Seymour elevation

Seymour (Indiana, Jackson County), United States elevation is 185 meters and Seymour elevation in feet is 607 ft above sea level ^{[src 1]}. Seymour is a populated place (feature code) with elevation that is 94 meters (308 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 162 to 201 m (531 to 659 ft) with average elevation of 176.7 meters (=580 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]}