Distance from Simpelveld to Spanaway
The shortest distance (air line) between Simpelveld and Spanaway is 5,008.74 mi (8,060.80 km).
How far is Simpelveld from Spanaway
Simpelveld is located in Zuid-Limburg, Netherlands within 50° 49' 59.88" N 5° 58' 59.88" E (50.8333, 5.9833) coordinates. The local time in Simpelveld is 11:23 (26.12.2025)
Spanaway is located in Washington, United States within 47° 5' 52.44" N -123° 34' 36.12" W (47.0979, -122.4233) coordinates. The local time in Spanaway is 02:23 (26.12.2025)
The calculated flying distance from Spanaway to Spanaway is 5,008.74 miles which is equal to 8,060.80 km.
Simpelveld, Zuid-Limburg, Netherlands
Related Distances from Simpelveld
Spanaway, Washington, United States