[KCDXC] Re: ARRL 160M contest

Wed, 6 Nov 2002 10:34:12 -0600

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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