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Republicans Choose Candidates for Plattsburgh City Seats

Plattsburgh city Republicans caucused late last week and chose their slate of candidates to challenge the Democrats vying for all City Hall seats in the November election.

All of Plattsburgh’s Common Council seats and the mayor’s position are open this year. City Republicans announced their choice for mayor and the six candidates for the Common Council late last week.
After saying he would not run because he planned to retire from city politics, independent James Calnon received the endorsement of city Republicans as their mayoral candidate to replace the retiring Don Kasprzak.

Calnon is currently the mayor pro tem and represents Ward Four on the Council. He joins Michael Drew as the only Republican candidates with previous experience on the Council. The Democrats chose a similar slate, with the mayoral candidate also a current city councilor and only one other candidate with previous experience.

And that new council faces, according to mayoral candidate James Calnon, continuing fiscal challenges.

Michael Drew served on the Common Council for six years. For two years, he was clerk of the council and was mayor pro tem for the other four. He was interim mayor in 2006 when then-mayor Dan Stewart left for a state position. Drew was defeated in a primary race for mayor that year by Kasprzak, the current mayor, who leaves office in December. The Ward Two candidate says he is returning to city politics, in part, because he is concerned about losing experienced councilors.

Drew says key issues for the city remain fiscally oriented.

Drew hopes that with a 60 million dollar budget, city voters will elect some individuals with experience dealing with Plattsburgh’s intricacies and issues.

Candidates must submit petitions with 5 percent of their ward’s registered voters to the Board of Elections by July 11th.
According to data from the Clinton County Board of Elections, 42 percent of registered city voters are Democrats, 24 percent Republican and 7 percent Independent.

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