Distance from Sissonville, West Virginia to Petersburg, North Dakota
Wondering how far is Petersburg, North Dakota from Sissonville, West Virginia? Here's our driving distance calculation from Sissonville to Petersburg.
The driving distance between Sissonville, West Virginia and Petersburg, North Dakota is 1256 mi or 2021 km . It takes 18 hr 37 min to reach Petersburg from Sissonville by road. The distance by air route is 1049 mi / 1688 km.
There are 2 routes to reach Petersburg, North Dakota from Sissonville, West Virginia.
1256 mi / 2021 km - 18 hr 37 min
1233 mi / 1984 km - 18 hr 49 min
Sissonville, West Virginia
| Time | 04:12:54 |
| Latitude | 38.5281485 |
| Longitude | -81.6309601 |
| State | West Virginia |
| Zip | |
| Timezone | America/Indiana/Vevay |
Petersburg, North Dakota
| Time | 03:12:54 |
| Latitude | 48.0111052 |
| Longitude | -98.0014884 |
| State | North Dakota |
| Zip | 58272 |
| Timezone | America/North_Dakota/Center |
Sissonville Distance to Popular Cities
| Route | Distance |
|---|---|
| Sissonville to Eskdale | 41 mi / 66 km |
| Sissonville to Pax | 61 mi / 98 km |
| Sissonville to Scarbro | 64 mi / 103 km |
| Sissonville to Oak Hill | 65 mi / 105 km |
| Sissonville to Hilltop | 66 mi / 106 km |
| Sissonville to Glen Jean | 66 mi / 106 km |
| Sissonville to Mt Hope | 68 mi / 109 km |
| Sissonville to Prosperity | 71 mi / 114 km |
| Sissonville to Bradley | 71 mi / 114 km |
| Sissonville to Gatewood | 72 mi / 116 km |
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