Distance from Boynton Beach to Jefferson City
Distance from Boynton Beach to Jefferson City is 1,751 kilometers. This air travel distance is equal to 1,088 miles.
The air travel (bird fly) shortest distance between Boynton Beach and Jefferson City is 1,751 km or 1,088 miles.
If you travel with an airplane (which has average speed of 560 miles) from Boynton Beach to Jefferson City, It takes 1.94 hours to arrive.
Boynton Beach
Boynton Beach is located in United States.
| GPS Coordinates (DMS) | 26° 31´ 31.2600'' N 80° 3´ 59.1480'' W |
|---|---|
| Latitude | 26.52535 |
| Longitude | -80.06643 |
| Altitude | 11 m |
| Country | United States |
Boynton Beach Distances to Cities
Jefferson City
Jefferson City is located in United States.
| GPS Coordinates | 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 |