# Why does a Boeing 747-8 take more runway length to take off, compared to an Airbus A380?

The Airbus A380 is the largest passenger airliner. That also includes the Boeing 747-8. Meaning, the Boeing 747-8 is smaller than the Airbus A380. As far as I know, smaller planes use less runway length to take off, compared to bigger planes. So, the Boeing 747-8 should consume less runway length than the Airbus A380 to take off, right?

I went to Google, and searched how much runway length is required for a Boeing 747 to take off. The top result showed 3040 m. I then searched how much runway length is required for an Airbus A380 to take off. The top result showed 2900 m.

How is this possible?

• The Airbus A380 is the largest commercial airliner in the world, but is not the largest of "all the planes" ... for that see Antonov A-255 Dec 7, 2021 at 12:20
• "smaller planes use less runway length to take off, compared to bigger planes" not absolutely true. The very small F104 (17m long, 6 m wide) needs 1.4 km of runway to take off. The avro RJ100 (30m long, 26m wide) needs1.2 km of runway. And the RJ100 weights 4 times the F104, but it has only 3 times the thrust (ea. engine provides 30kN, while the single engine in the F104 provides 44kN, 65 kN with afterburner) Dec 8, 2021 at 10:16
• @CGCampbell that very much depends on which metric you use - the H-4 Hercules for example has a larger wingspan than the AN-225.
– Moo
Dec 8, 2021 at 20:39

This table compares some relevant numbers for the two aircraft:

Aircraft Airbus A380 Boeing 747-8i
MTOW 575 000 kg 447 700 kg
Wing Area 845 m² 554 m²
Thrust 4x 348 kN 4x 296 kN
Thrust / Weight 2.42 N/kg 2.64 N/kg

(All values from Wikipedia (A380, 747-8), all at MTOW, standard sea level conditions)

As you can see, the 747-8 has a slightly higher thrust to weight ratio (about 9% higher), which means it can accelerate a bit faster (at MTOW). However, the A380's wing is much larger resulting in a smaller wing loading (about 14% lower) at MTOW. This results in a lower takeoff speed and therefore reduces the required runway for takeoff.

If you want to compare the takeoff distances at different takeoff weights, I recommend to look at the Airport Planning Documents. These will contain graphs like these:

(Airbus Aircraft Characteristics document for the A380, ISA conditions, Trent 900 engines)

(Boeing Airplane Characteristics for Airport Planning document for the 747-8)

The second graph shows that at MTOW the 747-8 requires a little over 3000 m of runway for takeoff (at sea level). At the same weight (about 448 t), the A380 only requires about 1750 m according to the first chart. This is because it has a much lower wing loading of only 530 kg/m² now. With a 3000 m runway, the A380 can lift off at over 570 t, more than 100 t heavier than the 747-8.

• This is key: lower wing loading means lower liftoff speed. Dec 7, 2021 at 14:57
• Since the existing version of the A380 is the one with the short fuselage, the "real" A380 would have a higher wing loading and, consequently, a higher takeoff distance. Pity it never went into production but hampered the version which did with a too large wing. Dec 7, 2021 at 21:22
• Then remove some wing surface area, preferably on the ends where it impedes access and flow at airports. Dec 8, 2021 at 4:05
• @Harper-ReinstateMonica then you have a plane that can access existing airports, but it needs a 7.5 km runway :) Dec 8, 2021 at 10:08
• And the two have similar displacement (?). When I start adding too much wing in KSP I can't get up to speed. Dec 8, 2021 at 11:42

If you play around with wing-loading, T/W ratio, and $$C_L$$, you can get the same or even better result. For instance, the A380 has less wing sweep, which lowers the takeoff distance (increases the $$C_L$$ in the relation above). Case in point, the A380's indicative performance V2 speed is 150 kts, compared to 175 kts for the 747-8.