1860 United States presidential election in Vermont
(Redirected from United States presidential election in Vermont, 1860)
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County Results
Lincoln 60-70% 70-80% 80-90%
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Elections in Vermont |
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The 1860 United States presidential election in Vermont took place on November 2, 1860, as part of the 1860 United States presidential election. Voters chose five electors of the Electoral College, who voted for president and vice president.
Vermont was won by Republican candidate Abraham Lincoln and his running mate Hannibal Hamlin They defeated Democratic candidate Stephen A. Douglas and his running mate Herschel V. Johnson. Lincoln won the state by a landslide margin of 56.45%.
With 75.86% of the popular vote, Vermont would be Lincoln's strongest victory in terms of percentage in the popular vote.[1]
Stephen A. Douglas was born in Brandon, Vermont.
Results
1860 United States presidential election in Vermont[2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Republican | Abraham Lincoln of Illinois | Hannibal Hamlin of Maine | 33,808 | 75.86% | 5 | 100% | ||
Democratic | Stephen Arnold Douglas of Illinois | Herschel Vespasian Johnson of Georgia | 8,649 | 19.41% | 0 | 0.00% | ||
Southern Democratic | John Cabell Brekinridge of Kentucky | Joseph Lane of Oregon | 1,866 | 4.19% | 0 | 0.00% | ||
Constitutional Union | John Bell of Tennessee | Edward Everett of Massachusetts | 217 | 0.49% | 0 | 0.00% | ||
N/A | Others | Others | 26 | 0.06% | 0 | 0.00% | ||
Total | 44,566 | 100% | 5 | 100% |
See also
References
- ^ "1860 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
- ^ "1860 Presidential General Election Results - Vermont".