What is more effective for braking?
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I am convinced he is wrong, however, to put all the braking down to "intake drag".
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Some thoughts...
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...
Easy answer...
Bottums Up
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).
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).
Brake fans
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?
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?
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Reverse Thrust Calculations.....
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…….
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…….
The drawing board
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.
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.
Last edited by Chris Scott; 24th Feb 2008 at 16:26. Reason: Typo
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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.
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