05-04-2011, 11:35 AM
Join Date: Aug 2010
Here's a formula for making twisted leaders
Until a few days ago, I never heard of a twisted leader. Some folks have been helping me get ready for a trip to Placencia (I leave for a week on 5/10). One of the things mentioned was the use of twisted leaders instead of spending time making biminis (Wasn't going for records, simply trying to reduce tarpon break-offs) I took him up on it and made a couple for my 5wt Sage that I use for LMB here in SoFL.
Made a single twist 6'leader of 20# fluoro. I felt that this single twist leader would perform like a straight 40# leader. And, undoubtedly, if I had made the leader with a butt section and a mid-section, it would ahve been even better. Tested it out this morning with a clouser, gurgler and a largish, wind-resistant bucktail fly. The leader was unreal: Turned over everything very nicely, casts appeared to be more accurate with softer landings. In three words, "It was great."
With my new found conversion from regular to twisted leaders, I figured that there must be a formula that will allow one to calculate how much starting line is needed for a given length of twisted leader. Could find one on the net, so I developed one based on the premise that this will be a two stage leader with a butt section that is twice as long as the mid-section...
Total length of line = Y
Total length of twisted leader = X
Given that we want a butt section that is twice the length of the mid-section, then
Length of butt section = 2/3 X and Length of mid-section = 1/3 X
Working with an already formed leader, the formula is:
X/3 [the mid-section] + 2 (2/3 X) [the full length of the butt section before it was folded at 1/3 the length of the single twisted line] = Y/2 [the length of the beginning full line folded in half], which reads
X/3 + 2 (2/3 X) = Y/2, solving for Y
Y = 10X/3: The length of line needed to form a twisted leader with a butt section twice the length of the mid-section is ten times the desired TL's length divided by 3. Or, 3.3 [10/3] times the TL's length
So, in order to find the length of line needed (Y) to form a 5' TL (X)...plugging in the numbers...
Y = 10(5)/3 = 50/3 = ~17'
Deriving a table from this formula (BTW, this formula is only good if you want the butt section to be 2/3s of the total length of the TL)
Length of line (Y) Length of TL (X)
17 ' 5'