  Fleurus Island Location in Antarctica  Fleurus Island Fleurus Island (Antarctic Peninsula) | |
| Geography | |
|---|---|
| Location | Antarctica |
| Coordinates | 64°34′S 62°13′W / 64.567°S 62.217°W |
| Administration | |
| Administered under the Antarctic Treaty System | |
| Demographics | |
| Population | Uninhabited |
Fleurus Island is an island lying 1 km (0.5 nmi) 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
- ↑ "Fleurus Island". Geographic Names Information System. United States Geological Survey, United States Department of the Interior. Retrieved 27 March 2012.
This article incorporates public domain material from "Fleurus Island". Geographic Names Information System. United States Geological Survey.
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