9

I will need to travel on an OEBB Railjet train, between Vienna and Graz, a service that runs once per hour. As I will be able to travel at any point during the day, I wonder if there is any way to get any data indicative of which trains (throughout a given day of the week) are normally the least busy, based on past travel statistics aggregated through various means.

Google Maps sometimes shows how busy a given tram/bus is - but that information is only given live as opposed to aggregated by time of the day and day of the week (as is the case with its Popular Times graphs for businesses); it's also only available for local public transport rather than for trains. OEBB's own Scotty app provides lots of information about train connections, but there too I've not been able to find this particular information regarding the crowding levels of trains.

  • There is a train before 6 o clock which is likely to be not busy – guest Oct 20 '20 at 15:42
10

You can get a rough estimate by looking at the ticket prices a week or two in advance. The departures with the cheapest tickets will likely be least crowded. If I for example check the coming Monday (October 26th), there are for most departures saver fares available ranging from €24.90 to €39.90. For one departure, no saver fares are available, but you need a full price ticket for €41.00.

The departures with tickets for €24.90 at 5:58, 20:58 and 21:58 will likely have the least number of passengers. The departures with only full price tickets for €41.00 at 18:58 will likely have the most.

BTW, the service runs every hour and not only every 2 hours.

  • Thanks! That can indeed be a useful rough metric to go by, but in the case of the Vienna-Graz link, it seems to too rough to be helpful: for all of tomorrow's services for instance, I get the same price (EUR 20.50, with a Vorteilscard), and the same for the day after tomorrow. Surely they won't all be at 100% capacity. It would be nice if any metric were available based on actual passenger numbers. I have no doubt OEBB themselves will not happily give those away, but perhaps some amateur/third-party websites that somehow manage to infer these data?! – z8080 Oct 20 '20 at 19:47
  • 4
    @z8080 That is simply because the 50% Vorteilscard tariff is cheaper than all the saver fares. If you choose to see the price without a Vorteilscard, the concept works on short notice, ie for tomorrow, as well. – Tor-Einar Jarnbjo Oct 20 '20 at 22:34
  • Ah yes, silly of me not to have realised :) Thanks again! – z8080 Oct 21 '20 at 8:09
9

ÖBB Auslastungsanzeige https://live.oebb.at is intended for this, but not all trains are covered.

This is what the crowding display looks like, if available:

crowding information example

  • Wow, what a great tool, shame it's not been advertised more. It's clearly in the early stages, and hopefully more features and trains will be added (currently not even all RJs are included). Funnily enough, the data shown seems to be mainly from past trains with limited info about currently-running trains, despite the site's name "live" – z8080 Oct 21 '20 at 8:39
  • 1
    This really looks suspiciously wrong. E.g. this train today has identical numbers on both halves of the train - varying from empty to full in adjacent cars. live.oebb.at/… – asdfex Oct 21 '20 at 16:13
  • 1
    I must admit that this sounds like a much better answer to the question than mine, but I must agree with asdfex that the results seem wrong. Even in the posted screenshot, it is not comprehensible why car 24 should be almost full when car 23 is almost empty. That is not how normal train passengers 'work' when they are looking for a free seat. It is also worth to notice that the service obviously only shows the estimated occupancy at the mentioned station (here Wien Hbf) and not during the entire trip. – Tor-Einar Jarnbjo Oct 21 '20 at 18:32
  • 1
    My guess is that the crowding display is based on seat reservations. Car 24 with the red symbol is where all the prebooked seats for Airail passengers are, see vagonweb.cz/razeni/… – user108733 Oct 22 '20 at 19:21
  • 1
    @user108733 That would make sense - but the crowded car shifts from day to day. If you check 18.10. - 21.10., it's either 22,23 or 24 and the same one in 32,33,34. – asdfex Oct 23 '20 at 8:19
0

So far, for what it's worth, my own preferred method is to look at the seat reservation schemes for a given train, and infer that it's representative of the day of the week and time. For instance, if I want to compare the business level of Sunday trains, I will look at these seat schemes for several trains on the upcoming Sunday. To compare across days of the week, one needs of course to look with an equal amount of days in advance for each 'candidate'.

  • I doubt that there is a strong correlation between the number of reservations and actual occupancy. Some departures are definitely more used for spontaneous travel without a reservation long time in advance than other departures. – Tor-Einar Jarnbjo Oct 23 '20 at 21:30
  • Right, but this noise factor should be equal across those candidates (across all trains in a day for instance), thus enabling this to be a potentially useful metric in choosing among them, no? – z8080 Oct 24 '20 at 9:41
  • No; I don't have any numbers to back my belief, but I assume that trains in the morning and early evening on weekdays are much more used by business travellers who either travel without a reservation or book on very short notice, while mid-day and late evening trains on weekdays and all trains on weekends are more used for leisure travel and therefore by people who plan and book much longer in advance. – Tor-Einar Jarnbjo Oct 24 '20 at 17:35
  • I see what you mean, that makes sense. As it stands, I suppose a combination of all three of the above presented methods should be able to at least roughly triangulate on the desired data :) – z8080 Oct 24 '20 at 18:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.