Latitude and longitude of David Island
- In Antarctica
Satellite map of David Island
David Island (66°25′S98°46′E) is an ice-covered island, 10 miles (16 km) long and 6 miles (10 km) wide, marked by rock exposures along its north and east sides, lying off Davis Peninsula in the Shackleton Ice Shelf. It was discovered in November 1912 by the Western Base party of the Australasian Antarctic Expedition (AAE) under Mawson, and named by him for Professor Sir T.W.
Latitude: -66° 24' 59.99" S
Longitude: 98° 45' 59.99" E