Distance from Petersburg to San Juan de Aznalfarache
The shortest distance (air line) between Petersburg and San Juan de Aznalfarache is 3,821.49 mi (6,150.09 km).
How far is Petersburg from San Juan de Aznalfarache
Petersburg is located in Virginia, United States within 37° 12' 15.48" N -78° 36' 31.32" W (37.2043, -77.3913) coordinates. The local time in Petersburg is 20:59 (12.03.2025)
San Juan de Aznalfarache is located in Sevilla, Spain within 37° 22' 0.12" N -7° 58' 59.88" W (37.3667, -6.0167) coordinates. The local time in San Juan de Aznalfarache is 02:59 (13.03.2025)
The calculated flying distance from San Juan de Aznalfarache to San Juan de Aznalfarache is 3,821.49 miles which is equal to 6,150.09 km.
Petersburg, Virginia, United States
Related Distances from Petersburg
San Juan de Aznalfarache, Sevilla, Spain