Converting Calculated headways to signal spaces.

[KenD]

i would like to know how you convert calulated headways to required signal types.

eg, A line has a civil line speed of 160km/h However the trains that use the line are only able to do a max of 120km/h.

Operationally thru trains require a headway of 3mins and stopping trains 6mins.

This working headway would be 125% of the calulated headway

I used the standard braking distance of 1189m, overlap 183m, 10sec for sighting. assume gradient of level.

For 4 aspect signalling I calulated a headway of 75 sec

For 3 aspect signalling 93 sec.

Would graphing help?

[PJW]

eg, A line has a civil line speed of 160km/h
I used the standard braking distance of 1189m,

What do you mean by "standard braking distance"? The IRSE exam generally gives the braking rate of 0.5m/s/s; 1189m seems a little short for the maximum permissible line speed so it looks like you are assuming higher brake rate as I make it 1998m (so round to 2000m).

Whatever the value is, this is the closest spacing of 3 aspect signals and therefore lowest possible headway distance is then derived by:
double signal spacing + sighting distance + train length + overlap length. Convert this to headway time by DIVIDING [thanks Ken!] by the timetabled speed of the train (i.e. could be the maximum train speed but may be less). Then, as you have said, you need to ensure that there is an appropriate % contingency to arrive at the timetabled headway.

Obviously other constraints dictate where signals can be placed so that in reality won't be able to put every signal at true minimum spacing, so unless your desired headway can still be achieved with signals spaced at 115% braking then it's not really a tenable solution and thus 4 aspects are needed.

Similar calculation for 4 aspects except instead of the braking from the Green aspect to the Red occuring over 2 signal sections, it occurs over 3 sections; hence multiply the spacing by 3 rather than 2, BUT don't forget that in the 4 aspect case the minimum signal spacing is 0.5 braking distance!

You need to do the sums for stopping headways (and that is where a graph can be helpful), but looking at the figures it seems that the non-stop is probably the more onerous. The reason that I say this: 120km/h = 75 mph = approx 34 m/s and thus will take just about 68s to stop at the usual brake rate and similar to accelerate, so compared to the non-stop train lose say 80sec (perhaps signals not a minimum spacing). If dwell time only 30s then total loss = 120s but there is 180s extra; seems highly likely that the signalling that gives 3 min non-stop headway can also give 6 min stopping.

[KenD]

Whatever the value is, this is the closest spacing of 3 aspect signals and therefore lowest possible headway distance is then derived by:
double signal spacing + sighting distance + train length + overlap length. Convert this to headway time by multiplying by the timetabled speed of the train (i.e. could be the maximum train speed but may be less). Then, as you have said, you need to ensure that there is an appropriate % contingency to arrive at the timetabled headway.

Shouldn't the value be divided by the speed of the train?

[KenD]

Also I can calculate Braking times and distances to halt but how would you calculate the time and distance required for a train to brake from say 100km/h to 40km/h for a turnout ?

[PJW]

Whatever the value is, this is the closest spacing of 3 aspect signals and therefore lowest possible headway distance is then derived by:
double signal spacing + sighting distance + train length + overlap length. Convert this to headway time by multiplying by the timetabled speed of the train (i.e. could be the maximum train speed but may be less). Then, as you have said, you need to ensure that there is an appropriate % contingency to arrive at the timetabled headway.

Shouldn't the value be divided by the speed of the train?

Oops- I changed the way this sentence worded and failed to re-read through. Yes headway distance (m) obviously must be DIVIDED by speed (m/s) in order to get a time (s)!

[PJW]

Also I can calculate Braking times and distances to halt but how would you calculate the time and distance required for a train to brake from say 100km/h to 40km/h for a turnout ?

If you can calculate braking times and distances to a halt from 100km/h and also from 40km/h, then the times you require are the difference between those values. OK this may be a bit of a long winded way to get the answer, but when you think about it, it must work. This of course assumes that there is no "brake build up delay"; in the real world things aren't quite so simple as there is an elapsed time when little or no braking occurs following the driver's braking request.

Actually for constant braking rate thoughout the speed range, you can calculate the time taken to decelerate from initial speed "u" to final speed "v" by making use of the equation v=u+at; t=(v-u)/a (with the acceleration "a" of course being negative).

One of the very useful things of the IRSE so often using a=-0.5m/s/s is that it takes 2 seconds to reduce speed by 1 m/s; Hence if the initial speed is say 25 m/s then to reduce speed to 15m/s is a reduction of 10m/s which will take 20s. The average speed during the braking period is 20m/s and hence the distance travelled is 20 m/s x 20 s = 400m; hence you can often do in your head.