Richard Harrison wrote:
Roy, W7EL wrote:
"What is the velocity factor, and how did you calculate it?"

Given:
length = 12 inches
diamwter = 6 in.
L = 38.6 microhenry

I used formula (37) from Terman`s Handbook to calculate 25 turns in the
coil. 471 inches of wire are needed in the coil.

The velocity of the EM wave traveling around the turns of the coil is
almost equal to the velocity in a straight wire. But, the time required
to travel 471 inches is 40 times the time required to travel 12 inches.
The velocity factor is the reciprocal of 40 or 0.025.

Not quite what I was expecting, but let's see if I understand what it
means. This means that if we put a current into one end of the inductor,
it'll take about 40 ns for current to reach the other end, right? So we
should expect a phase delay in the current of 180 degrees at 6.15 MHz,
or about 30 degrees at 1 MHz, from one end to the other?

Roy Lewallen, W7EL

"Roy Lewallen" wrote:
Richard Harrison wrote:
The velocity of the EM wave traveling around the turns of the coil is
almost equal to the velocity in a straight wire. But, the time required
to travel 471 inches is 40 times the time required to travel 12 inches.
The velocity factor is the reciprocal of 40 or 0.025.

Not quite what I was expecting, but let's see if I understand what it
means. This means that if we put a current into one end of the inductor,
it'll take about 40 ns for current to reach the other end, right? So we
should expect a phase delay in the current of 180 degrees at 6.15 MHz,
or about 30 degrees at 1 MHz, from one end to the other?

Dr. Corum's VF equation predicts a VF of approximately double
Richard's with corresponding delays of 1/2 of your calculated
values.
--
73, Cecil, W5DXP

Cecil, W5DXP wrote:
"Dr. Corum`s VF equation predicts a VF of approximately double
Richard`s----."

I wonder why? Dr. Terman wrote that the wave follows the turns in a
coil. My recollection of common solid-dielectric coax VF is about 2/3
that of free-space due to the fense plastic.

Twice the velocity factor in a coil requires a wave traveling faster
than light or taking a short-cut around the turns.

I often learn from my mistakes. Where did I err?

Best regards, Richard Harrison, KB5WZI

"Richard Harrison" wrote:
Twice the velocity factor in a coil requires a wave traveling faster
than light or taking a short-cut around the turns.

I often learn from my mistakes. Where did I err?

The current does take a short-cut due to adjacent coil coupling.
But please note the velocity factor only approximately doubles
from the "round and round the coil" calculation. Even though a
VF of 0.04 is ~double the "round and round the coil" approximation,
it is still 96% away from the VF=1.0 originally asserted by W8JI
which assumes that all the coils couple 100% to all the other coils.
--
73, Cecil, W5DXP

Cecil, W5DXP wrote:
"The current does take a short-cut due to adjacent coil coupling."

R.W.P. King wrote on page 81 of Transmission Lines, Antennas, and Wave
Guides:
"The electromagnetic field in the near zone is characterized by an
inverse-square law for amplitude and by quasi-instantaneous action."

I still don`t know what to make of King`s assertion regards
instantaneous action.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
Cecil, W5DXP wrote:
"The current does take a short-cut due to adjacent coil coupling."

R.W.P. King wrote on page 81 of Transmission Lines, Antennas, and Wave
Guides:
"The electromagnetic field in the near zone is characterized by an
inverse-square law for amplitude and by quasi-instantaneous action."

I still don`t know what to make of King`s assertion regards
instantaneous action.

From the IEEE Dictionary: "instantaneous - A qualifying term
indicating that no delay is purposely introduced in the action
of the device."

Does anyone have a formula for the coupling factor between
turns in a coil?
--
73, Cecil http://www.qsl.net/w5dxp

I'm going with Drs. Corum on this one. Solve equation 28 for tau, get
beta from equation 4. The phase velocity along the axis of the coil is
omega/beta.

The velocity factor in question is that phase velocity over the speed
of light in a vacuum.

The coil modes are surface waves in a weird coordinate system. Note
that the paper is very explicit in saying they're not TEM.

Throw equation 28 into Mathematica or Matlab or something and solve for
tau. The cases given after equation 28 with all the limitations
appear(ed?) to be a point of some contention, but equation 28 seems
*only* to have the limitation of circumferential symmetry of the
surface waves on the coil.


At the junctions between the wire and the coil, there is a transfer of
energy between the surface wave modes on the coil and the usual antenna
mode (I guess it's TEM?)

The coil is like G-line in that it guides surface waves, but the coil
modes are modes specific to the helical geometry; the G-line surface
waves are specific to the straight-wire geometry.

There is a mode on the helix where the waves go round and round the
turns, but the example given is a traveling wave tube for microwave
amplification, and it seems to me that there are a few turns over a few
inches for *microwave* frequencies.

I am not one to argue with a solution to Maxwell's equations.

-Dan

FWIW, tau=sqrt(L*C); Z0=sqrt(L/C); L=3.86e-5; tau=4e-8
(Please note: That is NOT!! a lumped model!)

implies that C= 41pF and Z0=965ohms

(v.f. = 0.058 is left as an exercise for the reader.)

But if the coil's axis is parallel to a ground plane, that 6" diameter
coil must be spaced about a quarter inch away from the ground plane
(axis 3.25" from ground plane) to get that 41pF capacitance, and that's
assuming a solid tube 3" in radius as a quick model .(An
approximation!)

Cheers,
Tom

Roy, you are allowing your imagination to stray.

Roy, W7EL wrote:
"This means that if we put a current into one end of the inductor, it`ll
take about 40 ns for current to reach the other end, right? So we should
expect a phase delay in current of 180 degrees at 6.15 MHz, from one end
to another?"

Hopper`s rule is one foot traveled per nanosecond. 40 feet of wire takes
40 nanoseconds.

The wavelength of 6.15 MHz is 48,8 or about 160 feet and in that space
the phase rotates 360-degrees. 40 feet is 1/4 of 360-degrees or
90-degrees at 6.15 MHz. At 1 MHz, the wavelength is 300 meters. 12,2
meters of wire is about 15-degrees of delay by my $1-dollar Chinese
calculator.

Best regards, Richard Harrison, KB5WZI