How do you calculate a ship's gross and net tonnage?

Formula, assumptions, and limits

Gross tonnage comes from one measured figure, the volume of every enclosed space in the ship, and a coefficient that grows slowly with that volume. Net tonnage starts from the cargo spaces, scales them by how deep the ship floats against how deep it is, and adds a passenger term. Both formulas are in Annex I to the International Convention on Tonnage Measurement of Ships, 1969 (ITC 69): Regulation 3 for gross tonnage, Regulation 4 for net tonnage.

Formula

• $V$: total volume of all enclosed spaces, m³ (Regulations 2(4), 2(5) and 6).
• $V_c$: total volume of the cargo spaces, m³ (Regulation 2(7)).
• $d$: molded draft amidships, m, taken from the load line Regulation 4(2) names.
• $D$: molded depth amidships, m, from the top of the keel to the underside of the upper deck at side (Regulation 2(2)).
• $N_1$: passengers in cabins with not more than 8 berths; $N_2$: all other passengers, both from the passenger certificate.
• $GT$ and $NT$ have no unit.

Regulation 4(1) attaches three conditions to the net tonnage formula: the factor $(4d/3D)^2$ is not taken as greater than 1; the cargo term $K_2 V_c (4d/3D)^2$ is not taken as less than $0.25\,GT$; and $NT$ is not taken as less than $0.30\,GT$. When $N_1 + N_2$ is under 13, both are taken as zero.

Derivation

The convention gives the two formulas as rules, not as results of a physical law, so the derivation is the order in which the calculator applies them.

1. $K_1$ from $\log_{10} V$, then $GT = K_1 V$, rounded down to a whole number under TM.5/Circ.6 interpretation R.3-2.
2. $K_2$ from $\log_{10} V_c$, and the draft factor $(4d/3D)^2$, capped at 1 by Regulation 4(1)(a).
3. The cargo term $K_2 V_c (4d/3D)^2$, raised to $0.25\,GT$ if it is less (Regulation 4(1)(b)).
4. The passenger term $K_3 (N_1 + N_2/10)$, zero when there are fewer than 13 passengers, added to the cargo term.
5. The sum raised to $0.30\,GT$ if it is less (Regulation 4(1)(c)), then rounded down under R.4-2.

The cap in step 2 bites when $d$ passes 75% of $D$, since $4d/3D = 1$ exactly there. That is why Regulation 4(2)(e) sets the draft of a ship with no load line at 75% of the molded depth: its factor is 1.

Assumptions

• $V$ and $V_c$ are the measured volumes from the tonnage survey. Regulation 7 takes every measurement to the nearest centimeter and calculates volumes by generally accepted methods, with an accuracy acceptable to the Administration.
• $K_1$ and $K_2$ come from the formula. TM.5/Circ.6 R.3-1 and R.4-1 let the Administration use the Appendix 2 table instead, which prints both to four decimals.
• The rounded-down gross tonnage, the figure on the certificate, goes into $K_3$ and the two floors. Regulation 4 calls for gross tonnage “as determined in accordance with the provisions of regulation 3” and does not say whether the floor of R.3-2 comes first; using the unrounded figure moves the net tonnage by at most 1 in the passenger cases tested here.

Worked example

A cargo ship with 50,000 m³ of enclosed spaces, 35,000 m³ of cargo spaces, a summer draft of 12.0 m, a molded depth of 17.0 m and no passengers.

1. $\log_{10} 50\,000 = 4.69897$, so $K_1 = 0.2 + 0.02 \times 4.69897 = 0.293979$.
2. $GT = 0.293979 \times 50\,000 = 14\,698.97$, rounded down to 14,698.
3. $\log_{10} 35\,000 = 4.54407$, so $K_2 = 0.290881$.
4. $4d/3D = 48/51 = 0.941176$, squared $0.885813$, under 1 and kept.
5. Cargo term $= 0.290881 \times 35\,000 \times 0.885813 = 9\,018.33$, above $0.25 \times 14\,698 = 3\,674.50$.
6. No passengers, so no passenger term; $9\,018.33$ is above $0.30 \times 14\,698 = 4\,409.40$.
7. $NT = 9\,018.33$, rounded down to 9,018.

With the Appendix 2 coefficients, 0.2940 and 0.2909, the same ship comes out at GT 14,700 and NT 9,018.

Edge cases and limits

• No cargo spaces. With $V_c = 0$ the cargo term is zero and the 0.25 GT floor takes its place.
• No passengers, small cargo term. The 0.30 GT floor always overrides the 0.25 GT floor, so on a ship with no passengers the 0.25 GT floor never decides the result. V = 20,000 m³, Vc = 9,000 m³, d = 7.2 m and D = 12.0 m give a cargo term of 1,607.53 and NT 1,716, which is 0.30 × 5,720.
• 12 or 13 passengers. Twelve count as none. Thirteen in cabins add $K_3 \times 13$ in full: 40.13 at GT 14,698.
• Draft past 75% of depth. At d = 13.5 m and D = 17.0 m the factor would be 1.121; it is taken as 1 and the cargo term becomes $K_2 V_c$, 10,180.85.
• A draft deeper than the depth would put the upper deck under water at the load line, and the calculator refuses it.
• Volumes past 1,000,000 m³ fall outside Appendix 2, which runs from 10 to 1,000,000 m³, and the calculator stops there.

Regulatory basis

The International Convention on Tonnage Measurement of Ships was adopted on 23 June 1969 and entered into force on 18 July 1982. Annex I holds the measurement rules: definitions in Regulation 2, gross tonnage in Regulation 3, net tonnage in Regulation 4, the change of net tonnage in Regulation 5, and volumes and measurements in Regulations 6 and 7. IMO TM.5/Circ.6 (19 May 2014), which superseded TM.5/Circ.5, carries the unified interpretations, the rounding rule among them. In the United States, 46 CFR Part 69, Subpart B, sets out the same formulas: section 69.57 for gross tonnage, section 69.63 for net tonnage and section 69.53 for the draft.

Common errors

1. Natural log in place of log10. Both coefficients use the base-10 log. With the natural log, K1 at 50,000 m³ comes out near 0.416 and the gross tonnage about 42% too high.
2. The wrong draft. A scantling or design draft, or the timber summer load line, in place of the draft Regulation 4(2) names. The draft factor then shifts every cargo-ship net tonnage that sits above the floors.
3. Net tonnage as gross minus deductions. Under Regulation 4 net tonnage is built up from the cargo spaces and the passengers; nothing is subtracted from the gross tonnage, and NT has its own floor of 0.30 GT.
4. Rounding to the nearest whole number. 14,698.97 is certified as 14,698, not 14,699.
5. Cargo spaces outside V. Regulation 2(7) counts a cargo space in Vc only if it is already in V, so Vc can never exceed V.

K1 and K2 by volume

The coefficient rises by 0.02 for every tenfold rise in volume: 0.22 at 10 m³, 0.30 at 100,000 m³ and 0.32 at 1,000,000 m³, the two ends of the convention’s Appendix 2 table. So a ship ten times the volume of another has a little over ten times its gross tonnage. The same function gives $K_2$ from the cargo volume.

K1 × V passes 500 between 1,883 and 1,884 m³: 499.93 rounds down to 499, and 1,884 m³ is the first whole volume certified at GT 500. Regulation 5.1.3 of the Maritime Labour Convention, 2006, requires a maritime labor certificate on ships of 500 gross tonnage or over engaged on international voyages, and MLC Article II(1)(c) defines gross tonnage by Annex I of the 1969 Tonnage Convention.

Draft, depth and the net tonnage factor

The factor $(4d/3D)^2$ ties net tonnage to how deep the ship floats on its assigned load line against its molded depth, and it reaches its cap of 1 at a draft of 75% of the depth. A ship with a deep freeboard pays less net tonnage on the same cargo volume.

Regulation 4(2) names the draft for each kind of ship:

• a ship under the International Convention on Load Lines: the summer load line draft, never the timber load line;
• a passenger ship: the deepest subdivision load line under SOLAS or another applicable agreement;
• a ship with a national load line: that summer load line;
• a ship with no load line but a draft restricted under national rules: the maximum permitted draft;
• any other ship: 75% of the molded depth amidships, which makes the factor 1.

Regulation 5(2) gives a ship assigned both a cargo and a passenger load line one net tonnage only, the one for the load line of the trade it is in.

What counts in V and in Vc

V is every enclosed space: any space bounded by the hull, by fixed or portable partitions or bulkheads, or by decks or coverings other than awnings (Regulation 2(4)). An opening in a deck or the hull does not take a space out. Regulation 2(5) excludes some open spaces, such as an erection open at the ends, but puts one back in if it has shelves or other means of securing cargo or stores, or any means of closing its openings.

Volumes are measured to the inner side of the shell or structural boundary plating in a ship constructed of metal, regardless of insulation, and appendages are included (Regulation 6). TM.5/Circ.6 R.6(2)-1 treats bulbs, fairwaters and shaft bossings as appendages; R.6-1 leaves out spaces above the upper deck, appendages and spaces open to the sea of 1 m³ or less.

Vc is narrower. Regulation 2(7) counts the enclosed spaces appropriated for cargo to be discharged from the ship, each marked CC in letters at least 100 mm high. TM.5/Circ.6 R.2(7)-1 to R.2(7)-11 sort the borderline spaces:

When net tonnage may be changed

An increase in net tonnage applies at once. Under Regulation 5(1), when a change in V, Vc, d, N1 or N2 raises the net tonnage, the new figure is determined and “applied without delay”.

A decrease waits. Regulation 5(3) bars a new International Tonnage Certificate (1969) with a lower net tonnage until twelve months after the current certificate was issued, except when the ship changes flag, undergoes alterations the Administration deems major, or is a passenger ship carrying large numbers of unberthed passengers in special trades such as the pilgrim trade. A ship cannot cut its net tonnage for one port call by declaring a lighter load line and switch back the next month. 46 CFR 69.71(c) carries the same twelve-month rule.

The Suez and Panama canals charge on their own tonnage. The Panama Canal Authority’s admeasurement regulations (Agreement No. 358, 12 December 2019) compute PC/UMS net tonnage as K4(V) + K5(V) from the ITC 69 enclosed volume, with spaces ITC 69 excludes added back for passenger vessels (Article 10); the Suez Canal Authority’s Rules of Navigation charge on the Suez Canal net tonnage of the 1873 Constantinople commission. Neither figure is the ITC 69 net tonnage this calculator returns.