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In passenger airplane how much thrust is required to takeoff? What factors does it depend on?

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    $\begingroup$ You want exact lbs.ft or % or for B747 or Cessna 172 or just principle behind it? $\endgroup$ – vasin1987 Jul 17 '15 at 4:31
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    $\begingroup$ 0, provided sufficient wind $\endgroup$ – Antzi Jul 17 '15 at 9:59
  • $\begingroup$ Elevation (pressure), temperature and weight are the three that come to mind. $\endgroup$ – Burhan Khalid May 24 '17 at 8:31
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You need at least enough thrust to keep the airplane flying at the speed of minimum drag. This is quite a bit higher than the minimum flight speed, so you should add something to quicken the acceleration to this point. Also, you want to climb eventually, so you better add some more thrust.

Normally, the static thrust of an airliner is at least a quarter of its weight. If the airliner is empty, this can become as much as half of the weight.

One reason is altitude capability: Since thrust goes down with the density of air, the static thrust in the cruise altitude is only a quarter of sea level thrust. The thrust of a modern high-bypass-ratio engine drops with speed, so at cruise speed and altitude, thrust is roughly a sixth of sea level static.

The second reason is safety: The take-off should be continued even after one engine fails in the late acceleration phase. Now a normally two-engined plane has only half as much thrust available and should still get into the air, so it doesn't crash into whatever follows at the end of the runway.

The thrust needed to sustain flight is about 1/18 of aircraft weight, and if you factor in the multiples given above, you will notice that if the aircraft can fly at full thrust in cruise, this fits nicely with a static sea level thrust force equivalent to one third of its weight force.

The factors for take-off thrust are:

  • Runway length: Short runways need more excess thrust for faster acceleration.
  • Runway elevation: Higher places have less dense air, so more speed is needed to get airborne, and less thrust is available from the engines than at sea level.
  • Runway slope. Taking off downhill is equivalent to having a little extra thrust.
  • Air temperature: Colder air is denser, so minimum speed is lower and the engines develop more thrust.
  • Wind speed: A headwind is equivalent to starting the takeoff run at this speed.
  • Minimum flight speed: A low wing loading and low-drag flap settings reduce the speed at which the aircraft can take off.
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  • $\begingroup$ The 1/18 is typical required thrust for airliners at best glide speed, but at $V_2$ it will be higher. I'd say perhaps 1/12–1/10, but I haven't seen those values quoted anywhere publicly, so that's just a guess based on simulator that estimates behaviour from geometry and some fudge-factors. And don't forget some reserve to be able to actually climb, not just sustain flight. $\endgroup$ – Jan Hudec Jan 29 '15 at 20:50
  • $\begingroup$ @JanHudec: True, especially with flaps set for takeoff. But I didn't want to imply that the 1/18th applies for takeoff. If the runway is long enough, though, the minimum thrust needed for takeoff would be indeed about 1/18th of weight. $\endgroup$ – Peter Kämpf Jan 29 '15 at 21:09
  • $\begingroup$ Only very theoretically. You'd need to lift off to the ground effect (where drag is lower) and accelerate in ground effect to best glide speed (that is something like 260-280 knots). $\endgroup$ – Jan Hudec Jan 29 '15 at 21:44
  • $\begingroup$ @JanHudec: I do all acceleration on the ground and lift off just when reaching minimum drag. Of course, climbing would require more thrust, but if the minimum is asked for, this is it. Only the runway needs to be long enough, and I need to disregard maximum tire speed. $\endgroup$ – Peter Kämpf Jan 29 '15 at 22:33
  • $\begingroup$ a) are you sure wheels would still provide less drag at 260 knots than wing-in-ground-effect (or at least wing out of ground effect) and b) are you sure the tires would not blow up long before you got to that speed? $\endgroup$ – Jan Hudec Jan 30 '15 at 19:41
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The amount of thrust necessary is the force required to accelerate the aircraft to take off speed, a speed which allows the wings to generate enough lift to sustain the aircraft in the air.

Factors include

  • aircraft weight
  • runway condition (length, slope, dry/wet)
  • flaps configuration
  • head wind component
  • If it's a multi-engine aircraft, the ability to safely get airborne in the event of a single engine failure is considered as well.

If by "amount of thrust", you mean the acceleration force: engine power ranges from 160hp on small propeller planes to thousands of pounds in turbine jets (Boeing 777's engine is rated at 417kN). In most GA planes, takeoff power = full power. In larger aircrafts, typically it is slightly below full power to save engine wear.


Note that "thrust" is not used to get an airplane off the ground. "Lift" is used to get it airborne and counteract the "weight". "Thrust" is produced by the engine(s) to propel the plane forward, "lift" is created by the wings.

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    $\begingroup$ Most airliners are actually in the tens of thousands of pounds per engine. The 777 is even more, with up to 115,000 lb. per engine (rated... it's been tested to something like 127,000 lb., IIRC.) Fun fact: A single engine from the 777-300ER is sufficiently powerful to fly a DC-9 straight up while still accelerating. $\endgroup$ – reirab Jan 29 '15 at 20:47
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    $\begingroup$ So essentially a 777-300ER is an overweight DC-9 VTOL? $\endgroup$ – Jon Story Jan 31 '15 at 3:12
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It all comes down to Newton's second law of motion in that:

$$F= m \cdot a$$

The trust required for a 70 m/sec. 10.000 kilos takeoff speed and weight with an acceleration of 2g is: 20 Kilo Newtons, it will take 35 seconds and 1225 meters of runway to become airborne .At the moment airplane manufacturers have to trade off Power and Speed to the best fuel economy. but with propellers a max speed is about 662 K hr ( SAAV 2000 ) and jets with a optimun speed is 880 k hrs. Both do not travel at the optimun speed for ANY airplane of around 700 kms per hour wich is where induced drag and atatic drag are at their lowest so a propeller airplane is slow and a jet is fast and the penalty is fuel consuption because the jet have travel at the speed of combustion in the engines and propellers can not be made bigger than they are now (speed of sound limits their size at the tips)

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    $\begingroup$ That's a lot of trust... :) $\endgroup$ – CGCampbell Jul 17 '15 at 11:37
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Notwithstanding exceptions and certain details the operator can decide about, generally the minimum thrust required to satisfy EASA (and FAA) regulations on takeoff performance for civilian jet aircraft is that which allows for a given weight, aircraft configuration, runway, and set of environmental conditions to

  • Accelerate to the value of V1 speed pertinent to that specific takeoff with all engines operating
  • Either reject the takeoff from V1 speed with either all engines
    operating or one engine inoperative within the allowed distance for
    the stop case or
  • Continue the takeoff from V1 speed with one engine inoperative, accelerating to VR and V2 speeds pertinent to that specific takeoff within the allowed distance for the takeoff case
  • Climb out from the runway at V2 speed while meeting a minimum climb gradient defined in the regulations (depending on the number of engines)
  • Clear all obstacles under the takeoff flight path by a certain predetermined margin
  • Reach a minimum height for flap retraction or attain clean configuration and associated speed within the maximum time limit for engine operation at takeoff thrust (usually either 5 or 10 minutes)

This is a rather simplified, condensed version of various paragraphs in EASA CS 25 (Certification Specification for Large Aircraft) and EASA Air Operations regulations, chapter POL (Performance and Operating Limitations - e.g. CAT.POL.A.205 details takeoff distance requirements). For more details, I'm afraid that actually reading the regs is as good as it gets...

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