Re: [bolger] Re: Physics?

> ARRRGGGH!

I agree! Lets give this one a rest.

Chuck

Not fair! Take those two buckets and fill one half full of water, and put
an equal weight of lead in the other, then put them in water and see which
sinks the farthest.

Chuck

>
> I thought of a new example. Take two 10-liter buckets and put
> 10 kilograms of lead in one and 10 liters of water in the other.
> The "dry" weight should be the same. Now go to the shore, a bucket
> in each hand, and wade until your hands are under water. Do the
> buckets still feel equally heavy?
>
>
> Sakari Aaltonen

Bravo Jim Pope.
My vote for the worlds best sub 40ft cruising monohull,
the Roger Martin designed Grey Wolf, holds a ton of water
in a tank just under the side-deck. Pumps it across to
a mirror tank when tacking. Sails flatter, and if racing the football
team on the rail can go below or home.This arrangement does take
72 cubic feet out of the boat, but the football teams bigger!
And half is a bouyancy tank
Jeff Gilbert.

----- Original Message -----
From: Jim Pope <jpope@...>
To: <bolger@egroups.com>
Sent: Tuesday, October 31, 2000 3:23 PM
Subject: Re: [bolger] Physics?


No soy estimado, I'm approximato
Pero for my two cents----
As long as the water filled compartment is submerged, its only effect is to
reduce the vessel's buoyancy. After all, it weighs the same as the rest of
the
water all around the boat. When, however, you lift that filled ballast tank
above the ocean's (or lake's, or river's) surface it weighs much more than
the
air and generates a gravity vector downward from its center of mass. If that
center is out on an arm from the boat's center of buoyancy, the water
ballast
adds to the total righting moment acting on the boat.

Bolger's boxes enjoy a huge movement of the center of buoyancy towards their
top deck as they heel over and submerge more and more of their square
topside
shape. That allows water ballast boxes located out towards the chine to
gain
effectiveness,
an advantage that low sided hulls, such as the traditional sharpie, can't
offer. To be self righting that type of boat needs to employ different
strategies such as lots of ballast. Traditional sharpies didn't, by the way,
and were not self righting.

Jim

Bill Jochems wrote:

> Estimado grupo Bolger
> I have a question about something which is related, I think, to David
> Ryan's question about 100 lbs. centered vs. 50 lbs. on each side. Bolger
> has written that water ballast is of no effect until it has been lifted
> above the water line. I don't understand this. It seems to me that
weight
> is weight, be it water, concrete or lead; and that a boat would behave the
> same whether ballasted with 500 lbs. of water, or 500 lbs of lead, if the
> centers of gravity of each were the same.
> Bill Jochems
> -----Original Message-----
> From: David Ryan <david@...>
> To:bolger@egroups.com<bolger@egroups.com>
> Date: Monday, October 30, 2000 10:17 AM
> Subject: Re: [bolger] Physics?
>
> >Hooray! Dr. Wah would be proud, even if he never did understand what
> >I was doing in his class.
> >
> >Moving on.
> >
> >In a flat bottomed boat, is there a difference between putting 100#
> >dead in the center of the center, or putting 50# on each chine?
> >
> >YIBB,
> >
> >David
> >
> >
> >
> >>That's about right. Let the weight of your 900# beam act in the middle;
> the
> >>lever arm is from fulcrum is then (15-3)=12', so the moment is
> 12*900=10800
> >>ft-lb. To balance the beam, you need to apply 10800/3 = 3600# on the
> short
> >>end (i.e. 3*3600=10800 => equilibrium).
> >>
> >>(To figure it much, much more simply, convert your units to SI metric,
> >>divide by 1000 and then once again by .001 ;-)
> >>
> >>Gregg Carlson
> >>
> >>At 09:58 AM 10/30/2000 -0500, you wrote:
> >>>FBBB --
> >>>
> >>>It's been 12 years since I cracked a physical text book (my degree is
> >>>in fine art, go figure.)
> >>>
> >>>Anyone care to check me on this:
> >>>
> >>>A 30 foot beam that weighs 30 pounds per foot is balanced on a
> >>>fulcrum 3 feet from one end (and 27 feet from the other.)
> >>>
> >>>By my (suspect) calculations, there is a torque of 10935 foot-pounds
> >>>on one side, and 135 foot-pounds on the other.
> >>>
> >>>Further (dubious) arithmetic says that if I apply 3600 pounds of
> >>>force directly on the end of the short side, the beam will be
> >>>balanced.
> >>>
> >>>Anyone wanna check my math?
> >>>
> >>>YIBB,
> >>>
> >>>David
> >>>
> >>>
> >>>CRUMBLING EMPIRE PRODUCTIONS
> >>>134 W.26th St. 12th Floor
> >>>New York, NY 10001
> >>>(212) 243-1636
> >>>
> >>>
> >>>Bolger rules!!!
> >>>- no cursing
> >>>- stay on topic
> >>>- use punctuation
> >>>- add your comments at the TOP and SIGN your posts
> >>>- add some content: send "thanks!" and "ditto!" posts off-list.
> >>>
> >>>
> >>
> >>
> >>
> >>Bolger rules!!!
> >>- no cursing
> >>- stay on topic
> >>- use punctuation
> >>- add your comments at the TOP and SIGN your posts
> >>- add some content: send "thanks!" and "ditto!" posts off-list.
> >
> >
> >CRUMBLING EMPIRE PRODUCTIONS
> >134 W.26th St. 12th Floor
> >New York, NY 10001
> >(212) 243-1636
> >
> >
> >Bolger rules!!!
> >- no cursing
> >- stay on topic
> >- use punctuation
> >- add your comments at the TOP and SIGN your posts
> >- add some content: send "thanks!" and "ditto!" posts off-list.
> >
> >
>
>
> Bolger rules!!!
> - no cursing
> - stay on topic
> - use punctuation
> - add your comments at the TOP and SIGN your posts
> - add some content: send "thanks!" and "ditto!" posts off-list.



Bolger rules!!!
- no cursing
- stay on topic
- use punctuation
- add your comments at the TOP and SIGN your posts
- add some content: send "thanks!" and "ditto!" posts off-list.

Same if its at the same level.
Jeff

----- Original Message -----
From: David Ryan <david@...>
To: <bolger@egroups.com>
Sent: Tuesday, October 31, 2000 4:16 AM
Subject: Re: [bolger] Physics?


Hooray! Dr. Wah would be proud, even if he never did understand what
I was doing in his class.

Moving on.

In a flat bottomed boat, is there a difference between putting 100#
dead in the center of the center, or putting 50# on each chine?

YIBB,

David

Spot on
Jeff

----- Original Message -----
From: David Ryan <david@...>
To: <bolger@egroups.com>
Sent: Tuesday, October 31, 2000 1:58 AM
Subject: [bolger] Physics?


FBBB --

It's been 12 years since I cracked a physical text book (my degree is
in fine art, go figure.)

Anyone care to check me on this:

A 30 foot beam that weighs 30 pounds per foot is balanced on a
fulcrum 3 feet from one end (and 27 feet from the other.)

By my (suspect) calculations, there is a torque of 10935 foot-pounds
on one side, and 135 foot-pounds on the other.

Further (dubious) arithmetic says that if I apply 3600 pounds of
force directly on the end of the short side, the beam will be
balanced.

Anyone wanna check my math?

YIBB,

David


CRUMBLING EMPIRE PRODUCTIONS
134 W.26th St. 12th Floor
New York, NY 10001
(212) 243-1636


Bolger rules!!!
- no cursing
- stay on topic
- use punctuation
- add your comments at the TOP and SIGN your posts
- add some content: send "thanks!" and "ditto!" posts off-list.

Yes.
Its a metal drogue.
Go high tech, get a colander.
Jeff

----- Original Message -----
From: <cliff25@...>
To: <bolger@egroups.com>
Hmm. Then I wonder if an empty water pail would be just as effective as
a sea anchor? The question pops to mind.
Cliff

I meant leave it open.
Jeff

----- Original Message -----
From: Cyber Rebel <secesh@...>
To: <bolger@egroups.com>
Sent: Saturday, November 04, 2000 5:31 PM
Subject: Re: [bolger] Re: Physics?


Well, if we are talking about a wooden tank filled with water, I would say
because the wood may expand, increasing pressure inside the tamk. I would
rather have a bung as the weak point, rather than blow out the seams, or
warp the panels. I'm probably waaaay off here, but it was the first thing
that came to mind.
____________________________________________________________________________
__
Human beings can always be counted upon to assert, with vigor, their
God-given right to be stupid.
-Dean Koontz
----- Original Message -----
From: Jeff Gilbert <jgilbert@...>
To: <bolger@egroups.com>
Sent: Friday, November 03, 2000 3:33 PM
Subject: Re: [bolger] Re: Physics?


> Heres one to drive you nuts.
>
> You have a boat thats form stable, but a bit too skittery.
> You decide to waterballast it below the sole, which is already below
> the waterline, just to settle it down a bit.
> Why bother with a bung?
>
> Jeff Gilbert
>
>
> Bolger rules!!!
> - no cursing
> - stay on topic
> - use punctuation
> - add your comments at the TOP and SIGN your posts
> - add some content: send "thanks!" and "ditto!" posts off-list.
>



Bolger rules!!!
- no cursing
- stay on topic
- use punctuation
- add your comments at the TOP and SIGN your posts
- add some content: send "thanks!" and "ditto!" posts off-list.

ARRRGGGH!
____________________________________________________________________________
__
Human beings can always be counted upon to assert, with vigor, their
God-given right to be stupid.
-Dean Koontz

----- Original Message -----
From: Chuck Leinweber <chuck@...>
To: <bolger@egroups.com>
Sent: Saturday, November 04, 2000 11:27 AM
Subject: Re: [bolger] Re: Physics?


> Not fair! Take those two buckets and fill one half full of water, and put
> an equal weight of lead in the other, then put them in water and see which
> sinks the farthest.
>
> Chuck
>
>
> >
> > I thought of a new example. Take two 10-liter buckets and put
> > 10 kilograms of lead in one and 10 liters of water in the other.
> > The "dry" weight should be the same. Now go to the shore, a bucket
> > in each hand, and wade until your hands are under water. Do the
> > buckets still feel equally heavy?
> >
> >
> > Sakari Aaltonen
>
>
>
>
> Bolger rules!!!
> - no cursing
> - stay on topic
> - use punctuation
> - add your comments at the TOP and SIGN your posts
> - add some content: send "thanks!" and "ditto!" posts off-list.
>

Well, if we are talking about a wooden tank filled with water, I would say
because the wood may expand, increasing pressure inside the tamk. I would
rather have a bung as the weak point, rather than blow out the seams, or
warp the panels. I'm probably waaaay off here, but it was the first thing
that came to mind.
____________________________________________________________________________
__
Human beings can always be counted upon to assert, with vigor, their
God-given right to be stupid.
-Dean Koontz

----- Original Message -----
From: Jeff Gilbert <jgilbert@...>
To: <bolger@egroups.com>
Sent: Friday, November 03, 2000 3:33 PM
Subject: Re: [bolger] Re: Physics?


> Heres one to drive you nuts.
>
> You have a boat thats form stable, but a bit too skittery.
> You decide to waterballast it below the sole, which is already below
> the waterline, just to settle it down a bit.
> Why bother with a bung?
>
> Jeff Gilbert
>
>
> Bolger rules!!!
> - no cursing
> - stay on topic
> - use punctuation
> - add your comments at the TOP and SIGN your posts
> - add some content: send "thanks!" and "ditto!" posts off-list.
>

<<....if the external bag is tethered to the boat you have an inertial
effect. it will slow the boat down.>>

Hmm. Then I wonder if an empty water pail would be just as effective as
a sea anchor? The question pops to mind.

Cliff

http://www.angelfire.com/mi/mmmkkk/
(Last of the Red Hot DJs)

Pippo:

Wouldn't the one with the water have more rotational inertia, as Gregg
pointed out?

Chuck

> Sakari, everything submerged in water has a net weight equal to its
> "dry" weight decreased by the weight of an equal volume of water. The
> water outside doesn't care whether it's lead or water or air.
> I don't like "visual" examples because they're generally misleading,
> however think about two ballast chambers, sealed, identical in shape.
> One totally filled (no sloshing) with X kg of water, the other one
> empty, with a lead sphere (or cube, or whatever) of mass X kg in the
> center of mass of the previous. The effects of the two will be, of
> course, indistinguishable from each other, and two such boats will
> behave exactly the same.
> Best, Pippo
>
> --- Inbolger@egroups.com, Sakari Aaltonen <sakaria@c...> wrote:
> >
> > The effect of a water ballast (rigidly connected to the boat) would
> > seem to depend on whether it is under water or not. Under water, it
> > "weighs" nothing; above, it does. So, if the water ballast is under
> > water, it does not contribute to the center of mass (only what takes
> > care of the rigid connection does). It's like walking around with a
> > balloon of air rigidly connected to you...
>
>
>
>
> Bolger rules!!!
> - no cursing
> - stay on topic
> - use punctuation
> - add your comments at the TOP and SIGN your posts
> - add some content: send "thanks!" and "ditto!" posts off-list.
>
>

Steve:

You have nailed it! When you add ballast to the outside of a boat or blimp,
you change the shape of the vessel, and therefore the form stability. If
you add a keel, it adds buoyancy. If you fill it with water, it does not
lower the CG, and you end up with no righting improvement. To do that you
need to lower CG, and if you want to do it with water, don't increase
buoyancy by adding a keel, instead, add the water to bottom of the existing
hull.

Chuck

> Jeff,
> I believe that the blimp filled with air would fall because of the weight

of

> the blimp. The helium filled blimp is a "lighter-than-air" craft, as the
> helium is lighter than the surrounding air, and in sufficient quantity to
> couteract the weight of the blimp it is in. The air inside the other does
> nothing to counteract it's weight and therefore is heavier in it's
> particular medium. Let's change the analogy to a steel boat, since steel

is

> heavier than water. A steel boat will float as long as it is filled with
> air, because the air is lighter than the displaced water. Fill the boat

with

> water, and it sinks. However, a tank of water on the outside of a steel

boat

> will not make it heavier, and therefore adds no actual ballast except the
> weight of the actual tank minus the water. To go back to your blimp

analogy,

> a helium filled blimp is in no way ballasted by the air inside the

gondola,

> but rather by the weight of the gondola itself. If, however, one places a
> tank of air inside the helium filled envelope of a gondola-less blimp, the
> corresponding amount of helium is displaced in that area of the envelope,
> and the blimp will rotate as if that side were ballasted. The same is true
> of a boat because the air inside is displaced by something heavier. Thus
> water in the boat = ballast, water outside the boat is not.
>
> Steve Lane
> Lansdowne, Maryland
>

____________________________________________________________________________

> __
> Human beings can always be counted upon to assert, with vigor, their
> God-given right to be stupid.
> -Dean Koontz
> ----- Original Message -----
> From: <freedem@...>
> To: <bolger@egroups.com>
> Sent: Thursday, November 02, 2000 8:45 AM
> Subject: [bolger] Re: Physics?
>
>
> > Ernie
> > If you check into the next post (xx35 I belive) Jeff Gilbert agrees
> > with you but it helped me see what Chuck Leinweber said earlyer about
> > a cilindar with a water tank attacked inside. the water is attached
> > to the boat and acts as part of the center of mass. it "hangs" like a
> > pendulum from the center of boyancy ( rather boyancy hangs upward from
> > the center of mass but whatever they are conected ) as a pendulum
> > weight is conected to the suporting point. when they get out of line
> > and righting force is developed that trys to move the centers back in
> > line vertically and restore equelibrum.. the effect is the same
> > weather the mass being lifted out of line with the center of boyancy
> > is denser ( lead) the same as ( water) or lighter than (wood) than the
> > medium it is floating in. it simply much be ( heavyer or denser )
> > than the boyancy of the vessel. the only difference is the value of
> > the force developed.
> >
> > Think of other fluid mediums say a blimp. the helium "flotation" wants
> > to stay above the metal gondala and engines . but if you had no
> > gondala or engines and simply had two blimpts tied side by side one
> > filled with air and the other with helium. they would rotate so the
> > one filled with air hangs below the helium. not because the air in the
> > one is heavyer than the air outside but because it is heavyer than the
> > helium filled blimp.
> > Please for give me long winded ness
> > Jeffery
> >
> >
> > --- Inbolger@egroups.com, "Ernie Murphy" <ernie@s...> wrote:
> > > "a boat would behave the same whether ballasted with 500 lbs. of
> > > water, or 500 lbs of lead"
> > >
> > > Well, not true, let me see if I can give you a feel for this. I
> > > concidered this while imagineering a catfish, which has a water-
> > > filled keel.
> > >
> > > Imagine you have any boat, and start filling baloons with water,
> > > knotting the end and tying off with string. Weigh each baloon, then
> > > dump it overboard. Tie the string to the gunnel and do another. How
> > > many baloons will it take to sink your boat?
> > >
> > > Well, not *quite* an infinite ammount. After all, the string and
> > > even the baloon itself have some weight. But close.
> > >
> > > Now start doing the same with fishing sinkers that have the same
> > > weight as a water filled baloon. As long as the weights stay off the
> > > bottom they will start to pull your boat down as soon as you add the
> > > first one.
> > >
> > > Sure, 500 pounds of water and 500 pounds of lead both weigh 500
> > > pounds *in air*. Put em underwater and the water weighs.... zero,
> > > nada, nothing. Just remember: all boats float cause they weigh
> > > nothing.
> > >
> > > It's displacement you see. A 500 pound boat displaces an volumn of
> > > water that if you could scoop out and take out on land and weigh it
> > > would weigh... exactly 500 pounds. If the weight of water your boat
> > > displaced weighed any less, your boat would be sinking.
> > >
> > > (Water inside the boat isn't displaced. It may be out of place,
> > but
> > > thats something different.)
> > >
> > > The lead will have weight underwater, as it is denser then the
> > > water. A thing is good for balast under water as long as it is
> > denser
> > > then water, but only by the difference in it's weight and the weight
> > > of water it displaces.
> > >
> > > For my 'water keel' catfish, the water filling the keel still does
> > > a few useful things: It still has mass, hence it will help damp out
> > > movement (it will take more time for it to heel all the way over wet
> > > over dry), and also help carry it thru tacks. But, since underwater
> > > water "don't weigh nothing," it will not add to the righting moment
> > > till the boat heels to it's side and the keel comes out dry. And
> > once
> > > outside it gets heavy FAST, helping to stop a rollover when it's
> > > getting disasterous.
> > >
> > > -Ernie
> > >
> > >
> > > --- Inbolger@egroups.com, "Bill Jochems" <wjochems@s...> wrote:
> > > > Estimado grupo Bolger
> > > > I have a question about something which is related, I think,
> > to
> > > David
> > > > Ryan's question about 100 lbs. centered vs. 50 lbs. on each side.
> >
> > > Bolger
> > > > has written that water ballast is of no effect until it has been
> > > lifted
> > > > above the water line. I don't understand this. It seems to me
> > > that weight
> > > > is weight, be it water, concrete or lead; and that a boat would
> > > behave the
> > > > same whether ballasted with 500 lbs. of water, or 500 lbs of lead,
> > > if the
> > > > centers of gravity of each were the same.
> > > > Bill Jochems
> > > > -----Original Message-----
> > > > From: David Ryan <david@c...>
> > > > To:bolger@egroups.com<bolger@egroups.com>
> > > > Date: Monday, October 30, 2000 10:17 AM
> > > > Subject: Re: [b
> >
> >
> >
> > Bolger rules!!!
> > - no cursing
> > - stay on topic
> > - use punctuation
> > - add your comments at the TOP and SIGN your posts
> > - add some content: send "thanks!" and "ditto!" posts off-list.
> >
>
>
>
> Bolger rules!!!
> - no cursing
> - stay on topic
> - use punctuation
> - add your comments at the TOP and SIGN your posts
> - add some content: send "thanks!" and "ditto!" posts off-list.
>
>

Fair enough, but if the external bag is tethered to the boat you
have an inertial effect. it will slow the boat down.
This is more like drag, or "handicap" or "inertial penalty"
than ballast in the classic sense.
I suppose the classic example is towing a dinghy full of water.
Its not ballast, its something best described by words not used here.
Jeff G

----- Original Message -----
From: Cyber Rebel <secesh@...>
To: <bolger@egroups.com>
Sent: Friday, November 03, 2000 5:07 PM
Subject: Re: [bolger] Re: Physics?


Jeff,
I believe that the blimp filled with air would fall because of the weight of
the blimp. The helium filled blimp is a "lighter-than-air" craft, as the
helium is lighter than the surrounding air, and in sufficient quantity to
couteract the weight of the blimp it is in. The air inside the other does
nothing to counteract it's weight and therefore is heavier in it's
particular medium. Let's change the analogy to a steel boat, since steel is
heavier than water. A steel boat will float as long as it is filled with
air, because the air is lighter than the displaced water. Fill the boat with
water, and it sinks. However, a tank of water on the outside of a steel boat
will not make it heavier, and therefore adds no actual ballast except the
weight of the actual tank minus the water. To go back to your blimp analogy,
a helium filled blimp is in no way ballasted by the air inside the gondola,
but rather by the weight of the gondola itself. If, however, one places a
tank of air inside the helium filled envelope of a gondola-less blimp, the
corresponding amount of helium is displaced in that area of the envelope,
and the blimp will rotate as if that side were ballasted. The same is true
of a boat because the air inside is displaced by something heavier. Thus
water in the boat = ballast, water outside the boat is not.

Steve Lane
Lansdowne, Maryland
____________________________________________________________________________
__
Human beings can always be counted upon to assert, with vigor, their
God-given right to be stupid.
-Dean Koontz
----- Original Message -----
From: <freedem@...>
To: <bolger@egroups.com>
Sent: Thursday, November 02, 2000 8:45 AM
Subject: [bolger] Re: Physics?


> Ernie
> If you check into the next post (xx35 I belive) Jeff Gilbert agrees
> with you but it helped me see what Chuck Leinweber said earlyer about
> a cilindar with a water tank attacked inside. the water is attached
> to the boat and acts as part of the center of mass. it "hangs" like a
> pendulum from the center of boyancy ( rather boyancy hangs upward from
> the center of mass but whatever they are conected ) as a pendulum
> weight is conected to the suporting point. when they get out of line
> and righting force is developed that trys to move the centers back in
> line vertically and restore equelibrum.. the effect is the same
> weather the mass being lifted out of line with the center of boyancy
> is denser ( lead) the same as ( water) or lighter than (wood) than the
> medium it is floating in. it simply much be ( heavyer or denser )
> than the boyancy of the vessel. the only difference is the value of
> the force developed.
>
> Think of other fluid mediums say a blimp. the helium "flotation" wants
> to stay above the metal gondala and engines . but if you had no
> gondala or engines and simply had two blimpts tied side by side one
> filled with air and the other with helium. they would rotate so the
> one filled with air hangs below the helium. not because the air in the
> one is heavyer than the air outside but because it is heavyer than the
> helium filled blimp.
> Please for give me long winded ness
> Jeffery
>
>
> --- Inbolger@egroups.com, "Ernie Murphy" <ernie@s...> wrote:
> > "a boat would behave the same whether ballasted with 500 lbs. of
> > water, or 500 lbs of lead"
> >
> > Well, not true, let me see if I can give you a feel for this. I
> > concidered this while imagineering a catfish, which has a water-
> > filled keel.
> >
> > Imagine you have any boat, and start filling baloons with water,
> > knotting the end and tying off with string. Weigh each baloon, then
> > dump it overboard. Tie the string to the gunnel and do another. How
> > many baloons will it take to sink your boat?
> >
> > Well, not *quite* an infinite ammount. After all, the string and
> > even the baloon itself have some weight. But close.
> >
> > Now start doing the same with fishing sinkers that have the same
> > weight as a water filled baloon. As long as the weights stay off the
> > bottom they will start to pull your boat down as soon as you add the
> > first one.
> >
> > Sure, 500 pounds of water and 500 pounds of lead both weigh 500
> > pounds *in air*. Put em underwater and the water weighs.... zero,
> > nada, nothing. Just remember: all boats float cause they weigh
> > nothing.
> >
> > It's displacement you see. A 500 pound boat displaces an volumn of
> > water that if you could scoop out and take out on land and weigh it
> > would weigh... exactly 500 pounds. If the weight of water your boat
> > displaced weighed any less, your boat would be sinking.
> >
> > (Water inside the boat isn't displaced. It may be out of place,
> but
> > thats something different.)
> >
> > The lead will have weight underwater, as it is denser then the
> > water. A thing is good for balast under water as long as it is
> denser
> > then water, but only by the difference in it's weight and the weight
> > of water it displaces.
> >
> > For my 'water keel' catfish, the water filling the keel still does
> > a few useful things: It still has mass, hence it will help damp out
> > movement (it will take more time for it to heel all the way over wet
> > over dry), and also help carry it thru tacks. But, since underwater
> > water "don't weigh nothing," it will not add to the righting moment
> > till the boat heels to it's side and the keel comes out dry. And
> once
> > outside it gets heavy FAST, helping to stop a rollover when it's
> > getting disasterous.
> >
> > -Ernie
> >
> >
> > --- Inbolger@egroups.com, "Bill Jochems" <wjochems@s...> wrote:
> > > Estimado grupo Bolger
> > > I have a question about something which is related, I think,
> to
> > David
> > > Ryan's question about 100 lbs. centered vs. 50 lbs. on each side.
>
> > Bolger
> > > has written that water ballast is of no effect until it has been
> > lifted
> > > above the water line. I don't understand this. It seems to me
> > that weight
> > > is weight, be it water, concrete or lead; and that a boat would
> > behave the
> > > same whether ballasted with 500 lbs. of water, or 500 lbs of lead,
> > if the
> > > centers of gravity of each were the same.
> > > Bill Jochems
> > > -----Original Message-----
> > > From: David Ryan <david@c...>
> > > To:bolger@egroups.com<bolger@egroups.com>
> > > Date: Monday, October 30, 2000 10:17 AM
> > > Subject: Re: [b
>
>
>
> Bolger rules!!!
> - no cursing
> - stay on topic
> - use punctuation
> - add your comments at the TOP and SIGN your posts
> - add some content: send "thanks!" and "ditto!" posts off-list.
>



Bolger rules!!!
- no cursing
- stay on topic
- use punctuation
- add your comments at the TOP and SIGN your posts
- add some content: send "thanks!" and "ditto!" posts off-list.

Heres one to drive you nuts.

You have a boat thats form stable, but a bit too skittery.
You decide to waterballast it below the sole, which is already below
the waterline, just to settle it down a bit.
Why bother with a bung?

Jeff Gilbert

On Fri, 3 Nov 2000, Giuseppe Bianco wrote:

> Sakari, everything submerged in water has a net weight equal to its
> "dry" weight decreased by the weight of an equal volume of water. The
> water outside doesn't care whether it's lead or water or air.
> I don't like "visual" examples because they're generally misleading,
> however think about two ballast chambers, sealed, identical in shape.
> One totally filled (no sloshing) with X kg of water, the other one
> empty, with a lead sphere (or cube, or whatever) of mass X kg in the
> center of mass of the previous. The effects of the two will be, of
> course, indistinguishable from each other, and two such boats will
> behave exactly the same.
> Best, Pippo

But the "empty" chamber will have air in it, no? Otherwise it cannot
be identical in shape. Air is very light compared to water, so its
"dry" weight decreased by the weight of an equal volume of water will,
in fact, be negative...

I thought of a new example. Take two 10-liter buckets and put
10 kilograms of lead in one and 10 liters of water in the other.
The "dry" weight should be the same. Now go to the shore, a bucket
in each hand, and wade until your hands are under water. Do the
buckets still feel equally heavy?


Sakari Aaltonen

Sakari, everything submerged in water has a net weight equal to its
"dry" weight decreased by the weight of an equal volume of water. The
water outside doesn't care whether it's lead or water or air.
I don't like "visual" examples because they're generally misleading,
however think about two ballast chambers, sealed, identical in shape.
One totally filled (no sloshing) with X kg of water, the other one
empty, with a lead sphere (or cube, or whatever) of mass X kg in the
center of mass of the previous. The effects of the two will be, of
course, indistinguishable from each other, and two such boats will
behave exactly the same.
Best, Pippo

--- Inbolger@egroups.com, Sakari Aaltonen <sakaria@c...> wrote:
>
> The effect of a water ballast (rigidly connected to the boat) would
> seem to depend on whether it is under water or not. Under water, it
> "weighs" nothing; above, it does. So, if the water ballast is under
> water, it does not contribute to the center of mass (only what takes
> care of the rigid connection does). It's like walking around with a
> balloon of air rigidly connected to you...

Jeff,
I believe that the blimp filled with air would fall because of the weight of
the blimp. The helium filled blimp is a "lighter-than-air" craft, as the
helium is lighter than the surrounding air, and in sufficient quantity to
couteract the weight of the blimp it is in. The air inside the other does
nothing to counteract it's weight and therefore is heavier in it's
particular medium. Let's change the analogy to a steel boat, since steel is
heavier than water. A steel boat will float as long as it is filled with
air, because the air is lighter than the displaced water. Fill the boat with
water, and it sinks. However, a tank of water on the outside of a steel boat
will not make it heavier, and therefore adds no actual ballast except the
weight of the actual tank minus the water. To go back to your blimp analogy,
a helium filled blimp is in no way ballasted by the air inside the gondola,
but rather by the weight of the gondola itself. If, however, one places a
tank of air inside the helium filled envelope of a gondola-less blimp, the
corresponding amount of helium is displaced in that area of the envelope,
and the blimp will rotate as if that side were ballasted. The same is true
of a boat because the air inside is displaced by something heavier. Thus
water in the boat = ballast, water outside the boat is not.

Steve Lane
Lansdowne, Maryland
____________________________________________________________________________
__
Human beings can always be counted upon to assert, with vigor, their
God-given right to be stupid.
-Dean Koontz

----- Original Message -----
From: <freedem@...>
To: <bolger@egroups.com>
Sent: Thursday, November 02, 2000 8:45 AM
Subject: [bolger] Re: Physics?


> Ernie
> If you check into the next post (xx35 I belive) Jeff Gilbert agrees
> with you but it helped me see what Chuck Leinweber said earlyer about
> a cilindar with a water tank attacked inside. the water is attached
> to the boat and acts as part of the center of mass. it "hangs" like a
> pendulum from the center of boyancy ( rather boyancy hangs upward from
> the center of mass but whatever they are conected ) as a pendulum
> weight is conected to the suporting point. when they get out of line
> and righting force is developed that trys to move the centers back in
> line vertically and restore equelibrum.. the effect is the same
> weather the mass being lifted out of line with the center of boyancy
> is denser ( lead) the same as ( water) or lighter than (wood) than the
> medium it is floating in. it simply much be ( heavyer or denser )
> than the boyancy of the vessel. the only difference is the value of
> the force developed.
>
> Think of other fluid mediums say a blimp. the helium "flotation" wants
> to stay above the metal gondala and engines . but if you had no
> gondala or engines and simply had two blimpts tied side by side one
> filled with air and the other with helium. they would rotate so the
> one filled with air hangs below the helium. not because the air in the
> one is heavyer than the air outside but because it is heavyer than the
> helium filled blimp.
> Please for give me long winded ness
> Jeffery
>
>
> --- Inbolger@egroups.com, "Ernie Murphy" <ernie@s...> wrote:
> > "a boat would behave the same whether ballasted with 500 lbs. of
> > water, or 500 lbs of lead"
> >
> > Well, not true, let me see if I can give you a feel for this. I
> > concidered this while imagineering a catfish, which has a water-
> > filled keel.
> >
> > Imagine you have any boat, and start filling baloons with water,
> > knotting the end and tying off with string. Weigh each baloon, then
> > dump it overboard. Tie the string to the gunnel and do another. How
> > many baloons will it take to sink your boat?
> >
> > Well, not *quite* an infinite ammount. After all, the string and
> > even the baloon itself have some weight. But close.
> >
> > Now start doing the same with fishing sinkers that have the same
> > weight as a water filled baloon. As long as the weights stay off the
> > bottom they will start to pull your boat down as soon as you add the
> > first one.
> >
> > Sure, 500 pounds of water and 500 pounds of lead both weigh 500
> > pounds *in air*. Put em underwater and the water weighs.... zero,
> > nada, nothing. Just remember: all boats float cause they weigh
> > nothing.
> >
> > It's displacement you see. A 500 pound boat displaces an volumn of
> > water that if you could scoop out and take out on land and weigh it
> > would weigh... exactly 500 pounds. If the weight of water your boat
> > displaced weighed any less, your boat would be sinking.
> >
> > (Water inside the boat isn't displaced. It may be out of place,
> but
> > thats something different.)
> >
> > The lead will have weight underwater, as it is denser then the
> > water. A thing is good for balast under water as long as it is
> denser
> > then water, but only by the difference in it's weight and the weight
> > of water it displaces.
> >
> > For my 'water keel' catfish, the water filling the keel still does
> > a few useful things: It still has mass, hence it will help damp out
> > movement (it will take more time for it to heel all the way over wet
> > over dry), and also help carry it thru tacks. But, since underwater
> > water "don't weigh nothing," it will not add to the righting moment
> > till the boat heels to it's side and the keel comes out dry. And
> once
> > outside it gets heavy FAST, helping to stop a rollover when it's
> > getting disasterous.
> >
> > -Ernie
> >
> >
> > --- Inbolger@egroups.com, "Bill Jochems" <wjochems@s...> wrote:
> > > Estimado grupo Bolger
> > > I have a question about something which is related, I think,
> to
> > David
> > > Ryan's question about 100 lbs. centered vs. 50 lbs. on each side.
>
> > Bolger
> > > has written that water ballast is of no effect until it has been
> > lifted
> > > above the water line. I don't understand this. It seems to me
> > that weight
> > > is weight, be it water, concrete or lead; and that a boat would
> > behave the
> > > same whether ballasted with 500 lbs. of water, or 500 lbs of lead,
> > if the
> > > centers of gravity of each were the same.
> > > Bill Jochems
> > > -----Original Message-----
> > > From: David Ryan <david@c...>
> > > To:bolger@egroups.com<bolger@egroups.com>
> > > Date: Monday, October 30, 2000 10:17 AM
> > > Subject: Re: [b
>
>
>
> Bolger rules!!!
> - no cursing
> - stay on topic
> - use punctuation
> - add your comments at the TOP and SIGN your posts
> - add some content: send "thanks!" and "ditto!" posts off-list.
>

Ernie:

This is a complicated concept, and you have to be very careful with
examples. With your balloon and fish weight test, you have to remember that
with the balloons, you are increasing the volume of the boat, thus the
buoyancy, That extra buoyancy is what floats all that water.

Think about it this way: If you put a 500 pound container of water in a
small boat, it will sink x amount. If you put 500 pounds of lead in a
similar boat, it will sink a similar amount. If the lead were blocked up so
that its center of gravity were the same as the one with the container of
water wouldn't they both have the same righting moment also.

The thing that is so confusing is that you would never put lead ballast in a
boat, then block it up a foot or two off the bottom, which is what you
effectively do when you put water ballast in.

Chuck

> "a boat would behave the same whether ballasted with 500 lbs. of
> water, or 500 lbs of lead"
>
> Well, not true, let me see if I can give you a feel for this. I
> concidered this while imagineering a catfish, which has a water-
> filled keel.
>
> Imagine you have any boat, and start filling baloons with water,
> knotting the end and tying off with string. Weigh each baloon, then
> dump it overboard. Tie the string to the gunnel and do another. How
> many baloons will it take to sink your boat?
>
> Well, not *quite* an infinite ammount. After all, the string and
> even the baloon itself have some weight. But close.
>
> Now start doing the same with fishing sinkers that have the same
> weight as a water filled baloon. As long as the weights stay off the
> bottom they will start to pull your boat down as soon as you add the
> first one.
>
> Sure, 500 pounds of water and 500 pounds of lead both weigh 500
> pounds *in air*. Put em underwater and the water weighs.... zero,
> nada, nothing. Just remember: all boats float cause they weigh
> nothing.
>
> It's displacement you see. A 500 pound boat displaces an volumn of
> water that if you could scoop out and take out on land and weigh it
> would weigh... exactly 500 pounds. If the weight of water your boat
> displaced weighed any less, your boat would be sinking.
>
> (Water inside the boat isn't displaced. It may be out of place, but
> thats something different.)
>
> The lead will have weight underwater, as it is denser then the
> water. A thing is good for balast under water as long as it is denser
> then water, but only by the difference in it's weight and the weight
> of water it displaces.
>
> For my 'water keel' catfish, the water filling the keel still does
> a few useful things: It still has mass, hence it will help damp out
> movement (it will take more time for it to heel all the way over wet
> over dry), and also help carry it thru tacks. But, since underwater
> water "don't weigh nothing," it will not add to the righting moment
> till the boat heels to it's side and the keel comes out dry. And once
> outside it gets heavy FAST, helping to stop a rollover when it's
> getting disasterous.
>
> -Ernie
>
>
> --- Inbolger@egroups.com, "Bill Jochems" <wjochems@s...> wrote:
> > Estimado grupo Bolger
> > I have a question about something which is related, I think, to
> David
> > Ryan's question about 100 lbs. centered vs. 50 lbs. on each side.
> Bolger
> > has written that water ballast is of no effect until it has been
> lifted
> > above the water line. I don't understand this. It seems to me
> that weight
> > is weight, be it water, concrete or lead; and that a boat would
> behave the
> > same whether ballasted with 500 lbs. of water, or 500 lbs of lead,
> if the
> > centers of gravity of each were the same.
> > Bill Jochems
> > -----Original Message-----
> > From: David Ryan <david@c...>
> > To:bolger@egroups.com<bolger@egroups.com>
> > Date: Monday, October 30, 2000 10:17 AM
> > Subject: Re: [bolger] Physics?
> >
> >
> > >Hooray! Dr. Wah would be proud, even if he never did understand
> what
> > >I was doing in his class.
> > >
> > >Moving on.
> > >
> > >In a flat bottomed boat, is there a difference between putting 100#
> > >dead in the center of the center, or putting 50# on each chine?
> > >
> > >YIBB,
> > >
> > >David
> > >
> > >
> > >
> > >>That's about right. Let the weight of your 900# beam act in the
> middle;
> > the
> > >>lever arm is from fulcrum is then (15-3)=12', so the moment is
> > 12*900=10800
> > >>ft-lb. To balance the beam, you need to apply 10800/3 = 3600# on
> the
> > short
> > >>end (i.e. 3*3600=10800 => equilibrium).
> > >>
> > >>(To figure it much, much more simply, convert your units to SI
> metric,
> > >>divide by 1000 and then once again by .001 ;-)
> > >>
> > >>Gregg Carlson
> > >>
> > >>At 09:58 AM 10/30/2000 -0500, you wrote:
> > >>>FBBB --
> > >>>
> > >>>It's been 12 years since I cracked a physical text book (my
> degree is
> > >>>in fine art, go figure.)
> > >>>
> > >>>Anyone care to check me on this:
> > >>>
> > >>>A 30 foot beam that weighs 30 pounds per foot is balanced on a
> > >>>fulcrum 3 feet from one end (and 27 feet from the other.)
> > >>>
> > >>>By my (suspect) calculations, there is a torque of 10935 foot-
> pounds
> > >>>on one side, and 135 foot-pounds on the other.
> > >>>
> > >>>Further (dubious) arithmetic says that if I apply 3600 pounds of
> > >>>force directly on the end of the short side, the beam will be
> > >>>balanced.
> > >>>
> > >>>Anyone wanna check my math?
> > >>>
> > >>>YIBB,
> > >>>
> > >>>David
> > >>>
> > >>>
> > >>>CRUMBLING EMPIRE PRODUCTIONS
> > >>>134 W.26th St. 12th Floor
> > >>>New York, NY 10001
> > >>>(212) 243-1636
> > >>>
> > >>>
> > >>>Bolger rules!!!
> > >>>- no cursing
> > >>>- stay on topic
> > >>>- use punctuation
> > >>>- add your comments at the TOP and SIGN your posts
> > >>>- add some content: send "thanks!" and "ditto!" posts off-list.
> > >>>
> > >>>
> > >>
> > >>
> > >>
> > >>Bolger rules!!!
> > >>- no cursing
> > >>- stay on topic
> > >>- use punctuation
> > >>- add your comments at the TOP and SIGN your posts
> > >>- add some content: send "thanks!" and "ditto!" posts off-list.
> > >
> > >
> > >CRUMBLING EMPIRE PRODUCTIONS
> > >134 W.26th St. 12th Floor
> > >New York, NY 10001
> > >(212) 243-1636
> > >
> > >
> > >Bolger rules!!!
> > >- no cursing
> > >- stay on topic
> > >- use punctuation
> > >- add your comments at the TOP and SIGN your posts
> > >- add some content: send "thanks!" and "ditto!" posts off-list.
> > >
> > >
>
>
>
> Bolger rules!!!
> - no cursing
> - stay on topic
> - use punctuation
> - add your comments at the TOP and SIGN your posts
> - add some content: send "thanks!" and "ditto!" posts off-list.
>
>

On Thu, 2 Nov 2000, Giuseppe Bianco wrote:

> Ernie, I think that Bill is right. As long as the ballast (be it lead,
> water, potato chips or air) remains rigidly connected to the boat, the
> effect is exactly the same for any material you use for the ballast
> itself, provided the center of mass remains the same. The flooded keel
> example is a bit misleading in this context...
> Best, Pippo

The effect of a water ballast (rigidly connected to the boat) would
seem to depend on whether it is under water or not. Under water, it
"weighs" nothing; above, it does. So, if the water ballast is under
water, it does not contribute to the center of mass (only what takes
care of the rigid connection does). It's like walking around with a
balloon of air rigidly connected to you...


Sakari Aaltonen

John
I belive the water keel has righting moment even below the waterline
because it has mass and there fore is forcing the center of boyancy
deeper in the fluid.

Think of it as a teter totter, or a leaver with the pivot in the
middle and have it resting on a pontoon picture the load / ballest/
on the left side make the pivot point high enough that the left end
of the lever cas swing with out hitting the pontoon. the right end
of the lever is the mastif you try to pull it down to the right it
will want to return to a vertical positin now if the ballest were
water and was in the water and could pivot on the fulcrum there would
not be a righting force( discounting the weight of the tank holding
the water) BUT since the tank is ridgedly conected to the boyancy
when you try to pull the lever down to the right it twists the pontoon
so that the right side of the pontoon is pushed deeper and the left
side lifts above the water because the center of boyancy shifts to the
right as the center of mass shifts to the left a righting force is
developed so that they try to re aline them selves vertically.

One more example if you'll forgive me imagion that the lead weight
were shaped into a closed cilindar so the combo of the air and metal
were the same density ( ie size and shape) as the tank of water even
though the density is the same as water were it to be fixed to a
pontoon by a plank and laid on the water the lead end would sink till
it hung straight down from the pontoon

I hope I'm helping this discussion
Jeffery


--- Inbolger@egroups.com, "John S Harper/Raleigh/IBM" <jsharper@u...>
wrote:

> > For my 'water keel' catfish, the water filling the keel still

does

> > a few useful things: It still has mass, hence it will help damp

out

> > movement (it will take more time for it to heel all the way over

wet

> > over dry), and also help carry it thru tacks. But, since

underwater

> > water "don't weigh nothing," it will not add to the righting

moment

> > till the boat heels to it's side and the keel comes out dry. And

once

> > outside it gets heavy FAST, helping to stop a rollover when it's
> > getting disasterous.
>
> This part really hurts my brain. (I consider that fun.)
>
> I think I've finally got a grasp of what's going on.
>
> At first glance, the water keel appears magical; it has mass when

you

> consider roll-damping and momentum when tacking but no mass when you
> consider righting moment!
>
> I think I've got the answer. It doesn't affect righting-moment

because it

> has neutral bouyancy and therefore the boat's static stability curve

(up to

> the point some part of the keel is above the waterline) would be the

same

> with or without this water-filled keel. When I say without the
> water-filled keel I don't mean the keel is still there but empty I

mean the

> whole keel is removed.
>
> I still have the sneaking suspicion that my analysis is wrong.
>
> Somebody st

> For my 'water keel' catfish, the water filling the keel still does
> a few useful things: It still has mass, hence it will help damp out
> movement (it will take more time for it to heel all the way over wet
> over dry), and also help carry it thru tacks. But, since underwater
> water "don't weigh nothing," it will not add to the righting moment
> till the boat heels to it's side and the keel comes out dry. And once
> outside it gets heavy FAST, helping to stop a rollover when it's
> getting disasterous.

This part really hurts my brain. (I consider that fun.)

I think I've finally got a grasp of what's going on.

At first glance, the water keel appears magical; it has mass when you
consider roll-damping and momentum when tacking but no mass when you
consider righting moment!

I think I've got the answer. It doesn't affect righting-moment because it
has neutral bouyancy and therefore the boat's static stability curve (up to
the point some part of the keel is above the waterline) would be the same
with or without this water-filled keel. When I say without the
water-filled keel I don't mean the keel is still there but empty I mean the
whole keel is removed.

I still have the sneaking suspicion that my analysis is wrong.

Somebody straighten out my thinking,
John

Ernie
If you check into the next post (xx35 I belive) Jeff Gilbert agrees
with you but it helped me see what Chuck Leinweber said earlyer about
a cilindar with a water tank attacked inside. the water is attached
to the boat and acts as part of the center of mass. it "hangs" like a
pendulum from the center of boyancy ( rather boyancy hangs upward from
the center of mass but whatever they are conected ) as a pendulum
weight is conected to the suporting point. when they get out of line
and righting force is developed that trys to move the centers back in
line vertically and restore equelibrum.. the effect is the same
weather the mass being lifted out of line with the center of boyancy
is denser ( lead) the same as ( water) or lighter than (wood) than the
medium it is floating in. it simply much be ( heavyer or denser )
than the boyancy of the vessel. the only difference is the value of
the force developed.

Think of other fluid mediums say a blimp. the helium "flotation" wants
to stay above the metal gondala and engines . but if you had no
gondala or engines and simply had two blimpts tied side by side one
filled with air and the other with helium. they would rotate so the
one filled with air hangs below the helium. not because the air in the
one is heavyer than the air outside but because it is heavyer than the
helium filled blimp.
Please for give me long winded ness
Jeffery

--- Inbolger@egroups.com, "Ernie Murphy" <ernie@s...> wrote:
> "a boat would behave the same whether ballasted with 500 lbs. of
> water, or 500 lbs of lead"
>
> Well, not true, let me see if I can give you a feel for this. I
> concidered this while imagineering a catfish, which has a water-
> filled keel.
>
> Imagine you have any boat, and start filling baloons with water,
> knotting the end and tying off with string. Weigh each baloon, then
> dump it overboard. Tie the string to the gunnel and do another. How
> many baloons will it take to sink your boat?
>
> Well, not *quite* an infinite ammount. After all, the string and
> even the baloon itself have some weight. But close.
>
> Now start doing the same with fishing sinkers that have the same
> weight as a water filled baloon. As long as the weights stay off the
> bottom they will start to pull your boat down as soon as you add the
> first one.
>
> Sure, 500 pounds of water and 500 pounds of lead both weigh 500
> pounds *in air*. Put em underwater and the water weighs.... zero,
> nada, nothing. Just remember: all boats float cause they weigh
> nothing.
>
> It's displacement you see. A 500 pound boat displaces an volumn of
> water that if you could scoop out and take out on land and weigh it
> would weigh... exactly 500 pounds. If the weight of water your boat
> displaced weighed any less, your boat would be sinking.
>
> (Water inside the boat isn't displaced. It may be out of place,
but
> thats something different.)
>
> The lead will have weight underwater, as it is denser then the
> water. A thing is good for balast under water as long as it is
denser
> then water, but only by the difference in it's weight and the weight
> of water it displaces.
>
> For my 'water keel' catfish, the water filling the keel still does
> a few useful things: It still has mass, hence it will help damp out
> movement (it will take more time for it to heel all the way over wet
> over dry), and also help carry it thru tacks. But, since underwater
> water "don't weigh nothing," it will not add to the righting moment
> till the boat heels to it's side and the keel comes out dry. And
once
> outside it gets heavy FAST, helping to stop a rollover when it's
> getting disasterous.
>
> -Ernie
>
>
> --- Inbolger@egroups.com, "Bill Jochems" <wjochems@s...> wrote:
> > Estimado grupo Bolger
> > I have a question about something which is related, I think,
to
> David
> > Ryan's question about 100 lbs. centered vs. 50 lbs. on each side.

> Bolger
> > has written that water ballast is of no effect until it has been
> lifted
> > above the water line. I don't understand this. It seems to me
> that weight
> > is weight, be it water, concrete or lead; and that a boat would
> behave the
> > same whether ballasted with 500 lbs. of water, or 500 lbs of lead,
> if the
> > centers of gravity of each were the same.
> > Bill Jochems
> > -----Original Message-----
> > From: David Ryan <david@c...>
> > To:bolger@egroups.com<bolger@egroups.com>
> > Date: Monday, October 30, 2000 10:17 AM
> > Subject: Re: [b

Ernie, I think that Bill is right. As long as the ballast (be it lead,
water, potato chips or air) remains rigidly connected to the boat, the
effect is exactly the same for any material you use for the ballast
itself, provided the center of mass remains the same. The flooded keel
example is a bit misleading in this context...
Best, Pippo

--- Inbolger@egroups.com, "Ernie Murphy" <ernie@s...> wrote:
> "a boat would behave the same whether ballasted with 500 lbs. of
> water, or 500 lbs of lead"
>
> Well, not true, let me see if I can give you a feel for this.

"a boat would behave the same whether ballasted with 500 lbs. of
water, or 500 lbs of lead"

Well, not true, let me see if I can give you a feel for this. I
concidered this while imagineering a catfish, which has a water-
filled keel.

Imagine you have any boat, and start filling baloons with water,
knotting the end and tying off with string. Weigh each baloon, then
dump it overboard. Tie the string to the gunnel and do another. How
many baloons will it take to sink your boat?

Well, not *quite* an infinite ammount. After all, the string and
even the baloon itself have some weight. But close.

Now start doing the same with fishing sinkers that have the same
weight as a water filled baloon. As long as the weights stay off the
bottom they will start to pull your boat down as soon as you add the
first one.

Sure, 500 pounds of water and 500 pounds of lead both weigh 500
pounds *in air*. Put em underwater and the water weighs.... zero,
nada, nothing. Just remember: all boats float cause they weigh
nothing.

It's displacement you see. A 500 pound boat displaces an volumn of
water that if you could scoop out and take out on land and weigh it
would weigh... exactly 500 pounds. If the weight of water your boat
displaced weighed any less, your boat would be sinking.

(Water inside the boat isn't displaced. It may be out of place, but
thats something different.)

The lead will have weight underwater, as it is denser then the
water. A thing is good for balast under water as long as it is denser
then water, but only by the difference in it's weight and the weight
of water it displaces.

For my 'water keel' catfish, the water filling the keel still does
a few useful things: It still has mass, hence it will help damp out
movement (it will take more time for it to heel all the way over wet
over dry), and also help carry it thru tacks. But, since underwater
water "don't weigh nothing," it will not add to the righting moment
till the boat heels to it's side and the keel comes out dry. And once
outside it gets heavy FAST, helping to stop a rollover when it's
getting disasterous.

-Ernie

--- Inbolger@egroups.com, "Bill Jochems" <wjochems@s...> wrote:
> Estimado grupo Bolger
> I have a question about something which is related, I think, to
David
> Ryan's question about 100 lbs. centered vs. 50 lbs. on each side.
Bolger
> has written that water ballast is of no effect until it has been
lifted
> above the water line. I don't understand this. It seems to me
that weight
> is weight, be it water, concrete or lead; and that a boat would
behave the
> same whether ballasted with 500 lbs. of water, or 500 lbs of lead,
if the
> centers of gravity of each were the same.
> Bill Jochems
> -----Original Message-----
> From: David Ryan <david@c...>
> To:bolger@egroups.com<bolger@egroups.com>
> Date: Monday, October 30, 2000 10:17 AM
> Subject: Re: [bolger] Physics?
>
>
> >Hooray! Dr. Wah would be proud, even if he never did understand
what
> >I was doing in his class.
> >
> >Moving on.
> >
> >In a flat bottomed boat, is there a difference between putting 100#
> >dead in the center of the center, or putting 50# on each chine?
> >
> >YIBB,
> >
> >David
> >
> >
> >
> >>That's about right. Let the weight of your 900# beam act in the
middle;
> the
> >>lever arm is from fulcrum is then (15-3)=12', so the moment is
> 12*900=10800
> >>ft-lb. To balance the beam, you need to apply 10800/3 = 3600# on
the
> short
> >>end (i.e. 3*3600=10800 => equilibrium).
> >>
> >>(To figure it much, much more simply, convert your units to SI
metric,
> >>divide by 1000 and then once again by .001 ;-)
> >>
> >>Gregg Carlson
> >>
> >>At 09:58 AM 10/30/2000 -0500, you wrote:
> >>>FBBB --
> >>>
> >>>It's been 12 years since I cracked a physical text book (my
degree is
> >>>in fine art, go figure.)
> >>>
> >>>Anyone care to check me on this:
> >>>
> >>>A 30 foot beam that weighs 30 pounds per foot is balanced on a
> >>>fulcrum 3 feet from one end (and 27 feet from the other.)
> >>>
> >>>By my (suspect) calculations, there is a torque of 10935 foot-
pounds
> >>>on one side, and 135 foot-pounds on the other.
> >>>
> >>>Further (dubious) arithmetic says that if I apply 3600 pounds of
> >>>force directly on the end of the short side, the beam will be
> >>>balanced.
> >>>
> >>>Anyone wanna check my math?
> >>>
> >>>YIBB,
> >>>
> >>>David
> >>>
> >>>
> >>>CRUMBLING EMPIRE PRODUCTIONS
> >>>134 W.26th St. 12th Floor
> >>>New York, NY 10001
> >>>(212) 243-1636
> >>>
> >>>
> >>>Bolger rules!!!
> >>>- no cursing
> >>>- stay on topic
> >>>- use punctuation
> >>>- add your comments at the TOP and SIGN your posts
> >>>- add some content: send "thanks!" and "ditto!" posts off-list.
> >>>
> >>>
> >>
> >>
> >>
> >>Bolger rules!!!
> >>- no cursing
> >>- stay on topic
> >>- use punctuation
> >>- add your comments at the TOP and SIGN your posts
> >>- add some content: send "thanks!" and "ditto!" posts off-list.
> >
> >
> >CRUMBLING EMPIRE PRODUCTIONS
> >134 W.26th St. 12th Floor
> >New York, NY 10001
> >(212) 243-1636
> >
> >
> >Bolger rules!!!
> >- no cursing
> >- stay on topic
> >- use punctuation
> >- add your comments at the TOP and SIGN your posts
> >- add some content: send "thanks!" and "ditto!" posts off-list.
> >
> >

Jim:

No pienso que tiene absolutamente razón.

There was a long discussion about this subject on the boatdesign discussion
group a while back. Before that, I thought that water ballast had to be
above the water line to have any effect, too. Here is the mental example
that showed me the light:

Imagine you have a "boat" that is a perfect cylinder. The weight is such
that it is 3/4 submerged in water. (More like a submarine, but OK for the
sake of this discussion) Now what we want to do is attach a bladder of
water to the INSIDE of the tube. First we rotate the "hull" until the
bladder is straight up and let it go. Notice that the tube rolls over and
comes to rest with the water ballast down. Next we rotate the tube until
the ballast is high on one side, but just below the water line. Now release
the tube and what happens? Usted está correcto, saltamontes. Again the
tube turns until the water is down. True, the effect of the water ballast
is slight, but it does work above the waterline.

Now try this with the bladder attached to the outside of the tube. In this
case, (assuming the bladder itself is weightless) the ballast drops below
the waterline and has no further effect. From this experiment, you can see
that the subject is complicated (form and weight distribution are factors),
but you can not say that water ballast is only effective above the water
line.

( Thanks to Lew Clayman, and Peter Vanderwaart for their inspiration)

Chuck

> No soy estimado, I'm approximato
> Pero for my two cents----
> As long as the water filled compartment is submerged, its only effect is

to

> reduce the vessel's buoyancy. After all, it weighs the same as the rest of

the

> water all around the boat. When, however, you lift that filled ballast

tank

> above the ocean's (or lake's, or river's) surface it weighs much more than

the

> air and generates a gravity vector downward from its center of mass. If

that

> center is out on an arm from the boat's center of buoyancy, the water

ballast

> adds to the total righting moment acting on the boat.
>
> Bolger's boxes enjoy a huge movement of the center of buoyancy towards

their

> top deck as they heel over and submerge more and more of their square

topside

> shape. That allows water ballast boxes located out towards the chine to

gain

> effectiveness,
> an advantage that low sided hulls, such as the traditional sharpie, can't
> offer. To be self righting that type of boat needs to employ different
> strategies such as lots of ballast. Traditional sharpies didn't, by the

way,

> and were not self righting.
>
> Jim

Agree - the only important thing is the position of the Center of
Mass, coincident with the barycenter in a constant gravitational
field ;-), around which a boat moves. Once you have that, as well as
the displacement of course, the shape of the boat dictates the
righting moment as a function of the heel angle.
Best, Pippo

--- Inbolger@egroups.com, GHC <ghartc@p...> wrote:
> True (makes no difference)

Bill
On the other hand if you have a cubic foot of lead that weights 708
pounds on a scale on dry land it will only weigh about 646 ounds ybder
the water because it will displace 62 pounds of water and thus be more
bouant than its weight.
Jeffery

--- Inbolger@egroups.com, "Bill Jochems" <wjochems@s...> wrote:
> Estimado grupo Bolger
> I have a question about something which is related, I think, to
David
> Ryan's question about 100 lbs. centered vs. 50 lbs. on each side.
Bolger
> has written that water ballast is of no effect until it has been
lifted
> above the water line. I don't understand this. It seems to me that
weight
> is weight, be it water, concrete or lead; and that a boat would
behave the
> same whether ballasted with 500 lbs. of water, or 500 lbs of lead,
if the
> centers of gravity of each were the same.
> Bill

No soy estimado, I'm approximato
Pero for my two cents----
As long as the water filled compartment is submerged, its only effect is to
reduce the vessel's buoyancy. After all, it weighs the same as the rest of the
water all around the boat. When, however, you lift that filled ballast tank
above the ocean's (or lake's, or river's) surface it weighs much more than the
air and generates a gravity vector downward from its center of mass. If that
center is out on an arm from the boat's center of buoyancy, the water ballast
adds to the total righting moment acting on the boat.

Bolger's boxes enjoy a huge movement of the center of buoyancy towards their
top deck as they heel over and submerge more and more of their square topside
shape. That allows water ballast boxes located out towards the chine to gain
effectiveness,
an advantage that low sided hulls, such as the traditional sharpie, can't
offer. To be self righting that type of boat needs to employ different
strategies such as lots of ballast. Traditional sharpies didn't, by the way,
and were not self righting.

Jim

Bill Jochems wrote:

> Estimado grupo Bolger
> I have a question about something which is related, I think, to David
> Ryan's question about 100 lbs. centered vs. 50 lbs. on each side. Bolger
> has written that water ballast is of no effect until it has been lifted
> above the water line. I don't understand this. It seems to me that weight
> is weight, be it water, concrete or lead; and that a boat would behave the
> same whether ballasted with 500 lbs. of water, or 500 lbs of lead, if the
> centers of gravity of each were the same.
> Bill Jochems
> -----Original Message-----
> From: David Ryan <david@...>
> To:bolger@egroups.com<bolger@egroups.com>
> Date: Monday, October 30, 2000 10:17 AM
> Subject: Re: [bolger] Physics?
>
> >Hooray! Dr. Wah would be proud, even if he never did understand what
> >I was doing in his class.
> >
> >Moving on.
> >
> >In a flat bottomed boat, is there a difference between putting 100#
> >dead in the center of the center, or putting 50# on each chine?
> >
> >YIBB,
> >
> >David
> >
> >
> >
> >>That's about right. Let the weight of your 900# beam act in the middle;
> the
> >>lever arm is from fulcrum is then (15-3)=12', so the moment is
> 12*900=10800
> >>ft-lb. To balance the beam, you need to apply 10800/3 = 3600# on the
> short
> >>end (i.e. 3*3600=10800 => equilibrium).
> >>
> >>(To figure it much, much more simply, convert your units to SI metric,
> >>divide by 1000 and then once again by .001 ;-)
> >>
> >>Gregg Carlson
> >>
> >>At 09:58 AM 10/30/2000 -0500, you wrote:
> >>>FBBB --
> >>>
> >>>It's been 12 years since I cracked a physical text book (my degree is
> >>>in fine art, go figure.)
> >>>
> >>>Anyone care to check me on this:
> >>>
> >>>A 30 foot beam that weighs 30 pounds per foot is balanced on a
> >>>fulcrum 3 feet from one end (and 27 feet from the other.)
> >>>
> >>>By my (suspect) calculations, there is a torque of 10935 foot-pounds
> >>>on one side, and 135 foot-pounds on the other.
> >>>
> >>>Further (dubious) arithmetic says that if I apply 3600 pounds of
> >>>force directly on the end of the short side, the beam will be
> >>>balanced.
> >>>
> >>>Anyone wanna check my math?
> >>>
> >>>YIBB,
> >>>
> >>>David
> >>>
> >>>
> >>>CRUMBLING EMPIRE PRODUCTIONS
> >>>134 W.26th St. 12th Floor
> >>>New York, NY 10001
> >>>(212) 243-1636
> >>>
> >>>
> >>>Bolger rules!!!
> >>>- no cursing
> >>>- stay on topic
> >>>- use punctuation
> >>>- add your comments at the TOP and SIGN your posts
> >>>- add some content: send "thanks!" and "ditto!" posts off-list.
> >>>
> >>>
> >>
> >>
> >>
> >>Bolger rules!!!
> >>- no cursing
> >>- stay on topic
> >>- use punctuation
> >>- add your comments at the TOP and SIGN your posts
> >>- add some content: send "thanks!" and "ditto!" posts off-list.
> >
> >
> >CRUMBLING EMPIRE PRODUCTIONS
> >134 W.26th St. 12th Floor
> >New York, NY 10001
> >(212) 243-1636
> >
> >
> >Bolger rules!!!
> >- no cursing
> >- stay on topic
> >- use punctuation
> >- add your comments at the TOP and SIGN your posts
> >- add some content: send "thanks!" and "ditto!" posts off-list.
> >
> >
>
>
> Bolger rules!!!
> - no cursing
> - stay on topic
> - use punctuation
> - add your comments at the TOP and SIGN your posts
> - add some content: send "thanks!" and "ditto!" posts off-list.

> As to centers of buoyancy, I think this is a bit of a red herring.

If

> you ballast with lead, your center of buoyancy *isn't* going to be
> the same. You can make lead as ineffective as water, but why would
> you?

Getting the correct answers about water ballast depends on being very
clear about what you are comparing. Are you considering the water
ballast as

1) placed on the inside of an otherwise unballasted hull

or

2) attached to the outside of an otherwise unballasted hull

or

3) used instead of metal ballast.

If you take viewpoint 2, then water ballast has no effect until
raised above the water level. Otherwise, it does.

Peter

True (makes no difference) - search the group archive for more than you
ever wanted to know 'bout this.

Gregg Carlson

At 10:53 AM 10/30/2000 -0700, you wrote:

>Estimado grupo Bolger
> I have a question about something which is related, I think, to David
>Ryan's question about 100 lbs. centered vs. 50 lbs. on each side. Bolger
>has written that water ballast is of no effect until it has been lifted
>above the water line. I don't understand this. It seems to me that weight
>is weight, be it water, concrete or lead; and that a boat would behave the
>same whether ballasted with 500 lbs. of water, or 500 lbs of lead, if the
>centers of gravity of each were the same.
>Bill Jochems
>-----Original Message-----
>From: David Ryan <david@...>
>To:bolger@egroups.com<bolger@egroups.com>
>Date: Monday, October 30, 2000 10:17 AM
>Subject: Re: [bolger] Physics?
>
>
>>Hooray! Dr. Wah would be proud, even if he never did understand what
>>I was doing in his class.
>>
>>Moving on.
>>
>>In a flat bottomed boat, is there a difference between putting 100#
>>dead in the center of the center, or putting 50# on each chine?
>>
>>YIBB,
>>
>>David
>>
>>
>>
>>>That's about right. Let the weight of your 900# beam act in the middle;
>the
>>>lever arm is from fulcrum is then (15-3)=12', so the moment is
>12*900=10800
>>>ft-lb. To balance the beam, you need to apply 10800/3 = 3600# on the
>short
>>>end (i.e. 3*3600=10800 => equilibrium).
>>>
>>>(To figure it much, much more simply, convert your units to SI metric,
>>>divide by 1000 and then once again by .001 ;-)
>>>
>>>Gregg Carlson
>>>
>>>At 09:58 AM 10/30/2000 -0500, you wrote:
>>>>FBBB --
>>>>
>>>>It's been 12 years since I cracked a physical text book (my degree is
>>>>in fine art, go figure.)
>>>>
>>>>Anyone care to check me on this:
>>>>
>>>>A 30 foot beam that weighs 30 pounds per foot is balanced on a
>>>>fulcrum 3 feet from one end (and 27 feet from the other.)
>>>>
>>>>By my (suspect) calculations, there is a torque of 10935 foot-pounds
>>>>on one side, and 135 foot-pounds on the other.
>>>>
>>>>Further (dubious) arithmetic says that if I apply 3600 pounds of
>>>>force directly on the end of the short side, the beam will be
>>>>balanced.
>>>>
>>>>Anyone wanna check my math?
>>>>
>>>>YIBB,
>>>>
>>>>David
>>>>
>>>>
>>>>CRUMBLING EMPIRE PRODUCTIONS
>>>>134 W.26th St. 12th Floor
>>>>New York, NY 10001
>>>>(212) 243-1636
>>>>
>>>>
>>>>Bolger rules!!!
>>>>- no cursing
>>>>- stay on topic
>>>>- use punctuation
>>>>- add your comments at the TOP and SIGN your posts
>>>>- add some content: send "thanks!" and "ditto!" posts off-list.
>>>>
>>>>
>>>
>>>
>>>
>>>Bolger rules!!!
>>>- no cursing
>>>- stay on topic
>>>- use punctuation
>>>- add your comments at the TOP and SIGN your posts
>>>- add some content: send "thanks!" and "ditto!" posts off-list.
>>
>>
>>CRUMBLING EMPIRE PRODUCTIONS
>>134 W.26th St. 12th Floor
>>New York, NY 10001
>>(212) 243-1636
>>
>>
>>Bolger rules!!!
>>- no cursing
>>- stay on topic
>>- use punctuation
>>- add your comments at the TOP and SIGN your posts
>>- add some content: send "thanks!" and "ditto!" posts off-list.
>>
>>
>
>
>
>Bolger rules!!!
>- no cursing
>- stay on topic
>- use punctuation
>- add your comments at the TOP and SIGN your posts
>- add some content: send "thanks!" and "ditto!" posts off-list.
>
>

Well, we didn't discover the source of the 1% error, but that's close enough!

Anyway, indeed there is!

By spreading the mass apart, you increase the polar moment of inertia,
which will dampen or slow the roll rate or response to an impulse like a
gust of wind or wave. (It will do nothing to improve ultimate righting
moment.)

470 sailors (and similar 2-man light dinghys) use this idea to adjust the
pitch response to waves - i.e crew spreads apart to dampen waves motion or
scoots together to make boat more "pitchy".

Gregg Carlson

At 12:16 PM 10/30/2000 -0500, you wrote:

>Hooray! Dr. Wah would be proud, even if he never did understand what
>I was doing in his class.
>
>Moving on.
>
>In a flat bottomed boat, is there a difference between putting 100#
>dead in the center of the center, or putting 50# on each chine?
>
>YIBB,
>
>David
>
>
>
>>That's about right. Let the weight of your 900# beam act in the middle; the
>>lever arm is from fulcrum is then (15-3)=12', so the moment is 12*900=10800
>>ft-lb. To balance the beam, you need to apply 10800/3 = 3600# on the short
>>end (i.e. 3*3600=10800 => equilibrium).
>>
>>(To figure it much, much more simply, convert your units to SI metric,
>>divide by 1000 and then once again by .001 ;-)
>>
>>Gregg Carlson
>>
>>At 09:58 AM 10/30/2000 -0500, you wrote:
>>>FBBB --
>>>
>>>It's been 12 years since I cracked a physical text book (my degree is
>>>in fine art, go figure.)
>>>
>>>Anyone care to check me on this:
>>>
>>>A 30 foot beam that weighs 30 pounds per foot is balanced on a
>>>fulcrum 3 feet from one end (and 27 feet from the other.)
>>>
>>>By my (suspect) calculations, there is a torque of 10935 foot-pounds
>>>on one side, and 135 foot-pounds on the other.
>>>
>>>Further (dubious) arithmetic says that if I apply 3600 pounds of
>>>force directly on the end of the short side, the beam will be
>>>balanced.
>>>
>>>Anyone wanna check my math?
>>>
>>>YIBB,
>>>
>>>David
>>>
>>>
>>>CRUMBLING EMPIRE PRODUCTIONS
>>>134 W.26th St. 12th Floor
>>>New York, NY 10001
>>>(212) 243-1636
>>>
>>>
>>>Bolger rules!!!
>>>- no cursing
>>>- stay on topic
>>>- use punctuation
>>>- add your comments at the TOP and SIGN your posts
>>>- add some content: send "thanks!" and "ditto!" posts off-list.
>>>
>>>
>>
>>
>>
>>Bolger rules!!!
>>- no cursing
>>- stay on topic
>>- use punctuation
>>- add your comments at the TOP and SIGN your posts
>>- add some content: send "thanks!" and "ditto!" posts off-list.
>
>
>CRUMBLING EMPIRE PRODUCTIONS
>134 W.26th St. 12th Floor
>New York, NY 10001
>(212) 243-1636
>
>
>Bolger rules!!!
>- no cursing
>- stay on topic
>- use punctuation
>- add your comments at the TOP and SIGN your posts
>- add some content: send "thanks!" and "ditto!" posts off-list.
>
>

>On Mon, 30 Oct 2000, David Ryan wrote:
>> In a flat bottomed boat, is there a difference between putting 100#
>> dead in the center of the center, or putting 50# on each chine?
>
>Yes.
>
>One difference, in heeling, the outboard weight will generate greater
>lever arm.
>

But won't this be counteracted by the weight on the other side?

-D

CRUMBLING EMPIRE PRODUCTIONS
134 W.26th St. 12th Floor
New York, NY 10001
(212) 243-1636

Estimado grupo Bolger
I have a question about something which is related, I think, to David
Ryan's question about 100 lbs. centered vs. 50 lbs. on each side. Bolger
has written that water ballast is of no effect until it has been lifted
above the water line. I don't understand this. It seems to me that weight
is weight, be it water, concrete or lead; and that a boat would behave the
same whether ballasted with 500 lbs. of water, or 500 lbs of lead, if the
centers of gravity of each were the same.
Bill Jochems
-----Original Message-----
From: David Ryan <david@...>
To:bolger@egroups.com<bolger@egroups.com>
Date: Monday, October 30, 2000 10:17 AM
Subject: Re: [bolger] Physics?

>Hooray! Dr. Wah would be proud, even if he never did understand what
>I was doing in his class.
>
>Moving on.
>
>In a flat bottomed boat, is there a difference between putting 100#
>dead in the center of the center, or putting 50# on each chine?
>
>YIBB,
>
>David
>
>
>
>>That's about right. Let the weight of your 900# beam act in the middle;

the

>>lever arm is from fulcrum is then (15-3)=12', so the moment is

12*900=10800

>>ft-lb. To balance the beam, you need to apply 10800/3 = 3600# on the

short

>>end (i.e. 3*3600=10800 => equilibrium).
>>
>>(To figure it much, much more simply, convert your units to SI metric,
>>divide by 1000 and then once again by .001 ;-)
>>
>>Gregg Carlson
>>
>>At 09:58 AM 10/30/2000 -0500, you wrote:
>>>FBBB --
>>>
>>>It's been 12 years since I cracked a physical text book (my degree is
>>>in fine art, go figure.)
>>>
>>>Anyone care to check me on this:
>>>
>>>A 30 foot beam that weighs 30 pounds per foot is balanced on a
>>>fulcrum 3 feet from one end (and 27 feet from the other.)
>>>
>>>By my (suspect) calculations, there is a torque of 10935 foot-pounds
>>>on one side, and 135 foot-pounds on the other.
>>>
>>>Further (dubious) arithmetic says that if I apply 3600 pounds of
>>>force directly on the end of the short side, the beam will be
>>>balanced.
>>>
>>>Anyone wanna check my math?
>>>
>>>YIBB,
>>>
>>>David
>>>
>>>
>>>CRUMBLING EMPIRE PRODUCTIONS
>>>134 W.26th St. 12th Floor
>>>New York, NY 10001
>>>(212) 243-1636
>>>
>>>
>>>Bolger rules!!!
>>>- no cursing
>>>- stay on topic
>>>- use punctuation
>>>- add your comments at the TOP and SIGN your posts
>>>- add some content: send "thanks!" and "ditto!" posts off-list.
>>>
>>>
>>
>>
>>
>>Bolger rules!!!
>>- no cursing
>>- stay on topic
>>- use punctuation
>>- add your comments at the TOP and SIGN your posts
>>- add some content: send "thanks!" and "ditto!" posts off-list.
>
>
>CRUMBLING EMPIRE PRODUCTIONS
>134 W.26th St. 12th Floor
>New York, NY 10001
>(212) 243-1636
>
>
>Bolger rules!!!
>- no cursing
>- stay on topic
>- use punctuation
>- add your comments at the TOP and SIGN your posts
>- add some content: send "thanks!" and "ditto!" posts off-list.
>
>

On Mon, 30 Oct 2000, David Ryan wrote:

> In a flat bottomed boat, is there a difference between putting 100#
> dead in the center of the center, or putting 50# on each chine?

Yes.

One difference, in heeling, the outboard weight will generate greater
lever arm.

Hooray! Dr. Wah would be proud, even if he never did understand what
I was doing in his class.

Moving on.

In a flat bottomed boat, is there a difference between putting 100#
dead in the center of the center, or putting 50# on each chine?

YIBB,

David

>That's about right. Let the weight of your 900# beam act in the middle; the
>lever arm is from fulcrum is then (15-3)=12', so the moment is 12*900=10800
>ft-lb. To balance the beam, you need to apply 10800/3 = 3600# on the short
>end (i.e. 3*3600=10800 => equilibrium).
>
>(To figure it much, much more simply, convert your units to SI metric,
>divide by 1000 and then once again by .001 ;-)
>
>Gregg Carlson
>
>At 09:58 AM 10/30/2000 -0500, you wrote:
>>FBBB --
>>
>>It's been 12 years since I cracked a physical text book (my degree is
>>in fine art, go figure.)
>>
>>Anyone care to check me on this:
>>
>>A 30 foot beam that weighs 30 pounds per foot is balanced on a
>>fulcrum 3 feet from one end (and 27 feet from the other.)
>>
>>By my (suspect) calculations, there is a torque of 10935 foot-pounds
>>on one side, and 135 foot-pounds on the other.
>>
>>Further (dubious) arithmetic says that if I apply 3600 pounds of
>>force directly on the end of the short side, the beam will be
>>balanced.
>>
>>Anyone wanna check my math?
>>
>>YIBB,
>>
>>David
>>
>>
>>CRUMBLING EMPIRE PRODUCTIONS
>>134 W.26th St. 12th Floor
>>New York, NY 10001
>>(212) 243-1636
>>
>>
>>Bolger rules!!!
>>- no cursing
>>- stay on topic
>>- use punctuation
>>- add your comments at the TOP and SIGN your posts
>>- add some content: send "thanks!" and "ditto!" posts off-list.
>>
>>
>
>
>
>Bolger rules!!!
>- no cursing
>- stay on topic
>- use punctuation
>- add your comments at the TOP and SIGN your posts
>- add some content: send "thanks!" and "ditto!" posts off-list.

CRUMBLING EMPIRE PRODUCTIONS
134 W.26th St. 12th Floor
New York, NY 10001
(212) 243-1636

That's about right. Let the weight of your 900# beam act in the middle; the
lever arm is from fulcrum is then (15-3)=12', so the moment is 12*900=10800
ft-lb. To balance the beam, you need to apply 10800/3 = 3600# on the short
end (i.e. 3*3600=10800 => equilibrium).

(To figure it much, much more simply, convert your units to SI metric,
divide by 1000 and then once again by .001 ;-)

Gregg Carlson

At 09:58 AM 10/30/2000 -0500, you wrote:

>FBBB --
>
>It's been 12 years since I cracked a physical text book (my degree is
>in fine art, go figure.)
>
>Anyone care to check me on this:
>
>A 30 foot beam that weighs 30 pounds per foot is balanced on a
>fulcrum 3 feet from one end (and 27 feet from the other.)
>
>By my (suspect) calculations, there is a torque of 10935 foot-pounds
>on one side, and 135 foot-pounds on the other.
>
>Further (dubious) arithmetic says that if I apply 3600 pounds of
>force directly on the end of the short side, the beam will be
>balanced.
>
>Anyone wanna check my math?
>
>YIBB,
>
>David
>
>
>CRUMBLING EMPIRE PRODUCTIONS
>134 W.26th St. 12th Floor
>New York, NY 10001
>(212) 243-1636
>
>
>Bolger rules!!!
>- no cursing
>- stay on topic
>- use punctuation
>- add your comments at the TOP and SIGN your posts
>- add some content: send "thanks!" and "ditto!" posts off-list.
>
>

FBBB --

It's been 12 years since I cracked a physical text book (my degree is
in fine art, go figure.)

Anyone care to check me on this:

A 30 foot beam that weighs 30 pounds per foot is balanced on a
fulcrum 3 feet from one end (and 27 feet from the other.)

By my (suspect) calculations, there is a torque of 10935 foot-pounds
on one side, and 135 foot-pounds on the other.

Further (dubious) arithmetic says that if I apply 3600 pounds of
force directly on the end of the short side, the beam will be
balanced.

Anyone wanna check my math?

YIBB,

David


CRUMBLING EMPIRE PRODUCTIONS
134 W.26th St. 12th Floor
New York, NY 10001
(212) 243-1636