Hard braking down hill blowouts

[Michael Press]

In article <[email protected]>,
Ben C <[email protected]> wrote:

> On 2008-03-31, Michael Press <[email protected]> wrote:
> > In article <[email protected]>,
> > Ben Kaufman <[email protected]> wrote:
> >
> >> On Sat, 29 Mar 2008 21:39:19 -0700 (PDT), Hank <[email protected]> wrote:
> >> >Why 10mph? Even if it's winding, surely 15 doesn't seem too out of the
> >> >realm of sanity, IMO. I guess I'd have to see the road before passing
> >> >judgment on that, but 10mph just seems you're trading one danger for
> >> >another, if you're having frequent blowouts.
> >>
> >> I could handle the road at 20 mph but my assumption has been that
> >> higher speeds generate greater rim temperatures due to the square of
> >> the velocity for the kinetic energy formula. If the hill wasn't so
> >> steep then taking it faster would probably require a lot less use of
> >> the brakes but based upon other safer hills that take me to about 40
> >> mph, I would say this should be a significantly faster decent.
> >
> > I explained this several times. The amount of energy dissipated
> > is the same for a given descent and initial and terminal speeds.
>
> Yes. But the power at which the rim is heated and at which it cools is
> not.
>
> > Forget kinetic energy as a function of speed.
>
> Why? It seems to me it is quite important.

You did not say why.

> > Concentrate on total energy dissipated. The goal is to reduce the
> > heating of the tire bead and the air in the tube. If the rim is always
> > hot as with constant braking you are guaranteed to heat the bead an
> > inflation air.
>
> If you heat the rim at the same rate at which it is also cooling, then
> you're going to be all right aren't you?

You describe an equilibrium where the rim remains hot.
If the rim remains hot, it heats the bead and inflation air.
I advocate a scheme where the rim gets very hot quickly,
then release the brakes and cool the rim. Most of the heat
dissipation is into the air, and not into the tire.
Steady braking maximizes heat transfer into the tire.

> > If you allow for free runs downhill with no braking, the rim will cool
> > quickly, very much more quickly if you allow the free runs to have
> > maximum speed before braking.
>
> Yes. But it might not be safe to descend with no braking.

Where did I advocate not braking?
I remember advocating coming to a full stop
and assessing the condition of the rims and tires.


> > Heavy braking decreases the amount of time to heat the rim.
> > Then you get off the brakes and allow the rim to cool. The
> > energy dissipation rate for a hot rim increases with the
> > temperature, so you are spending more time dissipating heat to
> > the air at a high rate; dissipating heat that will not go to
> > heating the bead and inflation air.
>
> I agree that non-continuous braking may help.
>
> > Another thing you have not considered is that when coasting downhill
> > at speed you dissipate more energy to wind drag the faster you go; the
> > rate of dissipation increases more than linearly with speed.
>
> This is all good advice, but it doesn't answer the question: is it
> better to brake continuously to maintain 10mph or continuously to
> maintain 20mph on a hill with a coasting speed of 50mph?
>
> If my tyre blows off either way, do I get further before it blows off if
> I go at 10mph or if I go at 20mph?

Continuous braking is the method most likely to overheat tires.

--
Michael Press

 

[Michael Press]

In article <[email protected]>,
Ben Kaufman <[email protected]> wrote:

> On Mon, 31 Mar 2008 11:29:49 -0700, Michael Press <[email protected]> wrote:
>
> >In article <[email protected]>,
> > Ben Kaufman <[email protected]> wrote:
> >
> >> On Sat, 29 Mar 2008 21:39:19 -0700 (PDT), Hank <[email protected]> wrote:
> >> >Why 10mph? Even if it's winding, surely 15 doesn't seem too out of the
> >> >realm of sanity, IMO. I guess I'd have to see the road before passing
> >> >judgment on that, but 10mph just seems you're trading one danger for
> >> >another, if you're having frequent blowouts.
> >>
> >> I could handle the road at 20 mph but my assumption has been that higher speeds
> >> generate greater rim temperatures due to the square of the velocity for the
> >> kinetic energy formula. If the hill wasn't so steep then taking it faster would
> >> probably require a lot less use of the brakes but based upon other safer hills
> >> that take me to about 40 mph, I would say this should be a significantly faster
> >> decent.
> >
> >I explained this several times. The amount of energy dissipated
> >is the same for a given descent and initial and terminal speeds.
> >Forget kinetic energy as a function of speed. Concentrate on
> >total energy dissipated. The goal is to reduce the heating of
> >the tire bead and the air in the tube. If the rim is always hot
> >as with constant braking you are guaranteed to heat the bead
> >an inflation air. If you allow for free runs downhill with no
> >braking, the rim will cool quickly, very much more quickly if
> >you allow the free runs to have maximum speed before braking.
> >Heavy braking decreases the amount of time to heat the rim.
> >Then you get off the brakes and allow the rim to cool. The
> >energy dissipation rate for a hot rim increases with the
> >temperature, so you are spending more time dissipating heat to
> >the air at a high rate; dissipating heat that will not go to
> >heating the bead and inflation air. Another thing you have not
> >considered is that when coasting downhill at speed you dissipate
> >more energy to wind drag the faster you go; the rate of dissipation
> >increases more than linearly with speed.
>
> As I have explained in previous messages, I cannot to go very fast due to
> concerns about cars.

Of course. Stay safe. Stop and check the condition of the rim and tires.

--
Michael Press

 

[Tom Sherman]

Bill Bushnell wrote:
> [email protected] wrote:
>> I instrumented a wheel with pressure sensor and temperature
>> thermocouple to monitor a blow-off but was unsuccessful getting the
>> tire to separate even though temperature reached 150 degC and pressure
>> 125 psi. The mechanism for blow-off is still unclear but it seams the
>> bead material softens and creeps of the hooked bead of the rim.
>
> I stopped getting blowoffs when I reduced cold inflation pressure from 85 to 75
> psi (Ritchey Tom Slick, 559x36 on Ritchey rim). I believe that the likelihood of
> blowoff may be especially sensitive to inflation pressure because even though
> I've been riding around lately with an extra 45 lbs. on the bike, I still don't
> get blowoffs, even on the roads where they are likely to occur (Hicks, Vista
> Verde/Ramona, etc.)
>

Any past blowoffs on the (ISO 406-mm?/451-mm?) front wheel?

--
Tom Sherman - Holstein-Friesland Bovinia
The weather is here, wish you were beautiful

 

[Garry Lee]

I've had about 3 such blowoffs. One in Gran Canaria, one on the
descent from Neuschwannstein Castle in Germany and one somewhere in
Ireland. Can't remember where.

 

[Ben Kaufman]

On Mon, 31 Mar 2008 18:05:04 -0700 (PDT), Ron George <[email protected]>
wrote:

>On Mar 27, 10:06 pm, Ben Kaufman <spaXm-mXe-anXd-paXy-5000-
>[email protected]> wrote:
>> Is it normal to have blow-outs on a road bike from prolonged hard braking going
>> about one mile down a steep hill or should superior wheels and tires be able
>> to deal with the generated heat? I have an old panasonic but keep the tires and
>> tubes up to date (PerfomanceBike GT2 Kevlar, rated at 105 lbs, 26TPI, which are
>> not the best in the world but a heck of a lot better than what my LBS sold me.
>> It is a 27 inch rim and I could not find any better quality tires). But is it
>> the tire/wheel quality at issue? I have been thinking about getting a new bike
>> rather than trying to upgrade this one for a number of reasons (I don't think
>> it's even possible to switch to the current wheel size) but the blow-outs are
>> my biggest justification of expenditure to my wife.
>>
>> Thanks.
>>
>> Ben
>
>Ben,
>
>Here are some tubulars popping off wheels. Famous accidents of the
>Tour de France. http://cozybeehive.blogspot.com/2007/11/tubulars-exploding-and-peeling-off.html
>
>This is an interesting discussion, I'm watching where its leading.
>
>Ron
>http://cozybeehive.blogspot.com

Thanks, especially the point that this racer has fear of a repeat after the
first time, and his motivation is to win a race, mine is to just get down
safely.

Ben

 

[[email protected]]

On Mar 31, 4:47 pm, Ben C <[email protected]> wrote:
> On 2008-03-31, Michael Press <[email protected]> wrote:
>
>
>
> > In article <[email protected]>,
> > Ben Kaufman <[email protected]> wrote:
>
> >> On Sat, 29 Mar 2008 21:39:19 -0700 (PDT), Hank <[email protected]> wrote:
> >> >Why 10mph? Even if it's winding, surely 15 doesn't seem too out of the
> >> >realm of sanity, IMO. I guess I'd have to see the road before passing
> >> >judgment on that, but 10mph just seems you're trading one danger for
> >> >another, if you're having frequent blowouts.
>
> >> I could handle the road at 20 mph but my assumption has been that
> >> higher speeds generate greater rim temperatures due to the square of
> >> the velocity for the kinetic energy formula. If the hill wasn't so
> >> steep then taking it faster would probably require a lot less use of
> >> the brakes but based upon other safer hills that take me to about 40
> >> mph, I would say this should be a significantly faster decent.
>
> > I explained this several times. The amount of energy dissipated
> > is the same for a given descent and initial and terminal speeds.
>
> Yes. But the power at which the rim is heated and at which it cools is
> not.
>
> > Forget kinetic energy as a function of speed.
>
> Why? It seems to me it is quite important.
>
> > Concentrate on total energy dissipated. The goal is to reduce the
> > heating of the tire bead and the air in the tube. If the rim is always
> > hot as with constant braking you are guaranteed to heat the bead an
> > inflation air.
>
> If you heat the rim at the same rate at which it is also cooling, then
> you're going to be all right aren't you?
>
> > If you allow for free runs downhill with no braking, the rim will cool
> > quickly, very much more quickly if you allow the free runs to have
> > maximum speed before braking.
>
> Yes. But it might not be safe to descend with no braking.
>
> > Heavy braking decreases the amount of time to heat the rim.
> > Then you get off the brakes and allow the rim to cool. The
> > energy dissipation rate for a hot rim increases with the
> > temperature, so you are spending more time dissipating heat to
> > the air at a high rate; dissipating heat that will not go to
> > heating the bead and inflation air.
>
> I agree that non-continuous braking may help.
>
> > Another thing you have not considered is that when coasting downhill
> > at speed you dissipate more energy to wind drag the faster you go; the
> > rate of dissipation increases more than linearly with speed.
>
> This is all good advice, but it doesn't answer the question: is it
> better to brake continuously to maintain 10mph or continuously to
> maintain 20mph on a hill with a coasting speed of 50mph?
>
> If my tyre blows off either way, do I get further before it blows off if
> I go at 10mph or if I go at 20mph?

I think another pertinent factor is the effect on the shape of the
tire due to the tangential braking force between the tire and the
road.

At the contact patch, during braking, the road is pulling backward
(i.e. uphill) on the tire tread at the contact patch. This probably
has the effect of distorting the bead shape where it interfaces with
the rim. I suspect there's a tendency to put more force radially
outward on the bead just in front of the contact patch, and to reduce
the radial outward force just behind the contact patch. And I'd bet
that this is important in determining whether the tire stays in
place. That is, if these forces distort the bead wire and its rubber
"hook" sufficiently, you get a blowout.

IOW, I think that a person who heated a rim, or over-pressured a tire,
or both, in a workshop "bench" test may not see a blowout; but that
same temperature and pressure combination might blow out when under
heavy braking force.

This just throws another variable into the mix. The curve described
(zero to peak to zero) for braking force vs. descending speed wouldn't
tell the whole story. A graph of blowout risk would need to include
this effect.

Oh, and my guess would be that steel bead wires would resist that
distortion better than kevlar beads. I'd also guess that it's
possible to compound rubber surrounding the bead wire to be less
susceptible to softening under high temperatures. Seems to me that
these measures would reduce hill-descent blowouts. (Not that I've had
more than one.)

Does anyone know, do these types of blowouts happen more often with
kevlar beads?

- Frank Krygowski

 

[sergio]

On Apr 1, 6:40 pm, [email protected] wrote:
> Does anyone know, do these types of blowouts happen more often with
> kevlar beads?

I have experienced three blowouts, all with HiLite Comp kevlar bead
tires.
One while descending, another at a shop when the mechanic was
installing a new tube (pinched?), the third one at night (!) after a
good mountain ride.
So, no wonder I very much prefer iron beads.

Going back to your observation of the effect ot the forces through the
contact patch, they may be important while taking a turn, not so much
when going straight.

Sergio
Pisa

 

[Bill Bushnell]

Tom Sherman <[email protected]> wrote:
> Bill Bushnell wrote:
> > [email protected] wrote:
> >> I instrumented a wheel with pressure sensor and temperature
> >> thermocouple to monitor a blow-off but was unsuccessful getting the
> >> tire to separate even though temperature reached 150 degC and pressure
> >> 125 psi. The mechanism for blow-off is still unclear but it seams the
> >> bead material softens and creeps of the hooked bead of the rim.
> >
> > I stopped getting blowoffs when I reduced cold inflation pressure from 85 to 75
> > psi (Ritchey Tom Slick, 559x36 on Ritchey rim). I believe that the likelihood of
> > blowoff may be especially sensitive to inflation pressure because even though
> > I've been riding around lately with an extra 45 lbs. on the bike, I still don't
> > get blowoffs, even on the roads where they are likely to occur (Hicks, Vista
> > Verde/Ramona, etc.)
> >

> Any past blowoffs on the (ISO 406-mm?/451-mm?) front wheel?

No. Just the rear wheel.

--
Bill Bushnell
http://pobox.com/~bushnell/

 

[Bill Bushnell]

Bill Bushnell <[email protected]> wrote:
> [email protected] wrote:
> > I instrumented a wheel with pressure sensor and temperature
> > thermocouple to monitor a blow-off but was unsuccessful getting the
> > tire to separate even though temperature reached 150 degC and pressure
> > 125 psi. The mechanism for blow-off is still unclear but it seams the
> > bead material softens and creeps of the hooked bead of the rim.

> I stopped getting blowoffs when I reduced cold inflation pressure from 85 to 75
> psi (Ritchey Tom Slick, 559x36 on Ritchey rim). I believe that the likelihood of
> blowoff may be especially sensitive to inflation pressure because even though
> I've been riding around lately with an extra 45 lbs. on the bike, I still don't
> get blowoffs, even on the roads where they are likely to occur (Hicks, Vista
> Verde/Ramona, etc.)

> Perhaps you would consider repeating your test with incrementally more air
> pressure in your tire.

One further thought. Shortly after I began inflating my tires to 75 psi and
observed no more blow offs, I switched from the tan wall version of the Tom Slick
to the black wall version. Could the black coating on the bead hold better to a
hot rim than the tan bead?

--
Bill Bushnell
http://pobox.com/~bushnell/

 

[Ben Kaufman]

On Mon, 31 Mar 2008 21:48:37 -0400, Ben Kaufman
<[email protected]> wrote:

<SNIIP>
>How about steel? The wheels on my old panasonic are pretty heavy.
>
>Ben

Just weighed the front one without tube or tire, 3.3 lbs.

Ben

 

[Luns Tee]

In article <01eda793-00dc-4bd1-815d-678f484ef68d@e10g2000prf.googlegroups.com>,
<[email protected]> wrote:
>For a 40 mph coasting-speed hill, the peak of the braking power curve
>was around 23 mph. Faster or slower than 23 mph meant less braking
>effort.

The relationship between 40mph and 23mph comes from assuming the
only forces at play are gravity vs. wind and braking, with gravity and
wind being linear and cubic with respect to velocity. The braking power
is then:

P= k1*v - k2*v^3

This being zero at 40mph tells us that k1/k2=(40mph)^2. The
maximum power is when the derivative of this WRT the velocity is zero,
or:
k1 - 3*k2*v^2 = 0 -> v=sqrt(k1/k2/3) = 40mph/sqrt(3) = 23mph.

>It would take considerable (and impractical) calculations to
>discover which two points on either side of the peak of the curve
>correspond to each other.

For any given velocity v, if there's another velocity v2 with
the same braking power, then

k1*v - k2*v^3 = k1*v2 - k2*v2^3
k1*(v-v2) = k2*(v^3-v2^3)
k1/k2= v^2 + v*v2 + v2^2 = (40mph)^2 (unless v=v2)

This is a simple quadratic, with solution of:

v2= sqrt(40mph^2 - 3/4*v^2) - v/2


>That is, for 15 mph on the hill, the same braking power is needed at
>some speed between 23 mph and 40 mph, but the precise speed is not
>practical.

sqrt(40^2 - 3/4 * 15^2) - 15/2 ~= 30.3mph

This is as precise as the underlying assumptions are.

-Luns

 

[[email protected]]

On Sun, 6 Apr 2008 21:23:43 +0000 (UTC), [email protected]
(Luns Tee) wrote:

>In article <01eda793-00dc-4bd1-815d-678f484ef68d@e10g2000prf.googlegroups.com>,
> <[email protected]> wrote:
>>For a 40 mph coasting-speed hill, the peak of the braking power curve
>>was around 23 mph. Faster or slower than 23 mph meant less braking
>>effort.
>
> The relationship between 40mph and 23mph comes from assuming the
>only forces at play are gravity vs. wind and braking, with gravity and
>wind being linear and cubic with respect to velocity. The braking power
>is then:
>
> P= k1*v - k2*v^3
>
> This being zero at 40mph tells us that k1/k2=(40mph)^2. The
>maximum power is when the derivative of this WRT the velocity is zero,
>or:
>k1 - 3*k2*v^2 = 0 -> v=sqrt(k1/k2/3) = 40mph/sqrt(3) = 23mph.
>
>>It would take considerable (and impractical) calculations to
>>discover which two points on either side of the peak of the curve
>>correspond to each other.
>
> For any given velocity v, if there's another velocity v2 with
>the same braking power, then
>
> k1*v - k2*v^3 = k1*v2 - k2*v2^3
> k1*(v-v2) = k2*(v^3-v2^3)
> k1/k2= v^2 + v*v2 + v2^2 = (40mph)^2 (unless v=v2)
>
> This is a simple quadratic, with solution of:
>
> v2= sqrt(40mph^2 - 3/4*v^2) - v/2
>
>
>>That is, for 15 mph on the hill, the same braking power is needed at
>>some speed between 23 mph and 40 mph, but the precise speed is not
>>practical.
>
> sqrt(40^2 - 3/4 * 15^2) - 15/2 ~= 30.3mph
>
> This is as precise as the underlying assumptions are.
>
>-Luns

Dear Luns,

It may be too much trouble, but can that be used to produce the graph
of speed 0-40 versus braking power from the old post, whose link is
now dead?

I think that Jim Smith was using your approach:
http://groups.google.com/group/rec.bicycles.tech/msg/ab705c3d6e475419

But his graph is gone:
http://groups.google.com/group/rec.bicycles.tech/msg/21c12e0e024dcf14

Cheers,

Carl Fogel