mod vador, have you ever contributed anything at all to this ng or any other in
any way related to sailing? refresh out memories, if you would.

mode (modervador)
Date: 9/28/2004 12:25 AM Eastern Daylight Time
Message-id:

(JAXAshby) wrote in message
...
I suggest that Jax post his equations

D = 16 T^2

S = T * A

A (one G) = 32 ft/sec^2

you can find these equations in a junior high school science class for
advanced
students.

Ah, I see your problem. You have already substituted the accelleration
constant g into d=16t^2 and multiples of g were apparently not dealt
with properly. Use d=(1/2)at^2 and you might get better results.

%mod%

"NOYB" wrote in message link.net...
"JAXAshby" wrote in message
...
**some** is a pedant use of the word. particularly in the context of
multi-G
shock loads. i.e. "reducing" by "some" measure from 4.898 G's to
4.89799983
G's.

A wave surge isn't causing multi-G shock loads, jackassby.

Ding! Give the man a cigar!

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


BWAAAHAAAA!!!! THIS gem from someone who just stated that anybody with
mathematical knowledge would know all about catenary behavior!!!!!
BWAAAHAAAA!


here ya go, dum-dum. you learned something today.

cat·e·nar·y ( P ) Pronunciation Key (ktn-r, k-tn-r)
n. pl. cat·e·nar·ies
The curve formed by a perfectly flexible, uniformly dense, and inextensible
cable suspended from its endpoints. It is identical to the graph of a
hyperbolic cosine.

cate·nary adj.


catenary


\Cat"e*na*ry\, n.; pl. Catenaries. (Geol.) The curve formed by a rope or chain
of uniform density and perfect flexibility, hanging freely between two points
of suspension, not in the same vertical line.

n : the curve theoretically assumed by a perfectly flexible and inextensible
cord of uniform density and cross section hanging freely from two fixed points

a properly sized chain rode of adequate scope DOES NOT *HANG* freely
from two fixed points.

really, junnie? skywaves prevent it?

If it ever does, it should be intermittently, as the temporary shape
of the (then temporary) catenary dampens any shock loading.

obviously, junnie, you have no idea the forces involved. It also appears you
do not have anything remotely connected to the mental candle power needed to
understand it. even hoary admitted his math ignorance with his 625 SAT score,
and you -- junnie, are not willing to brag you are as stew ped mathematically
as hoary.

give it up, junnie. anchoring the way you suggest will drag anchor in wind and
waves. injure someone that way and no stupid frickin claim of "it was an act
of god" will save you from going to jail if I hear about it. it is as criminal
as driving a car drunk. "reasonable men" don't do that, and "reasonable men"
is the legal standard applied by the courts.

Asshole, I full well understand what a catenary is

obviously not, for you post the stupidity below.

Now, YOU said that
a line, chain, piece of string ALWAYS has a catenary. Prove that
point, please. I can certainly prove you wrong!!!

go back an read the definition again. here, let me help you out and repost it.
this really is easy stuff, vase kisser.

here ya go, dum-dum. you learned something today.

cat·e·nar·y ( P ) Pronunciation Key (ktn-r, k-tn-r)
n. pl. cat·e·nar·ies
The curve formed by a perfectly flexible, uniformly dense, and inextensible
cable suspended from its endpoints. It is identical to the graph of a
hyperbolic cosine.

cate·nary adj.


catenary


\Cat"e*na*ry\, n.; pl. Catenaries. (Geol.) The curve formed by a rope or
chain
of uniform density and perfect flexibility, hanging freely between two
points
of suspension, not in the same vertical line.

n : the curve theoretically assumed by a perfectly flexible and
inextensible
cord of uniform density and cross section hanging freely from two fixed
points

Asshole, I full well understand what a catenary is. Now, YOU said that
a line, chain, piece of string ALWAYS has a catenary. Prove that
point, please. I can certainly prove you wrong!!!

jaxashby tells the world how the issue is that he gets things exactly backwards.

(JAXAshby) wrote in message ...
nice cut and paste mod vader IV. too bad you didn't address the issue.

go stand in the corner. no sharp objects for you.

(modvador)
Date: 9/27/2004 12:24 PM Eastern Daylight Time
Message-id:

(JAXAshby) wrote in message
...
the accelleration needed is 1/2 g

1/2 G (note the capitization) means a 11,000# strain on a 22,000 # boat.

Capital G is the universal gravitational constant. Lowercase g is the
gravitational accelleration constant, the constant describing the
force which accellerates a mass towards the Earth's center as measured
at the Earth's surface. g is not constant everywhere on Earth because
Earth is spinning and because Earth is not a perfect sphere.

When one discusses forces on a body, one may compare these forces to
the force due to gravity, because gravity is familiar. A 2 lb mass
experiences a 2 lbf downward force from gravitational attraction on
Earth, thus a 1/2 g force (note lower case) would equate to 1 lbf. It
would not be possible to know what a 1/2 G force would be without
knowing the mass of the planet and the distance from the center, which
is why g is typically used instead of G since g already has that
information factored in.

you
got 11,000# chain/chocks/anchor on your boat.
btw, yo-yo, the G-loads can be one hell of a lot higher than 1/2.

The issue is indeed over the forces and whether the anchoring is
designed to withstand them and/or to minimize them.

%mod%

jaxashby tells the world how only sailboats have anchors.

(JAXAshby) wrote in message ...
in other words, odor vader, you contributed not a thing to the discussion about
dangerously lazy sailors trying to injury other sailors.

not surprising, for you have never posted anything remotely related to sailing
in the past.

(modervador)
Date: 9/27/2004 12:54 PM Eastern Daylight Time
Message-id:

(JAXAshby) wrote in message
...
Starting with boat travelling at 8 ft/sec, a 2 g force will require 6
inches to stop the boat, and the boat will come to a stop in 1/8th of
a second.

1/4 second, not 1/8th. one hell of a difference.

but thanks for googling for hours trying to make an unproveable point.

Oh, quite provable indeed. I'm not aware of any specific links on
Google which would lead one to the equations involving accelleration,
velocity, distance and time, but it seems like you're interested
enough to research it for yourself. I remember them from high school
physics class many years ago, so there was no need.

The relevant relations here are v=at and d=(1/2)at^2. In this case,
we're specifying a=2g and solving to get an 8 ft/sec change in
velocity. Plugging in 32 ft/sec^2 for g and my (correct) answer of
1/8th second for t, we get

v= 2*32 ft/s^2 *1/8 sec = 64/8 ft/sec = 8 ft/sec, which was the
velocity given as the starting condition.

d= (1/2)*2*32 ft/sec^2 *(1/8 sec)^2 = 32/64 ft = 1/2 ft = 6 inches.

You had originally stated that for 2 g and 8 ft/sec, stopping distance
would be 4 inches; that math was questioned. I have provided not only
the correct stopping distance but the correct time. I stand by my
math.

%mod%

jaxashby tells the world how he is satisfied by his performance of a
simple physics problem, a performance so poor it would make a junior
high student wince with embarassment, yet he still feels qualified to
lecture others.

(JAXAshby) wrote in message ...
mod vador, have you ever contributed anything at all to this ng or any other in
any way related to sailing? refresh out memories, if you would.

mode (modervador)
Date: 9/28/2004 12:25 AM Eastern Daylight Time
Message-id:

(JAXAshby) wrote in message
...
I suggest that Jax post his equations

D = 16 T^2

S = T * A

A (one G) = 32 ft/sec^2

you can find these equations in a junior high school science class for
advanced
students.

Ah, I see your problem. You have already substituted the accelleration
constant g into d=16t^2 and multiples of g were apparently not dealt
with properly. Use d=(1/2)at^2 and you might get better results.

%mod%