Seymourville, West Virginia

From Wikipedia, the free encyclopedia

CountryUnited States
Elevation1,030 ft (310 m)
Seymourville, West Virginia
Seymourville is located in West Virginia
Seymourville
Seymourville
Seymourville is located in the United States
Seymourville
Seymourville
Coordinates: 39°03′25″N 79°06′32″W / 39.05694°N 79.10889°W / 39.05694; -79.10889
CountryUnited States
StateWest Virginia
CountyGrant
Elevation1,030 ft (310 m)
Time zoneUTC-5 (Eastern (EST))
  Summer (DST)UTC-4 (EDT)
Area codes304 & 681
GNIS feature ID1555594[1]

Seymourville is an unincorporated community in Grant County, West Virginia, United States. Seymourville is 4.5 miles (7.2 km) north of Petersburg.

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