How much weight does Palaniuk need?

If you hang around enough tournament weigh-ins, you hear a lot of talk about how much weight the leader will need to have on the final day to put things away — to all but guarantee (there are no true guarantees) victory.


Here at the St. Lawrence River, it would be easy to say that Palaniuk needs 25 pounds to guarantee victory here, since a lot of locals seem to think that a 25-pound limit is attainable.


I think there's a better way to calculate this, and it uses information we already have.


The two numbers we need to know are the weight of the heaviest catch in the tournament so far, and the lead Palaniuk has going into the final round.


The first number is 23-14. That's what Steve Kennedy had yesterday, and it propelled him into the finals. The second number is 3-12. That's the margin between Palaniuk, in first, and Jonathon VanDam, in second, at the end of the weigh-in yesterday.


Take the first number and subtract the second and you get 20-2, and that's what I think Palaniuk needs today to virtually ensure victory.


Why those numbers?


Well, the biggest bags of the tournament typically come on the first two days. That's when the fish were the freshest, the least picked over and the least harassed. No, it's not always true, but it's true far more often than not, and it means that the biggest catch of the event is very unlikely to come on the final day, even though only the most successful anglers are on the water.


The second number — the lead — merely tells us how close to the first number the leader needs to be if he's to beat the second best angler having a great day.


Could JVD catch 25 pounds and pass Palaniuk? Of course he could, but it's very unlikely.


According to BASSTrakk, Palaniuk has a limit weighing 21-10 already in his livewell. If his number is close to accurate, JVD will need every bit of 25 pounds to catch him.


Of course, BASSTrakk is not always accurate.

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