On 18 Sep 2004 23:14:45 GMT, (JAXAshby) wrote:

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 what would the forces be using your example?

I don't have a trig calc handy, but do this. divide 50 feet by 20 feet, then
10 feet, then 5 feet, then 1 foot, then 1 inch, then 1/10th inch. that will
give you the tangent of each angle.

look up each tangent, then divide each number into 1#. that will give you the
#'s force on the end points of the line.

a catenary is worse and much, much, much more difficult to calculate, but the
above will give you an idea of the HUGE forces involved once the chain starts
to pull tight

Thanks - that's pretty much what I figured.

Later,

Tom

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?

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

"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".

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?

Later,

Tom

whoring, once you lose sight of a trick, you lose sight of anthing real.

try again, this time without anything in your mouth.

From: JohnH
Date: 9/18/2004 8:34 PM Eastern Daylight Time
Message-id:

On 18 Sep 2004 23:14:45 GMT, (JAXAshby) wrote:

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 what would the forces be using your example?

I don't have a trig calc handy, but do this. divide 50 feet by 20 feet,
then
10 feet, then 5 feet, then 1 foot, then 1 inch, then 1/10th inch. that will
give you the tangent of each angle.

look up each tangent, then divide each number into 1#. that will give you
the
#'s force on the end points of the line.

a catenary is worse and much, much, much more difficult to calculate, but
the
above will give you an idea of the HUGE forces involved once the chain
starts
to pull tight

50/20=2.5
50/10=5
50/5=10
50/1=50
50/1/12=600
50/1/120=6000

If each of these are divided into 1, the results would be, respectively:
0.4
0.2
0.1
0.02
0.001666...
0.0001666...

These numbers don't look so big. Could you have made an error?


John H

On the 'Poco Loco' out of Deale, MD,
on the beautiful Chesapeake Bay!

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

genie, shut up.

Subject: Why need anchor chain?
From: 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

"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.

"JAXAshby" wrote in message
...
genie, shut up.

Subject: Why need anchor chain?
From: 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.



http://www.johnsboatstuff.com/Articles/anchor.htm

" Having a lot of sag in the rode reduces shock loads "

Isn't that what started this whole argument? Anchor chain introduces sag in
the rode. Sag in the rode reduces shock loads. So I guess I was right when
I said that anchor chain acts as a shock absorber, eh jaxie?

"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?

Later,

Tom

The end points are not at the same elevation.