Distance from Searcy to Sand Springs
Distance from Searcy to Sand Springs is 408 kilometers. This air travel distance is equal to 254 miles.
The air travel (bird fly) shortest distance between Searcy and Sand Springs is 408 km or 254 miles.
If you travel with an airplane (which has average speed of 560 miles) from Searcy to Sand Springs, It takes 0.45 hours to arrive.
Searcy
Searcy is located in United States.
| GPS Coordinates (DMS) | 35° 15´ 2.3040'' N 91° 44´ 10.5000'' W |
|---|---|
| Latitude | 35.25064 |
| Longitude | -91.73625 |
| Altitude | 82 m |
| Country | United States |
Searcy Distances to Cities
| Searcy | Distance |
|---|---|
| Distance from Searcy to Tulsa | 398 km |
| Distance from Searcy to Starkville | 334 km |
| Distance from Searcy to Farmers Branch | 541 km |
| Distance from Searcy to Dallas | 542 km |
| Distance from Searcy to Oklahoma City | 526 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 |
Sand Springs Distances to Cities
| Sand Springs | Distance |
|---|---|
| Distance from Sand Springs to Tahlequah | 106 km |
| Distance from Sand Springs to Amarillo | 528 km |
| Distance from Sand Springs to Sapulpa | 16 km |
| Distance from Sand Springs to Wasco Ca | 2,094 km |
| Distance from Sand Springs to Tulsa | 11 km |