Distance from Aleksandrovskoye to Bloomfield
Distance from Aleksandrovskoye to Bloomfield is 8,656 kilometers. This air travel distance is equal to 5,379 miles.
The air travel (bird fly) shortest distance between Aleksandrovskoye and Bloomfield is 8,656 km or 5,379 miles.
If you travel with an airplane (which has average speed of 560 miles) from Aleksandrovskoye to Bloomfield, It takes 9.6 hours to arrive.
Aleksandrovskoye
Aleksandrovskoye is located in Russia.
| GPS Coordinates (DMS) | 44° 42´ 51.0120'' N 43° 0´ 2.9880'' E |
|---|---|
| Latitude | 44.71417 |
| Longitude | 43.00083 |
| Altitude | 295 m |
| Country | Russia |
Aleksandrovskoye Distances to Cities
| Aleksandrovskoye | Distance |
|---|---|
| Distance from Aleksandrovskoye to Chesterfield | 9,668 km |
| Distance from Aleksandrovskoye to Cerritos | 11,096 km |
| Distance from Aleksandrovskoye to Bloomfield | 8,656 km |
| Distance from Aleksandrovskoye to Hanford | 10,861 km |
| Distance from Aleksandrovskoye to Rapid City | 9,618 km |
Bloomfield
Bloomfield is located in United States.
| GPS Coordinates | 40° 48´ 24.3720'' N 74° 11´ 7.5120'' W |
|---|---|
| Latitude | 40.80677 |
| Longitude | -74.18542 |
| Altitude | 51 m |
| Country | United States |