Lewis Island (Antarctica)
  Location in Antarctica | |
| Geography | |
|---|---|
| Location | Antarctica |
| Coordinates | 66°6′S 134°22′E / 66.100°S 134.367°ECoordinates: 66°6′S 134°22′E / 66.100°S 134.367°E |
| Highest elevation | 30 m (100 ft) |
| Administration | |
|
None | |
| Demographics | |
| Population | Uninhabited |
| Additional information | |
| Administered under the Antarctic Treaty System | |
Lewis Island is a small rocky island rising to 30 metres (100 ft), marking the east side of the entrance to Davis Bay in Antarctica. It was delineated from air photos taken by U.S. Navy Operation Highjump (1946–47) and named by the Advisory Committee on Antarctic Names for James B. Lewis, Passed Midshipman on the sloop Peacock of the U.S. Exploring Expedition (1838–42) under Charles Wilkes.[1]
See also
References
- ↑ "Lewis Island". Geographic Names Information System. United States Geological Survey. Retrieved 2013-06-13.
This article incorporates public domain material from the United States Geological Survey document "Lewis Island (Antarctica)" (content from the Geographic Names Information System).
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