Comment by jld
11 days ago
I saw a talk a long time ago about the structural aspects of runway design. The most interested fact I remember was that the stresses on the runway generated by departures was higher than those of arrivals, as departures repeatedly stress the same part of the runway, while jets land on a much more distributed area of the runway.
Plus jets weigh a lot less at arrival than at departure.
When I worked at Boeing, I talked about autoland systems with my lead engineer. He said the autoland was too perfect, as the airplanes would touch down at the same place every time.
This caused that place in the runway to suffer severe fatigue damage.
IIRC, there was a similar problem on aircraft carrier flight decks, where they had to induce some randomized amount of dispersion to keep the tailhook from hitting the same spot over and over again.
I work at a self-driving car company and we observed a similar problem when we did some off-road testing on dirt tracks. The cars were too precise and they were cutting deep ruts into the soil. We too solved it by adding a pseudo-random offset to the track.
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Hmm in distributed computer systems similar patterns exist, e.g. adding jitter to avoid thundering herd effects.
This feels like an essential pattern of the universe or something…
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Similar problem with catenary wires on electric trains wearing into the contacts. The wires zigzag to distribute the wear.
Citation please. Doesn't pass the sniff test.
I suspect the ocean in its various states provides quite a bit of dispersion. Replacing deck plates on a ship is a normal part of maintenance. I find it very hard to believe they'd induce randomness rather than having just that one plate get a different hardness (I know some people will screech about that but trust me, the warship industry is well practiced at such things).
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Do you know how they keep the concrete from cracking? All the pads in general are in way better shape than my driveway, and the driveway has decent support underneath and is subjected much, much less load.
Maybe they use plentiful jagged interlocking sharp granite as the base l? (like railroad track foundation)
Next time you're at SFO, SJC, or any other major hub sitting in the plane before it backs out of the gate take a second to gaze upon and admire how pristine all the concrete pads are, it's really impressive.
Mostly the extensive subgrade work, as I understand. Similar to a road, there’s a bed of sand and aggregate under the concrete surface.
The concrete they use is very precisely mixed to a specification and then it’s tested for adherence to the spec.
A runway is also going to be 3-4x thicker than a 4-6” thick driveway slab. Probably they also use fiberglass or PVC coated rebar instead of plain steel rebar.
Definitely not an expert here but I can read a civil site plan and hire civil site work subcontractors frequently.
Here’s some information on concrete testing: https://www.concrete.org/frequentlyaskedquestions.aspx?faqid...
Also, Grady is one of the best creators on YouTube, I can’t help but watch his full videos whenever they pop up. I always learn something, even if I’m familiar with the subject.
Edit: Granite is not one of the listed materials in Part 4 - Base Courses of the FAA runway construction guide, here’s the entire thing for reference - https://www.faa.gov/airports/engineering/construction_standa...
Most concrete cracking you will see in residential construction and private driveways are either because the ground wasn't compacted well enough before the pour, or more often they didn't put a thick enough layer of stone to prevent the ground from moving. Cutting out depth from the base of crushed stone is often the easiest way to cut costs because it means less material brought in and less material to dig out.
Granted private driveways don't need to be absolutely perfect, but if you want it to last for a really long time you need deeper base layers.
Same with any roadway. The base is everything. I visited some European contries and noticed that the roads seemed to have fewer cracks and potholes than many roads in the US. I had assumed it was better maintenance, but the reason I was told is that they spend a lot more on preparing the base than is typical in the US.
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If I had to guess, all things equal, its probably much thicker than your average driveway.
Yes, for example taxiways and aprons designed to take the weight of large aircraft like A380s can be ~470mm (almost half a metre) thick, and that's only half of the structure with the subgrade and sub-base together being about as thick! Whereas the standard for driveways where I live is 125mm thick.
Not sure if that's a serious question, but your driveway might lack a proper foundation, so the surface is moving and cracks. Also, it's likely not concrete, but tarmac (which is much softer).
A significant number of American driveways are concrete. I'm not going to look up numbers, but I would have to believe that more are concrete than asphalt/tarmac. Unpaved driveways could outnumber both, who knows, but most people with paved driveways have concrete.
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"stress" (in engineering terms) has a particular meaning and is not a generic term. It is not really a synonym for "forces" or "what makes other stuff break"!
Let's look at just the downward forces:
I need some quick figures 1 - an early Boeing 747: 330 tonnes (metric) fully loaded and 160 tonnes empty. A tonne is 1000 Kg.
According to 2: 240 feet per minute vertical is a hard landing which about 1.2m/s. 60 - 180 is considered ideal, so let's go for about 150fpm which is about 0.7m/s.
We have to estimate the maximum downward force on take off. At the point of just before lift off, the plane has rotated to say, let's say 45 degrees, and its engines are delivering enough force and its wings are delivering enough force to push it into the air. Surely at take off, that vertical force is simply the weight of the aircraft, which has remained the same all the time. It doesn't suddenly push down harder than its weight, that's just what it feels like for a passenger.
So let's allow our jet to be empty on landing and also let the acceleration due to gravity be 10m/s/s
So what is the instantaneous downward force of a mass of 160 tonnes dropping at 0.7 m/s compared to a dead weight load of 330 tonnes. Both are in a gravitational field of 10 m/s/s (or m^s-2).
Now this is where I get a bit lost because force = mass x acceleration and the landing plane is descending at a constant velocity of 0.7 m/s. Mind you, the ascending plane is also ... ascending, or will do but it does not have an instantaneous upward velocity so at wheels off it has a vertical acceleration of zero.
Help!
1 https://measuringly.com/how-much-does-boeing-747-weigh/ 2 https://aviation.stackexchange.com/questions/47422/what-is-t...
You might find these points helpful:
1) when an airliner lands, the undercarriage legs, which are telescopic sprung and damped struts, spread the vertical deceleration over a finite period (I cannot say how long it lasts, but I would say of the order of a second or so.)
2) At the point of touchdown, the wings are generating lift about equal to the aircraft’s weight. This decreases quite rapidly, largely on account of the decease in angle of attack as the nosewheel comes down and from the deployment of spoilers, but it would be mistaken to think that the runway is immediately supporting the full weight of the airliner after touchdown.
3) On takeoff, until the nosewheel is lifted to initiate rotation, a significant fraction of an airliner’s weight is being supported by the runway. During rotation, as the angle of attack increases, the lift increases [1] until it exceeds the weight, at which point the airliner lifts off.
4) If we ignore the fact that the undercarriage is sprung, then the airliner has no vertical velocity until it lifts off. Right at that point, however, when the lift exceeds the weight, it gains a vertical acceleration.
I hope this helps!
[1] Plus a vertical component of the engine thrust, but no airliner rotates to anything like 45 degrees - in fact, if it has not left the ground at a rotation angle equal to the angle of maximum lift coefficient (~10 - 15 degrees), it is not going to do so without going faster.
OK so let's look at a second. I think I can get away with this analysis:
On take off f = 330 x 10 = 3300 (units etc)
On landing f = 160 x 10.7 = 1712
So, if you are gentle enough on landing and the aircraft is nearly half the weight it was on take off then the downward force on landing is very much less than that on take off.
That 160 tonnes empty also implies I've thrown the passengers, crew and luggage out too, which is a bit rough. Let's try total fuel at "about 180 to 213 tonnes" and allow that we need a factor of safety, so let's say 40 tonnes of fuel left over on landing.
On landing f = 200 x 10.7 = 2140
So, I'm still going to need some convincing about landing aircraft causing more damage than those taking off.
I was only waffling about rotation angles whilst trying to get to grips with what is going on. I now don't think the engines have anything to do with this analysis. Mind you I am just about old enough to remember watching Lightnings (https://en.wikipedia.org/wiki/English_Electric_Lightning) taking off. Imagine a large silver firework ...
Wouldn’t the force be the force required to decelerate the plane’s vertical velocity to 0 m/s over whatever small amount of time?
Isn’t the acceleration here just the difference in velocity between the plane and the runway so 0.7 m/s/s?
IF the velocity would change from 0.7m/s to 0m/s over one second, the acceleration would be 0.7m/s/s. But if the time span over which that velocity change is (much) shorter, the acceleration would be (much) higher.
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Yeah, the higher departure stress due to greater fuel weight at takeoff was mentioned in this video.
I'm now curious about the engineering of the displaced threshold. This is a portion of the runway that aircraft can taxi onto and use for takeoff but not for landing. I thought (assumed) that the landing was harder on the runway surface than takeoffs, hence the displaced threshold wasn't designed for that force.
The displaced threshold could also be used to ensure obstacle and terrain clearance on landing - simply disallow that portion from being used in order to create an offset from the obstacle. But I don't know whether this is a very common reason for displaced threshold usage.
-- Video also mentions https://skybrary.aero/ which I'd not heard of previously. Looks neat. I'll have to check it out.
Isn't it the opposite? Landing stress a sub-section of the runway while departures stress a larger portion?
I'd be surprised that a heavier plane on takeoff exerts more force on the runway than a lighter plane landing.
And as the departing plane goes faster, doesn't the lift take stress off the runway?
> And as the departing plane goes faster, doesn't the lift take stress off the runway?
Only for a short period between rotation and liftoff. Most of the takeoff roll is spent building up horizontal speed; the pilot doesn't command the aircraft to pitch up before it's ready to lift off.
There will be lift almost as soon as the plane begins moving forward, reducing the weight of the plane, which would seem to reduce downward stress.
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It's the same principle as walking on snow in normal shoes vs. snow shoes. Taking off is normal shoes, a lot of pressure concentrated at the very first part of the runway. Landing is snow shoes because it's distributed across more of the physical surface, and the plane weighs a lot less when it lands anyway.
Planes all start their take off from basically the same position and stress the whole runway, slowly lowering as lift increases, but at their highest weight.
And this is because pilots are trained to keep their nose gear on the centerline, and there are relatively few aircraft types in use which receive the "heavy" after their flight number over ATC. So wheels are going to roll over the exact same tracks repeatedly.
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Watch the video. He says for long range flights, fuel is half of the total weight of the plane.
According to the video the much larger weight is the main or even only cause of takeoff exerting more load on the runway than landings.