Distance between Alexandrov, Vladimir Oblast and Chekhov
Distance between Alexandrov, Vladimir Oblast and Chekhov is 158.74 km. This distance is equal to 98.64 miles, and 85.66 nautical miles.
The distance line on map shows distance from Alexandrov, Vladimir Oblast to Chekhov between two cities.
If you travel with an airplane (which has average speed of 560 miles per hour) between Alexandrov, Vladimir Oblast to Chekhov,
It takes 0.18 hours to arrive.
| Alexandrov, Vladimir Oblast | 56.3947309 | 38.712037 |
|---|---|---|
| Chekhov | 55.1526207 | 37.4604756 |
| Distance | 158.74 km | 98.64 miles |
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