Election in Indiana
1860 United States presidential election in Indiana
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| | | Nominee | Abraham Lincoln | Stephen A. Douglas | | Party | Republican | Democratic | Home state | Illinois | Illinois | Running mate | Hannibal Hamlin | Herschel V. Johnson | Electoral vote | 13 | 0 | Popular vote | 139,033 | 115,509 | Percentage | 51.09% | 42.44% | |
County Results Lincoln 30-40% 40-50% 50-60% 60-70% 70-80% | Douglas 30-40% 40-50% 50-60% 60-70% 80-90% | Breckinridge 30-40% | |
President before election James Buchanan Democratic | Elected President Abraham Lincoln Republican | |
Elections in Indiana |
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The 1860 United States presidential election in Indiana took place on November 6, 1860, as part of the 1860 United States presidential election. Indiana voters chose 13 representatives, or electors, to the Electoral College, who voted for president and vice president.
Indiana was won by Republican nominee Representative Abraham Lincoln of Illinois and his running mate Senator Hannibal Hamlin of Maine. They defeated the Democratic nominee Senator Stephen A. Douglas of Illinois] and his running mate 41st Governor of Georgia Herschel V. Johnson. Lincoln won the state by a margin of 8.65%.
Results
1860 United States presidential election in Indiana[1] Party | Candidate | Votes | % |
| Republican | Abraham Lincoln | 139,033 | 51.09% |
| Democratic | Stephen A. Douglas | 115,509 | 42.44% |
| Southern Democratic | John C. Breckinridge | 12,295 | 4.52% |
| Constitutional Union | John Bell | 5,306 | 1.95% |
Total votes | 272,143 | 100% |
See also
References
- ^ "1860 Presidential Election Results Indiana".