Determine the free surface constant for a fuel oil tank 30 ft. long by 40 ft. wide by 15 ft. deep. The specific gravity of the fuel oil is .85 and the ship is floating in saltwater (S.G. 1.026).

Correct Answer: Option C — 3787

Free surface constant (FSC) equals (l * b^3) / 12 multiplied by the ratio of liquid specific gravity to saltwater specific gravity (1.026). For a 30x40x15 ft tank with 0.85 SG oil, FSC computes to 3787.