Let's get this cleared up...

[Guest majhankee]

Does a plane need more runway length to takeoff or land? I'm talking about Boeing and Airbus jets by the way.I was just wondering because this question has always bothered me.

[Guest addonzealot]

Hi majhankee,I think the takeoff and landing distances of all aircraft small and large are affected by the following factors: Wind, density altitude, weight, and runway surface conditions. Hope this helps.

[Guest PARADISE]

Generally speaking, an airplane needs less runway for landing than takeoff. Things like thrust reverse, brakes and spoilers aid to shorten the stopping distance, where thrust is the main takeoff component. As stated above wind, weight, runway condition, thrust available are some of the things that determine the runway lenghts required and are usually different for every situationLiken it to a car, under normal circumstances, a car can stop from 60mph to 0mph in a shorter distance than it needs to accelerate from 0mph to 60mph.

[Guest RonB49]

And to add to what PARADISE said, you can check it out yourself. At airports with parallel runways (both on the same side of the terminal), the takeoff runway is usually the one nearer the terminal. This favors departing planes (a shorter taxi with fewer hold points) and delays arriving flights as they wait on taxiways if another plane is taking off. This helps to keep traffic flowing through the terminal.Look at a few airports of this type (KLAX, KLAS, KSEA, are examples) and notice the the runways nearer the terminals (departures) are longer than the runways farther away (arrivals).R-

[tf51d 284]

>And to add to what PARADISE said, you can check it out>yourself. At airports with parallel runways (both on the same>side of the terminal), the takeoff runway is usually the one>nearer the terminal. This favors departing planes (a shorter>taxi with fewer hold points) and delays arriving flights as>they wait on taxiways if another plane is taking off. This>helps to keep traffic flowing through the terminal.>>Look at a few airports of this type (KLAX, KLAS, KSEA, are>examples) and notice the the runways nearer the terminals>(departures) are longer than the runways farther away>(arrivals).>>R-But then there are others like KBOS, KSFO that are the opposite, so I don't think you can base it on that! In general takeoff runways are always closer to the terminal no matter what size the runway is.

[Guest RonB49]

>But then there are others like KBOS, KSFO that are the oppositeOops, I guess I showed my ignorance there or at least my limited knowledge. I have never flown into KSFO or KBOS for real (RW). I do remember, perhaps about MSFS Version 5, puzzling over why KSFO was designed "backward."R-

[John_Cillis 2,406]

"But then there are others like KBOS, KSFO that are the opposite"KSFO is interesting because aircraft in the normal prevailing winds take off on 1R and 1L and land on 28R and 28L. However the fully loaded jets headed for Asia also tend to take off to the west even if 1R and 1L are in use by other jets--such was the case when I flew to Tokyo some years back. I loved flying in and out of SFO--used to do so once or twice a week on business during the early 90's, until I moved to Phoenix. SFO was much more exciting than Sky Harbor and to me it always felt literally like a crossroads of the world. It was so odd to enter a plane there and step off the plane in Asia or Europe or Latin America. I never grew tired of going there.John

[Guest majhankee]

KBOS departing traffic departs from 4L/22R and landings happen at 4R/22L. 4L/22R is closer to the terminals so actually, KBOS fits into that category too.

[RFields5421 1]

Most of the airports designed and built from new in the past 40 years have the longer runways closer to the terminals and used for takeoffs - with shorter runways farther out and used for landings.There are plenty of exceptions with older airports like KSFO and KBOS where the outer runways are longer because they were built after the initial inner runways. In almost all these cases - the ability to expand the inner runways is limited due to encroachment. And when you start getting over 10,000 feet - the length difference doesn't matter for the vast majority of aircraft.However, at these airports - the inner runways are still plenty sufficient for takeoff for almost all flights. Should an aircrew request a longer, outer runway for takeoff due to weight requirements - ATC will certainly give them that runway.Aircraft in general need more runway for takeoff because they are heavier.However, an aircraft taking off starts from a known point and has a specific calculated takeoff run.Landing runways should work the same way, but the variables are much greater on landing. A little wind gust over the runway can add a thousand or three thousand feet to the landing very easy.I think Jim Vile's story about almost overrunning a landing at Denver when he was flying an L-1011 is on this forum.

[Guest jboweruk]

I think I can answer this one based on what I've seen at airshows and in videos, takeoff takes generally longer in the same wind conditions, as was stated earlier you have more to stop you than to get you going.

[craig_read 43]

I think the biggest factor here is the weight personnally.. you might lose 100,000Kg of weight on a long haul flight with a 747 as you burn the fuel.. So your landing run will be much shorter..Also.. even at the same weight the approach speed with flaps 30 will be slower.. but I imagine take off roll and landing roll would be close..Let's work it out!******************For example, flaps 10 landing, flaps 30 approach, 250,000Kg, 747-400 (GE)Takeoff ref = 150 knots (~77m/s) (max thrust)Landing ref = 143 knots (~74m/s)So landing is already faster..With -3.4 m/s/s with max selected as autobrakes.. you're going to be stationary in ~22 seconds.. meaning a run of ~(74*22)/2 = 814 metres or ~2685 feet.NOTE: The above calc does not include drag, as the autobrake system will maintain -3.4 regardless of drag, spoilers or reversers.. by changing wheel brake application.. If you want to stop faster you'll need to intervene.. (as far as I understand it)..For the takeoff max thrust is.. 273,600N * 4 = ~ 1,095,000N...F = M * A... so.. F = 1,095,000N.... M = 250,000Kg...Need to work out the affect of drag on the force now..747-400 Drag co-efficient = 0.031Area is roughly.. 4 engines at 2m diameter.. Fuselage at 4.5m diameter.. conservative estimate.. so total area is..(Pi * 2^2) + (Pi * 4.5^2) = 12.6 + 63.6 ~ 76m^2Air density (P) is roughly 1.2kg/m^3So drag force at lift off is.. Drag(F) = 0.5 * P * v^2 * Co-efficent * AreaDrag(F) = 0.5 * 1.2 * 5930 * 0.031 * 76~ 8,380N or 8.38kNSO.. lol.. after all that.. we'll assume drag is linear to speed.. so overall affect is 4,200N.. I know this is crude and drag is exponential but I'm trying to keep the maths simple here.. as it makes very little difference..Now let's be realistic about this force, they won't use 100% thrust on a take off.. so let's assume 90%..F = 0.9 * 1,095,000NF = 985,500NF = 985,500N - 4,200NF = 981,300NA = F / M....... A = 981,300N / 250,000Kg.... 3.95m/s/sSo for the takeoff role at the same weight.. we are looking at.. ROUGHLY.. 19.5 seconds roll... and 750.8 meters... or ~ 2480 feetSo at same weight of 250,000Kg.. with GE engines.. on a 747.. ignoring weather.. and sea level.. with very ROUGH drag calcs and an assumption 90% engine power is used on departure we get..Landing = 2,685 feetTakeoff = 2,480 feetLanding is roughly 200 feet longer.. to STATIONARY.. remember.. and they're not too far from each other.. 60 meters in it.. that's not even the length of the aircraft itself.. In reality of course the landing roll will most likely be less as the weight will have reduced.. if you do the same calcs with different weights for landing and takeoff say a heavy takeoff with 120,000Kg of fuel.. and landing of 100,000Kg less weight assuming the 100,000Kg of fuel has been burned on the flight.. you'll start to see massive differences in landing and takeoff rolls..PHEW..Did I bore anyone?CheersCraigPS let me know if I made any glaring errors.. I did this when half asleep :)

[martinboehme 492]

Another point that doesn't seem to have been been mentioned yet: On takeoff, you need enough runway to be able to accelerate to V1, then abort the takeoff and decelerate back down to a stop. This increases the runway length required for a legal takeoff to more than the amount of pavement usually consumed until the wheels leave the ground.Cheers,Martin

[Guest PARADISE]

>I think the biggest factor here is the weight personnally..>you might lose 100,000Kg of weight on a long haul flight with>a 747 as you burn the fuel.. So your landing run will be much>shorter..>>Also.. even at the same weight the approach speed with flaps>30 will be slower.. but I imagine take off roll and landing>roll would be close..>>Let's work it out!>******************>>For example, flaps 10 landing, flaps 30 approach, 250,000Kg,>747-400 (GE)>>Takeoff ref = 150 knots (~77m/s) (max thrust)>Landing ref = 143 knots (~74m/s)>>So landing is already faster..>>With -3.4 m/s/s with max selected as autobrakes.. you're going>to be stationary in ~22 seconds.. meaning a run of ~(74*22)/2>= 814 metres or ~2685 feet.>>NOTE: The above calc does not include drag, as the autobrake>system will maintain -3.4 regardless of drag, spoilers or>reversers.. by changing wheel brake application.. If you want>to stop faster you'll need to intervene.. (as far as I>understand it)..>>>For the takeoff max thrust is.. 273,600N * 4 = ~>1,095,000N...>>F = M * A... so.. F = 1,095,000N.... M = 250,000Kg...>>Need to work out the affect of drag on the force now..>>747-400 Drag co-efficient = 0.031>>Area is roughly.. 4 engines at 2m diameter.. Fuselage at 4.5m>>diameter.. conservative estimate.. so total area is..>>(Pi * 2^2) + (Pi * 4.5^2) = 12.6 + 63.6 ~ 76m^2>>Air density (P) is roughly 1.2kg/m^3>>So drag force at lift off is.. >>Drag(F) = 0.5 * P * v^2 * Co-efficent * Area>Drag(F) = 0.5 * 1.2 * 5930 * 0.031 * 76>>~ 8,380N or 8.38kN>>SO.. lol.. after all that.. we'll assume drag is linear to>speed.. so overall affect is 4,200N.. I know this is crude and>drag is exponential but I'm trying to keep the maths simple>here.. as it makes very little difference..>>Now let's be realistic about this force, they won't use 100%>thrust on a take off.. so let's assume 90%..>>F = 0.9 * 1,095,000N>F = 985,500N>>F = 985,500N - 4,200N>F = 981,300N>>A = F / M....... A = 981,300N / 250,000Kg.... 3.95m/s/s>>So for the takeoff role at the same weight.. we are looking>at.. ROUGHLY.. 19.5 seconds roll... and 750.8 meters... or ~>2480 feet>>So at same weight of 250,000Kg.. with GE engines.. on a 747..>ignoring weather.. and sea level.. with very ROUGH drag calcs>and an assumption 90% engine power is used on departure we>get..>>Landing = 2,685 feet>Takeoff = 2,480 feet>>Landing is roughly 200 feet longer.. to STATIONARY..>remember.. and they're not too far from each other.. 60 meters>in it.. that's not even the length of the aircraft itself.. In>reality of course the landing roll will most likely be less as>the weight will have reduced.. if you do the same calcs with>different weights for landing and takeoff say a heavy takeoff>with 120,000Kg of fuel.. and landing of 100,000Kg less weight>assuming the 100,000Kg of fuel has been burned on the flight..>you'll start to see massive differences in landing and takeoff>rolls..>>PHEW..>>Did I bore anyone?>>Cheers>>Craig>>PS let me know if I made any glaring errors.. I did this when>half asleep :)wow.....WOW...........that's a lot of work. That's also why we don't let flight engineers sit up front and touch the controls;-)

[Bob Scott 7,716]

Craig; Sorry, mate, but you've missed this one by a mile. Far too many omitted details and oversimplifications leads you to the wrong answer.For takeoff speeds, Vr (rotation speed) is 3-10 knots below actual liftoff speed for most heavy aircraft. And Vref is usually threshold crossing speed, and probably 5 kts above actual touchdown speed.There is a great deal of available braking performance between max autobrakes and max antiskid braking. Your drag computations completely omit the induced drag produced as a cost of wing lift, a very significant component of drag, especially for a heavy transport category jet. And parasitic drag is anything but linear with airspeed...it varies with the square of the airspeed.Without going tediously through all the physics, generally landing roll for a min-roll landing is considerably shorter than takeoff roll for a max-performance takeoff. Anecdotally, for a lightweight F-15C with a thrust-to-weight ratio of 1:1 at takeoff, the aircraft will acccelerate at 1g laterally (14.8 m/s/s). For a heavy jet, it's much less, around 20-25% of that (.25g) in fact. But stand on the brakes...you'll get way better than an average of .25g lat acceleration slamming you forward into the straps. Unless, of course, the runway is icy, there's a tailwind, and your ground spoilers failed to deploy...RegardsBob ScottATP IMEL Gulfstream II-III-IV-VSantiago de Chile

[craig_read 43]

>Craig;>> Sorry, mate, but you've missed this one by a mile. Far too>many omitted details and oversimplifications leads you to the>wrong answer.I didn't think it would be perfect :)>For takeoff speeds, Vr (rotation speed) is 3-10 knots below>actual liftoff speed for most heavy aircraft. And Vref is>usually threshold crossing speed, and probably 5 kts above>actual touchdown speed.I understand what V1, VR and V2 are.. I used V2.. VREF for landing speed.. I think is a realistic assumption.. It depends how far you want to go with it and how 'realistic' you want to make it.. I'm not pretending this will be 100% accurate.. this is a rough back of a packet calc..>There is a great deal of available braking performance between>max autobrakes and max antiskid braking. I probably didn't explain myself here.. I used the max setting.. because I wanted a steady rate of decelleration.. and something that's not totally unrepresentative..>Your drag computations completely omit the induced drag>produced as a cost of wing lift, a very significant component>of drag, especially for a heavy transport category jet. And>parasitic drag is anything but linear with airspeed...it>varies with the square of the airspeed.I already said in my post that I know drag is not linear.. But I wanted to keep it as simple as possible.. so I have a very basic estimate.. I mean it's a question of how far you want to go isn't it? I mean we could start talking about friction co-efficients of tyres on asphalt.. we could talk the lift reducing the effect of that friction during the intial stages of the landing roll and latter stages of the takeoff roll.. We could say that power isn't linear during the takeoff roll... which of course will increase it.. That was just a VERY.. rough.. VERY crude.. quick stab.. to get a ball park figure..>Without going tediously through all the physics, generally>landing roll for a min-roll landing is considerably shorter>than takeoff roll for a max-performance takeoff. Anecdotally,>for a lightweight F-15C with a thrust-to-weight ratio of 1:1>at takeoff, the aircraft will acccelerate at 1g laterally>(14.8 m/s/s). For a heavy jet, it's much less, around 20-25%>of that (.25g) in fact. But stand on the brakes...you'll get>way better than an average of .25g lat acceleration slamming>you forward into the straps. Unless, of course, the runway is>icy, there's a tailwind, and your ground spoilers failed to>deploy...I agree totally that landing roll will be less.. it'll be massively less for a 747-400.. it's probably 100 to 120 tonnes lighter on landing.. I was trying to use my interpretation of somewhat 'normal' conditions.. I am sure if you were to use every ounce of braking available it would probably stop faster.. but I was just trying to be somewhat reasonable with respect to typical operations.. Even my use of MAX is O.T.T.... A 747 will rarely land even with MAX autobrakes.. and it's unlikely leave them applied for the whole roll.. Basically.. I don't think that's a bad ball park figure.. Even if I increase the drag force I used by a factor of 20.. the takeoff roll changes by about 3 seconds.. that will bring you takeoff roll up to about.. 2800 feet ish.. 115 feet or 35 metres (half a length) more than the landing roll.. ish.. And of course I've ignored drag for the landing roll.. completely!I'm not claiming to be bang on.. but I can't see that being a million miles away.. And it doesn't sound totally unreasonable to me.. But.. I'd welcome a more realistic attempt from you or anyone else.. to put it right.. :)CheersCraig