Hauling, Rod-loading.

On Nov 10, 2:22*pm, rb608 wrote:
It's my contention, however, the the force terms should be grouped,
and that the unbalanced force causing the line acceleration is
actually (Frt - Flt). *This allows your model to fit into the F=ma
equation as:

(Frt - Flt) = Fi * Fa


Having said this however, I'd clarify that in reality, the line
tension Flt and the force on the rod tip (parallel to the line) Frt
are the same force (in opposite directions), because it's the rod tip
that imparts that force to the line. One needs to examine exactly
where one wishes to consider the free-body diagram. Because we're
looking at how the line accelerates, we can leave the rod out of the
equation altogether. The unbalanced force imparted as line tension by
the rod tip F will cause the mass of the line m to accelerate at a
rate a. Because m is constant, more F causes faster a. It's not a
static model because a0. It's a lot simpler mathematically than it
is with a rod in your hands (well, my hands anyway.)

rb608 wrote:
On Nov 10, 2:22 pm, rb608 wrote:
It's my contention, however, the the force terms should be grouped,
and that the unbalanced force causing the line acceleration is
actually (Frt - Flt). This allows your model to fit into the F=ma
equation as:

(Frt - Flt) = Fi * Fa


Having said this however, I'd clarify that in reality, the line
tension Flt and the force on the rod tip (parallel to the line) Frt
are the same force (in opposite directions), because it's the rod tip
that imparts that force to the line. One needs to examine exactly
where one wishes to consider the free-body diagram. Because we're
looking at how the line accelerates, we can leave the rod out of the
equation altogether. The unbalanced force imparted as line tension by
the rod tip F will cause the mass of the line m to accelerate at a
rate a. Because m is constant, more F causes faster a. It's not a
static model because a0. It's a lot simpler mathematically than it
is with a rod in your hands (well, my hands anyway.)


What I basically want to do is to find out the proportions and relations
of line tension to rod loading for any given rod and any given mass of
line. One is of course obliged to assume an optimum cast as well, as
without the correct and constant, ( or steadily increasing) line
tension, acceleration is erratic, and losing tension at any point in the
cast causes it to fail, ( or be less than optimal).

Given that the basic theory is correct, then I only need to graph the
math and I should be able to plug in any data I want and achieve the
optimum by modeling.

Unfortunately, I seem to be having more trouble with the math modeling
than I should be.

It would seem obvious that the force required to tension the line is not
used to accelerate it.

Therefore, the force Lt = the force Rt minus the force Fa. Using the
actual data one can work out the rod loading for any given mass, line
tension etc, for any mass of line within the rod capabilities, among
other things. That´s really the point of the exercise, and also to
explain why and how a haul works.

This should also explain why a haul gives a much longer cast, as the
line tension is converted into momentum. But I would like to prove this
mathematically, not just theoretically and empirically.

wrote:

This should also explain why a haul gives a much longer cast, as the
line tension is converted into momentum. But I would like to prove this
mathematically, not just theoretically and empirically.

Anyway, I am going to give up for the nonce, and go through it all again
later. I have concentrated so hard on it that I am merely confusing
myself now, and going round in circles. Right now I can´t even
substitute equations properly.

On Nov 10, 3:20*pm, "
wrote:
It would seem obvious that the force required to tension the line is not
used to accelerate it.

It is the same force. Ignore gravity and imagine a fly line suspended
horizontally. Add a force to one end. the line will accelerate in
that direction (F=ma) for as long as the force is applied. The longer
the duration of the force, the faster the line moves. The velocity at
any time is the acceleration times the time (v = at).

If you further acclerate the line with a haul, that adds to the
velocity imparted by the rod tip, and V = v(rod tip) + v(hauling
speed). Walla, greater line speed, longer casting distance. That's
a gross simplification without considering the flexure of the rod tip,
but it's the basic model.

As near as I can speculate based on my knowledge of the physics and my
ineptitude at actually doing it, my guess is there's a "magic" timing
and speed for which the haul effect is maximized without being mostly
attenuated by sudden tip flexure from the added load; but I'm
convinced in this endeavor, the art is of far more beauty than the
mathematics.

On Nov 10, 9:51*pm, rb608 wrote:
On Nov 10, 3:20*pm, "
wrote:

It would seem obvious that the force required to tension the line is not
used to accelerate it.

It is the same force. *Ignore gravity and imagine a fly line suspended
horizontally. *Add a force to one end. *the line will accelerate in
that direction (F=ma) for as long as the force is applied. *The longer
the duration of the force, the faster the line moves. *The velocity at
any time is the acceleration times the time (v = at).

If you further acclerate the line with a haul, that adds to the
velocity imparted by the rod tip, and *V = v(rod tip) + v(hauling
speed). * *Walla, greater line speed, longer casting distance. *That's
a gross simplification without considering the flexure of the rod tip,
but it's the basic model.

As near as I can speculate based on my knowledge of the physics and my
ineptitude at actually doing it, my guess is there's a "magic" timing
and speed for which the haul effect is maximized without being mostly
attenuated by sudden tip flexure from the added load; but I'm
convinced in this endeavor, the art is of far more beauty than the
mathematics.

The main problem with that is, that line "acceleration" is impossible
without line tension. Line tension is also what makes the line stay
straight, and also holds it in the air. Without tension it immediately
begins to fall or deform.

The force required to maintain tension in the line can not be used to
accelerate it. The added acceleration only appears when the line is
released at the correct moment, and that tension is also then
converted to momentum.

If you go back to the fence post model for a moment. Lay a line on the
ground and tie one end to the top of the post. Slowly pull the line
taut. The first thing that happens is that the line lifts into the air
as a result of the tension on it.

The same thing happens if you accelerate the line without it being
fixed, but in this case against the inertia of the line itself.

The second thing that happens is that the line exerts a force on both
ends, on the post, and on your hand. This force is directly
proportional to the tension in the line. The tension in the fence
post model does not accelerate the line in the accepted sense, it is
fixed between two points, it is merely a force present at all points
parallel to the line. If you let go of the line suddenly, it will
spring away from you as tension is released as the the tension is
converted to momentum. The quicker you release it, the faster it jumps
away.

The rod spring allows you to force the release of tension more or less
immediately, also adding the force of the rod spring itself, and when
casting with a fixed line length, it is this tension which allows you
to cast at all. Without it, the line simply falls to the ground.

It is the tension which causes the line to roll out in a loop. The
line as a whole does not actually go anywhere independently, as if you
were throwing a stone for instance, it simply rolls out under
tension. If there is sufficient momentum left, after the main roll
out, and you release the line, it will keep on going for a while until
it loses tension and momentum.( Shooting). Even after you have
released the line on a shoot, there is still some tension on the rear
end of the line as it pulls the backing. Without this tension, ( which
is also the reason for rear tapers on shooting heads etc) the line
would not turn over in a loop, it would ( does!) simply crumple and
start to fall.

The reason the haul gives such a massive increase in distance, or
forces turnover into a wind etc, is that the greater the line tension,
the straighter the line, and the less fluid friction, and the faster
the line rolls out. The line is not "accelerated" by the haul at all
in the accepted sense, and the concept of "line speed" is a complete
misnomer in my opinion. Even a small fast haul adds a lot of tension
to the line which makes it straighter, and also adds considerably to
the momentum which eventually accrues if the line is released. ( When
shooting, casting heads etc). When casting "normally", the line is not
released at all, ( and should not be, as this decreases tension), but
simply rolls out to it´s end and then falls out of the air as tension
decreases.

Once you know that you are trying to "roll the line out" under
tension, then a lot of difficulties seem to simply fade away. using
this concept I have taught people to cast very well in a couple of
hours, and several other instructors I know now use the method
exclusively.

You don´t have to think about a lot of "steps" in the cast ( there are
none anyway it is a complete fluid motion), you don´t learn a lot of
mistakes, and you don´t have to think about much at all. Most people
do it instinctively once they grasp the concept.

Give it a try, I am sure you will be most pleasantly surprised. It is
harder for a caster with ingrained faults and muscle memory to do,
becaue he has to "unlearn" or forget everything else first, but it is
still a lot easier than trying to "improve" poor casting by adding
more steps and rules etc.

Think of the cast as one fluid action, where you are trying to roll
the line out under tension. Forget everything else, rod positions,
hand positions etc etc. Just do it.

Lastly, the maths are not at all necessary, but I would like
mathematical proof as well, I don´t tell beginners any math or
anything like that. I basically tell them what I have just told you.
Most are casting pretty well within an hour.

TL
MC

This guy, who is also a champion caster in a number of disciplines,
uses the same basic teaching technique;

http://www.ukswff.co.uk/Casting4.shtml

But don´t start trying to analyse lots of things before you have got
the simple easy casting stroke down. Just roll the line out under
tension.

TL
MC

rb608 wrote:
On Nov 10, 3:20 pm, "
wrote:
It would seem obvious that the force required to tension the line is not
used to accelerate it.

It is the same force. Ignore gravity and imagine a fly line suspended
horizontally. Add a force to one end. the line will accelerate in
that direction (F=ma) for as long as the force is applied. The longer
the duration of the force, the faster the line moves. The velocity at
any time is the acceleration times the time (v = at).

If you further acclerate the line with a haul, that adds to the
velocity imparted by the rod tip, and V = v(rod tip) + v(hauling
speed). Walla, greater line speed, longer casting distance. That's
a gross simplification without considering the flexure of the rod tip,
but it's the basic model.

As near as I can speculate based on my knowledge of the physics and my
ineptitude at actually doing it, my guess is there's a "magic" timing
and speed for which the haul effect is maximized without being mostly
attenuated by sudden tip flexure from the added load; but I'm
convinced in this endeavor, the art is of far more beauty than the
mathematics.






















There are other ways to explain it. Consider a steel cable. Everybody
knows what happens when a steel cable under tension breaks, or is
released. It snakes away with very considerable force. We have all heard
horror stories of such a cable cutting a man in half, or decapitating
someone.

This is purely the result of tension on the cable. As soon as it is
released, the tension is converted to kinetic energy.

It is my contention that precisely the same thing happens to a fly-line.
Of course the fly-line is on a smaller scale, and has less mass than a
steel cable, but in my opinion the principle is exactly the same.

Here thr main point is that the tension alone is not sufficent, once
under tension, the line and rod must be accelerated while maintaining
that tension.

All faults can be traced directly to the lack of, or erratic line
tension. Tailing loops are caused by concave movements of the rod tip,
whihc causes a curve in the line, a curved line is not under tension
like a straight one, Indeed, being staright is a prerequisite of
obtaining the tension, and thus allowing acceleration of the line and
rod tip.

The other points, most of which I have not yet mentioned to a wider
public, all confirm the theory, and allow one to easily explain what
happens while casting, and also explain all the faults, and why they occur.