Distance from Martinsburg to Sayreville Junction
Distance from Martinsburg to Sayreville Junction is 330 kilometers. This air travel distance is equal to 205 miles.
The air travel (bird fly) shortest distance between Martinsburg and Sayreville Junction is 330 km or 205 miles.
If you travel with an airplane (which has average speed of 560 miles) from Martinsburg to Sayreville Junction, It takes 0.37 hours to arrive.
Martinsburg
Martinsburg is located in United States.
| GPS Coordinates (DMS) | 39° 27´ 22.3560'' N 77° 57´ 50.0040'' W |
|---|---|
| Latitude | 39.45621 |
| Longitude | -77.96389 |
| Altitude | 139 m |
| Country | United States |
Martinsburg Distances to Cities
| Martinsburg | Distance |
|---|---|
| Distance from Martinsburg to Los Angeles | 3,616 km |
| Distance from Martinsburg to Framingham | 635 km |
| Distance from Martinsburg to Steubenville | 250 km |
| Distance from Martinsburg to Pittsburgh | 205 km |
| Distance from Martinsburg to Dix Hills | 422 km |
Sayreville Junction
Sayreville Junction is located in United States.
| GPS Coordinates | 40° 27´ 55.3680'' N 74° 19´ 49.5480'' W |
|---|---|
| Latitude | 40.46538 |
| Longitude | -74.33043 |
| Altitude | 35 m |
| Country | United States |