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Hauling, Rod-loading.



 
 
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  #31  
Old November 10th, 2008, 05:09 PM posted to rec.outdoors.fishing.fly
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Posts: 785
Default Hauling, Rod-loading.

Just in case you still haven´t grasped it;

http://img73.imageshack.us/my.php?image=nonameyt7.jpg

  #32  
Old November 10th, 2008, 05:18 PM posted to rec.outdoors.fishing.fly
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Posts: 785
Default Hauling, Rod-loading.

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?
  #33  
Old November 10th, 2008, 05:19 PM posted to rec.outdoors.fishing.fly
rw
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Posts: 1,773
Default 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.


Hey, he's using Newtons! Can't you read, dumbo?

--
Cut "to the chase" for my email address.
  #34  
Old November 10th, 2008, 05:19 PM posted to rec.outdoors.fishing.fly
Tim Lysyk
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Posts: 179
Default Hauling, Rod-loading.

wrote:
On Nov 10, 5:50 pm, Scott Seidman wrote:
wrote in news:af6faa37-7c18-480a-ba93-
:

Probably a good idea, teachers should have open minds, and actually
look at theories and equations before they simply trash them out of
hand.
You still have not said a single sensible word about the theory, or
the equations. The only possible conclusions are, that you either don
ït want to do so because you are stupid and biased, or you are just
too stupid to so so at all.
I feel sorry for your students.

I haven't looked through your equations, because I don't know why I
should. If a student handed me a manifesto like that, I'd hand it back
and ask him or her to do a better job.

Step one would be to tell your reader WHY you are doing this exercise.
What do you expect to show us? A model that you are asking a question is
a valuable tool, but a model in a vacuum is masturbation.

I gave up even trying to figure this out when I saw acceleration in units
of "ms^2" instead of ms^(-2)

--
Scott
Reverse name to reply


That is the standard European unit notation for Newtons. That should
be quite obvious, and as you correctly translated the unit you
mentioned to the standard American notation ms^(-2), ( which is also
merely a variation of standard mathematical notation, you can´t be
completely stupid.

http://de.wikipedia.org/wiki/Newton_(Einheit) ( Doubtless you can find
it in English as well).

It makes no difference to me whether you look at it or not, or whether
you try to trash it without looking at it, doing so merely
demonstrates your stupidity, inertia, and bias. Those are not very
good traits for somebody who purportedly teaches graduate engineering.

The reason for the exercise is to discover the optimum casting stroke
and demonstrate it mathematically for any given parameters. There are
other reasons as well, testing various combinations of rod and line,
and building a graphic simulation of the process.

This is already in hand.

Nobody is forcing you to look at or believe anything at all. If you
merely wish to demonstrate how stupid you are, by trashing something
without either looking at it, or understanding it, that is just fine
with me.

You STILL have not written a single sensible word in regard to either
the theory, or the equations. Just more silly bull****.


I think Scott was refering your phrase: "Fa = the acceleration of the
line in ms²". Acceleration is usually expressed as change in disrance
per some unit of time, such as m/(s^2), or meters divided by seconds
squared. You have it as meters times seconds squared.

Tim Lysyk
  #36  
Old November 10th, 2008, 05:24 PM posted to rec.outdoors.fishing.fly
[email protected]
external usenet poster
 
Posts: 785
Default Hauling, Rod-loading.

On Nov 10, 6:19*pm, Tim Lysyk wrote:
wrote:
On Nov 10, 5:50 pm, Scott Seidman wrote:
wrote in news:af6faa37-7c18-480a-ba93-
:


Probably a good idea, teachers should have open minds, and actually
look at theories and equations before they simply trash them out of
hand.
You still have not said a single sensible word about the theory, or
the equations. *The only possible conclusions are, that you either don
ït want to do so because you are stupid and biased, or you are just
too stupid to so so at all.
I feel sorry for your students.
I haven't looked through your equations, because I don't know why I
should. *If a student handed me a manifesto like that, I'd hand it back
and ask him or her to do a better job.


Step one would be to tell your reader WHY you are doing this exercise. *
What do you expect to show us? *A model that you are asking a question is
a valuable tool, but a model in a vacuum is masturbation.


I gave up even trying to figure this out when I saw acceleration in units
of "ms^2" instead of ms^(-2)


--
Scott
Reverse name to reply


That is the standard European unit notation for Newtons. That should
be quite obvious, and as you correctly translated the unit you
mentioned to the standard American notation ms^(-2), *( which is also
merely a variation of standard mathematical notation, you can´t be
completely stupid.


http://de.wikipedia.org/wiki/Newton_(Einheit) ( Doubtless you can find
it in English as well).


It makes no difference to me whether you look at it or not, or whether
you try to trash it without looking at it, doing so merely
demonstrates your stupidity, inertia, and bias. Those are not very
good traits for somebody who purportedly teaches graduate engineering.


The reason for the exercise is to discover the optimum casting stroke
and demonstrate it mathematically for any given parameters. There are
other reasons as well, testing various combinations of rod and line,
and building a graphic simulation of the process.


This is already in hand.


Nobody is forcing you to look at or believe anything at all. If you
merely wish to demonstrate how stupid you are, by trashing something
without either looking at it, or understanding it, that is just fine
with me.


You STILL have not written a single sensible word in regard to either
the theory, or the equations. Just more silly bull****.


I think Scott was refering your phrase: "Fa = the acceleration of the
line in ms²". *Acceleration is usually expressed as change in disrance
per some unit of time, such as m/(s^2), or meters divided by seconds
squared. You have it as meters times seconds squared.

Tim Lysyk


That is not the case, you obviously use different notation.

The notation I use is the standard SI notation, as shown here;
http://img73.imageshack.us/my.php?image=nonameyt7.jpg

If you wish to use your notation, then go ahead, it is simple to
convert to whatever notation you please, merely substitute the units
you do not agree with. Makes no difference at all to the equations.

  #38  
Old November 10th, 2008, 05:41 PM posted to rec.outdoors.fishing.fly
rw
external usenet poster
 
Posts: 1,773
Default Hauling, Rod-loading.

wrote:
On Nov 10, 6:19 pm, Tim Lysyk wrote:

wrote:

On Nov 10, 5:50 pm, Scott Seidman wrote:

wrote in news:af6faa37-7c18-480a-ba93-
:


Probably a good idea, teachers should have open minds, and actually
look at theories and equations before they simply trash them out of
hand.
You still have not said a single sensible word about the theory, or
the equations. The only possible conclusions are, that you either don
ït want to do so because you are stupid and biased, or you are just
too stupid to so so at all.
I feel sorry for your students.

I haven't looked through your equations, because I don't know why I
should. If a student handed me a manifesto like that, I'd hand it back
and ask him or her to do a better job.


Step one would be to tell your reader WHY you are doing this exercise.
What do you expect to show us? A model that you are asking a question is
a valuable tool, but a model in a vacuum is masturbation.


I gave up even trying to figure this out when I saw acceleration in units
of "ms^2" instead of ms^(-2)


--
Scott
Reverse name to reply


That is the standard European unit notation for Newtons. That should
be quite obvious, and as you correctly translated the unit you
mentioned to the standard American notation ms^(-2), ( which is also
merely a variation of standard mathematical notation, you can´t be
completely stupid.


http://de.wikipedia.org/wiki/Newton_(Einheit) ( Doubtless you can find
it in English as well).


It makes no difference to me whether you look at it or not, or whether
you try to trash it without looking at it, doing so merely
demonstrates your stupidity, inertia, and bias. Those are not very
good traits for somebody who purportedly teaches graduate engineering.


The reason for the exercise is to discover the optimum casting stroke
and demonstrate it mathematically for any given parameters. There are
other reasons as well, testing various combinations of rod and line,
and building a graphic simulation of the process.


This is already in hand.


Nobody is forcing you to look at or believe anything at all. If you
merely wish to demonstrate how stupid you are, by trashing something
without either looking at it, or understanding it, that is just fine
with me.


You STILL have not written a single sensible word in regard to either
the theory, or the equations. Just more silly bull****.


I think Scott was refering your phrase: "Fa = the acceleration of the
line in ms²". Acceleration is usually expressed as change in disrance
per some unit of time, such as m/(s^2), or meters divided by seconds
squared. You have it as meters times seconds squared.

Tim Lysyk



That is not the case, you obviously use different notation.

The notation I use is the standard SI notation, as shown here;
http://img73.imageshack.us/my.php?image=nonameyt7.jpg


Uh, unless I'm hallucinating, that web site contradicts your "notation."

--
Cut "to the chase" for my email address.
  #39  
Old November 10th, 2008, 05:47 PM posted to rec.outdoors.fishing.fly
rw
external usenet poster
 
Posts: 1,773
Default Hauling, Rod-loading.

wrote:
On Nov 10, 5:50 pm, Scott Seidman wrote:

wrote in news:af6faa37-7c18-480a-ba93-
:


Probably a good idea, teachers should have open minds, and actually
look at theories and equations before they simply trash them out of
hand.


You still have not said a single sensible word about the theory, or
the equations. The only possible conclusions are, that you either don
ït want to do so because you are stupid and biased, or you are just
too stupid to so so at all.


I feel sorry for your students.


I haven't looked through your equations, because I don't know why I
should. If a student handed me a manifesto like that, I'd hand it back
and ask him or her to do a better job.

Step one would be to tell your reader WHY you are doing this exercise.
What do you expect to show us? A model that you are asking a question is
a valuable tool, but a model in a vacuum is masturbation.

I gave up even trying to figure this out when I saw acceleration in units
of "ms^2" instead of ms^(-2)

--
Scott
Reverse name to reply



That is the standard European unit notation for Newtons.


Quite right. In a little-known act, the European Union, in the spirit of
notational simplification and standardization, decreed that the s^2 be
moved from the denominator to the numerator, saving one (or in some
cases three) characters.

Unfortunately, an amendment to repeal the Law of Gravity was tabled.

--
Cut "to the chase" for my email address.
  #40  
Old November 10th, 2008, 05:47 PM posted to rec.outdoors.fishing.fly
rb608
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Posts: 681
Default Hauling, Rod-loading.

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.
 




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