Nice work, but I don't believe you can capture relative polar path
differences by using just the maximum latitude of the path. It's more
complicated than that. There can be dramatic differences in relative path
performance depending on propagation conditions. For example, when
conditions are good, W1 will have excellent polar paths on 40m to UA9, UA0,
etc. But when there's an auroral disturbance, the path can literally
shutdown. A station with a lower latitude path could have a huge advantage
in that case. Time of year also has an effect on when paths open and for how
Further, there are cases where the difference is so large that a simple
factor doesn't reflect it. For example, one year when I was a 20m op at
KC1XX in CQ WW SSB, I sat down at the radio at 0000z and tuned what was
essentially a rapidly fading band -- just a few weak stations to the south.
Meanwhile, the 20m ops at K3LR were running JAs at well over 100+/hr (I
think they logged something like 136 JAs that hour, to a handful or random
contacts I logged.) I ran the corrdinates for KC1XX and K3LR through your
spreadsheet and got points for JA of 5.56 for KC1XX and 5.36 for K3LR. Even
if I had been able to work a few JAs, the point factor wouldn't have made a
dent in the rate difference.
73, Dick WC1M
From: Pat Rundall [mailto:email@example.com]
Sent: Friday, July 01, 2011 3:45 PM
Subject: [CQ-Contest] Distance + Path DX contest scoring method (Excel proof
After reading many posts on this, I decided to create a scoring system that
would incorporate both distance and polar path. To do this, I simply used
Excel with the data from the CTY.DAT file to get lat/lon and then calculate
distance from a known point and the polar path. Users can test it with their
own QTH and tweak the effect of distance and polar path as they wish.
The default scoring method gives a score of:
QSO points = 1 + distance points + polar path points
where distance points = point to point distance in km / 5000
and polar path points = 3 * (max latitude of the path/90) (for QSOs where
the longitude differs by more than 90 degrees)
A bit complex? Perhaps.
If you have MS Excel, please take a look at this and see if you think this
would make sense. Before you complain about it, note my comments at the top
of the sheet.
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