Distance from Lexington Park to San Giorgio di Piano
The shortest distance (air line) between Lexington Park and San Giorgio di Piano is 4,342.59 mi (6,988.72 km).
How far is Lexington Park from San Giorgio di Piano
Lexington Park is located in Maryland, United States within 38° 15' 10.08" N -77° 33' 27.36" W (38.2528, -76.4424) coordinates. The local time in Lexington Park is 23:16 (04.03.2025)
San Giorgio di Piano is located in Bologna, Italy within 44° 38' 60" N 11° 22' 59.88" E (44.6500, 11.3833) coordinates. The local time in San Giorgio di Piano is 05:16 (05.03.2025)
The calculated flying distance from San Giorgio di Piano to San Giorgio di Piano is 4,342.59 miles which is equal to 6,988.72 km.
Lexington Park, Maryland, United States
Related Distances from Lexington Park
San Giorgio di Piano, Bologna, Italy