[KCDXC] Re: ARRL 160M contest

Wed, 6 Nov 2002 14:01:12 -0600

If I can get an antenna up, I'll try to play.  Probably doing QRP so I can
get a certificate!

-----Original Message-----
From: tombaugh [mailto:[email protected]] 
Sent: Wednesday, November 06, 2002 1:52 PM
To: [email protected]
Subject: Re: [KCDXC] Re: ARRL 160M contest

Absolutely! It's only once a year Jim. And Stateside contacts are plentiful.

Tom Baugh
AE9B
----- Original Message -----
From: "Reicher, James" <[email protected]>
To: <[email protected]>
Sent: Wednesday, November 06, 2002 1:43 PM
Subject: RE: [KCDXC] Re: ARRL 160M contest

> I don't have an antenna for 160 yet.  I suppose I could shunt-feed my
tower.
> Is there a low-power category for the contest?
>
> 73 de N8AU, Jim in Raymore, MO
>
> -----Original Message-----
> From: tombaugh [mailto:[email protected]]
> Sent: Wednesday, November 06, 2002 1:36 PM
> To: [email protected]
> Subject: Re: [KCDXC] Re: ARRL 160M contest
>
>
> Can you get on 160M Jim?
>
> Tom Baugh
> AE9B
> ----- Original Message -----
> From: "Reicher, James" <[email protected]>
> To: <[email protected]>
> Sent: Wednesday, November 06, 2002 10:34 AM
> Subject: RE: [KCDXC] Re: ARRL 160M contest
>
>
> > Actually, you should be using a different formula.
> >
> >
> > F(X) * Q = F(Y)* Q
> > If F is frequency in megahertz, Q is QSOs, X is 1.8 and Y is 27 then
> > 27MHz(QSOs) = 1.8 MHz(QSOs)
> > Solving for QSO's, it follows that F(X)/F(Y) = 1 QSO
> > If F(X)=27MHZ and F(Y)=1.8MHZ, then
> > 27MHz / 1.8MHz = 15 QSOs
> > Thus, (15 QSOs * 1.8MHz) = (1 QSO * 27MHz) or, to put it simply, It 
> > takes 15 160meter contacts to make 1 CB contact!
> >
> > 73 de N8AU, Jim in Raymore, MO
> > (Putting on the earplugs and body armor! Hi hi)
> >
> >
> >
> >
> > -----Original Message-----
> > From: Alan KI7WO [mailto:[email protected]]
> > Sent: Wednesday, November 06, 2002 10:18 AM
> > To: [email protected]
> > Subject: [KCDXC] Re: ARRL 160M contest
> >
> >
> > Hi Tom,
> >
> > I have been thinking about this 160 Meter thing.  Since I have no 
> > antenna for this band I have been using the old slide rule to 
> > calculate some figures.  Here is what I have come up with:
> >
> > 160 meters divided by 27 Mhz equals 5.9 roughly so if I get 1 point 
> > per
> SSB
> > Q then I need 6 Q's to equal one 160 SSB contact and if I get 2 
> > points per CW Q this requires 3 CW Q's and surely I should get  4 
> > points per RTTY Q
> so
> > 2 RTTY Q's on 27 Mhz would make one "GOOD ONE" on 160.
> >
> > I know that I can count on you & AK0A to help make these two Q's for 
> > my "first ever" 160 effort.
> >
> > Alan  [email protected]
> >
> >
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