Distance from Jefferson Valley-Yorktown to Seguin
The shortest distance (air line) between Jefferson Valley-Yorktown and Seguin is 1,575.15 mi (2,534.96 km).
How far is Jefferson Valley-Yorktown from Seguin
Jefferson Valley-Yorktown is located in New York, United States within 41° 19' 4.8" N -74° 11' 57.12" W (41.3180, -73.8008) coordinates. The local time in Jefferson Valley-Yorktown is 08:10 (05.01.2026)
Seguin is located in Texas, United States within 29° 35' 20.76" N -98° 1' 56.64" W (29.5891, -97.9676) coordinates. The local time in Seguin is 07:10 (05.01.2026)
The calculated flying distance from Seguin to Seguin is 1,575.15 miles which is equal to 2,534.96 km.
Jefferson Valley-Yorktown, New York, United States