Distance from Alexandria Bay, New York to Seymour, Indiana

Wondering how far is Seymour, Indiana from Alexandria Bay, New York? Here's our driving distance calculation from Alexandria Bay to Seymour.

Distance
750 mi
1207 km
Travel Time
11 hr 23 min

The driving distance between Alexandria Bay, New York and Seymour, Indiana is 750 mi or 1207 km . It takes 11 hr 23 min to reach Seymour from Alexandria Bay by road. The distance by air route is 634 mi / 1020 km.

There are 2 routes to reach Seymour, Indiana from Alexandria Bay, New York.

Via I-90 W and I-71 S
750 mi / 1207 km - 11 hr 23 min
Via I-90 W
813 mi / 1308 km - 12 hr 3 min

Alexandria Bay, New York

Time 09:10:11
Latitude 44.3358836
Longitude -75.9177309
State New York
Zip
Timezone America/New_York

Seymour, Indiana

Time 09:10:11
Latitude 38.9592201
Longitude -85.8902547
State Indiana
Zip 47274
Timezone America/Indiana/Vevay

Nearby Cities

Village Green

105 mi / 169 km

Jordan

113 mi / 182 km

Weedsport

117 mi / 188 km

Port Byron

121 mi / 195 km

Cayuga

133 mi / 214 km

Clyde

141 mi / 227 km

Sharon Springs

150 mi / 241 km

Fultonville

154 mi / 248 km

Fonda

155 mi / 249 km

Tribes Hill

159 mi / 256 km

Alexandria Bay Distance to Popular Cities

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