Seymour elevation

Seymour (Wisconsin, Outagamie County), United States elevation is 243 meters and Seymour elevation in feet is 797 ft above sea level ^{[src 1]}. Seymour is a populated place (feature code) with elevation that is 36 meters (118 ft) smaller than average city elevation in United States.

Below is the Elevation map of Seymour, which displays elevation range with different colors. Scale of the first map is from 231 to 279 m (758 to 915 ft) with average elevation of 249.3 meters (=818 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 Seymour elevation map:^{[src 2-3]}

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