Wire Length Discussion
Date: Sun, 11 Jun 2006 11:47:10 -0600 Subject: wire length - Original poster: "Gates"
How do you calculate the length of wire I use to make my secondary coil? What's the formula? Thanks gang
Date: Sun, 11 Jun 2006 11:55:44
-0600 - Subject: Re: wire length - Original poster: Vardan
Hi,
Today, the "length of the wire" has been found not to matter as far as streamer
length or anything. There are a few who still think it is based on 1/4
wavelength or something, but I totally disagree with them as do most.
You want a coil say 4 inches in diameter and maybe two plus feet long. About
900-1200 turns works fine.
The length of the wire is:
3.14159 x secondary diameter x number of turns
So a 4 inch secondary with 1000 turns uses 1050 feet of wire. Much wire data is
here in the back:
http://hot-streamer.com/temp/FormulasForTeslaCoils.pdf
Also see John's notes here:
http://hometown.aol.com/futuret/page5.html
Cheers, Terry
Date: Mon, 12 Jun 2006 00:17:28 -0600 - Subject: Re: wire length - Original poster: "Dr. Resonance"
The wire length is not that critical as long as it falls within a certain range
for the number of turns. The larger the dia. of the secondary coil the more
potential (voltage) your coil will achieve although plasma based on input power
is really a better indicator of spark length (Freau equation). Most
experimenters plan on using 1000 to 1400 turns on either a 4 inch or 6 inch dia.
(ID) white PVC tube for excellent performance with a copper tubing subdivided
sparkgap.
Contact me off list and I can email you the plans for a nice performing coil
using a 4 inch ID PVC tube. It runs on a 12 kV 60 mA neon sign transformer and
produces sparks over 4 feet long. A nice lightweight portable coil for home or
school use.
Dr. Resonance
Date: Mon, 12 Jun 2006 00:17:50
-0600 - Subject: Re: wire length - Original
poster: "Dr. Resonance"
It should be noted that Terry's sample calculation is for a 4 inch OD (outside
dia.) tube. If you are using standard PVC schedule 40 water pipe (like Home
Depot stuff) then the calculation would be for 4.5 inch OD tubing with 1,000
turns and becomes:
sec wire length = (3.14159 x 4.5 x 1000 turns)/12 in/ft. = 1,178 feet of
wire.
If you are using 24 AWG double-build 200 degree C. magnet wire then the 1,000
turns would work out to a lineal winding length of 22.5 inches. Allow 3/4 inch
on top and 3/4 inch on bottom of your tube so tube length (total) before winding
is 24 inches.
A .0188 uF 32 kV MMC cap design (16 pcs of 0.15 uF 2 kV MMC caps in series)
with a copper tube spark gap (200 CFM fan on each end --- one pushing and one
pulling air through a 6 inch dia. pvc tube) will produce 58 inch long sparks
with a 12/60 NST. Very compact and good performance for a small portable coil.
Be sure to use a strike rail around the primary coil.
Dr. Resonance
Date: Mon, 12 Jun 2006 09:02:17 -0600
- Subject: Re: wire length - Original
poster: "Scott Hanson"
DC -
Is something wrong with your cap calcs here?
Eight of the C-D 15uF 2KV caps in series will provide the specified .0188uF
value, but at a voltage rating of only 16KV.
Sixteen of the same caps in series will provide only .0094uF at a more
conservative 32KV.
What is your recommended value: .018uF, or .009uF?
Did you mean TWO parallel strings of 16 caps each?
Regards, Scott Hanson
Date: Mon, 12 Jun 2006 11:19:11
-0600 - Subject: Re: wire length - Original
poster: "Barton B. Anderson"
Hi Scott,
I use 2 parallel strings in my .0188 uF MMC. It's only 32 caps, so it's low cost
and still robust. I think D.C. meant to call out 2 strings.
Take care, Bart
Date: Mon, 12 Jun 2006 14:14:41 -0600 - Subject: Re: wire length - Original poster: "Dr. Resonance"
Thanks for bringing this matter to my attention. Working too late at night again!. It is 16 in series for .0094 uF 32 kV total capacitance per string --- then, two strings connected in parallel for the .0188 uF value total.
Date: Mon, 12 Jun 2006 14:14:50
-0600 - Subject: Re: wire length -
Original poster: "Dr. Resonance"
Two strings in parallel is correct. Working too late at night again!
Date: Mon, 12 Jun 2006 14:15:04
-0600 - Subject: Re: Wire Length -
Original poster: Jared E Dwarshuis
The classic equation for an air cored inductor, derived with Maxwell's equations
is:
L = u Nsqrd Area / length
However the numerator and denominator can be multiplied by 4pi, yielding:
L = u (2pi R N) sqrd / 4pi l
Since: 2pi R N is wire length , we can write:
L = u (wire length)sqrd / 4pi l*
I put a star next to the length because solenoids in the real world do not have
a perfectly uniform magnetic field. We then need to make our solenoid length
just a little bit longer to get the correct inductance.
Now we can talk about standing wave resonance in a Tesla coil. We will use a
simple version of capacitance in the lc equation.
We can use a sphere for our top end capacitor. The capacitance of a sphere is: c = 4pi e R*
I put a star next to the radius because a Tesla coil inductor has self
capacitance that must be accounted for. We find that R, in real life is going
to be a bit smaller due to the self capacitance of the inductor.
Now we examine Tesla's equation:
C/4 Wire length = 1/ 2pi sqrt (lc)
Substituting from above for L and c, we get:
C/4 Wire = C/ Wire' x 1/2pi sqrt (l*/R*)
Set: 2pi = sqrt (l*/R*)
Then:
C/ 4 Wire = C/Wire'
Inverting frequency gives us the period:
4 Wire /C = Wire'/C
A casual inspection shows that this equation can only be satisfied if Wire' = 4
Wire.
Now we apply the Lorentz equation, as we are observing time dilation and
distance contraction.With a Gamma of 4, we predict that the velocity of waves down the length of wire
in our inductor will measure the sqrt(15/16)C
Thus, the actual wave velocity is reference frame dependant.
Tesla must not have understood all of this (in the late 1880's) or he would have
found the general form of the equation which describes node formation.
n/2 C/w = 1/2pi sqrt ( u x (w/2n)sqrd x 2n/ 4pi l* x 4pi e R*)
Where n = 1/2, 2/2, 3/3...........
The nodality equation also yields a gamma of 4
Sincerely: Jared Dwarshuis (and by proxy), Lawrence Morris