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Re: [PSUBS-MAILIST] Weighing

To the weighers out there,

     Carsten has a really good idea there!  Yes, it  would work.  When the 
sub is balanced on the fulcrum the Center of Mass (CM) is directly above 
the pivot point of the fulcrum.  As I mentioned, you can hold a three ton 
mass balanced with just a tiny force from one finger on the bow or 
stern.  If you mark the position of the fulcrum and then move it, say two 
inches aft of the CM, the bow would then exhibit a downward force because 
of the offset of the mass.  We would then have to apply a force upward on 
the bow (through the bathroom scale) to bring the sub back into 
balance.  In my physics classes, we call this balance Rotational 
Equilibrium.  It means that the right-hand torque is equal to the left-hand 
torque, just as it was when the fulcrum was under the CM.
     For those interested in this technique you should be aware if the 
potential error.  If, say that the sub weighs 5500 pounds.  When you move 
the fulcrum aft by two inches it would set up a torque equal to (5500 
Lbs)x(2 Inches) or 11,000 In-Lbs.  If the distance between the scales and 
the fulcrum is say 60 inches, then the scale would read 183.3 pounds.  Thus 
the torque generated by the force through the scale would be (183.3 
Lbs)x(60 Inches) or 11,000 In-Lbs.  If we divide the 11,000 In-Lbs. by 2 
inches, we get 5500 pounds, the weight of the sub.  But say that you moved 
the fulcrum 1.9 inches by error.  Then the torque would be 10,450 
In-Lbs.  The scale would read (10,450)/(60 Inches) or 174.2 
Lbs.  Calculating the weight of the sub would yield 5,225 Lbs.  You would 
be in error by 225 Lbs.
     One would have to be extremely accurate with the placement of the 
fulcrum at the 2-inch offset.  Comparing this to Ray's suggestion of using 
a truck scale, the 20 Lb. error of such a scale would be surpassed with 
this method if the fulcrum were out just 0.00364 inche from its 2-inch 
displacement point.  To lower the error one could move the displacement 
from say 2 inches to a foot or more, but then the common bathroom scale 
would not work, as it would be greatly overloaded.
     A person could move the fulcrum say a foot to reduce error, allow the 
bow of the sub to rest on a block, then add weights to the stern at a given 
distance until equilibrium was once again reached.  The weights (wrenches, 
bars of steel, the wife, family dog, etc) could be individually weighed on 
a very accurate scale.  Second thought, leave the wife out of the weighing 
picture.  NOT a good idea!
     It is easier to have more accuracy with a 12-inch fulcrum move, than a 
2-inch fulcrum move.
     One other IMPORTANT thought for these techniques...   When adding 
weights to bring the sub into equilibrium do not add weights until the sub 
begins to move from its then bow-low position.  The amount of weight needed 
should be determined with he sub completely level.  In other words, add 
weight force the sub to level, determine if you have to add or take off 
weight, then repeat.  It is also VERY essential to have the sub level 
through all the final determinations.

G. Boucher




At 01:42 AM 8/4/2002, you wrote:
>Gary R. Boucher schrieb:
> >
> >   My thoughts on W&B
> >
>.....
> >      To determine the exact position for center of mass of my boat, I took
> > a simple piece of heavy angle iron, turning it so that the 90-degree
> > pointed edge faced upward.  I then used hoists to lift my boat and set it
> > down on the angle.  Repeated balancing efforts allowed me to zero in on the
> > exact position of the CM.  Empty, my sub weighs over 5000 pounds and I
> > could hold it balanced with one finger on this steel fulcrum.  Of course
> > calculations are fine two, but nothing gives you the exact center of mass
> > like a fulcrum.
>.....
> > Gary Boucher
>
>Hello Gary - based on this we should able to meassure the exact weight
>of
>the sub with a normal personal home scale !
>
>1.) Did your meassuring.. call this point A.
>2.) After mark the point - put the angle iron 2 inch aft
>     call this point B. This is our new turning point.
>     Add the personal scale with another angle bar
>     under the bow just 100 inch from point B = called point C.
>3.) Read the meassuring (M1)at the scale and subtract the weight of the
>     iron bar of the scale. (M1y)
>
>4.) Calculate : Weight sub (W) x AB balance  M1y x CB
>     or W =  (M1y x CB) / AB
>
>Example =
>
>4.1)
>
>Meassuring say: 120 pounds
>Iron bar on scale say 3 pound
>AB =   2 inch
>CB = 100 inch
>
>Weight of sub = ((120 pound - 3 pound) x 100 inch ) / 2 inch = 5850
>pound.
>
>No need for a road scale.
>
>Just find the longitundinal center of graphity and than use a small
>personal scale. If the scale is to small - just make the distance
>AB smaller - (but allways positiv - never zero).
>
>Carsten