Question 4805Deck 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. What is TCG?
AI Explanation
The correct answer is A) 0.00 feet.
To calculate the transverse center of gravity (TCG), we need to use the formula: TCG = (LM - FSML) / (VM - FSMT).
Given the values in the problem:
VM = 974,441 foot-long tons
LM = 40,301 foot-long tons
FSML = 30,572 foot-long tons
FSMT = 18,732 foot-long tons
Plugging these values into the formula:
TCG = (40,301 - 30,572) / (974,441 - 18,732)
TCG = 9,729 / 955,709
TCG = 0.00 feet
The other options are incorrect because they do not match the calculated value of 0.00 feet for the TCG.
Related Questions
#4803 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 TCG? #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? #4806 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? #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?