Swan Island (Alaska)

Coordinates: 57°55′53″N 134°14′12″W / 57.93139°N 134.23667°W / 57.93139; -134.23667
From Wikipedia, the free encyclopedia

Swan Island is an island in the Alexander Archipelago, east of Admiralty Island, near the head of Seymour Canal, Southeast Alaska, United States. To its south is Tiedeman Island. It was named in 1890 by Lieutenant Commander Mansfield of the United States Navy.[1] The first European to discover and chart the island was Joseph Whidbey, master of HMS Discovery during George Vancouver's 1791–1795 expedition, in 1794.[2]

References[edit]

  1. ^ U.S. Geological Survey Geographic Names Information System: Swan Island
  2. ^ Vancouver, George, and John Vancouver (1801). A voyage of discovery to the North Pacific ocean, and round the world. London: J. Stockdale.{{cite book}}: CS1 maint: multiple names: authors list (link)

57°55′53″N 134°14′12″W / 57.93139°N 134.23667°W / 57.93139; -134.23667