Although the formulae did not copy properly, if one wants to help me with
this, they can go to the link below for the proper fomatting of these
formulae. I would really appreciate it.

Thank you.

j/b


d= .0808 (12 gauge stranded copper)
S= 6 inches
r=Effective dielectric constant (Air = 1.00054)


Here is the Wikipedia link for these formulae:
http://unblockmoney.info/browse.php?...xhZGRlcl9saW5l

The formula to calculate the impedance of ladder line is:





The formula to calculate the impedance of open (air dielectric) ladder line
is:



The formula to calculate the distance between conductors of open ladder line
is:



Whe
Z0 = Impedance.
S = Center to center distance between wires.
d = Diameter of the wire.
r = Effective dielectric constant (Air = 1.00054).
The unit of measure for S and d are not critical as long as they are the
same.

I GET DIFFERENT RESULTS WHEN COMPARING TO A GRAPH.

"justme" wrote in :

Although the formulae did not copy properly, if one wants to help me
with this, they can go to the link below for the proper fomatting of
these formulae. I would really appreciate it.

Thank you.

j/b

....

I read all three posts, and it was a bit hard to understand your problem.

Firstly, the formulae on the Wiki page that use the log term are
approximations of the function using the cosh term, and the approximation
is poor for small S/d.

Have a look at the calculator at http://www.vk1od.net/calc/tl/twllc.htm ,
it uses the cosh term in its calcs. BTW, the metric equivalent of your
example is 2mm wires spaced 150mm.


Owen

Owen Duffy wrote in
:

....
Firstly, the formulae on the Wiki page that use the log term are
approximations of the function using the cosh term, and the
approximation is poor for small S/d.

Have a look at the calculator at
http://www.vk1od.net/calc/tl/twllc.htm , it uses the cosh term in its
calcs. BTW, the metric equivalent of your example is 2mm wires spaced
150mm.

cosh above should be acosh!

Owen

.... not doing too well thismorning, too eager to get out to breakfast.

I also meant to note that the log formula is 276log(S/d) using the same
notation as the acosh formula.

The authors have cobbled formulas from different places, and they do not
use a consistent meaning for S and d.

But, use the acosh formula for close spacings.

Owen

Owen Duffy wrote in news:Xns9C913EF99FBC5nonenowhere@
61.9.191.5:

The authors have cobbled formulas from different places, and they do not
use a consistent meaning for S and d.

Sorry, I was wrong there, they are 'correct'... 276log(2*6/0.08) gives 600
ohms which is the correct answer.

Check that the graph you are referencing uses the same meaning for the two
dimensions.

Owen

On Sep 24, 3:46*pm, Owen Duffy wrote:
Owen Duffy wrote in news:Xns9C913EF99FBC5nonenowhere@
61.9.191.5:

The authors have cobbled formulas from different places, and they do not
use a consistent meaning for S and d.

Sorry, I was wrong there, they are 'correct'... 276log(2*6/0.08) gives 600
ohms which is the correct answer.

Check that the graph you are referencing uses the same meaning for the two
dimensions.

Owen

Close one end and then measure the impedance. If you are making your
own you will eventually have to do it anyway to determine your
matching system