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!
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