1864 United States presidential election in Indiana

1864 United States presidential election in Indiana

November 8, 1864
 
Nominee Abraham Lincoln George B. McClellan
Party National Union Democratic
Home state Illinois New Jersey
Running mate Andrew Johnson George H. Pendleton
Electoral vote 13 0
Popular vote 149,922 130,223
Percentage 53.5% 46.5%

County Results

President before election

Abraham Lincoln
Republican

Elected President

Abraham Lincoln
National Union

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.[1] Lincoln defeated McClellan in the national election with 212 electoral votes.[2]

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.[3] This table reflects the statewide popular vote as calculated by Walter Dean Burnham in his influential study, Presidential Ballots, 1836–1892.[1]

1864 United States presidential election in Indiana[1]
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

See also

Notes

  1. ^ a b c Burnham 1955, p. 249.
  2. ^ "1864 Electoral College Results". National Archives. Retrieved November 1, 2025.
  3. ^ Dubin 2002, p. xi.

Bibliography