# Fifty Plus (50+) - A cycling math problem - What's your answer?

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Kurt Erlenbach
07-13-12, 05:27 PM
Today on my morning ride my mind began to wander, and I started thinking about my shoes. I've had this pair for about three years, and they've taken me over 10,000 miles. Thinking about those numbers, I began to to think about how far the shoes travel to take me a mile. The pattern in space the shoe follows is something of a sine wave, but it accelerates and decelerates as the pedals rotate around the bottom bracket. Given those facts, how does one calculate the distance a shoe travels when the bike goes one mile?

I think I have the answer, and I'd like to ask for your's. Here are what I think are the relevant facts: Average speed: 17.3 miles per hour. Average cadence 78 rpm. Crankarm length 175 mm. If you think additional facts are relevant, please so state. Answer must be in feet, and show your work.

cyclinfool
07-13-12, 05:43 PM
a mile

contango
07-13-12, 05:46 PM
Crankarm length of 175mm means your pedals would describe a circle of diameter of just about 1100mm, or 43.3", or 3.61 feet

78rpm equals 4680 rotations per hour. So for 4680 rotations to take you 17.3 miles each rotation would take you 19.52 feet

I'm going to take a bit of a leap of faith and speculate that the distance the shoe travels per rotation is the same whether you pedal for the full rotation normally, or you were in a theoretical situation where you lifted the back wheel, rotated the pedals once while stationary, and then rolled forward 19.52 feet with the pedals stationary. That may not be entirely accurate but given you probably aren't doing the same cadence the entire time, the same speed the entire time, or necessarily even in the same gear the entire time, it's probably a fair enough approximation (that said it's late, so maths geeks feel free to tell me I'm wrong)

That would make the distance the pedals travelled per rotation 19.52 + 3.61 = 23.13 feet.

So over the course of your 17.3 miles in an hour those 4680 rotations would have the pedals moving 23.13 x 4680 = 108,248 feet, so over the course of one mile they would travel 108248/17.3 = 6257 feet. So the pedals are moving 18.5% further than the bottom bracket.

locolobo13
07-13-12, 05:51 PM
I was thinking more cycloid like. Or a "Curtate Cycloid"?

http://en.wikipedia.org/wiki/Cycloid

Have no idea how to calculate distance.

Kurt Erlenbach
07-13-12, 06:44 PM
Tom - But the shoe is moving up and down in addition to forward, so it's distance will be greater than a mile, won't it? The bike as a whole goes one mile, but the circular motion of the pedals, just like the circular motion of the wheels, means those parts travel farther. Put simply, the wheel hub goes one mile, but a point on the tire moves more than a mile, just like the shoes.

TomD77
07-13-12, 06:57 PM
Tom - But the shoe is moving up and down in addition to forward, so it's distance will be greater than a mile, won't it? The bike as a whole goes one mile, but the circular motion of the pedals, just like the circular motion of the wheels, means those parts travel farther. Put simply, the wheel hub goes one mile, but a point on the tire moves more than a mile, just like the shoes.

I thought about it more after the post and deleted the post to think more. I've come to the conclusion that there is no correlation of pedal distance to forward distance. Say you have a bike with an infinite gear ratio, then any given pedal speed can have from a zero forward velocity to infinite forward velocity. Therefore the answer changes relative to the crank distance, gear ratio and wheel size.

cyclinfool
07-13-12, 07:03 PM
I stick with my original answer, that is unless you tell me the exact direction of travel, the date and time of travel and your longitude and lattitude because we all know that we are spinning through space on a rotating ball in a moving universe - and that poor little shoe is traveling with us...

Kurt Erlenbach
07-13-12, 07:07 PM
Tom - OK, assume a 700 x 25 tire. The gear inches ratio could be calculated from that with the speed and cadence number.

Kurt Erlenbach
07-13-12, 07:25 PM
Here's my answer. Assume you're on a trainer. The bike moves a distance of zero, but the pedals and shoes move pi(crank length) * # of revs. Using that formula, going 17.3 mph means 1522.4 ft/min = 3.4682 min per mile. 78 rpm cadence thus = 270.5 revs per mile. pi(175 mm) * 270.5 revs = 148,715.49 mm per mile. 148,715.49 mm = 487.912 feet. The answer thus, I think, is 5280 + 487.9 = 5767.9 feet. But after reading the wikipedia article lobolobo links above, I think the correct answer requires calculus that this lawyer, who barely passed Calculus for Babies in college, no longer knows.

Am I right or wrong?

Dudelsack
07-13-12, 07:44 PM
N+1. The answer always seems to work with cycling related questions.

DnvrFox
07-13-12, 08:05 PM
Err . .

Does this change your calculations any?

http://static.ddmcdn.com/gif/bungled-personal-flight-attempt-1.jpg

Bearhawker
07-13-12, 08:13 PM
Another wrinkle... if the outer edge of the tire goes around faster than the hub does, does the center of the hub actually move?

Kurt Erlenbach
07-13-12, 08:35 PM
My answer is wrong because the distance the shoe travels on a trainer is pi(2 * 175 mm), not pi(175mm).

Kurt Erlenbach
07-13-12, 08:36 PM
Another wrinkle... if the outer edge of the tire goes around faster than the hub does, does the center of the hub actually move?

On a trainer, the answer is no. In my problem, the hub moves exactly one mile.

Retro Grouch
07-13-12, 08:42 PM
When you're out on a morning ride, you need to focus more on what you're doing.

Kurt Erlenbach
07-13-12, 09:25 PM
OK, here's an new part of the answer. The length of the arc of a cycloid is 8r. But the movement of shoes on a bike describe a curtate cycloid, not a cycloid. The movement of a point on the outside of the tire describes a cycloid. The radius of the tire is about 340 miilimeters, or 13.39 inches, making the tire circumference 2* 3.1416 * 340 = 2,136 millimeters = 7 feet, or 753.3 revolutions per mile. The distance a point on the tire moves thus is 8r(753.4) or 8* 13.39 * 753.3 = 6733 feet. So the valve stem goes 6733 feet (or a little less, because it's not on the outside) when the bike goes a mile. Now I need the find the formula for calculating the arc of a curtate cycloid, which would describe the distance the shoes move. I learned something tonight.

Kurt Erlenbach
07-13-12, 09:26 PM
And R-Grouch, you are completely right.

zonatandem
07-13-12, 09:53 PM
The shoe does not tragel anywhere.

TomD77
07-13-12, 11:06 PM
It's not that hard. Say you are in a 50:18 gear and (for simplicity) your wheel circumference is exactly 6.5 feet and your crank length is exactly 7 inches. Each rotation of the crank (1.82 feet) will yield 18 feet of forward movement for a ratio of very close to 10:1 of crank distance to forward distance. So the answer is 528 feet per mile. If you're asking about the more complicated case of pedal movement relative to a stationary reference, it is additive.

Look at your watch as you are in a car at 60 mph, how far does the second hand travel in one minute if it is perpendicular to the direction of travel? One mile plus the circumference of the dial.

TomD77
07-13-12, 11:38 PM
When you're out on a morning ride, you need to focus more on what you're doing.

Dunno about that. My best rides are when I have some thought train going and I'll temporarily surface and wonder where the last 5 miles got off to.

07-14-12, 05:30 AM
Tom - But the shoe is moving up and down in addition to forward, so it's distance will be greater than a mile, won't it? The bike as a whole goes one mile, but the circular motion of the pedals, just like the circular motion of the wheels, means those parts travel farther. Put simply, the wheel hub goes one mile, but a point on the tire moves more than a mile, just like the shoes.

If you have a cadence counting computer take the distance traveled in miles by the bike and compare it to the distance derived from the circumference the shoe travels and the total pedal strokes. No matter how you visualize the "sine" wave or other foot motion, all it does is travel around a circle x number of times for y minutes. I believe that all the other perceived motions cancle out.

As a quick example, assume a gear ratio that results in the foot going forward relative to the ground at twice bike speed at the top of the pedal circle. That would also result in a foot speed relative to the ground of 0 at the bottom of the circle. I contend that the overall result is that your shoes don't know the difference between riding a stationary bike and riding down the trail. They only know the cicle distance and the time.

That could be completely wrong, but it's my intuitive "guess". :innocent:

BeastRider
07-14-12, 06:05 AM
Today on my morning ride my mind began to wander, and I started thinking about my shoes. I've had this pair for about three years, and they've taken me over 10,000 miles. Thinking about those numbers, I began to to think about how far the shoes travel to take me a mile. The pattern in space the shoe follows is something of a sine wave, but it accelerates and decelerates as the pedals rotate around the bottom bracket. Given those facts, how does one calculate the distance a shoe travels when the bike goes one mile?

I think I have the answer, and I'd like to ask for your's. Here are what I think are the relevant facts: Average speed: 17.3 miles per hour. Average cadence 78 rpm. Crankarm length 175 mm. If you think additional facts are relevant, please so state. Answer must be in feet, and show your work.

REALLY??.....That's what you are concerned about??......

Garfield Cat
07-14-12, 06:45 AM
The shoes only travel when you get off the bike and walk around in them. The rest is meaningless. Take a look at your heel cushion and the scuff marks.

Kurt Erlenbach
07-14-12, 07:47 AM
The path of the shoe is not a curtate cycloid because the chain and gearing intervene. The path a shoe traces on a child's tricycle is a curtate cycloid, when the pedal is connected directly to the drive wheel. Knowing the tire radius is 340 mm allows calculation of the gain ratio - 270.5 turns of the crank = 270.5(2pi(175mm)) =297,431 mm = 975.8 feet. So to move a point on the outside of the wheel 6733 feet, which is what's required to roll a wheel of r=340 mm one mile, requires turning the crank a distance of 975.8 feet. Somewhere in the ratio of those numbers, about 6.9 to 1, is an important fact.

Dudelsack
07-14-12, 08:23 AM
My answer is wrong because the distance the shoe travels on a trainer is pi(2 * 175 mm), not pi(175mm).
Someone always drags pi into threads on this forum.
http://i180.photobucket.com/albums/x40/TWLBA/1c7c6886.jpg

Retro Grouch
07-14-12, 10:14 AM
Someone always drags pi into threads on this forum.

Bikey Mikey
07-14-12, 12:13 PM
I would never drag pi; I'd eat it--well, the one ending with "e."

Phil85207
07-14-12, 08:06 PM
There must be a pill you can take for that affliction. Thinking to much on a bike ride.

jalbri
07-14-12, 09:14 PM
In my opinion the answer definitely depends on the gear the bicycle is in. Also, I don't think it matters whether the bike is moving forward at all unless you are trying to find the distance traveled relative to a particular point on the ground.

BluesDawg
07-14-12, 10:05 PM
42

Bob Nichols
07-15-12, 05:36 AM
Did you take into account that the earth surface speed due to rotation is about 1000 miles per hour at the euqator, and the earth is going around the sun at a speed of 67,000 miles per hour, and who knows how fast our solar system is travelling through the universe.

Bikey Mikey
07-15-12, 08:27 AM
Just remember that you're standing on a planet that's evolving
And revolving at nine hundred miles an hour,
That's orbiting at nineteen miles a second, so it's reckoned,
A sun that is the source of all our power.
The sun and you and me and all the stars that we can see
Are moving at a million miles a day
In an outer spiral arm, at forty thousand miles an hour,
Of the galaxy we call the 'Milky Way'.
Our galaxy itself contains a hundred billion stars.
It's a hundred thousand light years side to side.
It bulges in the middle, sixteen thousand light years thick,
But out by us, it's just three thousand light years wide.
We're thirty thousand light years from galactic central point.
We go 'round every two hundred million years,
And our galaxy is only one of millions of billions
In this amazing and expanding universe.

The universe itself keeps on expanding and expanding
In all of the directions it can whizz
As fast as it can go, at the speed of light, you know,
Twelve million miles a minute, and that's the fastest speed there is.
So remember, when you're feeling very small and insecure,
How amazingly unlikely is your birth,
And pray that there's intelligent life somewhere up in space,
'Cause there's bugger all down here on Earth.

Monty Python, The Galaxy Song (http://www.lyricsdepot.com/monty-python/galaxy-song.html)

Retro Grouch
07-15-12, 02:59 PM
42

Kurt Erlenbach
07-15-12, 03:23 PM
In my opinion the answer definitely depends on the gear the bicycle is in. Also, I don't think it matters whether the bike is moving forward at all unless you are trying to find the distance traveled relative to a particular point on the ground.

Of course, when you talk about distance ridden, it's always about the distance you and the bike travel, which really is the only relevant number - this calculating is just a way to pass the time on the bike.. Regarding gear, that factor is taken into account by the rpm/speed ratio. A higher rpm with equal speed equals a lower gear and vice versa. I calculate a gain ratio of 1/6.9 - for each turn of the crank, the wheels rotate 6.9 times using a 78 rpm cadence and 17.3 average speed on wheels with a 340 mm radius and a 175 mm crank. That's how gearing come into the equation. A higher gear with the same rpm results in a higher speed, thus a higher gain ratio. So there's no need to know just what gear you're in. I am sure that the four factors - cadence, crank length, average speed, and wheel radius - are what's needed to get an answer. I also sure that rotating distance plus forward distance is close to, but not exactly, correct.

BeastRider
07-15-12, 08:42 PM
waste of time. there are MUCH better things to think about on a ride.

BikinPotter
07-15-12, 10:03 PM
261697

Kurt Erlenbach
07-16-12, 06:02 PM
We have a winner, and the answer is not nearly as intuitive as I would have thought. From hamster over at Road Cycling:

"Actually, the path of the shoe is still one of the cycloids, even though there's gearing in between. If the ratio is 6.9 to 1, it is a curtate cycloid, but, with sufficiently low gearing, it could become a prolate cycloid.

Equations of motion are

x = v*t + r*cos(w*t)
y = r*sin(w*t)

They describe a curtate cycloid when v > w*r, a regular cycloid when v = w*r, and a prolate cycloid when v < w*r. On a child's tricycle, the ratio v/wr is fixed and equal to the ratio of the wheel radius and the crankarm length. On a regular bicycle, it is variable depending on gearing."

From davids:

"The wikipedia article on cycloids shows you how to do it: http://en.wikipedia.org/wiki/Cycloid

I think the only difference is that the bike is travelling more quickly along the x-direction than a true cycloid (your feet are spinning just the same, but you're moving forward more quickly). So you have to modify the equation in the x-direction:

x = r(t-sin(t)) + (v-r)t,

where v is the average velocity in the x-direction.

y stays the same:

y = r(1-cos(t))

The arc length is now more difficult to compute, because the dx/dt derivative has an extra (v-r) term. You can plug the integral into Mathematica and it can probably do it for you (I could probably do it by hand by collecting like terms but I'm lazy as hell):

It's the integral from t = [0,2*pi] of sqrt((r*sin(t))^2 + (r*(1-cos(t))+(v-r))^2).

That integral gives you the distance your foot travels when your bike moves a distance v*2pi."

From spunky: "Doing that integral numerically (for the OPs parameters) I get d_shoe/d_bike = 1.008, so not even 1% more distance covered. Cool."

From me: "So the answer is 5,322 feet?"

And from spunky: "The answer is then 1 mile * 1.008 = 1.008 miles = 5322 ft just like you stated :) So yes, I believe this is the correct answer!"

So, what is to me a very surprising answer - When the bike goes 5280 feet at 17.3 mph with a cadence of 78 rpm, the rotating movement of your feet results in your feet traveling only 42 feet farther than the bottom bracket and the rest of the bike. A point on outside of the wheel moves 6733 feet to move the wheel 5280 feet.

I know some of you think this exercise was silly, but I found it very enlightening. Thank you for your answers.

TomD77
07-16-12, 06:58 PM
Your answer seemed counterintuitive until I thought about this way. Try a thought experiment using your hypothetical mile but with a pedal arrangement that has a pure vertical movement rather than circular. It would take 3.46 minutes @ 17.3 mph resulting in 540 pumps (2 pumps = one full cycle, down and up for 270 pump cycles) of 14 vertical inches each. Using this simplified case it is plain that the feet will have traveled 540 X the hypotenuse created by legs of 9.78 feet (5280/540) and 1.17 feet (14/12). A few seconds with a calculator yields 5318.8 feet, close enough to your answer to be a rounding error.

Kurt Erlenbach
07-16-12, 07:33 PM
Your answer seemed counterintuitive until I thought about this way. Try a thought experiment using your hypothetical mile but with a pedal arrangement that has a pure vertical movement rather than circular. It would take 3.46 minutes @ 17.3 mph resulting in 540 pumps (2 pumps = one full cycle, down and up for 270 pump cycles) of 14 vertical inches each. Using this simplified case it is plain that the feet will have traveled 540 X the hypotenuse created by legs of 9.78 feet (5280/540) and 1.17 feet (14/12). A few seconds with a calculator yields 5318.8 feet, close enough to your answer to be a rounding error.

A superb insight, Tom. In fact, after thinking about it, I feel silly that I originally went with the 2pi(175)(270) + 5280 kind of calculation. For a while I even thought that was close to right. Today while thinking about the right answer I could visualize the shoe's path as a long, low sinusoidal arc, whose path clearly was not going to be much longer than 5280 feet. Seeing it as a series of narrow right triangles obviously, in retrospect, would have been the way to ballpark it.

jalbri
07-16-12, 09:25 PM
From a former (retired) math teacher, that is a great problem! If I were still working I would have turned my AP Calculus students loose on it because it is exactly the type of problem that appears on that exam. Thank you!

z90
07-16-12, 09:39 PM
The path of the shoe is not a curtate cycloid because the chain and gearing intervene. The path a shoe traces on a child's tricycle is a curtate cycloid, when the pedal is connected directly to the drive wheel. Knowing the tire radius is 340 mm allows calculation of the gain ratio - 270.5 turns of the crank = 270.5(2pi(175mm)) =297,431 mm = 975.8 feet. So to move a point on the outside of the wheel 6733 feet, which is what's required to roll a wheel of r=340 mm one mile, requires turning the crank a distance of 975.8 feet. Somewhere in the ratio of those numbers, about 6.9 to 1, is an important fact.
Kurt, I think it effectively is a curtate cycloid if instead of using the wheel's actual diameter for the large radius you use its effective diameter. In other words, calculate how far the bike moves for one turn of the crank, then use that as the effective circumference of the wheel, and get the large radius from that.

By the way, I got as far as figuring out how many arcs there had to be, and then balked when I saw the formula for the length of the arc. Calculus was too long ago!

skycomag
07-17-12, 07:13 AM
measure th e length of the crank arm from the crank bolt center to the end of the to the pedal at the most forward part of the pedal that you are using from pt.a to pt. a . for 1 rev.
next roll the bike till your crank make one full rotation using the a/a pt. to how many feet the bike traveled.
next multiply that distance by how feet in a mile, 5280 ft.

so if the crank is 20" round x roll of the bike 10ft( 1 rev) x 1 mile/5280 ft = 10,56,000 feet traveled in 1 mile.

this my guess

DnvrFox
07-17-12, 07:14 AM
"10,56,000 feet traveled in 1 mile"

10,56,000

Is that the new - new math? :)

skycomag
07-17-12, 07:40 AM
after thinking onthis, the crank arm doesn't matter . it's how many rev the foot makes in one mile.
i'm changing my ans.
crank rotates let's say if the crank arm rotates 10 feet in 1 rev.then x this by 5280 feet in one mile =52800 feet

measure th e length of the crank arm from the crank bolt center to the end of the to the pedal at the most forward part of the pedal that you are using from pt.a to pt. a . for 1 rev.
next roll the bike till your crank make one full rotation using the a/a pt. to how many feet the bike traveled.
next multiply that distance by how feet in a mile, 5280 ft.

so if the crank is 20" round x roll of the bike 10ft( 1 rev) x 1 mile/5280 ft = 10,56,000 feet traveled in 1 mile.

this my guess

BeastRider
07-17-12, 07:45 AM
This thread has become so over-the-top ridiculous that I believe I'll unsubscribe and go ride my bike....All the while not really caring how many revolutions my foot makes in a mile.....

DnvrFox
07-17-12, 08:52 AM
Empirical evidence shows

On my mountain bike, in middle chain ring and middle cog, over a measured mile

440 revolutions

Length from center of crank to top of pedal = 7.25 inches (radius) D = 14.5

(pi x d) 3.14 x14.5 = 45.53 inches of circumference

45.53 x 440 revs = 20032 inches over one mile of pedal rotation, or 1669 feet of pedal rotation, plus the mile - 5280 feet - (if you like, but it has nothing to do with the pedal rotation) = 6949 feet.

David Bierbaum
07-17-12, 09:51 AM
We have a winner, and the answer is not nearly as intuitive as I would have thought. From hamster over at Road Cycling:

"Actually, the path of the shoe is still one of the cycloids, even though there's gearing in between. If the ratio is 6.9 to 1, it is a curtate cycloid, but, with sufficiently low gearing, it could become a prolate cycloid.

Equations of motion are

x = v*t + r*cos(w*t)
y = r*sin(w*t)

They describe a curtate cycloid when v > w*r, a regular cycloid when v = w*r, and a prolate cycloid when v < w*r. On a child's tricycle, the ratio v/wr is fixed and equal to the ratio of the wheel radius and the crankarm length. On a regular bicycle, it is variable depending on gearing."

From davids:

"The wikipedia article on cycloids shows you how to do it: http://en.wikipedia.org/wiki/Cycloid

I think the only difference is that the bike is travelling more quickly along the x-direction than a true cycloid (your feet are spinning just the same, but you're moving forward more quickly). So you have to modify the equation in the x-direction:

x = r(t-sin(t)) + (v-r)t,

where v is the average velocity in the x-direction.

y stays the same:

y = r(1-cos(t))

The arc length is now more difficult to compute, because the dx/dt derivative has an extra (v-r) term. You can plug the integral into Mathematica and it can probably do it for you (I could probably do it by hand by collecting like terms but I'm lazy as hell):

It's the integral from t = [0,2*pi] of sqrt((r*sin(t))^2 + (r*(1-cos(t))+(v-r))^2).

That integral gives you the distance your foot travels when your bike moves a distance v*2pi."

From spunky: "Doing that integral numerically (for the OPs parameters) I get d_shoe/d_bike = 1.008, so not even 1% more distance covered. Cool."

From me: "So the answer is 5,322 feet?"

And from spunky: "The answer is then 1 mile * 1.008 = 1.008 miles = 5322 ft just like you stated :) So yes, I believe this is the correct answer!"

So, what is to me a very surprising answer - When the bike goes 5280 feet at 17.3 mph with a cadence of 78 rpm, the rotating movement of your feet results in your feet traveling only 42 feet farther than the bottom bracket and the rest of the bike. A point on outside of the wheel moves 6733 feet to move the wheel 5280 feet.

I know some of you think this exercise was silly, but I found it very enlightening. Thank you for your answers.

For some reason, this explanation makes me want to find my old Spirograph set!

cranky old dude
07-17-12, 10:57 AM
We have a winner, and the answer is not nearly as intuitive as I would have thought. From hamster over at Road .......

So, what is to me a very surprising answer - When the bike goes 5280 feet at 17.3 mph with a cadence of 78 rpm, the rotating movement of your feet results in your feet traveling only 42 feet farther than the bottom bracket and the rest of the bike. A point on outside of the wheel moves 6733 feet to move the wheel 5280 feet.

I know some of you think this exercise was silly, but I found it very enlightening. Thank you for your answers.

Hey Dude! Are you resting?

Nah. I'm just waiting for my bike to catch up with me! :D

Kurt Erlenbach
07-17-12, 02:41 PM
Jalbri, let us know how the AP students do with this brain squeezer. I'd love to see this at a Math Counts competition in some form. My old Engineering Cal professor is smiling down on me and saying "I told you to listen and to show ALL of your work"

Bill

My daughter coaches the Mu Alpha Theta math competition team at Rickards HS in Tallahassee, and they just returned from their national competition in Boston (5th place in the US - Go Rickards!). I've asked her to use this problem for their competition next year. We'll see how they do. She recognized it as a curtate cycloid problem right away, but was confounded by the gearing issue, which I think is correctly dealt with by calculating a proper radius for the inner circle, then running the standard formula. A more interesting problem than I thought it was, that's for sure.

fietsbob
07-18-12, 08:37 AM
Did you include the shipping to get the shoes to your door, from exporting factory?