Walter...
Problem is that the, "point of contact" to a moving loop is an inconstant, since the loop is always unrolling as well as moving forward....so I don't think using it as a references has merit.
The formula I showed relates the forward motion of the end of the loop with respect to a presumed stationary rod tip. Of course, if the rod tip does not remain stationary, as with a forward followthrough, then the movement of the loop would slightly change.
These measurements have actually been made by Bruce Richards and physicist Noel Perkins of the U. of Mich.....but it's easy to set up your own physics experiment to prove it (which is one of the things I did.)
All you have to do is to take a thread spool and loop a piece of line around it . One end (the tag end) is the, "fly". The other is anchored at what we'll call the, "rod tip". Now the spool (loop) is moved a given distance and then stopped. It's all in the same time frame. We see that the, "fly" moved twice as far as the, "loop". Since it was in the same time frame, it had to do so at twice the loop speed.
To do the experiment with the, "shoot", we marked the line at the, "rod tip" which was a screw eye in a board. That way we could determine the distance that both the, "fly" and the, "rod leg" moved as well as the distance moved by the end of the, "loop". All in the exact same time frame. With all those figures it was simple arithmatic to come up with the answer.
I, then, used a larger spool (bigger, "loop") and was surprised to find that the relative figures were the same.
Gordy
From: WALTER/SUE SIMBIRSKI <simbirsw@xxxxxxx>
To: Gordon Hill <hillshead@xxxxxxx>
Subject: Re: FLYCASTING MATH ANSWERS
Date: Thu, 25 May 2006 09:54:05 -0600
Gordy - I'm a bit confused by both answers to question "2.) Now you have a cast made with a line shoot. You know the speed of the fly leg and the speed of the rod leg. What formula is used for determining loop speed ?" .
I think both answers are correct depending on the frame of reference. I think we can all agree that if the fly leg is moving 10 miles/hour faster than the rod leg then the loop is moving 5 miles/hour faster than the rod leg. If the chosen frame of reference is the something stationary (like the ground or the rod tip) and the speed of the rod leg is x miles/hour then the speed of the loop can be determined by:
(x + (x+10))/2 = x + 5
i.e. (the ground speed of the rod leg + the ground speed of the fly leg) divided by 2. [Gordy's formula]
On the other hand if we define a moving frame of reference, i.e. the point of contact between the rod leg and the loop, we then define speed in relation to the moving frame of reference (i.e. the speed of the rod leg). We could think of the rod leg and fly leg speeds as a form of "air speed". The "air speed" of the rod leg is 0 and the "air speed" of the fly leg is 10. This makes the air speed of the loop 10/2 = 5. Now to determine the ground speed of the loop we add the ground speed of the frame of reference, i.e. x, and arrive at a ground speed for the loop of
x + 5
i.e. ground speed of the rod leg + (relative speed of the fly leg divided by 2) [Jerry's formula]
Cheers
Walter
From: Gordon Hill <hillshead@xxxxxxx>
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Subject: Re: FLYCASTING MATH ANSWERS
Date: Wed, 24 May 2006 08:24:20 -0400
Hi, Group...
For those who got the hard to read HTML version of Tom's answers to, "FLY CASTING MATH", this is cleaned up by Walter.
My comments appear in text following Tom's answers which are in red. Mine are in BOLD CAPS and in green .
Gordy
From: Walter Simbirski <simbirsw@xxxxxxx>
To: Gordon Hill <hillshead@xxxxxxx>
Subject: Re: RE: text
Date: Wed, 24 May 2006 00:24:03 -0600
Hey Gordy!HTML stands for HyperText Markup Language. It was originally used to format web pages but is also usedto format emails as well. Somehow part of the html formatting got lost so that the email program didn't realizeit was supposed to use all of the gobbledygook as html. I've repaired the message and appended it to this.You should be able to view it okay now.CheersWalterTom...
Check out my additional notes below our answers to # 9.) in green. (Scroll way down for details.)
Gordy
From: "Gordon Hill" <hillshead@xxxxxxx>
To: cooper@xxxxxxxxxxxxxxxxxx
Subject: Re: Fly Casting, "Math"
Date: Mon, 22 May 2006 10:28:23 -0400
Tom...
My comments in BOLD CAPS.
You did really well !!!!
Gordy
From: "North Fork Flies - Tom Cooper" <cooper@xxxxxxxxxxxxxxxxxx>
Reply-To: "North Fork Flies - Tom Cooper" <cooper@xxxxxxxxxxxxxxxxxx>
To: "Gordon Hill" <hillshead@xxxxxxx>
Subject: Re: Fly Casting, "Math"
Date: Mon, 22 May 2006 03:16:12 -0400
My answers are in red----- Original Message -----From: Gordon HillSent: Sunday, May 21, 2006 8:14 AMSubject: Fly Casting, "Math"Jerry...
My server may have prevented some of the Group from getting my questions on, "Fly Casting Math." Would you help by sending this out ? Of these 9 questions, I'd expect an MCI candidate to correctly answer 6 of them (*) An exceptional candidate to be correct on 8.
1.)* You are given the speed of the fly leg of a loop for a cast with no shooting of line. What is the formula for calculating the loop speed?
Loop speed is 1/2 of the fly leg speed.
YES.
2.) Now you have a cast made with a line shoot. You know the speed of the fly leg and the speed of the rod leg. What formula is used for determining loop speed ?
Loop speed is 1/2 of the fly leg speed + the rod leg speed
IT'S: FLY LEG SPEED + ROD LEG SPEED DIVIDED BY 2.
3.)* You are casting a fly tied on a # 12 hook. What formula or, "rule" might you sue to help determine the approximate, "X" size for your class tippet ?
Rule of 3's. Divide fly size by 3 (or 4) or a 4X tippet
YES.
4.)* What formula or, "rule" would you use for determining the diameter of your 5X tippet ?
Subtract 5, The X size, from 011 (the diameter of 0X tippet) or 006.
YES. (ANOTHER WAY TO PUT IT IS TO SUBTRACT THE X NUMBER FROM 11 AND DIVIDE BY 1000.)
5.)* You know your tippet size to be 4X. What formula might you use to determine its pound breaking strength ?
Sorry to say I don't know this one, but with the wide range of test breaking strengths does a rule of thumb work?
"RULE OF 9'S"...SUBTRACT THE X NUMBER FROM NINE AND YOU GET AN APPROXIMATE VALUE FOR BREAKING STRENGTH. (LESS RELIABLE THAN IT USED TO BE PRIOR TO THE NEW TIPPET MATERIALS MANY OF WHICH ARE STRONGER/DIAMETER.)
6.)* What, "rule" or formula might you use to come up with the weight in grains for the first 30' of your #6 designated fly line ? (A rough approximation)
Line weight X 30 or 6 X 30 = 180. I like to use ((Line Weight X 25) + 25) or 6X25 = 150 +25 = 175 It is more accurate on lighter weights.
I AGREE. USING 30 GETS YOU CLOSE IN RANGE FOR A 4 WT. LINE. IT IS LESS ACCURATE AS YOU GO IN EITHER DIRECTION FROM THAT.
7.) What is the so-called, "IGFA Formula" for determining the weight in pounds of a game fish ? You have only the length of the fish and the girth of the fish measured in inches.
Girth Squared X Length/800 = Weight
YES.
8.)* Can you come up with a very general, "rule of thumb" to describe a typical fly leader in terms of percentage for butt section, tapered section, and tippet section ?
60 - 20 - 20
YES.
9.) You are trying for maximum distance. can you come up with a simple formula for determining the approximate length of your shoot when knowing the amount of line carried on your last back cast ?
Back cast length x 1.333 = total forward cast length.
I HAVE NOT HEARD OF THAT FORMULA.....I'LL TRY IT AND SEE. (DO YOU HAVE A REFERENCE FOR IT ?)
I'VE USED: 50% OF THE LINE CARRIED FOR THE LENGTH OF THE SHOOT. (THE LENGTH OF THE SHOOT IS THE VALUE I ASKED FOR IN THE QUESTION.)
After typing the above comment, I went out and we tested it. I took note of the fact that the 1.333 formula means this:
The caster can carry 60' on his last back cast. His total distance for his presentation would then be 60 X 1.333 which is 70.9 feet.
Now most CCI candidates who pass make a 75' distance cast when carrying about 50'. Your 1.333 formula would mean that they could achieve only 1.333 X 50 = 66.5'
I can make a 90' cast routinely by carrying 60'. This makes the 1.333 formula short of the mark, as I see it. (Most MCI's I've coached can do the same thing once they reach test level.)
Gordy
Tom Cooper
* expected of an MCI candidate.
Gordy