Distance from San Giuseppe Vesuviano to Walnut Park
Distance from San Giuseppe Vesuviano to Walnut Park is 10,416 kilometers. This air travel distance is equal to 6,472 miles.
The air travel (bird fly) shortest distance between San Giuseppe Vesuviano and Walnut Park is 10,416 km or 6,472 miles.
If you travel with an airplane (which has average speed of 560 miles) from San Giuseppe Vesuviano to Walnut Park, It takes 11.56 hours to arrive.
San Giuseppe Vesuviano
San Giuseppe Vesuviano is located in Italy.
| GPS Coordinates (DMS) | 40° 49´ 49.2240'' N 14° 30´ 12.3120'' E |
|---|---|
| Latitude | 40.83034 |
| Longitude | 14.50342 |
| Altitude | 106 m |
| Country | Italy |
San Giuseppe Vesuviano Distances to Cities
| San Giuseppe Vesuviano | Distance |
|---|---|
| Distance from San Giuseppe Vesuviano to Walnut Park | 10,416 km |
| Distance from San Giuseppe Vesuviano to Inver Grove Heights | 8,092 km |
| Distance from San Giuseppe Vesuviano to Watauga | 9,267 km |
| Distance from San Giuseppe Vesuviano to Irving | 9,250 km |
| Distance from San Giuseppe Vesuviano to Casper | 9,002 km |
Walnut Park
Walnut Park is located in United States.
| GPS Coordinates | 33° 58´ 5.0520'' N 118° 13´ 30.2520'' W |
|---|---|
| Latitude | 33.96807 |
| Longitude | -118.22507 |
| Altitude | 48 m |
| Country | United States |