What is the simplest formula for predicting the changes in freeboard of
a floating cube when weight is added to one edge?

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Clarification of Question by sir-ga on 20 May 2002 19:53 PDT

hedgie-ga,

SIMPLE is the key word here.  The intention is to find a simple means
to predict a change in waterline of a small barge when weight is added
to one edge.  The weight would not be enough to alter the floating
condition of the barge by more than fifteen degrees.  It would not
topple.  The center of the weight would be at the longitudinal center
and would only cause a change in heel or roll. The barge is
rectangular with straight, vertical sides.

EXAMPLE:  A barge of 40' length, 20' width and 5' height has a draft
(immersed depth) all the way around of 4' and 1' of freeboard (ie.
distance from the water to the top of the barge) when floating in salt
water that has a density of 64pounds per cubic foot.  1000 pounds is
hung by a wire over the starboard side at the longitudinal center.

How far will the oppasite side rise.
How far will the weighted side go down.

If this is the only thing we need to know in every case and the only
variables that will ever change are
draft
length
width
weight applied
what is the SIMPLEST formula to solve for the change in the port and
starboard waterlines.

Sir

do you know where the center of mass of the barge is?

if you did this with a cube, of course, the cube would just spin so
that the point where the weight was attached was on the bottom....  so
the critical thing for stability is that the center of mass be as low
as possible compared to the center of buoyancy....

1000 pounds will lower the barge as a whole 1000/(40x20x64) = 0.0195
feet = .235 inches (assuming that the 1000-pound weight is very dense
compared with water, otherwise you need to correct for that, of
course)

the barge is currently displacing 40x20x4x64 = 204800 pounds of water,
i.e. that's how much it weighs.

if we assume that the center of mass is originally a distance X below
the center of buoyancy (which is currently two feet below the water),
then the barge will rotate until the new center of mass is below the
center of the barge.

the new center of mass is offset by 1000 pounds x 10 feet / 204800
pounds = 0.0488 feet = .586 inches.

so the angle the barge will rotate is arctan(.586 inches / X)... if
the center of mass is one foot from the bottom of the boat, say, that
would be 12 inches below the COB, so the angle is arctan(.586/12)=2.8
degrees....  at the edge of the boat, that's a movement of 10 feet x
sin(angle) = 5.85 inches.

for small angles, this simplifies to vertical = weight * offset *
offset / (bargeweight * X)

perhaps a simpler way to put it is (offset^2) * (ratio of added weight
to barge weight) * (X)

where X, again, is the distance between the center of mass and the
center of buoyancy.

remember to add back in the .235 inches the whole barge sinks into the
water from the added weight, so that's just over six inches total.

(again, this is only an approximation, but it gives a simple rule of
thumb about how much you'll tip.  please pay somebody more than $4 to
figure this out for you if you're about to actually sink a boat!)

I enjoyed working out this answer, I hope you find this helpful. 
Thanks for using Google Answers!

good answer, well written, clear.

I spent 4 hours last night looking through Navy engineering practice
tests, pulled out my oldschool physics book, and then came to the
conclusion that without taking a course in fluid dynamics, I wouldn't
be able to answer this question... nice stumper!

I feel that the question is not fully defined:

LET US SAY
 cube has a side of length A  and density D [in g per cubic cm = g*(cm)^(-3) ]
 weight is a wire of length A, attached to one unsubmerged edge,
 which has total mass M and crossection C.

 THEN,
 depending on M and D and A, the cube may topple (wire in the water) 
 or just tilt (wire  still in the air).

 I do not see a clear measure of FREEBOARD. Do you want formula
 for low and high  water-edge distance or just formula for stability - i.e.
 for what values  the cube will topple?

Thanks for the clarification  Sir.

   The approximate solution, valid for small deviation from
   the unperturbed situation can be aproximated by a simple
   formula. The general case can only be solved numerically,
   (by finding roots some trigonometric expressions) and
   cannot be exactly expressed by an algebraic formula.



     You came up with a nice problem, Sir, but
     you did throw the barge out  with the barge-water :-):

   By representing the barge by a cube, you oversimplified
   (as pointed out by  davidmaymudes-ga ) too much.  Floating
   cube is only stable for very low (<.2) or very high (.8)
   densities. (These are in gram per cubic cm, water being 1.)

   The ratio W/H is essential. Pencil will float on the side, but
   not be stable when vertical, right? There is a  simple stability
   criterion tying W/H to D.

   So lets restate the problem as follows:

Lets have a rectangular block of wood
   (with volume Width*Height*Length and density D) floating
   in the water. Lets place weight of Mass M on one edge,
   (either at the center or distributed uniformly).
Then
    Tilt = X * (M*g)  , where M*g is weight of added load,

    tilt is angle in radians, or (which as about the same for
    small angles) sin of that angle.

    WHAT IS THE X?  (for small values of M, Tilt < 15 * 3.1 /180
radians).

---------------------------------------------------------------------

    That can be answered (but may take more then $5 ;-).

    Did davidmaymudes answered the question?

    I am not sure if I  can fully follow his reasoning: When he says:

>    if the center of mass is one foot from the bottom of the boat,
say, that..

    Is that an assumption or is that so for this shape?

  The quantity X should have dimension of inverse force (since tilt is
dimesionless)

 The simplified formula he gives : vertical = offset^2 *
(Wsmall/Wbarge)

 does not seem to have consistent dimensions.  May be he can clarify
this?

OK for those who wonder...

I just came up with a simpler method that works for small angles of
heel and rotation around an axis near the waterline which covers most
ballasting type situations.

Disregard the righting arm stuff, finding center of bouyancy-center of
gravity etc.

when a barge shaped object is heeled some of the barge comes out of
the water and some goes in.  These areas are triangular in section. 
the force they produce through lost bouyancy on one side and gained
bouyancy opn the other is equal to their volume times the density of
the medium.  If you flip one over to the other side the two triangluar
sections make one rectangular section the voolume of which is easy to
calculate.

seawater weighs 64 pounds per cubic foot.  

if the combined sections weigh 500 pounds and their outer edge is 10
feet from the center of the barge multiply 2/3 of the 10 foot radius
by the weight to get the rotational force in foot pounds.

When you know how much force it takes to sink one side of the barge 1
inch you can multiply by 6 to find out how much it takes to sink one
side of the barge six inches.

If you go more than 3-4 degrees this system probably goes to hell
because the barge begins rotating around an axis that is no longer in
the center and the heavy side goes down progressivley more than the
light side goes up.  But hey I can do this math in a few minutes
without a scientific calculator..

I'm going to try this out and see if it really works.
Can anyone run a proper calculation and tell me if I'm on the right
track?