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reversers and a/c speed

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reversers and a/c speed

Old 24th Dec 2010, 10:38
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reversers and a/c speed

Very straightforward question: why jet engine reversers are more effective at high speeds?

I have a couple of reasons in mind, but I couldn't find an "official" explaination!
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Old 24th Dec 2010, 11:01
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I think that is due to characteristics of jet engine.
At high speed we have large amount of air mass, so engine works better and it produces more thrust.
You should also remember that on contaminated rwy, efficiency of wheel brakes decreases and the only proper way to brake are reversers.

cheers !
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Old 24th Dec 2010, 12:06
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In fact the Jet Engine thrust is a little bit less at high speed, so the reverse thrust as well.
We can assume the produced reverse thrust is more or less same througout the landing roll. However at high speed due to lift, the efficiency of brakes are less. For the desired fixed deceleration rate at landing roll (e.g MED Autobrake Selection on an Airbus FBW), most of the deceleretion job is done by T/R at high speed moments.

The following figure is not based on speed but based on stopping distance, it may give a good idea the role of deceleration methods.



Edit reason: editorial

Last edited by JABBARA; 24th Dec 2010 at 13:35.
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Old 24th Dec 2010, 13:54
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JABBARA

Thanks for that post

I love data.

Now since that stopping distance question keeps coming up on the forum your curve will always stick in my mind if.......... I could understand all parts of it.

Questions

The left axis appears to be braking effect

What are the units?

Since speed is not in the curve, I assume that the bottom axis of distance is for a typical landing for the specified aircraft?

I don't understand the aero drag being greater at the 3000m mark (end of roll)? so I probably have missed the understanding of the curves

Help
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Old 24th Dec 2010, 14:12
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Also see http://flightsafety.org/files/alar_bn8-4-braking.pdf
Fig 3 shows stopping force vs speed, i.e. the contribution of each retarding device to the decelerating force vs speed.
Fig 4 (similar to #3) shows the % of stopping energy vs distance.
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Old 24th Dec 2010, 15:23
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What indeed is the 'official explanation'? It is an oft-quoted mantra that 'reversers are more effectiv/efficient at high speed' but I think perhaps we should insert than wheelbrakes in there? Jabbara's 'weight on wheels v braking force' has some merit.
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Old 24th Dec 2010, 15:33
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I have to check my literature on this. However, I believe that re-ingestion is one of the things influencing this. As speed reduces the reverse thrust plume re-ingests into the inlet affecting the efficiency of the thrust reversers. Reverse thrust decreases with decreasing speed.
Will have to check this when I am back in the office.
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Old 24th Dec 2010, 16:12
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I have to check my literature on this. However, I believe that re-ingestion is one of the things influencing this. As speed reduces the reverse thrust plume re-ingests into the inlet affecting the efficiency of the thrust reversers. Reverse thrust decreases with decreasing speed.
Will have to check this when I am back in the office.
I think you will find that the reingestion is at a cross-over point and not linear until then. Affected by engine power setting (air out the reversers) vs forward speed.

Just above no-effect,
below a chance for some ingestion mostly through the fan (FOD) and increasing chance for some in the core and a surge
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Old 24th Dec 2010, 16:28
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Originally Posted by BOAC
What indeed is the 'official explanation'? It is an oft-quoted mantra that 'reversers are more effectiv/efficient at high speed' but I think perhaps we should insert than wheelbrakes in there? Jabbara's 'weight on wheels v braking force' has some merit.
I was going to write exactly this as a response, because this is exactly the reason I wrote the initial question!

Knowing that:
1) reversers are more effective than brakes in the inital part of the landing roll

2) reversers are more effective than brakes if landing on a slippery rwy

So "reversers are more effective/efficient at high speed" implies that they are MORE EFFECTIVE THAN BRAKES and NOT more effective themselves for some particular reason that I am missing??

so...still looking for a definite answer!

ps: BOAC, the quoting on your profile is my favorite one!!
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Old 24th Dec 2010, 16:41
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One would think that it could vary by aircraft type. For instance, the
DC-9 and 737-100/200 with their big external buckets would seem to be likely to provide more drag at high speed.
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Old 24th Dec 2010, 16:54
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Some thoughts; by no means exact science.
Assuming that the same mass of air can be thrown forward by the engine reverse during the landing (unlikely to be true but good enough for the range of speeds on ground), then the change in momentum of that airmass relative to the static air is greater at high speed than at lower speeds. There should be a difference between; ‘v’ and ‘v squared’ amongst all that. Also, any ‘bucket effect’ - ‘v squared’ (as # 10).
Must stop thinking – back to the Christmas pud (now adding the brandy).
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Old 24th Dec 2010, 17:06
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thread from 2003 on these boards:

Why are thrust reversers ineffective at low speed? [Archive] - PPRuNe Forums


'Checkboard' seems to give a very plausible and detailed answer near the bottom of the thread.
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Old 24th Dec 2010, 20:18
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Net thrust of the engine is always gross thrust minus ram drag. Take away the net thrust, and you're left with ram drag, also referenced as inlet drag or intake drag. Spool the engine to greater power settings, and the drag rise increases.

Whether a straight turbojet or a turbofan, the inlet drag is considerable; the engine produces much more thrust than what is usable to propel the aircraft, but a certain portion of that thrust is used to overcome the drag produced by the engine. Higher velocities and higher power settings produce more inlet drag.

When reverse thrust is used, the thrust produced by the engine (by fan or by jet exhuast, depending on the powerplant), is diverted. Generally not directly against the direction of travel. The contribution to slowing the aircraft during landing, by re-directed exhaust gasses or fan flow is relatively small. The redirected gas path is often only slightly forward or at right angles to the direction of travel, and doesn't produce significant retarding action or deceleration (acceleration, to be correct).

Some aircraft will produce enough reverse airflow to back the aircraft, some won't. In either case, the contribution of this airflow to stopping the aircraft during landing is relatively slight.

The drag rise in the engine, specifically at the inlet, accounts for most of the retarding action used during landing. The faster the airplane, and the higher the power setting (the faster the engine is spooled, the more thrust produced, but take away that thrust, and the more drag remains. It's this drag that accounts for the effectivity if reverse thrust during landing.

Re-ingestion affects engine longevity, but doesn't account for a change in effective reverse thrust during landing.
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Old 24th Dec 2010, 23:44
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Lomapaseo

I guess the answer to your question about the unit of vertical axis is in the Figure 4 of the link given by Safetypee.

About the second part of your question, I guess there is a bit misunderstanding. The following examples may explain better:

At given airplane and given spesific conditions,

1. if pilot wants to stop airplane at 2000m with manual brake and Max Reverse, the Kinetic Energy which airplane has at touchdown is killed by
10% Tire Brake and Rolling drag combination
40% Max Reverse
50% Aerodynamic Drag (= parasit Drag)
Total 100%

2. If the pilot uses only Max Reverse but never wheel brake, the airplane stops near 3000m, and this time the Kinetic Energy which airplane has at touchdown is killed by
3% Only by Rolling drag
42% Max Reverse
55% Aerodynamic Drag (= parasit Drag)
Total 100%


A few word about dragging force of the reverse thrust vs. speed:

I believe Figure 3 at the link is not precisely drawn for engineering purposes but only to give an idea to pilots about the effect of tools they are using for deceleration. But still, the curve for Max Reverse (darker dashed line) shows practically the the dragging force produced by Max Reverse changes only a little bit vs. speed till the moment it is deliberately closed to idle below 80Kts
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Old 25th Dec 2010, 10:03
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I've always understood that the airflow from the reverser buckets can be likened to an aerodynamic "barn door" shoved out into the airflow. By that I mean that the plume of air from the reversers is "seen" by the ambient airflow as a (semi) solid obstruction and hence creates drag which varies directly with airspeed.

The catastrophic destruction of the circulation around the wing of the Lauda B767 which had the uncommanded reverse at high altitude all those years ago would seem to support this simplistic theory.

OTOH, I may well be completely on the wrong tack - please add learned comments as necessary.
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Old 25th Dec 2010, 10:15
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I've always understood that the airflow from the reverser buckets can be likened to an aerodynamic "barn door" shoved out into the airflow. By that I mean that the plume of air from the reversers is "seen" by the ambient airflow as a (semi) solid obstruction and hence creates drag which varies directly with airspeed.
This is not the case, at all.

The catastrophic destruction of the circulation around the wing of the Lauda B767 which had the uncommanded reverse at high altitude all those years ago would seem to support this simplistic theory.
No, it doesn't.

Disruption of lift doesn't imply that a wall of air creates drag. Multiple factors contribute to the hazard of a deployed reverser in flight, among them lift disruption.
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Old 25th Dec 2010, 13:48
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I may be picking at too fine a detail here but;

I had thought that the loss of lift was over a very small portion of wing and that the loss of thrust itself resulted in the aircraft rolling into that engine.

This roll could not be easily corrected by instinct since the available inboard airelons at that flight regime behind the engine were disrupted by the reverser.

happy to be corrected if I'm confused
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Old 25th Dec 2010, 14:15
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I don't think it is any different than the effectiveness of a reversing propellor, which is greater at high speeds than at low speeds.

For instance, on a turboprop aircraft, the flight idle blade angle of the controllable pitch propellor normally produces zero thrust at the aircraft's stall speed. That same idle blade angle produces positive thrust when the aircraft is static, but a negative thrust component at speeds above stall.

Immediate selection of reverse at high speed (on a flapless touchdown, say) is very effective, but as the forward speed decays the effectiveness of reverse also decays.
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Old 26th Dec 2010, 08:40
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Thank you Guppy. Please explain.
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Old 26th Dec 2010, 14:58
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I already did. Read.
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