I see this stated from time to time, for example in an answer to Can large airliners do aerobatics such as loops?

A barrel roll is one thing; they can be performed at constant load factor of 1 G (assuming sufficient control authority).

In a properly executed barrel roll, however, I do not think there is any point in the trajectory where the vertical velocity is constant. In particular, the entry and exit of the maneuver require transitions from level to ascending, and from descending to level trajectories, respectively. In these two cases, at least, I cannot see how a load factor greater than 1 can be avoided.

This IAC guide states a maximum in the range 2.5 - 3 (and a minimum of about 0.5), though it is probably not intended to be a minimum feasible value.

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    $\begingroup$ +1 for thinking while reading and bringing this to our attention. $\endgroup$ Commented Jan 12, 2018 at 17:25
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    $\begingroup$ Mandatory Bob Hoover ice tea video: youtube.com/watch?v=V9pvG_ZSnCc $\endgroup$
    – Adam
    Commented Jan 12, 2018 at 17:59
  • $\begingroup$ Clearly, when the nose is below the horizon and rising, the G-load must be greater than one. But it doesn't have to be much greater than one. $\endgroup$ Commented Jan 12, 2018 at 22:13
  • $\begingroup$ @CharlesBretana I think you have a point, though there is also the issue of maintaining 1G while inverted yet still keeping the axis of the entire maneuver horizontal. Perhaps a pear-shaped (narrow at the top) barrel would allow that, and in the video of Tex Johnston rolling the Boeing -80, that appears to be what he does, though it is difficult to tell whether the camera is panning. $\endgroup$
    – sdenham
    Commented Jan 13, 2018 at 2:19
  • $\begingroup$ @sdenham, Actually, I misspoke slightly in my earlier comment, you can remove the phrase " "nose is below the horizon' Any time the nose is rising, (or to be perfectly accurate, when the flight path angle is increasing) the vertical component of lift must be greater than the aircraft weight, which is impossible if aircraft G is less than or equal to one. F=ma, and if flight path angle is increasing, vertical velocity is increasing, so vertical acceleration is positive. Can't have acceleration without force, so total vertical Force must be positive i.e., greater than aircraft weight. $\endgroup$ Commented Jan 13, 2018 at 14:36

4 Answers 4


You are absolutely right, a load factor of greater than 1 is impossible to avoid in a proper barrel roll. The barrel part of its name comes from the spiral path the aircraft needs to perform in order to add a centrifugal acceleration which is greater than gravitational acceleration at the top of the roll. This is the condition to ensure a still positive normal load factor in the reference system of the aircraft through the whole roll. Therefore, at the bottom of the roll the normal acceleration needs to be greater than 2 g.

I took the liberty to edit the wrong answer. Given that it has received 14 upvotes at this time, it should teach everyone around here a lesson how much to rely on the number of votes to assess the correctness of an answer ;-).

  • $\begingroup$ I see that you have not repeated my mistake of assigning a unit to a dimensionless ratio. $\endgroup$
    – sdenham
    Commented Jan 12, 2018 at 18:20
  • $\begingroup$ You say "proper" barrel roll. Well then, how about improper? I.e. counteract (or enhance, when overhead) all centrifugal forces by adequate acceleration, effectively (I think) creating a pretty spiral-like, curved trajectory. Or does that not count as one? $\endgroup$ Commented Jan 12, 2018 at 23:37
  • $\begingroup$ @PeterA.Schneider: Perhaps as opposed to a simple roll, aka aileron roll. That wikipedia article has a diagram comparing various kinds of rolls. Thanks to the "do a barrel roll" meme from Starfox, many people don't know what a real barrel roll is. In the Starfox games, you can only do an aileron roll, even though the game calls it a barrel roll, and the meme gets applied to any kind of rotation, which of course mostly aren't barrel rolls. $\endgroup$ Commented Jan 13, 2018 at 5:53

If you do, it would look sloppy as the idea of a barrel roll is the aircraft rotates at a constant radius about a point off one of the wings, creating a helical or spiral flightpath. The erect portion of the maneuver would require around a 2g load factor to perform correctly, the inverted section reducing to 1g or so.


The claim is true in theory (spare some mathematical nitpicking), but false in practice. The mathematically sound version of the claim is:

A barrel roll can be performed at a constant load factor of 1+ϵ, where ϵ is an arbitrarily small number greater than zero.

For easier analysis, we can reduce a barrel roll to a recovery from a dive:

  1. Start in straight and level flight. Use the elevator to maintain a load factor of one at all times.

  2. Push the stick hard over and perform an aileron roll while maintaining a load factor of one.

  3. During the roll, the vertical component of the lift vector will be less than gravity (sometimes even negative). The aircraft inevietably starts to descend.

  4. Once you brought your wings back to level attitude, you will find yourself in a dive, still maintaining a load factor of one.

  5. Recover from the dive.

Two methods to reduce the required acceleration become immediately obvious:

  • Recover slowly, employing a load factor only slightly larger than unity. In practice, the recovery must be complete before the aircraft overspeeds or impacts the ground, both of which happen rather quickly.

  • Perform the aileron roll quickly to minimise the time in non-level attitude. The less time you spend in the roll, the slower the resulting descent will be. If you can perform the roll in zero time, there will be no dive to recover from. In practice, the aircraft's roll rate is limited.

  • $\begingroup$ I think you have made a good case for the feasibility of making some sort of roll maneuver at a load factor close to 1 throughout, I am not so sure it could be called a barrel roll, at least by the IAC definition.Maybe arguably a degenerate form of a barrel roll in the limit as the barrel radius goes to zero and the roll rate goes to infinity, but why not call that an aileron roll? $\endgroup$
    – sdenham
    Commented Jan 13, 2018 at 19:59
  • $\begingroup$ @sdenham - A degenerate barrel roll can look like either a loop (with some slight sideways offset) or like an aileron roll (with some slight corkscrewing), but it is not identical to these as the instantaneous accelerations and roll rates are different. Nevertheless, the key to a real world low-g barrel roll is to perform it very much like an aileron roll and not like a loop. $\endgroup$
    – Rainer P.
    Commented Jan 13, 2018 at 22:49

In a classic moment of airliner history, Tex Johnson barrel rolled the Boeing Dash 80 (the 707 prototype) over a crowd of potential customers. Not once, but twice. And this was the Dash 80's first flight in front of the public.

Johnson reported that he kept the plane at 1 g throughout the maneuver, thus posing no unexpected stresses on it. True that it probably wasn't of competitive aerobatic quality...

but, true to Johnson's intent ("selling airplanes") Boeing did end up selling a lot of 707's.

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    $\begingroup$ Could you expand on the physics of how the entry and exit of that maneuver could have been accomplished at 1G? While it may not have been a textbook barrel roll, it has all the essential features, including the entry and exit accelerations I mention in the question. $\endgroup$
    – sdenham
    Commented Jan 12, 2018 at 18:08
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    $\begingroup$ You can cancel gravity during the barrel roll by superimposing it on a parabolic arc. Sometime before and/or after the roll you will have to exceed 1G to transition from/to level flight. $\endgroup$
    – amI
    Commented Jan 12, 2018 at 20:07
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    $\begingroup$ The question was can an airliner execute a barrel roll at 1G. I believe Tex Johnson answered that question in the most practical way, by doing it. Note the photo in the linked article, with the dash 80 inverted over Seattle. How is that not an answer to the question? $\endgroup$
    – tj1000
    Commented Jan 13, 2018 at 0:24
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    $\begingroup$ @aml That is an ingenious solution, though I think it is pretty clear from the video that this is not the trajectory that Mr. Johnston took the 367-80 on. On the other hand, I think it is quite possible that in his comment, he was tacitly excluding the entry and exit phases. $\endgroup$
    – sdenham
    Commented Jan 13, 2018 at 1:45
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    $\begingroup$ Really? Tex Johnson, test pilot extraordinare, Bill Allen (Boeing's president who really flipped when he saw that), and just about every other aviation person who witnessed this called it a barrel roll. That seems like rather expert opinion. $\endgroup$
    – tj1000
    Commented Jan 13, 2018 at 5:36

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