Distance from Cuyahoga-Falls to Jacksonville
Driving route: -- ( - )
Driving route: -- ( - )
Extended route information
The flight distance between the nearest airports Cuyahoga-Falls (CLE) and Jacksonville (JAX) is 733.65 mi (1,180.70 km). This corresponds to an approximate flight time of 1h 53min. Similar flight routes: CLE → DAB , CLE → SFB , CLE → SAV , CLE → MCO , PIT → JAX
Bearing: 180.73° (S)The initial bearing on the course from Cuyahoga-Falls to Jacksonville is 180.73° and the compass direction is S.
Midpoint: 35.73309,-81.57592The geographic midpoint between Cuyahoga-Falls and Jacksonville is in 373.20 mi (600.60 km) distance between both points in a bearing of 180.73°. It is located in United States of America, North Carolina, Burke County.
Distance: 746.39 mi (1,201.20 km)The shortest distance (air line) between Cuyahoga-Falls and Jacksonville is 746.39 mi (1,201.20 km).
Driving route: -- ( - )The shortest route between Cuyahoga-Falls and Jacksonville is according to the route planner. The driving time is approx. . Half of the trip is reached in .
Time difference: 0hCuyahoga-Falls and Jacksonville are both in the same time zone America/New_York.
How is the distance calculated?
To calculate the distance between Cuyahoga-Falls and Jacksonville, the place names are converted into coordinates (latitude and longitude). The respective geographic centre is used for cities, regions and countries. To calculate the distance the Haversine formula is applied.
Similar routes:
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