Wheelspin

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Hi All,

Just for laughs today on my 60km ride, I started out with the valve
stems in sync. This was a casual ride on rolling fills with fixed gear.
No front braking, and only about 4 short downhill "slowings" with the
rear. No discernible wheelslip. Any guesses as to what the difference
was at the end of the ride?

Joseph

Dear Joseph,

Unpredictable would be my guess.

The answer is "no change". Exactly the same as when I left. I wonder
how many times it went 360 deg? I wonder how much was slip and how much
was diameter difference? I suppose I could test by using 2 identical
cycle computers calibrated by weighted roll-out, one on the front, one
on the back. If the rear measures more distance it slipped.

Joseph

Dear Joseph,

Consider the practical problem of getting an "exact" rollout.

Let's say that both wheels are measured carefully at 2000 mm rollout.

(Real 700c tires are around 2100, but this makes the arithmetic
simple.)

But one wheel is actually 2000.5 mm.

The other wheel is only 1999.5 mm.

Pretend that temperature, inflation, f/r weight ratio, bouncing,
turns, and sand on the road can be ignored.

We have an error of 1 mm /2000 mm = 0.0005, or 0.05% error. That's
awfully precise.

But 60 km = 60,000 meters = 30,000 two-meter wheelspins.

And 30,000 x 0.0005 = an error of 15 two-meter wheelspins.

Cheers,

Carl Fogel

Joseph Santaniello said:

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Let's assume that the two wheels are identical in size. Assume the
path of the front wheel is a sinusoid with wavelength Lw and
amplitude A. We can approximate the distance traveled by the wheel
as

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(1) L = Lw*(1 + (pi*A/Lw)^2).

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From Bicycling Science, 3rd edition, p. 269, the amplitude of the
rear wheel is

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(2) B = A/sqrt(q + (2*pi*Lb/Lw)^2).

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N.B. My Lw is D in the book, and my Lb is Lw in the book.

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From these we can compute the difference in travel between front
and rear wheels for a given straight distance, set this equal to
the wheel circumference, and solve for the total distance. I get

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(3) Lt = Dw*Lw^2*(Lw^2 + 4*pi^2*Lb^2)/4/pi^3/A^2/Lb^2

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Assume that

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Dw = wheel diameter = 27 inch
Lb = wheel base = 36 inch
A = amplitude of wobble = 2 inch
Lw = wavelength of wobble (guess) = 240 inch

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then the distance required for the front wheel to turn one more
revolution than the rear is about 4 miles. Note, however, that
this is a fourth-power of Lw (wavelength of wobble) so your mileage
may vary.

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That is very interesting. I wonder if this can be quantified in
terms of "wasted" effort. In other words, how much time can one (if
at all) save by minimizing amplitude and wavelength of wobble. I'm
thinking more in terms of resistance from scrub or other factors
rather than increased distance covered (unless the object is
RAAM...).

I suppose you have noticed that rear tires wear substantially faster
than front ones, even though the front is where heavy braking occurs
on descents. That wear occurs from slip of tire to road and has also
been observed in motor vehicles, some of which are more heavily loaded
on the non drive wheels.

If you need more evidence, listen to light weight tubulars during
acceleration where the slip is audible in a sprint.

Jobst Brandt

In article
<[email hidden]>
,

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Hi All,

Just for laughs today on my 60km ride, I started out with the valve
stems in sync. This was a casual ride on rolling fills with fixed gear.
No front braking, and only about 4 short downhill "slowings" with the
rear. No discernible wheelslip. Any guesses as to what the difference
was at the end of the ride?

Joseph

Dear Joseph,

Unpredictable would be my guess.

The answer is "no change". Exactly the same as when I left. I wonder
how many times it went 360 deg? I wonder how much was slip and how much
was diameter difference? I suppose I could test by using 2 identical
cycle computers calibrated by weighted roll-out, one on the front, one
on the back. If the rear measures more distance it slipped.

Roll out distance will vary 20 mm within reasonable
inflation pressure range. The difference in loading of
the front tire and rear will likely produce a 1 mm
difference in roll out. Supposing the wheel has a roll
out distance of 2100 mm. Then the difference in the
number of revolutions over one kilometer is
(1/2099 - 1/2100) * 1000000 ~ 0.22 revolutions.

--
Michael Press

On 18 Jan 2007 06:30:52 -0800, [email hidden] may have

Quoted message said:

Hi All,

Just for laughs today on my 60km ride, I started out with the valve
stems in sync. This was a casual ride on rolling fills with fixed gear.
No front braking, and only about 4 short downhill "slowings" with the
rear. No discernible wheelslip. Any guesses as to what the difference
was at the end of the ride?

Without measurements in between, at a frequency great enough to see if
there was a continuous or discontinous variance pattern, the final
position doesn't really mean much. It could be a simple coincidence.

--
My email address is antispammed; pull WEEDS if replying via e-mail.
Typoes are not a bug, they're a feature.
Words processed in a facility that contains nuts.

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Joe Riel said:

Let's assume that the two wheels are identical in size.
Assume the path of the front wheel is a sinusoid with
wavelength Lw and amplitude A. We can approximate
the distance traveled by the wheel as

(1) L = Lw*(1 + (pi*A/Lw)^2).

From Bicycling Science, 3rd edition, p. 269, the
amplitude of the rear wheel is

(2) B = A/sqrt(q + (2*pi*Lb/Lw)^2).

N.B. My Lw is D in the book, and my Lb is Lw in the book.

From these we can compute the difference in travel
between front and rear wheels for a given straight distance,
set this equal to the wheel circumference, and solve for
the total distance. I get

(3) Lt = Dw*Lw^2*(Lw^2 + 4*pi^2*Lb^2)/4/pi^3/A^2/Lb^2

Assume that

Dw = wheel diameter = 27 inch
Lb = wheel base = 36 inch
A = amplitude of wobble = 2 inch
Lw = wavelength of wobble (guess) = 240 inch

then the distance required for the front wheel to
turn one more revolution than the rear is about
4 miles. Note, however, that this is a fourth-power
of Lw (wavelength of wobble) so your mileage may vary.

That is very interesting. I wonder if this can be quantified in terms
of "wasted" effort.

Did you mean the riding or the computation? (OOh, sorry, I just
couldn't resist).

Mark J.

tiborg said:

Did you get off periodically to check that they weren't about to go out
of sync by a full revolution?

My first thought. A friend/coach once quipped that I go twice as far
when we went riding (an exageration since we're riding a lot of
paceline), so I made core/pilates a priority, and mentally focus on
going straight all the time, and holding the bike stable and center.
Yesterday I rode through 2" snow/ice up a rolling logging road gaining
about 1k ft, trying to keep myself as straight as possible, the plan
being to observe in my track how badly I failed, and how aimlessly I
really do appear to meander, esp. when the going get's tough, on the
return trip. Quite an education.

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Hi All,

[snip]

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A related issue: My wife and I have different brands of cyclometers on
our bikes. I always calibrate them by rollout measurements at proper
tire pressure with proper weight on the bikes. Somehow, hers
invariably comes out measuring less mileage than mine, by a percent or
two. I have to fiddle with her calibration so she doesn't feel like
she was cheated!

- Frank Krygowski

The only fair way to calibrate is to actually ride between mile
markers. If it says you went more than a mile increase circumference
and try again, and vice-versa. Also, compare multiple mile-marker
spans and average. You can pretty well nail it that way.

Bill Westphal

Slightly different angle to this but it sounds like we should be
putting our computer sensors on the rear wheel, not the front to get a
truer distance traveled measurement. Yet either wheel will be rolling
farther than the rider actually travels on the road.

Bill Westphal said:

The only fair way to calibrate is to actually ride between mile
markers. If it says you went more than a mile increase circumference
and try again, and vice-versa. Also, compare multiple mile-marker
spans and average. You can pretty well nail it that way.

It's a bit discouraging, though, because whenever I've taken
calibration factors out of a "manual" or roll-out measurement, then
compared it to mile markers, I've always been long. You mean my 50
mile ride was really only 49 miles? Darn!

But, "I calibrated it to within 1% between mile markers on a downhill
run on U.S. 431" sounds sufficiently technical to trump anybody else's
odometer measurements.

Pat

Email address works as is.

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Hi All,

Just for laughs today on my 60km ride, I started out with the valve
stems in sync. This was a casual ride on rolling fills with fixed gear.
No front braking, and only about 4 short downhill "slowings" with the
rear. No discernible wheelslip. Any guesses as to what the difference
was at the end of the ride?

Joseph

Dear Joseph,

Unpredictable would be my guess.

The answer is "no change". Exactly the same as when I left. I wonder
how many times it went 360 deg? I wonder how much was slip and how much
was diameter difference? I suppose I could test by using 2 identical
cycle computers calibrated by weighted roll-out, one on the front, one
on the back. If the rear measures more distance it slipped.

Joseph

You could set both of the cycle computers to 10% of the actual wheel
diameter, this would multiply your difference by ten fold.

Dave