MU reading and aircraft deceleration

[pstan]

My buddy was in Twin Falls Idaho KTWF a month or so ago. He was given a friction value of .4 for the runway I think he said, maybe .3. Not being familiar with what that meant, I looked in the AIM where 4−3−9. Runway Friction Reports and Advisories, para c states

"MU (friction) values range from 0 to 100 where zero is the lowest friction value and 100 is the
maximum friction value obtainable. For frozen
contaminants on runway surfaces, a MU value of
40 or less is the level when the aircraft braking
performance starts to deteriorate and directional
control begins to be less responsive. The lower the
MU value, the less effective braking performance
becomes and the more difficult directional control
becomes. "

So..

1. hypothetically, does this mean a MU of 100 would mean the brake pucks, if able to provide enough force, should be able to decelerate an aircraft on dry pavement (just sort of wheel skid) at about .7G where G = 32 feet per second squared?

2. on a slippery runway, a MU reading of .5 should mean the deceleration force from the tires on the runway is twice as much as with a MU reading of .25?

thanks for your corrections/inputs

PStan

 

[MauleSkinner]

There's a lot of variation in readings from various types of friction meters. The Canadian Runway Friction Index puts the different meter types in an interesting context...you might Google that if you're so inclined.

 

[Kritchlow]

Common in the US is a Tapley meter. most commercial operations cease at nil, which is <20.
There are charts available describing the different mu values.

 

[pstan]

Thanks guys

Any opinions on the question of halving the the mu reading? Should this double the minimum braking distance and half the deceleration rate once braking commenced?
Assuming braking done at just below the locked wheel condition.

Stan

 

[petrolero]

Thanks guys

Any opinions on the question of halving the the mu reading? Should this double the minimum braking distance and half the deceleration rate once braking commenced?
Assuming braking done at just below the locked wheel condition.

Stan

Well if absolutely everything else were equal and the coefficient of static friction halved, then yeah your braking distance would double. A mu of zero is physically impossible but would mean that NONE of the airplane's kinetic energy can get converted into braking force. Conversely, a mu of 100 would be like a cog. It essentially cannot slip no matter how much force is applied. That's impossible too in the sense that at some amount of force the tire will enter sliding friction mode.

The coefficient of sliding friction is lower than that of static friction, which is of course why we have anti-lock or anti-skid brakes.

I tried writing the equation but it looks ugly. This shows it better than I can...

http://hyperphysics.phy-astr.gsu.edu/hbase/crstp.html

 

[MauleSkinner]

Given that a .4 on a Tapley meter is different than a .4 on other brands, and the fact that the FAA has always taken the stance that mu readings were unusable without other data, I'd say that those kinds of assumptions probably aren't valid.

If you look at the CRFI, you'll find that linear equations really don't work there, either. The CRFI used to include a table that converted other friction readings to CRFI numbers, but I don't see that anymore, either.

The amount of braking power available for the same tires on the same surface changes with aircraft speed as well...at slower speeds, you can apply more pressure to the brakes without skidding. I think its more complex than a simple algebra problem.

 

[petrolero]

I think its more complex than a simple algebra problem.

It absolutely is more complex than a simple equation. Stopping distance is a function of coefficient of friction as well as kinetic energy (which contains the square of velocity) and weight on the tires. Weight on the tires is not a constant in aircraft calcs like it is assumed to be in automobiles. That's why if you tag the brakes too hard too early, you'll lock the wheels up regardless of the coefficient of static friction.

So thinking of this another way... if you double your speed (all else equal) you quadruple your braking distance. But the OP just asked about halving mu.... which would theoretically double braking distance if all else were equal.

That's why I included the classic rhetorical IF... if all else is absolutely identical.