1860 United States presidential election in New York

The 1860 United States presidential election in New York took place on November 6, 1860, as part of the 1860 United States presidential election. Voters chose 35 electors of the Electoral College, who voted for president and vice president.

1860 United States presidential election in New York

← 1856 November 6, 1860 1864 →
Turnout95.5%[1] Increase 5.6 pp
  Abraham Lincoln O-26 by Hesler, 1860 (cropped).jpg DemocraticLogo.svg
Nominee Abraham Lincoln Democratic electors
Party Republican Fusion
Home state Illinois Illinois / Tennessee / Kentucky
Running mate Hannibal Hamlin Herschel Johnson /
Edward Everett /
Joseph Lane
Electoral vote 35 0
Popular vote 362,646 312,510
Percentage 53.71% 46.29%

President before election

James Buchanan
Democratic

Elected President

Abraham Lincoln
Republican

New York was won by Republican candidate Abraham Lincoln, who defeated the Democratic fusion ticket.

Lincoln won the Empire State by a margin of 7.42%.

New York in the election was one of the 4 states that had a fusion ticket for the Democratic Party. The other three states were New Jersey, Pennsylvania and Rhode Island.

ResultsEdit

1860 United States presidential election in New York[2]
Party Candidate Votes Percentage Electoral votes
Republican Abraham Lincoln 362,646 53.71% 35
Fusion[a] Stephen A. Douglas / John Bell / John C. Breckinridge 312,510 46.29% 0
Totals 675,156 100.0% 35

NotesEdit

  1. ^ The slate of electors were pledged to 3 different candidates: 18 to Douglas, 10 to Bell, and 7 to Breckinridge.[3]

ReferencesEdit

  1. ^ Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
  2. ^ "1860 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved 3 August 2012.
  3. ^ Dubin, Michael J., United States Presidential Elections, 1788–1860: The Official Results by County and State, McFarland & Company, 2002, p. 187