Re: Exploding tires II

Joe Riel said:

Frank Krygowski said:

While I'm not willing to guarantee Joe modeled convective cooling
exactly, I'd bet that radiant cooling is much less of a factor given
the relatively low temperatures and the relatively low emissivity of
aluminum. But I admit, I'm not expert at that stuff.

Nor am I. In fact, I'll guarantee the converse, I didn't model it
*exactly* 8-). Radiation cooling may not be insignicant; however, I'm
having a hard time selecting an appropriate emissivity. I'm guessing
that the sidewalls should be that of sandblasted aluminum (0.2), but
the spoked surface (is there a better term?) might be quite a bit
higher since the emissivity of anodized aluminum is around 0.8.

Hmmm. Yes, that's higher than I would have guessed.

Quoted message said:

See, for example, http://www.infrared-thermography.com/material.htm.
Any suggestions will be appreciated. With those values, a rim
at 125C with ambient at 25C will radiate about 40W.

Still not very large.

Quoted message said:

Quoted message said:

In any case, if we say he should add radiant cooling to his
calculations, and if we say he should treat the convection as forced
and turbulent, then the fundamental problem is made worse: the
temperatures, and thus pressures, that he calculates will be even
lower. And his calculated pressure is not enough to solely account
for tires blowing off rims.

I agree here. Adding radiation and turbulent cooling will only
decrease the temperature. However, one effect that might increase
it somewhat is a non-isothermal surface. I've written the
differential equations that model this, including both radiation
and forced convection and will solve them numerically today.

What alloy of aluminum is typically used for rims? Specifically, I'm
looking for its thermal conductivity, density, and coefficient of
specific heat (Cp).

There are many extrudable alloys - 2024, 5086, 6061, 6063, 7075 and many
others - but I don't know which go into rims.

I'd bet MATWEB would have much of the info on those alloys. You could
get an idea how variable those properties are between alloys. Density
would be almost uniform, and maybe Cp. Conductivity, I'll bet, varies.

Also, what is a reasonable number for the

Quoted message said:

cross-sectional area of the material? From a web search I'm currently
using 82mm^2; however it wasn't clear what rim that was for. I'd
prefer a number for an MA-40 type rim. An electronic drawing of the
cross section, so that I can include it in the paper, would be ideal.

I once figured cross sectional areas of a few rims, using two methods.
First is just to take the rim mass and assume the center of mass is at
the bead seat diameter.

mass = area * (circumference at center of mass) * density

and again, density's pretty uniform, about 2.77 g/cm^3

My second method was to use a hacksawed section of rim as a rubber stamp
and break the resulting cross section down into simple shapes. Tedious.

The two methods agreed acceptably. A Super Champion clincher rim (Mod
58? Not sure) gave 0.152 square inches by one method, 0.153 by another.
That's 9.84 *10^-5 m^2, or 98.4 mm^2. That's a 530 gram rim.

--
Frank Krygowski [To reply, remove rodent and vegetable dot com.
Substitute cc dot ysu dot
edu]

Frank Krygowski said:

There are many extrudable alloys - 2024, 5086, 6061, 6063, 7075 and
many others - but I don't know which go into rims.

I'd bet MATWEB would have much of the info on those alloys. You could
get an idea how variable those properties are between alloys. Density
would be almost uniform, and maybe Cp. Conductivity, I'll bet, varies.

Thanks for the reference, I wasn't aware of that site <www.matweb.com>.
Checking those gives

Type k (W/m/K)
------- ----------
2024-XX 121-193
5086-XX 125
6061-XX 154-180
6063-XX 193-218
7075-XX 130-155

The range depends on the XX (O, T1, T2, etc) which I assume indicates
heat treatment. What would be appropriate for a rim?

Quoted message said:

Also, what is a reasonable number for the

Quoted message said:

cross-sectional area of the material? From a web search I'm currently
using 82mm^2; however it wasn't clear what rim that was for. I'd
prefer a number for an MA-40 type rim. An electronic drawing of the
cross section, so that I can include it in the paper, would be ideal.

I once figured cross sectional areas of a few rims, using two
methods. First is just to take the rim mass and assume the center of
mass is at the bead seat diameter.

mass = area * (circumference at center of mass) * density

and again, density's pretty uniform, about 2.77 g/cm^3

My second method was to use a hacksawed section of rim as a rubber
stamp and break the resulting cross section down into simple shapes.
Tedious.

I have a planimeter, so measuring the area is straightforward,
however, I don't have a rim lying around to sacrifice---I tossed
my old rims when I moved, alas.

Quoted message said:

The two methods agreed acceptably. A Super Champion clincher rim (Mod
58? Not sure) gave 0.152 square inches by one method, 0.153 by
another. That's 9.84 *10^-5 m^2, or 98.4 mm^2. That's a 530 gram rim.

Joe

Frank Krygowski said:

Quoted message said:

This does not live up to reality. My experience with tire
blow-offs is on roads where one must brake nearly continuously
because the road is curvy or rough, as in trail, so more than 20mph
is illusory. I don't think contributors to this thread have
experienced enough failures of this kind if any, to come up with a
valid model. Certainly this doesn't occur when tucked in and
descending at 45mph.

Quoted message said:

Quoted message said:

We need a better model to analyze.

Quoted message said:

Enough experience??

Quoted message said:

The model is based on known physics. Maybe the model can be
improved, but I don't see that firsthand experience is necessary to
do the physics. After all, nobody seems to doubt your accounts of
it happening. We're just trying to figure out _why_ it happens.
And I think it's clear by now that it's not mere overpressure doing
it.

To approach the subject reasonably, those parameters that are
understood to be involved need to be included, others not. To do this
is easier if the analyst understands the circumstances and dynamic
effects under which the phenomenon occurs. That is what I mean by
experience. For instance, I have had a tire blow off after braking to
a stop on a moderate ~8% grade, noticing a lump-lump-lump... as the
bicycle approached zero speed. This was a relatively short heating
period so it is evident to me, from this and the steaming rim event,
that we haven't gotten much closer to a good model.

The idea that the tire bead gets soft occurred to me but I later
rejected it because I have been in many situations where high rim
temperatures occurred only for a short duration, not long enough to
heat the air in the tube. There was no residual effect over many
miles in which the tires wore out while repeating the rim heating. I
get to review this every summer in riding over many mountain roads and
have reduced the tire blow-off to air temperature in the tire alone.

Jobst Brandt
[email hidden]

Joe Riel said:

Quoted message said:

Quoted message said:

If you want to calculate this it should include radiation cooling
(that is independent of speed but temperature dependent, rim and
ambient) and forced convention (which is greater than at ground
speed because the top of the rim is going twice that fast) based
possibly on a 200lb load on a (let's say) 15% grade on one wheel
brake.

Quoted message said:

Quoted message said:

While I'm not willing to guarantee Joe modeled convective cooling
exactly, I'd bet that radiant cooling is much less of a factor
given the relatively low temperatures and the relatively low
emissivity of aluminum. But I admit, I'm not expert at that stuff.

Quoted message said:

Nor am I. In fact, I'll guarantee the converse, I didn't model it
*exactly* 8-). Radiation cooling may not be insignificant; however,
I'm having a hard time selecting an appropriate emissivity. I'm
guessing that the sidewalls should be that of sand blasted aluminum
(0.2), but the spoked surface (is there a better term?) might be
quite a bit higher since the emissivity of anodized aluminum is
around 0.8. See, for example,

http://www.infrared-thermography.com/material.htm

Quoted message said:

Any suggestions will be appreciated. With those values, a rim at
125C with ambient at 25C will radiate about 40W.

At these temperatures, I believe emissivity is closer to 1 or a black
body. Emissivity is temperature dependent and at low temperatures,
where long wavelength infrared is the carrier, most surfaces are black
bodies. This is more obvious with sunlight heating where absorptivity
is drastically different from emissivity.

Quoted message said:

Quoted message said:

In any case, if we say he should add radiant cooling to his
calculations, and if we say he should treat the convection as forced
and turbulent, then the fundamental problem is made worse: the
temperatures, and thus pressures, that he calculates will be even
lower. And his calculated pressure is not enough to solely account
for tires blowing off rims.

Quoted message said:

I agree here. Adding radiation and turbulent cooling will only
decrease the temperature. However, one effect that might increase
it somewhat is a non-isothermal surface. I've written the
differential equations that model this, including both radiation
and forced convection and will solve them numerically today.

Quoted message said:

What alloy of aluminum is typically used for rims? Specifically,
I'm looking for its thermal conductivity, density, and coefficient
of specific heat (Cp). Also, what is a reasonable number for the
cross-sectional area of the material? From a web search I'm
currently using 82mm^2; however it wasn't clear what rim that was
for. I'd prefer a number for an MA-40 type rim. An electronic
drawing of the cross section, so that I can include it in the paper,
would be ideal.

I don't know what the alloy is but 2024 or so would probably be OK,
the differences being insignificant once we get to the temperature.
The two measurements of interest are a plot of rim temperature and air
temperature in the tire. If we had a transmitting sensor in the tire
and a optical pyrometer on the rim I would gladly purchase the needed
tubes and explode them on a steep hill on the back wheel. We have a
few good test courses around here. I think Joaquin Rd (aka walking)
Los Trancos Woods, might do the job as would Sierra Road near Alum
Rock Park (SJ).

http://tinyurl.com/46w8z (Joaquin)
http://tinyurl.com/49vyn (Sierra)

Jobst Brandt
[email hidden]

Quoted message said:

The idea that the tire bead gets soft occurred to me but I later
rejected it because I have been in many situations where high rim
temperatures occurred only for a short duration, not long enough to
heat the air in the tube. There was no residual effect over many
miles in which the tires wore out while repeating the rim heating.
I get to review this every summer in riding over many mountain roads
and have reduced the tire blow-off to air temperature in the tire
alone.

My high school physics class was a long time ago, so I can't answer
this question. But it seems to cut to the chase.

Let's say your tires have 100 psi in them at rest- say 70 deg F. You
descend, oh let's say the Col de Tende and the rims heat up from the
braking. Let's suppose the air temperature inside the inner tubes
reaches 200 def F. How much has the pressure increased? Enough to
blow the tire off?

FWIW, Rivendell Bicycle Works published an informal attempt to blow
tires off rims, in Issue 31 (p. 9). They got a tire (Rivendell brand
tire on a Bontrager rim) up to 200 psi without it blowing off the rim
(they snipped the bead in 10 places and repeated, but it blew off the
rim at 120 psi. That's neither here nor there, though).

As I mentioned in another post, the use of temperature-indicating
labels on tandem rims has shown rim temps as high as 250 deg F, and
temperature increases to 240 deg F in twenty seconds of hard braking.
With enough time, rims at those temperatures could significantly
increase the temperature of the air in the tubes, I would think.

Tim McNamara said:

Quoted message said:

The idea that the tire bead gets soft occurred to me but I later
rejected it because I have been in many situations where high rim
temperatures occurred only for a short duration, not long enough to
heat the air in the tube. There was no residual effect over many
miles in which the tires wore out while repeating the rim heating.
I get to review this every summer in riding over many mountain roads
and have reduced the tire blow-off to air temperature in the tire
alone.

My high school physics class was a long time ago, so I can't answer
this question. But it seems to cut to the chase.

Let's say your tires have 100 psi in them at rest- say 70 deg F. You
descend, oh let's say the Col de Tende and the rims heat up from the
braking. Let's suppose the air temperature inside the inner tubes
reaches 200 def F. How much has the pressure increased? Enough to
blow the tire off?

FWIW, Rivendell Bicycle Works published an informal attempt to blow
tires off rims, in Issue 31 (p. 9). They got a tire (Rivendell brand
tire on a Bontrager rim) up to 200 psi without it blowing off the rim
(they snipped the bead in 10 places and repeated, but it blew off the
rim at 120 psi. That's neither here nor there, though).

As I mentioned in another post, the use of temperature-indicating
labels on tandem rims has shown rim temps as high as 250 deg F, and
temperature increases to 240 deg F in twenty seconds of hard braking.
With enough time, rims at those temperatures could significantly
increase the temperature of the air in the tubes, I would think.

Dear Tim,

Going from 70 degrees F to 200 degrees F inside the tube is
going from 294K to 366K, a roughly 25% increase in absolute
temperature, so it's reasonable to assume that the 100 psi
would increase to only 125 psi.

http://members.aol.com/javawizard/tture.html

To double a 70F 100 psi pressure to 200 psi would seem to
require the air in the tube to rise from 70F to an unlikely
600F.

Of course, some of us run higher pressures--my chubby 700c x
26 touring bike tires are usually about 120 psi because
that's what they end up with after the air compressor has.
its way with them.

A 25% increase from 120 psi would bring my tires up to 150
psi. I'd think twice about riding with 25% more pressure
than usual on a rim hot enough to boil water.

Assuming that Jobst is quenching his hot rims at 5280 feet,
the rims are at least warmed up to 201 degrees F:

http://www.csgnetwork.com/h2oboilcalc.html

This page has links to a number of other odd calculators. At
10,000 feet, Jobst's hissing rims are still at least 192
degrees F.

Carl Fogel

Joe Riel said:

...
I have a planimeter, so measuring the area is straightforward,
however, I don't have a rim lying around to sacrifice---I tossed
my old rims when I moved, alas....

The Velocity [1] site has cross-sections as GIF images that you could
enlarge and print out.

[1] <http://www.velocityusa.com/>.

--
Tom Sherman

Tim McNamara said:

Quoted message said:

The idea that the tire bead gets soft occurred to me but I later
rejected it because I have been in many situations where high rim
temperatures occurred only for a short duration, not long enough to
heat the air in the tube. There was no residual effect over many
miles in which the tires wore out while repeating the rim heating.
I get to review this every summer in riding over many mountain
roads and have reduced the tire blow-off to air temperature in the
tire alone.

Quoted message said:

My high school physics class was a long time ago, so I can't answer
this question. But it seems to cut to the chase.

Quoted message said:

Let's say your tires have 100 psi in them at rest- say 70 deg F.
You descend, oh let's say the Col de Tende and the rims heat up from
the braking. Let's suppose the air temperature inside the inner
tubes reaches 200 deg F. How much has the pressure increased?
Enough to blow the tire off?

Quoted message said:

FWIW, Rivendell Bicycle Works published an informal attempt to blow
tires off rims, in Issue 31 (p. 9). They got a tire (Rivendell
brand tire on a Bontrager rim) up to 200 psi without it blowing off
the rim (they snipped the bead in 10 places and repeated, but it
blew off the rim at 120 psi. That's neither here nor there,
though).

I didn't see how these tests were conducted but I am certain that I
did not reach 200psi in the blow-offs that I experienced.

Quoted message said:

As I mentioned in another post, the use of temperature-indicating
labels on tandem rims has shown rim temps as high as 250 deg F, and
temperature increases to 240 deg F in twenty seconds of hard
braking. With enough time, rims at those temperatures could
significantly increase the temperature of the air in the tubes, I
would think.

I think I have made clear why I think this can only be resolved by a
road test. I propose using a thermocouple attached to the bed of the
rim and a pressure sensor connected to a valve stem extender. Data
collection vs. time could be collected by a flash memory on a PC-board
in the rear wheel so that no moving contacts or telemetry is
necessary. The timer can be started at the top of the hill and
recording continues while braking with only the rear brake until the
bottom or when the tire blows off, at which time the front brake can
stop the bicycle.

What would be known is the gradient and the constant speed at which
the test was performed (the rider controls that with his bicycle
speedometer and writes it on the data card that contains the
temperature and pressure profile vs time. The burst pressure is then
obvious if it occurred.

Jobst Brandt
[email hidden]

Quoted message said:

... I have had a tire blow off after braking to
a stop on a moderate ~8% grade, noticing a lump-lump-lump... as the
bicycle approached zero speed. This was a relatively short heating
period so it is evident to me, from this and the steaming rim event,
that we haven't gotten much closer to a good model.

The idea that the tire bead gets soft occurred to me but I later
rejected it because I have been in many situations where high rim
temperatures occurred only for a short duration, not long enough to
heat the air in the tube. There was no residual effect over many
miles in which the tires wore out while repeating the rim heating. I
get to review this every summer in riding over many mountain roads and
have reduced the tire blow-off to air temperature in the tire alone.

^^^ ^^^^^^^^^^^

I'm assuming you mean, the higher air pressure caused by the higher air
temperature. Correct me if I'm wrong in that assumption.

But if you believe that, you have to somehow reconcile that the
resulting air pressure during a descent is much less than what tires
have routinely been shown to hold, when cold. How do you explain this
discrepancy?

You also have to find some significant fault with Joe's nicely-done
calculations. So far, it looks as if any plausible lack of accuracy
would tend to overstate, not understate, the air pressure.

If you're basing your guess on the idea that the blowoff happens only
after longer-term braking, there are other plausible explanations
besides gradual heating of the air. For example, longer-term contact
between the hot rim and the tire bead could change the properties of the
rubber - say, perhaps softening it and allowing it to distort.
Simultaneously, longer-term braking force could gradually induce
distortion and circumferential creep in the tire at the bead seat.
Perhaps the combination of these factors eventually unseats the bead.
(I wouldn't expect circumferential creep, if it exists, to be absolutely
uniform around the rim; perhaps the motion "piles up" at one spot, where
the blowout occurs.)

You note that you've blown tires on steep, very slow speed descents.
Note that in those cases, braking force on the tire is high, so any
tendency to creep may be high. Power input to the rim is not
particularly high, and air pressure certainly doesn't get as high as in
the room temperature tests that have been described as successfully passed.

--
--------------------+
Frank Krygowski [To reply, remove rodent and vegetable dot com,
replace with cc.ysu dot edu]

Tom Sherman said:

Quoted message said:

...
I have a planimeter, so measuring the area is straightforward,
however, I don't have a rim lying around to sacrifice---I tossed
my old rims when I moved, alas...

Stop that! We also needn't get into metallurgy or tire emissivity.
What may not be apparent is that the main entry and exit of heat to
the air in the tire is through the rim, the rest of the tube being
fairly well insulated in the tire.

Quoted message said:

The Velocity [1] site has cross-sections as GIF images that you
could enlarge and print out.

Quoted message said:

[1] <http://www.velocityusa.com/>.

This does not require damaging any rims. I merely costs a tube for
each successful blow-off. As I just described, this is not difficult
to instrument. It just takes some time and a little money. I'll
gladly do the riding because I know ho benign such an event can be if
not cornering hard.

With all instrumentation being in the rear wheel, this becomes fairly
simple. There must be someone here connected with a laboratory where
such measurements are routine.

Jobst Brandt
[email hidden]

Quoted message said:

Tim McNamara said:

Quoted message said:

The idea that the tire bead gets soft occurred to me but I later
rejected it because I have been in many situations where high rim
temperatures occurred only for a short duration, not long enough
to heat the air in the tube. There was no residual effect over
many miles in which the tires wore out while repeating the rim
heating. I get to review this every summer in riding over many
mountain roads and have reduced the tire blow-off to air
temperature in the tire alone.

My high school physics class was a long time ago, so I can't answer
this question. But it seems to cut to the chase.

Let's say your tires have 100 psi in them at rest- say 70 deg F.
You descend, oh let's say the Col de Tende and the rims heat up from
the braking. Let's suppose the air temperature inside the inner
tubes reaches 200 def F. How much has the pressure increased?
Enough to blow the tire off?

<snip>

Quoted message said:

Going from 70 degrees F to 200 degrees F inside the tube is going
from 294K to 366K, a roughly 25% increase in absolute temperature,
so it's reasonable to assume that the 100 psi would increase to only
125 psi.

Well, now you have me surfing the net in an effort to conclusively
display my ignorance. Consider it done. According to

http://maps.unomaha.edu/Peake/3510/pressure.html,

Charles's Law states that "If volume is constant the pressure of a gas
is directly related to temperature. Pressure increases with increase
in temperature at rate of 1/273 of value at 0 C for each 1deg. C
change in temp. p=po+1/273 tpo where p is pressure po is pressure at 0
C and t is change in temp in celsius. Another version of the equation
is p=po ( 1+ 1/273 t)"

So, if we posit a 65 deg C increase in temperature, then... hmmmm. Is
the formula valid if we're starting not at 0 deg C but at roughly 22
deg C? Does

p= 100 psi * (1 + (1/273 * 65)) = 100 * 1.26 = 126 psi

work out correctly? That hardly seem like enough in itself to blow a
tire off a rim. Or is the factor different than 1/273 if the starting
temperature isn't 0 C?

Quoted message said:

To double a 70F 100 psi pressure to 200 psi would seem to require
the air in the tube to rise from 70F to an unlikely 600F.

My algebra never was good and is worse now for decades of disuse.
Danged embarrassing at times like this. I can't even solve Charles's
law for t. Sheesh. I get t = 1.992 when seeking a p = 200 psi
starting with a po = 100 psi. That doesn't work. What was it
Heinlein said about people who can't do algebra?

Anyway, it is seeming to me that increased air pressure from heating
is not the only component of the phenomenon. Perhaps someone literate
in math can enlighten me.

Frank Krygowski said:

Quoted message said:

... I have had a tire blow off after braking to a stop on a
moderate ~8% grade, noticing a lump-lump-lump... as the bicycle
approached zero speed. This was a relatively short heating period
so it is evident to me, from this and the steaming rim event, that
we haven't gotten much closer to a good model.

Quoted message said:

Quoted message said:

The idea that the tire bead gets soft occurred to me but I later
rejected it because I have been in many situations where high rim
temperatures occurred only for a short duration, not long enough to
heat the air in the tube. There was no residual effect over many
miles in which the tires wore out while repeating the rim heating.
I get to review this every summer in riding over many mountain
roads and have reduced the tire blow-off to air temperature in the
tire alone.

Quoted message said:

I'm assuming you mean, the higher air pressure caused by the higher
air temperature. Correct me if I'm wrong in that assumption.

What else? Let me say it this way. The tire blow-off occurs from
excess pressure in the tire cause by brake heating of the air in the
tube. The principal heat path is the bed of the rim while heat
exchange through the tire casing is relatively low. Therefore the
heating and cooling path is between tube and rim.

Quoted message said:

But if you believe that, you have to somehow reconcile that the
resulting air pressure during a descent is much less than what tires
have routinely been shown to hold, when cold. How do you explain
this discrepancy?

These "routine" tests were, as far as has been reported, static tests.
I think adding a bit of bead squirm as the casing alters its tension
angle in the load affected zone would help dislodge the tire.

Quoted message said:

You also have to find some significant fault with Joe's nicely-done
calculations. So far, it looks as if any plausible lack of accuracy
would tend to overstate, not understate, the air pressure.

I don't car because I know it occurs and that I have repeatedly had
rims amazingly hot without tire failure and no subsequent failure.
This indicates to me that the tire does not gradually creep off, there
being no way for it to retreat from such a position.

Quoted message said:

If you're basing your guess on the idea that the blowoff happens
only after longer-term braking, there are other plausible
explanations besides gradual heating of the air.

I never said that. These blow-offs have occurred after a short hard
braking at times.

Quoted message said:

For example, longer-term contact between the hot rim and the tire
bead could change the properties of the rubber - say, perhaps
softening it and allowing it to distort.

As I said, if this were the case, how would it recover once misshapen
and close to separation. You questioned why experience might be
valuable in defining a failure model. I think your line of
questioning makes my perception of that problem clear. You have not
observed enough of these to detect a trend and possible cause. This
makes analysis difficult... but as you can see in this thread, I have
proposed a method by which this can be resolved experimentally to most
peoples satisfaction.

Quoted message said:

Simultaneously, longer-term braking force could gradually induce
distortion and circumferential creep in the tire at the bead seat.
Perhaps the combination of these factors eventually unseats the
bead. (I wouldn't expect circumferential creep, if it exists, to be
absolutely uniform around the rim; perhaps the motion "piles up" at
one spot, where the blowout occurs.)

I hope you aren't suggesting that longitudinal creep of the tire on the
rim has any play in this. I have not has a tilted valve stem since I
got off tubulars, where this was typical on descents.

Quoted message said:

You note that you've blown tires on steep, very slow speed descents.
Note that in those cases, braking force on the tire is high, so any
tendency to creep may be high. Power input to the rim is not
particularly high, and air pressure certainly doesn't get as high as
in the room temperature tests that have been described as
successfully passed.

I guess you missed it but I have had and observed in others failure
after a short steep descent that required near stopping speed in a
curve or for obstacles. You are creating a model on a false
hypothesis in this case.

Jobst Brandt
[email hidden]

Quoted message said:

This does not require damaging any rims. I merely costs a tube for
each successful blow-off. As I just described, this is not difficult
to instrument. It just takes some time and a little money. I'll
gladly do the riding because I know ho benign such an event can be if
not cornering hard.

With all instrumentation being in the rear wheel, this becomes fairly
simple. There must be someone here connected with a laboratory where
such measurements are routine.

Why not use the temperature-sensitive strips mentioned earlier, like
those at:
http://www.omega.com/Temperature/pdf/TL-10.pdf

You get 10 strips for $13.50 that each indicate within a 10 degree F
band how hot the rim got. The ones that go from 190 F - 280 F seem
likely to cover the range of interest. That would provide
useful data without requiring any special instrumentation.

Tim McNamara said:

Quoted message said:

Tim McNamara said:

[email hidden] writes:

> The idea that the tire bead gets soft occurred to me but I later
> rejected it because I have been in many situations where high rim
> temperatures occurred only for a short duration, not long enough
> to heat the air in the tube. There was no residual effect over
> many miles in which the tires wore out while repeating the rim
> heating. I get to review this every summer in riding over many
> mountain roads and have reduced the tire blow-off to air
> temperature in the tire alone.

My high school physics class was a long time ago, so I can't answer
this question. But it seems to cut to the chase.

Let's say your tires have 100 psi in them at rest- say 70 deg F.
You descend, oh let's say the Col de Tende and the rims heat up from
the braking. Let's suppose the air temperature inside the inner
tubes reaches 200 def F. How much has the pressure increased?
Enough to blow the tire off?

<snip>

Quoted message said:

Going from 70 degrees F to 200 degrees F inside the tube is going
from 294K to 366K, a roughly 25% increase in absolute temperature,
so it's reasonable to assume that the 100 psi would increase to only
125 psi.

Well, now you have me surfing the net in an effort to conclusively
display my ignorance. Consider it done. According to

http://maps.unomaha.edu/Peake/3510/pressure.html,

Charles's Law states that "If volume is constant the pressure of a gas
is directly related to temperature. Pressure increases with increase
in temperature at rate of 1/273 of value at 0 C for each 1deg. C
change in temp. p=po+1/273 tpo where p is pressure po is pressure at 0
C and t is change in temp in celsius. Another version of the equation
is p=po ( 1+ 1/273 t)"

So, if we posit a 65 deg C increase in temperature, then... hmmmm. Is
the formula valid if we're starting not at 0 deg C but at roughly 22
deg C? Does

p= 100 psi * (1 + (1/273 * 65)) = 100 * 1.26 = 126 psi

work out correctly? That hardly seem like enough in itself to blow a
tire off a rim. Or is the factor different than 1/273 if the starting
temperature isn't 0 C?

Quoted message said:

To double a 70F 100 psi pressure to 200 psi would seem to require
the air in the tube to rise from 70F to an unlikely 600F.

My algebra never was good and is worse now for decades of disuse.
Danged embarrassing at times like this. I can't even solve Charles's
law for t. Sheesh. I get t = 1.992 when seeking a p = 200 psi
starting with a po = 100 psi. That doesn't work. What was it
Heinlein said about people who can't do algebra?

Anyway, it is seeming to me that increased air pressure from heating
is not the only component of the phenomenon. Perhaps someone literate
in math can enlighten me.

Dear Tim,

Ya got me.

I used English major physics, which has the advantage of
simplicity, if not accuracy, and assumed (remember my "if"😉
that pressure per square inch and absolute temperature for a
confined gas are roughly proportional.

That is, double the absolute temperature in degrees K and
the pressure per square inch also doubles. I kinda-sorta
think that this is what Charles Law is saying, 0 degrees C
being about 273 degrees K and a degree on either scale being
the same.

I daringly rounded a degree or two

Secretly, I long to use degrees Rankine, which is to
Farenheit as Kelvin is to Celsius--460 degrees Rankine is
about 0 degrees F.

Here's a page that not only offers to convert Rankine and
the other scales, but even calculates temperatures below
absolute zero, a trick sometimes helpful when my theories
stray even further from reality than normal:

http://cryowwwebber.gsfc.nasa.gov/introduction/Temp_Calc.html

Carl Fogel

In article <[email hidden]>,
[email hidden] says...

Quoted message said:

What would be known is the gradient and the constant speed at which
the test was performed (the rider controls that with his bicycle
speedometer and writes it on the data card that contains the
temperature and pressure profile vs time.

A small GPS on the handlebars or in the jersey pocket will give a
tracklog with time and position at each log point. If you have an
aneroid assist GPS the altitude in the tracklog and waypoints will be
pretty accurate too. A waypoint at the start and blow off would complete
the record (to allow easy verification of the altitudes from a topo map).

Not necessary but nice.

Bruce Graham

Quoted message said:

Frank Krygowski writes:

.... you have to somehow reconcile that the

Quoted message said:

resulting air pressure during a descent is much less than what tires
have routinely been shown to hold, when cold. How do you explain
this discrepancy?

These "routine" tests were, as far as has been reported, static tests.
I think adding a bit of bead squirm as the casing alters its tension
angle in the load affected zone would help dislodge the tire.

I agree with this. IOW, I'm guessing this effect of the braking force
is important.

Quoted message said:

Quoted message said:

You also have to find some significant fault with Joe's nicely-done
calculations. So far, it looks as if any plausible lack of accuracy
would tend to overstate, not understate, the air pressure.

I don't car because I know it occurs and that I have repeatedly had
rims amazingly hot without tire failure and no subsequent failure.

This sounds suspiciously like "I don't care what your stress
calculations show, I know I stretched my chain by pedaling so hard on
that last hill."

Sure sounds like myth and lore!

Quoted message said:

This indicates to me that the tire does not gradually creep off, there
being no way for it to retreat from such a position.

Quoted message said:

If you're basing your guess on the idea that the blowoff happens
only after longer-term braking, there are other plausible
explanations besides gradual heating of the air.

I never said that. These blow-offs have occurred after a short hard
braking at times.

OK. Was there time for the heat generated in the rim to transfer to the
air in the tire and raise it's pressure above, say, 120 psi? Joe's
computation was steady state. The situation you just described would
have to look at transients - and once again, that means Joe's would be
conservative (i.e. overstating the air pressure) in yet another way.

Quoted message said:

Quoted message said:

For example, longer-term contact between the hot rim and the tire
bead could change the properties of the rubber - say, perhaps
softening it and allowing it to distort.

As I said, if this were the case, how would it recover once misshapen
and close to separation. You questioned why experience might be
valuable in defining a failure model. I think your line of
questioning makes my perception of that problem clear. You have not
observed enough of these to detect a trend and possible cause. This
makes analysis difficult... but as you can see in this thread, I have
proposed a method by which this can be resolved experimentally to most
peoples satisfaction.

I think that experiment would be valuable. I'm not allowed to ship you
our infra-red thermometer, though! ;-)

Quoted message said:

Quoted message said:

Simultaneously, longer-term braking force could gradually induce
distortion and circumferential creep in the tire at the bead seat.
Perhaps the combination of these factors eventually unseats the
bead. (I wouldn't expect circumferential creep, if it exists, to be
absolutely uniform around the rim; perhaps the motion "piles up" at
one spot, where the blowout occurs.)

I hope you aren't suggesting that longitudinal creep of the tire on the
rim has any play in this. I have not has a tilted valve stem since I
got off tubulars, where this was typical on descents.

I don't think any longitudinal creep has to involve the whole tire. As
a temporary simplification, imagine a tire's bead heated, softened and
expanded in length (i.e. circumference). [Hold on, bear with me.]
Braking force on any particular contact patch occurs once per revolution.

If at one spot around the circumference, the tire/rim contact is slighly
less strong due to manufacturing tolerances, that inch or two of tire
might creep a slight amount while the rest of the bead remained in
place. If the tire bead rubber is softened by heat, the tire might lose
its grip and be unable to retain the air pressure.

To me, that's easiest to visualize if we've got a longitudinally
expanded tire bead - like a rug with a wrinkle - but if the material
properties are badly affected by heat, it might happen without that
[doubtful] expansion of bead length.

I'm curious - have you blowouts being more, or less, common with steel
vs. kevlar bead wires?

Quoted message said:

Quoted message said:

You note that you've blown tires on steep, very slow speed descents.
Note that in those cases, braking force on the tire is high, so any
tendency to creep may be high. Power input to the rim is not
particularly high, and air pressure certainly doesn't get as high as
in the room temperature tests that have been described as
successfully passed.

I guess you missed it but I have had and observed in others failure
after a short steep descent that required near stopping speed in a
curve or for obstacles. You are creating a model on a false
hypothesis in this case.

I'm looking for a model that involves more than the simplistic notion
that air pressure _alone_ is the cause, because I think we've proven the
necessary air pressure isn't there. I haven't seen a rebuttal for that
last idea.

Hmmm. If it _were_ just over-pressure caused by heat, then the solution
could be to deflate tires before a descent. Alternately, we could
design pressure-limiting relief valves for the valve core, that would
bleed off excess air - and inconveniently, stop to pump tires once
things had cooled.

If the problem were related to tire squirm, initial deflation might make
the problem worse.

--
Frank Krygowski [To reply, remove rodent and vegetable dot com.
Substitute cc dot ysu dot
edu]

Tim McNamara said:

Quoted message said:

Tim McNamara said:

[email hidden] writes:

> The idea that the tire bead gets soft occurred to me but I later
> rejected it because I have been in many situations where high rim
> temperatures occurred only for a short duration, not long enough
> to heat the air in the tube. There was no residual effect over
> many miles in which the tires wore out while repeating the rim
> heating. I get to review this every summer in riding over many
> mountain roads and have reduced the tire blow-off to air
> temperature in the tire alone.

My high school physics class was a long time ago, so I can't answer
this question. But it seems to cut to the chase.

Let's say your tires have 100 psi in them at rest- say 70 deg F.
You descend, oh let's say the Col de Tende and the rims heat up from
the braking. Let's suppose the air temperature inside the inner
tubes reaches 200 def F. How much has the pressure increased?
Enough to blow the tire off?

<snip>

Quoted message said:

Going from 70 degrees F to 200 degrees F inside the tube is going
from 294K to 366K, a roughly 25% increase in absolute temperature,
so it's reasonable to assume that the 100 psi would increase to only
125 psi.

Well, now you have me surfing the net in an effort to conclusively
display my ignorance. Consider it done. According to

http://maps.unomaha.edu/Peake/3510/pressure.html,

Charles's Law states that "If volume is constant the pressure of a gas
is directly related to temperature. Pressure increases with increase
in temperature at rate of 1/273 of value at 0 C for each 1deg. C
change in temp. p=po+1/273 tpo where p is pressure po is pressure at 0
C and t is change in temp in celsius. Another version of the equation
is p=po ( 1+ 1/273 t)"

So, if we posit a 65 deg C increase in temperature, then... hmmmm. Is
the formula valid if we're starting not at 0 deg C but at roughly 22
deg C? Does

p= 100 psi * (1 + (1/273 * 65)) = 100 * 1.26 = 126 psi

work out correctly? That hardly seem like enough in itself to blow a
tire off a rim. Or is the factor different than 1/273 if the starting
temperature isn't 0 C?

You are safe assuming a value of 1/273. As you may know, this is where
the idea of "absolute zero" got started. If this constant remains
constant, then at -273 C ones gas would have zero volume. This doesn't
happen, so we know it isn't always 1/273, but for all but the very
coldest temperatures we don't have to worry about it.

Quoted message said:

Quoted message said:

To double a 70F 100 psi pressure to 200 psi would seem to require
the air in the tube to rise from 70F to an unlikely 600F.

My algebra never was good and is worse now for decades of disuse.
Danged embarrassing at times like this. I can't even solve Charles's
law for t. Sheesh. I get t = 1.992 when seeking a p = 200 psi
starting with a po = 100 psi. That doesn't work. What was it
Heinlein said about people who can't do algebra?

Maybe simpler to look at ideal gas law:
pressure * volume = constant * temperature
This makes it clear that a doubling of temperature leads to a doubling
of pressure, increasing temperature by a factor of 1.25 raises pressure
by a factor of 1.25 etc. Just remember you have to measure temperature
above absolute zero.

Tim McNamara said:

Charles's Law states that "If volume is constant the pressure of a gas
is directly related to temperature. Pressure increases with increase
in temperature at rate of 1/273 of value at 0 C for each 1deg. C
change in temp. p=po+1/273 tpo where p is pressure po is pressure at 0
C and t is change in temp in celsius. Another version of the equation
is p=po ( 1+ 1/273 t)"

To me, those verbal descriptions are much harder to understand than the
formulas. I've given some below.

Quoted message said:

So, if we posit a 65 deg C increase in temperature, then... hmmmm. Is
the formula valid if we're starting not at 0 deg C but at roughly 22
deg C? Does

p= 100 psi * (1 + (1/273 * 65)) = 100 * 1.26 = 126 psi

work out correctly?

Yes, that's correct.

Quoted message said:

That hardly seem like enough in itself to blow a
tire off a rim.

Exactly.

Quoted message said:

[CF:]

Quoted message said:

To double a 70F 100 psi pressure to 200 psi would seem to require
the air in the tube to rise from 70F to an unlikely 600F.

Actually, about 532 deg F. But still highly unlikely.

Quoted message said:

My algebra never was good and is worse now for decades of disuse.
Danged embarrassing at times like this.

Here are the equations solved out so you can plug in pressures as
meaured by the gage, in psi, and temperatures in degrees Fahrenheit:

To find the pressure due to a certain temperature change:

P2= (T2+460)*(P1+14.7)/(T1+460) - 14.7

To find the temperature required for a certain pressure change:

T2 = (P2+14.7)*(T1+460)/(P1+14.7) - 460

Quoted message said:

Anyway, it is seeming to me that increased air pressure from heating
is not the only component of the phenomenon.

That's certainly what I think.

--
Frank Krygowski [To reply, remove rodent and vegetable dot com.
Substitute cc dot ysu dot
edu]

Peter Rathman said:

Quoted message said:

This does not require damaging any rims. I merely costs a tube for
each successful blow-off. As I just described, this is not
difficult to instrument. It just takes some time and a little
money. I'll gladly do the riding because I know ho benign such an
event can be if not cornering hard.

Quoted message said:

Quoted message said:

With all instrumentation being in the rear wheel, this becomes
fairly simple. There must be someone here connected with a
laboratory where such measurements are routine.

Quoted message said:

Why not use the temperature-sensitive strips mentioned earlier, like
those at:

http://www.omega.com/Temperature/pdf/TL-10.pdf

I guess you didn't see the method I propose because the heat strip
doesn't have the resolution required to answer the blow-off questions
and we know nothing about the pressure or the rate of temperature
rise.

Quoted message said:

You get 10 strips for $13.50 that each indicate within a 10 degree F
band how hot the rim got. The ones that go from 190 F - 280 F seem
likely to cover the range of interest. That would provide useful
data without requiring any special instrumentation.

OK, do it and let us know what you found.

Jobst Brandt
[email hidden]

Frank Krygowski said:

Quoted message said:

Quoted message said:

... you have to somehow reconcile that the resulting air pressure
during a descent is much less than what tires have routinely been
shown to hold, when cold. How do you explain this discrepancy?

Quoted message said:

Quoted message said:

These "routine" tests were, as far as has been reported, static
tests. I think adding a bit of bead squirm as the casing alters
its tension angle in the load affected zone would help dislodge the
tire.

Quoted message said:

I agree with this. IOW, I'm guessing this effect of the braking
force is important.

Quoted message said:

Quoted message said:

Quoted message said:

You also have to find some significant fault with Joe's
nicely-done calculations. So far, it looks as if any plausible
lack of accuracy would tend to overstate, not understate, the air
pressure.

Quoted message said:

Quoted message said:

I don't care because I know it occurs and that I have repeatedly
had rims amazingly hot without tire failure and no subsequent
failure.

Quoted message said:

This sounds suspiciously like "I don't care what your stress
calculations show, I know I stretched my chain by pedaling so hard
on that last hill."

No it doesn't. It mans I don't care what analytical results we get if
it is not shown to cause a blow-off. That they occur and easily, has
been demonstrated to my satisfaction. To calculate that it does not
occur doesn't help. If you review the results of calculated pressures
and that tires withstand pressures up to 160 psi, a pressure not
attained in the analytical model, then that is proof that it doesn't
occur in so many words.

Quoted message said:

Sure sounds like myth and lore!

Well it isn't reality.

Quoted message said:

Quoted message said:

This indicates to me that the tire does not gradually creep off,
there being no way for it to retreat from such a position.

Quoted message said:

Quoted message said:

Quoted message said:

If you're basing your guess on the idea that the blowoff happens
only after longer-term braking, there are other plausible
explanations besides gradual heating of the air.

Quoted message said:

Quoted message said:

I never said that. These blow-offs have occurred after a short
hard braking at times.

Quoted message said:

OK. Was there time for the heat generated in the rim to transfer to
the air in the tire and raise it's pressure above, say, 120 psi?
Joe's computation was steady state. The situation you just
described would have to look at transients - and once again, that
means Joe's would be conservative (i.e. overstating the air
pressure) in yet another way.

That is exactly my point. We need a graph of rim temperature and tore
pressure vs time and possibly distance.

Quoted message said:

Quoted message said:

Quoted message said:

For example, longer-term contact between the hot rim and the tire
bead could change the properties of the rubber - say, perhaps
softening it and allowing it to distort.

Quoted message said:

Quoted message said:

As I said, if this were the case, how would it recover once
misshapen and close to separation. You questioned why experience
might be valuable in defining a failure model. I think your line
of questioning makes my perception of that problem clear. You have
not observed enough of these to detect a trend and possible cause.
This makes analysis difficult... but as you can see in this thread,
I have proposed a method by which this can be resolved
experimentally to most peoples satisfaction.

Quoted message said:

I think that experiment would be valuable. I'm not allowed to ship
you our infra-red thermometer, though!

Quoted message said:

Quoted message said:

Quoted message said:

Simultaneously, longer-term braking force could gradually induce
distortion and circumferential creep in the tire at the bead seat.
Perhaps the combination of these factors eventually unseats the
bead. (I wouldn't expect circumferential creep, if it exists, to
be absolutely uniform around the rim; perhaps the motion "piles
up" at one spot, where the blowout occurs.)

Quoted message said:

Quoted message said:

I hope you aren't suggesting that longitudinal creep of the tire on
the rim has any play in this. I have not has a tilted valve stem
since I got off tubulars, where this was typical on descents.

Quoted message said:

I don't think any longitudinal creep has to involve the whole tire.
As a temporary simplification, imagine a tire's bead heated,
softened and expanded in length (i.e. circumference). [Hold on,
bear with me.] Braking force on any particular contact patch occurs
once per revolution.

If there is any at all, it would involve the whole tire, the tire
having a steel bead wire as a core. There has been no indication of
longitudinal creep in any tires that I have seen. The stems on blown
tubes were radial in position.

Quoted message said:

If at one spot around the circumference, the tire/rim contact is
slightly less strong due to manufacturing tolerances, that inch or
two of tire might creep a slight amount while the rest of the bead
remained in place. If the tire bead rubber is softened by heat, the
tire might lose its grip and be unable to retain the air pressure.

Quoted message said:

To me, that's easiest to visualize if we've got a longitudinally
expanded tire bead - like a rug with a wrinkle - but if the material
properties are badly affected by heat, it might happen without that
[doubtful] expansion of bead length.

Quoted message said:

I'm curious - have you blowouts being more, or less, common with
steel vs. Kevlar bead wires?

I don't know. I use steel bead tires except when a spare must be used
on a tour where I can't readily get another tire.

Quoted message said:

Quoted message said:

Quoted message said:

You note that you've blown tires on steep, very slow speed
descents. Note that in those cases, braking force on the tire is
high, so any tendency to creep may be high. Power input to the
rim is not particularly high, and air pressure certainly doesn't
get as high as in the room temperature tests that have been
described as successfully passed.

Quoted message said:

Quoted message said:

I guess you missed it but I have had and observed in others failure
after a short steep descent that required near stopping speed in a
curve or for obstacles. You are creating a model on a false
hypothesis in this case.

Quoted message said:

I'm looking for a model that involves more than the simplistic
notion that air pressure _alone_ is the cause, because I think we've
proven the necessary air pressure isn't there. I haven't seen a
rebuttal for that last idea.

I explained why I believe pressure is the main and probably only cause
and how this can be ascertained. I don't know what you find
"simplistic" about that.

Quoted message said:

Hmmm. If it _were_ just over-pressure caused by heat, then the
solution could be to deflate tires before a descent. Alternately,
we could design pressure-limiting relief valves for the valve core,
that would bleed off excess air - and inconveniently, stop to pump
tires once things had cooled.

Yes. how often would you like to do that? As you can see, I do much
touring in mountains and would therefore do a lot of pumping and
letting out of air. A pressure relief valve is a difficult device to
tune properly and not one that I would like to sell to the average
rider considering liability.

Quoted message said:

If the problem were related to tire squirm, initial deflation might
make the problem worse.

I doubt it.

Jobst Brandt
[email hidden]