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I've read several times that the navy versions of jet aircraft need to have a strengthened undercarriage. Here is one example, and another.

I've always just automatically assumed this was needed because aircraft landings are "rough". That is, the aircraft smacks down hard on the carrier deck (or so went my assumption).

Now I find myself questioning this. Naval jets land by catching a wire that brakes them hard. They may also takeoff with a catapult, which is some running device that pulls the nose gear forward at high g's.

So, what is the real reason that naval versions of jets need a strengthened undercarriage?

Federico
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DrZ214
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    I am pretty sure this was explained somewhere on this site, including picture of the aircraft skin (of E-2C or something like that) buckling under the load on touch-down before even catching the wire, but the search does not seem to work for me lately. – Jan Hudec May 23 '16 at 08:40
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    @JanHudec - I addressed this question here: http://aviation.stackexchange.com/a/25090/7394

    "An F/A-18 touches down around 720 fpm (12 ft/s). It's rated to twice that. CTOL fighters typically do about half that. I believe airliners average under 200 fpm (3 ft/s)."

     – Hephaestus Aetnaean May 23 '16 at 12:42

3 Answers3

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The landing on carrier is indeed hard. The reason is not the deceleration (which is handled by the hook), but the touch-down. Since the deck is short, the wires can't be spaced very far apart, so the aircraft must touch down very precisely. Since the precision is better at steeper angle, the aircraft landing on carrier do not flare. At all. So they hit the deck at more than twice the vertical speed compared to typical landing on decent runway.

Jan Hudec
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    They also land with higher speed, enough to go around on a "bolter", powering up as as soon as they hit without having to wait for spool up. Carrier landings are indeed a lot harder than normal ones. – Simon May 23 '16 at 10:48
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    @Simon, as far as I could find, airliners normally touch down with vertical speed below 300 ft/min (and I suppose it is similar with fighters as the speeds are similar too) while given the speed and approximately normal glide-slope, my guess would be the carrier landing would be at 600–800 ft/min. But it's just a guess. – Jan Hudec May 23 '16 at 10:58
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    @JanHudec you're pretty much dead right - F-18's typically land at 700fpm, and are rated up to 1,500. Airliners will typically aim to touchdown at <300fpm and most (/all?) have a maximum of somewhere around 600-700fpm before they need a "hard landing" inspection – Jon Story May 23 '16 at 11:05
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    @Simon To be very clear, they in fact unconditionally power up to full throttle as a matter of procedure as soon as they hit the deck. If they catch the wire then it will stop them no matter what. If they don't catch the wire then they're already roaring at full power for a touch-and-go. – J... May 23 '16 at 21:47
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    Worth noting that vertical speed can also be unpredictable - the carrier deck is not fixed and will have motion of its own. It seems probable that one unfortunately-timed swell could substantially increase that relative velocity... – Andrew is gone May 24 '16 at 18:19
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I'm not sure why "Naval jets land by catching a wire that brakes them hard. They may also takeoff with a catapult, which is some running device that pulls the nose gear forward at high g's." would make you question your assumption that carrier "landings are 'rough'".

In addition to Jan Hudec's description of the landing process, your statement about takeoffs is reasonably accurate, as well. According to Wiki, the C-13-1 catapult (used on many Nimitz class carriers) can shoot 80,000 lbs to 140 knots in 310 feet generating 2.81g with a total force of 225,140 lb (Thanks reirab!). All that stress goes through the nose gear.

Between the launch and the landing, there are considerably higher forces on the undercarriage of the plane, thus it needs to be considerably stronger than that of an equivalent land-based aircraft.

FreeMan
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    'g's are a measure of acceleration. 1 g = 9.8 m/s^2 = 32 ft/s^2 1 knot = 1.688 ft/s, so 140 kt = 236.3 ft/s. So, we have 310 ft = (1/2) a t^2 and a t = 236.3 ft/s. This gives us t = 2.63 s and a = 90 ft/s^2, which is 2.81 'g's. Accelerating 80,000 lb at 2.81 g requires 2.81 * 80,000 = 225,140 lb of force. – reirab May 23 '16 at 19:39
  • As I said, @reirab, I'm sure someone can calculate the G-force. :) Thanks! – FreeMan May 23 '16 at 19:41
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    @FreeMan, better than that - I was once involved in measuring it. The problem we had was actually engine mounting points cracking, particularly on aircraft used for training. We didn't know g levels to expect, so we instrumented one plane and an instructor flew a sequence of increasingly heavy landings, simulating a "worst case" student. Based on that, we set up the system to record up to 30g maximum. The first set of data from real student pilots was fairly useless, because about half the data points were off the scale. On bad hair day, it's not so much "landing" as "controlled crashing". – alephzero May 23 '16 at 21:09
  • @alephzero How in the world were the students walking away from that kind of g-force? – TomMcW May 23 '16 at 23:08
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    Those were the accelerations on the "solid" structure of the plane when the wheels hit the deck, not the same as the accelerations at the pilot's seat. The plane will pivot around the wheels, reducing the acceleration the pilot feels. But they are a measure of what the airframe has to withstand. Incidentally, after the instructor flew a landing that clocked at 10g, his comment was "I'm not being paid extra to take any more of this - that's as much data as you are going to get from me!" – alephzero May 23 '16 at 23:19
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    The lesson being that the back-of-the-envelope can tell you pretty accurately the average force involved in flinging something into the air or bringing it to a halt. Whether it snaps or not is more critically dependent on the max force (or perhaps the max deformation) :-) – Steve Jessop May 24 '16 at 10:10
  • @SteveJessop alephzero was talking about vertical force on landing. My calculation was for the horizontal force applied by the catapult. That said, you're still, of course, absolutely correct about the need to design for maximum expected forces rather than just average ones. According to p12 of this paper, the shocks on an F/A-18 compress up to 12.3 inches. If the acceleration were constant (it isn't,) going from 1,500 fpm to 0 over 12.3" would require an acceleration of 305 ft/s^2 = 9.53 g. – reirab May 24 '16 at 17:03
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It's the first reason you list: the aircraft hit the deck hard. It's not just the undercarriage; the whole airframe has to be ruggedized to withstand the greater shock of carrier landings.

Tyler Durden
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