Re: Tire burn-out during landings

From:         yarvin-norman@CS.YALE.EDU (Norman Yarvin)
Organization: Yale Computer Science Department
Date:         29 Dec 92 22:53:48 PST
References:   1 2 3
Followups:    1
Next article
View raw article
  or MIME structure (Robert Dorsett) writes:
> [...] (3) the *additional* wear and tear on the brakes, as they 
>must absorb the spinning energy, in addition to performing their normal 
>task of slowing down the airplane.  [...]

This additional energy is negligible.  Consider just the energy of the wheel
itself.  For a wheel which is rolling along the ground there is the relation:

	(rotational energy) = (some constant) * (energy of forward motion)

where the constant is independent of speed.  I seem to recall that for a
cylindrical wheel of uniform consistency, the constant is 2/7.  At absolute
worst the constant will be 1.  (This would occur if the entire mass of the
wheel were on the tread of the tire.)

Furthermore the energy of forward motion of the wheel is an insignificant
portion of the energy of the entire aircraft.  This goes by weight; if the
airliner weighs 100,000 pounds and a wheel weighs 300, the proportion of
energy in that wheel would be 3/1000 of the aircraft's energy.  Then, using
the 2/7 figure, the spinning energy of the wheel would be 3/1000*2/7 = .08%
of the energy of forward motion of the aircraft.

Assuming constant deceleration force, stopping distance would be lengthened
by that same .08%.  Even the weight of the mechanism required to speed up the
tires might be a bigger factor.

In any case, the practicality of preventing tires from disintegrating
depends on how fast tires presently disintegrate.  How much matter really is
there in that cloud of smoke?  Perhaps a gram per cubic meter of smoke?  And
how much tire is left on the runway?  Do they have to go out and scrape it
off now and then?  (I imagine not.)  Seems to me the loss of tire material
is negligible also.  Compared, that is, with the other costs of running the

Norman Yarvin