Re: Space winding question

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Terry Fritz" <twftesla-at-uswest-dot-net>

Hi,

At 08:45 PM 3/9/2001 -0600, you wrote: 
>I was wondering who could answer my question about space wound secondarys. 
>Lets say that you have a 12 inch secondary form, and you wind it with # 18 
>wire,"non space wound" with 1000 turns. The H/D ratio turns out to be 4.5 to 
>1. Then lets say that you decide to wind another 12 inch "space wound" with 
>1000 turns, and the H/D ratio comes out to 6 to 1. Would the two coils 
>resonate at the same frequency? In other words would the tank circuit see 
>the two coils as having the same characteristics ? I am really stumped on 
>this one.
>                                                                 Thanks 
>everyone.
> 

Your questions is straight forward and we can figure it out.

The coil's resonant frequency is determined basically by it's inductance
and effective capacitance.  "Sort of" like a simple LC circuit.  "I" don't
think wire length has anything at all to do with it and wire gauge does not
have much affect either.  An excellent site that tells of the calculations
plus a lot more is at:

http://home.earthlink-dot-net/~electronxlc/


So we have two 1000 turn coils 12 x 54 and 12 x 60.

The inductance of a coil is given by the following to with about 1% accuracy:

L = (N x R)^2 / (9 x R + 10 x H)


L = inductance of coil in microhenrys (µH)
N = number of turns
R = radius of coil in inches (Measure from the center of the coil to the
middle of the wire.)
H = height of coil in inches

Putting in our numbers:

L = (1000 x 6)^2 / (9 x 6 + 10 x 54) == 60600uH = 60.6mH

L = (1000 x 6)^2 / (9 x 6 + 10 x 60) == 55000uH = 55.0mH

So now we know the inductance of both coils...

For the capacitance, we can use the famous Medhurst equation also at the
site above that is about 1% accurate:

C = 0.29 x L +0.41 x R + 1.94 x SQRT(R^3/L)

C = self capacitance in picofarads
R = radius of secondary coil in inches
L = length of secondary coil in inches 

for the first coil I get C = 22.0pF
for the second coil I get C = 23.54pF

So now we know the effective capacitances for both coils.

Using the resonant circuit formula (also at the site above):

F = 1 / (2 x pi x SQRT(L x C)

F = frequency in hertz
L = inductance in henrys
C = capacitance in farads

The first coil gives:

F = 1 / (2 x 3.14159 x SQRT(0.0606 x 22 x 10^-12) = 137.84KhZ

and the second coil gives:

F = 1 / (2 x 3.14159 x SQRT(0.0550 x 23.54 x 10^-12) = 139.87KhZ

Thus, the two coils have very close to the same frequency around 139kHz.

See how easy that is :-))  Ok, there are a bunch of computer programs
around that will do all this easily but this is the "stuff" behind how
those programs work.

There are other programs that will find the effective capacitance with a
toroid in place too which is not a simple formula.

http://hot-streamer-dot-com/TeslaCoils/Programs/E-Tesla6.zip

There are super sophisticated programs and techniques like Paul Nicholson's
wonderful TSSP project uses that can tell all kinds of wonderful things
about resonating secondaries.  A nice paper about the work so far is below
in a zipped PDF file:

http://hot-streamer-dot-com/temp/pn2551-10c.pdf
Note: I had to save this file to my disk and then open it with acrobat.  It
is converted from postscript so it has gone through a lot of conversion ;-)

The originals are at:
http://www.abelian.demon.co.uk/tssp/

Pretty heavy reading but it does show where the art is at.


Cheers,

	Terry