An upright vessel has negative GM. GM becomes positive at the angle of loll because the __________.
A. free surface effects are reduced due to pocketing
B. KG is reduced as the vessel seeks the angle of loll
C. effective beam is increased causing BM to increase
D. underwater volume of the hull is increased
A. free surface effects are reduced due to pocketing
B. KG is reduced as the vessel seeks the angle of loll
C. effective beam is increased causing BM to increase
D. underwater volume of the hull is increased
Tricky.... D.
ReplyDeletesharp21 you are correct about this question being "Tricky". I would say this question is worth protesting for a few reasons, however if you really think about it I would have to say for no other choice given "C." is the best answer. Remember to always chose the "best answer" offered!
ReplyDeleteThe following is from my book, STABILITY AND TRIM FOR THE SHIP'S OFFICER, 4th Edition, page 104 to 105, if you have a copy:
"Effect of Negative GM on Vessels
If the center of gravity, G, lies above the transverse metacenter, M, (G above M), the vessel is in a state of unstable equilibrium, that is, she possesses a negative metacentric height, GM. Once disturbed bay an outside force, there is no tendency for the vessel to right itself at small angles of inclination. An upsetting moment is formed, and the vessel will incline, actually flop to either the port or starboard side, from the upright position. This is know as an angle of loll. An angle of loll should never be confused with list, a vessel’s inclination due to an off the centerline G.
A negative GM does not mean that the vessel will capsize. It merely means that the vessel does not have any initial stability, and that she will incline to an angle where B has moved far enough toward the low side of the vessel to be once more in the same vertical line as G. Figure 5-12 illustrates this situation. A vessel with a negative GM can actually flop from an angle of loll on one side to the other while the vessel is turning to change course while underway or during cargo operations. This condition can happen at the end of a voyage when the bunkers are burned out and ballast has not been taken to maintain a positive GM. A negative GM should always be avoided. All stability criteria indicates the vessel’s master maintain a positive GM at all times."
A NEGATIVE GM SHOULD ALWAYS BE AVOIDED!!!!!!!!!!! That said what actually happens when a vessel is in still water at an Angle of Loll. I believe this question addresses the fact that when a vessel has an angle of loll it is in equilibrium G aligned with B there is no righting arm or righting moment to restore it to a upright condition. To analyze this properly a new set of hydrostatic data is needed for this conditon. Choice "C." basically is saying for a wall sided ship the beam will increase at the waterplane which increased the moment of inertia of the waterplane. For a rectangular barge the moment of inertial of the waterplane will be similar to moment of inertia for a rectangular tank = lb^3/12. So if Beam increases slightly the moment of inertial will increase greatly!
BM, the metacentric radius, = moment of inertia of the waterplane/displaced volume.
KM = KB + BM, so if new hydrostatic data, KM, was calculated for the vessel at an angle of loll, you could find the it is perfectly stable in still water at its angle of loll and will in fact roll about it for very small angles. Keep in mind it could also flop over the the other side if the roll angle was big enough.
This video may help you to understand this better:
https://www.youtube.com/watch?v=eRgwGunG1LI
Finally let's look at the choices that can be eliminated from this question:
A. Pocketing to reduce free surface just does not apply here.
B. KG is not going to move.
D. The vessel's displacement will not change, however the drafts can be reduced due to the increased value of TPI or TPC due to the inclined waterline.
FINAL THOUGHT: A VERY SMALL NEGATIVE GM WILL PRODUCE A TREMENDOUS ANGLE OF LOLL!!!!!!!!!
I was leaning toward charley; due to the fact the KB would move toward the low side even though there is no righting couple it has reached equilibrium. Really good thought provoking question!
ReplyDeleteTo me, it is a very poorly worded question and I am wondering what you think of it.
ReplyDeleteRichard