Learning objectives
By the end of this chapter you will be able to:
- define the tonnes per centimetre immersion and derive it from the waterplane area using TPC = (AW × ρ) ÷ 100;
- calculate the sinkage or rise from loading or discharging a weight using sinkage or rise = w ÷ TPC;
- handle several weights at once through the net weight loaded or discharged;
- explain why the TPC increases with draught, and read it from the hydrostatic table;
- convert the TPC between salt, dock and fresh water using TPCDW = (TPCSW ÷ 1025) × ρDW;
- calculate the weight to load to reach a target draught, and judge when the TPC method is safe and when the table must be used;
- read metric draught marks to the centimetre.
Chapter 2 handled big changes of weight through the hydrostatic table. But most of a deck officer's daily arithmetic concerns small changes: two hundred tonnes of bunkers, a hundred tonnes of fresh water, a crane lift here and a slop discharge there. For these the table is heavier machinery than the job needs. This chapter supplies the light tool: one number, read straight off the table or the deadweight scale, that converts tonnes into centimetres of draught.
4.1 What the TPC is
Ask the simplest possible question: how many tonnes must be loaded to sink the ship exactly one centimetre deeper? When the ship sinks one centimetre, the extra water she displaces is a slice one centimetre thick with the shape and area of her waterplane. The volume of that slice is AW × 1 cm, which is AW ÷ 100 cubic metres, and its mass, by the Law of Flotation, is the weight that pushed her down. That mass is the tonnes per centimetre immersion:
In Chapter 3 we found the waterplane area of MV Ninja at a draught of 6.00 m to be 3225 m². Find her TPC in salt water at this draught, and compare it with the hydrostatic table.
TPC = (AW × ρ) ÷ 100 = (3225 × 1.025) ÷ 100 = 33.06 t
The hydrostatic table gives exactly 33.06 t at 6.00 m. The TPC column of the table is nothing more than the waterplane area column in disguise: the two are locked together by this formula, and checking one against the other, as we have just done, is a habit worth keeping.
4.2 Sinkage and rise
With the TPC in hand, small changes of draught become one line of arithmetic. Load a weight w and the ship sinks; discharge it and she rises; and because the TPC is tonnes per centimetre, the answer arrives in centimetres:
MV Ninja floats at 7.00 m in salt water, where her TPC is 33.82 t. She loads 265 t of bunkers. Find her new mean draught.
sinkage = w ÷ TPC = 265 ÷ 33.82 = 7.84 cm = 0.078 m
new draught = 7.000 + 0.078 = 7.078 m
MV Ninja lies at 8.00 m in salt water (TPC 34.64 t). She loads 420 t of cargo and discharges 180 t of ballast. Find her new mean draught.
net weight loaded = 420 − 180 = 240 t
sinkage = 240 ÷ 34.64 = 6.93 cm = 0.069 m
new draught = 8.000 + 0.069 = 8.069 m
Several weights are handled through their net effect: one addition or subtraction first, one division after. Had the net figure been negative the ship would have risen instead.
The centimetre is also the working unit of the draught marks themselves. Metric marks place numerals 10 cm high with their bottoms on every 20 cm of draught, and the officer reads the waterline against them, estimating the final centimetre by eye. TPC and draught marks speak the same language.
4.3 The TPC changes with draught
Chapter 3 showed the waterplane growing as the ship sinks deeper into her fuller body; the TPC, being the waterplane area in tonnes, grows with it. For MV Ninja the figure runs from 30.30 t at the lowest tabulated draught of 2.60 m to 35.28 t at the summer load line and 35.54 t at 10.40 m. The hydrostatic table tabulates it at every draught, and the curve below is that column drawn out:
Key point
Always take the TPC from the table at the draught you are actually at, and treat it as constant only while the draught change stays small, a few tens of centimetres at most. For large changes the TPC at the start is wrong by the finish; then the honest method is the difference of displacements from Chapter 2. Section 4.5 shows the two methods side by side.4.4 The TPC in dock water and fresh water
The one centimetre slice has the same shape and volume whatever the ship floats in, provided she floats at the same draught, but its mass depends on the density of the water. The TPC therefore scales in direct proportion to density, and the MCA sheet writes the conversion with density in kilograms per cubic metre:
At a draught of 6.00 m the salt water TPC of MV Ninja is 33.06 t. Find her TPC in dock water of density 1012 kg/m³.
TPCDW = (TPCSW ÷ 1025) × ρDW = (33.06 ÷ 1025) × 1012 = 32.64 t
Fresh water is simply the case ρDW = 1000: the fresh water TPC is always TPCSW ÷ 1.025.
MV Ninja lies at 6.00 m in fresh water and discharges 200 t of cargo. Find her rise.
TPCFW = (33.06 ÷ 1025) × 1000 = 32.25 t
rise = w ÷ TPC = 200 ÷ 32.25 = 6.20 cm = 0.062 m
The same discharge in salt water would have raised her only 200 ÷ 33.06 = 6.05 cm. In lighter water every tonne moves the ship a little further. The 6.00 m draught was read in fresh water, so it is the fresh water waterplane at that draught, and the fresh water TPC, that the discharge acts on.
Changing the water, not the weight
So far the draught has been held fixed and the density has entered only through the TPC. The other case is the ship whose weight is fixed while the water changes, as when she leaves the sea for a fresh water dock. Nothing has been loaded or discharged, so her displacement Δ is unchanged; it is the volume she must displace, ∇ = Δ ÷ ρ, that grows in lighter water, and she sinks until the extra volume is under water. The common error is to read the fresh water column of the hydrostatic table at the old draught and call that her new displacement; that column says what mass a given draught supports in fresh water, not what draught a given mass floats at.
MV Ninja at 6.000 m in salt water (Δ = 18064 t, TPC 33.06 t) moving into fresh water: by the table, enter the salt water displacement column with 18064 × 1.025 ÷ 1.000 = 18515.6 t, which lies 0.680 of the 664 t step between 6.00 m (18064 t) and 6.20 m (18728 t), so the new draught is 6.136 m. By the TPC, the volume displaced at the old waterline, 18064 ÷ 1.025 = 17623.4 m³, supports only 17623.4 t in fresh water, a shortfall of 440.6 t, and she sinks 440.6 ÷ 32.25 = 13.7 cm to 6.137 m. The two routes agree to the millimetre of linear interpolation; the ship has sunk about 14 cm without a tonne coming aboard. This TPC route is the fresh water allowance of Chapter 5, Δ ÷ (4 × TPC) millimetres: 137 mm here and 216 mm at the summer displacement.
4.5 Loading to a target draught
The commonest TPC question on board runs the other way: the draught we want is known, and the tonnage to reach it is not. Multiply the centimetres to go by the TPC and the answer appears. But this is also where the TPC must be handled with judgement, so we work the same problem by both methods and watch them meet:
MV Ninja floats at 9.40 m in salt water and is to complete loading to her summer draught of 9.60 m. Find the weight to load (a) by difference of displacement and (b) by the TPC, and compare the answers. (Table: Δ at 9.40 m = 29751 t, TPC 35.22 t; Δ at 9.60 m = 30456 t, TPC 35.28 t.)
(a) w = 30456 − 29751 = 705 t
(b) centimetres to go = 20 cm; mean TPC = (35.22 + 35.28) ÷ 2 = 35.25 t; w = 20 × 35.25 = 705 t
The two agree because over 20 cm the TPC scarcely changes, and using the mean of the two tabulated values absorbs what little change there is. Over a large change the two methods part: from 6.00 m (Δ = 18064 t, TPC 33.06 t) to the summer draught the table gives 30456 − 18064 = 12392 t, while method (b) with the TPC held at 33.06 t gives 360 × 33.06 = 11902 t, 490 t short, and even the mean of the two end values, 34.17 t, gives 12301 t, 91 t short. Over a change of that size only the table gives the correct figure.
Interactive: the TPC curve explorer
Slide the draught and read the TPC from the MV Ninja table, with the waterplane area recovered alongside it.
Interactive: the sinkage and rise simulator
Set the present draught, then load (positive) or discharge (negative) a net weight. The simulator takes the TPC from the table at your draught, converts for the water density, and moves the ship. Keep the weight modest: the tool, like the method, is for small changes.
Interactive: the TPC density converter
Enter a salt water TPC and slide the dock water density: the converter applies TPCDW = (TPCSW ÷ 1025) × ρDW.
Chapter summary
- TPC = (AW × ρ) ÷ 100: the mass of a one centimetre slice of water shaped like the waterplane.
- Sinkage or rise = w ÷ TPC, in centimetres. Several weights act through their net total.
- The TPC grows with draught because the waterplane grows; take it from the table at the actual draught.
- TPCDW = (TPCSW ÷ 1025) × ρDW; fresh water is the case ρDW = 1000.
- Weight to a target draught = centimetres to go × TPC, safe only while the change is small (from 9.40 to 9.60 m both routes give 705 t; from 6.00 to 9.60 m a single TPC is 490 t short); otherwise use the difference of displacements.
- When the water changes and the weight does not, Δ is fixed and ∇ = Δ ÷ ρ changes: find the new draught from the salt water column at Δ × 1.025 ÷ ρDW, or from the buoyancy shortfall at the old waterline divided by the TPC in the new water. Never read the new displacement off the fresh water column at the old draught.
Self test questions
Work each question with pencil, paper and the MV Ninja data booklet first. Your score appears in the bar below.