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Re: [PSUBS-MAILIST] Weighing
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
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