On Jun 10, 8:09*am, "Vito" <v...@cfl.rr.com> wrote:
> ? wrote:
> > That depends on whether you're riding the street or on the race track.
> Racing tires are designed to have max stick in a certain temp
> range. *One gets that temp by adjusting air pressure: down = hotter, up =
> colder.
>
> It is important to keep the pressure/temp relationship linear (don't ask me
> ask the tire engineers). Air, especially from a compressor, contains water.
> At racing speeds the tires get hot enough to turn it to steam, causing a
> nonlinear pressure increase, especially on cages where tires run hotter.
As the sulfur that bonds molecular chains of rubber begins to melt at
234°F~246°F the rubber compound begins to devulcanize, and the rubber
on the surface of rear tire begins to slough off, turn into "smileys,"
and then begin to roll into little black balls...
http://www.csgnetwork.com/prescorh2oboilcalc.html
Add 29.4 PSI *gauge* inflation pressure to the 14.7 PSI *absolute*
pressure which your pressure gauge doesn't indicate and you get 44.1
PSI Absolute in your tire, which is 89.76 Inches Hg.
The BP of H2O is 322.810°F at 44.1 PSIA. If the rider isn't getting
his tires up to
323°F, he isn't boiling the condensation inside his tires and doesn't
*need* nitrogen.
When water boils, it expands by a factor of 1600:1...
The amount of water in compressed air can vary widely, though. I won't
dig into my steam engineers books to get into the effect of this
pressure rise on the
*volume* of the tire, even though volumetric increase due to pressure
rise *does* affect the *size* of the tire's contact patch.
If you *ass*-ume that air (or nitrogen) is a perfect *dry* gas, you
can calculate the *linear* rise in tire temperature by using the
formula
P1/T1 = P2/T2
Where pressure is absolute and temperature is expressed in degrees
Kelvin for the metric system or degrees Rankine for the English
system.
With an initial inflation pressure of 36 PSIG at 59°F and a final
(hot) pressure of 40 PSIG, the calculation is as follows:
( 36 + 14.696 ) / ( 59 + 459.67 ) = ( 40 + 14.696 ) / ( X + 459.67 ) ;
50.696 / 518.67 = 54.696 / ( X + 459.67 ) ;
0.0977423 = 54.696 / ( X + 459.67 ) ;
54.696 / 0.0977423 = ( X + 459.67 ) = 559.5935° Rankine
559.5935° Rankine - 459.67 = 99.92° Fahrenheit.
If you want to work this problem out *including* the inscrutable
factor of *volume,* the formula is
( P1 * V1 ) / T1 = ( P2 * V2 ) / T2
Ya didn't know you were talking to a former rocket scientist and steam
engineer, didja?