Distance from Jefferson City to Tinton Falls
Distance from Jefferson City to Tinton Falls is 1,565 kilometers. This air travel distance is equal to 972 miles.
The air travel (bird fly) shortest distance between Jefferson City and Tinton Falls is 1,565 km or 972 miles.
If you travel with an airplane (which has average speed of 560 miles) from Jefferson City to Tinton Falls, It takes 1.74 hours to arrive.
Jefferson City
Jefferson City is located in United States.
| GPS Coordinates (DMS) | 38° 34´ 36.1200'' N 92° 10´ 24.6720'' W |
|---|---|
| Latitude | 38.57670 |
| Longitude | -92.17352 |
| Altitude | 195 m |
| Country | United States |
Jefferson City Distances to Cities
| Jefferson City | Distance |
|---|---|
| Distance from Jefferson City to Sacramento | 2,544 km |
| Distance from Jefferson City to Boston | 1,834 km |
| Distance from Jefferson City to San Francisco | 2,640 km |
| Distance from Jefferson City to Los Angeles | 2,386 km |
| Distance from Jefferson City to Dallas | 767 km |
Tinton Falls
Tinton Falls is located in United States.
| GPS Coordinates | 40° 18´ 15.4080'' N 74° 6´ 1.5120'' W |
|---|---|
| Latitude | 40.30428 |
| Longitude | -74.10042 |
| Altitude | 14 m |
| Country | United States |