Lake Alfred elevation

Lake Alfred (Florida, Polk County), United States elevation is 57 meters and Lake Alfred elevation in feet is 187 ft above sea level ^{[src 1]}. Lake Alfred is a populated place (feature code) with elevation that is 222 meters (728 ft) smaller than average city elevation in United States.

Below is the Elevation map of Lake Alfred, which displays elevation range with different colors. Scale of the first map is from 26 to 64 m (85 to 210 ft) with average elevation of 43.9 meters (=144 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 Lake Alfred elevation map:^{[src 2-3]}

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

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