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Old 03-31-2008, 09:45 PM   #1
spence
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Quote:
Originally Posted by Redsoxticket View Post
lets say that you have your drag set at 5 pds at the reel.

Now, how many pounds will it take (more or less) if the measurement was taken with with the rod flexing at about 4 ft from the rod given the same criteria above, that is 5 pds at the reel ?
Even less I'd imagine. Unless the drag numbers at the reel assumed some flex from the rod.

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Old 03-31-2008, 09:56 PM   #2
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I say more because there will be a Force vectors (Fx and Fy) at each guide which absorb the high loads.
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Old 03-31-2008, 10:01 PM   #3
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If you had a pool cue rod with only one guide then the measurement would be the same that is 5 pds.
Now with multiple guides, flexing rod then the weight will be absorbed on the guides.
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Old 03-31-2008, 10:05 PM   #4
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I think the question here is about the net force on the drag. Certainly the flex on the rod will absorb force, but are the drag settings from the manufacturer assuming fishing conditions or just a raw measurement?

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Old 03-31-2008, 10:08 PM   #5
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I don't know Spence but I'd assume raw measurement because every rod is different slow, fast etc.
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Old 03-31-2008, 10:21 PM   #6
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Friend was holding scale 6-8ft below where I was standing, so rod had full bend. Tried a 7lb drag too, your just asking for pulled hooked hooks and equipment failure there imo. But to each their own I guess.
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Old 03-31-2008, 10:59 PM   #7
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Tony the 7 pd taken at the reel ? If yes then what is the measurement if you tied a loop in the line at the 6-8 ft distance and recorded the pounds of force when the line moves.
If the 7 pds. is to much weight for the equipment then what is it at 5 pds ?

I'm thinking that 7 pds. at the reel can be 12 pds I guess at the loop.
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