Re: BPS Formula?

Original poster: "Barton B. Anderson by way of Terry Fritz <twftesla-at-qwest-dot-net>" <tesla123-at-pacbell-dot-net>

Hi Nolan,

I would think less than 346 BPS. Here's how I go about it.

First off, although you have set the static gap to an arc voltage of 17000
volts, you have to look at the charging system. Eg., assume the charging system
is a typical pole pig (240, 14.4kv, 10kva) and the cap is 60nF. You get:
: 20362Vp = (1.414 x Vrms) = charging voltage.
: 12.44 j = (.5 x C x Vp^2) = cap energy.
: 20734 ohms  = V(rms) / I(rms) = Z impedance.
This cap will fully charge in Z * Cp * 5 = 6.22ms (in a perfect world).

Now lets set a static gap for an arc voltage of 17000v with 2 gaps at .375" at
.42121" spacing. Varc = (3000000*4*d/(d/r+1+sqrt((d/r+1)^2+8)) where d is gap
spacing in meters (0.010699) and r is electrode radius in meters (0.004763)
(referencing North).

You want to find how fast the 60nF cap will charge to the arc voltage of 17000
volts. To find the time in ms, we can rearrange the formula Varc =
Vcharge(1-e^(-t/zc) where e is Eulers constant (2.7182818.), zc is the time
constant, and t is time.

Rearranging for time, we get t = zc*-LN(1-Varc/Vcharge):
: zc = 60nF x 20734 ohms = 0.0012s = time constant.
: Varc = 17000 volts.
: Vcharge = 20362 volts.
: time(ms) = (.0012 x -LN(1-17000/20362)) x 1000 = 2.241ms.

This means it will take 2.241ms to reach the arc voltage, at which time the gap
conducts. This is the firing time. If we assume this was a constant value, then
BPS is 1/firing time x 1000 = 446.3 BPS. Now remember, this is a 10kva example.
Going through the same procedure for 3kva (acheived by reducing the transformer
current), you should reach a value of 134 BPS. Note, I didn't change the input
voltage to get 134, as that would have required readjusting the gap spacing. I
played this one out as if I was adjusting the ballast (current) on a pig.

For a rotary, the voltage at the time of conduction across the gap is governed
by the time between electrode alignment, the time constant, and the charging
current. If full charge is 6.22ms and we are running at 346 BPS, take
1/346x1000 to get the alignment time = 2.89ms. Obviously, the breakrate didn't
allow the 6.22ms to fully charge, so the voltage at conduction will be less.
Varc = Vcharge(1-e^(-t/zc)) = 18367 volts which is 90.2% of full charge. At
3kva, this will drop down to 10219 volts (about 50%) because there's less
current to charge the cap between electrodes.

Anyway, that's how I would go about it. Also note, if you play with numbers in
a static gap looking at bps, you may find that bps can go way up there
depending on inputs. I suspect in this application, breakrates way out in
wonderland take on a whole new concept and I wouldn't put a whole lot of faith
in the numbers.

Take care,
Bart

Tesla list wrote:
>
> Original poster: "tmoore by way of Terry Fritz <twftesla-at-qwest-dot-net>"
> <tmoore-at-erie-dot-net>
>
> Hello all,
>
>     I have been messing around with excel and trying to figure out new
> things. So I am wondering if I got this right for a static spark gap, Joules
> in tank circuit=1/2*(0.000000001*Cap)*Volts^2 where cap is in nano farads.
> and volts is what the spark gap is set for.  So say you had a tank cap of
> 60nf and the gap was set at 17000V you would get Joules=8.67 is this right.
> so to figure out the bangs per second it would be power input/joules. so say
> you have a power in put of 3kva would you get about 346BPS right????? So any
> ways if you can tell me if this is right or tell me how to figure it out
> then thanks! Also if you had a rotary spark gap with barley any space
> between the flying electrode and the stationary electrode that fired 350BPS
> with a tank cap of 60nf and a power input of 3Kva you would get 17kv in the
> tank circuit right? Thanks for any help!
>
> Sincerely,
> Nolan Moore