1860 United States presidential election in Vermont
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.
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 County Results
Lincoln 60-70% 70-80% 80-90%
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Vermont was won by Republican candidate Abraham Lincoln, who won the state by a 56.45% margin.
With 75.86% of the popular vote, Vermont would be Lincoln's strongest victory in terms of percentage in the popular vote.[1]
Northern Democratic presidential candidate Stephen A. Douglas was born in Vermont, more specifically in the town of Brandon.
Results
| 1860 United States presidential election in Vermont[2] | ||||||||
|---|---|---|---|---|---|---|---|---|
| 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 A. Douglas of Illinois | Herschel Vespasian Johnson of Georgia | 8,649 | 19.41% | 0 | 0.00% | ||
| Southern Democratic | John C. Breckinridge 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% | ||||
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References
- "1860 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- "1860 Presidential General Election Results - Vermont".
State results of the 1860 U.S. presidential election | ||
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