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

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

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

VCIVIL_LIMIT = [({E + U}/ G) (g R)]^1/2  = 182 ft/sec = 124 mph = 62% VCRUISE 

(360 x 182) / (2  x 5,280) = 2 degrees/second

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

Accordingly, our solution is...

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