Distance between Villanova and Sayreville Junction
Distance between Villanova and Sayreville Junction is 98.67 km. This distance is equal to 61.31 miles, and 53.24 nautical miles.
The distance line on map shows distance from Villanova to Sayreville Junction between two cities.
If you travel with an airplane (which has average speed of 560 miles per hour) between Villanova to Sayreville Junction,
It takes 0.11 hours to arrive.
| Villanova | 40.0375832 | -75.3491813 |
|---|---|---|
| Sayreville Junction | 40.4653838 | -74.3304256 |
| Distance | 98.67 km | 61.31 miles |
Related Distances
- Distance between Califon and Villanova (87,41 km.)
- Distance between Newark Liberty International Airport and Villanova (122,89 km.)
- Distance between George Washington's Mt Vernon Estate, Museum & Gardens and Villanova (209,90 km.)
- Distance between Teterboro Airport and Villanova (141,46 km.)
- Distance between Denmar and Villanova (473,20 km.)
- Distance between Villanova and By Way (184,38 km.)