Distance from Junin to San Fernando del Valle de Catamarca
Distance between Junin and San Fernando del Valle de Catamarca is 820 kilometers (509 miles).
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Distance Map Between Junin and San Fernando del Valle de Catamarca
Junin, La Plata, Argentina ↔ San Fernando del Valle de Catamarca, Argentina = 509 miles = 820 km.
How far is it between JunÃn and San Fernando del Valle de Catamarca
Junin is located in Argentina with (-34.5838,-60.9433) coordinates and San Fernando del Valle de Catamarca is located in Argentina with (-28.4696,-65.7852) coordinates. The calculated flying distance from Junin to San Fernando del Valle de Catamarca is equal to 509 miles which is equal to 820 km.
| City/Place | Latitude and Longitude | GPS Coordinates |
|---|---|---|
| Junin | -34.5838, -60.9433 | 34° 35´ 1.7520'' S 60° 56´ 35.9520'' W |
| San Fernando del Valle de Catamarca | -28.4696, -65.7852 | 28° 28´ 10.4520'' S 65° 47´ 6.8640'' W |
Junin, La Plata, Argentina
Related Distances from Junin
| Cities | Distance |
|---|---|
| Junin to Azul | 300 km |
| Junin to Campana | 228 km |
| Junin to Dolores | 425 km |
| Junin to Lincoln 4 | 66 km |
| Junin to Lujan | 198 km |
| Junin to Mar Del Plata | 554 km |
| Junin to Mercedes | 173 km |
| Junin to Moron | 245 km |
| Junin to Necochea | 551 km |
| Junin to Nueve De Julio | 107 km |
San Fernando del Valle de Catamarca, Argentina