Distance from Clarendon to Sand Springs
Distance from Clarendon to Sand Springs is 2,740 kilometers. This air travel distance is equal to 1,703 miles.
The air travel (bird fly) shortest distance between Clarendon and Sand Springs is 2,740 km or 1,703 miles.
If you travel with an airplane (which has average speed of 560 miles) from Clarendon to Sand Springs, It takes 3.04 hours to arrive.
Clarendon
Clarendon is located in Jamaica.
| GPS Coordinates (DMS) | 17° 57´ 20.5920'' N 77° 14´ 25.8720'' W |
|---|---|
| Latitude | 17.95572 |
| Longitude | -77.24052 |
| Altitude | 76 m |
| Country | Jamaica |
Clarendon Distances to Cities
| Clarendon | Distance |
|---|---|
| Distance from Clarendon to Oklahoma City | 2,788 km |
| Distance from Clarendon to Pampa | 3,044 km |
| Distance from Clarendon to Amarillo | 3,092 km |
| Distance from Clarendon to Minnesota | 3,570 km |
| Distance from Clarendon to Lubbock | 3,002 km |
Sand Springs
Sand Springs is located in United States.
| GPS Coordinates | 36° 8´ 23.3160'' N 96° 6´ 32.0040'' W |
|---|---|
| Latitude | 36.13981 |
| Longitude | -96.10889 |
| Altitude | 212 m |
| Country | United States |