OCIMF_Mooring_Equipment_Guidelines_(Second_Edition)

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.