Distance from San Angelo to Lyaskovets
The shortest distance (air line) between San Angelo and Lyaskovets is 6,262.75 mi (10,078.93 km).
How far is San Angelo from Lyaskovets
San Angelo is located in Texas, United States within 31° 26' 32.64" N -101° 32' 58.56" W (31.4424, -100.4504) coordinates. The local time in San Angelo is 02:21 (25.02.2025)
Lyaskovets is located in Veliko Tarnovo, Bulgaria within 43° 6' 0" N 25° 43' 0.12" E (43.1000, 25.7167) coordinates. The local time in Lyaskovets is 10:21 (25.02.2025)
The calculated flying distance from Lyaskovets to Lyaskovets is 6,262.75 miles which is equal to 10,078.93 km.
San Angelo, Texas, United States
Related Distances from San Angelo
Lyaskovets, Veliko Tarnovo, Bulgaria