With the journey planner API. Some routes can have a duration value of 0. For example some stations which are directly connected.

Is there a way of getting a duration value which includes seconds?

With the journey planner API. Some routes can have a duration value of 0. For example some stations which are directly connected.

Is there a way of getting a duration value which includes seconds?

Hello, thank you for your query.

Unfortunately, Journey Planner does not return durations more granular than whole minutes.

Very unfortunate. Thanks

As I noted at the foot of my post at

the bus timetables in the Datastore Journey Planner zip file used to show times between stops in seconds but that seemed to stop, at least for new files, a few weeks after the system changed to the blessed Adiona. It was spurious accuracy of course but it did seem to recognise that some timings between stop pairs were much greater than others.

I asked then why this change (along with some others) had been made. Nobody has replied but I suppose it is not important in the scheme of things; my post was for interest as much as anything else.

I seem to recall that tube working timetables were traditionally worked in fractions of a minute, in which case it would be odd if it is still the practice) if the Datastore tube files did not use that. Times between tube stops are likely to vary much less than times between bus stops.

@Ja

Having written my own Journey Planner software (“Aubin”) from first principles, the station interchanges time can be computed in a number of ways, but the basic way is to sum up

- The walking time from the walking speed (of 1.68 metres per second) multiplied by the distance. Go faster: how London got hooked on speed | London Evening Standard | Evening Standard.
- Half the service interval, because if you arrive at a “platform” at a random time then this is average time you will wait using Integral maths. Integral - Wikipedia. If there are 4 trains per hour, the average wait is 7.5 minute.
- Note same-platform (such as Baker Street to Liverpool Street, Uxbridge to Rayners Lane, Queens Park to H&W) and cross-platform interchanges (Mile End, Finsbury Park, and many Victoria Line stations - Template:Victoria Line RDT - Wikipedia )
- If possible take account of slower speeds for stair climbs and people in crowds move more slowly especially when it’s busy and/or crowded.

Note that TfL has researched walking distances but not speeds

Thanks for the response, I won’t delve into it because it isnt important to me really. However, I’m curious what the “integral” maths being used is. Since saying just integral maths is a bit vague.

I did the maths about 15 years ago, so I don’t have a note of it.

But basically, if you arrive at a platform at any time then the next train is between 0 and 3600 ÷ tph.

If you sum up these calculations and average them, you get **1800÷ tph** seconds as an athematic mean.

The proper way to express this is with an integral calculation.

Why do you need integral maths? How can it be anything other than half the headway if the headway is absolutely consistent?

It gets more complicated if you factor in bunching. I vaguely remember reading some years ago that the average waiting time on a high frequency bus service was actually the headway itself. I wonder whether the tech available to help manage the service has reduced the average waiting time to less than the headway?

Because it the way to solve this kind of thing? I’ve got lots of maths qualifications and I feel it necessary to refer to them from time to time… My six-form class was four of us double triple-maths and to this day I have a love of mental maths and huge dislike for the one-trick-pony that is mechanics.

@briantist

It may well be the way to approach a more generalised problem but if a special case can be dealt with more simply, that’s the way to go. In this case, uniform distribution of arrival/waiting times, so symmetry. For every passenger arrival X minutes above half the headway there is another one X minutes below, so they cancel out. That’s true for all X so as all the variations from half the headway cancel out you are left with half the headway.

I can remember a uni exam question on a Markov process which I started to answer in the standard way, finished the first page of workings then looked at the question again and realised it was such a special case that it could be done in a few lines. Crossed out the page, wrote “I must be an idiot” and knocked it off!

Erm, OK. I’m a mathematician so I do things the hardest way possible…