Distance from Paso-Robles to Alexandria
Driving route: -- ( - )
Driving route: -- ( - )
Extended route information
The flight distance between the nearest airports Paso-Robles (SJC) and Alexandria (DCA) is 2,392.69 mi (3,850.67 km). This corresponds to an approximate flight time of 5h 1min. Similar flight routes: SJC → IAD , SJC → BWI , SJC → RIC , SJC → PHF , SFO → DCA
Bearing: 84.78° (E)The initial bearing on the course from Paso-Robles to Alexandria is 84.78° and the compass direction is E.
Midpoint: 39.28573,-99.33779The geographic midpoint between Paso-Robles and Alexandria is in 1,193.59 mi (1,920.90 km) distance between both points in a bearing of 84.78°. It is located in United States of America, Kansas, Rooks County.
Distance: 2,387.18 mi (3,841.79 km)The shortest distance (air line) between Paso-Robles and Alexandria is 2,387.18 mi (3,841.79 km).
Driving route: -- ( - )The shortest route between Paso-Robles and Alexandria is according to the route planner. The driving time is approx. . Half of the trip is reached in .
Time difference: 3hThe time difference between Paso-Robles (America/Los_Angeles) and Alexandria (America/New_York) is 3 hours. This means that it is now 08:58 (27.04.2024) in Paso-Robles and 11:58 (27.04.2024) in Alexandria.
How is the distance calculated?
To calculate the distance between Paso-Robles and Alexandria, 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:
Similar distance (± 0.5%)
Alexandria is just as far away from Paso-Robles as Paso-Robles from Havana (3,931 km), Philadelphia (3,984 km), Guatemala City (3,801 km), Jacksonville (3,662 km), Ottawa (3,898 km), Cancún (3,653 km), Baltimore (3,871 km), Washington (3,841 km), San Salvador (3,978 km).
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