Different landing distance with/without rev thrust
Hi,
This is a performance question. I've looked up FCOM 1 for the different landing distance based on landing with/without reverse thrust. My thought is the rate of deceleration after landing is the same with or without reverse thrust with the use of Autobrakes. If so, why does FCOM directs me to add an adjustment to the landing distance with autobrake 1, No Reverser on a dry runway and also on a Good Braking Action runway with varying autobrakes settings with No Reverser? The rate of deceleration should be the same and if so, shouldn't the landing distance be the same too? Thanks for the enlightenment. |
I would think: Friction of the tires on the runway is proportional to normal force (the weight of the plane pressing down on them). At the beginning of the landing run, there is still substantial lift from at least part of the wing, even with spoilers deployed, reducing the effective weight on the wheels, and thus the friction available for braking.
Therefore, a rate-based autobrake system just won't have enough braking to work with, until the plane slows enough and puts enough weight on the wheels. Reverse gets you down to a speed where the brakes become fully effective, in less time and thus in less distance. At which point the autobraking will give a constant rate of deceleration (if so designed), based on the combination of reverse thrust (if used) and friction braking. |
Autobrakes can give you less braking than is available by the conditions of the day, it can't give you more. If you are in a state where the max available braking action is less than that which the autbrakes is trying to achieve, then you will be limited by the conditions. In which circumstances changing the conditions - by, for example, using TRs - should change your stopping performance.
|
Depending on the autobrake setting on your type, you should have reversers slowing you before the autobrake kicks in ( eg 4 seconds on the A320 with autobrake low) |
|
Originally Posted by pattern_is_full
(Post 10266706)
I would think: Friction of the tires on the runway is proportional to normal force (the weight of the plane pressing down on them). At the beginning of the landing run, there is still substantial lift from at least part of the wing, even with spoilers deployed, reducing the effective weight on the wheels, and thus the friction available for braking.
Therefore, a rate-based autobrake system just won't have enough braking to work with, until the plane slows enough and puts enough weight on the wheels. Reverse gets you down to a speed where the brakes become fully effective, in less time and thus in less distance. At which point the autobraking will give a constant rate of deceleration (if so designed), based on the combination of reverse thrust (if used) and friction braking. Thanks for all the replies above. Please share more |
Whilst on ground school for the 146 some 20+ yrs ago we were told that regulatory requirements did not take into account reverse thrust for landing distance requirements, it was an unaccounted stopping bonus. Thus on the 146 the landing distance required was increased by 10% as it didn't have RT.
https://flightsafety.org/files/alar_bn8-3-distances.pdf When discussing landing distance, two categories must be considered:
|
All times are GMT. The time now is 09:43. |
Copyright © 2024 MH Sub I, LLC dba Internet Brands. All rights reserved. Use of this site indicates your consent to the Terms of Use.