Current load calculations
Step I: Determine vessel cbaracteristics.
Vessel particulars for a 250 kdwl tanker in a fully loaded condition are
LJlp = 329.3 ID
T == 19,2 m
Step 2: Obtain current coefficient ar 8, J 70 a
WDjT == 1.1:1
and that the tanker has a "conventional" bow
cx~ = 0
CYM = -0.175
CUc == +0.13
Figure 2.6
Figure 2. 5
Figure 2.5
Seep 3: Compute the current forces
where:
v~
C
Pc ,
= l'c --V ; TLBP
, 7600
F I'M = C YAr ~ 7600 VlT'L 'c 8P
F '(F,' = c~ V2T'L , 8P
7600.
kg
1025 - salt water at 20°C
ID~
= 2 knots
=0
FYAr = (+ 0.175) (--^1025 ) (2)2(19,2)(329.3) == 597 kN
7600
(
Fyp~ = (+ 0,45) 7600 1025) (2)2(19.2)(329.3):, 1534 kN
BalIas(cd 250 kdwl (anker
For a bailas(ed 250 kdwt tanker, the following (anker characteristics are used:
AT (Transverse or Head-On Wind Area) = 1380 m^2
AL (Longirudinal or Broadside Wind Area) = 7622 m^1
L8P (Length between perpendiculars) = 329.3 ml
T(A verage Dra fl)" = 6. I m
Trim (Down by the stern) = 1°
The same equations used for the loaded case apply to tbe balIasted case also. Only tbe peninenl
characteristks and the wind and currentcoefficienls vary.
*TUis draft is less tha'" the IMO draft (abL g.6 m) a~surned rar fannulae listed in Seclion 2,4 since the vessel iQvestiga(ed
is a pre-MARPQL Iypc.