Distance from Seymour to San Cesareo
The shortest distance (air line) between Seymour and San Cesareo is 4,880.59 mi (7,854.56 km).
How far is Seymour from San Cesareo
Seymour is located in Indiana, United States within 38° 56' 51.36" N -86° 6' 32.04" W (38.9476, -85.8911) coordinates. The local time in Seymour is 09:14 (19.03.2025)
San Cesareo is located in Roma, Italy within 41° 49' 0.12" N 12° 48' 0" E (41.8167, 12.8000) coordinates. The local time in San Cesareo is 15:14 (19.03.2025)
The calculated flying distance from San Cesareo to San Cesareo is 4,880.59 miles which is equal to 7,854.56 km.
Seymour, Indiana, United States