Re: [PSUBS-MAILIST] Weighing

[MerlinSub@t-online.de (Carsten Standfuss)]

Nice.. you have to work more acurate than 
with the bathroom scale experiment sample :-)

For the reason than the most parts of a subs hull like the cylinder, the
frame 
and the endcaps, stern conus, front dome etc. have there CM and CB in
the middle 
line of the cylinder - I prefer to calculate the vertical CM. Its quick
done 
- just some 14 days on Cssx.. Smaller items on board have not a so great
influence 
on it. 

I am not so affraid about the vertial CM to CB because this can be
calculate 
in a very early project stage. Will not move much during building. 

The longitudinal CM to CB is a much bigger deal
because of the long lever any mass at the ends of the boats have. 

The stern of torpedo shapped long Sgt.Peppers for example is almost free
flooding,
no bouancy/lift there. I calculate the stern engine weight with 7 kg (14
pounds) but the real things was more in the 14 kg (28 pounds) area. Put
all the lead in the bow area.. was boat was
still to heavy on the stern.. remove the plastic bow thruster pipes and
made 
some from steel. Works better but for the reason that the bow is much
closer to CB than the 
stern didn't reach zero.. Final solution was to remove stern battery
No.4
Lost 25 % of endurance - just because the main stern engine was a little
to heavy..   

Back to the bathroom scale - just told your neighbours  that you want to
make a 
SWB : a Submarine-Weighing-Barbecue - and they have to come with there
scales.. 
use 3-4 with a longer angle iron should fix the 2 inch problem to a 6-8
inch..  

How can I make the angel iron experiments on a 57 ts boat - without to
cut the boat 
in two - or lower/sunk the iron into the street ? :-) 

For the other guys.. note that the CB on surface station can be
different form the CB
on diving station. 

Sample 1
Imagine a sub with a big pressuretight tower in the bow area. 
Imagine that this boat is perfect in trim horizontal underwater (CB on
top of CM at zero degree) 
- if this boat surfaced the CB moved aft and the boat stick the nose
deeper into the 
water than the stern - create a trim to the nose on surface station. 

Sample 2
The same boat - but in perfect horizontal trim on the surface. If the
boat dive the tower creates
bouancy forward of CM and the nose raised. 
For this reason boats with a pressuretight big tower forward of CM
should be in a trim stage 
of laying some degrees to the stern on the surface.

Carsten

Gary R. Boucher schrieb:
> 
>      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