Distance from Sissonville, West Virginia to Oxbow, North Dakota
Wondering how far is Oxbow, North Dakota from Sissonville, West Virginia? Here's our driving distance calculation from Sissonville to Oxbow.
The driving distance between Sissonville, West Virginia and Oxbow, North Dakota is 1128 mi or 1815 km . It takes 16 hr 52 min to reach Oxbow from Sissonville by road. The distance by air route is 952 mi / 1532 km.
There are 3 routes to reach Oxbow, North Dakota from Sissonville, West Virginia.
1128 mi / 1815 km - 16 hr 52 min
1105 mi / 1778 km - 17 hr 4 min
1191 mi / 1917 km - 17 hr 31 min
Sissonville, West Virginia
Time | 17:29:29 |
Latitude | 38.5281485 |
Longitude | -81.6309601 |
State | West Virginia |
Zip | |
Timezone | America/Indiana/Vevay |
Oxbow, North Dakota
Time | 16:29:29 |
Latitude | 46.6699634 |
Longitude | -96.8003601 |
State | North Dakota |
Zip | 58047 |
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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