OpenStreetMap
I recently added “area roads” to downtown Durham, NC. I made the surface of the roads visible on Openstreetmap. I was wondering if their is any specific way to map road areas. I currently map them as “area = highway” and “highway = service”. Feel free to check it out!
Downtown Durham osm.org/#map=19/35.99361/-78.89968
Downtown Chapel Hill osm.org/#map=19/35.91627/-79.05645
Chapel Hill Public Library osm.org/#map=19/35.93251/-79.03549
All three links lead to a place in Openstreetmap where I mapped roads as areas.
I anyone else mapped roads as areas, please reply a link to the location!
Дискусија
Коментар корисника Hjart у 21 Септембар 2017 у 19:46
In Denmark (the country in north Europe) I know of one case of roads mapped as areas: The area around osm.org/way/441584563.
This doesn’t show on our standard layer, but imho has a number of advantages.
See more at osm.wiki/Proposed_features/area:highway
Коментар корисника Jual Knalpot у 21 Септембар 2017 у 20:19
aaaaaaaaaaaaaaa mantapppppppppppppp
Коментар корисника maxerickson у 21 Септембар 2017 у 21:02
Something may have gone wrong, several streets are mapped as part of a highway=service area:
osm.org/relation/7546630#map=19/35.93045/-79.03707
Коментар корисника maxerickson у 21 Септембар 2017 у 21:03
Oh I see, I didn’t read closely enough. I think if you are going to make areas, they should match the type of the road they cover, not just be universally service.
Коментар корисника Nakaner у 21 Септембар 2017 у 21:30
Hi BjornRassmussen,
adding
highway=serviceto an area a secondary roads runs through as you did two days ago is tagging for the renderer and discouraged. In addition, applications which do other things based on OSM data than rendering maps, will treat the outer ring of this polygon as a circular service road along the outer boundary of the road area! Please don’t do it. I therefore reverted this change.If the road is
highway=secondary, its area has to havearea:highway=secondary. If the roads ishighway=tertiary, its area has to havearea:highway=tertiary.Best regards
Michael