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Vessbook

Port-to-port sea distance

Lake Charles to Rotterdam distance

The sea distance from Lake Charles (United States) to Rotterdam (Netherlands) is about 4,953 nautical miles (9,173 km), via the Dover Strait. At 12 knots that is about 17 days 5 hours at sea.

Distance

4,953 nm

9,173 km

Sea days @ 12 kn

17.2

17 days 5 hours

ECA miles

2,338 nm

47% of the route

Canals & straits

1

Dover Strait

Sea days by speed

SpeedOne wayRound voyage
10 knots20.6 days41.3 days
11 knots18.8 days37.5 days
12 knots17.2 days34.4 days
13 knots15.9 days31.7 days
14 knots14.7 days29.5 days
15 knots13.8 days27.5 days

Steaming time only, with no allowance for weather, canal waiting or port time. The round voyage assumes the ship ballasts back on the same route.

Routes compared

Shortest route

4,953 nm

17.2 days at 12 kn · Dover Strait · 2,338 nm in ECA

See it on the map

Voyage estimates by ship size

About this trade

US Gulf clean products. Diesel and gasoline exports from the Texas and Louisiana refining belt to Latin America, the Caribbean and Europe, the core MR trade in the Atlantic.

Typical ship sizes
MR, LR1
Related benchmark routes
TC14 (USG–UKC MR), TC21 (USG–Caribbean MR)

Questions

How far is Lake Charles from Rotterdam by sea?

The sea distance from Lake Charles to Rotterdam is about 4,953 nautical miles (9,173 km), via the Dover Strait.

How long does a tanker take to sail from Lake Charles to Rotterdam?

About 17 days 5 hours at 12 knots, or 14.7 days at 14 knots. That is steaming time only; add time for port calls, canal transits and weather.

Which canals and straits does the Lake Charles–Rotterdam route pass?

The shortest route passes the Dover Strait.

How much of the voyage is inside an Emission Control Area?

About 2,338 nm, or 47% of the voyage, is inside an ECA, where ships must burn 0.10% sulphur fuel or use a scrubber.

Terms on this page

Distances are calculated by Vessbook on a modelled global shipping-lane network for a tanker at 12 m draft, from berth area to berth area.

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