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- rdd@cactus.org (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
- airplane.
-
- --
- Norman Yarvin yarvin@cs.yale.edu
-
-