Sample solution for righting moment problem

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)

A. 10,656 foot-tons (Correct)  B. 12,340 foot-tons (Incorrect) C. 13,980 foot-tons (Incorrect) D. 17,520 foot-tons (Incorrect)

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'

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)