Exploding tires II

I used to wrench for a dept. store in Alberta...the "shop" was an unused
storeroom in back behind the recieving dock. Busy season was mid-November
to mid December, as Canadian parents enjoy being unconsionably cruel by
buying their children an Xmas present that they aren't going to be able to
use for another THREE MONTHS...

Anyway...

The air line for the shop was taken from the compressor on the refuse
compactor unit outside the dock.

We had in a new line of "roadies" with 27x 1 1/8 tyres. I began building
them just as the outside temp headed for -15 degrees centigrade -- and
stayed there.

We would hang them from a bar suspended from the roof of the (heated) dock.

....and the tyres began blowing out mysteriously...

I was only 18 at the time, and a fair wrench despite my youth, but evidently
physics was not my strong suit...

air at -15 degrees, pumped in at 90 ppsi, then allowed to warm to +20
degrees or warmer? Gee, THAT shouldn't cause a problem, should it???

I admit, it took me awhile to figure it out...

Brian wrote:


air at -15 degrees, pumped in at 90 ppsi, then allowed to warm to +20
degrees or warmer? Gee, THAT shouldn't cause a problem, should it???

Actually, I'm very surprised it did. The resulting pressure should have
been about 103 psi. That's usually not high enough to cause a problem.

It's the ratio of the _absolute_ temperatures that matters, not the
ratio of the Celsius temperatures.


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

air at -15 degrees, pumped in at 90 ppsi, then allowed to warm to +20
degrees or warmer? Gee, THAT shouldn't cause a problem, should it???

Actually, I'm very surprised it did. The resulting pressure should have
been about 103 psi. That's usually not high enough to cause a problem.

It's the ratio of the _absolute_ temperatures that matters, not the ratio
of the Celsius temperatures.

For those of us who have forgotten our high-school physics, could you give a
few examples of absolute vs Celsius (or Fahrenheit) temperatures? In the
real world, we might conceivably see a tire/tube start the day at 40 degrees
(F) and reach a peak of well over 100 (F), possibly 150? To tell you the
truth, I don't honestly know just how hot the air in a tube might get on a
very long, very steep descent with ambient air temp at 90 degrees. However,
that's a real-world example that some of us experience.

--Mike-- Chain Reaction Bicycles
www.ChainReactionBicycles.com

"Mike Jacoubowsky" wrote in message
om...
air at -15 degrees, pumped in at 90 ppsi, then allowed to warm to +20
degrees or warmer? Gee, THAT shouldn't cause a problem, should it???

Actually, I'm very surprised it did. The resulting pressure should have
been about 103 psi. That's usually not high enough to cause a problem.

It's the ratio of the _absolute_ temperatures that matters, not the
ratio
of the Celsius temperatures.

For those of us who have forgotten our high-school physics, could you give
a
few examples of absolute vs Celsius (or Fahrenheit) temperatures? In the
real world, we might conceivably see a tire/tube start the day at 40
degrees
(F) and reach a peak of well over 100 (F), possibly 150? To tell you the
truth, I don't honestly know just how hot the air in a tube might get on a
very long, very steep descent with ambient air temp at 90 degrees.
However,
that's a real-world example that some of us experience.

Roughly speaking and assuming the tire volume doesn't change and air is
neither added to nor removed from the tire then the pressure is proportional
to the temperature expressed in degrees Kelvin. 0 degrees Celsius is 273
deg.K. Therefore we're talking about a tire going from 258 deg. K to 293
deg. K. If the pressure at 258K is 90 psi then the pressure at 293K will be
approximately 293/258 X 90 = 102 psi.

--Mike-- Chain Reaction Bicycles
www.ChainReactionBicycles.com

Kelvin = 273.15 + Celcius
(Farenheit -32) * 5/9 gives celcius
Pressure increases proportionally with Kelvin tempature (unless near
condensation point of the gases involved).

Desert, 32F 0C 273K dawn pump to 100 PSI
104F 40C 313K afternoon shade to 114 PSI
212F 100C 373K afternoon in sun with black tires getting soft to 136 PSI

a couple of years ago, at the Springfield IL iron man triathalon,
everyone pumps up their tires in the early morning dawn when it is
cool. After swimming 1/2 miles and running 5 miles, they got back,
after heating in the sun, hopped on their bikes and several tires blew
in from the increased pressure. Once person rode on his flat 45 miles
instead of putting on another tire.


Mike Jacoubowsky wrote:

air at -15 degrees, pumped in at 90 ppsi, then allowed to warm to +20
degrees or warmer? Gee, THAT shouldn't cause a problem, should it???

Actually, I'm very surprised it did. The resulting pressure should have
been about 103 psi. That's usually not high enough to cause a problem.

It's the ratio of the _absolute_ temperatures that matters, not the ratio
of the Celsius temperatures.

For those of us who have forgotten our high-school physics, could you give a
few examples of absolute vs Celsius (or Fahrenheit) temperatures? In the
real world, we might conceivably see a tire/tube start the day at 40 degrees
(F) and reach a peak of well over 100 (F), possibly 150? To tell you the
truth, I don't honestly know just how hot the air in a tube might get on a
very long, very steep descent with ambient air temp at 90 degrees. However,
that's a real-world example that some of us experience.

--Mike-- Chain Reaction Bicycles
www.ChainReactionBicycles.com

"Ian S" wrote: (clip) If the pressure at 258K is 90 psi then the pressure
at 293K will be approximately 293/258 X 90 = 102 psi.
^^^^^^^^^^^^^
I have one more little nit to pick, but since we're talking physics here,
you need to remember that 90 PSIG = 105 PSIA*. So 293/258 x 104 = 118 PSIA,
which is 103 PSIG.

* PSIG = Gauge pressure. PSIA = Absolute pressure = Gauge pressure + 15.

Not very important to a bicyclist, but TERRIBLY important to someone grading
a high school physics exam. :-)

"Leo Lichtman" wrote in message
...

"Ian S" wrote: (clip) If the pressure at 258K is 90 psi then the
pressure
at 293K will be approximately 293/258 X 90 = 102 psi.
^^^^^^^^^^^^^
I have one more little nit to pick, but since we're talking physics here,
you need to remember that 90 PSIG = 105 PSIA*. So 293/258 x 104 = 118
PSIA,
which is 103 PSIG.

* PSIG = Gauge pressure. PSIA = Absolute pressure = Gauge pressure + 15.

Not very important to a bicyclist, but TERRIBLY important to someone
grading
a high school physics exam. :-)

So where'd the 104 in your calculation come from?

"Ian S" wrote: So where'd the 104 in your calculation come from?
^^^^^^^^^^^^^
Sorry--my error. The first time I wrote it, I treated atmospheric pressure
as 24 psi. Then, when I edited it, I overlooked that entry. Take off five
points.

Mike Jacoubowsky wrote:

For those of us who have forgotten our high-school physics, could
you give a
few examples of absolute vs Celsius (or Fahrenheit) temperatures? In the
real world, we might conceivably see a tire/tube start the day at 40 degrees
(F) and reach a peak of well over 100 (F), possibly 150? To tell you the
truth, I don't honestly know just how hot the air in a tube might get on a
very long, very steep descent with ambient air temp at 90 degrees. However,
that's a real-world example that some of us experience.


OK, the relationship is (P1*V1)/T1 = (P2*V2)/T2 where P and T refer to
absolute pressures and temperatures, V is volume. If volume is
(reasonably) assumed constant, it cancels out, and you can rearrange to:

P2 = P1 *(T2/T1)

But again, P & T have to be measured on absolute scales. They're
usually not, so you have to convert.

For metric measurements like the originally quoted Celsius, the absolute
temperature scale is Kelvin. To convert Celsius to Kelvin, add 273.15

If you're starting with degrees Fahrenheit, the corresponding absolute
scale is Rankine. To convert Fahrenheit to Rankine, add 459.67 (or
alternately, convert Fahrenheit to Celsius and use the Kelvin scale.)

For pressures, your gage measures "gage pressure," psig. To convert to
absolute pressure (psia) add the pressure of the atmosphere, 14.7 psia.

So, 40 deg. F = 499.67 deg R 150 deg F = 609.67 deg R
90 psig = 104.7 psia

and P2 = 104.7 psia*(609.67/499.67) = 127.7 psia

But now we have to get that absolute pressure converted back to gage
pressure. Subtract 14.7 and you get P2 = 113 psig.

So in general, it's not as bad as you might think.



There are commercially available stick-on temperature indicators that
record maximum temperatures of the surface to which they're stuck.
http://www.tempil.com/Tempilabel.htm

ISTR someone checking tandem rims on mountain descents with these
things, but I don't recall what the maximum temperature was.


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

Ian S wrote:
"Mike Jacoubowsky" wrote in message
om...

air at -15 degrees, pumped in at 90 ppsi, then allowed to warm to +20
degrees or warmer? Gee, THAT shouldn't cause a problem, should it???

Actually, I'm very surprised it did. The resulting pressure should have
been about 103 psi. That's usually not high enough to cause a problem.

It's the ratio of the _absolute_ temperatures that matters, not the
ratio
of the Celsius temperatures.

For those of us who have forgotten our high-school physics, could you give
a
few examples of absolute vs Celsius (or Fahrenheit) temperatures? In the
real world, we might conceivably see a tire/tube start the day at 40
degrees
(F) and reach a peak of well over 100 (F), possibly 150? To tell you the
truth, I don't honestly know just how hot the air in a tube might get on a
very long, very steep descent with ambient air temp at 90 degrees.

However,
that's a real-world example that some of us experience.


Roughly speaking and assuming the tire volume doesn't change and air is
neither added to nor removed from the tire then the pressure is proportional
to the temperature expressed in degrees Kelvin. 0 degrees Celsius is 273
deg.K. Therefore we're talking about a tire going from 258 deg. K to 293
deg. K. If the pressure at 258K is 90 psi then the pressure at 293K will be
approximately 293/258 X 90 = 102 psi.

Might be easier to invoke the Rankine scale, which also gives absolute
temperatures but is easier to relate to the common Farenheit scale
(degrees R = degrees F + 459). Mike's example would then start at 40F =
499R and heat up to 150F = 609R. Assuming a 90 psi tire in the morning,
the long hot downhill ride would hit (609/499 x 90 psi) = 110 psi.
(Note Ian was calculating the pressure change for Brian's Canadian
store, not Mike's example. The answers should be different!)

I don't question Brian's claim that the department store tires popped
when they came inside. But given the pressure change in Mike's example,
I wouldn't want to ride one of those bikes down a long hill on a summer
afternoon!

Pat