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United States presidential election in New York, 1860

From Wikipedia, the free encyclopedia

United States presidential election in New York, 1860

← 1856 November 6, 1860 1864 →
Abraham Lincoln by Alexander Hesler.jpg
Nominee Abraham Lincoln Democratic electors
Party Republican Fusion
Home state Illinois Kentucky / Illinois
Running mate Hannibal Hamlin Joseph Lane /
Herschel Johnson
Electoral vote 35 0
Popular vote 362,646 312,510
Percentage 53.71% 46.29%

President before election

James Buchanan

Elected President

Abraham Lincoln

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.

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 four state that had a fusion ticket for the Democratic Party. The other three states were New Jersey, Pennsylvania and Rhode Island.

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United States presidential election in New York, 1860[1]
Party Candidate Votes Percentage Electoral votes
Republican Abraham Lincoln 362,646 53.71% 35
Fusion John C. Breckinridge / Stephen A. Douglas 312,510 46.29% 0
Totals 675,156 100.0% 35


  1. ^ "1860 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved 3 August 2012.
This page was last edited on 27 November 2018, at 02:44
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