Distance from Sissonville, West Virginia to Elliott, North Dakota
Wondering how far is Elliott, North Dakota from Sissonville, West Virginia? Here's our driving distance calculation from Sissonville to Elliott.
The driving distance between Sissonville, West Virginia and Elliott, North Dakota is 1170 mi or 1883 km . It takes 17 hr 38 min to reach Elliott from Sissonville by road. The distance by air route is 985 mi / 1585 km.
There are 3 routes to reach Elliott, North Dakota from Sissonville, West Virginia.
1170 mi / 1883 km - 17 hr 38 min
1146 mi / 1844 km - 17 hr 50 min
1242 mi / 1999 km - 18 hr 27 min
Sissonville, West Virginia
| Time | 08:55:04 |
| Latitude | 38.5281485 |
| Longitude | -81.6309601 |
| State | West Virginia |
| Zip | |
| Timezone | America/Indiana/Vevay |
Elliott, North Dakota
| Time | 07:55:04 |
| Latitude | 46.4021898 |
| Longitude | -97.8148228 |
| State | North Dakota |
| Zip | 58054 |
| 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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