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      Distance from Dent to San Pancrazio Salentino

      The shortest distance (air line) between Dent and San Pancrazio Salentino is 5,086.48 mi (8,185.90 km).

      How far is Dent from San Pancrazio Salentino

      Dent is located in Ohio, United States within 39° 11' 29.4" N -85° 20' 23.64" W (39.1915, -84.6601) coordinates. The local time in Dent is 11:23 (31.08.2025)

      San Pancrazio Salentino is located in Brindisi, Italy within 40° 25' 0.12" N 17° 49' 59.88" E (40.4167, 17.8333) coordinates. The local time in San Pancrazio Salentino is 17:23 (31.08.2025)

      The calculated flying distance from San Pancrazio Salentino to San Pancrazio Salentino is 5,086.48 miles which is equal to 8,185.90 km.

      Dent, Ohio, United States

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      Dent to Baltimore (Maryland)430.43 mi (692.71 km)
      Dent to Philadelphia (Pennsylvania)510.03 mi (820.81 km)
      Dent to Brooklyn (New York)576.04 mi (927.05 km)
      Dent to New York (New York)577.69 mi (929.70 km)
      Dent to Queens (New York)584.77 mi (941.10 km)
      Dent to Minneapolis (Minnesota)594.29 mi (956.41 km)

      San Pancrazio Salentino, Brindisi, Italy

      Related Distances to San Pancrazio Salentino