Distance from San Narciso to Harker Heights
Distance from San Narciso to Harker Heights is 13,511 kilometers. This air travel distance is equal to 8,395 miles.
The air travel (bird fly) shortest distance between San Narciso and Harker Heights is 13,511 km or 8,395 miles.
If you travel with an airplane (which has average speed of 560 miles) from San Narciso to Harker Heights, It takes 14.99 hours to arrive.
San Narciso
San Narciso is located in Philippines.
| GPS Coordinates (DMS) | 15° 0´ 51.4800'' N 120° 4´ 49.0800'' E |
|---|---|
| Latitude | 15.01430 |
| Longitude | 120.08030 |
| Altitude | 16 m |
| Country | Philippines |
San Narciso Distances to Cities
| San Narciso | Distance |
|---|---|
| Distance from San Narciso to Tigard | 10,819 km |
| Distance from San Narciso to Webster Groves | 13,282 km |
| Distance from San Narciso to Harker Heights | 13,511 km |
| Distance from San Narciso to Ridgeland | 13,854 km |
| Distance from San Narciso to Savannah | 14,328 km |
Harker Heights
Harker Heights is located in United States.
| GPS Coordinates | 31° 5´ 0.6360'' N 97° 39´ 35.0640'' W |
|---|---|
| Latitude | 31.08351 |
| Longitude | -97.65974 |
| Altitude | 236 m |
| Country | United States |