<<Units Units Units

KE=1/2 MV**2
M= lbs Mass or Weight/ 32 ft/sec**2
V= speed in ft/sec or knts*1.7
1000 lbs weight or force =  31.25 lbs mass
100 kts = 170 ft/sec>>

Okay, I can't stop myself - got to jump in.  Pounds "weight" is actually a
force, pounds "mass" is a convenience term that is the mass of an object
that exerts 1 pound of force on a scale at a standard gravity.  We engineers
commonly use the shortcut of talking about lbm (pounds mass).  The term
"slug" is defined as the mass for which one lbf will accelerated it at one
ft/sec^2.  The commonly accepted acceleration of "standard gravity" (at 45
degrees latitude) is 32.174 ft/sec^2 and hence a slug is about 32.2 lbm, not
31.25.

All that doesn't mean anything much, so let's compute the temperature of the
brake rotor after a maximum-energy stop.  For a 3400 lbm aircraft stopping
from 80 kts the kinetic energy to be absorbed by the brakes is almost
474,000 ft-lb, or 609 BTU's (some, but not a lot of energy is absorbed by
aerodynamic drag and tire losses, while the residual thrust is working the
other way).  I don't know what a brake rotor weighs, but I'll guess 5
pounds.  One thing I couldn't find is the specific heat of cast iron, but
most steel alloys have a specific heat of between 0.1 and 0.14, so I'll pick
0.12.  The temperature rise during a braking event is then BTU/(sp.ht.Xlbm),
or about 1,000 degrees F.  The brake temperature at the beginning could be
100, but if you drag the brakes a lot during taxi, heating them up to maybe
500 and then do a rejected takeoff, you could see 1500 degrees.  The brake
pad material will probably continue to work at 1,000 degrees, but probably
not at 1,500.  You might think that some of the heat gets absorbed in the
caliper and wheel, but it is not so - almost all the heat goes directly into
the rotor and very little is rejected into the air during that short time.
Metallic linings will keep working at higher temperatures than conventional
organics so that would help.

Gary Casey