Distance from Portoviejo to Savage
Distance from Portoviejo to Savage is 5,235 kilometers. This air travel distance is equal to 3,253 miles.
The air travel (bird fly) shortest distance between Portoviejo and Savage is 5,235 km or 3,253 miles.
If you travel with an airplane (which has average speed of 560 miles) from Portoviejo to Savage, It takes 5.81 hours to arrive.
Portoviejo
Portoviejo is located in Ecuador.
| GPS Coordinates (DMS) | 1° 3´ 16.4880'' S 80° 27´ 16.0200'' W |
|---|---|
| Latitude | -1.05458 |
| Longitude | -80.45445 |
| Altitude | 44 m |
| Country | Ecuador |
Portoviejo Distances to Cities
| Portoviejo | Distance |
|---|---|
| Distance from Portoviejo to Prospect Heights | 4,838 km |
| Distance from Portoviejo to Bothell | 6,750 km |
| Distance from Portoviejo to Savage | 5,235 km |
| Distance from Portoviejo to Castro Valley | 6,056 km |
| Distance from Portoviejo to Pinole | 6,092 km |
Savage
Savage is located in United States.
| GPS Coordinates | 44° 46´ 44.8680'' N 93° 20´ 10.8240'' W |
|---|---|
| Latitude | 44.77913 |
| Longitude | -93.33634 |
| Altitude | 216 m |
| Country | United States |