1860 United States presidential election in Rhode Island

none June 2025 September 2023 1860 United States presidential election 1860 United States presidential election in Rhode Island Rhode Island presidential no 1856 United States presidential election in Rhode Island 1856 1864 United States presidential election in Rhode Island 1864 November 2, 1860 x200px Abraham Lincoln by Alexander Hesler, 1860-restored (cropped).png Abraham Lincoln Republican Party (United States) Illinois Hannibal Hamlin 4 12,244 61.37% 100x100px 100x100px100x100px Stephen Douglas (Dem) John C. Breckinridge (SD) John Bell (CU) Fusion Party (US) DemocraticSouthern DemocraticConstitutional Union Illinois (Douglas) Kentucky (Breckinridge) Tennessee (Bell) Herschel V. Johnson (Dem) Joseph Lane (SD) Edward Everett (CU) 0 7,707 38.63% 1860 United States presidential election in Rhode Island results map by county.svg 250px County results Lincoln #e27f90 50–60% #cc2f4a 60–70% President James Buchanan Democratic Party (United States) Abraham Lincoln Republican Party (United States) 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. Rhode Island was won by Republican candidate Abraham Lincoln, who won by a margin of 22.74%. Results 1860 United States presidential election in Rhode Island1860 Presidential General Election Results - Rhode Island http://uselectionatlas.org/RESULTS/state.php?year=1860&fips=44&f=1&off=0&elect=0 U.S. Election Atlas 3 August 2012PartyCandidateVotesPercentageElectoral votes Republican Abraham Lincoln 12,244 61.37% 4 Fusion Stephen A. Douglas / John C. Breckinridge / John Bell 7,707 38.63% 0 Totals 19,951 100.00% 4 Results by county +1860 United States Presidential Election in Rhode Island (by county)https://books.google.com/books?id=Jz4KAQAAMAAJ&pg=PA47 The Tribune Almanac and Political Register 1861 Tribune Association. enCountyAbraham LincolnRepublicanStephen A. Douglas/John C. Breckinridge/John BellFusionTotal votescast#%#%Bristol 66759.08%46240.92%1,129Kent 1,24665.48%65734.52%1,903Newport 1,61064.68%87935.32%2,489Providence 7,20259.63%4,87540.37%12,077Washington 1,51964.56%83435.44%2,353Total12,24461.37%7,70738.63%19,951 Analysis 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, neighboring Massachusetts and Maine.1860 Presidential Election Statistics https://uselectionatlas.org/RESULTS/stats.php?year=1860&f=1&off=0&elect=0 2018-03-05 Dave Leip’s Atlas of U.S. Presidential Elections 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 not only 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. See also United States presidential elections in Rhode Island References Category:1860 Rhode Island elections 1860 Rhode Island