1840 United States presidential election in New York

1840 United States presidential election in New York
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Turnout	91.9% 21.4 pp
Harrison 40–50% 50–60% 60–70%	Van Buren 40–50% 50–60% 60–70% 70–80%
President before election; Martin Van Buren; Democratic	Elected President ; William Henry Harrison; Whig

A presidential election was held in New York from November 2–4, 1840 as part of the 1840 United States presidential election.^[2] Voters chose 42 representatives, or electors to the Electoral College, who voted for President and Vice President.

New York voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won New York by a narrow margin of 3.00%.

Results

1840 United States presidential election in New York^[3]
Party		Candidate	Running mate	Popular vote		Electoral vote
Party		Candidate	Running mate	Count	%	Count	%
	Whig	William Henry Harrison of Ohio	John Tyler of Virginia	226,001	51.18%	42	100.00%
	Democratic	Martin Van Buren of New York	Richard M. Johnson of Kentucky	212,733	48.18%	0	0.00%
	Liberty	James G. Birney of New York	Thomas Earle of Pennsylvania	2,809	0.64%	0	0.00%
Total				441,543	100.00%	42	100.00%

^ Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
^ Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson, NC: McFarland & Co. p. xvi.
^ "1840 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved December 23, 2013.

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