Distance between Pereslavl-Zalessky and Alexandrov, Vladimir Oblast
Distance between Pereslavl-Zalessky and Alexandrov, Vladimir Oblast is 38.59 km. This distance is equal to 23.98 miles, and 20.82 nautical miles.
The distance line on map shows distance from Pereslavl-Zalessky to Alexandrov, Vladimir Oblast between two cities.
If you travel with an airplane (which has average speed of 560 miles per hour) between Pereslavl-Zalessky to Alexandrov, Vladimir Oblast,
It takes 0.04 hours to arrive.
Pereslavl-Zalessky | 56.7333333 | 38.85 |
---|---|---|
Alexandrov, Vladimir Oblast | 56.3947309 | 38.712037 |
Distance | 38.59 km | 23.98 miles |
Related Distances
- Distance between Pereslavl-Zalessky and CityName=Semkhoz; RegionName=Sergiyevo-Posadskiy rayon~[0x1d]50; CountryName=Rossiyskaya Federatsiya (67,47 km.)
- Distance between Klin and Pereslavl-Zalessky (138,15 km.)
- Distance between Pereslavl-Zalessky and Rzhev (282,45 km.)
- Distance between Pereslavl-Zalessky and Kubinka (184,58 km.)
- Distance between Yaroslavl and Pereslavl-Zalessky (116,99 km.)
- Distance between Pereslavl-Zalessky and Alexandrov, Vladimir Oblast (38,59 km.)