Folded Dipole Calculator Help

[Phil Allison]

"amdx"

Those "non existent compications" are what I'm trying to eliminate

** Then just STOP inventing them !!!!!

Yes, I understand that


Would 72 ohms still be the correct number if d1 is twice the diameter of
d2?

** Read point 2 over and over - dickhead !!

I would think that is the reason the author put that variable in the
program.

** I *would* like to see any sign of actual THINKING from you at all !!!

But will not hold my breath waiting for it.



..... Phil

 

[amdx]

"amdx"

** Then just STOP inventing them !!!!!




** Read point 2 over and over - dickhead !!

Your not stating your opinion very clearly.
But I think you mean to say, the relative diameters of d1 and d2 would not
have
any effect on the underlying 72 ohms used in the calculations for the
program.
Thanks, Mike
If I'm wrong, please don't be shy, let me know.

 

[Phil Allison]

"amdx"

Your not stating your opinion very clearly.

** Yes I am.

But I think you mean to say, the relative diameters of d1 and d2 would not
have any effect on the underlying 72 ohms used in the calculations for the
program.

** The calculator does not use ANY particular impedance for the computation
of a RATIO !!

Once that RATIO is computed, it then multiplies with the impedance figure
YOU supply to get the final figure.



..... Phil

 

[amdx]

"amdx"


** Yes I am.



** The calculator does not use ANY particular impedance for the
computation of a RATIO !!

Once that RATIO is computed, it then multiplies with the impedance figure
YOU supply to get the final figure.

.... Phil

Yes Phil, I worded that incorrectly.
I understand the calculator finds the ratio of the
(impedance of a folded dipole) / (impedance of a dipole )
and then multiplies this ratio times the impedance of a dipole in free
space. (72 ohms)
I also found in the 1985 ARRL Handbook a graph* that shows a change of
conductor diameter from 10 to 10,000 only changes the impedance by 30 %.
Extrapolating from the graph, I see doubling the diameter of the conductor
only
increases the resistance by about two ohms. So if we only doubled half** of
the dipole
it seems the effect would be even less.
Thanks, Mike
* graph is non linear
** as in making d1 twice the diameter of d2