Distance from Seaford, New York to Fontainebleau, Florida

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

Distance
1324 mi
2131 km
Travel Time
19 hr 25 min

The driving distance between Seaford, New York and Fontainebleau, Florida is 1324 mi or 2131 km . It takes 19 hr 25 min to reach Fontainebleau from Seaford by road. The distance by air route is 1102 mi / 1773 km.

There are 2 routes to reach Fontainebleau, Florida from Seaford, New York.

Via I-95 S
1324 mi / 2131 km - 19 hr 25 min
Via I-81 S and I-95 S
1419 mi / 2284 km - 20 hr 49 min

Seaford, New York

Time 05:33:12
Latitude 40.6659344
Longitude -73.4881809
State New York
Zip
Timezone America/New_York

Fontainebleau, Florida

Time 05:33:12
Latitude 25.7728774
Longitude -80.3478301
State Florida
Zip
Timezone America/Kentucky/Monticello

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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