Re: Antenna Length Calculation...was: "your mail"

[DeWitt Harrison <de@aztek.com>]

On Tue, 13 May 97 10:26:51 EDT, Dan Masi wrote:

>On May 13,  7:32am, Kurt Wesseling wrote:
>> [ ... ]
>> 2808/FmHz   ...that is divide 2808 by the frequency in megahertz. 
>> This gives us the length in inches. (You metric guys can extrapolate)
>> 
>> So, with 93.0 mHz we get: 2808/93.0 =  30.1935483871
>> 
>> I think it will be safe to drop most of the digits after the 
>> decimal...hehheh.  So, for 93.0 mHz we're talking about just over 30 
>> inches.
>
>Kurt, what was wrong with Phil's calculation?  He calculated,
>correctly, that 1/4 wavelength of a 93Mhz signal, in free space,
>is about 20 inches.  In real life, we'd slow down the propagation
>speed a tad below 300Mm/s, but that would only serve to show the
>effective wavelength as a bit *less* than 20". Where does the number
>that you use, that gives us 30", come from?

Dan, I think Kurt may be right for once.  :-^)  My figger'n is as follows:

(3*10^8 m/s)/(93*10^6 Hz) = 3.226     m / wavelength

3.226 m / 4 = 0.8065     m / quarter wave

Phil said about 80 cm which is close enough. However,

0.8065 m / 0.0254   m / inch  =   31.75 inches

I don't know why Kurt's method gives an answer which is a bit
smaller. Must be a real world antenna,  radio guy kind of thing.
Cheers,

DeWitt Harrison   de@aztek.com
Boulder, CO
88 5kcstq