Distance from Sissonville, West Virginia to Pillsbury, North Dakota
Wondering how far is Pillsbury, North Dakota from Sissonville, West Virginia? Here's our driving distance calculation from Sissonville to Pillsbury.
The driving distance between Sissonville, West Virginia and Pillsbury, North Dakota is 1201 mi or 1933 km . It takes 17 hr 49 min to reach Pillsbury from Sissonville by road. The distance by air route is 1011 mi / 1627 km.
There are 2 routes to reach Pillsbury, North Dakota from Sissonville, West Virginia.
1201 mi / 1933 km - 17 hr 49 min
1177 mi / 1894 km - 18 hr 1 min
Sissonville, West Virginia
| Time | 04:16:39 |
| Latitude | 38.5281485 |
| Longitude | -81.6309601 |
| State | West Virginia |
| Zip | |
| Timezone | America/Indiana/Vevay |
Pillsbury, North Dakota
| Time | 03:16:39 |
| Latitude | 47.2069306 |
| Longitude | -97.7906569 |
| State | North Dakota |
| Zip | |
| 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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