Distance from Jacksonville to San Fernando de Monte Cristi
Distance between Jacksonville and San Fernando de Monte Cristi is 1539 kilometers (957 miles).
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Distance Map Between Jacksonville and San Fernando de Monte Cristi
Jacksonville, Tallahassee, United States ↔ San Fernando de Monte Cristi, Dominican Republic = 957 miles = 1539 km.
How far is it between Jacksonville and San Fernando de Monte Cristi
Jacksonville is located in United States with (30.3322,-81.6557) coordinates and San Fernando de Monte Cristi is located in Dominican Republic with (19.8483,-71.646) coordinates. The calculated flying distance from Jacksonville to San Fernando de Monte Cristi is equal to 957 miles which is equal to 1539 km.
| City/Place | Latitude and Longitude | GPS Coordinates |
|---|---|---|
| Jacksonville | 30.3322, -81.6557 | 30° 19´ 55.8480'' N 81° 39´ 20.3400'' W |
| San Fernando de Monte Cristi | 19.8483, -71.646 | 19° 50´ 53.7360'' N 71° 38´ 45.4920'' W |
Jacksonville, Tallahassee, United States
Related Distances from Jacksonville
| Cities | Distance |
|---|---|
| Jacksonville to Las Vegas | 3552 km |
| Jacksonville to East Lake | 330 km |
| Jacksonville to Clearwater | 356 km |
| Jacksonville to Boynton Beach | 475 km |
| Jacksonville to Boston | 1840 km |
San Fernando de Monte Cristi, Dominican Republic