World's Fastest Train

Copyright 2010 by Paul Niquette. All rights reserved.
As postulated in the puzzle, if the California High Speed Rail (CHSR) project were to adopt the World's Fastest Train for the route from Sacramento to Los Angeles, a non-stop service for the whole 412 miles would be covered in about 1:09, assuming its record-setting speed... 
#1. Maximum Train Velocity ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ vMAX  = 357 mph
...but stopping at a station along the route will call for adding at least...
#7. Station Dwell Time ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ tDWELL = 5 minutes that for a one-station stop, the time enroute increases to 1:14, which reduces the overall train speed to 333 mph.  That matches the corresponding entry in the table for Solution {a}.  Adding stops at all eight stations would increase the overall time enroute by 40 minutes to 1:49 and reduce the speed enroute to about 227 mph, which also matches the correspond entry for Solution {a}.

Most solvers will notice that the entries in Solution {a} do not take into account braking time going into each intermediate station and the acceleration time following the dwell time.  Those intervals can be calculated using...
#5. Maximum Acceleration ~~~~~~~~~~~~~~~~~~~~~ aMAX = 0.1 g = 2.2 mphps 
#6. Maximum Braking Deceleration ~~~~~~~~~~~  - aMAX = - 0.1 g = - 2.2 mphps 
...both being limited by passenger comfort.  For each intermediate stop we might calculate as follows:
Acceleration Time = vMAX  / aMAX =  357 mph / 2.2 mphps = 162 sec = 2.7 minutes
Braking Time = vMAX  / - aMAX =  357 mph / - 2.2 mphps = - 162 sec = - 2.7 minutes
...and adding their absoute magnitudes (|2.7| = |-2.7|) to tDWELL  gives us a total of about 10.4 minutes, resulting in a speed for the route of 310 mph.  That matches the entry for Solution {b}.  Checking the entry for eight enroute stations, we find an added time of 8 * 10.4 minutes or 1:23 for a total of 2:32 for the whole 412 miles at an average speed of 162 mph as offered in Solution {b}.

Solvers of the Express vs Local puzzle will see something amiss in Solution {b}.  All during deceleration and acceleration, the train is actually moving some distance along the trackway toward the destination.  That distance is given by a familiar formula..
(vMAX)2 / aMAX = (357 * 5280/3600 fps)2 / (2.2 * 5280/3600 fpsps) =  85,000 ft
...which amounts to 16 miles, with the station located at about the halfway point.  If the World's Fastest Train bypassed that station, it would cover the 16 miles a whole lot faster than a train that is stopping at the station  -- but not in zero time -- as apparently assumed in Solutions {a} and {b}. 

At 357 mph, the V150 takes 2.7 minutes to cover 16 miles, which should be deducted from the 10.4-minute time penalty for stopping at each station, giving a better estimate of 7.7 minutes per station.  The average speed with one enroute station stop is given by 412 / 357 + 7.7 / 60 = 1.28 hours for an average speed of 321 mph, which matches the first entry in Solution {c};  Likewise for eight stops 412 / 357 + 8 * 7.7 / 60 = 1.28 hrs and 189 mph. 

However, as we shall see, Solutions {a}, {b}, and {c} are all incorrect.

Solvers of the World's Fastest Dragster will be quite familar with effect of aerodynamic drag on acceleration at high speeds. The World's FastestTrain has the same problem along with another one -- adhesion.  The phrase "steel wheels on steel rails" celebrates the efficiency of railroads attributed to low rolling resistance compared to "rubber wheels on paved roads."  Well, that same property sets a limit on tractive effort as recognized in...

#3. Maximum Tractive Effort ~~~~~~~~~~~~~~~~~~~ eMAX = 27,600 pounds-force
Operating near the upper limit of its speed, the V150 will be facing its maximum in aerodynamic drag and using all of its...
#2. Maximum Propulsive Power ~~~~~~~~~~~~~~~~~~ pMAX  = 26,300 horsepower do so, with almost no power left over for acceleration.  Solvers can easily confirm that by reviewing the video.  As the train approaches 574.8 kph (357.0 mph), the acceleration has noticeable slowed to less than 1.0 kph/sec (0.6 mphps or about 1/100th of a g).  When the train is pulling out of a station, nearly all of that 26,300 hp would be available for acceleration.  And the V150 with...
#4. Maximum Train Weight ~~~~~~~~~~~ wMAX = 100 tons = 200,000 pounds-force
 ...could theoretically accelerate at eMAX/ wMAX = 27,600 / 200,000 = 0.14 g or 3.1 mphps.  That would be excessive for passengers who are not strapped into their seats, as mandated for airliners.  People of all ages may be standing or walking in the aisles or sipping coffee.  It is customary to set a limit on accelerations to aMAX = 0.1 g or about 2.2 mphps.  The graph below gives a normalized depiction in the time domain of acceleration and speeds on a V150 class train.
Acceleration is limited by passenger comfort at low speeds to aMAX no matter how fast the train might ultimately travel, vMAX This limit will prevail for as long as 1.5 minutes after leaving the station.  After that interval, acceleration begins to decline, as more and more of the train's power pMAX is required to overcome aerodynamic drag. 
With constant acceleration aMAX, the V150's speed would increase linearly along the dashed line toward vMAX  in about 3 minutes, as assumed in Solutions {a}, {b}, and {c}. 
Reality is something else.  Aerodynamic drag increases according as the square of train-speed, and the required horsepower to overcome aerodynamic drag increases with the cube of train-speed.  No matter how much of the V150's power is devoted to tractive effort eMAX, with increasing speed, more and more of that tractive effort is devoted to overcoming aerodynamic drag and thus unavailable for acceleration.

Some solvers may notice that at a speed below about 180 mph (50% of the 357 mph) aerodynamic drag has not taken effect to reduce the acceleration below aMAX. The numbers suggest that any train with aerodynamic properties comparable to the V150's will show the same pattern.  Since the present TGVs and Aerostars operate with a vMAX < 200 mph, the linear acceleration model applies to them.

Exclamation Point Alert: At speeds faster than about 180 mph, all of the V150's 32 drive wheels will be on the brink of losing their frictional grip on the rails!

Solvers of Trail Braking , Station Stop, and Trip Time puzzles will have found the "phase-plane plot" (derivative of a variable dx/dt plotted against the variable x) to be a useful tool.  The graph below plots train speed against distance from the station and will come in handy for solving the World's Fastest Train puzzle.  It shows the effect of aerodynamic drag on velocity in the space domain, with the V150 taking 28 miles to reach the record-setting speed of 357 mph following a station stop.  Deceleration is another matter as can be seen in the asymmetry of the phase-plane plot.

In the space domain above, constant deceleration takes the shape of a parabola.  Inasmuch as aerodynamic drag adds to the braking effort, a comfortable, constant deceleration of  0.1 g can be readily assured.  However, constant deceleration does require the use of gradually decreasing horsepower. Indeed, a sudden loss of catenary power from the wayside while traveling at 357 mph would instantaneously remove the tractive effort in the face of 27,600 lb of aerodynamic drag.  The result would be a jolting deceleration of 27,600 / 200,000 = 0.13 g.

The pattern for deceleration and acceleration can be readily calculated in the time domain using an elementary piecewise model on a spreadsheet.  Those time intervals added to the 5 minutes for tDWELL sum to 14.4 minutes.  The total distance, 8 + 28 = 36 miles, would be covered by a V150-class train traveling at 357 mph in 6.0 minutes. 

Thus, for the V150,  tDELTA = 14.4 - 6.0 = 8.4 minutes. That 8.4 minutes is the estimated allowance in the timetable required for each intermediate station and is the basis for Solution {d}...



The model assumes a fixed value for tDWELL = 5 minutes, which is much longer than the typical value of tDWELL = 20 seconds used in full metro services for daily commuters on short segments.  By way of justification, each duplex vehicle in the V150 consist is equipped with only one stairway and only one door at the platform level.  In contrast, BART vehicles  offer more than twice the number of doors per commuting passenger than do TGV train-sets for the cross-country travelers. Moreover, high-speed trains going long distances between stations require more time for loading and unloading passengers, since many are required to lug -- well, luggage. Finally, extra time embedded in tDWELL assures "feathered stopping" and reduction of "starting jerk." 

Superimposed on the phase-plane plot above are tDELTA values for hypothetical trains operating at lower vMAX  than the World's Fastest Train.  Solvers of the Express vs Local puzzle will not be surprised that tDELTA  increases with vMAX along with tDWELL .  One might ask, Does the increase in vMAX overcome the potential impact of tDELTA on trip-time?  Yes, and then some (but not much).

Here is a phase-plane plot for the V150 with a distinctive feature added that will be useful for estimating average speed between stations. During Acceleration phase, the blue line shows that 16 miles outside the station, the train reaches 350 mph.  Now, a train-speed within 2% of vMAX  = 357 mph ought to qualify for Cruise phase as shown, even though the train is still gradually accelerating. The model has been set up so that the World's Fastest Train gets put into Brake phase just as it reaches its vMAX  = 357 mph.  Taking all three phases into account, the train in the model comes to a stop at the next station about 45 miles beyond the previous station.  That is the shortest segment that can take advantage of the record-setting speed of the V150.

Whereas the blue line depicts instantaneous speed, the superimposed green line shows the average speed 'made good' to that location along the trackway.  The average is calculated in the model by the ratio of distance to time-of-arrival at that distance. Solvers are invited to make eight observations...
  1. End of Acceleration Phase: The train is traveling at 350 mph but has averaged only 225 mph.
  2. End of Cruise Phase: The train is traveling at 357 mph but has averaged only 280 mph.
  3. End of Brake Phase: On arrival at the station 45 miles away, the train has averaged 255 mph.
  4. Maximum Average Speed is 290 mph, where the blue line crosses the green line.
  5. Longer Segments will allow the train to continue cruising, with the blue line level at 357 mph.
  6. Longer Segments will also allow the green line to approach 357 mph asymptotically.
  7. Shorter Segments are represented by both dashed blue line and dashed green line.
  8. Shorter Segments are typified by a 16-mile distance between stations, during which the V150 does no better than the Train Grande Vitesse (TGV), which set the speed record 17 years earlier at 320 mph , both trains averaging 155 mph for the segment.
Back to the planned California High Speed Rail project with its eight intermediate stations on the 412-mile route between Sacramento and Los Angeles: As mentioned in the puzzle, the segment distances range from 13 miles to 113 miles and average 46 miles (412 / 9).  By happenstance of puzzle-making, that average fits the case in the graph above and would allow the World's Fastest Train to cruise at its world-record speed of 357 mph between stations -- but for no more than 48 seconds, on average  Exclamation point optional.

Six years after the publication of the World's Fastest Train puzzle, French manufacturer Alstom produced a 45-minute documentary entited Fastest Train in the World Ever Made about the AGV Automotrice Grande Vitesse being put into operation. 

Solvers may notice several references to the AGV's pantograph, which confirms discoveries made in another puzzle entitle Pantograph Design,

Meanwhile, the residence of your puzzle-master is now a mere 20 minutes away from the station in Lamballe, which will soon offer non-stop AGV service to and from Paris at 300 km/hr (200 mph).  Hoo-hah.

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