FERESTRĂU-OITUZ

Bacău County · ROMANIA (RO) · 46.20°N / 26.58°E

ADM1 · ID GeoNames 678,337

Ferestrău-Oituz is a city in Romania with a population of 1,036. It sits at 46.20/26.58, runs on Europe/Bucharest, is 3 km from Oituz.

Population 1,036

Place Type Populated Place

Elevation n/a

Timezone Europe/Bucharest

All facts

fields

Country: Romania
Subdivision: Bacău County
Coords: 46.20°N / 26.58°E
UTC offset: +2
DST: Observed

Proximity

nearest 10 · within 50 km

E Oituz Bacău County 9,596 3.1 km

E Marginea Bacău County 684 4.9 km

N Cireșoaia Bacău County 1,811 5.5 km

NW Cerdac Bacău County 1,571 6.1 km

NE Bahna Bacău County 594 6.3 km

NE Nicorești Bacău County 902 6.7 km

NE Satu Nou Bacău County 1,699 7.6 km

NE Pârgăreşti Bacău County 4,798 7.8 km

W Slănic-Moldova Bacău County 4,955 8.2 km

E Bogdăneşti Bacău County 2,740 8.4 km

Air access

nearby

LRCC

Oituz (PA&CO) Heliport

Heliport

2.3 km

BCM

Bacău George Enescu International Airport

Large airport

43.8 km

FAQ

3 questions

What is the population of Ferestrău-Oituz?: 1,036 people, per GeoNames.
Which timezone is Ferestrău-Oituz in?: Europe/Bucharest.
Which country is Ferestrău-Oituz in?: Romania (RO).

Get this page as JSON

no auth

cURL 200 OK

$ curl https://api.cityapi.org/v1/cities/678337

{
  "data": {
    "id": 678337,
    "name": "Ferestrău-Oituz",
    "ascii_name": "Ferestrau-Oituz",
    "country_code": "RO",
    "population": 1036,
    "elevation": null,
    "timezone_id": "Europe/Bucharest",
    "feature_class": "P",
    "feature_code": "PPL",
    "latitude": 46.20249,
    "longitude": 26.57629
  }
}