Seymour Hill
Seymour Hill is a peak in Ferry, Washington and has an elevation of 3,330 feet.| Tap on a place to explore it |
- 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″ northLongitude
-118.34722° or 118° 20′ 50″ westElevation
3,330 feet (1,015 metres)Open location code
85W35M43+X4OpenStreetMap ID
node 356541233OpenStreetMap feature
natural=peakGeoNames ID
5810021
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Satellite Map
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About Mapcarta. Data © OpenStreetMap contributors and available under the Open Database License". Text is available under the CC BY-SA 4.0 license, except for photos, directions, and the map. Photo: Lumpytrout, CC BY-SA 3.0.