Hi Art,
Do you mean a round inner conductor & square tube outer?
See the lower left corner of page 24-22 of Reference Data for Radio
Engineers, 6th edition, by ITT.
It is beyond my typing limit... and I don't have a scanner. Oh, what
the heck, let's see if I have enough screen ink...
Zo approx= [138 log10 (p) + 6.48 - 2.34(1+0.405p^-4)/(1-0.405p^-4)
-.48(1+0.163p^-8)/(1-0.163p^ -8) - .12
(1+0.067(p)^-12)/(1-0.067(p)^-12)]e^-0.5
Where D is the square dimension and d is the inner diameter and p= D/d
I suggest entering that formula into a spread sheet or a Basic program
to avoid wearing out the slide rule.
Here it is in Quick Basic ---
DQ = 1.4465147 : REM -- inside dimension of square outer conductor
D = 1 : REM -- diameter of inner conductor
P = DQ / D
Zo = (138 * LOG(P) * LOG(10) + 6.48 - 2.34 * (1 + .405 * P ^ -4) / (1 -
.405 * (P) ^ -4) - .48 * (1 + .163 * (P) ^ -8) / (1 - .163 * (P) ^ -8) -
.12 * (1 + .067 * (P) ^ -12) / (1 - .067 * (P) ^ -12)) * (2.718281828#) ^
-(1 / 2)
PRINT Zo; "ohms"
73 ohms,
Marv WC6W
On Wed, 20 Feb 2002 00:26:36 +0000 Arthur Moe <kb7ww@easystreet.com>
writes:
>
> Could some one pass along the formula for calculating the equivalent
> round
> section of a square tube so I can calculate line impedance.
>
>
> Thank you
>
> Art
> KB7WW
>
>
> --
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>
>
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