Brain Teaser - Riding against the wind

[CR Haeger]

Here is a short brain teaser that you can cogitate on next time you are out on a bicycle...

Question 1: Solve the puzzle "Riding against the wind" given here

Sam Lloyd puzzle - courtesy of Math is Fun

Question 2: Solve the puzzle again, but now with some additional assumptions included

- No acceleration or deceleration
- No up or down grade slope
- Riders power (watts) is constant in both directions and weight with bicycle is 100kg
- Wind is constant speed with air density = 1.28 kg/m^3
- Coeff of rolling resistance = 0.004
- Coeff of wind * Frontal area = 0.4 m^2
- Drivetrain efficiency = 95%

Question 3: What else can you determine about the rider, bicycle and/or wind?

Enjoy and remember your units...

[Gerson W. Barbosa]

#1 only:

r = rider's speed
w = wind speed

r + w = 1/3 mi/min (1)
r - w = 1/4 mi/min (2)

From (1), w = 1/3 - r

Replacing this in (2) it follows that

r - 1/4 = 1/3 - r
2*r = 1/3 + 1/4
2*r = 7/12

r = 7/24 [mi/min]

t = 1 [mi]/(7/24 [mi/min]

t = 24/7 [min]

[CR Haeger]

Yep, that is correct for "r".

However, I'm not sure if "w" is wind speed. That's what prompted me to pose the other questions.

[Gerson W. Barbosa]

(05-24-2015 04:49 PM)CR Haeger Wrote:  Yep, that is correct for "r".

However, I'm not sure if "w" is wind speed. That's what prompted me to pose the other questions.

Modulus of the component of the rider's speed due to the horizontal component of the wind speed might have been better. This would match the horizontal component of the wind speed if there were no friction involved, I think.

[CR Haeger]

Yeah, solution 1 seems to assume that the effect of wind on rider speed is exactly equal and opposite for the two directions.

[CR Haeger]

I worked #2 a few ways and on a few calculators.

I get t = 3' 26.8" or about 1 second longer than the answer for #1...mainly due to the "gentle wind".

[Claudio L.]

(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

V2=4/3V1

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?

Claudio

EDIT: Should be "at least one of the roots should be POSITIVE", obviously if one is real, the other one too.

[Claudio L.]

(05-28-2015 02:10 PM)CR Haeger Wrote:  I worked #2 a few ways and on a few calculators.

I get t = 3' 26.8" or about 1 second longer than the answer for #1...mainly due to the "gentle wind".

My solution above takes you exactly to your answer: 3 min. 26.8 seconds.

[CR Haeger]

Thanks Claudio for you clear worked example which appears correct for the assumptions you included. I also solved for W, P then V3 in a similar way. I forgot about using V2=4/3V1 as a substitution in my work.

I did work it with additional friction losses due to rolling resistance and (rider to wheel) drive-train losses. Perhaps these end up being negligible effects or drop out of the equations. Maybe these become more apparent when the difference in times is greater, say 3 and 5 minute 1-mile trips...

[Claudio L.]

(05-28-2015 08:33 PM)CR Haeger Wrote:  I did work it with additional friction losses due to rolling resistance and (rider to wheel) drive-train losses. Perhaps these end up being negligible effects or drop out of the equations. Maybe these become more apparent when the difference in times is greater, say 3 and 5 minute 1-mile trips...

You are right, I think all other effects are probably negligible for the given rider speed. The quadratic term can grow very quickly and render all other losses insignificant for the few digits we measure time with.
I didn't try to solve #2, I was actually trying to solve #1, but thought constant power was a more correct assumption, all others seemed arbitrary and without substance.
Somebody should contact the author of the puzzle and tell him his assumption is not a good example for kids learning physics. They shouldn't pull wild assumptions out of thin air like that.