quarter wave coils (fwd)

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: Gerry Reynolds <greynolds@xxxxxxxxxx>



---------- Forwarded message ----------
Date: Fri, 15 Dec 2006 06:17:23 -0500
From: Jared Dwarshuis <jdwarshuis@xxxxxxxxx>
To: tesla@xxxxxxxxxx
Subject: quarter wave coils

Hi: Mike

Sorry about the delay, I have been working long hours lately. Yes I would
like to see your math!

Wheelers is a lot like the form: L = u Nsqrd A / H

Multiplying top and bottom by 4pi to get wire length

L = u (2pi R N)sqrd/ 4pi H

L = 4pi X 10^-7  ( 2pi R N) sqrd/ 4pi H

L =          10^-7 (2pi R N)sqrd/ H

L =           10^-7  4 (pi)sqrd (R) sqrd (N)sqrd/ H

L(metric) =  10^-6   39.48  (N)sqrd (R)sqrd / 10H

Since there are 39.37 inch per meter, we can write a non metric version of
the classic formulae, accurate to 1/3 percent of the metric version.

L( from inches) = (N)sqrd (R)sqrd / [10H]         Micro Henry

The wheeler formulae has the addition of radius in the denominator.

 L( from inches) = (N)sqrd (R)sqrd / [9R + 10H]         Micro Henry

As the solenoid becomes long, the R term in the denominator of
Wheeler becomes negligible and we converge on the classic solution to within
1/3 percent.

Neat stuff.


Sincerely: Jared Dwarshuis