Sand Hill (Herkimer County, New York)
| Sand Hill | |
|---|---|
  Sand Hill Location of Sand Hill within New York | |
| Highest point | |
| Elevation | 1,542 feet (470 m) |
| Coordinates | 43°15′31″N 75°00′53″W / 43.25861°N 75.01472°WCoordinates: 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
- ↑ "Sand Hill". Geographic Names Information System. United States Geological Survey. Retrieved 2017-12-08.
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