- The ship must be in a light, near-condition with all weights (cargo, bunkers, ballast, stores, fresh water, loose gear) accounted for and positioned; all such weights must be lined up on the centreline or paired, and their positions accurately recorded (or removed and weighed).
- All tanks must be completely empty, pressed up (to prevent free surface) or their free-surface effect computed; no free liquid surfaces or sloshing should be present in the tanks; fuel/water/ballast tanks should be emptied or pressed up and kept gas-free.
- Deck: no undischarged items, all cargo on board weighed and centred; the ship must be upright, at even keel or known initial trim, and floating freely (not touching piers, hangars, moorings, lines or holding fast - always moored with loose ropes and no gangway contact).
- All personnel, apart from the inclining team and observers, should be docked? tensed, off the deck? person on board should not move about; boats and cranes set so as not to interfere.
- Decide the draft marks (recording of forward, aft, midships mean) and water density (sample) at each set of readings, and assess the waterplane area and displacing conditions from hydrostatic data/docking information.
- Prepare the inclining weights (movable ballast of known size) and their movement plan; ensure calm water, no currents or waves, and stable reference (plumb lines, clinometer/pendulum) freely available.
- The ship should be at midships age at a suitable inclination, generally less than about 3 to 4 degrees, and the reference pendulum (plumb line or spirit level) checked for free movement before starting.
- All weights should be moved smoothly amidships to starboard/port in a planned sequence (e.g. pendulum deflects), readings taken at each step.
Given speed increased 18% above normal (V1=1.18 Vn) for 7.5 h, then reduced to 9% below normal (V2=0.91 Vn) for 10 h, then a final speed V3 for the remainder of the day such that the total fuel for the 24 h day equals the normal day's consumption.
Fuel consumption is proportional to V^3 (per unit time). Normal day consumption = k x Vn^3 x 24.
Consumption in the three parts = k[V1^3 x 7.5 + V2^3 x 10 + V3^3 x 6.5], where 6.5 = 24 - 17.5 h.
Set equal to the normal consumption k x Vn^3 x 24:
1.18^3 x 7.5 + 0.91^3 x 10 + V3^3 x 6.5 = 1 x 24.
1.643 x 7.5 = 12.32; 0.7536 x 10 = 7.536; so 12.32 + 7.536 + 6.5 V3^3 = 24.
6.5 V3^3 = 24 - 19.86 = 4.14, so V3^3 = 0.6369, V3 = 0.8606 Vn.
Distances travelled in the day:
D = V1 x 7.5 + V2 x 10 + V3 x 6.5 = 1.18 x 7.5 + 0.91 x 10 + 0.8606 x 6.5
= 8.85 + 9.1 + 5.59 = 23.54 (as a multiple of distance per hour at Vn? Units: this is (Vn x hours) normalised by Vn, so in units of (Vn hours), normal daily distance = 24 x Vn).
Normal daily distance = 24 (in those units). Today's distance = 23.54.
Percentage difference = (23.54 - 24)/24 x 100 = -1.9%.
Answer: the distance travelled that day is about 1.9% less than the normal daily distance (approximately 0.46 Vn hours less). (Using the exact cubic relation the result is about -1.9%.)