I've been told before that the runways for large commercial aircraft have to be built to withstand a large amount of force. But I'm wondering exactly how much it is? When a 747 touches down, how much force is being imparted from the wheels to the tarmac? And, further, how much force is imparted into the runway while the plan is braking right after touchdown?

Bonus Points:

  • How much weight does the gear + the wheels absorb and dissipate?
  • I'm sure there's a range from a "greaser" landing with light braking all the way to whatever the highest the main gear is rated for, it would be nice to know both extremes and the mean.
  • 2
    $\begingroup$ Here's an interesting document about the forces on a 747 landing gear, but note a lot of the sources cited within have broken links, so I've not added as an answer. freewebs.com/josfrijmann/files/pro/… $\endgroup$
    – CGCampbell
    Jul 24, 2014 at 21:24
  • $\begingroup$ Also, I think a single strut should be able to take the weight / impact from the whole plane, since that may just be the strut to hit first, so the runway in turn should be able to take the impact of a single strut. $\endgroup$
    – SQB
    Jul 25, 2014 at 9:17
  • $\begingroup$ Yes, certifications require single strut landing (in case of cross-winds, etc). $\endgroup$
    – CGCampbell
    Jul 25, 2014 at 13:04

2 Answers 2


I do not know the exact numbers for oleo stroke and so on, but this is how you would calculate this. A 747-400 weighs 400 tons on take-off and 296 tons on landing, maximum. See here for the source of those figures.

Next is the landing speed, this is approx. 160 knots = 82 m/s. Now let's assume the pilot has misjudged the height and does not flare, but slams the aircraft with the 3° approach angle into the runway. It is supposed to survive this, so let's just continue. This gives you a vertical speed of 4.3 m/s, and at 296 tons this is an energy of 2,750 kWs = 0.76 kWh, which needs to be dissipated by the landing gear. Now I make the assumption that the gear stroke is 0.5 m (those who know please put it in the comments, and I will correct the calculation). We have 0.5 m to decelerate a mass of 296 tons from 4.3 m/s to zero. If we assume a constant deceleration, the force will also be constant and the sink speed will decrease linearly.

The average sink speed during this process is 2.15 m/s, so it will take 0.23 s and an acceleration of 18.5 m/s$^2$. Force is mass times acceleration, so the force is 5,473 kN or 1.23 million pounds. This is just the inertial force to stop the descent. While taxiing to the take-off position, the aircraft will press with 400 tons = 878,000 lbs on the runway, since the wings do not yet produce any lift. This shows that even a hard landing does not stress the landing gear so much - after all, the acceleration is just shy of 2 g, acting on a much lighter plane.

In reality, the landing gear of a 747 is staggered, so the inner main gears will touch the ground first. Also, I expect that the force will not be constant along the whole gear stroke. This will change the details of this approximation, but the general magnitude should not be different.

This answer goes into more details on how to calculate the damage an airplane will do to a given runway or apron.

Now for the braking stresses. The landing field length of a 747-400 is 2175 m, and let's just assume that the pilot forgot to use thrust reversers, aerodynamic drag was switched off that day, and all the braking power had to be supplied by the 16 main wheels. Let's also assume that the pilot uses 1200 m of that field length for braking (I'm just making this up to get to an upper limit of what force will act on the tarmac). Now we need to decelerate from 82 m/s to zero within 1200 m. Linear deceleration means an average speed of 41 m/s, so the whole process takes 29.27 s. Dividing speed by time yields a deceleration of 2.8 m/s$^2$.

To bring the aircraft to a full stop, we have an energy of 995,152 kWs = 276.4 kWh to dissipate along a distance of 1200 m. Using again Newton's second law, we see that this requires a horizontal force of 829 kN = 186,322 lbs, which translates into 51,8 kN = 11,645 lbs per wheel. This is certainly more than what happens in reality, but to put it into proportion: The static load per main wheel at the maximum landing mass is 174 kN = 39,150 lbs (assuming that 4% of the mass is carried by the nose gear). This extreme (horizontal) braking force is still less than 30% of the (vertical) static load, which is well below of the maximum braking coefficient of an aircraft wheel on a dry runway.

EDIT: CGCampell correctly remarked that emergency procedures at take-off weight will produce the biggest braking loads. Now I will calculate the highest possible braking loads, and for this I need this plot of a polynominal for the braking coefficient, which is the ratio between vertical and horizontal forces before the tire skids. I do not know the source; I collected it somewhere in the past and never found a reason to doubt its validity. tire friction coefficient

Shortly before the aircraft comes to a stop, the highest friction coefficient is reached, and then little lift is produced by the wings, so the vertical tire loads are those of the static case. At 96% of 400 tons acting on 16 wheels, this is 24 tons = 235,344 N = 52,907 lbs of downward force per wheel. Since the friction coefficient is 1 at low speed, the same load is transferred horizontally from each wheel to the ground, almost five times more than what I approximated for the landing above. Clearly, going to the limits produces much higher loads.

Formulas used in the calculation:

How to calculate the deceleration of a mass $m$ from a speed $v_0$ to $v_1$

Energy has no absolute value, we just add or subtract energy. The amount of energy change $∆E$ between the aircraft at a speed $v_0$ and at a speed $v_1$ is: $$∆E = \frac{m}{2}\cdot (v_0^2 - v_1^2)$$

The unit of energy is the Joule, which equals the Watt-second or the Newton-meter. Thus 1 J = 1 Ws = 1 Nm = 1 $\frac{\text{kg}\cdot \text{m}^2}{\text{s}^2}$. Metric units are neat, right?

In physics, energy $E$ equals work $W$, and work is performed when a force $F$ travels along a distance $s$. As does the braking force, acting on the moving aircraft: $$E = W = F_{brake}\cdot s$$

How do we get the rollout distance $s$?

We use the fact that the energy change due to deceleration equals the brake energy: $$\frac{m}{2}\cdot (v_0^2 - v_1^2) = F_{brake}\cdot s \Leftrightarrow s = \frac{m}{2\cdot F_{brake}}\cdot (v_0^2 - v_1^2)$$

How can we find the time $t$ it takes to slow the aircraft down?

If the force is constant, the acceleration $a = \frac{m}{F}$ is constant, too. So the speed over time $v(t)$ is: $$v(t) = v_0 + a\cdot t$$

We decelerate, so $a$ is a negative number. But the same works for a positive acceleration, too. To find the time $∆t = t_1 - t_0$ we make use of the fact that v varies linearly over time, such that the average speed is the arithmetic mean between $v_0$ and $v_1$. And time is just distance divided by speed: $$∆t = \frac{2\cdot s}{v_0 + v_1} = \frac{m\cdot s}{F_{brake}\cdot(v_0 + v_1)}\cdot (v_0^2 - v_1^2)$$

With such formulas it is always a good idea to compare the units on both sides. Yup, seconds, so the result is really a time.

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    $\begingroup$ Wouldn't you use takeoff weight, since an emergency could required return to field at max takeoff weight (less fuel used in return). $\endgroup$
    – CGCampbell
    Jul 25, 2014 at 13:06
  • $\begingroup$ @CGCampbell on the aircraft in question you would dump fuel to get to the proper landing weight... $\endgroup$
    – Jae Carr
    Jul 25, 2014 at 13:27
  • $\begingroup$ I know this question is 747, and the video here is a 757, but "ThomsonFly 757 bird strike & flames captured on video", youtu.be/9KhZwsYtNDE this plane returns to the airport with a full load. $\endgroup$
    – CGCampbell
    Jul 25, 2014 at 15:42
  • 1
    $\begingroup$ @CGCampbell Ah, good point, I keep forgetting sometimes procedures have to be put off in the face of an emergency. Apologies. The original question was supposed to be about ranges of normal operation though, I probably should have just started by saying that. $\endgroup$
    – Jae Carr
    Jul 25, 2014 at 18:12
  • $\begingroup$ @CGCampbell: Good point! An aborted take-off should produce the biggest braking forces. More load on the wheels means more braking power. I could also use actual friction coefficients for rubber and concrete to get to the maximum load possible. I will edit the post accordingly. $\endgroup$ Jul 25, 2014 at 19:58

I just saw this question and wanted to add my perspective from the point of view of a structural engineer who flies for fun as a private pilot.

I think the whole energy of impact of the airplane has to be dealt with as a non-elastic collision between the plane and the concrete slab of the runway! Assuming we can ignore the damping effect of landing wheels! The plane hits the runway and becomes one with it and starts to press and drag it down with the total mass of the plane and the mass of concrete slab ( part of it which is affected by impact) and a trapezoid influence area of crushed and compacted aggregate underlayment. In the absence of structural data on the properties of runway slab it is not possible to come up with even a rough numerical estimate of how much the flexibility and dynamic response of slab helps. A vibrating trail of the slab will follow the landing gear lie a long tidal wave that is powered by the impact of the tires and takes away from the shock of impact. This means the deceleration length is more than 0.5 meters, The kinetic energy of landing is dissipated by a large amount by slab and its substructure. So the impact is softer.
As far as braking forces the undesired heating of tires on touch down smoking has an incidental benefit of raising the coefficient of friction and helping with the braking.

As a side note, piloting small lightweight planes is fun and has many educational benefits too. The fact that brakes are not automatic allows you to burn them in wet runways a bit for better grip!

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    $\begingroup$ Why on earth would you neglect the landing wheels' influence? Especially for a 747 which has large stroke suspension. Really, the more logical choice would be to neglect the runway properties and assume it's a rigid body (even when assuming rigid suspension, the deformation of a tyre is a magnitude more than the runway deformation) $\endgroup$
    – Sanchises
    Sep 2, 2016 at 11:01
  • $\begingroup$ I meant ignoring tires' pneumatic deflection only, not landing gear! I said the stroke would be greater even. If you want to be a bit more careful you have to consider the fact that the wings are designed to be very flexible and in a hard landing they bend and gently deflect breaking the impact. In fact the entire fuselage and empennage is designed to flex longitudinally like a dolphin. So much so that passengers at the rear complain of vibration. Analysis of landing should consider the dynamic interaction of plain, air lift, and runway, being mindful even of the harmonics resonances. $\endgroup$
    – kamran
    Sep 2, 2016 at 14:50
  • 1
    $\begingroup$ Still - do you think the displacement of the tarmac will be significant compared to the 0.5m stroke? By completely ignoring the tarmac, any calculation of the runway-tyre force will be off by ~10% tops. Of course, the stresses may be a lot more than you would expect based on the forces, but that's not what the question asks. $\endgroup$
    – Sanchises
    Sep 2, 2016 at 15:07

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