I will be flying to Birmingham in a few days. I will go somewhere else near Banbury for a few days and then return to the city.

If I buy an off-peak return ticket BHI-BAN-BHI or BHM-BAN-BHM for that journey will I be able to leave in BHM resp. get onto the platform in BHI? And wil that be a legitimate way of travelling? Or do I need to buy two separate tickets for BHI-BAN and BAN-BHM?


It depends on which train ticket you buy.

The Off-Peak Saver Return on Birmingham Stations to Banbury permits a break-of-journey on the outbound portion, so you could join the train at Birmingham Airport.

I suspect the cheaper off-peak saver return ticket on Birmingham Airport to Banbury would not allow you to go so far as Birmingham Stations, but I haven't checked in the routeing guide. If it did allow this, it would be a cheaper alternative.

  • By the fact that there is no separate Off-Peak Non-Saver Return listed on brfares, I assume that what National Rail lists as “Off-Peak Return” (Crosscountry) is an SVR, is that right?
    – Anaphory
    Mar 13 '16 at 20:56
  • If I used it correctly, a Routeing point calculator tells me 'Travel was NOT allowed at NFM64: SDS from BAN to BHM on route 00000: £8.20 is more expensive than SDS from BAN to BHI on route 00211: £7.30.', so no?
    – Anaphory
    Mar 13 '16 at 21:03

National Rail Conditions of Carriage (PDF)


16. Starting, breaking or ending a journey at intermediate stations You may start, or break and resume, a journey (in either direction in the case of a return ticket) at any intermediate station, as long as the ticket you hold is valid for the trains you want to use. You may also end your journey (in either direction in the case of a return ticket) before the destination shown on the ticket

  • 2
    This only applies if the ticket conditions don't say otherwise. Some OFFPEAK R tickets do not allow it on the outbound journey.
    – Calchas
    Mar 13 '16 at 20:19
  • @ Calchas Yes, you have to go from BHI to BHM first, then go to BAN.
    – Him
    Mar 13 '16 at 20:28

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.