Distance Calculator

Distance from Seymour to Lawrence

Distance between Seymour and Lawrence is 98 kilometers (61 miles).
Driving distance from Seymour to Lawrence is 110 kilometers (68 miles).

air 98 km
air 61 miles
car 110 km
car 68 miles

Distance Map Between Seymour and Lawrence

Seymour, Indianapolis, United StatesLawrence, Indianapolis, United States = 61 miles = 98 km.

How far is it between Seymour and Lawrence

Seymour is located in United States with (38.9592,-85.8903) coordinates and Lawrence is located in United States with (39.8387,-86.0253) coordinates. The calculated flying distance from Seymour to Lawrence is equal to 61 miles which is equal to 98 km.

If you want to go by car, the driving distance between Seymour and Lawrence is 109.79 km. If you ride your car with an average speed of 112 kilometers/hour (70 miles/h), travel time will be 00 hours 58 minutes. Please check the avg. speed travel time table on the right for various options.
Difference between fly and go by a car is 12 km.

City/PlaceLatitude and LongitudeGPS Coordinates
Seymour 38.9592, -85.8903 38° 57´ 33.1920'' N
85° 53´ 24.9000'' W
Lawrence 39.8387, -86.0253 39° 50´ 19.1400'' N
86° 1´ 30.9360'' W

Estimated Travel Time Between Seymour and Lawrence

Average SpeedTravel Time
30 mph (48 km/h) 02 hours 17 minutes
40 mph (64 km/h) 01 hours 42 minutes
50 mph (80 km/h) 01 hours 22 minutes
60 mph (97 km/h) 01 hours 07 minutes
70 mph (112 km/h) 00 hours 58 minutes
75 mph (120 km/h) 00 hours 54 minutes
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