1860 United States presidential election in Maine
Election in Maine
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County Results Lincoln 50–60% 60-70% 70-80%
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The 1860 United States presidential election in Maine took place on November 2, 1860, as part of the 1860 United States presidential election. Voters chose eight electors of the Electoral College, who voted for president and vice president.
Maine was won by Republican candidate Abraham Lincoln, who won by a margin of 32.82%.
With 62.24% of the popular vote, Maine would prove to be Lincoln's fourth strongest state in terms of popular vote percentage after Vermont, Minnesota and Massachusetts.[1]
Results
1860 United States presidential election in Maine[2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Republican | Abraham Lincoln of Illinois | Hannibal Hamlin of Maine | 62,811 | 62.24% | 8 | 100.00% | ||
Democratic | Stephen A. Douglas of Illinois | Herschel Vespasian Johnson of Georgia | 29,693 | 29.42% | 0 | 0.00% | ||
Southern Democratic | John Cabell Breckinridge of Kentucky | Joseph Lane of Oregon | 6,368 | 6.31% | 0 | 0.00% | ||
Constitutional Union | John Bell of Tennessee | Edward Everett of Massachusetts | 2,046 | 2.03% | 0 | 0.00% | ||
Total | 100,918 | 100.00% | 8 | 100.00% |
See also
References
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