1864 United States presidential election in Vermont
(Redirected from United States presidential election in Vermont, 1864)
The 1864 United States presidential election in Vermont took place on November 8, 1864, as part of the 1864 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for president and vice president.
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County Results
Lincoln 60-70% 70-80% 80-90% 90-100%
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Vermont voted for the National Union candidate, incumbent Republican President Abraham Lincoln and his running mate Andrew Johnson. They defeated the Democratic candidate, George B. McClellan and his running mate George H. Pendleton. Lincoln won the state by a wide margin of 52.20%.
With 76.10% of the popular vote, Lincoln's victory with in the state would be his second strongest victory in the country in terms of percentage in the popular vote after Kansas.[1]
Results
edit1864 United States presidential election in Vermont[2] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
National Union | Abraham Lincoln (incumbent) | 42,420 | 76.10% | 5 | |
Democratic | George B. McClellan | 13,322 | 23.90% | 0 | |
Totals | 55,742 | 100.0% | 26 |
See also
editReferences
edit- ^ "1864 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
- ^ "1864 Presidential General Election Results - Vermont".