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.