Is it possible to use ONLY thrust reversers to steer and bring a smaller size commercial jet aircraft, with a minimum amount of fuel onboard, to a complete stop, after touching down on a longer than average length commercial airport runway?
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$\begingroup$ possible duplicate of For large jets, what is the primary means of slowing down after landing? $\endgroup$– UnrecognizedFallingObjectMay 3, 2015 at 1:27
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$\begingroup$ If you have enough runway, you can stop without the use of any stopping mechanism except drag. Is there a maximum distance you're thinking of? $\endgroup$– cpastMay 3, 2015 at 1:47
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$\begingroup$ I was thinking in the distance of 8,000 feet, and at sea level...thanks $\endgroup$– garyv440May 3, 2015 at 1:53
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3 Answers
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Steer: No. Stop: Yes.
Thrust reversers are either in or out. They cannot be engaged proportionally to some control setting in the cockpit. Once they are unlocked, the airflow through them sucks them out completely. The only way to vary their braking effect would be to throttle the engine up or down, and airliner engines react rather sluggishly to control inputs. Their responsiveness will not suffice to steer the aircraft; only rudder inputs (and those only at higher speed), nose wheel steering and differential braking with the wheel brakes are suitable for directional control on the ground.
Their braking effect is biggest at high speed and tapers off as the aircraft decelerates. For the first third of the landing run they will help noticeably in slowing the aircraft, but you will need a long runway when you want to land without wheel brakes. Details depend on landing weight, airfield elevation and wind speed.
answered May 3, 2015 at 16:24
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From a physics perspective, we can calculate the amount of force needed to slow a plane down within a given distance. The stopping distance equation is:
$$d_f=\frac{v_i^2}{2a}$$
If we assume the plane has 7000 ft to stop with only reverse thrust, and the touchdown speed is 130 knots = 220 ft/s, we can solve for acceleration.
All calculations will be approximate
$$7000=\frac{220^2}{2a}$$ $$a=3.5 ft/s^2$$
Using the second law of motion, we can find the force required to decelerate a 120,000 lb airplane at that rate.
$$F=ma$$ $$F=120,000\times3.5=415,000ftlb/s^2=13000lbf$$
This is about 25% of the max forward thrust of a 737-700. The engines have a bypass ratio of over 5:1, so it seems reasonable that the engines should be able to average this amount of reverse thrust force. This estimate also ignores all effects of drag or friction that would also help to stop the aircraft.
For steering, the pilots could use differential thrust. It would be more difficult than using the rudder and nose wheel, but it should be sufficient. A sideways wind component would make it more difficult to steer though.
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I'm not a commercial pilot but I've just been researching this for other reasons: to look at brake wear mechanisms.
Thrust reversers are normally supposed to be engaged on Airbus and Boeing commercial jets (that have them) when you are landing. Engaging them just means opening/moving the various flaps that divert the flow.
However due to noise regulations, you normally are not allowed to actively use them: so you put the engines into idle thrust. However this has several advantages: - the system is set up to allow a much bigger reverse thrust if you need it in a hurry - you also don't have any positive thrust from the idle thrust since you have it reversed (which is not negligible, it varies from 3-6% of total max thrust)
The FAR/JAR landing calculations do not assume any credit for reverse thrusters, so in ideal conditions you shouldn't need them. But if you have a system malfunction or other problem (bad braking, poor flap control, slippery runway, short runway, very heavy plane etc) then they can help you.
I believe most big commercial jet engines that have reverse thrust, can go up to 80%-90% of N1 (fan speed). That's because most work by opening up flaps that only divert the fan air output (which is about 80% of the thrust). These are for the "translating cowl type".
Airbus says that on the A330 with the GE CF6-80E1A2 engine, at 150kts and at 90% N1 you get 16,000 lb/engine reverse thrust (sea level). Which is 22% of max thrust. It drops to 8,000 lbs at 0 kts
The reversers that open after the rear of the engine ("target type") probably go to higher percentages, but I don't know how much. They are often on military planes.
I suspect that because of the strict noise abatement rules now in place at most big airports, reverse thrusters are not much used in standard landings. And braking capacity has improved a lot. So if you cannot use them, then you don't have them in th design: which is why the Airbus 380 only has them on two engines. More for emergencies I'd guess.
If you look at the efficiencies of wheel braking and air drag, the air drag works best at high air speeds: which is good, because this is where kinetic energy is the highest. As you slow down it becomes less effective, but then wheel braking can take over and control the deceleration at slower speeds, and where you might want to maneuver more.
Incidentally, you can see how much idle thrust occurs from Airbus and Boeing planes by looking at the charts they provide for Ground Towing Requirements. They tell you how much pullback is on the tow bar for the number of engines that are at idle. These charts are in the files they provide on the following (Section 5):
Airbus Aircraft Characteristics - Airpot and Maintenance Planning
Boeing Aircraft Characteristics - Airpot and Maintenance Planning
