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Brain Teaser - Riding against the wind
05-28-2015, 06:24 PM (This post was last modified: 05-28-2015 06:46 PM by Claudio L..)
Post: #7
RE: Brain Teaser - Riding against the wind
(05-24-2015 06:34 PM)CR Haeger Wrote:  Yeah, solution 1 seems to assume that the effect of wind on rider speed is exactly equal and opposite for the two directions.

Assuming that the wind will change the rider's speed the same both ways seems just as arbitrary as assuming that the wind will cause identical delay in time both ways. In my opinion, taking the average is just as good as the solution they proposed.

If we neglect any friction, and say that the only resistance is due to wind:

Going against the wind:

F1 = C * (V1+W)^2

and in favor of wind:
F2 = C * (V2-W)^2

and no wind:

F3 = C * V3^2

Where V1,2,3 are the actual speeds of the rider, W= wind speed and C = drag coefficient with density and all other constants included.

V1 and V2 are known. Assuming constant power (now this is a reasonable assumption)

P = F1 * V1 = F2 * V2 = F3 * V3


F1 = 4/3 F2

(V1+W)^2 = 4/3 (4/3V1-W)^2

The roots of this quadratic expression would give the actual wind speed (I didn't do the numbers, at least one should be real).

Once W is known,

P = F1 * V1 = C * (V1+W)^2 *V1

Since C is unknown, just compute P/C = V1 * (V1+W)^2.

Then P = F3 * V3 = C * V3^3

and V3 = (P/C)^(1/3)

This is similar to #2, except:

* Neglects friction, drivetrain efficiency and all other effects but wind (which aren't part of the original problem).
* It assumes no other parameters than the ones given in the problem.

So I see this as an alternative to solution #1, with a less arbitrary assumption (constant power instead of constant speed change).

Did I make any mistakes here?


EDIT: Should be "at least one of the roots should be POSITIVE", obviously if one is real, the other one too.
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RE: Brain Teaser - Riding against the wind - Claudio L. - 05-28-2015 06:24 PM

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