Sevastopol, Indiana
From Wikipedia, the free encyclopedia
| Sevastopol | |
|---|---|
| — Town — | |
 Sevastopol
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| Coordinates: 41°07′45″N 86°01′08″W / 41.12917°N 86.01889°WCoordinates: 41°07′45″N 86°01′08″W / 41.12917°N 86.01889°W | |
| Country | United States |
| State | Indiana |
| County | Kosciusko |
| Township | Franklin |
| Elevation[1] | 879 ft (268 m) |
| ZIP code | 46510 |
| FIPS code | 18-68760[2] |
| GNIS feature ID | 443221 |
Sevastopol is an unincorporated town in Franklin Township, Kosciusko County, Indiana. The community is located in the far southwest corner of the county, five miles from Mentone, seven from Burket, and 15 from Warsaw, the county seat.
[edit] Geography
Sevastopol is located at 41°07′45″N 86°01′08″W / 41.12917°N 86.01889°W.
[edit] References
- ^ "US Board on Geographic Names". United States Geological Survey. 2007-10-25. http://geonames.usgs.gov. Retrieved 2008-01-31.
- ^ "American FactFinder". United States Census Bureau. http://factfinder.census.gov. Retrieved 2008-01-31.
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