Re: [PSUBS-MAILIST] Weighing

["Gary R. Boucher" <protek@shreve.net>]

Doc,
     All of the calculations are for the sub outside of the water.  The 
Center of Mass is independent of being inside or outside of the 
medium.  The Center of Buoyancy is different.  The CB is usually 
calculated.  I had a spread sheet that I used where I listed EVERYTHING 
that went into the water.  For example; even a steel bar used for support 
had it's volume calculated.  It was given an X, and Y position.  Each 
volume multiplied by the weight density of water gave a small upward 
force.  Each force was summed as a torque about the X=0 and Y=0 axis of my 
sub.  The result is the CB.  The CB MUST be located above the CM or the sub 
will roll inverted.  This is good for excitement, but not good for the crew 
that has to right the mistake.
     Also, remember that nothing inside the hull contributes to the 
CB.  The Hull itself is the largest component effecting buoyancy.  Thought, 
only things that go into the water can generate buoyancy.
     Payload in many airplanes is a large percent of the aircraft's 
weight.  Moderate payloads in a sub are usually a much smaller ratio to the 
weight of the sub.
     Remember that the CM will always swing down to be located below the 
CB.  If you add payload and the CM changes by going forward, that new CM 
will rotate the sub bow downward until the CM is directly below the CB.  In 
Most subs the CB much higher than their CM.  Therefore a small longitudinal 
change in CM will cause only a small rotation.  In my sub the CB is very 
close to the CM.  Therefore if I change my CM by just an inch or so I have 
a very noticeable angle up or down as a result.

G Boucher



At 09:03 AM 8/4/2002, you wrote:
>Howdy Gary
>
>UUUUH, pardon my mathematical ignorance, but is the calculation of the CM
>outside the operational medium (H2O) a moot point given the locations of 
>static
>buoyancy and variable payload ballast?
>
>  Generally, the scale we're working on means that weight and balance may
>be greatly affected by considerably small variations in payload weight.
>
>Or am I still thinking too much in terms of airplanes?
>
>Just curious
>
>Doc
>
>"Gary R. Boucher" wrote:
>
> >      Sometimes we in the States refer to Trigonometry as "Trig."  As to
> > finding the vertical CM, imagine that position directly above the fulcrum
> > when the sub is level and balanced.  Call this vertical distance to the CM
> > location (Ycm).  If you allow the sub to pivot forward just a few degrees
> > this CM moves forward, actually rotating around the fulcrum pivot with
> > constant radius Ycm.  This distance is what we are looking for.  The CM of
> > the boat is always a Ycm away from the fulcrum.  Now when the sub has
> > rotated slightly (just a few degrees) the CM would project downward on a
> > true horizontal plane slightly in front of the fulcrum.  Call this forward
> > horizontal projection (H).  Call the bow down angle of the sub 
> (Ang).  Thus:
> >
> > Cos Ang = H / Ycm
> >
> > or
> >
> > Ycm = H / Cos Ang
> >
> > Now a forward torque equal to W (weight of the sub) times H (moment arm we
> > just defined) is generated holding the bow down firmly.
> >
> > To lift the bow we go to the stern and add weights until the point where it
> > just barely "Begins" to move toward level from its bow down angle Ang.  The
> > weights are placed on a vertical line attached to the stern where that
> > line's distance to the fulcrum in a true horizontal direction is (D).  The
> > amount of weight added lets call (WA).
> >
> > Just as the bow starts to rise from its resting position the WA x D has
> > created a torque equal, but opposite, to the W x H shown above.  Therefore:
> >
> > W x H = WA x D
> >
> > or
> >
> > H = ( WA x D ) / W
> >
> > and finally
> >
> > Ycm = [ (WA x D ) / W ]  / Cos Ang
> >
> > Now you don't want to measure the angle (Ang).  It should be calculated
> > from drop of the bow and distance from the fulcrum.  It requires great
> > accuracy and a protractor is not sufficient.  The angle may be only a few
> > degrees.
> >
> > This technique allowed me to find the CM in the vertical plane.  My sub
> > behaved exactly as predicted.
> >
> > Gary Boucher
> >
> > >Gary I did not understand what you excact did.
> > >Can you explain the little "trig" to
> > >meassure and calculate the vertical CM ?
> > >
> > >best regards Carsten