Sample Righting Moment Problem Tankship Northland
Use the following Stability Data Reference Book pages for information on the Cargo Ship S.S. American Mariner or the Tank Ship S.S. Northland to determine the correct answers to some of the stability problems.
Download Blue pages Download White pages Download Salmon pages
Your vessel’s drafts are: FWD 17′-05″, AFT 20′-01″; and the KG is 22.4 feet. What is the righting moment when the vessel is inclined to 15˚ ? (Use the reference material in Section 1, the blue pages, of the Stability Data Reference Book)
- Answers
- A. 10,656 foot-tons
- B. 12,340 foot-tons
- C. 13,980 foot-tons
- D. 17,520 foot-tons
- Step 1:
- Find Mean Draft.
- Fwd Draft 17′-05″ + Aft Draft 20′-01″ = 37′-06″
- Mean Draft = 37′-06″/2
- Mean Draft = 18′-09“
- Step 2:
- Find the displacement of the vessel by using the mean draft and the Deadweight Scale (Blue Pages).
- Displacement = 9450 tons
- Step 3:
- Find the change in the center of gravity between the actual KG and the assumed KG for the vessel..
- Actual KG: 22.4′
- Table KG: -20.0′
- GG1: = 2.4′
- Step 4:
- Find the corrected righting arm due the change in the center of gravity.
- correction = GG1 x Sin Angle (angle of list)
- = 2.4′ x Sin 15˚
- = 2.4′ x .2588 (.2588 is Sin 15˚ to four decimal places. Not .25 stated in the Cross Curves.
- = 0.62′
- Corrected GZ = GZ – correction
- for GZ, find the intersection of the curved line for inclination (15˚) and displacement tonnage
- (9,450) in the Cross Curves (Blue Pages)
- = 1.75′ – 0.62′ = 1.13′
- Find the corrected righting arm due the change in the center of gravity.
- Step 5: Find the righting moment.
- Righting Moment = GZ x Weight (displacement)
- = 1.13′ x 9,450 tons
- = 10,656 foot-tons (correct answer is A)