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Re: Tesla coil formula

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

Shawn wrote
"it is the inductance of the coil itself that determines this".
Well said! I usually stay away from these posts, but I'm bored (just uploaded 3 pcb's for 4pcb.com to quote). Nothing else to do.
The length of the wire has no ability to account for the interturn capacitance or interturn inductance which is formed by currents within the wire and currents affected by other currents in adjacent turns. This is "rediculously" obvious. If one must consider these "real" situations, then wire length theory hits a dead end.
The bottom line is wire length thoery is limited, and we are beyond that limit (but not everyone has traveled down that path). There was a time when coils were built with wire length theory, they worked fine, and the theory was accepted.
This list has battled these issues in the past (ref the archives). The fact is, as we strive to understand the physics encompassed with our coils, we realized wire length theory could not provide the answers. If you want to "ball park" these physics, fine. But, if you really want to know, follow the currents. 

Take care,
Bart

Tesla list wrote:


Original poster: Shaun Epp <scepp@xxxxxxx>

Peter,
This short path that you talk about known as self inductance, is exactly why the wire length is not the determining factor for resonance wavelength of a coil, it is the inductance of the coil itself that determines this. The inductance includes the magnetic flux that crosses adjacent conductors of the coil. 

Shaun


----- Original Message ----- From: "Tesla list" <tesla@xxxxxxxxxx>
To: <tesla@xxxxxxxxxx>
Sent: Wednesday, December 07, 2005 2:21 PM
Subject: Re: Tesla coil formula



Original poster: "Peter Terren" <pterren@xxxxxxxxxxxx>

Jared
Your reasoning seems to be at odds with the rest of the world and scientific observation despite repeated posts from other members. 
Electricity is not water or string and analogies can only take you so far.
Can you appreciate the possibility that current flow in the first turn of a coil can affect the last turn of a coil. Just like two transformers with one turn each? This can happen "at the speed of light". Now consider the current that is taking "the long way there" by travelling through many turns of a coil also at "the speed of light" but will take a lot longer to get there. This is where your analogy founders because it only accounts for "the long way there". 

Sorry I don't know any maths.

Peter
http://tesladownunder.com/


----- Original Message ----- From: "Tesla list" <tesla@xxxxxxxxxx>
To: <tesla@xxxxxxxxxx>
Sent: Wednesday, December 07, 2005 12:42 PM
Subject: Re: Tesla coil formula


Original poster: Jared E Dwarshuis <jdwarshui@xxxxxxxxx>


Hi Steve

The equations we developed are all based on the mathematics of
standing waves.

(1) A standing wave always partitions the length of a uniform
medium in even intervals.

(2) A Standing wave is the superposition of a forward and
reflected waves.

(3) Velocity / wavelength = frequency,  is maintained with
standing waves.

Examination of the left side of our equation stipulates that:

 n/2 C/ Wire length = frequency.     n = 1/2, 2/2, 3/2, 4/2...

In order to fit standing waves into a length of  wire we must
partition it into quarter wave segments.

We could write this as an angular frequency (w)

(w) = 2pi n/2 C/wire     n= 1/2, 2/2,....

The right side of our equation is:

 (w) = 1/ square root of (LC)

We have modified our inductance to reflect the fact that for an
inductor the voltage A to B = - L di/dt

The mathematics of standing wave resonance guarantee that our voltage
from A to B is between quarter wave segments. ( if it were not so then
the inductance sums to zero for a half wave coil)

So for example, when we put a full wave length on an inductor, for the
sake of angular frequency we are only interested in the inductance
from one quarter of the windings.

We can generalize this as:

 L = u / 4pi  x ( wire / 2n)sqrd x (2n)/length   n = 1/2, 2/2, 3/2,
4/2...

Now we can say that:

2pi  n/2 C/w = 1/ sqrt ( (u/ 4pi) x ((wire length)sqrd / 2n length) x
Capacitance )

n = 1/2, 2/2, 3/2..

We must subtract the  self capacitance of the coil. So we use
Medhurst's  equation to sum the capacitance between the voltage
potentials of the quarter wave regions.

Fiddles also operate according to the conditions of standing wave
resonance, and must satisfy these conditions:

(1) A standing wave always partitions the length of a uniform
medium in even intervals.

(2) A Standing wave is the superposition of a forward and
reflected waves.

(3) Velocity / wavelength = frequency,  is maintained with
standing waves.



Respectfully: Jared Dwarshuis, Larry Morris