Imagine a third world war or a major war involving significant use of carriers, something akin to a modern day variant of the battle of Midway involving carrier battles.

It isn't unthinkable that carriers could sink or be outright destroyed.

In a scenario where multiple allied nations are contributing carriers of different types, say for this particular battle, a couple carriers from the US, and a couple in total coming from the UK and France, the HMS Queen Elizabeth and Charles de Gaulle, respectively.

Being able to share resources throughout the battle (and war) appears to be a useful resource in and of itself.

In particular, what would the aircraft already launched do if their carrier is struck hard and is sinking? Aircraft are valuable assets, ejecting and ditching them in the sea I would imagine is the last option. Left with no other US carriers, no refueling, and no friendly airstrip nearby, is it part of the SOP (Standard Operating Procedure) to request landing on foreign friendly carriers? Even if it isn't part of the SOP, could it be done?

Moreover, should it be done?

My own reasoning:

I imagine that it could be done on at least the Charles de Gaulle.

On the Charles de Gaulle:

I assume the arrestor hook is compatible enough, although it wouldn't surprise me if there are implementation differences between nations on these systems that would render it useless. It would be interesting to know, if anyone knows for sure.

In any case, the Charles de Gaulle can (according to Wikipedia) take the E-2 Hawkeye, so I assume this is a strong indication it would be able to take an American E-2 Hawkeye as well. I don't assume they modify the arrestor hardware on the E-2 that the French use?

As for the F/A-18 and F-35C, I imagine they're using the same arrestor hardware as the E-2, hence they should also be able to land.

As for launching: Again the E-2 can be launched from the Charles de Gaulle, so it can probably launch a US Navy E-2 as well? The catapult link is the same? Same for the F/A-18 and F-35C? There's also the question of the power required to launch them, the calibration thereof, and the profile of how that power is delivered, perhaps this would be a challenge or somehow inflict undue stress to components.

On the Queen Elizabeth (and same with the HMS Prince of Wales):

The HMS Queen Elizabeth uses F-35B, but I believe these perform slow rolling landings.

Is it accurate that the Queen Elizabeth features no arrestor cabling at all? Does it have any nets that can capture aircraft? If not, I believe it would near impossible to safely land any non-compatible aircraft.

If they cannot even land, then launching is a moot question.

  • $\begingroup$ Although aircraft are expensive, pilots and carriers are much more so. It doesn't make a lot of sense to risk pilot and carrier for the sake of the plane with a landing that you aren't sure you can execute safely. With that said, I wouldn't be surprised if people had at least run the numbers about what's possible. $\endgroup$
    – Cadence
    Commented Sep 1, 2023 at 12:32
  • $\begingroup$ @Cadence Assuming that that arrestor gear works the same on the Charles de Gauller as a USN carrier, do you imagine the landing characteristics would be so different that it would be unsafe, in terms of risking the pilot and/or carrier? $\endgroup$ Commented Sep 1, 2023 at 13:37
  • $\begingroup$ This question has multiple issues. It's about naval strategy, not aviation per se, and it is opinion-based. I am voting to close. $\endgroup$ Commented Sep 3, 2023 at 23:46
  • 1
    $\begingroup$ Agree it is opinion based and involves naval strategy. There are some parts of it that are on-topic and factual, but they are extremely broad. Perhaps it could be narrowed in focus to something like "could aircraft A from country B land on carrier C from country D". I am voting to close. $\endgroup$
    – Daniel K
    Commented Sep 4, 2023 at 1:12
  • $\begingroup$ P.S. Most aircraft carriers just don't have enough deck or hangar space to add a full cycle of airplanes from another ship. The real answer is probably no, simply because they wouldn't all fit. $\endgroup$ Commented Sep 5, 2023 at 21:49


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