Distance from Jefferson City to Reynoldsburg
Distance from Jefferson City to Reynoldsburg is 822 kilometers. This air travel distance is equal to 511 miles.
The air travel (bird fly) shortest distance between Jefferson City and Reynoldsburg is 822 km or 511 miles.
If you travel with an airplane (which has average speed of 560 miles) from Jefferson City to Reynoldsburg, It takes 0.91 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 |
Reynoldsburg
Reynoldsburg is located in United States.
| GPS Coordinates | 39° 57´ 17.2440'' N 82° 48´ 43.6320'' W |
|---|---|
| Latitude | 39.95479 |
| Longitude | -82.81212 |
| Altitude | 266 m |
| Country | United States |
Reynoldsburg Distances to Cities
| Reynoldsburg | Distance |
|---|---|
| Distance from Reynoldsburg to Jamaica Plain | 1,015 km |
| Distance from Reynoldsburg to Lincolnia | 504 km |
| Distance from Reynoldsburg to Boardman | 217 km |
| Distance from Reynoldsburg to Dublin Oh | 30 km |
| Distance from Reynoldsburg to Janesville | 601 km |