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Wire Length

 

Date:  Sun, 11 Jun 2006 11:47:10 -0600

From:  Tesla list <tesla@pupman.com>

To:  tesla@pupman.com

Subject:  wire length

 

Original poster: "Gates" <ryker@isys.ca>

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

From:  Tesla list <tesla@pupman.com>

To:  tesla@pupman.com

Subject:  Re: wire length

Original poster: Vardan <vardan01@twfpowerelectronics.com>

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

From:  Tesla list <tesla@pupman.com>

To:  tesla@pupman.com

Subject:  Re: wire length

Original poster: "Dr. Resonance" <resonance@jvlnet.com>

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

From:  Tesla list <tesla@pupman.com>

To:  tesla@pupman.com

Subject:  Re: wire length

 Original poster: "Dr. Resonance" <resonance@jvlnet.com>

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

From:  Tesla list <tesla@pupman.com>

To:  tesla@pupman.com

Subject:  Re: wire length

 Original poster: "Scott Hanson" <huil888@surfside.net>

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

From:  Tesla list <tesla@pupman.com>

To:  tesla@pupman.com

Subject:  Re: wire length

 Original poster: "Barton B. Anderson" <bartb@classictesla.com>

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

From:  Tesla list <tesla@pupman.com>

To:  tesla@pupman.com

Subject:  Re: wire length

 Original poster: "Dr. Resonance" <resonance@jvlnet.com>

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.

Dr. Resonance
----- Original Message ----- From: "Tesla list" <tesla@pupman.com>
To: <tesla@pupman.com>
Sent: Monday, June 12, 2006 10:02 AM
Subject: Re: wire lengnth


>Original poster: "Scott Hanson" <huil888@surfside.net>
>
>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
>----- Original Message ----- From: "Tesla list" <tesla@pupman.com>
>To: <tesla@pupman.com>
>Sent: Sunday, June 11, 2006 11:17 PM
>Subject: Re: wire lengnth
>
>
>>Original poster: "Dr. Resonance" <resonance@jvlnet.com>
>>
>>
>>
>>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 for .009 uF -- two series circuits in parallel) 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.

 

 

 Date:  Mon, 12 Jun 2006 14:14:50 -0600

From:  Tesla list <tesla@pupman.com>

To:  tesla@pupman.com

Subject:  Re: wire length

 Original poster: "Dr. Resonance" <resonance@jvlnet.com>

Two strings in parallel is correct.  Working too late at night again!

Dr. Resonance



>Original poster: "Barton B. Anderson" <bartb@classictesla.com>
>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
>Tesla list wrote:
>
>>----- Original Message ----- From: "Tesla list" <tesla@pupman.com>
>>To: <tesla@pupman.com>
>>Sent: Sunday, June 11, 2006 11:17 PM
>>Subject: Re: wire lengnth
>>
>>
>>>Original poster: "Dr. Resonance" <resonance@jvlnet.com>
>>>
>>>
>>>
>>>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
>>>----- Original Message ----- From: "Tesla list" <tesla@pupman.com>
>>>To: <tesla@pupman.com>
>>>Sent: Sunday, June 11, 2006 12:55 PM
>>>Subject: Re: wire lengnth
>>>
>>>
>>>>Original poster: Vardan <vardan01@twfpowerelectronics.com>
>>>>
>>>>Hi,
>>>>
>>>>Today, the "length of the wire" has been found not to matter as
>>>>far as st....
>

 

 

Date:  Mon, 12 Jun 2006 14:15:04 -0600

From:  Tesla list <tesla@pupman.com>

To:  tesla@pupman.com

Subject:  Re: Wire Length

Original poster: Jared E Dwarshuis <jdwarshui@emich.edu>

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