As I have previously
posted...

because *you* have posted something -- anything -- proves nothing at all.

what a self-centered turd you are for even suggesting such.

I see no evidence that any ground tackle found on any boat other that
something bordering a ship will support anything remotely close to
40,000#

therefore, genei clown, if *you* anchor on all chain and drag down on someone
injuring them you deserve to go to jail.

you need stretch, yo-yo. don't use all chain. unless of course you like you
sex with a 300 pound drug dealer locked in a 8' x 10' cell.

nobody has posted any documented evidence of deck hardware being
damaged solely by using an all chain rode,


they most certainly have. not here among the junior high school set, but it
has be documented far and wide, and with specific pictures.

this ain't rocket science, little boy. just ordinary knowlege for those with
an upper two digit or low three digit IQ.

JAXAshby wrote:
I see no support for Jax's immovable object (anchor) theory,


there you go, genei. you WANT an anchor to move. It is called "dragging" and
it is NOT an Act of God. It is criminal negilgence if you injure someone with
your deliberate attempts to hit them by anchoring in an obviously irresponsible
way.

jail time, dood, for you.

but, genie, you don't ever anchor longer than a short to time fish, do you? In
other words, you zero point zero experience anchoring.



Doodles, the odds of YOU having any great degree of experience anchoring
in bad weather, with ANY particular set up, are in the realm of "slim to
none".

otn

Gene Kearns wrote:
On 24 Sep 2004 03:04:19 GMT, (JAXAshby) wrote:

nobody has posted any documented evidence of deck hardware being
damaged solely by using an all chain rode,


they most certainly have. not here among the junior high school set, but it
has be documented far and wide, and with specific pictures.


You know you have 'Ole Jax by the Speedos when he posts to himself no
less than 9 times and still hasn't provided proof... or any link to
proof.....



Uh...I'll leave having Jax by the Speedos to you, thankyewverymuch.

--
We today have a president of the United States who looks like he is the
son of Howdy Doody or Alfred E. Newman, who isn't smarter than either of
them, who is arrogant about his ignorance, who is reckless and
incompetent, and whose backers are turning the United States into a pariah.

What, me worry?

(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) 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%

On 27 Sep 2004 09:54:28 -0700,
(modervador) wrote:

(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.

Not that you need my support, but you are correctomundo.

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

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) 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%