Where is the catenary on a yo-yo hanging from a finger? It is a line
between 2 points.

"JAXAshby" wrote in message
...
clown, check the meaning of the word.

Gene Kearns
Date: 9/18/2004 7:22 PM Eastern Daylight Time
Message-id:

On 18 Sep 2004 20:49:57 GMT, (JAXAshby) wrote:

spell it any way you want, you still do not have any idea what it is.

yo-yo, a string strung between two points has, and always has, a
catenary.
go
look it up.

And a yo-yo is about the depth of your comprehension, here....

--



Grady-White Gulfstream, out of Southport, NC.

http://myworkshop.idleplay.net/cavern/
Homepage
http://www.southharbourvillageinn.com/directions.asp Where
Southport,NC
is located.
http://www.thebayguide.com/rec.boats
Rec.boats
at Lee Yeaton's Bayguide

On Sat, 18 Sep 2004 22:04:06 -0400, "NOYB" wrote:


"Short Wave Sportfishing" wrote in message
.. .

On Sat, 18 Sep 2004 19:46:04 -0400, Gene Kearns

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.

Thanks for the link Gene, by the way, I missed that the first time
around.

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.

Well, yes, but the static load page is "under construction" and that's
what I was looking for, but hey - at least I have a better picture of
what's what.

Interesting about that kellet thing - that makes a lot of sense. I
knew the term, but I didn't know the term, know what I mean?

Thanks.

Later,

Tom

On Sun, 19 Sep 2004 02:24:40 GMT, "Calif Bill"
wrote:


"Short Wave Sportfishing" wrote in message
.. .
On Sat, 18 Sep 2004 20:14:34 -0400, "NOYB" wrote:


"Gene Kearns" wrote in message
.. .
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.


the forces get out of hand ********VERY******** quickly. Even worse,
is
that
the weight in the middle (or chain) has momentum as the boat rocks, so
the
"natural" position of the weight overshoots and makes for seriously
high
g-loads.


There is no weight "in the middle" (other than the weight of the rode)
.... so you put two anchors on the same rode? Odd.

Also, in jaxassby's example, the points can't always be 100' apart if the
weight is hanging further and further down each time...unless he has an
extremely elastic line and there's a large amount of stretch. I assume
that
jaxassby meant to say "using a 100' rope".

The main definition of catenary is that of a curve formed by a
perfectly flexible, uniformly dense, and inextensible cable suspended
from its endpoints. It would look a lot like hyperbolic cosine if you
graphed it out. Which, now that I think about it, wouldn't look a lot
like an anchor rode as much as a tow line.

I'm more curious about strain towards the middle of the curve. That
would be fairly easy to measure at either end, but if you have two
opposing forces of two different weights, say as in a barge tow, the
center strain would constantly move towards one or the other depending
on the weights.

How would you determine that mathematically?

The end points are not at the same elevation.

True, but then a hyperbolic curve does not necessarily have to have
equal level end points (in this case, height). It only has to have a
90º tangent at some point along the curve.

I think I'm getting one of my headaches again. I retired to get away
from all this stuff. :)

Later,

Tom
-----------
"Angling may be said to be so
like the mathematics that it
can never be fully learnt..."

Izaak Walton "The Compleat Angler", 1653

gene, go stand in the corner and keep quiet. the man asked a valid question
and got a valid answer. the fact that you are not capable of understanding
either the question or the answer means not a thing except that you are too
stupid to understand. now, be quiet. adults are having a conversation.

Gene Kearns
Date: 9/18/2004 7:46 PM Eastern Daylight Time
Message-id:

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.


the forces get out of hand ********VERY******** quickly. Even worse, is
that
the weight in the middle (or chain) has momentum as the boat rocks, so the
"natural" position of the weight overshoots and makes for seriously high
g-loads.


There is no weight "in the middle" (other than the weight of the rode)
.... so you put two anchors on the same rode? Odd.

Using that concept, most people use kellets and think it is a good and
useful idea.

--



Grady-White Gulfstream, out of Southport, NC.

http://myworkshop.idleplay.net/cavern/ Homepage
http://www.southharbourvillageinn.com/directions.asp Where Southport,NC
is located.
http://www.thebayguide.com/rec.boats Rec.boats
at Lee Yeaton's Bayguide

tom, you got it close enough. when the chain pulls tight the forces become
huge *********VERY********** quickly.

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?

the forces get out of hand ********VERY******** quickly. Even worse, is
that
the weight in the middle (or chain) has momentum as the boat rocks, so the
"natural" position of the weight overshoots and makes for seriously high
g-loads.


There is no weight "in the middle" (other than the weight of the rode)
.... so you put two anchors on the same rode? Odd.

Using that concept, most people use kellets and think it is a good and
useful idea.

What's a kellet?


Later,

Tom

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.

If the wind pulled so hard that the line was perfectly straight, it wouldn't
be a catenary any longer

a chain (or nylon rode, or a piece of cord string) ALWAYS has a catenary.
ALWAYS.

to pull the line straight requires infinite effort. and to pull it "just a
little bit more" when it is already stressed with a 1,000# pull is HUGELY more
than 1000#.

That is what breaks boats loose, or at least boats without 15 or 20 feet of
stretch left in the rode.

the vertical component of force exerted by the boat on the
anchor would be so high that the anchor would pull loose.

no ****. why do you think boats pull loose.

Thanks for the link Gene, by the way, I missed that the first time
around.

don't thank him, Tom. The site is like a Texan fraud. "Big hat, no cattle."

While the site uses a lot of math symbols, it is the work of someone who
understands damned little about catenaries AND thinks boats anchor in lite
winds always.

if you wish to anchor on mostly or all chain in a blow, you MUST hang LARGE
weights (plural) in the middle of the rode. otherwise, the forces on the
anchor and deck chocks get huge very quickly as the chain goes tight, and you
will break loose (which is not an Act of God, but rather the consequence of an
act of neglegence).

boy, you guys are stucking fupid. give you an example understandable by a
sophomore in high school and you STILL insist on anchoring in such a fashion as
to injure boaters around you. Did you ever consider some district attorney
might charge you with criminal neglegence if you hurt someone anchoring your
silly, lazy ass way (which, in fact, is your justification as to why you --
utterly unwilling to pull up an anchor by hand -- install an electric anchor
puller that works only with chain).

catenaires, and they strains imposed on the end points, is a freshman physics
discussion understood by tens of millions of 18 year old boys every year.

From: "NOYB"
Date: 9/18/2004 8:14 PM Eastern Daylight Time
Message-id:


"Gene Kearns" wrote in message
.. .
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.


the forces get out of hand ********VERY******** quickly. Even worse, is
that
the weight in the middle (or chain) has momentum as the boat rocks, so
the
"natural" position of the weight overshoots and makes for seriously high
g-loads.


There is no weight "in the middle" (other than the weight of the rode)
.... so you put two anchors on the same rode? Odd.

Also, in jaxassby's example, the points can't always be 100' apart if the
weight is hanging further and further down each time...unless he has an
extremely elastic line and there's a large amount of stretch. I assume that
jaxassby meant to say "using a 100' rope".