Distance from Aleksandrovskoye to Hanford
Distance from Aleksandrovskoye to Hanford is 10,861 kilometers. This air travel distance is equal to 6,749 miles.
The air travel (bird fly) shortest distance between Aleksandrovskoye and Hanford is 10,861 km or 6,749 miles.
If you travel with an airplane (which has average speed of 560 miles) from Aleksandrovskoye to Hanford, It takes 12.05 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 |
Hanford
Hanford is located in United States.
| GPS Coordinates | 36° 19´ 38.8200'' N 119° 38´ 44.4480'' W |
|---|---|
| Latitude | 36.32745 |
| Longitude | -119.64568 |
| Altitude | 80 m |
| Country | United States |