[Date Prev][Date Next]
[Chronological]
[Thread]
[Top]
Re: [PSUBS-MAILIST] Weighing
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