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KazohiroNakamura
20th Feb 2008, 14:07
does anybody know what is more effective for braking? Is it reverse trust, speed brakes or normal braking?

nakamura

vp1
20th Feb 2008, 14:50
Normally wheel braking is most effective throughout the entire speed range.Reverse trust is most effective at higher speeds,typically speeds above 100kts. Reverse trust can sometimes be more effective than wheel brakes on wet and slippery runways. Like reverse trust,speed brakes provide aerodynamic braking at higher speeds.They also assist wheel braking by dumping lift,thereby putting the entire aircraft weight on the wheels, earlier in the landing roll.

Dream Land
20th Feb 2008, 15:30
I would say spoilers are used to improve wheel braking by killing lift.

KazohiroNakamura
20th Feb 2008, 17:04
So, it safe to assume that @ high speeds, revers thrust is more effective than normal braking. And at low speeds it's the way around

Chris Scott
20th Feb 2008, 17:06
Hi KasohiroNakamura,
You didn't tell us which aeroplane you are discussing, but, if its a modern jet airliner with high-bypass-ratio turbofan engines, you already have good advice.

The wing devices you describe as "speed-brakes" may need some clarification. In flight, some or all of them double as roll-spoilers. On touchdown, usually after main-wheel spin-up, they extend to a higher angle than possible in flight. In this mode, they are called "ground-spoilers" or "lift-dumpers". And that is their chief job, as vp1 clearly wrote. At high speed, with the trailing-edge wing-flaps fully down, the wing generates a lot of lift even when the nose is down, making the brakes less effective. Ground-spoiler drag effect is roughly proportional to the airspeed squared, so they are pretty useless below 80 kts. [Don't forget, the wing flaps also provide drag.]

Fan-thrust reversers are nothing like as effective as the old-fashioned "bucket" or "clamshell" thrust reversers of the 1960s. There is a very good reason for this. In the 1960s, the fans were much smaller, or even non-existent. So it was structurally possible to catch the entire exhaust output of the engine (N1 and N2 air), and deflect it forwards and about 45-degrees outwards. So, even at a restricted power setting, you got useful reverse thrust right down to a standstill. This was invaluable on slippery runways.

Fan reversers do not deflect the air very much forward - much more sideways. This means they are reasonably effective at high speed, because the speed of the aircraft vectors the air forward. But they are less and less effective as you slow down. The other snag is that ALL the non-fan exhaust is still generating forward thrust. By the time you are down to 60kts, the only advantage is that you are not getting forward thrust. So, at low airspeed, idle reverse is probably just as effective as full reverse.

To summarise:
On modern jets, particularly with carbon brakes, anti-skid brakes completely outperform the reversers, except on an icy runway (don't go there). And the ground-spoilers/lift-dumpers are mainly to help the brakes by putting all the weight on the wheels. For most landings, full reverse is noisy, fuel-consuming, can lead to debris being ingested, and wears the engine - for little benefit. Brake units are cheaper than engines. Idle reverse is usually the best compromise.

Turbo-props, like the Hercules, may be a different matter...

Chris Scott
20th Feb 2008, 17:12
Quote:
So, it safe to assume that @ high speeds, revers thrust is more effective than normal braking. And at low speeds it's the way around

Not quite. On a dry runway, modern brakes should always be more effectve than fan reversers, even at high speed.

Chris

KazohiroNakamura
20th Feb 2008, 17:44
Hi there chris, thanks for you detailed explanation. I was talking about a normal jet, lets say the B737.

The reason why I posted this question is the fact that I learned on my ATPL course that brakes are more effective at high speeds. But when I opened other books like for instance 'Ace the technical pilot interview', I read that reverse thrust is more effective than normal brakes. :confused::confused:

So I am a bit confused, also because I know this book contains lots of mistakes. But is this one of those mistakes?

Many Thank

Nakamura

Wizofoz
20th Feb 2008, 17:57
Kaz,

One more example of "Ace the Pilot Technical Quiz" being a load of bollocks!!

What you are being told here is good gen. Wheel brakes are by far the most effective stopping tool (Which is why, in an RTO, ensuring max braking comes before speed brakes or reverse thrust).

Do a search and you'll find numerous references to ATPTQ being filled with inaccuracies.

Chris Scott
20th Feb 2008, 18:01
It depends how old the book is, Nakamura... And you have not mentioned which B737 you are talking about. If it is an old 737-200 with JT8D engines, it has a bucket reverser. If it is a 737-300 or later, it will have CFM-56 with a big fan (and a bypass-ratio of about 5:1), with fan-reversers only, but more modern brakes. This makes for a different ball-game.

But perhaps I should say that, if you are on a limiting and/or contaminated runway, USE EVERYTHING AVAILABLE to stop. In the recent A340-600 runway over-run at Quito, it is said the R/W was so slippery that the anti-skid brakes did not give a good enough deceleration. If they had been able to use reverse, even at idle power, it would have made a big difference. [They could not, because L/G damage fooled the A/C into 'thinking' that it was still airborne.]

FougaMagister
20th Feb 2008, 18:32
When talking about reverse thrust, most posters seem to imply jet engines. On turboprop aircraft, reverse thrust stays effective a lower speeds. Actually, if selecting full reverse as soon as the nosewheel is down, we would stop in a few hundred meters - then soon start taxying backwards! But unless we have to take the first RET, we will also only use idle reverse. On aircraft fitted with spoilers not working as lift dumpers, a short burst of reverse thrust upsets laminar airflow on the wings and so helps to put weight on wheels.

Carbon brakes are more effective when hot; steel brakes when cold.

Cheers :cool:

SNS3Guppy
20th Feb 2008, 21:03
The most effective braking comes from the wheel brakes. The wheel brake effectiveness is dependent on a number of factors ranging from the availability of antiskid to runway condition, brake condition, brake temperature and aircraft weight, etc. All else being equal, brakes are still the single most reliable, and most effective means of stopping the airplane.

Reverse thrust is most effective at high speed when the internal drag of the engine is combined with the faster landing speed of the aircraft to produce the most total drag. Reverse thrust is a function of internal engine drag (intake or inlet drag). The reversing action of cascades, blockers and clamshells isn't there to direct thrust forward...that's fairly minimal and of little effect. Where the true braking action comes in reverse thrust is by removing the forward thrust...all you're left with is drag. The higher the power setting the greater the drag. Net engine thrust is output thrust minus intake drag. Take away the output thrust, and you're left with drag. The higher the airspeed, the greater the effect, and the higher the power setting the greater the effect.

This still doesn't make the retarding action of referve thrust more effective or greater than wheel brakes. It also means that the actual values vary with air density, landing speed, etc, and can't always be relied upon. For this reason, landing distances are calculated without the availability of reverse thrust (in most, but not all cases). Brakes are essential, reverse is gravy.

Ground spoilers aid in braking effectiveness by spoiling left and allowing more weight on the brakes; more weight on the brakes mean more brake energy. Aerodynamic drag on spoilers is greatest at higher airspeeds.

Spoilers and reverse are more effective at high speeds than at at low speeds, but this does not mean they are the most effective components in stopping the aircraft at a high speed or low speed. Just that they're at their most effective stage at high speeds.

hawk37
21st Feb 2008, 12:37
"Fan reversers do not deflect the air very much forward - much more sideways. This means they are reasonably effective at high speed, because the speed of the aircraft vectors the air forward. But they are less and less effective as you slow down."

Let's take a simplified case of the engine with the thrust reverser deployed producing a constant thrust during a portion of aircraft's deceleration (pilot sets and maintains appropriate N1). Perhaps not 100% true, but for aircraft speeds involved during landing seems ball park.

Assume the thrust angle of the exhaust is 60 degrees off of the direction of travel. Then surely the deceleration force is cos(60) * thrust, or .5 * thrust, regardless of the aircraft's forward speed. No? Is my example unrealistic for the aircraft deceleration once reverse thrust is established?

SNS3Guppy
21st Feb 2008, 13:21
The deflected exhaust gasses do not contribute appreciably to slowing the aircraft. Some aircraft generate adequate vectored thrust in this manner to perform a push back, but even in these cases, it's not the thrust vector from the buckets, blockers, or cascades that's producing the retarding effect you feel with reverse thrust. It's the intake drag.

When you remove the forward thrust, your'e not doing so for the purposes of directing it opposite the direction of motion. You're removing it so that all you're left with is intake drag. The effect of exhaust gasses vectored opposite the direction of motion is minimal.

BOAC
21st Feb 2008, 14:45
The 'old chestnut' again!

KN - the best bits of advice I can offer are:-

Refer to the Boeing 737-700 QRP where you will see an addition of up to 1100m in landing run required for 'no reversers' and 415m for one. Draw your own conclusions.

If the surface is very slippery, the wheel brakes can be virtually ineffective, so at ANY speed reversers will probably be better than brakes. Hence the Boeing recommendation to use up to max reverse if required down to 60kts

The best thing reversers normally do is put less energy (=heat) into the brakes which can help with quick t/rounds.

SNS3Guppy
21st Feb 2008, 15:09
Which "old chestnut" is that?

Of course the recommendation is to use reverse; it's not harming the engine and it reduces brake energy. In the case of autobrakes, the function of the autobrake is a rate of acceleration, and it's a combination of whatever forces are stopping the airplane. In a given autobrake setting, a higher reverse thrust means less braking, less brake energy, less heat, quicker turns...but this doesn't imply that the primary means of stopping the airplane, or the most effective or most critical component, is not the wheel brake.

chornedsnorkack
21st Feb 2008, 15:51
If the surface is very slippery, the wheel brakes can be virtually ineffective, so at ANY speed reversers will probably be better than brakes. Hence the Boeing recommendation to use up to max reverse if required down to 60kts


And below those 60 knots?

60 knots is still a speed where you do not like to slide into snowdrifts, ravines, buildings et cetera.

BOAC
21st Feb 2008, 15:56
They say below ok in an emergency to a standstill, of course. I assumed KN was referring to non-emergency ops.

Chris Scott
21st Feb 2008, 17:12
Quote from hawk37 (Chris, can you explain a bit further?):
Assume the thrust angle of the exhaust is 60 degrees off of the direction of travel. Then surely the deceleration force is cos(60) * thrust, or .5 * thrust, regardless of the aircraft's forward speed. No?


Hi, I'll give it a go. :} You are spot-on for the case when the aircraft is stationary. As to why it's different with forward speed, I'm going to try and simplify the applied-maths, but can't avoid using the concept of vectors.

Let's stick at first with your figures, for sake of argument. Imagine the aircraft is pointing North (000 degrees), and the wind is calm.

In the stationary case, let us consider a portion of the mass of the reverser exhaust gas that is coming out horizontally from the R/H side of the thrust-reverser assembly.
The gas is travelling on a [U]heading and track of 060 at an unknown speed (say, 100 kts), thus creating a reaction force (vector) in a direction of 240 degrees. This would tend to push the aeroplane backwards (provided the nosewheel steering remained straight).

Now let's have the aeroplane rejecting a take-off at 150 kts TAS and G/S (wind still calm). Compared with the first case - because it is now coming from a moving aircraft - the exhaust air already has a forward (000 degrees) VECTOR of 150 kts in relation to the earth before it leaves the reverser assembly, i.e., 000deg/150kts.
The reverser will impart to it the same VECTOR as before (060deg/100kts), in relation to the aircraft.
If you combine these two vectors, the RESULTANT in relation to the earth is roughly (in the absence of any trig. tables, I'm afraid) 025deg/200kts.
This produces a reaction force in a direction of 205 degrees, better than in the stationary case.

In addition to the improved direction of the reaction force, its strength is much higher (proportional to the square of the gas speed). As the gas speed is doubled to 200kts - compared with 100 kts - the strength of the deceleration force vector is 4-times as strong as in the stationary case.

And although I asked if we could consider only one portion of the exhaust gas, it is clear that all of it will work in a similar fashion.

I think SNS3Guppy is going to demolish the figures we are using, however. He will point out that your hypothetical figure of the exhaust gas travelling "60 degrees off the direction of travel" is quite unrealistic, particularly for a cascade fan-reverser. And the core air is not reversed at all, remember.

But I hope he will acknowledge that a conservative figure of 90 degrees "off the direction of travel", becomes much lower when you apply the forward vector generated by the aircraft's speed. That is why it's so important to select reverse at high speed. I am convinced he is wrong, however, to put all the braking down to "intake drag".

Propellers can generate reverse thrust vastly more efficiently than jets, particularly at low-speed/stationary, as in the Hercules. But the other reason we do not reverse jets off the stand is the problem of dangerous flying debris and ingestion of same into the fan intake.

Hope this helps.

BelArgUSA
21st Feb 2008, 22:05
Nakamura-san -
xxx
I could not do better than Chris Scott's explanation given to you, which is very accurate.
xxx
The only thing I would like to add about the reversers, is that they block most of the residual thrust of the jet engines. Residual thrust is quite sizeable on modern high-bypass fan engines, and being able to cancel the idle thrust after landing is a further help, to the use of wheel brakes and (spoilers) speed brakes. Many airports request not to use reversers for noise reasons, obviously, and carbon brakes are excellent at doing their job... but I never fail to use reversers, even if only at idle, to block the residual thrust.
xxx
In the 747s that I fly (the old ones), the reversers are installed on the fan. In the initial 747 airplanes, with JT9D-3As, there were turbine reversers as well, installed on the exhaust of the engine, but they were a source of problems, so these were eventually removed.
xxx
Bear in mind that a typical 747 engine has amost a 5:1 bypass ratio. By deflecting the fan exhaust air, you deflect quite a bit of the residual thrust. The remaining "core thrust" accounts for very little of the idle thrust.
xxx
:)
Happy contrails

KazohiroNakamura
22nd Feb 2008, 05:19
Guys, thank you for your info. I got more than I asked for :ok::ok:It is really clear to me now.

This is what I like about our job. On one hand you know what you should know about your aircraft systems in order to operate the aircraft. And on the other hand, if you are more interested of course, you can go deeper and deeper in the technical aspect of the systems. It's like peeling off the layers of an onion.

Thanks again and happy landings :):)

Nakamura

SNS3Guppy
22nd Feb 2008, 06:45
I am convinced he is wrong, however, to put all the braking down to "intake drag".


I said no such thing. I never put all the braking down to intake drag. Just most of it. Anything mechanical experiences internal drag. Your car engine expends a great deal of it's energy not to power your car, but to overcome it's own internal resistance. Same for a jet engine. The net thrust that remains as useful energy for moving the aircraft is what's left over after overcoming the inefficiencies of the powerplant. Take away that thrust by deflecting it elsewhere, and while you may get some benifit from that thrust, the primary benifit is what's left when that thrust is removed. What's left is drag, and that's the chief component of reverse thrust.

ssg
23rd Feb 2008, 17:26
If you want to narrow the parameters a little bit to get to the truth...let's assume that we are landing on a short field runway, at high altitude, max landing weight..and our runway is 100 feet longer then the our allowable numbers.and at the end of the runway is a burning lake of Lava. In the planes that I fly, corporate jets...typicaly, landing distances are not calculated, and not certified with the use of Thrust Reversers...That might tell you something. Since speed brakes on these planes are supposed to be stowed prior to landing...That might tell you something. If you dig a little deaper, you might start looking at how antilock brakes work...early models just stopped the skid, with constant pressure applied, newer designs actualy decellerated the aircraft..and remember, no NASCAR vehicles have antilock brakes..that might tell you something...previous post to icy runways brought up the idea of how to stop a plane without wheel brakes, that might tell you something... Since landing a plane on a short field without an overrun is much more affected by approach speed and touchdown placement, that might tell you something. No doubt all this is affected by pilot technique and experience...not everyone brakes the same, or pulls the TRs the same...but if you really asking this question. If I asked you to only use one system, brakes, TRs, or spoilers, which one would it be...

Easy answer...

Capt Claret
24th Feb 2008, 03:13
I would have thought that if thrust reversers were more effective than wheel brakes, then the MEL would allow dispatch with U/S wheel brakes but functioning reversers. ...

As one who usually lands and uses only reverse to somewhere between about 100 and 80 kts, to keep the brakes cool for the next departure, I know that I can stop a whole lot shorter using brakes than I can reverse. (49,895 kg MLW).

Chris Scott
24th Feb 2008, 14:47
It occurs to me that, with the accountants demanding increasingly short turnrounds, high brake temps must be an increasing factor for short-haul departures.

Use of full reverse theoretically reduces engine life and increases fuel cost, so my feeling has been that, on long runways, idle reverse is the best compromise. But if the brake temps are above limits on the next scheduled departure, I would kick myself for not having used full reverse.

In my company, despite opposition from the accountants, our early A320s were fitted with brake fans. As well as the initial cost and extra fuel cost, their weight occasionally had a slight effect on available payload.

My question is: how many low-cost operators are currently fitting brake fans?

hawk37
24th Feb 2008, 15:13
Well Chris, your math is pretty darn good if you really don’t have access to trig. I get the same 200 Kts (obviously, from 100 * cos(60) + 150) and 23.4 degrees (vice your 25 degrees) using your method, and it does make some sense to me.
I’ll perhaps add a caveat though. What if we used the same 150 Kts aircraft TAS and 100 Kts exhaust gas speed, but choose not to deploy the thrust reversers? In this case, the gases start out at 0 Kts relative to the Earth, and after exiting the aircraft end up with 50 kts forward speed relative to the Earth. This means the aircraft would slow down without deploying the thrust reversers at all!!
However, perhaps more realistic figures such as 300 Kts gas speed, and abort from 100 kts would mean your calculation does show, based on a my chosen 60 deg off reverse direction, that indeed there would be more reverse thrust at higher speeds than lower speeds. Maybe when I’m more bored I’ll run a spreadsheet on it.

So I understand your comment about the improved direction of the reaction force. I’ll have to think more about your comment about the strength of the reaction force, you said “its strength is much higher (proportional to the square of the gas speed)”. But if we used 100 Kts aircraft speed, the strength would be 150 Kts, so I’m not seeing the squared relationship. Again, maybe my spreadsheet will address this. Let me know if you’d like to see it when/if I complete it.

Thanks for your response, it sent me in the right direction (pun intended). Of course, We’re causing great anguish to SNS3Guppy…….

Chris Scott
24th Feb 2008, 16:25
Quote from hawk37:
What if we used the same 150 Kts aircraft TAS and 100 Kts exhaust gas speed, but choose not to deploy the thrust reversers? In this case, the gases start out at 0 Kts relative to the Earth, and after exiting the aircraft end up with 50 kts forward speed relative to the Earth. This means the aircraft would slow down without deploying the thrust reversers at all!!
[Unquote]

Maybe that's what SNS3Guppy means by "intake drag" ! Your first figure [0 kts] is not right, though.

Stick with my example of the wind-calm case, aircraft pointing NORTH.
In the unlikely event that the gases were moving at only 100 kts AFT in relation to the jet pipe, they would be moving FORWARD at 50 kts NORTH (000 deg) in relation to the Earth as they left the jet pipe. Despite that, I accept that they would still be giving forward thrust, not reverse thrust.

So you have identified a probable flaw in my basic argument. Fan reversers are definitely most effective at high speed. I have got the right result, but maybe for the wrong reason... [Don't mention it to SNS3Guppy, will you?] ;)

Back to the drawing board. :{

PS My reference to the "reaction force being proportional to the square of the gas speed" was a simplistic idea based on a reaction force, "R", being proportional to the kinetic energy of the mass of gas coming continuously from the fan cascades:
Kinetic Energy = Mass x Velocity-squared.
Afraid my Physics (Dynamics) is too rusty to go further at the moment.

PBL
24th Feb 2008, 20:37
There is an Airbus document entitled "Getting to Grips with Approach and Landing Accident Reduction" which includes two graphs; "typical" stopping forces distribution and "typical" stopping energy distribution versus stopping distance for a dry runway at sea level and max landing weight. These are Figures 3 and 4 on pp4-5 of the section on "Optimum Use of Braking Devices", which are on pp230-1 of the PDF copy of the manuscript.

PBL