Tire Pressure & relationship to punctures

On 18 Aug 2005 20:56:38 -0700, "peter"
wrote:

But higher pressures in the range that you mention often
change the contact patch from a long, thin, oval to a
shorter (and wider) oval:

low high
pressure pressure

/XX\
XXXXXX \XX/

The ASCII diagram above is exaggerated, but illustrates the
idea. The higher pressure tire has less area, but sweeps a
wider path and therefore rolls over more debris.
The high-pressure tire sweeps a debris path 2X wide,
compared to the 1x of the lower pressure tire ...

Why do you think such an effect occurs? That would imply that
increasing the pressure would result in a more obvious bulge of the
sidewall in the vicinity of the contact patch. But the actual effect is
the opposite; at 120 psi there is much less sidewall deformation at the
bottom of the tire than there is with 80 psi. Assuming the same tire
and loading, the contact patch at 80 psi will be both longer and wider
than the one at 120 psi.

Dear Peter,

Hmmm . . . my first thought was that you're right, so I
assumed that I was thinking of smaller wheels with wider
tires versus larger wheels with narrower tires, or just
confused.

Then I decided to prove that I was wrong and amused my
neighbors by pouring a few glasses of water on my driveway
and repeatedly riding my bicycle through the puddle and into
the well-lit garage, with its smooth concrete floor.

After all, it's width that we're interested in, so a long,
wet front-tire track is easier than a stationary contact
patch.

My measurements didn't strike me as convincing, but it
didn't seem as if I was seeing the increase in width that I
expected when I dropped the pressure from 120 psi to 80 psi.

But the tracks were shrinking even as I looked at them, the
reduction is actually quite small, and I think that the
angle of the belts affects things.

Unfortunately, it's too hot and dry in Pueblo right now to
do this with much confidence. At 11 p.m., the damned wet
tire-tracks evaporated so fast that they shrank visibly
while I was kneeling on the cement floor and trying to get
my calipers on them.

The reduction in any case is going to be small. Taking an
idealized 2.0" x 0.5" rectangular front contact patch an
inch square supporting 80 pounds at 80 psi, pumping the tire
up to 120 psi would theoretically reduce it a touch less
than 20% in each direction to 1.61" x 0.41"--so I might have
been trying to measure a tenth of an inch of damp concrete.

As for the angled belt business, I rummaged around and found
a vaguely remembered passage in Sharp's 1896 "Bicycles &
Tricycles." Translating his section bb and aa, he says at
the end of section 336 that the tension on a tire the long
way (fore and aft) is twice as much as it is sideways. (This
is along the lines of the inflation of the correctly belted
tire shortening the tire's circumference and causing the
tire to constrict onto the rim.)

This is for what Sharp calls the Palmer-type tire, with
angled belts. I'm not sure if this is the same as modern
tires, nor how it affects the contact patch as pressure
rises or falls, but it suggests that the contact patch may
be trickier than I thought at first, or than I thought after
reading your reply, which struck me as plausible.

So I thought at first that I was mistaken, but now I'm not
even sure of that. If anyone has a link or some pictures
that show contact-patch width changes with pressure, I hope
that they'll be interested enough to post them.

I expect that there are more modern calculations than
Sharp's, but if anyone is interested, I'll copy them out.

Carl Fogel

Peter Rathman writes:

But higher pressures in the range that you mention often change the
contact patch from a long, thin, oval to a shorter (and wider)
oval:

low high
pressure pressure

/XX\
XXXXXX \XX/

The ASCII diagram above is exaggerated, but illustrates the
idea. The higher pressure tire has less area, but sweeps a wider
path and therefore rolls over more debris. The high-pressure tire
sweeps a debris path 2X wide, compared to the 1x of the lower
pressure tire ...

Why do you think such an effect occurs? That would imply that
increasing the pressure would result in a more obvious bulge of the
sidewall in the vicinity of the contact patch. But the actual effect
is the opposite; at 120 psi there is much less sidewall deformation
at the bottom of the tire than there is with 80 psi. Assuming the
same tire and loading, the contact patch at 80 psi will be both
longer and wider than the one at 120 psi.

I think that boils down to this subject:

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

Jobst Brandt

On Fri, 19 Aug 2005 04:30:24 GMT, Paul Kopit
wrote:

On Thu, 18 Aug 2005 19:00:25 -0400, "caaron"
wrote:

I've noticed that when I keep my tires inflated to 80 or 90 psi I almost
never get flats (maybe 1 flat in 6 months), but when I keep them at
110-120psi I get flats very frequently (maybe one flat per week).

Can anyone explain why that is, and if there is any disadvantage to keeping
them at 80-90psi? I'm riding on 700x23c michelins and I ride mostly on the
highway shoulders and city streets.

Perhaps the tubes are stretched thinner at higher pressures? Tubes
that are sized the same on the box seem to have different thickness of
rubber and different cross section size. Less rubber between the road
and air space and more flats.

Dear Paul,

That's an idea, but . . .

I don't think that the tire (much less the rim) constraining
the inner tube actually expands even 1% when the pressure
increases from 80 to 120 psi.

The problem is that once the tube meets the inside of the
tire, there just isn't any place for it to go if it wants to
become thinner.

(Inner-tube rubber is pretty much incompressible--it remains
the same thickness unless there's some place for it to go.
Unless it can squish out somewhere else, solid rubber is
about as stubborn as water when it comes to compression.)

Thicker thorn-resistant inner tubes work in two ways.

First, piercing the extra-thick rubber on the outer curve of
the tube simply requires a slightly longer thorn. That's why
Kevlar tire belts actually help a bit in thorn country, even
though the thorns go through the Kevlar threads like needles
through steel wool--the tips on a goathead thorn are short
enough that just a little more padding can help.

Second, many thorn punctures are pin-pricks that you find by
inflating the tube, holding it underwater in a sink, and
looking for slow, tiny bubbles. With a thicker
thorn-resistant tube pressed against the tire by 80 to 120
psi, the extra rubber tends to close the pin-prick--the
rubber under pressure can squish into the tiny void and fill
it if the layer of rubber is thick enough.

Carl Fogel

Carl Fogel writes:

As for the angled belt business, I rummaged around and found a
vaguely remembered passage in Sharp's 1896 "Bicycles & Tricycles."
Translating his section bb and aa, he says at the end of section 336
that the tension on a tire the long way (fore and aft) is twice as
much as it is sideways. (This is along the lines of the inflation of
the correctly belted tire shortening the tire's circumference and
causing the tire to constrict onto the rim.)

I think the term angled belt refers to casing plies.

I'm not sure what that text says but the stress in a typical bias ply
tire is analyzed in "the Bicycle Wheel":

7. Rim Compression from Spoke Tension
C = N * T / (2 * pi) Compression at rim joint
N = 36 Number of spokes
T = 900 N Tension in one spoke
C = 36 * 900 / (2 * pi) = 5156 N
8. Constriction Force of Inflated Tire
T = C - E Tire tension from inflation
P = 0.8 MPa Tire pressure
d = 25 mm = 0.025 m Diameter of tire cross section
r = d/2 = 0.0125 m Radius of tire cross section
a = 45 deg Cord angle of casing (45 typical)
A = pi * r^2 Area of tire cross section
C = P * A * 2 * tan^2(a) Consticting component
E = P * A Expanding component
A = pi * 0.0125 = 4.909e-4 m^2
C = 0.8e6 * 4.909e-4 * 2 * 1 = 785.4 N
E = 0.8e6 * 4.909e-4 = 392.7 N
T = 785.4 - 392.7 = 392.7 N
For T = 0 the cord angle must be 35.27 degrees as it is in most hoses.

This is for what Sharp calls the Palmer-type tire, with angled
belts. I'm not sure if this is the same as modern tires, nor how it
affects the contact patch as pressure rises or falls, but it
suggests that the contact patch may be trickier than I thought at
first, or than I thought after reading your reply, which struck me
as plausible.

Today we call them casing plies and they were always 45 degrees until
radial tires were introduced by Michelin. The method for manufacture
in Sharp's days was that of tubulars:

http://www.sheldonbrown.com/brandt/making-tubulars.html

So I thought at first that I was mistaken, but now I'm not even sure
of that. If anyone has a link or some pictures that show
contact-patch width changes with pressure, I hope that they'll be
interested enough to post them.

I expect that there are more modern calculations than Sharp's, but
if anyone is interested, I'll copy them out.

Jobst Brandt

Wow, thanks for all of the replies to my question. Few of my flats are from
thorns. Most are from glass or small shards of metal. From my experience
the explanation that the tire is more compliant at lower pressure and forms
around the objects resisting punctures whereas at higher pressure the
objects have more ability to penetrate the tire seems correct. The tires do
seem faster at higher pressure (perhaps due to less road resistance). But
if I have to decide between several flats a week at high pressure or an
extra mile per hour at lower pressure I'll go for the lower pressure.

Thanks for all of the replies and comments.

Chuck

wrote:
no
significant difference between old and new tire other than a puncture
re sisting name lime Armadillo or the like.

I've noticed you have a low opinion of the flat resistance of this line
of tire. My experience with them is limited to the following:

1. I've felt them in the store and noticed how incredibly stiff and
hard to bend they are, like no other tire I've ever felt.
2. I checked the manufacturer's info to see why this was so. I found
that they have not a simple Kevlar belt, but instead have filled the
weave holes in that belt with a Kevlar-based elastomer. There are also
a few other reinforcements.
3. The following test backs up the idea that they are more puncture
resistant:
http://biomech.me.unr.edu/wang/abstracts/puncture.htm
Although their analysis contains at least one shocking statistical
error, and you will notice in addition some odd statements about the
dynamics of the situation, the actual tests themselves seem
straightforward and to lead to the conclusion that these tires really
do have extra flat resistance.

"Armadillo" seems like an entirely appropriate name to me: heavy, slow,
and armored.

George King writes:

Often enough riders will report here that the new brand of tire
they are using greatly reduced the incidence of flats even when
there is no significant difference between old and new tire other
than a puncture resisting name like Armadillo or the like.

I've noticed you have a low opinion of the flat resistance of this line
of tire.

I don't recall ever mentioning that, only that some of the names are
clever ploys.

My experience with them is limited to the following:

1. I've felt them in the store and noticed how incredibly stiff and
hard to bend they are, like no other tire I've ever felt.

That would give them the highest rolling resistance of any similar
tire. I have not had one in hand or seen one on a bicycle.

2. I checked the manufacturer's info to see why this was so. I found
that they have not a simple Kevlar belt, but instead have filled the
weave holes in that belt with a Kevlar-based elastomer. There are
also a few other reinforcements.

Ooh! That sounds fishy as hell. What is a "Kevlar-based elastomer",
considering that Kevlar is a filament. And what are weave holes? How
do you impregnate a cloth with Kevlar? In any event, this may be a
good reason not to ride Armadillos that seem to approach solid tires
in their mode of operation.

3. The following test backs up the idea that they are more puncture
resistant:

http://biomech.me.unr.edu/wang/abstracts/puncture.htm

Although their analysis contains at least one shocking statistical
error, and you will notice in addition some odd statements about the
dynamics of the situation, the actual tests themselves seem
straightforward and to lead to the conclusion that these tires
really do have extra flat resistance.

This is not the way puncture tests are done in the industry. The test
bed described defies quantitative definition and repeatability and no
bicycle rider should be involved in the test.

Tire tests are normally made using a calibrated sharp object applied
with a standard force to the area of interest (tread center in this
case). I don't have much faith in the results this web page presents,
both from the described method and the lack of technical content in
the text.

"Armadillo" seems like an entirely appropriate name to me: heavy,
slow, and armored.

I don't think I would give that name to my tire.

I find interesting the extremes that this subject brings up here on
wreck.bike with some contributors riding 19mm cross section tires at
140psi and reporting great results, while at the other extreme we see
many flat tires, heavy tires, Kevlar bands, slime and even solid
tires.

Although I carry a patch kit, it is not something I open often and as
I mentioned, tires wearing to the cords don't make that change
noticeably. I still believe the people who get many flats (riding on
the same streets that I do) don't find tools and money on the road,
let alone ever see an elliptical oil ring on the pavement. Flats seem
to be more a matter of awareness and looking at the road now and then.

Jobst Brandt

On Fri, 19 Aug 2005 08:37:16 -0400, "caaron"
wrote:

Wow, thanks for all of the replies to my question. Few of my flats are from
thorns. Most are from glass or small shards of metal. From my experience
the explanation that the tire is more compliant at lower pressure and forms
around the objects resisting punctures whereas at higher pressure the
objects have more ability to penetrate the tire seems correct. The tires do
seem faster at higher pressure (perhaps due to less road resistance). But
if I have to decide between several flats a week at high pressure or an
extra mile per hour at lower pressure I'll go for the lower pressure.

Thanks for all of the replies and comments.

Chuck

Dear Chuck,

Here's a calculator that predicts the time lost due to
rolling resistance at various pressures for various tires:

http://www.analyticcycling.com/ForcesTires_Page.html

If you use the defaults and compare a premium clincher at
120 psi and 80 psi, it predicts that a pair of 80 psi tires
will cost the default rider 70 seconds in a 40 km (24.9
mile) ride that lasts 83 minutes, 20 seconds at 8 m/s (17.9
mph).

tire seconds
psi lost 40km
@ 5000 seconds
---- ---------
150 -17.5 (faster)
140 -14
130 -9
120 0 (standard)
110 13
100 26
90 43
80 70 (slower)
70 100
60 140
50 200

The calculator is based on some old metal-drum
rolling-resistance data from Jobst Brandt and doesn't
address the question of how much highly-inflated tires lose
to increased bouncing on real-world pavement.

But it suggests that we're fooling ourselves when we think
that we can sense any speed difference in rolling resistance
between 120 psi and 90 psi--the extra 43 seconds is less
than 1% of 5,000 seconds.

Carl Fogel

Carl Fogel writes:

The calculator is based on some old metal-drum rolling-resistance
data from Jobst Brandt and doesn't address the question of how much
highly-inflated tires lose to increased bouncing on real-world
pavement.

Bounce losses are base on an invalid model, one that is not random but
rather like a cattle guard taken at the worst possible speed, one in
which the next bump is encountered at the steepest incline and never
landing on the back side to get that energy back.

First, that model only works with large (cattle guard) roughness that
causes tire lift-off and only with ones of regular spacing to act as a
saw tooth hindrance. Roads are made of random roughness in all
instances except washboard where the cattle guard syndrome is
guaranteed because the divots are made by exactly the action that
causes the dread bounce losses.

But it suggests that we're fooling ourselves when we think that we
can sense any speed difference in rolling resistance between 120 psi
and 90 psi--the extra 43 seconds is less than 1% of 5,000 seconds.

I'll agree with that fully.

Jobst Brandt

I find fewer flats at higher pressure. I have attributed this to the
opposite effect of what has been described here. That is, the softer
the tire, the easier it is for a piece of glass or metal to embed in
the rubber initially, so after a few hundred rotations of getting
worked in, the object pierces the tire and punctures the tube. With a
tire made harder by higher pressure, the pieces of glass or metal sluff
off the hard tire more often than the soft tire. That's my story and
I'm sticking to it. Glenn