Sixty (60) tons of cargo are raised with a heavy lift boom 45 feet from the centerline. The vessel's displacement including the weight lifted is 18,400 tons. The angle of list caused by the suspended weight is 1.5°, KM is 28.75 ft., and BM is 17.25 ft. What is the KG?
A. 11.65 feet
B. 22.85 feet
C. 23.15 feet
D. 23.82 feet
A. 11.65 feet
B. 22.85 feet
C. 23.15 feet
D. 23.82 feet
Hmm, it's been a while and I don't have my Stability book with me. I can get the shift of G and KB but I'm drawing a blank after that. I'd love to see a walkthrough as refresher.
ReplyDeleteGoogled the formula, which I've been trying to avoid.
ReplyDeleteC - final answer!
You need the "Inclining Experiment Formula" for this one.
ReplyDeleteTan (angle of heel) =
Weight in LT x Distance from Centerline in feet / GM in feet x Displacement in LT
Solving for GM we get:
GM = Weight x Distance/ tan (angle of list) x Displacement
GM = 60 x 45 / tan (1.5 degrees) x 18,400 = 5.603 feet
KG = KM - GM = 28.75 feet - 5.60 feet = 23.146 feet or Ans. C. 23.15 feet
Now where did the Inclining Experiment Formula come from? It is derived in my book STABILITY AND TRIM FOR THE SHIP's OFFICER, 4th Edition on page 115:
Derivation of Inclining Experiment Formula
If any weight is shifted on board a vessel, the center of gravity of the vessel will shift in a direction parallel to the weight movement. The distance G moves is found by the formulas:
1. GG' = (w x d) / Displ. (this is the transverse shift in G)
2. tan (angle of list) = GG'/ GM
[See Figure 6-2 D in the book or image the triangle formed]
3. GG' = GM tan (angle of list)
[transposing equation (2.) to solve for GG']
4. GM tan (angle of list) = (w x d) / Displ.
[(setting equation (3.) equal to (1.)]
5. GM = (w x d) / (Displ.)(tan (angle of list))
[ Solving for GM, we get the Inclining Experiment Formula!]
CHARLEY IS CLOSEST.
ReplyDeleteGreat review of this subject!
Appreciate the work through!
ReplyDelete