Hauling, Rod-loading.

rb608 wrote:
On Nov 10, 6:22 am, 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


Now, I admittedly can't haul worth a ****, but I do know my physics
(maybe that's my problem.) I'm having a bit of difficulty reconciling
your theory above with that of Mr. Newton who opined a much simpler
equation, F=ma.

Force equals mass times acceleration. That's it.

The extra terms you included (Ff and Flt) are misplaced IMO. Under no
circumstances would they be multiplication terms in the equation. The
line tension is just another Force that would be part of the net force
on the tip of the rod, not a separate term.

The air resistance term (what units are you thinking for that one?)
also does not belong there. Any resisting forces due to fluid
friction would similarly be a part of the net value of the Force at
the rod tip. It does not belong in the equation as a multiplicative
term.

As I said, I can't haul; but I fear the inaccuracies of your
mathematical equation may be detracting from your practical
instruction.

Line tension is a "special" force, considered to be parallel to the
string under tension. However, as the line acceleration is dependent on
this tension, it must be proportional to it.

The force on the rof tip, and the force on the end of the line are
"normal" forces and can be measured directly.

Friction and line tension both affect the acceleration of the line when
it is moving, and have to be included in the dynamic equation.

For a static system they can be ignored.

Try again;

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 m/s², 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/s² = 30g * Fa * Ff * Flt

Second equation;

Flt = 0,01 kgm/s² / 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 * 1m/s² * 0.3 = Flt / 0.01kgm/s²

30g * 1m/s² = 0,03kgm/s²

* 0,3 = 0.09 kgm/s²

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 * 1m/s² * 0.3 * 9kgm/s²

Second equation;

Flt = Frt / fi * Fa * Ff

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

Third equation;

Fi * Fa * Ff = Flt / Frt

0,03 kgm/s² * 1 m/s² * 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

wrote:
Try again;

Please, stop. I thought your poetry was bad enough.

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

rw wrote in
m:

Women are the product of time and money:

women = time * money

A friend of mine authored a JIR paper on "cliche contradiction",
mathematically resolving "absence makes the heart grow fonder" with "out of
sight, out of mind". It made one of the "best of" books.



--
Scott
Reverse name to reply

Scott Seidman wrote:
rw wrote in
m:


Women are the product of time and money:

women = time * money


A friend of mine authored a JIR paper on "cliche contradiction",
mathematically resolving "absence makes the heart grow fonder" with "out of
sight, out of mind". It made one of the "best of" books.




I stole that from Wikipedia.

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

Ignoring friction for the moment;

Frt = Fi * Fa * Flt

Which gives;

0,01 kgm/s² = 30g *Fa *Flt

Therefore,

Flt=0,01 kgm/s² * Fa

And,

30g+Fa=Flt/0,01kgm/s²

Assuming an acceleration of 1m/s²

30g*1m/s²=Flt/0,001kgm/s²

30g*1m/s²=0,03kgm/s²

Therefore,

Flt= 0,03kgm/s²

That seems to be correct. I will try to graph that and see if it
correlates with my other stuff.

wrote:
Some considerations on casting and rod loading.

A simple calculation for casting in air is;

Frt = Fi * Fa * Ff * Flt

snipped mucho geeko

So, if I memorize all that will I no longer get wind knots?

Ever hopeful,

Russell

Who is mathematically challenged.

Russell D. wrote:
wrote:
Some considerations on casting and rod loading.

A simple calculation for casting in air is;

Frt = Fi * Fa * Ff * Flt

snipped mucho geeko

So, if I memorize all that will I no longer get wind knots?

Ever hopeful, ...

You better hold off on the memorizing for awhile there, Russell.
Turns out The Loony's Hauling and Loading treatise is a work in
progress. At least that's the latest story out of Loonyville. At
first all of roff was stupid, then he was using European notation
(LOL ;-), then his computer didn't do the formatting correctly,
now finally he admits it's full of errors and miscalculations but
that's exactly why he posted it here in the first place. To subject
it to peer review. But it's already been published to much acclaim,
he just won't say where.

There's no one as good at shameless lying on the fly than His Loony
Mikeness. And now he's frantically canceling most of his posts in
this thread so that in the future any remembrance of this pathetic
episode will be a baldfaced lie. Quite the piece of work our Loony.

--
Ken Fortenberry

In article ,
says...
rw wrote:
wrote:
Do you understand E=mc² or is it too simplistic for you?

Of course! Now I get it! You're casting with relativistic velocity and
converting the energy of the haul into mass, resulting in much better
shooting. Brilliant!

Casting at the speed of light. What an elegantly simple solution
to the double haul.

I'm gonna need to hit the gym. ;-)



And you will need to buy the new Winston Plasma Rod (US$ 1,230,895.95
retail), hand-made in Twin Bridges, MT by workers with thick asbestos
gloves.

--
Kevin Vang
reply to kevin dot vang at minotstateu dot edu

On Mon, 10 Nov 2008 03:22:51 -0800 (PST),
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

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


Um, Dave, shouldn't you be sorting all of this...?

HTH,
R