Distance from Seymour to Providence
Distance from Seymour to Providence is 1,268 kilometers. This air travel distance is equal to 788 miles.
The air travel (bird fly) shortest distance between Seymour and Providence is 1,268 km or 788 miles.
If you travel with an airplane (which has average speed of 560 miles) from Seymour to Providence, It takes 1.41 hours to arrive.
Seymour
Seymour is located in United States.
| GPS Coordinates (DMS) | 38° 57´ 33.1920'' N 85° 53´ 24.9000'' W |
|---|---|
| Latitude | 38.95922 |
| Longitude | -85.89025 |
| Altitude | 185 m |
| Country | United States |
Seymour Distances to Cities
| Seymour | Distance |
|---|---|
| Distance from Seymour In to New York City | 1,035 km |
| Distance from Seymour In to Mansfield Ma | 1,288 km |
| Distance from Seymour In to Fairland | 773 km |
| Distance from Seymour In to Waverly | 433 km |
| Distance from Seymour In to Farmington Mn | 872 km |
Providence
Providence is located in United States.
| GPS Coordinates | 41° 49´ 26.3640'' N 71° 24´ 46.1880'' W |
|---|---|
| Latitude | 41.82399 |
| Longitude | -71.41283 |
| Altitude | -17 m |
| Country | United States |
Providence Distances to Cities
| Providence | Distance |
|---|---|
| Distance from Providence to Boston | 66 km |
| Distance from Providence to Smithfield | 16 km |
| Distance from Providence to Phoenix | 3,671 km |
| Distance from Providence to Virginia | 790 km |
| Distance from Providence to Palmdale | 4,117 km |