Short Cut?

Copyright ©2003 by Paul Niquette.  All rights reserved.

A certain bicyclist commutes on a level road. 

One day he/she discovers a shortcut over a hill. 

The wheelperson estimates that his/her speed 
pedalling up the hill will be half that of coasting 
down the hill.

Should he/she take the shortcut?

his puzzle has a problem.  A puzzle is expected to have an unknown, and the problem is to figure out what the unknown is.  That's called the solution.  That's not the problem in this puzzle.  The problem in this puzzle is that nothing is known. Well, almost nothing is known.  The thing has one of two solutions: "yes" or "no."   As a sophisticated solver, you do your solving using what is known, of course.  If nothing is known, how will you ever find the solution?  For this puzzle, maybe your solution will have to be "maybe."  Naah.

The unknowns are plentiful.  Here they are -- all ten of them...

One-way commuting distance 
on the level road
Up-hill distance on "short cut"
Down-hill distance on "short cut"
Speed on the level road
Up-hill speed
Down-hill speed
Height of the hill
Weight of bike and rider
Power output of the rider
Coefficient of drag

About all we learn from the statements in the puzzle can be summarized in two relationships among the unknowns...

[1] The length of the "short cut" = XU + XD
[2] The bicyclist's estimate is that SD = 2 SU
The sophisticated solver will surely recall the essential property of wind resistance as described, for example, in A Certain Bicyclist...
The invisible enemy of the bicyclist is wind resistance. Even in still air, the bicyclist creates wind resistance, or drag. It is a force to be reckoned with.

A certain bicyclist is pedalling at 9 MPH on a calm day. There's a significant relative wind to overcome at that speed. (It is far more ornery, by the way, than the rolling friction of the wheels.)  If he/she were to speed up to 18 MPH, the drag would increase -- but not just double. It would go up by a factor of four! Wind resistance varies as the square of the relative wind speed.

...from which another relevant relationship among the unknowns can be derived...
[3] Wind resistance experienced on the level road = K SL2
...and as described further in the passage referenced above...
To overcome this increased force, the bicyclist must expend energy at a faster rate. That's called "power." But since power output is computed as the product of wind speed and drag, it goes up as the cube of speed. For 18 MPH, therefore, a bicyclist's output is eight times greater than for 9 MPH.
... such that yet another relationship among the unknowns can be derived...
[4] The power required for commuting on the level road P = K SL3
By the way, in that mathematical expression, we have made an assumption, haven't we.  As is our practice, let us make that assumption explicit...
Assumption #1 That requisite pedaling power on the level road at SL is dominated by wind resistance.
...and thus rolling resistance and other sources of friction are negligible.  Since those forces are nearly constant, they can surely be neglected at higher speeds, including SD > SL.  Likewise at lower speeds, SU < SL, Assumption #1 may be taken to apply, along with...
Assumption #2 That wind resistance is negligible going up-hill at SU.
ew persons who ride bicycles will dispute this assumption.  It does favor the "short cut," so if our solution turns out to be "no" (don't take the short cut), then we can let it go at that.  If, on the other hand, the solution is "yes" (take the short cut), then we will surely want to reconsider Assumption #2.

During the up-hill portion of the "short cut," all of the bicyclist's power is devoted to climbing -- to lifting the weight of bike and rider, W, to height H over distance XU at speed SU.  Sophisticated solvers know that power is the product of a force times the distance through which it is applied divided by the time required.  Thus,...

[5]  The power required for climbing the hill = W (H / XU) SU
...and to do that with same pedaling power as for commuting on the level road, from [4], we have that...
[6]   P = W (H / XU) SU = K SL3
The sophisticated solver will doubtless take note of that word "coasting" in the puzzle statement, for it means that on the down-hill part of the "short cut" the bicyclist is not pedaling.  The bicycle and rider, apparently, come coasting down the hill at some "terminal velocity" limited by wind-resistance.  Accordingly, a balance of forces applies...
[7]   K SD2 = W (H / XD)
Is there anything else to be seen in the puzzle statements?  Oh right, the word "commutes," which implies a daily round trip model for the puzzle.  If relationship [2] SD = 2 SU is to apply in both morning and evening trips then that requires...
[8]   XU = XD.
From relationship [6] and [7], the sophisticated solver readily sees that
[9]   SD = 2(K / W)(XD / H) SL3 and XD = (W / K) H / SD2
...such that...
[10]   SD = 21/3 SL = 1.256 SL and SU = 0.63 SL matter what the relative distances are.  Exclamatory punctuation might be permissible for such an observation.

ow, back to the objective of the puzzle, to decide whether to take the "short cut."  We have learned elsewhere that making judgments based on speeds can be deceiving and thus we deliberately keep time in the numerator.  The elapsed time for the one-way commuting alternatives are given by...

[11] Level road commuting time TLR = XL / SL
[12] Short cut commuting time TSC = XU / SU + XD / SD = 3 XD / SD
...and from [10] and [12] we see that for TSC < TLR...
[13]   2 XD < 2 (21/3 XL) / 3 < 0.84 XL.
    The short cut better be less than 84% 
    as long as the level-road route or forget it.
Case in Point

A certain bicyclist commutes four and a half miles on a level road, taking a half hour each way.  One day he/she discovers a shortcut over a hill that measures a mile and a quarter shorter.  The wheelperson finds that his/her speed pedaling up the hill is about half that of coasting down the hill.  How much time does he/she save each day?

Apparently, XL = 4.5 mi and XU + XD = 4.5 - 1.25 = 3.25 mi
By [8]  XU = XD we determine that XD= 1.625 mi
By [11]  TLR = XL / SL we confirm that 0.5 hr = 4.5 mi / 9.0 mph
By [10]  SD = 21/3 SL we calculate that SD = 11.3 mph
By [12]  TSC = 3 XD / SD = 3 * 1.625 / 11.3 = 0.43 hr = 26 minutes...
...a savings of four minutes each way, eight minutes per day.

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