Seymourville, West Virginia | |
|---|---|
  Seymourville  Seymourville | |
| Coordinates: 39°03′25″N 79°06′32″W / 39.05694°N 79.10889°W | |
| Country | United States |
| State | West Virginia |
| County | Grant |
| Elevation | 1,030 ft (310 m) |
| Time zone | UTC-5 (Eastern (EST)) |
| • Summer (DST) | UTC-4 (EDT) |
| Area code(s) | 304 & 681 |
| GNIS feature ID | 1555594[1] |
Seymourville is an unincorporated community in Grant County, West Virginia, United States. Seymourville is 4.5 miles (7.2 km) north of Petersburg.
References
- ↑ "US Board on Geographic Names". United States Geological Survey. October 25, 2007. Retrieved January 31, 2008.
Municipalities and communities of Grant County, West Virginia, United States | ||
|---|---|---|
| City |  Map of West Virginia highlighting Grant County | |
| Town | ||
| Unincorporated communities | ||
| Footnotes | ‡This populated place also has portions in an adjacent county or counties | |
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