PROBLEM and SOLVING Number 6:
A ship with a fore draft of 18 feet 06 inches
and aft draft of 20 feet 06 inches
has a displacement of 18,000 tons
and a TPI of 62 in sea water.
Find her new draft in
water of 1,015 oz. density,
also the new draft in fresh water.
C in D = (Displacement x 1,025 - GD) / (TPI x 1,000)
C in D = (18,000 x 1,025 – 1,015) / (62 x 1,000)
C in D = (18,000 x 10) / (62,000)
C in D = (180,000) / (62,000)
C in D = 180,000 / 62,000
C
in D = 2.9 inchesTherefore, Change in Draft in water of 1,015 oz density is 2.9 inches
Fore Draft in Seawater = 18 feet 06 inches
+ Increase in Fore Draft = 2.9 inches
Fore Draft in 1,015 oz density = 18 feet 08.9 inches
Aft Draft in Seawater = 20 feet 06 inches
+ Increase in Aft Draft = 2.9 inches
Aft Draft in 1,015 oz density = 20 feet 08.9 inches
Let us now solve her new draft in Fresh Water:
C in D = (Displacement x 1) / (TPI x 40)
C in D = (18,000 x 1) / (62 x 40)
C in D = (18,000) / (2,480)
C in D = 18,000
/ 2,480C in D = 7.26 inches
Fore Draft in Sea Water = 18 feet 06 inches
+ increase in Fore Draft = 7.26 inches
Fore Draft in Fresh Water = 18 feet (13.26 inches – 12 inches and add to feet)
Fore Draft in Fresh Water = 19 feet 01.26 inches
Aft Draft in Sea Water = 20 feet 06 inches
+ increase in Aft Draft = 7.26 inches
Aft Draft in Fresh Water = 20 feet (13.26 inches – 12 inches and add to feet)
Aft Draft in Fresh Water = 21 feet 01.26 inches
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Note:
Effect Density on Draft:
From Sea Water to Fresh Water the draft increases (sink +)
From Fresh Water to Sea Water the draft decreases (rise -)
From higher density to lower density draft increases (sink +)
From lower density to higher density draft decreases (rise -)