Trip Time

Copyright ©2008 by Paul Niquette. All rights reserved.

Sophisticated solvers discovered in the Station-Stop puzzle that cruise speed vC does not play a part in determining minimum headway tH.  The Trip Time puzzle is a different matter.  We will need to ascertain the total time it takes to depart one station then travel to -- and depart from -- the next station.  Let's call that tS.   Now, if the distance between stations xS is large, as with long-haul or commuter rail service, tS will approach xS / vC.  Typical passenger transit systems in urban environments, called full metro service, require xS to be less than two miles, which brings into consideration parameters other than vC for calculating Trip Time.
Pictured in our Trip Time puzzle is a "Baby Bullet" train operated by Caltrain as an SRO commuter-rail service in San Francisco.   For this analysis, let us use New York City Transit trains to illustrate a typical full-metro service.
We turn again to the phase-plane plot to estimate the impact of a Station Stop -- this time on Trip Time not on headway.  The lead car of a train, shown in green, approaches a station at cruise speed vC.  The New York Transit system does not have automatic train control. Upon reaching Point 1, the operator sees a wayside sign that calls for slowing to a platform operating speed vP.   At Point 2, the train is stabilized at vP.
At Point 3, if the previous train is detected still berthed in the platform, the operator sees a red light and will issue a full-service braking command, stopping the train at Point 4b, outside the station by a buffer distance xB.

If the platform is not occupied, the train operator sees a green light, which allows continuation at vP into the platform all the way to Point 4a where the operator performs a feathered stop at Point 5.

Stopped in the station platform, a train waits out its dwell time tD, during which doors are opened for unloading and loading of passengers.  Typically 20 seconds in duration, tD must be added directly to the Trip Time.

We now shift our attention to Point 1', the location of the rear-most extreme of the same train, with its phase-plane depicted in red in the sketch above.  When tD has elapsed, the conductor closes the doors and the operator commands the train to depart the station.  The train accelerates at a constant rate aS to vP, the platform operating speed, at Point 2'.

As the tail car reaches the far end of the platform at Point 3', the train is released to continue accelerating at aS from vP to cruise speed vC, which occurs well clear of the platform at Point 4'.

Here is a Base Case, using typical values...

  • Train Length:  xL = 510 ft (10-car consist of R142 vehicles, each 51 ft long)
  • Cruise Velocity: vC = 50 mph (73.3 fps)
  • Platform Velocity: vP = 30 mph (44.0 fps)
  • Normal Acceleration: aS = 2.5 mphps (3.7 fpsps)
  • Service Braking Deceleration: dS = 3.0 mphps (4.4 fpsps)
  • Feathered Stop Decleration: dF = 1.5 mphps (2.2 fpsps)
  • Dwell Time: tD = 20 seconds
The indicated distances and times can be calculated as follows:
xCP = (vC2 - vP2) / 2 dS
xP = 100 ft (assumed)
xPO = vP2 / 2 dS
xB = 210 ft (assumed)
xOP = vP2 / 2 aS
xFO = vP2 / 2 dF
xPC = (vC2 - vP2) / 2 aS
tCP = (vC - vP) / dS
tP = xP / vP
tPO = vP / dS
tB = xB / vP
tOP = vP / aS
tFO = vP / dF
tPC = (vC - vP) / aS
In accommodating a Station-Stop, the train must operate slower than cruise speed vC between Point 1 and Point 4' for a total time of...

tSS = tCP + tP + (xPO / vP) + tB + (xL - xF0) / vP + tF0 + tD + t0P + (xL - x0P) / vP + tPC

...and during that time, the lead car of the train will have traveled a total distance of...

xSS = xCP + xP + xP0 + xB + xL + xPC + xL

If the station in the Base Case were bypassed, say, by an express train, that distance xSS would have been covered at cruise speed vC, such that tBYPASS = xSS / vC.  Thus, the difference in Trip Time tDELTA enjoyed by an express train in bypassing a local station can be calculated as follows:

Sophisticated solvers will notice in those expressions that (a) both tSS and tBYPASS increase with all the same variables and (b) both tSS and tBYPASS decrease with increases in acceleration and decelerations aS, dS and dF.  By plugging in the Base Case numbers, we find that tSS = 96 seconds and tBYPASS = 36 seconds.  Accordingly, tDELTA = 60 seconds.  Thus, for every bypassed local station, an express train saves one minute in Trip Time.  For operational implications of that, see the Express vs Local puzzle.

Perturbations of the Base Case show some interesting interactions, if not ironic trade-offs...

  1. Increasing acceleration and decelerations aS, dS, and dF by 10% improves tSS by about 4 seconds at every station and reduces the benefit of express service tDELTA by 2 seconds for each bypassed station.
  2. Reducing buffer distances xP and xB by 10% will have neglible effects on tSS and tDELTA.
  3. Overcrowding on a station platform can result in extending dwell time tD by some time we shall call tCRUSH, which will be added directly to both tSS and tDELTA.
  4. Having both express and local services, an express station will likely reduce tCRUSH.  However, the same benefit would accrue to all stations by running more trains with closer headways.
  5. Maximum cruise speed vC for R142 vehicles is 55 mph, 10% over the Base Case value.  That would make an improvement in the benefits of express service tDELTA by 3.5 seconds while penalizing station stops by 4.1 seconds!  You may regard the exclamation point as signifying a counter-intuitive consequence.
Keeping time in the numerator where it belongs and using the New York City Transit system to typify full metro train services, we may offer as our soluton for the Trip Time puzzle...
Stopping for passengers adds about a minute of Trip-Time between stations.


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