Roy Lewallen wrote:
I did a search quite some time ago and failed completely in finding the
formula you describe, in Terman or any other "bible". The formula for
the capacitance of an isolated sphere is common, but not a cylinder. The
formula for a coaxial capacitor is common also, but the capacitance
calculated from it approaches zero as the outer cylinder diameter gets
infinite.


Roy, I can't answer the question you put to Reg, but the capacitance of
an isolated conducting cylinder is approximately that of an isolated
conducting sphere of the same surface area. Any unbalanced charge on a
conductor resides on the surface of the conductor and so the greater the
surface area, the greater the charge you can place on it to raise its
potential by some amount, etc. Obviously the cylinder's electric field
would depart from the radial field of a point charge at the center of
the sphere but I don't think that's relevant in this case. I hear bells
ringing. ;-)

Chuck



Maybe you could take a look after the wine wears off, and see if you can
locate the formula. By your earlier posting, it sounds like you've used
it frequently, so it shouldn't be too hard to find. I'd appreciate it
greatly if you would. And yes, I would make use of the formula -- I'm
very curious about how well a coil can be simulated as a transmission
line. The formula you use would be valid only in isolation, so
capacitance to other wires, current carrying conductors, and so forth
would have an appreciable effect. I showed not long ago that capacitance
from a base loading coil to ground has a very noticeable effect. Do you
have a way of taking that into account also?

Roy Lewallen, W7EL

Reg Edwards wrote:

How do you calculate the coil C to use in the transmission line

formulas?

Roy Lewallen, W7EL

===================================

I'm surprised a person of your knowledge asked.

Go to Terman's or other bibles, I'm sure you'll find it somewhere, and
find the formula to calculate the DC capacitance to its surroundings
of a cylinder of length L and diameter D.

Then do the obvious and distribute the capacitance uniformly along its
length.

The formula will very likely be found in the same chapter as the
inductance of a wire of given length and diameter.

I have the capacitance formula I derived myself somewhere in my
ancient tattered notes but I can't remember which of the A to S
volumes it is in.

I'm 3/4 ot the way down a bottle of French Red plonk. But Terman et
al should be be quite good enough for your purposes.

And its just the principle of the thing which matters. It's simple
enough. I don't suppose you will make use of a formula if and when
you find one.
----
Reg.

Roy Lewallen wrote:

What's it's velocity factor, and how did you calculate it?

I can't believe I did that! It must be from spending too much time
reading Internet postings. Of course I meant:

What's its velocity factor, and how did you calculate it?
^^^
I knew better than that by the time I'd finished grade school. Hope it
isn't all downhill from here.

Roy Lewallen, W7EL

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.

Best regards, Richard Harrison, KB5WZI

Tom Ring wrote:

Cecil Moore wrote:
Let's say we have a 1/2WL dipole in free space driven by a
self-contained source at the center. If we float a florescent
light bulb around the ends of the dipole, are you saying the
electric fields won't fire the bulb like it does on earth?

Stop acting like an idiot Cecil.

It was a technical question. I was just wondering what keeps
the wire from transferring energy when it is located in free
space.
--
73, Cecil http://www.qsl.net/w5dxp

Roy Lewallen wrote:

That's all very nice. Let's see if it's useful for anything.

A while back, Cecil posted a model of a base loaded vertical antenna. It
has an inductor which is vertically oriented. The bottom of the inductor
is 1 foot from the ground and the inductor is 1 foot long and six inches
in diameter. Inductance is 38.5 uH and it's self resonant at 13.48 MHz.
(Moving it very far from ground changes the resonant frequency to 13.52
MHz.)

What's it's velocity factor, and how did you calculate it?

13.48 MHz is not exactly the self-resonant frequency of the
coil. At 13.48 MHz, the one foot bottom section is 0.0137
wavelengths long, i.e. 4.9 degrees. So the coil occupies
85.1 degrees, i.e. 0.236 wavelength. The coil length is
coincidentally also one foot so the velocity factor is
4.9/85.1 = 0.058.
--
73, Cecil http://www.qsl.net/w5dxp

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.

13.48 MHz is not exactly the self-resonant frequency of the
coil. At 13.48 MHz, the one foot bottom section is 0.0137
wavelengths long, i.e. 4.9 degrees. So the coil occupies
~85.1 degrees at self-resonance. The coil length is coincidentally
also one foot so the velocity factor is 4.9/85.1 = 0.058.

I don't have the Terman Handbook. Does he take adjacent coil
coupling into account in that formula? If not, that's the
difference in the two results.

In either case, the velocity factor is not anywhere near 1.0
as the lumped circuit model would have us believe.

Does anyone have a formula for what percentage of current is
induced in coils farther and farther away from the primary
coil? I haven't found such a formula in my references but it's
got to exist.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil, W5DXP wrote:
"I don`t have the Terman Handbook."

Formula (37) on page 55 of the 1943 "Radio Engineers` Handbook is:

Lo = (r sq) (n sq) / 9(r) + 10(l)
Lo = approximate low-frequency inductance of a single-layer solenoid in
microhenries where r is the radius and l is the length of the coil in
inches.

Terman attributes the formula to H.A. Wheeler, "Simple Inductance
Formulas for Radio Coils", Proc. I.R.E., Vol 16, P1398, October 1928.

Best regards, Richard Harrison, KB5WZI

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

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