short bus, the example I gave is accurate, if understated. the forces on a
massless line with weight in center become huge **********VERY************
quickly at the line becomes tighter.

The forces on a catenary are even more huge and become worse even faster.

The math is easier on a massless line than a catenary, that is all.

NOYB"
Date: 9/18/2004 10:04 PM Eastern Daylight Time
Message-id:


"Short Wave Sportfishing" wrote in message
.. .
On Sat, 18 Sep 2004 19:46:04 -0400, Gene Kearns
wrote:

On 18 Sep 2004 21:15:33 GMT, (JAXAshby) wrote:

What happens during the interaction of forces on the rode would be
most fascinating.

a way to simplified look at it is to consider the chain/rode/line to
have zero
weight pulled between two points (say 100 feet apart), then hang a 1#
weight in
the center point and check how much strain it put on the end points when
the
weight hangs 20 feet, then 10 feet, then 5 feet, then 1 foot, then 1
inch, then
1/10th inch. Just use trig to figure the forces.

So.... we just used intuitive trig to figure out why (1) we use scope
with an anchor and (2) why we don't tie boats to the dock with chain.
Now *that* is some real science......

And your "simplified look" does not apply.... an anchor rode does not
employ both ends at the same "Y" value.... therefore assumptions of
Y=Y'=0 do not obtain and is, therefore, the root cause of your lack
of understanding in this area. There isn't *anything* *attached* to
the middle.

But wouldn't the strain be equal at the arthimetical center and can be
equated to weight? It's really just another to figure energy
transfer, right?

I'm not totaly familiar with this so if I mess this up, it's an
electronic engineer with a math degree playing at mechanics, but
catenary defined means the shape of the line (or in this case rode) as
a curve. A funciton of strain would be weights at either end. Strain
can be defined as stored energy which is, I would think, distributed
evenly along the line to the end points. One way to define how much
strain is being applied would be to add weight to the middle and
measure the deflection.

At that point, it becomes a trig function - yes/no?


Yes, assuming the line itself has negligible mass compared to the weight
pulling on it...which is not a practical assumption when comparing it to an
anchor line and chain rode. I don't know why jackassby even used this
example. Jackassby's example describes a straight line...not a catenary.
It is not even close to replicating what is happening to an anchor line and
chain rode. The only time that Jax's example *may* similate an anchor line
is when the force of the wind is so great that the line and chain is
perfectly straight...which *never* happens in a real world situation anyhow.
If the wind pulled so hard that the line was perfectly straight, it wouldn't
be a catenary any longer...it would be a straight line. And if it were a
straight line, the vertical component of force exerted by the boat on the
anchor would be so high that the anchor would pull loose. Go look at the
website Gene posted:
http://alain.fraysse.free.fr/sail/ro...ic/sta_hom.htm

It explains it all.