For a 75,000 m' gas carrier:
S )( MBL == 2.228 AT + 0.170 TLop
BR )( MBL == 2.012 AL + 1.291 Tlop
For a 125,000 mJ gas carrier (spheri.cal tanks):
S )( MBL == 2.388 AT + 0.170 TLsp
BR )( MBl == 1.898 AL + 1.291 TLop
2.6.5 Example Calculations
The following examples show the application of the approximate formulae.
- For an oil tanker Dj 90 kDWT with segregated ballast tanks
(a) Ship particulars:
Lop == 234.0 m
B = 39.6 m
D = 20.6 m
d == 14.0 m
(26)
(27)
(28)
(29)
ATI' == 755 m^1 (at design draft d); ATo = 1044 m< (at balJast draft)
AL8 == 4200 m^2 (at 6.7 m draft = 0.02 Lop + 2m); ALF == 2492 m^1 (at design draft d)
(b) Maximum forces on the ship, by use of formulae (7), (8), (9), (13), (14) and (15)
FuB load:
F, max == 415 + 139 = 554 kN
FYF mll)( == 375 + 760 = 1135 kN
FYA max == 707 + 813 == 1520 kN
Ballast (IMO draft):
F, max 520 + 72 = 592 kN
FYF [!lax 1193 + 41 = 1234 kN
FYA ma)( 1340 + 41 = 1381 kN
(c) Spring line requirements, by use of formula (22):
S x MBL == 2080 + 288 == 2368 kN
Assuming the ship will have a total of four spring lines (5 == 4), the min.imum breaking
strength required for each line will be:
2368
MBL = -- == 592 kN (== 60 tonnes force)
4
(d) Breast line requirements, using formula (23):
BR )( MBI. == 3678 + 4229 == 7C)07 kN
Assuming that the ship has a total of eight breast lines (B == 8) symme1rically arranged
(four forward and four aft), the minimum breaking strength required for each line will be:
7907
MBL == == 988 kN (= 101 tonnes force)
8