Fast Track

Copyright 2011 by Paul Niquette. All rights reserved.

 
The following conditions were prescribed for a curve in high-speed trackway:
Cruise Speed ~~~~~~~~~~~~~~~~ VCRUISE = 200 mph = 293 ft/sec
Track Gauge ~~~~~~~~~~~~~~~~~~G = 1 metrum asinorum = 4 ft 8.5 in = 4.7 ft
Height of Center of Mass ~~~~~~~~~~ H = 11/2 G =7.1 ft
Superelevation ~~~~~~~~~~~~~~~~~~~ E = 7 in = 0.58 ft
Unbalance ~~~~~~~~~~~~~~~~~~~~~~~~U = E / 2 = 4 in = 0.333 ft
Radius of Curvature ~~~~~~~~~~~~~~~~~~ R = 1 mile = 5,280 ft
Change in Alignment ~~~~~~~~~~~~~~~~~~~ A = 20 degrees 
Spirals (in + out) ~~~~~~~~~~~~~~~~~~~~~~~~ S = A / 2 = 10 degrees 
Acceleration/Deceleration ~~~~~~~~~~~~~~~~~~~ a = d = g / 10 = 3.2 ft/sec/sec
Car Length ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~C = 75 ft
Maximum Train Length ~~~~~~~~~~~~~~~~~~~~~~~ L = 20 x C = 1,500 ft
How much time will be lost negotiating the curve?

Six phases of operations are required for negotiating a curve... 

Phase 1 Deceleration from VCRUISE to VCIVIL_LIMIT 
Phase 2 Spiral into Curve
Phase 3 Curve at Radius R
Phase 4 Spiral out of Curve
Phase 5 Tangent to Clear End-of-Train
Phase 6 Acceleration from VCIVIL_LIMIT  to VCRUISE
Phase 3 Curve at Radius R

We begin our solution in Phase 3 by considering the train to be established in a curve with R = 1 mile operating at its civil limit VCIVIL_LIMIT.  In the puzzle formulation, we derived...

VBALANCED = [(E / G) (g R)]1/2
...in which the value of E was chosen to produce a balance of forces in the track.  To derive the 'civil limit' for the curve, we can simplify our treatment of unbalance U so that the height of the center of mass H need not be considered.  Let {E + U} be applied to synthesize a balance of forces in the rails...
VCIVIL_LIMIT = [({E + U}/ G) (g R)]1/2  = 182 ft/sec = 124 mph = 62% VCRUISE 
Operating in Phase 3 at 182 ft/sec, the train is steadily changing its direction of travel at...
(360 x 182) / (2  x 5,280) = 2 degrees/second
...and is specified to change direction of travel during Phase 3 by A - S = 10 degrees, which will require 5 seconds along the arc of a circle that measures 182 x 5 = 910 feet in length.

Phase 2 Spiral into Curve
Phase 4 Spiral out of Curve

As the train transitions from traveling on straight track to the curve at radius R  in Phase 2, its rate of turning in direction of travel must linearly increase from zero deg/sec to 2 deg/sec, thus averaging 1 deg/sec.  Likewise in Phase 4, the train will be decreasing its rate of turning back to zero and averaging 1 deg/sec. 

Both spirals are specified to change the train's direction by S = A / 2 = 10 degrees, which will each take 5 seconds along its spiraling arc for 182 x 5 = 910 feet in length.

Phase 1 Deceleration from VCRUISE to VCIVIL_LIMIT
Phase 6 Acceleration from VCIVIL_LIMIT  to VCRUISE

For safety, the train must be slowed to VCIVIL_LIMIT before entering the spiral into the curve. Accordingly Phase 1 will require (293 - 182) / 3.2 = 34.7 second and cover a total distance of 34.7 (293 + 182) / 2 = 8,238 ft.  The same times and distances apply to Phase 6.

Phase 5 Tangent to Clear End-of-Train

When the lead car returns to straight tangent track at the beginning of Phase 5, the rest of the consist is still operating in the spiral out of the curve.  The train must therefore continue at VCIVIL_LIMIT for the length of the train L = 20 x C = 1,500 ft, which requires 8.2 seconds.

 
Summary
   Time 
  sec 
Distance
     ft 
Phase 1 Deceleration to VCIVIL_LIMIT 
34.7
8,238
Phase 2 Spiral into Curve (5 deg)
5.0
910
Phase 3 Curve at Radius R (10 deg)
5.0
910
Phase 4 Spiral out of Curve (5 deg)
5.0
910
Phase 5 Tangent to Clear End-of-Train
8.2
1,500
Phase 6 Acceleration to VCRUISE
  34.7
8,238
Totals: 
   92.6
20,706
While negotiating a change of direction of 20 degrees, the train in the Fast Track puzzle will cover a total distance along the track alignment toward its destination of 20,706 feet (3.9 miles) over a period of 92.6 sec for an average speed of 224 ft/sec = 152 mph.  If there had been no such curve in the trackway, the train would be able to operate at VCRUISE = 200 mph = 293 ft/sec over the same distance in only 70.7 sec. 
Accordingly, our solution is...
92.6 - 70.7 = 21.9 seconds

 


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