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Hornellsville, New York

Hornellsville, New York
Town
Hornellsville, New York is located in New York
Hornellsville, New York
Hornellsville, New York
Location within the state of New York
Coordinates: 42°21′N 77°40′W / 42.350°N 77.667°W / 42.350; -77.667Coordinates: 42°21′N 77°40′W / 42.350°N 77.667°W / 42.350; -77.667
Country United States
State New York
County Steuben
Area
 • Total 43.6 sq mi (113.0 km2)
 • Land 43.5 sq mi (112.6 km2)
 • Water 0.2 sq mi (0.5 km2)
Elevation 1,198 ft (365 m)
Population (2010)
 • Total 4,151
 • Density 95/sq mi (37/km2)
Time zone Eastern (EST) (UTC-5)
 • Summer (DST) EDT (UTC-4)
FIPS code 36-35683
GNIS feature ID 0979079

Hornellsville is a town in Steuben County, New York, United States. United States. The population was 4,151 at the 2000 census. The name is taken from a prominent pioneer family.

The Town of Hornellsville is at the west border of the county, and surrounds the city of Hornell.

The town was first settled around 1790 in the vicinity of the City of Hornell. The Town of Hornellsville was formed from part of the Town of Canisteo in 1820. Territory from Hornellsville was used to form the Towns of Hartsville (1844) and part of Fremont (1854).

In 1852, the community of Hornellsville set itself apart as an incorporated village and became a city (Hornell) in 1888.

According to the United States Census Bureau, the town has a total area of 43.7 square miles (113 km2), of which, 43.5 square miles (113 km2) of it is land and 0.2 square miles (0.52 km2) of it (0.41%) is water.

The Southern Tier Expressway (Interstate 86 and New York State Route 17) passes through the north part of the town. New York State Route 21 and New York State Route 36 intersect by Hornell.

The Canisteo River flows southward through the town.

The west town line is the border of Allegany County, New York.


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