Distance from Arganzuela to Saint-Petersburg
Driving route: -- ( - )
Driving route: -- ( - )
Extended route information
The flight distance between the nearest airports Arganzuela (MAD) and Saint-Petersburg (LED) is 3,187.16 km. This corresponds to an approximate flight time of 4h 14min. Similar flight routes: MAD → HEL , MAD → TLL , MAD → RIX , MAD → SVO , ALC → LED
Bearing: 47.56° (NE)The initial bearing on the course from Arganzuela to Saint-Petersburg is 47.56° and the compass direction is NE.
Midpoint: 51.36867,9.69712The geographic midpoint between Arganzuela and Saint-Petersburg is in 1,595.60 km distance between both points in a bearing of 47.56°. It is located in Germany, Lower Saxony, Landkreis Göttingen, Hann. Münden (Deutschland, Niedersachsen, Landkreis Göttingen, Hann. Münden).
Distance: 3,191.20 kmThe shortest distance (air line) between Arganzuela and Saint-Petersburg is 3,191.20 km.
Driving route: -- ( - )The shortest route between Arganzuela and Saint-Petersburg is according to the route planner. The driving time is approx. . Half of the trip is reached in .
Time difference: 1hThe time difference between Arganzuela (Europe/Madrid) and Saint-Petersburg (Europe/Moscow) is 1 hours. This means that it is now 14:00 (12.05.2024) in Arganzuela and 15:00 (12.05.2024) in Saint-Petersburg.
How is the distance calculated?
To calculate the distance between Arganzuela and Saint-Petersburg, the place names are converted into coordinates (latitude and longitude). The respective geographic centre is used for cities, regions and countries. To calculate the distance the Haversine formula is applied.
Similar routes:
Similar distance (± 0.5%)
Saint-Petersburg is just as far away from Arganzuela as Arganzuela from Cairo (3,350 km), Saint Petersburg (3,191 km), Alexandria (3,177 km), Ankara (3,084 km), Dakar (3,154 km), Giza (3,349 km), Kharkiv (3,262 km), Bamako (3,114 km), Ouagadougou (3,124 km), Dnipro (3,166 km).