Distance from Jacksonville to Sibanicu
Distance between Jacksonville and Sibanicu is 1092 kilometers (679 miles).
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679 miles
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Distance Map Between Jacksonville and Sibanicu
Jacksonville, Tallahassee, United States ↔ Sibanicu, Camaguey, Cuba = 679 miles = 1092 km.
How far is it between Jacksonville and Sibanicú
Jacksonville is located in United States with (30.3322,-81.6557) coordinates and Sibanicu is located in Cuba with (21.235,-77.5264) coordinates. The calculated flying distance from Jacksonville to Sibanicu is equal to 679 miles which is equal to 1092 km.
| City/Place | Latitude and Longitude | GPS Coordinates |
|---|---|---|
| Jacksonville | 30.3322, -81.6557 | 30° 19´ 55.8480'' N 81° 39´ 20.3400'' W |
| Sibanicu | 21.235, -77.5264 | 21° 14´ 6.0000'' N 77° 31´ 35.0040'' W |
Jacksonville, Tallahassee, United States
Related Distances from Jacksonville
| Cities | Distance |
|---|---|
| Jacksonville to Boca Raton | 496 km |
| Jacksonville to Boynton Beach | 475 km |
| Jacksonville to Buenaventura Lakes | 262 km |
| Jacksonville to Fruit Cove | 40 km |
| Jacksonville to Doral | 563 km |
Sibanicu, Camaguey, Cuba
Related Distances to Sibanicu
| Cities | Distance |
|---|---|
| Esmeralda to Sibanicu | 121 km |
| Nuevitas to Sibanicu | 57 km |
| Guaimaro to Sibanicu | 30 km |
| Jimaguayu to Sibanicu | 39 km |
| Florida to Sibanicu | 90 km |