Hauling, Rod-loading.

On Nov 10, 5:19*pm, Scott Seidman wrote:
wrote in news:29b4afb1-d366-4ff8-9f76-
:

Which being the case, you will doubtless have no trouble at all
refuting it scientifically eh?

No real need. *I can't even follow your plain text equations because of
formatting issues, but I don't see any derivatives in time or space, so I
know whatever point approximations you're making are probably overly
simplistic.

--
Scott
Reverse name to reply

If you are unable to follow it, then how do you know it is wrong?

You are just bull****ting again. If you don´t know what kgm/s² means,
then you are too dumb to follow the equations anyway. The unit
contains the space time derivatives. In longhand;

kilogrammes per meter divided by seconds squared. ( Aka, "Newtons")

Obviously you are even too dumb to tie a line to a fencepost.

Makes no difference to me what you dumbos write, but for anybody
interested, and capable of understanding it, it may be useful.

Also, the more you dumb assholes react without actually knowing or
saying anything at all germane, purely for "personal" reasons,
independent of the facts at issue, the more sensible people assume
that it is worth looking at.

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.

On Nov 10, 6:04*pm, 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.

The coefficient of fluid friction, usually indicated by " µk " is a
dimensionless scalar value.

The rest of what you have written makes no sense in regard to the
equations and explanations I gave.

The units used for tension = Newtons. The units used for force on the
rod tip = Newtons. The rest is self explanatory. As indeed is the
theory itself after a little thought.

I am not here to give people basic algebra, nor to explain
mathematical concepts.

The theory itself has already proven extremely valuable in teaching
people to cast.

If you don´t understand it, then that is your problem not mine.

Do you understand E=mc² or is it too simplistic for you?

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!

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

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. ;-)

--
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 Nov 10, 12:18*pm, wrote:

The coefficient of fluid friction, usually indicated by " µk " is a
dimensionless scalar value.

That being the case, the units will not match on both sides of the
equation, at least not as you've described the variables. And if it's
dimensionless, how do you account for its variability with velocity?
Is it friction drag or pressure drag?

The units used for tension = Newtons. *The units used for force on the
rod tip = Newtons. *The rest is self explanatory. As indeed is the
theory itself after a little thought.

What should be self explanatory is that the rod tip Newtons on the
left are cancelled by the tension term Newtons on the right, leaving
the left side unitless with kg.m^(-2) on the right. That doesn't
work.

I am not here to give people basic algebra, nor to explain
mathematical concepts.

Okay.


The theory itself has already proven extremely valuable in *teaching
people to cast.

I have not discussed your theory, only with your mathematical modeling
of it. Anybody who could teach me to double haul would have my
respect; but not by insulting my intelligence.,


Do you understand E=mc² * or is it too simplistic for you?

No, I'm down wit dat. As my #20 Griffith's Gnat approaches c, it
approaches the size of a #6 and the time it takes to reach the trout
slows down, giving me more time for an upstream mend; but it doesn't
matter because my 3 wt rod now weighs 12,000,000 Newtons.

rb608 wrote:
On Nov 10, 12:18 pm, wrote:

The coefficient of fluid friction, usually indicated by " µk " is a
dimensionless scalar value.

That being the case, the units will not match on both sides of the
equation, at least not as you've described the variables. And if it's
dimensionless, how do you account for its variability with velocity?
Is it friction drag or pressure drag?

The units used for tension = Newtons. The units used for force on the
rod tip = Newtons. The rest is self explanatory. As indeed is the
theory itself after a little thought.

What should be self explanatory is that the rod tip Newtons on the
left are cancelled by the tension term Newtons on the right, leaving
the left side unitless with kg.m^(-2) on the right. That doesn't
work.

I am not here to give people basic algebra, nor to explain
mathematical concepts.

Okay.


The theory itself has already proven extremely valuable in teaching
people to cast.

I have not discussed your theory, only with your mathematical modeling
of it. Anybody who could teach me to double haul would have my
respect; but not by insulting my intelligence.,


Do you understand E=mc² or is it too simplistic for you?

No, I'm down wit dat. As my #20 Griffith's Gnat approaches c, it
approaches the size of a #6 and the time it takes to reach the trout
slows down, giving me more time for an upstream mend; but it doesn't
matter because my 3 wt rod now weighs 12,000,000 Newtons.

One of the reasons for making the theory/equations publicly available
was to check for any errors. I don´t know what you are seeing on your
screens. I have checked it several times, and it seems OK to me, also in
a word processor, and as a normal text file.

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 1m/s²

That gives;

30g * 1m/s² * 0.3 = Flt / 0.01kgm/s²

30g * 1ms² = 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 * 1ms² * 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 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

The model arises from the basic theory, and nobody has yet disputed the
equations. Nevertheless, I will have another close look at them.

The point is, that the static rod loading is directly proportional to
the line tension when the line is fixed. As tension is what keeps the
line in the air in a dynamic ( or a static ) system, ( line not fixed
at a point, such as a fence post), then the dynamic line tension is a
function of the rod tip acceleration against the mass ( inertia) of the
line.

Initial target is to find out the line tension for any given set of
variables.

In a static system, ( which is easily measured), the line tension is
directly proportional to the force on the line. The force on the line (
tension in Netwons, kgm/s²) is equal and opposite to the force on the
rod. ( No acceleration as such, in the sense that there is no or very
little movement, but of course the unit of acceleration may be applied).
That is to say, the line does not go anywhere, even though force is
applied, that is why the result of the equation is simply mass in grams.

So, if the force on the end of the line is (say) 10 grams/meter/s², then
the force on the rod tip is also 10 grams/meter/s², and total rod
loading is also 10 grams/meter/s².

Line tension increases drastically ( in the static system) when the rod
is loaded further by bending it against the fixed line. If the point is
not fixed, then line tension still increases, but is then also related
to the acceleration of the rod tip.

Tension can only increase if the rod tip continues to accelerate.
However, the rod loading remains constant at 10 grams/meter/s²

The line moves faster the greater the tension, and it can only move
efficiently when the rod is accelerated constantly throughout the
acceleration phase. Any slack or faltering in the application of force
results in less line tension, which results in the line slowing down.

Anyway, if you tell people that a haul is designed to increase line
tension, then they do it instinctively more or less immediately.

They don´t need to know anything about rod loading, or acceleration,
they just need to know that they have to accelerate the rod smoothly and
constantly, and the haul should also be smooth and constantly accelerating.

This has worked for everybody I have shown it to. I teach most people to
double haul in about two to three hours. I show it to them, explain it,
and they do the rest themselves.

Whether the mathematical model is entirely accurate, I am no longer
sure. I will have to check it again. Also, there are obviously problems
with various formats and notations. Complicated further because I simply
don´t know what you are seeing.

One last try at the basic equation.

The line is attached to a fixed point, the rod is bent until the force
on the fixed point is 10grams/meter/s²

The tension on the static rod, and on the static line are then
identical. Bending the rod further against the line causes an increase
in line tension which is proportional to the rod and line tension, and
directly proportional to it.

If the force on the fixed point is 100 grams/meter/s² then the rod
loading is the same, and so is the line tension.

Hauling the line at this point causes a massive increase in line
tension, this happens even without a rod, simply by pulling on the line
directly. In this case, the rod loading ( bend) is also proportional to
the angle of the rod and line, when the line is pulled.

All this gets very complex indeed, but the equation seems simple enough.
If you have a better idea to model it, then go ahead.

On Mon, 10 Nov 2008 19:35:49 +0100, "
wrote:

rb608 wrote:
On Nov 10, 12:18 pm, wrote:

The coefficient of fluid friction, usually indicated by " µk " is a
dimensionless scalar value.

That being the case, the units will not match on both sides of the
equation, at least not as you've described the variables. And if it's
dimensionless, how do you account for its variability with velocity?
Is it friction drag or pressure drag?

The units used for tension = Newtons. The units used for force on the
rod tip = Newtons. The rest is self explanatory. As indeed is the
theory itself after a little thought.

What should be self explanatory is that the rod tip Newtons on the
left are cancelled by the tension term Newtons on the right, leaving
the left side unitless with kg.m^(-2) on the right. That doesn't
work.

I am not here to give people basic algebra, nor to explain
mathematical concepts.

Okay.


The theory itself has already proven extremely valuable in teaching
people to cast.

I have not discussed your theory, only with your mathematical modeling
of it. Anybody who could teach me to double haul would have my
respect; but not by insulting my intelligence.,


Do you understand E=mc² or is it too simplistic for you?

No, I'm down wit dat. As my #20 Griffith's Gnat approaches c, it
approaches the size of a #6 and the time it takes to reach the trout
slows down, giving me more time for an upstream mend; but it doesn't
matter because my 3 wt rod now weighs 12,000,000 Newtons.

One of the reasons for making the theory/equations publicly available
was to check for any errors. I don´t know what you are seeing on your
screens. I have checked it several times, and it seems OK to me, also in
a word processor, and as a normal text file.

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 1m/s²

That gives;

30g * 1m/s² * 0.3 = Flt / 0.01kgm/s²

30g * 1ms² = 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 * 1ms² * 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 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

The model arises from the basic theory, and nobody has yet disputed the
equations. Nevertheless, I will have another close look at them.

The point is, that the static rod loading is directly proportional to
the line tension when the line is fixed. As tension is what keeps the
line in the air in a dynamic ( or a static ) system, ( line not fixed
at a point, such as a fence post), then the dynamic line tension is a
function of the rod tip acceleration against the mass ( inertia) of the
line.

Initial target is to find out the line tension for any given set of
variables.

In a static system, ( which is easily measured), the line tension is
directly proportional to the force on the line. The force on the line (
tension in Netwons, kgm/s²) is equal and opposite to the force on the
rod. ( No acceleration as such, in the sense that there is no or very
little movement, but of course the unit of acceleration may be applied).
That is to say, the line does not go anywhere, even though force is
applied, that is why the result of the equation is simply mass in grams.

So, if the force on the end of the line is (say) 10 grams/meter/s², then
the force on the rod tip is also 10 grams/meter/s², and total rod
loading is also 10 grams/meter/s².

Line tension increases drastically ( in the static system) when the rod
is loaded further by bending it against the fixed line. If the point is
not fixed, then line tension still increases, but is then also related
to the acceleration of the rod tip.

Tension can only increase if the rod tip continues to accelerate.
However, the rod loading remains constant at 10 grams/meter/s²

The line moves faster the greater the tension, and it can only move
efficiently when the rod is accelerated constantly throughout the
acceleration phase. Any slack or faltering in the application of force
results in less line tension, which results in the line slowing down.

Anyway, if you tell people that a haul is designed to increase line
tension, then they do it instinctively more or less immediately.

They don´t need to know anything about rod loading, or acceleration,
they just need to know that they have to accelerate the rod smoothly and
constantly, and the haul should also be smooth and constantly accelerating.

This has worked for everybody I have shown it to. I teach most people to
double haul in about two to three hours. I show it to them, explain it,
and they do the rest themselves.

Whether the mathematical model is entirely accurate, I am no longer
sure. I will have to check it again. Also, there are obviously problems
with various formats and notations. Complicated further because I simply
don´t know what you are seeing.

One last try at the basic equation.

The line is attached to a fixed point, the rod is bent until the force
on the fixed point is 10grams/meter/s²

The tension on the static rod, and on the static line are then
identical. Bending the rod further against the line causes an increase
in line tension which is proportional to the rod and line tension, and
directly proportional to it.

If the force on the fixed point is 100 grams/meter/s² then the rod
loading is the same, and so is the line tension.

Hauling the line at this point causes a massive increase in line
tension, this happens even without a rod, simply by pulling on the line
directly. In this case, the rod loading ( bend) is also proportional to
the angle of the rod and line, when the line is pulled.

All this gets very complex indeed, but the equation seems simple enough.
If you have a better idea to model it, then go ahead.


What about this, Dave...should it be sorted...?

HTH,
R

rb608 wrote:

No, I'm down wit dat. As my #20 Griffith's Gnat approaches c, it
approaches the size of a #6 and the time it takes to reach the trout
slows down, giving me more time for an upstream mend; but it doesn't
matter because my 3 wt rod now weighs 12,000,000 Newtons.

And it would collapse into a spinning black hole, emitting superheated
gas jets with a momentum approaching a nanoconnor.

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