Re: 1.5 wave coil

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Original poster: Jared E Dwarshuis <jdwarshui@xxxxxxxxx>

I was asked by several people at Dr. R's  Teslathon to describe how
our 1.5 wave coil worked.

Each coil measures 6.625 inch by 120 inch. Each coil is wound with
7,400 turns 26 gauge wire. That's 3,912 meters of wire for each coil.
Since it is a 1.5 wave coil we multiply 3912 x .6666666 giving us the
full wave length of 2608 meters.

Now the speed of light divided by the wavelength (2608 meters)gives us
115,000 Hz. This is the frequency that you drive the primary at.

A 1.5 wave coil represents six quarter wave sections, so you calculate
the inductance using 1/6 th  of the wire length or 3,912 / 6 = 652
meters

Since:    L = u  (wire length)sqrd / ((4 pi) x solenoid length)
L then equals .083 Henry. Since the distance between the current node
and voltage node is .51meter we can calculate the total capacitance as
45.07 Pf per meter x .51 meter and get 22.9 pf

Now you calculate the Medhurst for a solenoid 6.625 inch by 20 inch
and get 9.8 Pf. This gets subtracted from the total capacitance of
22.9 Pf and we get 13.1 Pf

Since the structure shares current at the current nodes (unlike a
quarter wave) we are only interested in half of this value or 6.55 Pf
for each capacitor.

We ran two identical coils side by side so we made 8 capacitor plates
this size

You can run individual coils and get sparks between the capacitors,
which we did for a much shorter coil earlier this summer. The ten foot
coils were far to long to get breakout between plates of a single coil
using a tiny NST, so we made two part resonant coils instead.

These are all the calculations that you need to produce a 1.5 wave
coil.  You have 6 choices for current nodes, take your pick.

The unsettling matter of  how energy is transferred between plates
simply indicates that the conventional notion of how capacitors work
is only partially correct.

Jared Dwarshuis and Larry Morris