I am idly curious whether any significant proportion of a landing plane's momentum is lost through wheel braking vs. everything else like thrust reversers, aero drag, etc?
If I have to pick a plane, let's go with a 737.
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$\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$– FedericoCommented Jul 29, 2022 at 15:59
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2 Answers
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The vast majority of the stopping power on a landing is the wheel brakes. Reverse thrust is a much smaller component, and as it happens is generally not taken credit for in published landing distance performance data. It's just a bonus. The pilot can choose to use reverse or not, and often won't, maybe only deploying them to idle reverse to have available just in case some braking power is lost for some reason.
Aerodynamic drag is significant at first, although still way less than wheel brakes, but quickly declines as speed comes down, whereas wheel brake power (as in, what you feel as deceleration, not so much actual mathematical energy conversion) is relatively constant all the way to being stopped.
Brakes on airliners and generally split into two separate circuits on each gear, leaving half braking if a sub-system fails. If that happens, landing distance goes way up.
Then there's anti-skid (ABS in a car). Anti-skid is the only thing that allows the pilot to use maximum braking. Without anti-skid, going hard on the brakes will likely lock wheels and blow tires, because airliner brakes are like truck airbrakes, with no feedback at the pedals, just spring pressure, and you can only modulate them based on your sensation of deceleration of the plane.
So, the large majority of an airliner's stopping power is in the brake system, with the anti-skid system making it possible to exploit all of the potential braking power.
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$\begingroup$ The OP specified in the comments that spin-up is not what they are asking about. $\endgroup$ Commented Jul 27, 2022 at 19:41
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2$\begingroup$ Well now that it's been revised... $\endgroup$– John KCommented Jul 28, 2022 at 0:00
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2$\begingroup$ Frictional braking power relies entirely on contact with good pavement. Ask a Minnesota driver about that. Whereas thrust reversers rely entirely on contact with good air. So unless God removes all air from the airport vicinity, thrust reversers will behave predicatably, if anything be higher performance in a rain squall due to higher thrust mass (the water). $\endgroup$ Commented Jul 28, 2022 at 1:01
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1$\begingroup$ Max reverse alone is like very lightly applying brakes. Even on a wet or compacted snow runway, most of the slowing power is coming from braking. If the plane has cascade style reversers that only direct air forward at a slight angle, they do almost nothing really. And you have to get the reverse off as the speed drops below 90-80 kt so it's only working during the initial part of the roll. $\endgroup$– John KCommented Jul 28, 2022 at 12:31
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$\begingroup$ Thanks - this is the kind of answer I was looking for. I had assumed that reverse thrust was the lion's share of braking but I guess not! $\endgroup$ Commented Jul 28, 2022 at 15:11
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Not much, as planes will never carry large heavy wheels into the air. The concept could be tested with models. Some energy will be required to overcome the angular inertia of the wheel.
In olden times, this was known as a "flywheel". This stored energy can be utilized, for example, to keep a piston engine turning smoothly between powerstrokes.
But with aircraft, weight savings is paramount. To scale, any energy going into spinning up wheels on landing is miniscule compared to the kinetic energy of the aircraft. And while we are at it, a bit of energy also goes into the "chirp" on touchdown from frictional heating, also found from braking.
answered Jul 28, 2022 at 9:01
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1$\begingroup$ I realize some bush pilots may disagree with the first sentence. $\endgroup$ Commented Jul 28, 2022 at 9:17
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3$\begingroup$ I don't think the question is about transferring energy to spin the wheel. It is about using the brakes mounted on the wheels. $\endgroup$ Commented Jul 28, 2022 at 15:04
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$\begingroup$ @FlorianF related, interesting enough to post. $\endgroup$ Commented Jul 28, 2022 at 15:27