j = sqrt(-1). Also, XC is
1/(i*W*C) that is again an imaginary number. You need to do
the math using complex arithmetic. If the circuit is in
exact resonance the current will approach infinity and the
voltages across each element will as well. There is
obviously series and effective shunt resistance in the
circuit elements to keep things within some reasonable
bounds at resonance.

David

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
--
Please remove one 'i' from my address when replying by email.

 

[dersh.z]

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.

 

[greenpjs]

On Thu, 28 Aug 2008 20:43:45 -0700, isw <isw@witzend.com> wrote:

<snip>

Hi all,
I am afraid I can't help the original poster, but I find this thread
very interesting. There is a pendulum at a local museum and I always
wondered how it worked. I assume that an electromagnet is used to
replace the small amount of energy lost during each swing of the
pendulum, but how is it actually connected and where is the
electromagnet? What does having an resonant circuit tuned to
slightly below the power line frequency have to do it? I guess I'm
asking for someone to explain the theory. I don't suppose there's a
"how stuff works" article on the subject, but I'll go check that out
now.

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

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

 

[Ancient_Hacker]

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.

 

[Franc Zabkar]

On Thu, 28 Aug 2008 06:01:52 +1000, Franc Zabkar
<fzabkar@iinternode.on.net> put finger to keyboard and composed:

It seems to me that a perfectly tuned LC circuit would require a
capacitance of 9.38uF at 60Hz.

2.pi.f = 1/sqrt(LC)

So as the capacitance falls from your initial value of 9.7uF, the
circuit becomes better tuned and the current increases. AISI, this
increase in current would accelerate the capacitor's failure. Would it
not be possible to detune the circuit to an initial frequency of 61Hz
and extend the life of the pendulum that way? Would the performance at
61Hz be as good as at 59Hz?

FWIW, I found this pendulum article (not the same as the OP's):
http://www.iop.org/EJ/article/0950-7671/27/3/301/siv27i3p57.pdf

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
--
Please remove one 'i' from my address when replying by email.

 

[Ross Herbert]

On Tue, 26 Aug 2008 08:54:50 -0700 (PDT), "dersh.z" <dersh.z@gmail.com> 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.

http://www.21stcenturysciencetech.com/2006_articles/Amplitude.W05.pdf

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.

 

[greenpjs]

On Fri, 29 Aug 2008 21:23:09 -0700, isw <isw@witzend.com> wrote:

In article <jm0gb4l8scr9pdrt9uis34o6rjf35scqea@4ax.com>,
greenpjs <greenpjs@neo.rr.com> wrote:

On Thu, 28 Aug 2008 20:43:45 -0700, isw <isw@witzend.com> wrote:

snip

Hi all,
I am afraid I can't help the original poster, but I find this thread
very interesting. There is a pendulum at a local museum and I always
wondered how it worked. I assume that an electromagnet is used to
replace the small amount of energy lost during each swing of the
pendulum, but how is it actually connected and where is the
electromagnet? What does having an resonant circuit tuned to
slightly below the power line frequency have to do it? I guess I'm
asking for someone to explain the theory. I don't suppose there's a
"how stuff works" article on the subject, but I'll go check that out
now.

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

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.

The two are similar in that both effect a transfer of momentum, but not
quite in the same way.

For the pendulum, the L-C circuit should have a high "Q", which means
that it will be slow to respond to changes. As the pendulum approaches,
the proximity of that large piece of iron alters the tuning, and
therefore the current through the coil (and amp-turns is what determines
the force that accelerates the pendulum). If you get it right, the
delayed response (due to the high Q; think of it as a phase shift) will
cause the peak of attraction to come just before the pendulum crosses
dead center (if not for the slow response due to the narrow bandwidth,
the peak of attraction would be at dead center and would fall off
equally whether the pendulum were approaching or receding).
Interestingly, the process tends to be self-limiting, too; if the
pendulum is moving faster (because of a larger swing), the attraction
will not be as great because there's not as much time for it to build
up, and vice-versa.

I'm an electroniker, not an orbital scientist; I was just using the
"slingshot" as another case of momentum transfer. I don't feel
comfortable addressing it at the level of explaining how it works in
detail. Very roughly, though, the spacecraft has to approach a planet
going the opposite way to the planet's motion around the sun, and leave
the planet going the same way. Properly done, the maneuver will add
twice the planet's velocity to the spacecraft's while very slightly
slowing down the planet.

Isaac

Isaac,
Thank you! Regarding the slingshot, as soon as you said "planet's
motion around the sun", it all made sense to me. I had been thinking
about a single isolated (stationary) planet rather than the entire
solar system.

Pat

 

[Jerry G.]

Most likely the caps are failing because they are being pulsed, or the
demand on them is high.

Try using film type capacitors, or high voltage rated ceramic
capacitors. You may have to put a number of them in parallel to get
the value you want.

Jerry G.


On Aug 26, 11:54 am, "dersh.z" <ders...@gmail.com> wrote:

Hi Everyone, first time post... Mr. Goldwasser suggested I run this
question by this group...

I work at a small college with a bottom electromagnet driven pendulum.

The electromagnet consists of a large coil (.75H) with an iron core
and a 10uf capacitor in series with the hot leg (110vac). The pendulum
is an 8lb. iron shotput.  Works great 24/7 except...

The caps, I've been using (GE 10uf @ 370VAC motor cap), last about 4
months before the circuit becomes "un-tuned" (measured capacitance
decreases - pendulum swing gradually increases then will eventually
stop and get stuck to the magnet).

I get the best swing using these params... l=.75H, C= 9.7uf, f= 59Hz.
(most of the 10uf caps have measured 9.6 - 9.8uf)

My question... can you recommend a higher reliability capacitor, one
that might last a few years instead of a few months?

Thank you for your time.

 

[PeterD]

On Fri, 29 Aug 2008 06:13:08 +1000, Franc Zabkar
<fzabkar@iinternode.on.net> wrote:

On Thu, 28 Aug 2008 09:12:30 -0400, PeterD <peter2@hipson.net> put
finger to keyboard and composed:

On Thu, 28 Aug 2008 16:12:32 +1000, Franc Zabkar
fzabkar@iinternode.on.net> wrote:

It seems to me that choosing a better spec cap is still only delaying
the inevitable. IMHO a better approach would be to choose a 9.1uF cap,
polypropylene film or otherwise. This assumes that the capacitance
required for resonance at 60Hz is 9.4uF. As it is now, the circuit is
slightly detuned above its resonant frequency. AFAICS this means that
when the 9.7uF cap inevitably degrades, the circuit drifts *toward*
resonance, in which case the capacitor's current and voltage both
*increase*, resulting in further degradation and even more drift
toward resonance. If, however, the circuit were to be detuned on the
other side of resonance, then any degradation would result in a drift
*away* from resonance, with a *reduction* in current and voltage, and
this would in turn would slow the rate of degradation of the cap.

Using that logic, then while the circuit drifts towards resonance, it
tends to degrade faster towards resonance, until it crosses over the
resonance point, then degradation slows. (or it fails at resonance!)

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

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