Re: Exploding tires II

Joe Riel said:

Quoted message said:

While the rest of your calculations are far beyond me, I'm curious
why your terminal velocity for your 200 lb rider/bike on a 10%
slope is less than predicted by the analytic cycling calculator:

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

Quoted message said:

Quoted message said:

When I plug in 0 watts for coasting, -0.10 for a 10% grade, and
90.72 kg for a 200 lb rider and bike, I get a terminal velocity of
23.61 kph, or about 52.8 mph, a 17% increase on your figure.

Quoted message said:

Quoted message said:

I hasten to add that I have no idea if this matters or if the
explanation would be obvious to someone who followed your
calculations.

Quoted message said:

Quoted message said:

It could be that the 0.5 frontal area and 0.5 drag coefficient
defaults at analytic cycling are for the kind of full tuck position
that you specifically mention is not being used in your model.

Quoted message said:

This is a good question. There are two reasons for the discrepancy.
First, I merely guessed at the 45mph (that was the only figure I did
so), so I wouldn't be surprised if there is some error. However,
the guess was partially educated, or at least thought about. The
relevant terminal velocity, as mentioned in the analysis, is not for
a cyclist descending in a full tuck, but rather for a cyclist in the
position required for continually braking, i.e. both hands on the
drops, and sitting up somewhat. This is significantly less
aerodynamic than a full tuck, so I expect that the terminal velocity
would be less. 45mph seemed a reasonable number to me.

Quoted message said:

It's easy enough to add a graph to show the max temperature for
various terminal velocities; I'll do that.

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 whith a valid model.
Certainly this doesn't occur when tucked in and descending at 45mph.

We need a better model to analyze.

Jobst Brandt
[email hidden]

Joe Riel said:

Quoted message said:

Quoted message said:

I'm curious why your terminal velocity for your 200 lb rider/bike
on a 10% slope is less than predicted by the analytic cycling
calculator:

Quoted message said:

[snipped]

Quoted message said:

Quoted message said:

This is a good question. There are two reasons for the
discrepancy. First, I merely guessed at the 45mph (that was the
only figure I did so), so I wouldn't be surprised if there is some
error. However, the guess was partially educated, or at least
thought about. The relevant terminal velocity, as mentioned in the
analysis, is not for a cyclist descending in a full tuck, but
rather for a cyclist in the position required for continually
braking, i.e. both hands on the drops, and sitting up somewhat.
This is significantly less aerodynamic than a full tuck, so I
expect that the terminal velocity would be less. 45mph seemed a
reasonable number to me.

Quoted message said:

Quoted message said:

It's easy enough to add a graph to show the max temperature for
various terminal velocities; I'll do that.

Quoted message said:

Note, from equation (12), that the maximum temperature rise is
proportional to the square root of the terminal velocity. So if the
terminal velocity were, say, 54mph rather than the 45mph used, the
computed maximum temperature rise would be 180degF*sqrt(54/45) =
197degF.

Terminal velocity probably has nothing to do with this because it
occurs at speeds where no significant energy is lost to wind drag.
Again, I think the analytic model is faulty.

Jobst Brandt
[email hidden]

Joe Riel said:

Quoted message said:

Dear Joe,

While the rest of your calculations are far beyond me, I'm
curious why your terminal velocity for your 200 lb
rider/bike on a 10% slope is less than predicted by the
analyctic cycling calculator:

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

When I plug in 0 watts for coasting, -0.10 for a 10% grade,
and 90.72 kg for a 200 lb rider and bike, I get a terminal
velocity of 23.61 kph, or about 52.8 mph, a 17% increase on
your figure.

I hasten to add that I have no idea if this matters or if
the explanation would be obvious to someone who followed
your calculations.

It could be that the 0.5 frontal area and 0.5 drag
coefficient defaults at analytic cycling are for the kind of
full tuck position that you specifically mention is not
being used in your model.

This is a good question. There are two reasons for the discrepancy.
First, I merely guessed at the 45mph (that was the only figure I did
so), so I wouldn't be surprised if there is some error. However, the
guess was partially educated, or at least thought about. The relevant
terminal velocity, as mentioned in the analysis, is not for a cyclist
descending in a full tuck, but rather for a cyclist in the position
required for contintually braking, i.e. both hands on the drops, and
sitting up somewhat. This is significantly less aerodynamic than a
full tuck, so I expect that the terminal velocity would be less.
45mph seemed a reasonable number to me.

It's easy enough to add a graph to show the max temperature for
various terminal velocities; I'll do that.

By definition the flow of air around the rim is not laminar because the
rim is constantly accelerating. Consider a point directly in front
of the axle: it is moving at a speed equal to the bicycle in a direction
pointed 45 degrees towards the ground, one half revolution later, when
this point is directly behind the axle, it is again moving at the
speed of the bicycle but pointing 45 degrees up, the velocity vector
for that point having rotated through 270 degrees.

Q_convection may be proportional to \sqrt{v}, but the flow is most
definitely not laminar.

Also, just a minior nitpick, but you state that the "s" in Fg = mgs
is the slope, when of course it is really sin(atan(slope)).

Quoted message said:

Terminal velocity probably has nothing to do with this because it
occurs at speeds where no significant energy is lost to wind drag.

I don't understand this. Terminal velocity is used only to provide
a means for determining the drag coefficient.

Hmmm [thinking]. What you are getting at, I believe, is that in the
low-speed region where maximum temperature arises (your proposition),
the drag on the bicycle does not follow the square law. That is, a
linear term becomes significant. Is that what you mean?

At 45mph/sqrt(5)=20mph (predicted worst-case speed with my model),
the braking loss is 800W and the air drag loss is 160W. What other
terms would you consider significant? At 20mph I believe that
air drag follows a square law reasonably well.

Quoted message said:

Again, I think the analytic model is faulty.

To be clear, do you mean in general, i.e. we aren't going to get a
reasonable result with any analytic model, or just this specific
model?

Joe Riel

Jim Smith said:

By definition the flow of air around the rim is not laminar because the
rim is constantly accelerating. Consider a point directly in front
of the axle: it is moving at a speed equal to the bicycle in a direction
pointed 45 degrees towards the ground, one half revolution later, when
this point is directly behind the axle, it is again moving at the
speed of the bicycle but pointing 45 degrees up, the velocity vector
for that point having rotated through 270 degrees.

I'm not sure that is sufficient to say that the flow is not locally
laminar (my term). Surely the flow across the blades of a slow speed
fan is modeled as laminar, even though the velocity vector is changing?

Quoted message said:

Q_convection may be proportional to \sqrt{v}, but the flow is most
definitely not laminar.

So what is it called? I assume it isn't turbulent. The significant
feature, of course, is whether convection is proportional to \sqrt{v},
at least for determining the worst-case velocity.

Quoted message said:

Also, just a minior nitpick, but you state that the "s" in Fg = mgs
is the slope, when of course it is really sin(atan(slope)).

Which is s/sqrt(1+s^2) and within 0.5% at a slope of 10%. A far
better approximation than the others, undoubtedly. But I'll make a
note of it. I did screw up the math there, I drew a vector diagram
and interpreted it incorrectly.

Thanks,

Joe

Tim McNamara said:

A Google search does turn up the Markal Company (under www.laco.com)
but no such product is listed on their Web site. "Heat sensing
sticker" is not a useful search phrase, either. "Self-adhesive
temperature indicator" is somewhat more successful I have no idea of
how accurately calibrated these things might be. However, anecdotal
reports of descending wet mountain roads and having the rims get hot
enough from braking to produce steam would be consistent with what
Rodriguez and Black report.

http://www.tempil.com/productdisplay.asp has a similar looking product
under the "Tempilabel" name.

Heat Indicating labels (some are reversible), crayons, cement, and lacquers:

http://www.omega.com/toc_asp/sectionSC.asp?section=F&book=temperature

--
Don McAllaster
Shrewsbury, Mass.

Quoted message said:

Quoted message said:

I have, and the tire didn't come off. I have a set of
Vredestein Fortezza Pista tires, which are spec.'ed with a
maximum pressure of 145 psi. When competing in time trials
of up to 40 km, I have been known to inflate them to maximum
pressure, or maybe a smidgen more. I also compete in roller
races where I inflate these tires to 160 psi. They are
mounted on a set of Campagnolo Shamal wheels. They have
never even given a hit of coming off.

Rims have blown apart at lower tyre pressures than this. At what tyre
pressure will a rim sustain with what amount of wear? Do rim manufacturers
publish such data? Are they too concerned about machined surfaces?

FWIW, Mavic specifies maximum recommended pressures for their rims by tire
size. Example:
http://www.mavic.com/servlet/srt/mavic/road-prod_fiche?product.id=55&lg=uk

Chris Neary
[email hidden]

"Science, freedom, beauty, adventure: what more could
you ask of life? Bicycling combined all the elements I
loved" - Adapted from a quotation by Charles Lindbergh

Chris Neary said:

Quoted message said:

Quoted message said:

I have, and the tire didn't come off. I have a set of Vredestein
Fortezza Pista tires, which are spec.'ed with a maximum pressure
of 145 psi. When competing in time trials of up to 40 km, I have
been known to inflate them to maximum pressure, or maybe a smidgen
more. I also compete in roller races where I inflate these tires
to 160 psi. They are mounted on a set of Campagnolo Shamal
wheels. They have never even given a hit of coming off.

Quoted message said:

Quoted message said:

Rims have blown apart at lower tyre pressures than this. At what
tyre pressure will a rim sustain with what amount of wear? Do rim
manufacturers publish such data? Are they too concerned about
machined surfaces?

Quoted message said:

FWIW, Mavic specifies maximum recommended pressures for their rims
by tire size. Example:

Quoted message said:

http://www.mavic.com/servlet/srt/mavic/road-prod_fiche?product.id=55&lg=uk

These numbers seem to arise from tire casing tension which increases
with cross section but is unrelated to blow-off pressure which is
entirely dependent on rim width. A free body diagram of pressure will
reveal that only air pressure, acting at the inside-to-inside width of
the rim, acts on the tire to affect lift-off force. I have no doubt
that casing failures are directly related to tire size (cross section)
but this is not related to lift-off force.

Jobst Brandt
[email hidden]

Joe Riel said:

Quoted message said:

Terminal velocity probably has nothing to do with this because it
occurs at speeds where no significant energy is lost to wind drag.

Quoted message said:

I don't understand this. Terminal velocity is used only to provide
a means for determining the drag coefficient.

Quoted message said:

Hmmm [thinking]. What you are getting at, I believe, is that in the
low-speed region where maximum temperature arises (your
proposition), the drag on the bicycle does not follow the square
law. That is, a linear term becomes significant. Is that what you
mean?

Not at all, the drag coefficient is useless because at the speeds this
occurs, wind drag is inconsequential. Rather than muddle the
computation, just assume there is no wind drag since we have little
idea of the rider weight to cross section ratio or any other
significant effects.

Quoted message said:

At 45mph/sqrt(5)=20mph (predicted worst-case speed with my model),
the braking loss is 800W and the air drag loss is 160W. What other
terms would you consider significant? At 20mph I believe that air
drag follows a square law reasonably well.

Sure it does but then blow-off problems arise on grades where such
speeds are usually not possible or at least reasonable.

Quoted message said:

Quoted message said:

Again, I think the analytic model is faulty.

Quoted message said:

To be clear, do you mean in general, i.e. we aren't going to get a
reasonable result with any analytic model, or just this specific
model?

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.

My last expert witness assessment was of a tandem that certainly never
exceeded 20mph on the route of the incident that blew the rear tire
off.

Jobst Brandt
[email hidden]

Peter wrote in message <[email hidden]>...

Quoted message said:

Trevor said:

It does not require the whole of the tyre to move. The fact that a tyre
blew off is indicative of the fact that some of the beading moved whilst
other parts stayed seated.

I think you'll need some evidence to convince others of this
theory. Why would this supposed creeping not also happen in the
vicinity of the valve stem where it would be noticed after the
blowout?

Tyre bead HAS to move to allow lift off. Only where the adhesion to the rim
is lowest will the tyre slip and so allow lift off. Low tyre pressure
ensures all of the tyre will slip and does show tyre dislocation.

Quoted message said:

Quoted message said:

Cleanliness of both rim and tyre is essential when using high pressures

to

Quoted message said:

Quoted message said:

reduce blow off situations.

My only experience with the phenomenon occurred when using only the
front brake on a small diameter wheel. Numerous other descents
have been troublefree by using both brakes and utilizing air
braking as much as possible (i.e. faster speeds).

Yes, I use air dam effect too.

Trevor

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

Quoted message said:

Trevor said:

[email hidden] wrote in message ...

Quoted message said:

Where do you descend 20% grades?

N.Wales

[snip]

Dear Trevor,

Can you tell us the names of these 20% grades in N. Wales?

Start as farm tracks off the Halkyn range.

Trevor

Frank Krygowski wrote in message <[email hidden]>...

Quoted message said:

Trevor said:

Frank Krygowski wrote in message <[email hidden]>...

Quoted message said:

I know that typical blowoffs don't rip the tire this way. Still, I
wonder if this gives further evidence that the problem is related to the
effect of temperature on the materials. I think most of us agree it's
not simply high pressure doing the damage.

Perhaps there are nicks on the hook of the rim, either escaping quality
control or subsequent damage due to mishandling, including the use of hard,
possibly steel, tyre levers. Because the tyre flexes with each rotation, a
small burr on the rim may cut the tyre casing. Eventually it rips a small
amount and when left in the boot the tubes pressure is sufficient to tear
the tyre apart.

Trevor

Why not glue wired on tyres? All the disadvantages of both tub's and wired
ons. Not true. No needle to put through tube.

Trevor
]

DRS wrote in message ...

Quoted message said:

"Trevor" <[email hidden]> wrote in message
news:[email hidden]

Quoted message said:

DRS wrote in message ...

[...]

Quoted message said:

Quoted message said:

FWIW, when Anna Meares was interviewed the other night at one point
she said she was running at 220psi, but obviously she was using
tubulars.

That pressure amazes me. Anna Meares, Australian, sprint gold medal?

Yes.

Quoted message said:

What makes it obvious tubulars were used?

Something she said about gluing the tyres to the rims.

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Notice subject heading has transferred to split rims. All your supposed
suggestion to calculating forces are irrelevant. Due to inaccuracies in
tolerance of manufacture, ongoing sample testing is really the only way to
go.

Trevor

[email hidden] wrote in message ...

Quoted message said:

Chris Neary said:

Quoted message said:

> I have, and the tire didn't come off. I have a set of Vredestein
> Fortezza Pista tires, which are spec.'ed with a maximum pressure
> of 145 psi. When competing in time trials of up to 40 km, I have
> been known to inflate them to maximum pressure, or maybe a smidgen
> more. I also compete in roller races where I inflate these tires
> to 160 psi. They are mounted on a set of Campagnolo Shamal
> wheels. They have never even given a hit of coming off.

Quoted message said:

Quoted message said:

Rims have blown apart at lower tyre pressures than this. At what
tyre pressure will a rim sustain with what amount of wear? Do rim
manufacturers publish such data? Are they too concerned about
machined surfaces?

Quoted message said:

FWIW, Mavic specifies maximum recommended pressures for their rims
by tire size. Example:

Quoted message said:

http://www.mavic.com/servlet/srt/mavic/road-prod_fiche?product.id=55&lg=uk

Quoted message said:

These numbers seem to arise from tire casing tension which increases
with cross section but is unrelated to blow-off pressure which is
entirely dependent on rim width. A free body diagram of pressure will
reveal that only air pressure, acting at the inside-to-inside width of
the rim, acts on the tire to affect lift-off force. I have no doubt
that casing failures are directly related to tire size (cross section)
but this is not related to lift-off force.

Jobst Brandt
[email hidden]

Trevor said:

Peter wrote in message <[email hidden]>...

Quoted message said:

Trevor said:

It does not require the whole of the tyre to move. The fact that a tyre
blew off is indicative of the fact that some of the beading moved whilst
other parts stayed seated.

I think you'll need some evidence to convince others of this
theory. Why would this supposed creeping not also happen in the
vicinity of the valve stem where it would be noticed after the
blowout?

Tyre bead HAS to move to allow lift off.

It certainly has to move radially outward. But your claim seemed to
be that there was an initial lateral shift of the bead moving parallel
to the rim until some bunching up occurred and then precipitated the
outward radial movement. In the only case I've experienced, the
blowout actually occurred a short time after I had braked to a stop.
I can see this happening because of increased temperature of the tire
and air since the hot rim would continue to heat the tire for awhile
after stopping. Less clear how your theory would apply since there
was no longer any force pushing the tire around the rim at the time
of the blowout.

Quoted message said:

Only where the adhesion to the rim
is lowest will the tyre slip and so allow lift off. Low tyre pressure
ensures all of the tyre will slip and does show tyre dislocation.

The tire and rim are rather uniform, so I'd expect that if one portion
of the tire has low enough adhesion to allow such slippage then the
slip would propogate to the next portion until the whole tire slips.

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.

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.

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.

So that brings us back to the question: What _does_ cause this to happen?

Quoted message said:

My last expert witness assessment was of a tandem that certainly never
exceeded 20mph on the route of the incident that blew the rear tire
off.

If we can work this out for a single bike, it won't take five minutes to
modify it for a tandem.

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

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 whith a valid model.
Certainly this doesn't occur when tucked in and descending at 45mph.

We need a better model to analyze.

Enough experience??

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.

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

Frank Krygowski said:

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.

Sounds a little premature to me. Better measure the temperature of an
actual rim under conditions which lead to blow off before making such
a declaration.

Frank Krygowski 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.

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.
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.

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). 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.

Thanks,

Joe Riel

"Trevor" <[email hidden]> wrote in message
news:[email hidden]

Quoted message said:

Why not glue wired on tyres? All the disadvantages of both tub's and
wired ons. Not true. No needle to put through tube.

Why are you top-posting all of a sudden?

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