Lewis Island (Antarctica)
  Lewis Island Location in Antarctica | |
| Geography | |
|---|---|
| Location | Antarctica |
| Coordinates | 66°6′S 134°22′E / 66.100°S 134.367°E |
| Highest elevation | 30 m (100 ft) |
| Administration | |
| Administered under the Antarctic Treaty System | |
| Demographics | |
| Population | Uninhabited |
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, with Anton Island 5 nmi to the south south-west[1]. 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.[2]
See also
References
- ^ "Antarctica Detail". geonames.usgs.gov. Archived from the original on 2021-06-02. Retrieved 2026-03-10.
- ^ "Lewis Island". Geographic Names Information System. United States Geological Survey, United States Department of the Interior. Retrieved 2013-06-13.
This article incorporates public domain material from "Lewis Island (Antarctica)". Geographic Names Information System. United States Geological Survey.