Dint Island
  Location in Antarctica | |
| Geography | |
|---|---|
| Location | Antarctica |
| Coordinates | 69°17′S 71°49′W / 69.283°S 71.817°WCoordinates: 69°17′S 71°49′W / 69.283°S 71.817°W |
| Length | 3 km (1.9 mi) |
| Country | |
|
None | |
| Demographics | |
| Population | Uninhabited |
| Additional information | |
| Administered under the Antarctic Treaty System | |
Dint Island is a rocky island, 3 kilometres (1.5 nmi) long. Probably first seen from the air by the United States Antarctic Service, 1939–41, it was first mapped from air photos taken by the Ronne Antarctic Research Expedition, 1947–48, by D. Searle of the Falkland Islands Dependencies Survey in 1960. It was so named by the UK Antarctic Place-Names Committee because a distinctive cirque makes a dent, or dint, on the south side of the island.[1]
Location
Dint Island is located at (69°17′S 71°49′W / 69.283°S 71.817°W) and lies 4 kilometres (2 nmi) off the west side of Alexander Island within Lazarev Bay. The island lies roughly 6 miles (10 km) southeast of Umber Island.
See also
References
- ↑ "Dint Island". Geographic Names Information System. United States Geological Survey. Retrieved 2012-01-21.
This article incorporates public domain material from the United States Geological Survey document "Dint Island" (content from the Geographic Names Information System).