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.

a mile

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.

I was thinking more cycloid like. Or a "Curtate Cycloid"?

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

Have no idea how to calculate distance.

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.

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...

Tom - OK, assume a 700 x 25 tire. The gear inches ratio could be calculated from that with the speed and cadence number.

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?

N+1. The answer always seems to work with cycling related questions.

Err . .

Does this change your calculations any?

Another wrinkle... if the outer edge of the tire goes around faster than the hub does, does the center of the hub actually move?

My answer is wrong because the distance the shoe travels on a trainer is pi(2 * 175 mm), not pi(175mm).

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

When you're out on a morning ride, you need to focus more on what you're doing.

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.

And R-Grouch, you are completely right.

The shoe does not tragel anywhere.

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.

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.

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".

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

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.

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.

Someone always drags pi into threads on this forum.