Jefferson elevation

Jefferson (Ohio, Ashtabula County), United States elevation is 294 meters and Jefferson elevation in feet is 965 ft above sea level ^{[src 1]}. Jefferson is a seat of a second-order administrative division (feature code) with elevation that is 15 meters (49 ft) bigger than average city elevation in United States.

Below is the Elevation map of Jefferson, which displays elevation range with different colors. Scale of the first map is from 252 to 319 m (827 to 1047 ft) with average elevation of 281.6 meters (=924 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 Jefferson elevation map:^{[src 2-3]}

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