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 | |
| Country | United States |
| State | New York |
| County | Steuben |
| Area | |
| • Total | 43.54 sq mi (112.78 km2) |
| • Land | 43.33 sq mi (112.22 km2) |
| • Water | 0.22 sq mi (0.56 km2) |
| Elevation | 1,198 ft (365 m) |
| Population | |
| • Total | 4,039  |
| • Estimate (2021)[2] | 4,002 |
| • Density | 93.24/sq mi (36.00/km2) |
| Time zone | UTC-5 (Eastern (EST)) |
| • Summer (DST) | UTC-4 (EDT) |
| FIPS code | 36-35683 |
| GNIS feature ID | 0979079 |
Hornellsville is a town in Steuben County, New York, United States. The population, not counting the city of Hornell, was 4,039 at the 2020 census.[2] The name is taken from a prominent pioneer family.[3]
The Town of Hornellsville is at the western border of the county, and surrounds the city of Hornell. Until 1906, the city of Hornell was named Hornellsville.
The name Hornellsville is used only in real estate or legal contexts, but rarely in conversation. This is because Hornell is much more frequently mentioned, and its boundaries are quite different.
- ^ "2016 U.S. Gazetteer Files". United States Census Bureau. Retrieved July 5, 2017.
- ^ a b c Bureau, US Census. "City and Town Population Totals: 2020—2021". Census.gov. US Census Bureau. Retrieved October 6, 2022.
{{cite web}}:|last1=has generic name (help) - ^ Gannett, Henry (1905). The Origin of Certain Place Names in the United States. Govt. Print. Off. pp. 161.