How to cast a fly more efficiently...

[danc]

Found this on another board. Believe it or not, the author is serious.

 

"Some considerations on casting and rod loading.

 

A simple calculation for casting in air is;

 

Frt = Fi * Fa * Ff * Flt

 

Where Frt = the force on the rod tip in kg.m/s², Fi = inertia (mass)

in grams, Ff = the coefficient of fluid kinetic friction µk ( air

resistance), Fa = the acceleration of the line in ms², and Flt = line

tension in kg.m/s²

 

As may be seen, the equation depends on line tension being greater

than zero to produce a positive value. The higher the line tension,

the greater the value. If the value is zero or less, then Frt=0

 

This equates to " No tension= "no force on the rod tip"

 

To find line tension, rod loading, acceleration,friction, etc. one may

simply substitute the equations, you can calculate all the variables,

and also prove that line tension is a major factor.

 

Frt = Fi * Fa * Ff* Flt

 

 

Flt = Frt / fi * Fa * Ff

 

 

Fi * Fa * Ff = Flt / Frt

 

So, as the rod loading at any given point is known, ( it is simply the

curve of the rod, can be measured statically for any weight).

 

We will assume a rod loading of 0.01 kgm/s² (10 grams ).

 

The line mass can be weighed. Assume 30g here.

 

As the static line tension is exactly equal to the static rod loading

this must also be 0.01kgm/s² but only when the rod and line are

static! What is left when the rod/line is moving must be the

acceleration. Tension is required to accelerate the line.

 

However, the actual acceleration of the line, and friction, are extra

variables we don?t know yet.

 

Plugged in to the first equation, we get;

 

0.01kgm/s? = 30g * Fa * Ff * Flt

 

Second equation;

 

Flt = 0,01 kgm/s? / 30g * Fa * Ff

 

Third equation;

 

30g + Fa * Ff = Flt / 0.01kgm/s²

 

We still need to know the coefficient of friction and the

acceleration. Unfortunately, as this coefficient is not a fundamental

force, it can not be derived from first principles, and must be

observed empirically. In this case we will simply assume it to be 0.3.

 

We don't know the acceleration either, but we will also simply assume

a value here, of 1ms²

 

That gives;

 

30g * 1ms² * 0.3 = Flt / 0.01kgm/s²

 

 

30g * 1ms² = 0,03kgm/s²

 

* 0,3 = 0.09 kgm/s?

 

Therefore,

 

Flt= 0,09kgm/s² / 0,01kgm/s²

 

The units cancel, and Flt = 9kgm/s²

 

 

Add the values to all equations;

 

Frt = Fi * Fa * Ff * Flt

 

0.01kgm/s² = 0,03kgm/s² * 1ms² * 0.3 * 9kgm/s²

 

Second equation;

 

Flt = Frt / fi * Fa * Ff

 

9kgm/s² = 0.01kgm/s² / 0,03kgm/s² * 1ms² * 0.3

 

Third equation;

 

Fi * Fa * Ff = Flt / Frt

 

0,03 kgm/s² * 1 ms² * 0.3 = 9kgm/s² / 0.01kgm/s²

 

This proves all equations.

 

Plugging in the values you have for any particular conditions will

tell you the line tension, the rod loading, the acceleration, and the

friction.

 

If you graph the information, you can read it off directly.

 

You can also see how changing the mass changes the tension and

acceleration, how friction affects the model, and a lot of other

things.

 

You can also plug in the force for a haul, and see how it affects the

setup.

 

Lots of things are possible.

 

These equations are rudimentary, but cover all major factors. I am

still working on equations for the conversion of line tension to line

momentum. The equation shown is also primarily designed to show what

happens on the forward stroke. From when the rod begins loading. One

may of course adjust it, and add other factors if desired.

 

There are a couple of points worthy of note. The fluid friction varies

according to the amount of line outside the rod tip, as of course does

the mass, and its velocity.

 

Once line has rolled out and is shot or released, the tension on the

line itself is governed by the momentum of the line pulling on the

backing. This retains some tension on the line. As long as the line

stays straight, as a result of this tension, it will fly further. Once

it starts to "crinkle" it collapses.

 

In order for the line to turn over completely, there must be

sufficient tension for it to do so.

 

The equation shown is just one of a series which I am trying to use to

set up a casting simulator, first as a mathematical model, and then

including programmed graphical elements. The target is a dynamic model

of casting, into which one may plug in any rod or line, and also show

the optimal length and weight for shooting heads etc etc. Hopefully it

will also show the effects of differences in rod tapers and action.

 

One of the main things of note here, is that it is rod and line

tension which keeps the line swinging back and forth when false

casting with a fixed line. The force applied to the butt only adds

sufficient force to account for "losses" to fluid friction.

 

Also, one does not "throw" or "cast" the line, one rolls it out. When

the line is rolling out, it is the tension on the bottom leg of the

line loop which causes this.

 

In order to convert the rod and line tension to line momentum, when

shooting line, the point at which tension is released, and how this is

done, is of major importance.

 

In order to "force" turnover for instance, more tension is required.

This can be done by "overpowering" the cast, or by using a "check

haul", pulling back on the line before it has unrolled will cause

tension to increase, and the line turns over faster.

 

Pulling back on a line which is already unrolled will of course merely

brake it.

 

This also demonstrates how hauling works, it does not accelerate the

line, or load the rod much, it increases system tension, mainly line

tension, which is converted to momentum.

 

This theory, and the related equations, are my original work, if you

use it, please credit where you got it from.

 

TL

MC"

 

Mike Connor-ROFF

[jwl]

wow I am glad mine just says, open bail and sling 'er out there

[BillM]

This is what happens with an engineer or someone with a similar mindset tries fly fishing

[kemper]

Interesting.

 

What scares me is I actually understood all of that.

 

But I couldnt finish my calculus midterm today. Go figgur

[Greencoachdog]

... just cast the damn line and catch the damn fish!!!

[the spirit of fishing]

thanks;

now I know why I gave up trying to flyfish-I was doing all my calculations in Imperial measure and didn't do a proper conversion to Metric....

[Rich]

Under a float is even more efficient.

[danbo]

Reminds me of my first ski lesson..bend zee knees, watch out for zee trees, 5 dollars please!

[douG]

Calculus is only good for those problems where you know there is enough info for the answer, but you just don't know how to get the answer.

 

Common sense is good for everything else.