A cycling math problem - What's your answer?
07-15-12 | 03:14 PM
#51
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From: The Path to Fredvana
Bikes: Long Haul Trucker 2010 , Felt Z90 2008, Rans Rocket 2001, Specialized Hardrock 1989
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.
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.
Last edited by z90; 07-15-12 at 03:26 PM.
07-15-12 | 03:22 PM
#52
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No, they can't.
Imagine a person walking through a grid of city streets. Suppose that he goes one block north, five blocks east, one block south, five blocks east, etc.
Next, imagine a bird that flies to every second corner visited by the person.
The distance traveled by the bird is less than the distance traveled by the man. He has to walk 1+5+1+5=12 blocks to complete one "period". The bird covers the same route in 2*sqrt(1*1+5*5)=10.2 blocks.
Likewise, the distance traveled by the rider's shoe with respect to the ground is less than the sum of the distance traveled by the bike + the distance traveled by the shoe with respect to the bike. This would be an exact analogy if the shoe were moving straight up and down (like on a stepper machine), but it also holds when the shoe is moving in a circle.
According to my estimates on the first page, if the bicycle travels 10,000 miles and the shoe travels 1,860 miles with respect to the bike, the total distance is approximately 10,000*(1+0.186^2/4) ~ 10,090 miles.
Imagine a person walking through a grid of city streets. Suppose that he goes one block north, five blocks east, one block south, five blocks east, etc.
Next, imagine a bird that flies to every second corner visited by the person.
The distance traveled by the bird is less than the distance traveled by the man. He has to walk 1+5+1+5=12 blocks to complete one "period". The bird covers the same route in 2*sqrt(1*1+5*5)=10.2 blocks.
Likewise, the distance traveled by the rider's shoe with respect to the ground is less than the sum of the distance traveled by the bike + the distance traveled by the shoe with respect to the bike. This would be an exact analogy if the shoe were moving straight up and down (like on a stepper machine), but it also holds when the shoe is moving in a circle.
According to my estimates on the first page, if the bicycle travels 10,000 miles and the shoe travels 1,860 miles with respect to the bike, the total distance is approximately 10,000*(1+0.186^2/4) ~ 10,090 miles.
07-15-12 | 04:24 PM
#53
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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.
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.
07-15-12 | 04:29 PM
#54
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From: The Path to Fredvana
Bikes: Long Haul Trucker 2010 , Felt Z90 2008, Rans Rocket 2001, Specialized Hardrock 1989
Ok, hamster's stair stepper did it for me. I can see now that the distance would be less than the sum of these two. So this is solvable for a fixed gear, but I think there isn't one answer for a bike that can coast and shift gears.
07-15-12 | 04:50 PM
#55
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I think hamster put it best when he said to imagine a stretched out sinusoid. Here's a plot using the OP's parameters (axis is in meters, with identical x- and y- scaling), except that I've tripled the OP's cadence to make the plot appear less stretched out (so imagine something 3x more stretched). [EDIT: now plot is using correct cadence).
In the limit of a very fast cyclist, with tons of stretching the vertical distance contributes negligibly to the total distance.
EDIT: I replaced original plot (still shown below) where I had tripled the OP's cadence with a true-to-form plot that shows the path traveled by a shoe, given the OP's parameters.
Last edited by spunkyj; 07-16-12 at 11:58 AM.
07-15-12 | 05:28 PM
#56
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That was why I said up to that point, I didnt think it was correct to just add the 5280 for the mile. There really can not be an "correct" answer because there are too many variable that can not be accounted for. As stated, no one holds exactly 78 rpms for one mile straight and if you take your foot off when you stop.
07-15-12 | 05:56 PM
#57
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That was why I said up to that point, I didnt think it was correct to just add the 5280 for the mile. There really can not be an "correct" answer because there are too many variable that can not be accounted for. As stated, no one holds exactly 78 rpms for one mile straight and if you take your foot off when you stop.
07-15-12 | 07:00 PM
#58
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From: Alpharetta, GA
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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.
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.
07-15-12 | 07:29 PM
#59
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From: Space Coast, Florida
07-15-12 | 08:24 PM
#60
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How do you get 5,322 feet? (I may have lost track of what specific question 5,322 is the answer to).
If the bike has travelled 10,000 miles, then each shoe has traveled about 10,000 * 1.008 = 10,080 miles (as per my last post). That means that each of your shoes have travelled 80 miles or 0.8% further then your bike.
Unless I made a mistake. But hamster's earlier estimate suggests not. At 17.3 mph with 78 rpm your are traveling 5.95 m/pedal revolution. that means that your foot travels 175 mm * 2 = 0.35 m vertically over 3 horizontal meters. The extra distance over just going the horizontal distance is then approximately:
sqrt(3.^2 + 0.35^2)/3. = 1.007
which is very close to the 1.008 I got from integrating over the cycloid.
EDIT: Ah, I see the original post asks for the distance traveled per shoe per 1 mile. 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!
If the bike has travelled 10,000 miles, then each shoe has traveled about 10,000 * 1.008 = 10,080 miles (as per my last post). That means that each of your shoes have travelled 80 miles or 0.8% further then your bike.
Unless I made a mistake. But hamster's earlier estimate suggests not. At 17.3 mph with 78 rpm your are traveling 5.95 m/pedal revolution. that means that your foot travels 175 mm * 2 = 0.35 m vertically over 3 horizontal meters. The extra distance over just going the horizontal distance is then approximately:
sqrt(3.^2 + 0.35^2)/3. = 1.007
which is very close to the 1.008 I got from integrating over the cycloid.
EDIT: Ah, I see the original post asks for the distance traveled per shoe per 1 mile. 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!
Last edited by spunkyj; 07-15-12 at 09:22 PM.
07-16-12 | 11:25 AM
#62
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Like the OP, the question has crossed my mind, but in a different context. As I spin up a steep hill, creeping forward at 5 mph or so despite a cadence in the 90s, I often wonder how far my feet are travelling.
Unlike the OP, I had no freakin' idea how to make the calculations.
Logically, the difference between d_feet and d_bike would be much larger in this context, but how much larger?
Unlike the OP, I had no freakin' idea how to make the calculations.
Logically, the difference between d_feet and d_bike would be much larger in this context, but how much larger?
07-16-12 | 11:43 AM
#63
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Like the OP, the question has crossed my mind, but in a different context. As I spin up a steep hill, creeping forward at 5 mph or so despite a cadence in the 90s, I often wonder how far my feet are travelling.
Unlike the OP, I had no freakin' idea how to make the calculations.
Logically, the difference between d_feet and d_bike would be much larger in this context, but how much larger?
Unlike the OP, I had no freakin' idea how to make the calculations.
Logically, the difference between d_feet and d_bike would be much larger in this context, but how much larger?
Note that the cycloid shape is much more rounded and less stretched out. In this case, integrating along the cycloid to determine its length I get:
d_shoe/d_bike = 1.16
This is very close to the value of d_shoe/d_bike=1.19 obtained when treating the circular motion of the pedals and linear motion of the bike independently. So it turns out that the OP's initial thinking gives a good approximation in the limit of a slow speed and high cadence (as one might expect).
Last edited by spunkyj; 07-16-12 at 11:47 AM.
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