Thumb test

In article <[email hidden]>,

Quoted message said:

Tim McNamara said:

Quoted message said:

Well, the larger tyre requires less pressure for a given casing
tension. I read that here recently and am still getting my head
around it. It's also mentioned here by Jobst Brandt:

http://www.sheldonbrown.com/brandt/rim-support.html

Quoted message said:

Quoted message said:

"[...] unit casing tension is equivalent to inflation pressure
times the radius of curvature divided by pi [...]".

Quoted message said:

Quoted message said:

I was a bit surprised by this at first, but then if you think
pressure is force per unit area, if you increase the area of the
inside of the casing, you need more force for a given pressure.
Not sure if this reasoning is bogus or not though.

Quoted message said:

If you have an inflation pressure of 100 psi, a tire with more
inside surface area will have a casing under greater tension because
there are more square inches. If I grok correctly.

Casing stress is arrived upon by cutting across the circular minor
diameter of the tire (the tire is a circular cross section having no
structural belt as radial tires do to change that) and take the two
halves as solid sections being pressed apart by inflation pressure.
That gives the lineal separation force which is the casing tension.

The above mentioned formula reduces to just that. For cord stress,
adjusting for 45 degree bias ply SQR(2) gets involved but this is
about casing tension which is the same regardless of fabric structure.

From the responses of people more knowledgeable in the topic than
myself, it's clear that I don't grok this correctly. So, let me ask:

1. In terms of riding a bike, what is the significance of casing
tension? Does it affect how the tire feels, rolling resistance,
traction, etc?

2. When I am on my bike, is it casing tension or inflation pressure
that holds my rims off the ground?

Tim McNamara said:

In article <[email hidden]>,

Quoted message said:

David L. Johnson said:

> The ordinal ranking of the top ten tires was (widths are actual):
> Deda Tre Giro d'Italia (700 x 24); Clement del Mondo (700 x 28
> tubular); Michelin Pro2 Race (700 x 25); Continental Ultra Gator
> (700 x 23), Mistuboshi Trimline (650B x 37), Panaracer Pasela (700
> x 35); Clement Criterium (700 x 21); Avocet Cross (700 x 35);
> Avocet Duro (700 x 28).

Quoted message said:

I wonder when this was done. The comparisons with most of these
tires just no longer matter. Certainly neither the Clement del
Mondo nor the Criterium are available any longer, and if they were,
they would cost a bleeding fortune. They were beautiful, handmade
silk tires, but their kind has gone extinct.

Quoted message said:

Or, it may be that someone has resurrected the names, though not
the tires.

It's not the tires themselves that are interesting but the relation
of inflation pressure to RR for tires with essentially smooth tread.

http://www.sheldonbrown.com/brandt/rolling-resistance-tubular.html

The RR curves for smooth tires are nearly identical but with a
multiplier. One could take one of the curves and multiply it to
generate the others. Exceptions are those that have significant
tread profile that causes tread squirm losses. Conspicuous are the
two Specialized tires with raised center ridge.

Interestingly, the BQ study found that tire pressure had less impact on
rolling speed than other factors (assuming the results are accurate).

Quoted message said:

Tubular tires, that have less RR than all the clinchers, are out of
place by the losses caused by elastomeric rim glue while their slopes
shows that they have inherently low RR due to their thin casings and
tread, and the thin latex inner tubes. The effect of inflation
pressure (slope) for them diminishes with less lossy material.

The BQ test tried some of the tires with butyl and latex tubes. The
tires rolled slower with latex tubes. They did note that these latex
tubes were not as thin as those in tubular tires.

As I've stated though, the results of the test are different enough that
I am somewhat skeptical, and wonder about the contribution of confounds.
Someone with a better understanding of this stuff than I would have to
read the test and its methodology.

Dear Tim,

Those coasting tests are well-meant, but unlikely to be accurate.

Real-world coasting tests have far too many variables to detect subtle
differences.

One run may roll over a slightly rougher section of the apparently
uniform road surface.

Or it may miss the undetectable half-inch high, ten-foot-long hump in
the apparently flat road.

Or it may shorten the course with slightly tighter turns.

Meanwhile, one rider cannot possibly duplicate his position and wind
drag second by second for two runs, much less two different riders.

And consider the effect of an imperceptible 0.5 mph headwind changing
to an imperceptible 0.5 mph tailwind on a "perfectly still day"
between runs for one rider:

http://www.kreuzotter.de/english/espeed.htm

Hands-on tops, no rpm, no watts, -2 for a 2% downhill slope:

wind bike
speed speed
mph mph
headwind 0.5 15.1
no wind 0.0 15.6
tailwind -0.5 16.1

Hell, even the temperature, barometric pressure, and humidity changes
between repeated runs can change the speed noticeably.

Sadly, calculators or controlled spin-down tests are usually needed to
detect small differences reliably. We can probably detect gross
differences (MTB knobbies versus narrow racing tires) by just coasting
next to friends or timing repeated runs, but smaller differences are
too easily swamped by the variables.

My maximum speed coasting down the steepest part of my daily ride has
varied from 36.8 to 43.0 mph so far this week.

Repeated runs minutes apart with great care on the approach and tuck
might narrow that figure.

But given normal wind variation and the slightly different line and
tuck each time for each run, I expect that there would still be a
surprising range of speeds for a dozen runs down my hill.

Spread that out over a long morning, with time spent changing to
different tires, tubes, and inflations, and small differences are just
not likely to show up.

Cheers,

Carl Fogel

"Ben C" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Does the "thumb test" (squeezing a tyre to see if it's hard enough)
measure tyre pressure or casing tension?

The answer has already been given by several posters here but seems to have
become lost in a sea of confusion. I'll offer the following in an attempt
at clarification.

The casing tension per unit length of casing can be calculated by
multiplying the internal air pressure times the cylindrical diameter divided
by 2x the wall thickness. This is commonly known as tangential stress or
hoop stress and can be verified by any modern text on thin-walled pressure
vessels.

This means that the casing tension is 'proportional' to the cylindrical
diameter. For example if two different size tires are inflated to 100psi,
say the cylindrical diameter of one tire is 12 inches and the diameter of
another tire is 1.20 inches, (assuming equal casing thickness) the tension
in the larger diameter tire will be 10x greater than the small tire.

Yet both tires will resist thumb indentation roughly the same because the
internal air pressure is equal in both cases. Therefore it would be more
accurate to say that the thumb test measures internal air pressure directly
and measures casing tension indirectly.

JL

Tim McNamara said:

In article <[email hidden]>,

Blair P. Houghton said:

I think the 2:1 ratio of Crr's in that table and the obvious mixing
of sizes in the list indicates that unless you buy exactly those
items that were tested you will not be able to tell by looking at the
tire specs whether it will have a high Crr or a low one.

I.e., get the ones that look cool, and go improve the engine.

The conclusion of the authors was that they would choose tires based not
only on rolling resistance but also other issues, such as comfort,
durability, etc. The authors are randonneurs and one author did find
that he rode significantly faster times on a 600K and 1000K brevet
(setting personal bests and course records) while his times on his
regular tires in other brevets were about typical. He felt that was an
indication that there was something to the results of their testing.

Surely.

But how many brands of tires label the packaging with
the Crr number?

It'd be neato-keen if we could have those at hand when
Froogling.

All we have are the ones in this one list comprising
a fraction of a percent of the market.

Anyone up for picketing at ISO headquarters Saturday
morning to demand a couple of extra digits on the molded-in
markings? 23-622-0038, perhaps?

--Blair
"Go right ahead. I'll be riding."

On 2006-10-03, [email hidden] <[email hidden]> wrote:

[snip]

Quoted message said:

The most intuitive way to see what happens may be to imagine an inner
tube or tire force-filled with water. The skin of the toroid is
obviously tight. Push inward on it locally anywhere, and you know that
the water is bulging it outward elsewhere and raising the tension.

Yes, although here we've gone to a stretchy skin containing an
incompressible fluid. An inner tube is a stretchy skin, but an outer
tyre is better imagined as a non-stretchy fabric casing containing a
compressible fluid (air).

In the water-balloon, it's the skin that provides the springiness; in
the tyre it's the air itself.

On the subject of the thumb test, I did some "back of the envelope"
calculations.

If an object full of compressed air has a radius of curvature R, and I
press part of it flat to form a circular contact patch, how does force
relate to penetration distance? By penetration distance, I mean the
distance in the direction of the pressing force between the point at the
centre of the contact patch and where that point was before I pressed it
in.

The force required is given by P*pi*r**2 where P is the air pressure
inside the object, and r is the radius of the contact patch.

I use ** to mean "raised to the power of".

r**2, the radius of the contact patch squared I make 2Rd - d**2 where d
is the penetration distance.

So F = P*pi*(2Rd - d**2)

It is sufficient to see here that F is proportional to R.

Now I have read it elsewhere in r.b.t that unit casing tension is given
by PR/pi.

So R cancels out-- whatever the radius of curvature, the force required
for a given penetration distance is proportional to unit casing tension.
If you want the same penetration for a given force at lower R, you will
require a higher pressure to keep the casing tension the same.

This "model" is an approximation, but I think the thumb test is close to
"squeeze the tyre with a given test force and see how far it depresses".
In that case, the amount it depresses depends on casing tension, and you
should expect the same depression for a fatter tyre at a lower pressure
as you'd get for a narrower tyre at a higher pressure.

jim beam said:

Blair P. Houghton said:

Quoted message said:

You seem determined to miss every point--including the nail. 🙂

Now press the nose of the balloon against your hand.

Notice what happens to the tension of the rubber against
your palm. It decreases. The rubber relaxes there.
It contracts towards its uninflated size.

That isn't increasing the tension in the rubber,
it's relieving it.

isn't that because because it's in contact with your hand? as i
understand it, if your hand was entirely frictionless, tension would
remain as the skin of the balloon has to be in equilibrium.

No it doesn't*.

If it were a spherical** balloon and the only forces on
it were the air pressures on either side, it might have
a constant relation between tension and pressure over
its whole surface. But once it contacts something else
outside, that no longer applies, and a similar slacking
on the flattened side should occur.

Try it on a slippery, flat, incompressible surface, if your
hand is too compliant and sticky.

--Blair
"I'll wait while you butter
your balloons..."

* - the opposite, in fact; if the skin of your hand sticks
to the balloon, the slackening in the balloon tries to
pull your skin towards the middle, so your skin is pulling
back outward, so as to stretch the balloon, not shrink it.
With slippery hands, the balloon can slacken more.

** - The fact that it's a baloney-shaped balloon was used
only to evoke your sense memory. I assume all of us have
squeezed a few in our younger days. The differing curvatures
in the balloon imply that the pressure/tension ratios vary
around the surface.

On Tue, 3 Oct 2006 13:40:52 -0700, "JL" <[email hidden]>

Quoted message said:

"Ben C" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Does the "thumb test" (squeezing a tyre to see if it's hard enough)
measure tyre pressure or casing tension?

The answer has already been given by several posters here but seems to have
become lost in a sea of confusion. I'll offer the following in an attempt
at clarification.

The casing tension per unit length of casing can be calculated by
multiplying the internal air pressure times the cylindrical diameter divided
by 2x the wall thickness. This is commonly known as tangential stress or
hoop stress and can be verified by any modern text on thin-walled pressure
vessels.

This means that the casing tension is 'proportional' to the cylindrical
diameter. For example if two different size tires are inflated to 100psi,
say the cylindrical diameter of one tire is 12 inches and the diameter of
another tire is 1.20 inches, (assuming equal casing thickness) the tension
in the larger diameter tire will be 10x greater than the small tire.

Yet both tires will resist thumb indentation roughly the same because the
internal air pressure is equal in both cases. Therefore it would be more
accurate to say that the thumb test measures internal air pressure directly
and measures casing tension indirectly.

JL

Dear JL,

The casing is motionless, forces balanced, internal air pressure
against outer air pressure and casing tension.

To push the casing in, you push against the tensioned casing.

Even a tiny pressure causes a tiny movement and an increase in the
casing tension of flexible, indented, deformed body.

But the internal pressure does not change.

Consider a flimsy sidewall 4.00 x 18 motorcycle trials tire, rolling
down the road at 20 mph in a wheelstand contest for ten minutes. (I've
trailed along behind on two wheels, hoping that the showoffs would
break their necks.)

The rider and motorcycle total 400 pounds, all loaded on the rear
axle. Even without wheelie contests, the front end is popped up
routinely in trials.

The tire pressure is only 4 psi, measured with low-pressure gauges
that read in half-pound increments.

The tire's contact patch starts out at 4 inches wide across these 5
rows of closely spaced knobs and never widens appreciably:

http://12.158.74.10/product_images/368_0620L.jpg

Mounted on an 18-inch diameter rim, the tire stands just under 26
inches high. The sidewalls are flimsy--a flat 1-ply trials tire
squashes flat under the parked machine with no rider.

According to the raw air pressure theory, we have 400 pounds to
distribute on a 4-inch wide contact patch at 4 psi.

That would mean a 25-inch long contact patch on a tire whose diameter
is just under 26 inches.

Pushing on a tense casing measures the casing tension that holds the
rim up. The thumb moves inward because the casing tension changes
locally. The air pressure stays the same.

Or so it seems to me. I do appreciate your nice exposition of casing
tension in an undeformed section.

Cheers,

Carl Fogel

In article <[email hidden]>,

JL said:

"Ben C" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Does the "thumb test" (squeezing a tyre to see if it's hard enough)
measure tyre pressure or casing tension?

The answer has already been given by several posters here but seems
to have become lost in a sea of confusion. I'll offer the following
in an attempt at clarification.

The casing tension per unit length of casing can be calculated by
multiplying the internal air pressure times the cylindrical diameter
divided by 2x the wall thickness. This is commonly known as
tangential stress or hoop stress and can be verified by any modern
text on thin-walled pressure vessels.

This means that the casing tension is 'proportional' to the
cylindrical diameter. For example if two different size tires are
inflated to 100psi, say the cylindrical diameter of one tire is 12
inches and the diameter of another tire is 1.20 inches, (assuming
equal casing thickness) the tension in the larger diameter tire will
be 10x greater than the small tire.

Yet both tires will resist thumb indentation roughly the same because
the internal air pressure is equal in both cases. Therefore it would
be more accurate to say that the thumb test measures internal air
pressure directly and measures casing tension indirectly.

OK. Here's where I may be confunded. If I take a piece of fabric and
stretch it, the tighter I stretch it for harder it feel to my thumb.
There's no inflation pressure to feel, since there's ambient pressure on
both sides.

In article <[email hidden]>,

Quoted message said:

Those coasting tests are well-meant, but unlikely to be accurate.

Real-world coasting tests have far too many variables to detect
subtle differences.

One run may roll over a slightly rougher section of the apparently
uniform road surface.

Or it may miss the undetectable half-inch high, ten-foot-long hump in
the apparently flat road.

Or it may shorten the course with slightly tighter turns.

At risk of sounding like I'm defending the report, I'll point out what
they wrote about their methodology.

The roll down test was done on a soapbox derby track near WoodlandPark
in Seattle. The course was 245 meters long starting at a 6% grade,
decreasing to a 4.5% grade and then reducing to a 0.5% grade over the
final 184 meters. The initial 16 meters were very smooth, the remainder
was uniform, moderately rough asphalt. There were no holes, ridged or
overlays. The course was swept prior to the testing. The rider coasted
from a standing start with no pedaling, holding the same position and
wearing the same clothing for all of the runs. The bike was timed over
184 meters. Speeds were between 10.6 mph and 16.9 mph. There was a set
of reference tires used to calibrate the test to try to compensate for
changing meteorological conditions. Two independent timers were used
and the times averaged. Multiple runs were made with each tire and the
measurements averaged.

All that being said, I remain concerned about the possibility of
unidentified confounds affecting the outcome. The testers made a
laudable effort at minimizing these. I see two possible major
compounds: using stopwatches to time the rider rather than a mechanical
trigger system, and the short timed coasting distance (184 m) which was
covered in 25.3 seconds (Deda Tre) to 30.6 seconds (Nifty Swifty). The
short distance may magnify the apparent differences caused by confounds
such as human error in timing and other conditions mentioned by Carl.

In article
<[email hidden]>,

Quoted message said:

The only way to reduce the tension in a local spot on an expanded
balloon surface is to contract it.

When you push in on an inflated balloon you decrease the
area of the local patch; hence you decrease the local
tension.

Another way to see this is that the local strain energy in
an elastic structure is proportional to the local
curvature. Decreasing the curvature decreases the strain
energy.

--
Michael Press

"Tim McNamara" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

In article <[email hidden]>,

Quoted message said:

Tim McNamara said:

> Well, the larger tyre requires less pressure for a given casing
> tension. I read that here recently and am still getting my head
> around it. It's also mentioned here by Jobst Brandt:

http://www.sheldonbrown.com/brandt/rim-support.html

Quoted message said:

> "[...] unit casing tension is equivalent to inflation pressure
> times the radius of curvature divided by pi [...]".

Quoted message said:

> I was a bit surprised by this at first, but then if you think
> pressure is force per unit area, if you increase the area of the
> inside of the casing, you need more force for a given pressure.
> Not sure if this reasoning is bogus or not though.

Quoted message said:

If you have an inflation pressure of 100 psi, a tire with more
inside surface area will have a casing under greater tension
because
there are more square inches. If I grok correctly.

Casing stress is arrived upon by cutting across the circular minor
diameter of the tire (the tire is a circular cross section having no
structural belt as radial tires do to change that) and take the two
halves as solid sections being pressed apart by inflation pressure.
That gives the lineal separation force which is the casing tension.

The above mentioned formula reduces to just that. For cord stress,
adjusting for 45 degree bias ply SQR(2) gets involved but this is
about casing tension which is the same regardless of fabric
structure.

From the responses of people more knowledgeable in the topic than
myself, it's clear that I don't grok this correctly. So, let me ask:

1. In terms of riding a bike, what is the significance of casing
tension? Does it affect how the tire feels, rolling resistance,
traction, etc?

2. When I am on my bike, is it casing tension or inflation pressure
that holds my rims off the ground?

Can't resist it.....the rim is standing on the casing cords :-)

Phil H

Carl Fogel said:

Basically, an inflated elastic skin automatically occupies the
lowest possible tension shape. If you force a deformation anywhere
locally, the tension rises.

Hold it. That is incorrect. Casing tension is inversely related to
curvature so that when you load a bicycle wheel, casing tension is
reduced in the area where the cross section bellies out, the free
standing radius of curvature being smaller in that area.

It is for this reason that a small cross section tire has less casing
stress than a larger one at the same pressure. aka small tires can be
inflated to higher pressures than large ones of similar quality. I
think that is or should be common knowledge.

Quoted message said:

It might be easier to work through the geometry than devise spring
devices inside short sections. Air and rubber are sufficient.

Tires are not rubber although their tread and "air chamber" is rubber.
I use that term because there are tubeless bicycle tires. The casings
are made of essentially inelastic fiber so any model constructed of
rubber is incorrect for demonstrating stress in tires.

Quoted message said:

The simplest tension figure that we can describe is a straight line,
say a rubber string stretched tight between two fixed points.

As I said, an elastic model has little if anything to do with a
pneumatic bicycle tire.

Quoted message said:

If we push the rubber string in any direction, the force increases
the tension.

Quoted message said:

Push the tight rubber string sideways at any point and the distance
obviously increases because a straight line is the shortest distance
between two points.

Quoted message said:

Instead of pushing the rubber string sideways, grab a point on the
rubber string and pull it toward either fixed point. Tension drops
on one side, but must rise to more than the original tension on the
other side--your pull has introduced a third fixed point and
stretched a section of the original rubber string further and
tauter.

Quoted message said:

Let's get rid of those pesky fixed points.

And lets get a real tire, not a rubber band.

Quoted message said:

An endless rubber string is just a rubber band. Imagine a rubber
band laid flat to form an air-tight seal between two plates.

Imagine a tire casing made of Kevlar cord.

Quoted message said:

Pump air into the sealed space and the rubber band will
automatically expand to form a circle with even pressure all around.

Quoted message said:

... and a raft of other inappropriate examples.

Jobst Brandt

Tim McNamara said:

Quoted message said:

Quoted message said:

> Well, the larger tyre requires less pressure for a given casing
> tension. I read that here recently and am still getting my head
> around it. It's also mentioned here by Jobst Brandt:

http://www.sheldonbrown.com/brandt/rim-support.html

Quoted message said:

Quoted message said:

Quoted message said:

> "[...] unit casing tension is equivalent to inflation pressure
> times the radius of curvature divided by pi [...]".

Quoted message said:

Quoted message said:

Quoted message said:

> I was a bit surprised by this at first, but then if you think
> pressure is force per unit area, if you increase the area of the
> inside of the casing, you need more force for a given pressure.
> Not sure if this reasoning is bogus or not though.

Quoted message said:

Quoted message said:

Quoted message said:

If you have an inflation pressure of 100 psi, a tire with more
inside surface area will have a casing under greater tension
because there are more square inches. If I grok correctly.

Quoted message said:

Quoted message said:

Casing stress is arrived upon by cutting across the circular minor
diameter of the tire (the tire is a circular cross section having
no structural belt as radial tires do to change that) and take the
two halves as solid sections being pressed apart by inflation
pressure. That gives the lineal separation force which is the
casing tension.

Quoted message said:

Quoted message said:

The above mentioned formula reduces to just that. For cord stress,
adjusting for 45 degree bias ply SQR(2) gets involved but this is
about casing tension which is the same regardless of fabric
structure.

Quoted message said:

From the responses of people more knowledgeable in the topic than
myself, it's clear that I don't grok this correctly. So, let me
ask:

Quoted message said:

1. In terms of riding a bike, what is the significance of casing
tension? Does it affect how the tire feels, rolling resistance,
traction, etc?

I think this puts the cart before the horse. Casing tension depends
on two parameters, inflation pressure and tire size. I think you know
the answer to that already. The harder the tire is inflated the
harsher the ride for any cross section and the larger the cross
section the softer the ride for any inflation pressure. Let the
manufacturer decide what casing stress is acceptable. It's not the
user's problem.

Quoted message said:

2. When I am on my bike, is it casing tension or inflation pressure
that holds my rims off the ground?

Inflation pressure does that but does so by casing tension. As I
said, it's only a problem for the engineer who wonders how loads get
from the rim to the ground. The item in the FAQ explains that.

http://www.sheldonbrown.com/brandt/rim-support.html

Jobst Brandt

Tim McNamara said:

In article <[email hidden]>,

Quoted message said:

Those coasting tests are well-meant, but unlikely to be accurate.

Real-world coasting tests have far too many variables to detect
subtle differences.

One run may roll over a slightly rougher section of the apparently
uniform road surface.

Or it may miss the undetectable half-inch high, ten-foot-long hump in
the apparently flat road.

Or it may shorten the course with slightly tighter turns.

At risk of sounding like I'm defending the report, I'll point out what
they wrote about their methodology.

The roll down test was done on a soapbox derby track near WoodlandPark
in Seattle. The course was 245 meters long starting at a 6% grade,
decreasing to a 4.5% grade and then reducing to a 0.5% grade over the
final 184 meters. The initial 16 meters were very smooth, the remainder
was uniform, moderately rough asphalt. There were no holes, ridged or
overlays. The course was swept prior to the testing. The rider coasted
from a standing start with no pedaling, holding the same position and
wearing the same clothing for all of the runs. The bike was timed over
184 meters. Speeds were between 10.6 mph and 16.9 mph. There was a set
of reference tires used to calibrate the test to try to compensate for
changing meteorological conditions. Two independent timers were used
and the times averaged. Multiple runs were made with each tire and the
measurements averaged.

All that being said, I remain concerned about the possibility of
unidentified confounds affecting the outcome. The testers made a
laudable effort at minimizing these. I see two possible major
compounds: using stopwatches to time the rider rather than a mechanical
trigger system, and the short timed coasting distance (184 m) which was
covered in 25.3 seconds (Deda Tre) to 30.6 seconds (Nifty Swifty). The
short distance may magnify the apparent differences caused by confounds
such as human error in timing and other conditions mentioned by Carl.

Dear Tim,

I agree that the testers made good-faith efforts. I hope that I'd be
just as willing as you are to point them out.

But I'm still awfully skeptical. I can't, for example, see how anyone
can adjust for random 0.5 mph wind variations during numerous runs over
an open 245 meter course.

I expected the speeds "between 10.6 and 16.9 mph" to correspond to the
"30.6 seconds (Nifty Swifty)" and the "25.3 seconds (Deda Tre)"--but
they don't seem to work out when I chuck them into a spreadsheet:

184 meters = 603.67 feet

603.67 feet / 10.6 mph = 38.8 seconds (10.6 mph -- minimum?)
603.67 feet / 13.45 mph = 30.6 seconds (30 .6 seconds Nifty Swifty)
603.67 feet / 16.27 mph = 25.3 seconds (25.3 seconds Deda Tre)
603.67 feet / 16.9 mph = 24.4 seconds (16.9 mph -- maximum?)

Were those two bikes not the fastest and slowest? Or am I just
misunderstanding you?

Cheers,

Carl Fogel

<[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Dear JL,

Unfortunately, the only thing that I can think of is to stretch an air
tight trampoline over a very deep chamber and raise the air pressure.

Somehow, I don't think that this is what you're after, but maybe I'm
wrong. It's really just the tire writ large.

The trampoline will bulge upward just like a tire bulging outward.

Place the weight representing the thumb press onto the bulging
trampoline representing the inflated tire.

Let's say that the air pressure bulging the trampoline upward is 100
psi.

The stretched trampoline's tension must provide 100 psi of force to
contain the air pressure--the forces must balance.

Place a 1 pound weight on the top of the bulging trampoline, a weight
on a flat, inch-square post.

I think that the trampoline will start to indent and that the local
tension will change. Even a tiny force will be acting at nearly right
angles to the tension surface, which is so faintly curved as to be
almost flat. The air pressure will remain 100 psi at all points.

In contrast, cover the opening to a valve stem from the inside with a
1-square-inch metal plate. The air pressure will force the plate up
against the underside of the trampoline with 100 pounds of force.
There's no side tension because the plate is loose. It won't budge
until we push down on it with a 101 pound force on a thin rod stuck
down through the valve hole.

r
| o |
| d |valve stem
| | |
| | |
| | |
___| | |__________trampoline/tire inflated to 100 psi
XXXXXXXXXX
loose metal 1-square-inch plate held up by 100 psi
shouldn't move until 101 pound weight rests on rod
no side tension on loose metal plate (?)

I think that this is just repeats what I've said before. I'm trying to
see flaws in the examples, but I'm going to need help. Maybe the
diagram above will let you spot a problem, or suggest something more
along the lines of what you had in mind with the large water balloon.

Cheers,

Carl Fogel

Carl, your valve stem example with a 1 sq in plate is a very close
approximation of a track pump. If the inside barrel diameter of the pump =
1.128 in. and plunger disc surface area is exactly 1 sq in, how much
downward force on the pump handle is required on the final stroke to bring a
tire up to 100 psi? What is causing the resistance?

JL

Carl Fogel said:

I agree that the testers made good-faith efforts. I hope that I'd be
just as willing as you are to point them out.

Quoted message said:

But I'm still awfully skeptical. I can't, for example, see how
anyone can adjust for random 0.5 mph wind variations during numerous
runs over an open 245 meter course.

Quoted message said:

I expected the speeds "between 10.6 and 16.9 mph" to correspond to
the "30.6 seconds (Nifty Swifty)" and the "25.3 seconds (Deda
Tre)"--but they don't seem to work out when I chuck them into a
spreadsheet:

Quoted message said:

184 meters = 603.67 feet

Quoted message said:

603.67 feet / 10.6 mph = 38.8 seconds (10.6 mph -- minimum?)
603.67 feet / 13.45 mph = 30.6 seconds (30 .6 seconds Nifty Swifty)
603.67 feet / 16.27 mph = 25.3 seconds (25.3 seconds Deda Tre)
603.67 feet / 16.9 mph = 24.4 seconds (16.9 mph -- maximum?)

Quoted message said:

Were those two bikes not the fastest and slowest? Or am I just
misunderstanding you?

The test shown on the web site:

http://www.sheldonbrown.com/brandt/rolling-resistance-tubular.html

These were made by a company that took the effort to build an
instrumented test stand for bicycle tires using a large diameter steel
drum normally used for automobile tires. By strain gauge
measurements, RR was directly measured at a controlled speed for a
series of tires with other parameters such as inner tube used, (except
tubulars) held constant. By measuring over a range of inflation
pressures, anomalies are visible and inflation effects separated from
tread pattern effects.

Coast down tests, on the road and on small drums (as Michelin showed
at InterBike last year) are flawed from the start as you point out,
the wind effects being larger than the differences to be measured.

Jobst Brandt

On Wed, 4 Oct 2006 01:38:31 -0700, "JL" <[email hidden]>

Quoted message said:

<[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Dear JL,

Unfortunately, the only thing that I can think of is to stretch an air
tight trampoline over a very deep chamber and raise the air pressure.

Somehow, I don't think that this is what you're after, but maybe I'm
wrong. It's really just the tire writ large.

The trampoline will bulge upward just like a tire bulging outward.

Place the weight representing the thumb press onto the bulging
trampoline representing the inflated tire.

Let's say that the air pressure bulging the trampoline upward is 100
psi.

The stretched trampoline's tension must provide 100 psi of force to
contain the air pressure--the forces must balance.

Place a 1 pound weight on the top of the bulging trampoline, a weight
on a flat, inch-square post.

I think that the trampoline will start to indent and that the local
tension will change. Even a tiny force will be acting at nearly right
angles to the tension surface, which is so faintly curved as to be
almost flat. The air pressure will remain 100 psi at all points.

In contrast, cover the opening to a valve stem from the inside with a
1-square-inch metal plate. The air pressure will force the plate up
against the underside of the trampoline with 100 pounds of force.
There's no side tension because the plate is loose. It won't budge
until we push down on it with a 101 pound force on a thin rod stuck
down through the valve hole.

r
| o |
| d |valve stem
| | |
| | |
| | |
___| | |__________trampoline/tire inflated to 100 psi
XXXXXXXXXX
loose metal 1-square-inch plate held up by 100 psi
shouldn't move until 101 pound weight rests on rod
no side tension on loose metal plate (?)

I think that this is just repeats what I've said before. I'm trying to
see flaws in the examples, but I'm going to need help. Maybe the
diagram above will let you spot a problem, or suggest something more
along the lines of what you had in mind with the large water balloon.

Cheers,

Carl Fogel

Carl, your valve stem example with a 1 sq in plate is a very close
approximation of a track pump. If the inside barrel diameter of the pump =
1.128 in. and plunger disc surface area is exactly 1 sq in, how much
downward force on the pump handle is required on the final stroke to bring a
tire up to 100 psi? What is causing the resistance?

JL

Dear JL,

Air pressure--no casing tension.

No movement from 0 to 100 psi, then a foot of movement at just over
100 psi (a little friction and the slight increase in air pressure).

And the pressure remains 100 psi during that foot of travel.

If we had a ten-story pump and a pressure-relief valve to keep the
tire pressure from rising over 100 psi, the handle would move a
hundred feet under the 101 pound weight.

Of course, in a thumb press, a widening contact patch will flatten out
some to provide more resistance as area increases.

But even pressing with something flat, circular, and non-expanding
like a nail head seems to provide a trampoline-like dent, not a
contact-patch-with-the-ground flattening.

That seems to suggest that the casing is being pulled into tension
with thumb-print dent.

Here's a picture that seems to show denting, not flattening:

http://server5.theimagehosting.com/image.php?img=225a%20dent.jpg

I put the details and two pictures in a new thread, since this one is
getting tangled. I thought that an actual picture might help, but
please don't take it as being presented as proof. The idea is for
people to have something concrete and visual to refer to in their
explanations.

Does it look as if there's a lot of local stretching and tension?

The question isn't facetious--the side of a toroid is trickier than my
feeble geometry can handle. It took me a while to notice that the
sidewall where most of us do our thumb pressing is much flatter than
the curved contact patch that most of us think of.

I'm also wondering what it would look like the other way, with a thumb
press from the inside denting the sidewall outward.

I think that the difference in air pressure would produce different
curves, but that both cases would require raising the local tension.

That is, I think that the inward dent would be sharper, with the air
pressure forcing the sides toward the denter, while an outward dent
would be more gradual:

narrow inward dent? wider outward dent?

Michael Press said:

In article
<[email hidden]>,

Tim McNamara said:

The ordinal ranking of the top ten tires was (widths are actual): Deda
Tre Giro d'Italia (700 x 24); Clement del Mondo (700 x 28 tubular);
Michelin Pro2 Race (700 x 25); Continental Ultra Gator (700 x 23),
Mistuboshi Trimline (650B x 37), Panaracer Pasela (700 x 35); Clement
Criterium (700 x 21); Avocet Cross (700 x 35); Avocet Duro (700 x 28).

Conspicuously absent is the Avocet Fasgrip.

The "Avocet Duro" *is* a Fasgrip, one that is 28mm. Different widths
have different names.

In article <[email hidden]>,

Quoted message said:

Tim McNamara said:

In article <[email hidden]>,

Quoted message said:

Those coasting tests are well-meant, but unlikely to be accurate.

Real-world coasting tests have far too many variables to detect
subtle differences.

One run may roll over a slightly rougher section of the
apparently uniform road surface.

Or it may miss the undetectable half-inch high, ten-foot-long
hump in the apparently flat road.

Or it may shorten the course with slightly tighter turns.

At risk of sounding like I'm defending the report, I'll point out
what they wrote about their methodology.

The roll down test was done on a soapbox derby track near
WoodlandPark in Seattle. The course was 245 meters long starting
at a 6% grade, decreasing to a 4.5% grade and then reducing to a
0.5% grade over the final 184 meters. The initial 16 meters were
very smooth, the remainder was uniform, moderately rough asphalt.
There were no holes, ridged or overlays. The course was swept
prior to the testing. The rider coasted from a standing start with
no pedaling, holding the same position and wearing the same
clothing for all of the runs. The bike was timed over 184 meters.
Speeds were between 10.6 mph and 16.9 mph. There was a set of
reference tires used to calibrate the test to try to compensate for
changing meteorological conditions. Two independent timers were
used and the times averaged. Multiple runs were made with each
tire and the measurements averaged.

All that being said, I remain concerned about the possibility of
unidentified confounds affecting the outcome. The testers made a
laudable effort at minimizing these. I see two possible major
compounds: using stopwatches to time the rider rather than a
mechanical trigger system, and the short timed coasting distance
(184 m) which was covered in 25.3 seconds (Deda Tre) to 30.6
seconds (Nifty Swifty). The short distance may magnify the
apparent differences caused by confounds such as human error in
timing and other conditions mentioned by Carl.

Dear Tim,

I agree that the testers made good-faith efforts. I hope that I'd be
just as willing as you are to point them out.

But I'm still awfully skeptical. I can't, for example, see how anyone
can adjust for random 0.5 mph wind variations during numerous runs
over an open 245 meter course.

I expected the speeds "between 10.6 and 16.9 mph" to correspond to
the "30.6 seconds (Nifty Swifty)" and the "25.3 seconds (Deda
Tre)"--but they don't seem to work out when I chuck them into a
spreadsheet:

184 meters = 603.67 feet

603.67 feet / 10.6 mph = 38.8 seconds (10.6 mph -- minimum?)
603.67 feet / 13.45 mph = 30.6 seconds (30 .6 seconds Nifty Swifty)
603.67 feet / 16.27 mph = 25.3 seconds (25.3 seconds Deda Tre)
603.67 feet / 16.9 mph = 24.4 seconds (16.9 mph -- maximum?)

Were those two bikes not the fastest and slowest? Or am I just
misunderstanding you?

I think that those were just the highest and lowest speeds seen during
the course of the testing.

In article <[email hidden]>,

Quoted message said:

Tim McNamara said:

Quoted message said:

>> Well, the larger tyre requires less pressure for a given casing
>> tension. I read that here recently and am still getting my head
>> around it. It's also mentioned here by Jobst Brandt:

http://www.sheldonbrown.com/brandt/rim-support.html

Quoted message said:

Quoted message said:

>> "[...] unit casing tension is equivalent to inflation pressure
>> times the radius of curvature divided by pi [...]".

Quoted message said:

Quoted message said:

>> I was a bit surprised by this at first, but then if you think
>> pressure is force per unit area, if you increase the area of the
>> inside of the casing, you need more force for a given pressure.
>> Not sure if this reasoning is bogus or not though.

Quoted message said:

Quoted message said:

> If you have an inflation pressure of 100 psi, a tire with more
> inside surface area will have a casing under greater tension
> because there are more square inches. If I grok correctly.

Quoted message said:

Quoted message said:

Casing stress is arrived upon by cutting across the circular minor
diameter of the tire (the tire is a circular cross section having
no structural belt as radial tires do to change that) and take the
two halves as solid sections being pressed apart by inflation
pressure. That gives the lineal separation force which is the
casing tension.

Quoted message said:

Quoted message said:

The above mentioned formula reduces to just that. For cord
stress, adjusting for 45 degree bias ply SQR(2) gets involved but
this is about casing tension which is the same regardless of
fabric structure.

Quoted message said:

From the responses of people more knowledgeable in the topic than
myself, it's clear that I don't grok this correctly. So, let me
ask:

Quoted message said:

1. In terms of riding a bike, what is the significance of casing
tension? Does it affect how the tire feels, rolling resistance,
traction, etc?

I think this puts the cart before the horse. Casing tension depends
on two parameters, inflation pressure and tire size. I think you
know the answer to that already. The harder the tire is inflated the
harsher the ride for any cross section and the larger the cross
section the softer the ride for any inflation pressure. Let the
manufacturer decide what casing stress is acceptable. It's not the
user's problem.

Quoted message said:

2. When I am on my bike, is it casing tension or inflation pressure
that holds my rims off the ground?

Inflation pressure does that but does so by casing tension. As I
said, it's only a problem for the engineer who wonders how loads get
from the rim to the ground. The item in the FAQ explains that.

http://www.sheldonbrown.com/brandt/rim-support.html

OK, that's where I'm baffled. On the one hand, casing tension is not
the rider's problem. Other the other hand, inflation pressure holds the
rim off the ground through casing tension. But on the gripping hand,
when I squeeze the tire with my thumb I am not feeling casing tension, I
am feeling inflation pressure. I must be missing something because 2
and 2 are not equaling 4. I dunno why my brain doesn't want to wrap
around this.