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

On Nov 10, 7:22*pm, wrote:
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


You might want to check your units, Mike. I get Frt would be in
Kg*g*m^2

If you mean acceleration of the line is in m/s^2, then the units of
Frt would be even worse: Kg*g*m^2/s^4


I think you need another factor in there...but Physics isn't my strong
suit.

--riverman
(unit analysis r us)

On Nov 10, 12:53*pm, riverman wrote:
On Nov 10, 7:22*pm, wrote:

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

You might want to check your units, Mike. I get Frt would be in
Kg*g*m^2

If you mean acceleration of the line is in m/s^2, then the units of
Frt would be even worse: Kg*g*m^2/s^4

I think you need another factor in there...but Physics isn't my strong
suit.

--riverman
(unit analysis r us)

There are several other factors, but for the purposes of demonstration
they can safely be ignored.

QUOTE
In physics String Tension is the magnitude of the pulling force
exerted by a string, cable, chain, or similar object on another
object. Tension is measured Newtons (kgm/s2) and is always parallel to
the string on which it applies. There are two basic possibilities for
systems of objects held by strings.[1] Either acceleration is zero and
the system is therefore in equilibrium or there is acceleration and
therefore a net force is present. Note that a string is assumed to
have negligible mass.
UNQUOTE

http://en.wikipedia.org/wiki/Tension_(physics)

TL
MC

As a rod is a spring, and spring tension may be equated to string
tension, one may use the same units. This also facilitates conversion
etc.

The basic premise is of course that every action provokes an equal and
opposite reaction ( Newton´s third law). This means that in the system
there is always equilibrium at any given point. The force on the rod
tip therefore must equal the tension in the line, except when hauling,
where the haul force primarily increases line tension.

This applies regardless of whether the system is moving or not. A
static equivalent would be attaching the line to a fence post, and
bending the rod against the tension of the line. The line tension is
directly proportional to the rod loading ( spring force). The force
on each end of the system is identical ( ignoring hysterysis* when
dynamically applied ) in equlibrium.


*hysteresis (hÄ*s'tÉ™rÄ“`sÄ*s), phenomenon in which the response of a
physical system to an external influence depends not only on the
present magnitude of that influence but also on the previous history
of the system. Expressed mathematically, the response to the external
influence is a doubled-valued function; one value applies when the
influence is increasing, the other applies when the influence is
decreasing. Magnetic hysteresis occurs when a permeable material like
soft iron is magnetized by being subjected to an external magnetic
field. The induced magnetization tends to lag behind the magnetizing
force. If a field is applied to an initially unmagnetized sample and
is then removed, the sample retains a residual magnetization (it has
become a permanent magnet). The graph of the magnetic induction B
versus the magnetic field H is called a hysteresis loop. The area of
the loop is proportional to the energy dissipated as heat when the
system goes through a cycle; this represents a considerable energy
loss in alternating-current machinery. Thermal hysteresis occurs when
the value of a given property of a body depends not only on the body's
temperature but also on whether the temperature is rising or falling.
An example is the dielectric constant versus temperature for certain
crystals. Another kind of hysteresis is a common feature of control or
cybernetic systems. A familiar example is a thermostat controlling a
source of heat and set at some temperature T0. When the room
temperature falls through T0 to some lower temperature T1, the heating
power is switched on. When the room temperature rises through T0 to
some higher temperature T2, the power is switched off. Thus, for
temperatures lower than T1, the heat is always on; for temperatures
higher than T2, the heat is always off; but for temperatures between
T1 and T2, the heat may be on or off (double-valued response),
depending on which of the two temperatures T1 and T2 occurred most
recently in the system's history. Unlike the previous examples, this
hysteresis effect is not naturally occurring; it is designed into the
control system to prevent the damage to the system that would arise
from switching on and off too frequently.

TL
MC

This incidentally proves the hauling theory, as when a rod and line
are attached to a fence post, a haul does not accelerate the line, but
it increases tension massively. This explains why even a small fast
haul increases line momentum very considerably. There has always been
very considerable controversy about this, some maintaining that the
haul increases rod loading ( it does to some extent), and other
maintaining that the line was accelerated. The line is accelerated to
some extent, but the primary factor is line tension.

Without the tension the line can not be accelerated, the greater the
tension, the greater the acceleration.

Imagine the traces of a dog sled. Before the sled can move, the traces
must be taut, ( = under tension), the greater the tension, the greater
the force on the sled, and the faster it moves. It is also necessary
to increase tension constantly in order to mainatin or increase it,
which further proves the theory, as this is also the ideal casting
stroke.

TL
MC

Lastly, tension is the force which allows fly casting to function at
all. Tension is what keeps the line in the air. that it also explains
why a haul works as it does, and the ideal casting stroke has not
hitherto been explored.

TL
MC

Using Newtons;

The newton is the unit of force derived in the SI system; it is equal
to the amount of force required to give a mass of one kilogram an
acceleration of one metre per second squared. Algebraically:

1N = 1kg*m/s²

To describe the force on the rod tip is also correct.

TL
MC

This incidentally also demonstrates why a cast with slack at any point
fails, and why the point of tension release ( conversion to momentum),
is of such paramount importance.

A smooth steadily accelerating cast, with the line under steadily
increasing tension, gives the best cast. A haul adds tension, and
increases rod tip speed.

Quite a few ramifications actually, I am presently graphing the
information, and trying to propgram a range of simulations to which I
may ad actual data from various rods, lines etc.

TL
MC

wrote:

This theory, and the related equations, are my original work, if you
use it, please credit where you got it from.

I suggest that you submit it for publication -- to The Journal of
Irreproducible Results.

http://www.jir.com/

--
Cut "to the chase" for my email address.

rw wrote:
wrote:
This theory, and the related equations, are my original work, if you
use it, please credit where you got it from.

I suggest that you submit it for publication -- to The Journal of
Irreproducible Results.

http://www.jir.com/

LOL !

The Loony was better off when he was refusing to post anything
about fly fishing. He came off as just a loony crank but now
that he's decided to grace us with his gibberish he comes across
as a loony crank and a laughable crackpot. It reminds me of the
time Gehrke posted his scientific treatise on the chemistry of
floatants and everybody who'd ever taken a chemistry class beyond
high school laughed their asses off.

--
Ken Fortenberry