Distance from Jacksonville to Ciego de Avila
Distance between Jacksonville and Ciego de Avila is 987 kilometers (614 miles).
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Distance Map Between Jacksonville and Ciego de Avila
Jacksonville, Tallahassee, United States ↔ Ciego de Avila, Cuba = 614 miles = 987 km.
How far is it between Jacksonville and Ciego de Ávila
Jacksonville is located in United States with (30.3322,-81.6557) coordinates and Ciego de Avila is located in Cuba with (21.84,-78.7619) coordinates. The calculated flying distance from Jacksonville to Ciego de Avila is equal to 614 miles which is equal to 987 km.
| City/Place | Latitude and Longitude | GPS Coordinates |
|---|---|---|
| Jacksonville | 30.3322, -81.6557 | 30° 19´ 55.8480'' N 81° 39´ 20.3400'' W |
| Ciego de Avila | 21.84, -78.7619 | 21° 50´ 24.0000'' N 78° 45´ 42.9840'' W |
Jacksonville, Tallahassee, United States
Related Distances from Jacksonville
| Cities | Distance |
|---|---|
| Jacksonville to Homestead | 613 km |
| Jacksonville to Deltona | 184 km |
| Jacksonville to Gainesville | 115 km |
| Jacksonville to Ives Estates | 541 km |
| Jacksonville to Boynton Beach | 475 km |
Ciego de Avila, Cuba
Related Distances to Ciego de Avila
| Cities | Distance |
|---|---|
| Camaguey to Ciego De Avila | 109 km |
| Bayamo to Ciego De Avila | 315 km |
| Artemisa to Ciego De Avila | 481 km |