Fleurus Island

Fleurus Island (64°34′S 62°13′W / 64.567°S 62.217°W / -64.567; -62.217Coordinates: 64°34′S 62°13′W / 64.567°S 62.217°W / -64.567; -62.217) is an island lying 0.5 nautical miles (1 km) south of Delaite Island in Wilhelmina Bay, off the west coast of Graham Land, Antarctica. it was shown on an Argentine government chart of 1950, and was named by the UK Antarctic Place-Names Committee in 1956 after the British ship Fleurus, which visited the area in 1928.[1]

See also

References

  1. ↑ "Fleurus Island". Geographic Names Information System. United States Geological Survey. Retrieved 2012-03-27.

 This article incorporates public domain material from the United States Geological Survey document "Fleurus Island" (content from the Geographic Names Information System).


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