Selinsgrove elevation

Selinsgrove (Pennsylvania, Snyder County), United States elevation is 139 meters and Selinsgrove elevation in feet is 456 ft above sea level ^{[src 1]}. Selinsgrove is a populated place (feature code) with elevation that is 140 meters (459 ft) smaller than average city elevation in United States.

Below is the Elevation map of Selinsgrove, which displays elevation range with different colors. Scale of the first map is from 119 to 300 m (390 to 984 ft) with average elevation of 163.4 meters (=536 ft) ^{[note 1]}
These maps also provides idea of topography and contour of this city, they are displayed at different zoom levels. More info about maps, scale and edge coordinates you can find below images.

#1 Selinsgrove elevation map:^{[src 2-3]}

#2 Zoom out 2x:^{[src 2-3]}

Selinsgrove Topographic map:^{[src 2-3]}