Airliner failing to execute flare during landing

Question

If an airplane such as Boeing-777 doesn't do a flare and instead continues "50, 40, 30, 20, 10..." at a steady rate, how hard will it hit the ground? (Added later) Specifically, will it be a "hard landing" or an outright crash? How many Gs will it feel like?

Answer 1

Too hard,... in some form or another, ranging from too hard for comfort to too hard to maintain structural integrity.

Feelwise for a passenger it's like jumping off the top of a household ladder. It is quite survivable, be it at a cost. It's a serious blow and it hurts unless you train for it.

It most definitely causes damage to the aircraft. The impact easily exceeds the tolerance range of the suspension, leaving quite some of it to be absorbed by the relatively fragile material of the fuselage.

A good example is the case of Speedbird 38 landing at Heathrow in 2008. Lack of energy made it impossible for this 777 to flare. Trying to flare would have made that aircraft stall, which would have been way more damaging. This crash made a lot of people at Boeing really proud. At the expense of the airframe, all survived, thus qualifying it as a 'good' landing.

This kind of accident is very valuable, because it is virtually impossible to put the aircraft's calculated stress resistance to the test that bluntly during certification. So apparently, you are far from the only one who would like to know the answer to that question.

Answer 2

It will be a hard landing, not a "crash". The normal decent rate at typical reference speeds of 130-140 knots is somewhere around 700 fpm. Aircraft carrier landings are pretty much that rate (you are basically descending into the water, but the deck gets in the way) but carrier airplanes are designed to take the forces.

Assuming you touch down on the runway (Speedbird 38 landed on grass, sinking at a very high rate, which ripped the gear out, and it's not really comparable), everybody will know it, you will probably blow tires, some passengers may have sore backs, and there is a good chance of bending something in the gear or in the structural attachments for the gear. But maybe not.

What will happen is the plane will taxi in or get towed in and go for a special hard landing inspection, specified for that type and which will have a dedicated AMM procedure.

This will involve going over the structure and all of the key attachment points that have possibly been overstressed, to find out if things are bent, elongated, distorted, cracked, etc. It could involve visual inspections, dimensional checks, and possibly specific Non Destructive Testing checks like X-ray, Ultrasound, Eddy Current, etc. to detect hidden overstress fractures.

If the plane passes the special inspection, it's good to go back on the line. One-time overloads on metal structures are not a problem if the overloads did not exceed the plastic deformation limits of the structure. In other words, if it's not permanently bent, or ripped, or cracked, it's fine.

Answer 3

If a standard 3 1/2° glidepath is used during final approach and Vref for the airplane is there around 150 knots, that equates to an approx 750-800 fpm descent, resulting in a very hard landing. The airplane could possibly even experience structural damage as a result

Answer 4

It will hit the ground with a vertical velocity equal to whatever its descent rate on the glideslope was. Descent rate, in ft/min, on a 2.5 degree glideslope is just under 5 times its ground speed (in knots). Simpler, take half the ground speed and multiply by 10. So an aircraft at a ground speed of 150 knots will have a descent rate of around 750 ft/min.

Actually doing the math, the numbers for various glide slope angles are as follows:

Glide Slope ---------- Descent rate (ft/min) (as multiple of Ground Speed)    
   2.5 deg ----------- 4.4 x Ground Speed (knots)   
   3.0 deg ----------- 5.3 x Ground Speed           
   3.5 deg ----------- 6.2 x Ground Speed      

By the way, these numbers come from having to convert knots (nm/hour) into feet per minute. 1 Knot is 1.69 ft/sec, and there are 60 seconds in each minute. so you have to multiply the sine of the glide path angle by 1.69 and then by 60....