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.

(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 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%