Distance from Seaford, New York to Faribault, Minnesota

Wondering how far is Faribault, Minnesota from Seaford, New York? Here's our driving distance calculation from Seaford to Faribault.

Distance
1227 mi
1975 km
Travel Time
18 hr 26 min

The driving distance between Seaford, New York and Faribault, Minnesota is 1227 mi or 1975 km . It takes 18 hr 26 min to reach Faribault from Seaford by road. The distance by air route is 1036 mi / 1667 km.

There are 3 routes to reach Faribault, Minnesota from Seaford, New York.

Via I-80 W and I-90 W
1227 mi / 1975 km - 18 hr 26 min
Via I-80 W, I-90 W and US-52 N
1229 mi / 1978 km - 18 hr 29 min
Via I-76 W, I-80 W and I-90 W
1275 mi / 2052 km - 19 hr 7 min

Seaford, New York

Time 06:01:49
Latitude 40.6659344
Longitude -73.4881809
State New York
Zip
Timezone America/New_York

Faribault, Minnesota

Time 05:01:49
Latitude 44.2949636
Longitude -93.268827
State Minnesota
Zip 55021
Timezone America/Menominee

Nearby Cities

East Atlantic Beach

19 mi / 31 km

Matinecock

21 mi / 34 km

Kings Point

25 mi / 40 km

Saltaire

25 mi / 40 km

Fire Island

29 mi / 47 km

Ocean Beach

28 mi / 45 km

Bronx

35 mi / 56 km

Fort Lee

37 mi / 60 km

New Rochelle

37 mi / 60 km

Leonia

38 mi / 61 km

Seaford Distance to Popular Cities

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