5 N Dime

I have been tying up flies for steelhead as of late. It was brought to
my attention that the proper ammount of trurns for the ribbing on these
flies is 5. Does anyone know the reason the number of turns should be
5?

I will be intrested to hear what the others have to say on this matter.


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theartoflee

He who asks is a fool for five minutes, but he who does not ask remains
a fool forever.
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"theartoflee" wrote in message
...

I have been tying up flies for steelhead as of late. It was brought to
my attention that the proper ammount of trurns for the ribbing on these
flies is 5. Does anyone know the reason the number of turns should be
5?

I will be intrested to hear what the others have to say on this matter.



Because it's prime. You do want your flies to be prime don't you?

It depends on if they are wild steelhead or hatchery steelhead. I've
found that the wild ones won't touch a fly with 4 turns, not even 4 and
a half.

When I give someone a fly recipe with ribbing, one of the common
questions they ask is; "How many turns on the ribbing?" If I say; "I
don't know 5 or 6 will do." They ask; "Well is it 5 or 6?" I now tell
everybody 5. I think it's become the standard because I catch myself
asking the same dumb question and getting the same answer.

I really doubt the fish cares.

But, be sure to start the ribbing from the top of the butt of the fly
or it's useless.

Cdog


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Corndog
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Just because - the world needs some standards!


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JeffK
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Jeff good answer I was guessing that it was so the commercial tiers and
illiterates could count the number of turns for the ribbing on there
free hand while the other hand handled the ribbing part.


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theartoflee

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a fool forever.
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I found this:

"As to flies tied on larger hooks(4-5/0), the number of turns needed to
provide the optimal torque, 3.24 x 10 to the fourth power negative
centripetal force accelerated toward the shank of the hook, is based in
McClingon’s constant. That constant dictates that one turn is needed
for a size “5” thread (based on the Buford Scale) for each 153 microns
of hook shank(or rotational axis) diameter. For example, a hook shank
that is 500 microns in diameter, would need a minimum of four wraps.
Of course, the material that is to be tied to the shank of the hook
increases the diameter of the wrap and therefore increasing the
centripetal force needed to secure it to the shank of the hook. Thus
the diameter of the material that is to be tied to the hook is to be
calculated using a factor of .5 of McClingon’s constant. It must be
added that if one increases the size of the thread usedby one Buford
unit, the number of wraps decreases by a factor of .347( and increases
at that same rate when one decreases the size of the thread by one
Buford unit)"


--
Future Fanatic

John
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the truth will set you free.
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Future Fanatic;96347 Wrote:
I found this:

"As to flies tied on larger hooks(4-5/0), the number of turns needed to
provide the optimal torque, 3.24 x 10 to the fourth power negative
centripetal force accelerated toward the shank of the hook, is based in
McClingon’s constant. That constant dictates that one turn is needed
for a size “5” thread (based on the Buford Scale) for each 153 microns
of hook shank(or rotational axis) diameter. For example, a hook shank
that is 500 microns in diameter, would need a minimum of four wraps.
Of course, the material that is to be tied to the shank of the hook
increases the diameter of the wrap and therefore increasing the
centripetal force needed to secure it to the shank of the hook. Thus
the diameter of the material that is to be tied to the hook is to be
calculated using a factor of .5 of McClingon’s constant. It must be
added that if one increases the size of the thread usedby one Buford
unit, the number of wraps decreases by a factor of .347( and increases
at that same rate when one decreases the size of the thread by one
Buford unit)"

You know,

I never thought of it that way.....but it DOES make a lot of sense!

My lofty take....
Maybe because an odd number has a more organic look or feel to it. Use
four or six wraps when building a robot, but five wraps on a Spey!
~James


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jcstikfish

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Future Fanatic wrote:
I found this:

"As to flies tied on larger hooks(4-5/0), the number of turns needed to
provide the optimal torque, 3.24 x 10 to the fourth power negative
centripetal force accelerated toward the shank of the hook, is based in
McClingon’s constant. That constant dictates that one turn is needed
for a size “5” thread (based on the Buford Scale) for each 153 microns
of hook shank(or rotational axis) diameter. For example, a hook shank
that is 500 microns in diameter, would need a minimum of four wraps.
Of course, the material that is to be tied to the shank of the hook
increases the diameter of the wrap and therefore increasing the
centripetal force needed to secure it to the shank of the hook. Thus
the diameter of the material that is to be tied to the hook is to be
calculated using a factor of .5 of McClingon’s constant. It must be
added that if one increases the size of the thread usedby one Buford
unit, the number of wraps decreases by a factor of .347( and increases
at that same rate when one decreases the size of the thread by one
Buford unit)"




Please keep your units standard. Hook shanks are referred to by gauge
(American Wire Gauge, AWG), not by diameter.

Diameters can be calculated by applying the formula:

D(AWG)=.005·92^((36-AWG)/39) inch.

For the 00, 000, 0000 etc. gauges you use -1, -2, -3, which makes more
sense mathematically than "double nought." This means that in American
wire gage every 6 gauge decrease gives a doubling of the wire diameter,
and every 3 gauge decrease doubles the wire cross sectional area.
Similar to dB in signal and power levels.

Conversion to microns is left to the reader.

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