Seymour elevation

Seymour (Tennessee, Sevier County), United States elevation is 289 meters and Seymour elevation in feet is 948 ft above sea level ^{[src 1]}. Seymour is a populated place (feature code) with elevation that is 10 meters (33 ft) bigger 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 270 to 461 m (886 to 1512 ft) with average elevation of 324.1 meters (=1063 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]}

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

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