View Full Version : Brake energy

I'm looking for a formula that would predict the amount of energy required to stop ; for example a 1000 lb aircraft at 60 knots and the amount of heat generated by the brakes in doing so.

Thanks for any replies.

Paul Wilson

31st Aug 2004, 21:31

Kinetic Energy = 0.5mass multiplied by velocity squared

using metric units - 1 Joule = 1kg *(metres/seconds) squared

chuck in a constant and the same equation will allow you to use any units you like, in theory you could have an equation reading something like

1 Mars bar = 1 double decker bus * (width of a 2p/a blink of an eye) squared * by constant C

Mad (Flt) Scientist

31st Aug 2004, 22:51

Of course, that's all the energy that has to be consumed to bring the aircraft to a halt - it might not all go to the brakes (in fact, certainly won't) so that is a conservative number. But if your energy distribution amongst the brakes is uneven it might NOT be conservative to divide by the number of wheels or bogies etc to get a "per brake" number - only testing will give you that kind of info reliably.

Centaurus

1st Sep 2004, 11:48

Somewhere back in one of the "Boeing Airliner" magazines I recall seeing a paragraph which stated that if you cannot touch the brake area without hurting your hand, then by definition the brakes are hot. But hot can mean ordinary hot, very hot or really really hot. You are not going to leave your hand there long enough to decide which.

Without a brake temperature indication system, it becmoes guesswork to determine how many million foot-pounds the brakes have absorbed during (say) a 100 knot abort on a long runway. Or a max weight landing and a high speed turn-off taxiway. I am discussing 737 ops here.

Can anyone give me advice on an intelligent way to apply the Brake Cooling Schedule in a 737 without having to delay every take off simply because the brakes were too hot to touch?

Tim Zukas

4th Sep 2004, 23:15

The example in the original question:

1000-lb a/c = 453.59 kg

60 knots = 30.87 meters/sec

So kinetic energy equals 216,080 joules. The conversion between heat and energy is pretty well fixed-- is it roughly 4.18 joules to a small calorie? If so, 52 kilocalories (do they still call them that?) have to be absorbed, mostly by the brakes. Enough to heat 10 lb of water 11.4 degrees C.

FlexibleResponse

7th Sep 2004, 07:26

Paul Wilson gave the answer to IHL’s very thought provoking question. Kinetic Energy = 0.5mass multiplied by velocity squaredActually, it comes from Einstein's famous formula:

E = m x c squared

In simple terms for us pilots to have a useful tool for determining braking energy, we can consider only these two major variables.

1 Aircraft weight

2 Groundspeed at brake application

As a result you can see that you can vary the a/c weight by quite a lot and not make too much difference to the energy required. But because the speed factor is squared, changes to brake application groundspeed becomes the most critical factor in determining brake energy required and how hot your brakes get.

For the Airbus A330/A340 series this information is available in a graph in the Minimum Equipment List for operation with unserviceable brake temperature indication. Other factors such as runway slope, number of reversers operating, etc,etc are allowed in this Airbus graph for but the major effects of weight and especially speed are very apparent.

Einstein right again? Buy that man a beer!

Centaurus,

I found some guidance at John Tullamarine’s:

Useful Website References (Tech Log ‘sticky’).

I have never been rated on the B737 and I don\'t know if the information is authoritive.

http://www.b737.org.uk/pilotnotes.htm

http://www.b737.org.uk/perf_brakecooling.gif

supercarb

7th Sep 2004, 17:18

Actually, it comes from Einstein's famous formula:

E = m x c squared

Er, no it doesn't. Try Isaac Newton.

FlexibleResponse

9th Sep 2004, 14:01

Thanks supercarb!

I think I agree with that Newtonian physics define the laws of motion.

And the mere refinement on their limitations by Einstein does not make the concept any more useful in this instance.

IHL,

This may be of interest:

http://www.infinityaerospace.com/Brake_Kinetic_Energy_Req.htm

http://www.risingup.com/fars/info/part23-735-FAR.shtml

maxy101

9th Sep 2004, 15:18

Don't wish to split hairs, but I think Einsteins theories describe the laws of motion more accurately re:relativity and Lorentz transforms, but for speeds well below the speed of light , "classical" physics approximates very well. i.e Newtonian physics.