Circling: Ground and Sky

Copyright ©2008 by Paul Niquette. All rights reserved.


et us start with the bicycle in the puzzle, assigning some value H as the height of the center of gravity of bike and rider in the vertical position.  For simplicity, it is preferable to take 'moment arms' about the point of contact of the surface with the wheels, which removes both L and FFRICTION from consideration.  Let  represent the angle by which the bicycle is leaning at any given moment.  The weight W develops a moment arm H sin  around the contact point of the wheels with the surface.  This is the 'torque' that tends to tip the bicycle over into a calamitous fall. 

From elementary physics, the centrifugal force FCENTRIFUGAL = (W / g)(V2 / R), where g = 32.2 ft/sec/sec, V = speed in ft/sec, and R = the local radius (in feet) of the curve being followed by the bicycle. Observe that FCENTRIFUGAL develops its moment arm from H cos , and we write...

R = V2 / g (sin  / cos ) = V2 / g tan 
...noting, after some algebra, the independence of R with respect to both W and H.

Turning now (no pun intended) to the commercial aircraft overhead at angle-of-bank , observe that, unlike in the bicycle calculations, there is no H with which to contend.  The vertical component of lifting force is given by L cos  = W, and the horizontal component must satisfy L sin  = FCENTRIFUGAL.  Perhaps amazingly, the expression for the radius of curvature for the commercial airliner... 

R = V2 / g tan 
...is not only independent of W but is also mathematically the same as that for the bicycle.  Exclamatory punctuation optional.

he puzzle calls for both the bike and the plane each to travel in a complete circle. Two equations are most relevant...

RBIKE= V 2BIKE/ g tan BIKE,
RPLANE= V2PLANE/ g tan PLANE.
Time intervals to complete the circular pathways are given by...
TBIKE= 2   RBIKE / VBIKE,
TPLANE= 2   RPLANE / VPLANE
...and following substitutions, we learn that...
TBIKE= 6 seconds, and 
TPLANE= 89 seconds 
...which means that...
 
The jet airliner arrives about a minute and a half after the bicycle and a quarter of a mile behind, having traveled an extra 4.5 miles, but will take only about 5 seconds to catch up.

...suggesting a slightly different case that may be of interest to some solvers.  If the bicycle were traveling at 12 mph and leans, say, 6 degrees, while a plane overhead traveling 10 times faster banks 45 degrees (a common practice for airliners), the two will arrive back at the starting point at about the same time.

Epilog

on't try this at home.  Depicted here is a rather extreme case of a left-hand turn being executed on a two-wheel vehicle.  Unlike Einstein's bike being ridden casually in a courtyard -- oh, and unlike a jet-plane maneuvering in the open sky -- the motorcycle is constrained to complete every turn within the confines of the racetrack. 

The objective, of course, is speed, higher the better.  The only variable is the lean-angle , shown here to be about 45 degrees. 

Assuming the requirement for a turn of 90 degrees ( / 2 radians), familiar relationships apply...

TMOTORCYCLE= ( / 2)  RMOTORCYCLE / VMOTORCYCLE , where...
RMOTORCYCLE = V 2MOTORCYCLE/ g tan MOTORCYCLE 
...and can be used to confirm that at 75 mph, the turn can be completed in 5 seconds.

 


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