With 61.37% of the popular vote, Rhode Island would prove to be Lincoln's fifth strongest state in terms of popular vote percentage in the 1860 election after Vermont, Minnesota, Massachusetts and Maine.[1]
Like New Jersey, New York and Pennsylvania, Rhode Island was one of the four states that had a fusion ticket for the Democrats, which was supported just by the Northern Democrats but also supporters of Southern Democrats and Constitutional Unionists. However, unlike in the other three states the electors on the Democratic ticket in Rhode Island were pledged solely to Douglas, therefore some sources credit the Democratic popular vote to the Northern Democratic nominee.
Results
1860 United States presidential election in Rhode Island[2]
1860, united, states, presidential, election, rhode, island, main, article, 1860, united, states, presidential, election, took, place, november, 1860, part, 1860, united, states, presidential, election, voters, chose, four, electors, electoral, college, voted,. Main article 1860 United States presidential election The 1860 United States presidential election in Rhode Island took place on November 2 1860 as part of the 1860 United States presidential election Voters chose four electors of the Electoral College who voted for president and vice president 1860 United States presidential election in Rhode Island 1856 November 2 1860 1864 Nominee Abraham Lincoln Stephen Douglas Dem John C Breckinridge SD John Bell CU Party Republican FusionHome state Illinois Illinois Douglas Kentucky Breckinridge Tennessee Bell Running mate Hannibal Hamlin Herschel V Johnson Dem Joseph Lane SD Edward Everett CU Electoral vote 4 0Popular vote 12 244 7 707Percentage 61 37 38 63 County Results Lincoln 50 60 60 70 President before electionJames BuchananDemocratic Elected President Abraham LincolnRepublicanRhode Island was won by Republican candidate Abraham Lincoln who won by a margin of 22 74 With 61 37 of the popular vote Rhode Island would prove to be Lincoln s fifth strongest state in terms of popular vote percentage in the 1860 election after Vermont Minnesota Massachusetts and Maine 1 Like New Jersey New York and Pennsylvania Rhode Island was one of the four states that had a fusion ticket for the Democrats which was supported just by the Northern Democrats but also supporters of Southern Democrats and Constitutional Unionists However unlike in the other three states the electors on the Democratic ticket in Rhode Island were pledged solely to Douglas therefore some sources credit the Democratic popular vote to the Northern Democratic nominee Results Edit1860 United States presidential election in Rhode Island 2 Party Candidate Votes Percentage Electoral votesRepublican Abraham Lincoln 12 244 61 37 4Fusion Stephen A Douglas John C Breckinridge John Bell 7 707 38 63 0Totals 19 951 100 0 4See also EditUnited States presidential elections in Rhode IslandReferences Edit 1860 Presidential Election Statistics Dave Leip s Atlas of U S Presidential Elections Retrieved 2018 03 05 1860 Presidential General Election Results Rhode Island U S Election Atlas Retrieved 3 August 2012 This Rhode Island elections related article is a stub You can help Wikipedia by expanding it vte Retrieved from https en wikipedia org w index php title 1860 United States presidential election in Rhode Island amp oldid 1127154308, wikipedia, wiki, book, books, library,
