Question 4806Deck General70% to pass
The DEEP DRILLER, at 60.0 feet draft in sea water, has VM = 974,441 foot-long tons, LM = 40,301 foot-long tons, TM = 3 foot-long tons, FSML = 30,572 foot-long tons, and FSMT = 18,732 foot-long tons, and FSMT = 18,732 foot-long tons. What is the LCG?
AI Explanation
The correct answer is C) 2.22 feet.
To calculate the LCG (Longitudinal Center of Gravity), we can use the formula: LCG = (LM + FSML) / (VM + TM).
Plugging in the given values, we get:
LCG = (40,301 + 30,572) / (974,441 + 3)
LCG = 70,873 / 974,444
LCG = 2.22 feet
The other options are incorrect because they do not match the calculated value of 2.22 feet. The LCG represents the longitudinal position of the vessel's center of gravity, and this calculation is a standard procedure for determining this important parameter in marine operations.
Related Questions
#4804 The DEEP DRILLER, loaded as shown in the Sample Load Form #4 (Drilling), discharges a non-liquid load of 275.8 long tons from a position 130 feet above the keel, 40 feet forward of amidships, and 30 feet to port of the centerline. What is the new KGT? #4805 The DEEP DRILLER, at 60.0 feet draft in sea water, has VM = 974,441 foot-long tons, LM = 40,301 foot-long tons, TM = 3 foot-long tons, FSML = 30,572 foot-long tons, and FSMT = 18,732 foot-long tons. What is TCG? #4807 The DEEP DRILLER is loaded as shown in Sample Load Form #4 (Drilling). What would be the new sum of transverse moments for the Stores & Supplies (Table 6) if paint weighing 3.48 tons is added to the paint locker? #4808 What is the new location of the longitudinal center of gravity if 200 short tons are discharged from 30 feet forward of amidships on a MODU with LCG 1.5 feet forward of amidships, and displacement 9,000 short tons? #4809 What is the shift in the longitudinal center of gravity if 200 short tons are discharged from 30 feet forward of amidships on a MODU with LCG 1.5 feet forward of amidships, and displacement 9,000 short tons?