The names of modern tankers in use by the USAF (and some other air forces) generally seem to have a "KC" prefix, e.g. KC-135 Stratotanker, KC-46 Pegasus, KC-767, and so forth.

  1. Does the "K" in "KC" stand for anything specific? According to this Air Force publication regarding designating and naming defense military aerospace vehicles (Attachment 3, page 24), "K" is used for tankers, but why? Is it just an arbitrary non-reserved letter, considering alternatives like "T" and "R" are already taken?
  2. According to the same publication, "C" stands for cargo, so "KC" would be for tanker/cargo aircraft. Are there any tankers that are not also cargo?
  • 1
    $\begingroup$ Did you try googling this before asking? First hit on google 'KC designation aircraft' $\endgroup$
    – ROIMaison
    Commented Apr 30 at 13:11
  • $\begingroup$ @ROIMaison I did but couldn't find any useful results with the terms I used. $\endgroup$
    – wolframw
    Commented Apr 30 at 13:17
  • $\begingroup$ Interesting - According to @ROIMaison's link, "K" for Kerosene and "C" for Cargo. i.e. it's a cargo transport dedicated to one very specific cargo. $\endgroup$
    – FreeMan
    Commented Apr 30 at 13:20
  • 2
    $\begingroup$ Please post that "K = 'kerosene'" as an answer. Wasn't aware of that before. $\endgroup$
    – Ralph J
    Commented Apr 30 at 14:08

2 Answers 2


With regard to the #2 question, yes: when the U.S. Navy used the A-6 in a tanker role for other carrier-based aircraft, it was designated the KA-6D.

Other carrier-based attack aircraft like the A-3 seem to have had similar designations.


The “KC” means Kerosene Cargo. (See comments section for source)

I did some research and I don’t think there’s any U.S. tanker name that doesn’t have the K on it.

Edit : I’ll just add the link here for convenience.

  • $\begingroup$ The DOD specification DoD 4120.15-L does not support your claim. It simply says K = Tanker and C = Transport. $\endgroup$ Commented May 1 at 16:04
  • $\begingroup$ @OrganicMarble, Hmm interesting. See the third comment in the original question. Is that explaining something different? $\endgroup$
    – Wyatt
    Commented May 4 at 15:54

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