to think this is a completely unreasonable bit of train fare policy?

[BertrandRussell]

I think this might need a diagram. But I’ll try without. We live near station T. Ds had work today near station H. H-T either direct or changing at F is £7. However, every second train means changing at S- and the fare is £13! It is a slightly longer journey- but It’s still to the same place. Is that ridiculous- or am I being unreasonable?

If the route via S is more popular - then it makes sense for it to cost more. Annoying, but not crazy.

I can take two routes to the airport - one of via London and the other is longer but not via London, London route more expensive because more people want to be on the trains in and out of London - so the train company can charge more.

Train pricing is completely unreasonable and illogical and infuriating.
I would use trains a lot more if they sorted this out.

Do different train companies operate the two routes?

Oh I think I can decode this.

Changing at S means going back down the line to S from the starting point, and then travelling back through T to your destination, right?

It’s what you get for having your local station not much more than a halt.

I used to live not far from there (if I’m right about where you are) and you get these quirks. Personally, I’d just time myself to catch the non-doubling-back route.

(Double checking it on google maps it seems to think doubling back all the way to R is the thing to do, which is odd, and even more irritating.)

There are many journeys where this can occur - and they don't necessarily involve doubling back. The £13 fare will be "any permitted" route and the £7 fare will be "not via S" or "via F". An example I am familiar with is Leeds to Sheffield where the "any permitted" single is £17.40 and the "not Doncaster" fare is £11.60. They even charge £17.40 on the few through trains from Leeds to Sheffield that stop at Doncaster - you don't have to change trains!