Pete Smith wrote:
> Even with the increased price of copper these days, 3/4" thinwall copper
> tubing is still pretty cheap. I recall that someone a few years ago
> had, on a web page, a RF grounding installation made entirely of 3/4" or
> 1" copper tubing.
>
> I'm a little surprised, though (not knowing any better) that the
> inductance of a tube at HF is only as good as the tube flattened
> (roughly equivalent to 1 1/4" strap  I would have thought it would be
> closer to strap the size of the circumference of the tube, or twice that.
>
>
I've been trying to come up with a good simple conceptual explanation
for why this is, and I'm going to draw some pictures and post it
eventually, but here goes, in words alone..
Imagine that the tube is a bunch of parallel wires arranged in a circle.
Say there's 8 of them, for now..
Start with one single wire.. it has some inductance L.
Now, our tube is represented by 8 parallel wires.. If all those wires
were a long distance apart, but in parallel, the overall inductance
would be L/8 (resistance too..)
But they're not a long distance apart. The magnetic field from one wire
interacts with all the other wires, so there's some mutual inductance
between them. That is, the inductance of two parallel wires (each of
inductance L) is not L/2, but somewhat more. (Ltotal = (L+M)/2) The
closer the wires, the bigger the mutual inductance, M, and the closer
the inductance gets to just L (if the two wires are on top of each
other.. not physically possible, but as an example)
OK, back to the tube...
Picking a given wire, say the one at the top (12 o'clock position).. It
has the least coupling to the one that's at the bottom (6 o'clock), and
gradually more to the ones closer on each side. Calculate all the
parallel inductances and you can work out what it is.
Now squash the tube (from the top), so 12 and 6 o'clock are closer to
each other.
The Mutual L for 12 and 6 is now greater, BUT, the mutual L for 3 and 9
o'clock is less. It doesn't exactly cancel, but the overall net effect
is "small"
(I would imagine that an exact analytical solution is straightforward,
assuming that elliptic integrals are your friend, and you KNOW, because
it has circular/elliptical symmetry, that the solution will be something
using a Bessel function)
As I recall, Terman or some contemporary of his has a description of
this too, where he talks about the geometric mean of the distances
between the wires. It also comes up when calculating the inductance of
"bundle" conductors for HV transmission lines, too (not only does using
a bundle reduce corona losses, it also reduces inductance, which is a
very good thing for power transmission lines)
(Grover's exact formula (page 31) for inductance between two filaments
separated by d, of length l, (in cm)
M = 0.002 *l *[ln(l/d+sqrt(1+l^2/d^2))sqrt(1+d^2/l^2)+d/l]
or for d<<l (the usual case), Grover recommends a series expansion:
M = 0.002 * l *[ln(2*l/d) 1 + d/l  1/4 *d^2/l^2 + ....)
but remember that Grover was working back in the days of slide rules and
log tables, when square roots were a pain, etc.)
Here's some quick excel data for a circle of 8 wires and a squashed tube
of 8 wires. You can see the total M doesn't change much
Length>> 100 cm
x y dist from 1st M (uH?)
1 0 3 o'clock
0.707 0.707 0.765309088 0.914686261
0 1 1.414213562 0.793167183 12 o'clock
0.707 0.707 1.84761955 0.740562029
1 0 2 0.725014038 9 o'clock
0.707 0.707 1.84761955 0.740562029
0 1 1.414213562 0.793167183 6 o'clock
0.707 0.707 0.765309088 0.914686261
sum>> 5.621844983
Flattened
M(uH)
1.571 0
0.785 0.1 0.23675715 1.148280776
0.000 0.1 1.004987562 0.860673365
0.785 0.1 1.788196466 0.746982368
1.571 0 2.570796327 0.675928883
0.785 0.1 1.788196466 0.746982368
0.000 0.1 1.004987562 0.860673365
0.785 0.1 0.23675715 1.148280776
sum>> 6.187801903
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