# 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, 2017 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. Feb 12, 2017 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 Feb 12, 2017 at 21:13
• A320 airborne in less than 800 meters. youtu.be/im32AHm0qq8 Jun 3, 2017 at 12:51
• "the required RWY length is twice as long." Yes, because in the event of an aborted takeoff they need to be able to decelerate to a stop (which will take roughly the same distance) before reaching the other end. Dec 25, 2019 at 17:18

## 1 Answer

At takeoff weight of 78.9 tonnes and -20°C temperature (best case scenario), the A320 needs 2,250 metres (7,400 feet). (This includes the distance to reject a takeoff.)

As for the forces, you forgot to take into account the aerodynamic drag and rolling friction, both will slow down the acceleration.

This PDF from Virginia Tech explains the forces at work and the equations needed.