1864 United States presidential election in Indiana
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A presidential election was held in Indiana on November 8, 1864, as part of the 1864 United States presidential election. The National Union ticket of the incumbent president Abraham Lincoln and the governor of Tennessee Andrew Johnson defeated the Democratic ticket of the former commanding general of the United States Army George B. McClellan and the U.S. representative from Ohio's 1st congressional district George H. Pendleton.[2] Lincoln defeated McClellan in the national election with 212 electoral votes.[3]
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November 8, 1864
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| Turnout | 82.9%[1]  6.5 pp | |||||||||||||||||||||||||
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 County Results
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General election
Summary
Indiana chose 13 electors on a statewide general ticket. Nineteenth-century election laws required voters to elect each member of the Electoral College individually, rather than as a group. This sometimes resulted in small differences in the number of votes cast for electors pledged to the same presidential candidate, if some voters did not vote for all the electors nominated by a party.[4] This table reflects the statewide popular vote as calculated by Walter Dean Burnham in his influential study, Presidential Ballots, 1836â1892.[2]
| Party | Candidate | Votes | % | ±% | |
|---|---|---|---|---|---|
| National Union | Abraham Lincoln Andrew Johnson |
149,922 | 53.52 |  2.38 | |
| Democratic | George B. McClellan George H. Pendleton |
130,223 | 46.48 | 4.11 | |
| Total votes | 280,145 | 100.00 | |||
