Distance from Shenandoah Junction, West Virginia to Petersburg, Nebraska

Wondering how far is Petersburg, Nebraska from Shenandoah Junction, West Virginia? Here's our driving distance calculation from Shenandoah Junction to Petersburg.

Distance
1227 mi
1975 km
Travel Time
18 hr 38 min

The driving distance between Shenandoah Junction, West Virginia and Petersburg, Nebraska is 1227 mi or 1975 km . It takes 18 hr 38 min to reach Petersburg from Shenandoah Junction by road. The distance by air route is 1073 mi / 1727 km.

There are 3 routes to reach Petersburg, Nebraska from Shenandoah Junction, West Virginia.

Via I-80 W
1227 mi / 1975 km - 18 hr 38 min
Via I-70 W
1259 mi / 2026 km - 19 hr 12 min
Via I-76 W and I-80 W
1250 mi / 2012 km - 19 hr 24 min

Shenandoah Junction, West Virginia

Time 14:28:43
Latitude 39.3587115
Longitude -77.8430506
State West Virginia
Zip 25442
Timezone America/New_York

Petersburg, Nebraska

Time 13:28:43
Latitude 41.8533427
Longitude -98.0786758
State Nebraska
Zip 68652
Timezone America/North_Dakota/New_Salem

Nearby Cities

Wilson-Conococheague

35 mi / 56 km

Brambleton

39 mi / 63 km

Arcola

41 mi / 66 km

McNair

45 mi / 72 km

Herndon

45 mi / 72 km

Reston

47 mi / 76 km

South Riding

48 mi / 77 km

Chantilly

48 mi / 77 km

Navy

52 mi / 84 km

Difficult Run

53 mi / 85 km

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