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.

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.

"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...

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.

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?

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.