Ferestrău-Oituz elevation

Ferestrău-Oituz (Bacău, Comuna Oituz), Romania elevation is 308 meters and Ferestrău-Oituz elevation in feet is 1010 ft above sea level ^{[src 1]}. Ferestrău-Oituz is a populated place (feature code) with elevation that is 56 meters (184 ft) bigger than average city elevation in Romania.

Below is the Elevation map of Ferestrău-Oituz, which displays elevation range with different colors. Scale of the first map is from 271 to 776 m (889 to 2546 ft) with average elevation of 505.9 meters (=1660 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 Ferestrău-Oituz elevation map:^{[src 2-3]}

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