Sand Hill (Herkimer County, New York)
| Sand Hill | |
|---|---|
  Sand Hill Location of Sand Hill within New York  Sand Hill Sand Hill (the United States) | |
| Highest point | |
| Elevation | 1,542 feet (470 m) |
| Coordinates | 43°15′31″N 75°00′53″W / 43.25861°N 75.01472°W[1] |
| Geography | |
| Location | ENE of Poland, New York, U.S. |
| Topo map | USGS Hinckley |
Sand Hill is a summit located in Central New York Region of New York located in the Town of Russia in Herkimer County, east-northeast of Poland.
References[edit]
- ^ "Sand Hill". Geographic Names Information System. United States Geological Survey, United States Department of the Interior. Retrieved 2017-12-08.