Seymour Hill
Seymour Hill is a peak in Ferry, Washington and has an elevation of 3,330 feet.- Type: Peak with an elevation of 3,330 feet
- Category: landform
- Location: Ferry, Washington, Pacific Northwest, United States, North America
- View on OpenStreetMap
Latitude
48.15739° or 48° 9′ 27″ north
Longitude
-118.34722° or 118° 20′ 50″ west
Elevation
3,330 feet (1,015 metres)
Open location code
85W35M43+X4
OpenStreetMap ID
OpenStreetMap feature
natural=peak
GeoNames ID
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Satellite Map
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