The way to check it is to plug your result into the equation, that is,
solve:

276*log(2*23.44/1.1)

What do you get?

Roy Lewallen, W7EL

Incidentally, the log formula for Z is actually an approximation. It's
no good when the wire spacing is very close compared to the wire
diameter, but it's fine for typical transmission lines.

yea right wrote:
Could someone check my math for me? It's been awhile since I used some of
these math skills.

I need to construct a short segment of 450 ohm open air xmittion line
between my antenna and a 9:1 balun mounted on a post. The formula given by
the 2003 ARRL handbook on page 19.4 gives the formula for calculating line
impedance as,
Z=276 Log(2S/d)

I had to re-arrange this to solve for S with Z and d as given.

Z = 450 desired xmission line imp.
d = 1.1mm the size in mm of the 18gauge wire I have.

Z/276 = Log(2S/d)
than
450/276 = Log2S - Log1.1

than
1.630 = log(2S) - .0414

than
1.630 = log 2 + log S - .0414

than
1.630 = .301 + log S - .0414

than
1.630 = .26 + log S

than
1.37 = log S

23.44 = S

23.44mm center to center

Does it check out?!
Thank You In Advance!