The present draft is 8.20 mtrs fwd and 8.60 mtrs aft. You want to load at your maximum load draft of 8.88 mtrs. If the TPC at this draft is 26, how much cargo remains to be loaded?
Given:
TPC = 23
Fwd draft = 8.20 m
Aft draft = 8.60 m
Maximum draft = 8.88 m
What is asked?
Tons of cargo to be loaded
Solution:
1. We have to solve for the present mean draft by adding forward and aft draft, and then divide the sum by two.
Given:
TPC = 23
Fwd draft = 8.20 m
Aft draft = 8.60 m
Maximum draft = 8.88 m
What is asked?
Tons of cargo to be loaded
Solution:
1. We have to solve for the present mean draft by adding forward and aft draft, and then divide the sum by two.
Mean draft = (fwd draft + aft draft) ÷ 2
= (8.20 m + 8.60 m) ÷ 2
= 16.80 m ÷ 2
= 8.40 m
2. Subtract the maximum load draft by the present mean draft.
Maximum draft = 8.88 m
Mean draft = - 8.40 m
0.48 m
3. Convert 0.48 meters to centimeters so that we could apply the TPC easily.
0.48 m × 100 cm/m = 48 cm
× 26 tons per centimeter immersion
1248 tons
A vessel has a summer draft of 11.184m, TPC 46.99tons and a FWA of 254mm. She is loaded in a berth where the water RD is 1.011 and her present mean draft is 10.98m. How much more cargo can she load in order to be at her summer load-line on reaching salt water?
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