| 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
- ↑ "Sand Hill". Geographic Names Information System. United States Geological Survey, United States Department of the Interior. Retrieved 2017-12-08.
 
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