Sand Hill (Herkimer County, New York)
From Wikipedia, the free encyclopedia
Mountain in New York, United States
Elevation1,542 feet (470 m)
| Sand Hill | |
|---|---|
  Sand Hill Location of Sand Hill within New York Show map of New York Sand Hill Sand Hill (the United States) Show map of 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 / 43.25861; -75.01472[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.
- ↑ "Sand Hill". Geographic Names Information System. United States Geological Survey, United States Department of the Interior. Retrieved December 8, 2017.
Related Articles
~