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International cycle routes don’t show in OSM cycling map

3

Hi. I have noticed several EuroVelo cycle routes missing in the standard OSM cycling map (for example, EV15 - Rhine Route). However, the relations do exist in the OSM database (EV15: https://www.openstreetmap.org/relation/2171555), but they seem only to be displayed on third-party maps such as https://cycling.waymarkedtrails.org/. I am new to the mechanics behind OSM, and have searched the wiki, the bulletin board, and this site to no avail, so please bear with my question: How is the set of relations displayed on the default cycling map chosen, and what can I do to incorporate the missing relations?

asked 10 Jun '18, 20:23

djefferies's gravatar image

djefferies
56114
accept rate: 0%


2 Answers:

3

There is no standard OSM cycling map, what you see as the "CycleMap layer" is just as third party as for example https://cycling.waymarkedtrails.org/

Or put differently you need to talk to the creator/operator of the layer in question for questions and support https://www.thunderforest.com/

answered 10 Jun '18, 23:13

SimonPoole's gravatar image

SimonPoole ♦
44.7k13326701
accept rate: 18%

1

International cycle routes are also displayed on https://cycle.travel/map (disclaimer - my site!).

(12 Jun '18, 10:22) Richard ♦

1

The issue with the default cycling map (the one you see as an option on OSM.org) not rendering cycling EuroVelo 15 may be that it is a "Super-Relation", or a relation of relations. I'm not sure why, but many renderers do not handle super-relations well. I suspect that is the issue here.

As already stated talking to the individual who renders the tiles for the default cycling layer would be more definitive.

I don't think there is a way around this issue, other than to hope that the renderer you are using begins to support super-relations.

answered 11 Jun '18, 07:52

keithonearth's gravatar image

keithonearth
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accept rate: 13%

OpenCycleMap doesn't support international cycling relations at all AIUI, it's not a super-relation issue.

(12 Jun '18, 10:22) Richard ♦