Distance from Painesville to Kalmthout
The shortest distance (air line) between Painesville and Kalmthout is 3,890.28 mi (6,260.80 km).
How far is Painesville from Kalmthout
Painesville is located in Ohio, United States within 41° 43' 26.4" N -82° 44' 47.04" W (41.7240, -81.2536) coordinates. The local time in Painesville is 01:46 (14.06.2025)
Kalmthout is located in Arr. Antwerpen, Belgium within 51° 22' 59.88" N 4° 28' 0.12" E (51.3833, 4.4667) coordinates. The local time in Kalmthout is 07:46 (14.06.2025)
The calculated flying distance from Kalmthout to Kalmthout is 3,890.28 miles which is equal to 6,260.80 km.
Painesville, Ohio, United States
Related Distances from Painesville
Kalmthout, Arr. Antwerpen, Belgium