TPC & MCTC Calculator

Calculate Tonnes Per Centimetre and Moment to Change Trim One Centimetre

Derive TPC and MCTC from waterplane geometry: LBP, beam, draught, waterplane and block coefficients, and water density. Naval architects and ship officers use this to verify hydrostatic tables and solve ballasting and trim problems. It runs free in your browser on Gera Tools, with nothing uploaded.

Last updated Source: Gera Tools

What is TPC?

TPC, tonnes per centimetre immersion, is the mass that must be loaded or discharged to change the mean draught by one centimetre. It equals the waterplane area times water density divided by 100, so a larger or fuller waterplane gives a higher TPC.

TPC and MCTC are two of the core hydrostatic quantities every ship officer and naval architect reaches for when loading, ballasting, or trimming a vessel. This calculator derives both from the waterplane geometry so you can verify a hydrostatic table, sanity-check a loading computer, or work a textbook problem.

How it works

The waterplane area sets TPC directly, while the longitudinal second moment of that area sets the trim stiffness:

A_w  = Cw × LBP × B                 (waterplane area, m²)
TPC  = A_w × ρ / 100                (tonnes per cm)
∇    = Cb × LBP × B × d             (displaced volume, m³)
W    = ∇ × ρ                        (displacement, t)
I_L  = k × LBP³ × B                 (longitudinal inertia of waterplane, m⁴)
BM_L = I_L / ∇
MCTC = W × GM_L / (100 × LBP) ≈ ρ × I_L / (100 × LBP)

Because longitudinal BM_L is so large, KB and KG are negligible against it, so GM_L ≈ BM_L is an accepted approximation for trim moments. The inertia coefficient k (typically 0.060–0.075) captures how the waterplane shape concentrates area toward the ends.

Worked example

For a vessel 150 m LBP, 22 m beam, at 8 m mean draught, with Cw = 0.82, Cb = 0.72, operating in sea water (ρ = 1.025 t/m³):

  • Waterplane area: 0.82 × 150 × 22 = 2,706 m²
  • TPC: 2,706 × 1.025 / 100 ≈ 27.7 t/cm — every 27.7 tonnes loaded or discharged changes mean draught by 1 cm
  • Displaced volume: 0.72 × 150 × 22 × 8 = 19,008 m³
  • Displacement: 19,008 × 1.025 ≈ 19,483 t
  • Longitudinal inertia (k = 0.070): 0.070 × 150³ × 22 = 52,987,500 m⁴
  • BM_L: 52,987,500 / 19,008 ≈ 2,788 m
  • MCTC: 1.025 × 52,987,500 / (100 × 150) ≈ 3,620 t·m/cm… but applying the full W × BM_L / (100 × LBP): 19,483 × 2,788 / 15,000 ≈ 3,622 t·m/cm

These are estimates within a few percent of the actual hydrostatic table values for a conventional vessel of this form. Always cross-check against the vessel’s official stability booklet for operational work.

How to use TPC and MCTC on passage

Adjusting mean draught with TPC: If you load 500 tonnes of bunkers, the change in mean draught is 500 / TPC = 500 / 27.7 ≈ 18 cm by the stern if loaded aft of the centre of flotation, or a bodily sinkage of 18 cm if loaded exactly at the centre of flotation.

Adjusting trim with MCTC: Trimming moment is mass × shift distance. If you shift 100 t of ballast 30 m aft, the trimming moment is 3,000 t·m. Trim change = 3,000 / MCTC. Using the example MCTC above, that is approximately 0.8 cm by the stern per metre of LBP — trivially small for this large vessel, confirming that large ships are very stiff in trim.

Why TPC and MCTC change with draught

Both quantities depend on the waterplane, which changes shape and area as a ship rises or sinks. A fuller hull form at a deeper draught will have a larger waterplane area (higher TPC) and a larger longitudinal second moment (higher MCTC). The ship’s hydrostatic tables provide these values at half-centimetre or centimetre draught intervals precisely because using a single value across a wide draught range introduces cumulative error in ballasting calculations.

For voyage planning calculations, always interpolate from the hydrostatic table at the departure and arrival draught rather than using a single figure from this estimator.