Distance from Sissonville, West Virginia to Max, North Dakota
Wondering how far is Max, North Dakota from Sissonville, West Virginia? Here's our driving distance calculation from Sissonville to Max.
The driving distance between Sissonville, West Virginia and Max, North Dakota is 1400 mi or 2253 km . It takes 20 hr 45 min to reach Max from Sissonville by road. The distance by air route is 1176 mi / 1893 km.
There are 1 routes to reach Max, North Dakota from Sissonville, West Virginia.
1400 mi / 2253 km - 20 hr 45 min
Sissonville, West Virginia
Time | 17:35:15 |
Latitude | 38.5281485 |
Longitude | -81.6309601 |
State | West Virginia |
Zip | |
Timezone | America/Indiana/Vevay |
Max, North Dakota
Time | 16:35:15 |
Latitude | 47.8211141 |
Longitude | -101.2987702 |
State | North Dakota |
Zip | 58759 |
Timezone | America/North_Dakota/Beulah |
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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