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 a cycloid or a curtate cycloid, according to one contestant over on 50+. 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.

I'll take a quick guess:
578.03 hours x 4680rph x 1.1meters per rev = 2,975,698 meters spinning or 9,762,790ft. Add in the overall motion of the bike (52,800,000ft) and we have 62,562,790ft or 11,849 miles of movement through space.

Your shoes travel one mile for every mile traveled, as does everything else attached to you or the bike.

The frame travels one mile, but the pedals and shoes, like a point on the tires, would travel farther because they're moving up and down and fore and aft relative to the frame. Not a trick question.

Sure, but using that thinking when you ride an out-and-back you've traveled 0 miles. That's not exactly what the OP was asking.

Because or variations in cadence I don't think this is a solvable problem.

OP. You need to ride harder to push such thought out of your head...

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 about cycloids, 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?

I understand the spirit of the question but yeah, if you travelled 10k , so did your shoes.

Its like saying you drove your car 10k but your hat went farther cause you were inside waving it all around.

Or how about the chain on the bike? Or the Spokes then? Or the pedals?

Dang, I think the value of all my used parts is going down the drain with this thread.

I found a mistake in my answer. The distance the shoes travel on a trainer is pi(2* 175mm) or 1,099.5 mm per rev. Recalculating . . .

First of all, you lost a factor of 2. The distance travelled per revolution on a trainer is 2*pi*crank length.

Even with a factor of 2, your feet travel at the speed of 2*pi*0.175*78*60*0.001 = 5.146 km/h = 3.2 mph with respect to the bike. Since the bike itself is moving at 17.3 mph, the trajectory of your feet looks like an extremely stretched sine wave, whose slope never exceeds sin-1(3.2/17.3) = 10.7°, and the total distance travelled by your foot in an hour is somewhere between 17.3 miles and 17.3/cos(10.7°) = 17.6 miles.

The wikipedia article on cycloids shows you how to do it: https://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.

A Curtate Cycloid:

https://mathworld.wolfram.com/CurtateCycloid.html

Right, and that's called "complete elliptic integral of the second kind", it is not expressible in elementary functions, but, to the first order of approximation, you should get something like v*(1+x^2/4), where x is the ratio of pedal speed to bike speed (~0.185).

Tried to do the series based on formulas from wolfram and wikipedia, I keep getting 1+x^2, which seems to be too large and incompatible with my upper bound from post #10. Either I'm making a mistake somewhere or one of the formulas is wrong.

Edit: found my mistake. To get the first nonzero term in x correctly, I had to take three terms from the expansion of E(k), not two. The answer is, in fact, 1+x^2/4.

42

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.

Well said sir, but I believe that's a slightly different question

Average speed is verboten in the 41...

....nope. Funny thing....bikes ride on tires the whole time and the tires cover the whole distance.

Some yahoo is going to use this thread in their physics class as a point of discussion.

OP....really? Oh well....who am I to judge...

Cycloid, schmecloid! OP was on the right path.

The total distance his shoes travel per mile consists of distance his shoes travel in circular motion around bottom bracket plus 1 mile of road traveled by said bottom bracket in forward motion. Times two - shoes normally come in pairs. You do not even need a calculator - any calculator! - to figure this out, let alone flexing your impressive trigonometry muscles.

Class dismissed.

Everyone seems to be thinking in relation to the Earth. Factor in the Earth's rotation about the sun and the Sun's movement in the galaxy and the expansion of the universe and you'll come up with a significantly higher number.

As we say in Flanders: "Ne kus van de juffrouw en een bank naar voor".
Good answer ... the terms "relative to the earth" were indeed never mentioned in the original question.

Tape your GPS to the shoe.

The total distance is incalculable unless you also factor in gear selection which would affect the number of revolutions, non?

For instance, if I ride 1km in 53/11 and 1 km in 39/25 the number of revolutions will be very different. Therefore the total distance traveled by my shoes (times 2) would also be wildly different.

More difficult to factor in is the left right wobble.

Non.
The OP stated his average cadence and that's all you need, I believe.