1860 United States presidential election in New York

1860 United States presidential election in New York
Turnout	95.5% 5.6 pp
Lincoln 50–60% 60–70% 70–80%	Douglas 50–60% 60–70% 70–80%
President before election; James Buchanan; Democratic	Elected President ; Abraham Lincoln; Republican

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 the tipping state in this election, and had Lincoln lost it there would have been a contingent election decided by Congress.

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

Lincoln won New York by a margin of 7.42 percentage points.

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

New York was home to a prominent Republican politician, William H. Seward who was widely expected to be nominated in the Republican Convention.

New York was key to the election; a Lincoln loss there would deadlock the Electoral College. Soon after his return from his Midwest tour, Seward embarked on another, speaking to large crowds across the state of New York. At his advisor Thurlow Weed's urging, he went to New York City and gave a patriotic speech before a large crowd on November 3, only three days before the election.^[2]

^ Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
^ Van Deusen, Glyndon (1967). William Henry Seward. New York: Oxford University Press. pp. 234–235. OCLC 426046.

[1] Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.

[2] Van Deusen, Glyndon (1967). William Henry Seward. New York: Oxford University Press. pp. 234–235. OCLC 426046.

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