Comment by gerdesj
14 days ago
"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.
Is it much shorter? An airplane does take a while to settle down. You land tail first and the nose generally takes a second or so to come down.
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