1866 Maine gubernatorial election explained

Election Name:1866 Maine gubernatorial election
Image1:File:Chamberlain_as_Governor_of_Maine.jpg
Before Election:Samuel Cony
Governor
Percentage2:37.59%
Popular Vote2:41,942
Party2:Democratic Party (United States)
Nominee2:Eben F. Pillsbury
Popular Vote1:69,636
Country:Maine
Percentage1:62.41%
Party1:Republican Party (United States)
Nominee1:Joshua Chamberlain
Map Size:300px
Election Date:September 10, 1866
Next Year:1867
Next Election:1867 Maine gubernatorial election
Previous Year:1865
Previous Election:1865 Maine gubernatorial election
Ongoing:No
Type:presidential
After Party:Republican Party (United States)
After Election:Joshua Chamberlain
Before Party:Republican Party (United States)

The 1866 Maine gubernatorial election was held on September 10, 1866. Republican candidate and war hero Joshua Chamberlain defeated the Democratic candidate Eben F. Pillsbury.[1]

General election

Candidates

Republican

During the American Civil War, Chamberlain played a crucial role at the Battle of Gettysburg.[3] This gave Chamberlain a war hero status.

Democratic

Results

The extremely popular Chamberlain was able to win election to a one-year term as governor. Chamberlain won a majority of 27,687 votes, trouncing his Democratic opponent.[4]

References

  1. Book: Dubin, Michael J.. United States Gubernatorial Elections, 1861_ÑÐ1911: The Official Results by State and County. 2010-06-11. McFarland. 978-0-7864-5646-8. en.
  2. Web site: Joshua Lawrence Chamberlain. 2021-09-01. National Governors Association.
  3. Web site: 2009-04-15. Defense of Little Round Top. 2021-09-01. American Battlefield Trust. en-US.
  4. Web site: Our Campaigns - ME Governor Race - Sep 10, 1866. 2021-09-01. www.ourcampaigns.com.