Hauling, Rod-loading.

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