How long of a runway does an A380 need? Created by Sal Khan.

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  1. Something is seriously wrong here. My calculations show that the takeoff run (point at which the A380 rotates) last 40 seconds. It travels about 1600m accelerating at 2.2 m/s2 and its rotation speed is 165 kts (305 km/h). This is calculated for MTOW 575 tonnes and total trust (4 engines) 1,244,000 N.

  2. Thanks a ton. I was a little shaky on constant acceleration, but it is definitely making more sense. It is also so nice to see how intuitive all of this really is

  3. Faster way of doing it is using motion equations, displacement=(velocity + initial v) * time / 2 = (78 m/s + 0 m/s) * 78 secs / 2 = 3042

  4. So much is incorrect here. First, where are you getting 1m/s square acceleration? The actual accelection of a fully loaded A380 is rather more rapid than this, and no A380 takeoff has ever taken 78 seconds. Based on the thrust to weight ratio of a loaded A380 at max thrust, you would be getting about 2m/sec sq. The official takeoff distance takes into account the distance to screen height (V2, 35ft AGL). According to Airbus, the official takeoff distance is 2,950 meters at MTOW. That means that the aircraft would have taken rather less than this distance to achieve V Rotate speed, because the official distance includes climb to screen height. The aircraft may cover as much as 2,000ft (650m) of distance between V Rotate and V2.

  5. question: why didn't you get the average velocity in getting the change in time? while you get the average velocity in getting displacement? is it because getting the change in time involves acceleration already? -.- please enlighten me…

  6. Hey I wondering if there is another way of solving this? My initial thought was to just count how far you travel if you accelerate at 1m/s up to 78 seconds. So at t=1, you have traveled 1 meter. At time t=2, you have traveled 2 more meters (total 3) etc. So would it be okay to just find the sum from 1 to 78? (This gives me a different number).

  7. Well, the real time that a fully loaded A380 takes to take-off is actually between 45 and 50 seconds, I did it many times and timed it. Not 78 … that would shrink the runway a bit, wouldn't it? 😉

  8. Everyone just calm down…this is an explanation of a physics concept to be used on the MCAT (it likely doesn't work IRL but no one is actually going to use these calculations IRL – It. Is. Just. For. The. MCAT.)

  9. The reality is an Airbus A380 from a beginning roll to rotation is 31-35 seconds (Vr @ 170+mph), and a total of 45 seconds till main landing gear is 30-feet above runway (V2) and does this in less than 6,000-ft. By 78 seconds, the aircraft is quite a distance from the airport and already switching to departure frequencies. This is a very large aircraft but by it's design, it can take-off and land in less space than is needed for a Boeing MD-86. The A380 is an incredible machine.

  10. plane takeoff and landing distances vary based on wind velocity, runway slope, and the wight of airplane. This was just simplified down for easy example guys.. calm down.

  11. Yes, but in this example he assumes that the necessary length of the runway is between two identical points on the plane.
    The length of the plane itself is very easy to add after you have calculated the runway itself, it's, like, Runway + Plane.
    No big deal.

  12. ANOTHER THING not taken into account is the fact that during takeoff, power is GRADUALLY increased to maximum … you don't just slam a jet engine into max thrust … that's a good way to destroy an engine. ALSO, the faster an aircraft travels on the ground, the more lift is generated, which means there's less weight (and friction) on the wheels, which means that ACCELERATION IS INCREASING for the ENTIRE takeoff run, and these calculations are therefore WRONG!.

  13. The thing this guy DOESN'T take into account is AIR DENSITY. COLD air is more dense than HOT air, and therefore provides MORE LIFT. Conversely, Warm or HOT air provides LESS lift. So the required runway length will CHANGE … DEPENDING UPON WEATHER CONDITIONS.
    The Youtube clip here -> /watch?v=hxeKmiFudFg shows an A380 taking off from Nunavut, Canada (VERY COLD AIR) from an 8600 ft (~2650m) runway with over 1000 ft of runway still available at liftoff.

  14. an a380 is larger than a 747. Its the biggest passenger plane in the world.. but ur right he must be using numbers that state the plane is at maximum takeoff weight because there are some runways bearly pushing 10,000 feet that 747 and a380 takeoff/land all the time

  15. Can sumone just sum up what he meant when he drew the average velocity on the graph and showed it took the same distance or something.. i was a bit confused there?? is the average velocity just bang in the middle of graph?

  16. Or, s=displacement or length of runway needed, u=initial velocity, t=time, a=acceleration: s=ut + 1/2(at^2): s=0 + 1/2(1 x 78^2) = 3042m

  17. OH GET A GIRL FRIEND! ,……………….who cares! and when u do get a girlfriend dont start talking this shit! CAUSE SHE IS GONNA RUN!

  18. @Omnilegence
    i don't think mentioning something that's more interesting then a airbus or at least useful in the exciting careers of forensics or astrology makes it a not pointless piece of knowledge for general use ,I am probably missing out on useful information in the career in airbus landing calculating tho?

  19. @AtomFA You are just too stupid.

    1. You have a picture of an Inquisitor from Warhammer 40K as your portrait, do you play Warhammer 40K? If so please explain the practical value of it, seeing as your what-I-assume-is-an objection to the video implies the feel the need for things to take interest in to have practical value.
    2. Calculating the runway length is a practical application, understanding how things move is of course practical. Foresics(blood splattering patterns), planetary orbits,…

  20. @AtomFA Hahahaha, I bet you thought that was going to sound like a clever little pedantic comment. This wasn't a demonstration of knowledge, this video was a walkthrough of a practical application of his knowledge of kinematics.

  21. Yeah, It doesn't take into account changing mass due to burning fuel, Friction, Drag, etc.
    It's a good place to start though. Thanks Sal!

  22. @oadekore as you said take-off distance depends on weight. but in this problem since velocity and acceleration are already give(assuming pilot calculated according to the weight of the craft) this becomes a simple kinematics problem.(note that he assumed at 280km/hr the aircraft will take off !! )

  23. The minimum distance for an A380 to takeoff is not as straight forward as you've portrayed in your calculations. the minimum distance for an A380 to take-off depends on weight/mass, which was never mentioned in your calculations. The lighter the weight the shorter the minimum distance to take off. An A380 computes the weight & then computes the minimum take off distance & the thrust input required.Without weight,your calculations would actually be catastrophic

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