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gengis
1st May 2006, 18:10
This may be flogging a dead horse somewhat, but I'd like to hear some diverse views on this subject. I am restricting this discussion to the 747-400 in specific.

The Maximum Takeoff Weight (MTOW) is 396 metric tonnes. Maximum Landing Weight (MLWT) - 302 tonnes (-400 Freighter) & 285 tonnes (-400 passengeer). Limiting load factors are +2.5g/-1.0g flaps UP or +2.0/-0g flaps DOWN.

For this discussion, let us define a hard landing as one in the region of 1.5g.

At the maximum landing weights permitted, simple arithmetic will show that this puts 427.5 tonnes on the main wheels (pax airplanes) & 453 tonnes (freighter). In both cases this is well above the maximum takeoff weight of the airplane. Clearly FAR 25 covers this in the required design Factor of Safety for the landing gear so this should still be a landing that the crew "walk away from".

Now let us consider "hard" landings at weights well below maximum landing weight ~ say 240 tonnes (pax) or 260 tonnes (freighter). Once again - at 1.5g on the concrete - this works out to 360 tonnes (pax) & 390 tonnes (freighter). This time however, the force on the main wheels is below the maximum takeoff weight of the airplane. In this event though, neither the limiting load on the landing gear nor the limiting load factor of the airplane would have been exceeded. Is it still incumbent on the crew to write this up in the book for a hard landing check or not? My personal take on this is..... yes. Better to err on the conservative side, for the sake of all, not least the next set of crew.

The permutations of possible weight vs vertical accelerations on landing can reveal a whole host of varying numbers on the wheels & airframe - such as very very light landing weights (such as during aircraft ferry)...... So would you say that 1.7g at a landing weight of 190 tonnes (amounting to 323 tonnes at touchdown) be less stressful to the airplane than a 1.5g landing at 240 tonnes (giving 360 tonnes) ....? Most guys i know think of hard landings in terms of number of "g"s on landing, but as the above suggests, it has surely got to be correlated with the weight as well.

Rainboe
1st May 2006, 18:34
I don't think it's purely providing a vertical acceleration to a large mass on the main undercarriage on touchdown. More comes into it- how about fuselage bending? The difference between take-off and landing weights is accounted for by fuel which is in the wings and centre fuselage which doesn't really affect the fuselage when the plane hits the deck. The fuselage still takes stress as well, the tailplane and engine mounts also.

The African Dude
1st May 2006, 19:28
It is my understanding that the landing gear is specified to absorb enough landing stresses, when operating within aircraft limits (mass and vertical acceleration), that the airframe is not subjected to more load than it has the ability (or design recommendation) to absorb. I also beleive, although do not have aerospace engineering experience outside Uni, that the maximum takeoff weight is more determined by the aerodynamics in terms of design lift available, and that as the aerodynamic design of the aircraft is refined an iterative process accomodates the landing gear weight (which correlates proportionally to some loose extent with the loads it is required to absorb) as supported by that aerodynamic lift.

The maximum takeoff weight therefore, is bound to be only a proportion of the loads which the landing gear can support. Landing at MLW at Maximum allowed ROD, plus the safety factor to which you refer, is much greater in terms of effective tonnage - as you showed with your calculations. Therefore I would write up a heavy landing if it exceeded the Mass-ROD limits specified in Std Operating Proc.'s for a normal landing. Otherwise it would be reasonable to trust that the standard loads from normal (SOP adherent) operation would be happily accomodated by the landing gear, and not fall into the 'heavy' category.

Hope that makes sense to somebody. :rolleyes:

Edit: Rainboe, comparing relatively light landings with ones involving greater fuel loads in the centre of the aircraft: I would imagine there might be an effect from inertia at high ROD, effectively introducing an additional downward bending moment to the centre of the fuselage (NLG and MLG before and aft of this point acting as 'simple' supports in the model). Whether this would introduce greater tensile (under fuselage) or compressive (top of fuselage) stresses and if so, what the magnitude of them would be, I wouldn't want to guess...

Rainboe
1st May 2006, 20:22
The problem is in a heavy landing, everybody is looking outside and nobody will glance at the VSI, so unless the aeroplane has a flight recorder than will spew out the figures, it takes an estimation of how hard ones backside was compressed to decide if a heavy may be on the cards.

I think inertia has little to do with a heavy landing. Light or heavy, the descent of the aeroplane is arrested in a few inches, so fuselage bending moments will be unaffected. A large amount of fuel will simply stress the wing structure. The major determinant of Max take-off weight has to be structural strength of the Gear, with wing strength and lift having a lot of effect too.

gengis
1st May 2006, 20:45
Rainboe:

"The problem is in a heavy landing, everybody is looking outside and nobody will glance at the VSI, so unless the aeroplane has a flight recorder than will spew out the figures, it takes an estimation of how hard ones backside was compressed to decide if a heavy may be on the cards."

The ACMS does provide an indication of the landing "g"s but i do feel that this should not be the primary determinant. I stand corrected, but the ACMS readout sampling rate is far lower than the Quick Access Recorder so at best it will provide one with a very approximate value. Notwithstanding this, i feel that the matter is not so straightforward, as i have attempted earlier to draw correlation between landing "g"s & landing weight.

"I think inertia has little to do with a heavy landing. Light or heavy, the descent of the aeroplane is arrested in a few inches, so fuselage bending moments will be unaffected. A large amount of fuel will simply stress the wing structure. The major determinant of Max take-off weight has to be structural strength of the Gear, with wing strength and lift having a lot of effect too."

This is a good point; nevertheless, would you say that the effects of "g" on the airframe (fuselage/engine mounts etc) during landing are the same as they are in turbulent flight? 1.5g at landing = 1.5g in flight? If so, then at the levels we are speaking of, the airplane is well inside its flaps down load factor limit of +2.0g. Unless of course we are all missing some other effect on the airframe that is peculiar only to the landing?

1st May 2006, 20:48
For this discussion, let us define a hard landing as one in the region of 1.5g.

This may be VERY low. You may want to think about instantanoeus accelerations closer to 2.5'g' than 1.5'g' - which drastically changes much of the argument relating hard landing loading cases and in-flight 'g' envelopes....

Bear in mind also that QAR/FDR data is usually sampled at quite low rates, so it's very easy to miss the true peak decelleration actually experienced by the aircraft.

The African Dude
1st May 2006, 21:17
Hi again Rainboe,
"The major determinant of Max take-off weight has to be structural strength of the Gear, with wing strength and lift having a lot of effect too."

Well, I'm not sure I agree, for the following reason. If you were to calculate the forces on the gear at Mtow and during an average weight 2.5g (Mad (Flt) Scientist) landing, the structural strength of the gear isn't really limiting at T/O.

Denti
1st May 2006, 21:31
On our fleet a hard landing is defined as > 2.00g, if we have a touchdown harder than that we get a printout stating the fact and requiring a tech log entry and hard landing inspection. But that is on 737s and not 747s, but would be surprised if it were different for the big birds as the limit is 2g with flaps down.

zlin77
1st May 2006, 22:45
Just to veer slightly off the topic, and please correct me if I'm wrong.
I believe that the certification requirements on Boeing A/C require the capability to land at Max T.O.W. providing the sink rate does not exceed 360f.p.m.
I'll let the experts work that out in G's!

The African Dude
2nd May 2006, 00:06
6ft/s = 1.829m/s
F = ma = 396000 x 9.81 = 3.884MN due gravitational loading only, i.e. "g"
Now, arresting ROD in half a second:
F = ma = 396000 x (1.829m/s)/(0.5s) = 1.448MN
This gives total force on aircraft of 3.884 ("g") + 1.448 = 5.332MN = 1.37g
Arresting ROD in a quarter of a second = 2.897MN = 1.75g

Why am I sitting here doing this instead of studying for my aircraft structures exam on Wednesday??

I don't even know if it's realistic, but seeing as we've come this far, assume a strut (well, oleo) length of about, I dunno, 1.5ft? = ~0.7m
Average speed of strut compression = half of original speed (vertically) = (1.829/2) = 0.9145m/s
so for a 0.7m oleo at that ROD it would reach full compression after 0.76 seconds.. so refining the original calculation with that time of deceleration would yield a g-force of about 1.25g..

edit to add back of envelope calc for entertainment purpoes only!

john_tullamarine
2nd May 2006, 01:32
Unfortunately gear certification justification and testing is a bit more complex than that .... MFS plays with that sort of stuff for his day job so I think I would take some note of his observations ...

XPMorten
2nd May 2006, 12:21
Inertia is definitely a consideration :ooh:
Hard landing;

Cheers,

M

gengis
2nd May 2006, 13:35
African: "6ft/s = 1.829m/s
F = ma = 396000 x 9.81 = 3.884MN due gravitational loading only, i.e. "g"
Now, arresting ROD in half a second:
F = ma = 396000 x (1.829m/s)/(0.5s) = 1.448MN
This gives total force on aircraft of 3.884 ("g") + 1.448 = 5.332MN = 1.37g
Arresting ROD in a quarter of a second = 2.897MN = 1.75g
Why am I sitting here doing this instead of studying for my aircraft structures exam on Wednesday??
I don't even know if it's realistic, but seeing as we've come this far, assume a strut (well, oleo) length of about, I dunno, 1.5ft? = ~0.7m
Average speed of strut compression = half of original speed (vertically) = (1.829/2) = 0.9145m/s
so for a 0.7m oleo at that ROD it would reach full compression after 0.76 seconds.. so refining the original calculation with that time of deceleration would yield a g-force of about 1.25g.."

All this sounds great but it is based on the premise of 6 ft/s or 360 fpm. On the -400, you are right: you would only get to 360 fpm after having flared the airplane - hence a very decent outcome. But for guys who fail to flare/flare late & you are not quick enough to catch it for them, i'm afraid you are looking at much more like 800 fpm on the wheels. That changes everything! :ugh:

In fact, Kinetic Energy = (mv2)/2; so at twice the vertical speed, the kinetic energy absorbed by the landing gear is 4 times!

I reckon it's very difficult to quantify what other effects something like this could have on the rest of the airplane. On the one hand if the airplane is certified for +2.0g in flight with flaps down, but on the other hand what additional problems would a 2.0g landing do to the airplane? Touching down without flaring or a late flare puts 4 times the energy on the wheels

The African Dude
2nd May 2006, 14:47
gengis, what you say is right - I'd guess that in situations where there was a bad/nonexistant flare, particularly when heavy, it would be fairly clear to the crew that the landing was sketchy and that the a/c should be checked over.

Perhaps it just comes down to a judgement call, given the difficulty involves in calculating accurately the energy and consequently stresses absorbed by the gear and through the load path of the airframe?