Inclining Result and a Part-Loaded Centre Tank on a Converted Coastal Tanker
A 92 m double-hull coastal tanker (B = 15.0 m) has been converted to carry two cargo grades. After the conversion an inclining experiment was run at a displacement of 4,860 t with KM of 6.42 m: two 5.0 t weights were shifted 8.0 m transversely and a 6.5 m pendulum showed 195 mm deflection, giving an uncorrected GM of 0.55 m. The report was issued without deduction for a slack slop tank (free-surface moment 730 t.m) or for 12 t of consumables left aboard at 5.0 m above the keel. The operator now intends to trade with a 400 t parcel in the No. 3 centre cargo tank (10.0 m wide, 24.0 m long, cargo SG 0.85), which will sit about 45% full for several days between discharge ports. KM at the loaded draft is 6.35 m and the parcel's centre of gravity is 3.2 m above the keel.
1. Recompute the lightship GM applying the corrections the report omitted, and say whether the IS Code 2008 initial-GM criterion is satisfied.
Read the worked answer
Measured GM from the experiment:
GM = w.d / (Δ . tan θ) = (2 x 5.0 x 8.0) / (4,860 x 0.030) = 80 / 145.8 = 0.55 m, so measured KG = 6.42 - 0.55 = 5.87 m.
Consumables (12 t at 5.0 m): the correct approach is to remove them from the light condition:
Δ = 4,848 t; KG = (4,860 x 5.87 - 12 x 5.0) / 4,848 = 5.872 m - a shift of 2 mm, i.e. negligible, because the weight sits close to the KG.
Slop tank free surface: FSC = FSM / Δ = 730 / 4,848 = 0.151 m.
GM (fluid) = 6.42 - 5.872 - 0.151 = 0.40 m.
The IS Code 2008 Part A criterion GM0 ≥ 0.15 m is met with a wide margin. But the free-surface term is 27% of the corrected GM and dwarfs the consumables correction - the lesson is that the two 'corrections' are not of comparable importance, and the report as issued (0.55 m) overstates the vessel's stiffness by nearly 40%.
2. Now assess the planned 400 t part-loaded parcel. What is the resulting fluid GM, and what does it tell you about relying on the 0.15 m rule?
Read the worked answer
Loading 400 t at Kg 3.2 m:
Δ = 4,848 + 400 = 5,248 t
KG (solid) = (4,848 x 5.872 + 400 x 3.2) / 5,248 = 29,747 / 5,248 = 5.668 m
GM (solid) = KM - KG = 6.35 - 5.668 = 0.682 m
Free surface, No. 3 centre tank at 45%:
i = l.b^3/12 = 24.0 x 10.0^3 / 12 = 2,000 m^4
FSM = ρ . i = 0.85 x 2,000 = 1,700 t.m → 1,700 / 5,248 = 0.324 m
Slop tank (scaled): 730 / 5,248 = 0.139 m
GM (fluid) = 0.682 - 0.324 - 0.139 = 0.22 m
GM0 is still above 0.15 m, but the margin is now thin and the whole reserve has been eaten by one slack tank. GM0 is a floor, not a design target: with 0.22 m the righting-lever curve will be low and late, and the Code's area criteria (0.055 m.rad to 30°, 0.09 m.rad to 40°, GZmax ≥ 0.20 m) are the criteria that will actually govern. The answer is that the 0.15 m rule passing is not evidence the loading is safe - the GZ curve must be computed for this exact condition.
3. What would you require before the loading plan is signed, and what operational limit would you set?
Read the worked answer
(1) Recompute, don't estimate. Run the approved loading computer for the actual voyage condition with every tank's real fill level, and compare the GZ curve against all six IS Code 2008 Part A 2.2 criteria - not just GM.
(2) Treat the slop tank and the cargo parcel as one design condition. The two free surfaces are additive, and the worst case is not necessarily 45% fill; check the filling sequence, because a wide shallow fill near 30-50% gives close to the maximum i and therefore the maximum FSM.
(3) Set a hard operational limit on the number of simultaneously slack tanks. As a working rule for this hull, no more than one cargo tank plus the slop tank slack at any time, and the parcel split into two smaller tanks filled in sequence if the loading computer shows the area criteria marginal.
(4) Fix the documentation. The inclining report must be reissued with the free-surface and consumables corrections before it is used as the basis of any loading instrument; the current 0.55 m figure should be withdrawn from the ship's stability booklet.
(5) Brief the master that GM0 ≥ 0.15 m being satisfied is not permission to sail; the loading computer output signed for the condition is.