[KCDXC] Re: ARRL 160M contest

Wed, 6 Nov 2002 13:36:01 -0600

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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