# What's the acceleration and absolute minimal RWY length for an A320 during take-off?

TL;DR: An A320 takeing-off with full thrust is able to reach $V_{TO}$ (77 m/s) in only 26 seconds and in as few as ~ 1.000 metres. Is that really true?

Last time I traveled with an A320 I was again impressed by the acceleration during takeoff. I think it's not outstanding (compared to others) but nonetheless impressive. A cause might be that passengers are somehow exposed to the powers and cannot influence it. But I enjoyed it and it made me doing some calculations.

I gathered the following data:

• usual take-off speed $V_{TO}=150$ knots (source)
• required runway length $2000 \text m$ (source)
• max. thrust per engine $\approx 120 \text{kN}$ (source)
• MTOW $78 \text{tons}$ (source)

For constant acceleration along a path the following rules apply:

$$\left. \begin{array}{} s = \frac{1}{2} at^2 & \Leftrightarrow & t = \sqrt{ \frac{2s}{a} } \\ v = a \cdot t & \Leftrightarrow & t = \frac{v}{a} \end{array} \right\} \Rightarrow a = \frac{1}{2}\frac{v^2}{s}$$

With the data from above this gives an acceleration from 0 to 100 km/h (or 62 mph or 28 m/s) in approximately 18.5 seconds and about 52 seconds for reaching the take-off speed. That sounded a bit lame compared to my 100hp station wagon.

So I investigated further, this time from Newton's point of view, $F=ma$, which gives an acceleration of $a = \frac{2 \cdot 120.000 N}{78.000 kg} \approx 3 \frac{m}{s^2}$. That gives 0…100 km/h in about 9 seconds. This far better coincides with my gut feeling as a pax.

But my second approach, $a \approx 3 \frac{m}{s^2}$, also resulted in the following conclusion and question:

An A320 takeing-off with full thrust is able to reach $V_{TO}$ (77 m/s) in only 26 seconds and in as few as ~ 1.000 metres. Is that really true? I was amazed at this because they all say the required RWY length is twice as long.

• "For constant acceleration along a path the following rules apply" - okay, but what about non-constant acceleration? Aircraft get fast, and fast means more drag, and more drag means lower acceleration per thrust force unit. – Nij Feb 12 '17 at 18:40
• That's exactly my dilemma. Constant acceleration says 18.5 secs from 0 to 100 km/h while Newton says 9 secs. But when I read @ ymb1's answer, it's even worse, given the drag and tyre friction. – PerlDuck Feb 12 '17 at 20:21
• 78.9T is a bit much, and in fact over MTOW on most A320s. Shortest runway we can operate per manual is 1600m. We have DOW routinely into the ~53-55T, so assuming an empty a/c with very little fuel, optimum flaps config and TOGA thrust, I assume as little as 1000m would be enough. Mathematically. There is a lot of extra safety margin to account for various scenarios (net vs gross engine performance, wind change, weight&balance errors) so you will end up with a much larger required take off run – Radu094 Feb 12 '17 at 21:13
• What about wind? Given enough wind, takeoff acceleration time can be exceptionally short. – J Walters Feb 13 '17 at 1:32
• A320 airborne in less than 800 meters. youtu.be/im32AHm0qq8 – pr1268 Jun 3 '17 at 12:51  