1864 United States presidential election in Michigan explained

See main article: 1864 United States presidential election.

Election Name:1864 United States presidential election in Michigan
Country:Michigan
Flag Year:1864
Type:presidential
Ongoing:no
Previous Election:1860 United States presidential election in Michigan
Previous Year:1860
Next Election:1868 United States presidential election in Michigan
Next Year:1868
Election Date:November 8, 1864
Image1:Abraham Lincoln November 1863.jpg
Nominee1:Abraham Lincoln
Party1:National Union Party (United States)
Home State1:Illinois
Running Mate1:Andrew Johnson
Electoral Vote1:8
Popular Vote1:79,149
Percentage1:53.60%
Nominee2:George B. McClellan
Party2:Democratic Party (United States)
Home State2:New Jersey
Running Mate2:George H. Pendleton
Electoral Vote2:0
Popular Vote2:68,513
Percentage2:46.40%
President
Before Election:Abraham Lincoln
Before Party:Republican Party (United States)
After Election:Abraham Lincoln
After Party:National Union Party (United States)
Map Size:300px

The 1864 United States presidential election in Michigan took place on November 8, 1864, as part of the 1864 United States presidential election. Voters chose eight[1] representatives, or electors, to the Electoral College, who voted for president and vice president.[2]

Michigan was won by National Union candidate, incumbent Republican president Abraham Lincoln and his running mate Andrew Johnson. They defeated the Democratic challenger George B. McClellan and his running mate George H. Pendleton. Lincoln won the state by a margin of 7.2%.[3]

, this is the last time Ottawa County voted for a Democratic presidential candidate.[4]

See also

Notes and References

  1. Web site: 1864 Election for the Twentieth Term (1865-1869). 2018-11-28.
  2. Web site: 1864 Presidential General Election Results - Michigan.
  3. Web site: 1864 Presidential General Election Results - Michigan.
  4. Menendez, Albert J.; The Geography of Presidential Elections in the United States, 1868-2004, pp. 222-227