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Re: Tesla Coil Formula
Original poster: Jared E Dwarshuis <jdwarshui@xxxxxxxxx>
Jared,
What is the point of putting 2pi n/2 C/ wire in there
Once again you are trying to state that the length of wire wound on
the coil is the wavelength of the coil. This has been proven by
qualified people on this group that it is not true, it was a old
fasion belief at one time but progress has been made!
Maybe it's time to accept the fact that you are WRONG!
The correct form for this is:
Fo = c c is the speed of light
----------- = 300 million meters per second
wavelength
Why do you put n/2 in your equation?.
Wavelength is determined by the inductance of the coil, dealing with
tesla coils that is, and capacitance is a combination of the medhurst
capacitance and the top load.
The stuff on the right hand side of the equals sign is just the L C
resonance formula that everyone has seen along with your wire length
garbage. It is not a new formula. You expanded on the inductance
part, why not expand on the capacitance too ?
The correct standard form for the resonance formula that we are all
familiar with is:
Fo = 1
--------------------
2 pi sqrt (L C)
There are no new formula's that you posted! Why would you write an
equation like this anyway? Most equation have one variable to be
determined on the left hand side of the equals sign and the rest on
the right hand side.
Shaun Epp
...................................................................
Hi Shaun:
(Part I)
Lets start with Amperes law and examine the magnetic field in a region
of an ideal solenoid. The closed integral B.ds = u I enclosed. Then B
= u I total/ length. We can also write this as: B = u i N /length
(where i N = I total)
Now we can go to the definition of inductance:
L = N flux / i
or:
L = N B A / i
Substituting in B we get:
L = u Nsqrd A / length
We can go a bit further and multiply the numerator and denominator by
4pi
L = u (2pi r N)sqrd/ 4pi length
Since (2pi r N) is the total length of wire in the inductor we get:
L = u (wire length)sqrd / 4pi length
Now we will use another inductor argument:
Voltage (a) to Voltage (b) = - L di/dt
Suppose we have a standing wave in our inductor. To qualify as a
standing wave it must partition the length of our inductor into
regions where voltage reaches a maximum and a minimum. So for example
if we were to fit a standing wave of one full wavelength to an
inductor, we would have four regions where voltage went from zero to a
maximum.
Now lets apply our added inductor argument. Oops the inductance is
zero going between points where the voltage is zero. Better try
something else, how about this:
Under standing wave resonance the inductance (for sake of angular
frequency) is defined between quarter wave regions.
Now we used the disclaimer (for sake of angular frequency), because
there are still four sections of the inductor storing energy.
So when we talk about standing wave resonance in an inductor. For the
sake of angular frequency, we will need to amend the wire length
inductance formula.
L = (u / 4 pi) x (wire length/2n)sqrd x (2n/l) ( where n =
1/2, 2/2, 3/2, 4/2, ??..)
So for the full wave (mentioned above) we would have two full nodes (
n = 4/2) and our inductance (for sake of angular frequency) comes from
one quarter of the length of the inductor.
(Part II)
F = C / wavelength = C/ wire length:
This holds true under standing wave resonance but only when we are
putting a full wave length on a coil. If we want to put half of a
wavelength on a solenoid coil, then we need to write:
F = C/ (2 x Wire length)
If we want to put a quarter wavelength on a coil then we write:
F = C/ (4 x Wire length)
Or we could simply write:
F = n/2 C/wire length (where n = 1/2, 2/2, 3/2???..)
By definition the angular frequency (w) = 2pi frequency
(w) also equals 1/ sqrt (LC)
So we can set our two expressions equal and we get:
2pi n/2 C/wire = 1/ sqrt [(u/ 4pi) x ( (Wire)sqrd / 2n l ) x
capacitance]
END.
With love and kisses: Jared Dwarshuis