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This is a follow on to an earlier discussion about pilot perception of G forces compared with actual aerodynamic loads on an airframe. Years ago, the Boeing 707 prototype famously did not one, but two 1G barrel rolls to impress potential buyers. I would imagine the expert pilot just kept the airframe at 1G positive load through the entire maneuver, which would put him weightless at the top of the roll (plane inverted). Is this what actually happened? How is this done?

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marked as duplicate by fooot, Pondlife, Ralph J, Sanchises, kevin Oct 18 '18 at 8:47

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  • $\begingroup$ He was not weightless at the top of the roll, he experienced 1G... Bob Hoover famously demonstrated this by pouring a glass of water while barrel rolling, the G force remained constant throughout the roll. I think you are trying to add the 1G aerodynamic load to the -1G of gravity to get to zero, but the force felt throughout the entire roll is 1G relative to the outside of the loop. $\endgroup$ – Ron Beyer Oct 17 '18 at 21:17
  • $\begingroup$ @RonBeyer Bob Hoover pours iced tea while doing a barrel roll in this video. $\endgroup$ – Gerry Oct 17 '18 at 22:26
  • $\begingroup$ The title could use some improvement. Do you really want to give the answer to the question in the title? PS is there any such thing as a 1G barrel roll meaning 1G from start of initiation to very end? Seems unlikely. $\endgroup$ – quiet flyer Oct 17 '18 at 22:37
  • $\begingroup$ @quietflyer The title seems fine to me: what G forces does the pilot perceive when the aircraft is subject to 1G of aerodynamic forces? $\endgroup$ – Timber Swett Oct 17 '18 at 22:55
  • $\begingroup$ G force on G-meter is same as G-force that pilot perceives, which is same as the G-load exerted on the aircraft structure. It's all the same. $\endgroup$ – quiet flyer Oct 18 '18 at 3:19
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You can't do a barrel roll at 1G. It's a loop pulled out into a coil, and you have to pull pitch to get it started so you are going to be 1.5 to 2 starting out, declining to 1G to .5 at the top, and pulling 1.5 to 2 on the pullout.

So the sensation at the start is like a hard climbing turn (which is what you're doing), pressed into your seat with about the force of a steep turn at the start, declining to normal gravity or bit less (giving a bit of that giddy becoming weightless sensation, but not coming out of your seat) as you ease off the pitch coming around the top, and about the same pressed into the seat force as at the start, or a little more, on the pullout.

A good barrel roll should be no more than 2Gs peak, but how much is a function of pilot skill and of the airplane's ability to keep up the energy through the first half. A jet with oodles of power to keep the speed up in the steep rolling climb can get away with a very gentle pull and will see only about 1.5 or less on the pull up and pull out.

But it's always going to be more than 1 starting out and finishing off and when Tex Johnston said in an interview it was a "1G maneuver", he really meant "not much more than" as in less than 2 (and no worse than a steep turn, unless you screw it up and fall out). Watch the film of the barrel roll and you see a firm pull up at the start with a fairly steep nose down attitude on the exit, which would have required a pull of at least 1.5 G in returning to level flight without over speeding.

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  • $\begingroup$ Outstanding answer John. So the airframe is always positively loaded around 1.5 - 2 G, with pilot "feeling" 0.5 - 1 G at the top (being head down). Perfect, thank you so much. Now I'll work on the control inputs. $\endgroup$ – Robert DiGiovanni Oct 18 '18 at 2:58
  • $\begingroup$ It's a weird sensation at the top of the roll, because "down" is blue sky, and "up" is green earth. Loops are similar, but the pull force and G load at the start and end is much higher because all the energy is directed in the vertical plane. A loop will be 3 Gs during the initial pull and 1G or less over the top. $\endgroup$ – John K Oct 18 '18 at 3:22
  • $\begingroup$ Ah, interesting, I didn't know the "1 G " reference came from an actual interview with the pilot. $\endgroup$ – quiet flyer Oct 18 '18 at 5:11
  • $\begingroup$ If longitudinal accelerations are ignored, one su-31 could barrel roll at 1g normal to plane's longitudinal axis(!) , finishing the roll slightly climbing. $\endgroup$ – qq jkztd Oct 18 '18 at 10:00
  • $\begingroup$ So you all see where we are going with this, using aerodynamic load to compensate for negative gravity over the top (inverted) flight, either with elevator, thrust, or both. I was fooled by the terminology "barrel roll". It is a pitch and roll (now able to make Gs with el). Bob Hoover was a genius. $\endgroup$ – Robert DiGiovanni Oct 18 '18 at 10:53
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By common terminology, when we speak of an "X G manuever" (say X=1 so a "1 G maneuver"), we mean that 1 G is the up-and-down acceleration indicated on the G-meter on the panel, and it also is the up-and-down acceleration (force per unit mass) acting to create stresses or strains on the aircraft's structure (and the pilot's body.) In other words, it is the "felt" acceleration or force per unit mass, or at least the up-and-down component thereof, in the pilot's reference frame. To get the ACTUAL up-and-down acceleration component (or force per unit mass) in the reference frame of the aircraft and pilot, we need to subtract the 1 G acceleration of gravity, or more precisely, the component of the the 1 G acceleration of gravity that acts in the up-and-down reference frame of the aircraft and pilot. Why? Because gravity exerts an equal acceleration (force per unit mass) on every molecule of the aircraft and pilot simultaneously and thus creates no stresses or strains within either, nor does it tend to accelerate the pilot toward or away from the aircraft and seat, so it is not a "felt" acceleration.

Two examples-- straight and level flight -- "felt" acceleration 1G, actual acceleration 0G. "Zero G" ballistic flight-- "felt" acceleration 0G, actual acceleration 1G downward (i.e. -1G).

So, in any maneuver that would commonly be described as a "1 G" maneuver, the G-force perceived by the pilot is +1 G, by definition.

Can a barrel roll be a "1 G" maneuver in this sense? Sure, as the aircraft goes over the very top of the roll. Can it be a "1G" maneuver in this sense from start to finish, including the initial pull-up and final round-out back to level flight? No it cannot. The barrel roll involves an arcing trajectory, and the G-load must be greater than 1-G during the initial pull-up and final round-out.

By the way, the "felt" acceleration as described above, is nothing more or less than the net aerodynamic force generated by the aircraft in whatever axis we are speaking of, or in all three axes if we want to be comprehensive. Normally we are most interested in the forces acting in the up-down direction in the aircraft's reference frame-- that's what the G-meter on the panel indicates-- and this is overwhelmingly dominated by the lift force generated by the wings. So when we speak of "feeling" 1 G (or "perceiving" 1 G), we are just saying that the wings are generating a net force equal to the mass of the (aircraft + contents) times the gravitational acceleration constant times two.

It may be sometimes a bit ambiguous as to whether the term "G-load" is being used to refer the TOTAL "felt" acceleration, or just the component that acts in the upward / downward direction in the pilot's reference frame. Either way, if we label a maneuver a "1 G maneuver", the logical conclusion is the the pilot is perceiving a 1-G acceleration-- i.e. the aircraft is exerting a 1 G force (per unit mass) on the pilot. Just as in straight and level flight.

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  • $\begingroup$ meant to comment in another place. $\endgroup$ – quiet flyer Oct 18 '18 at 5:10

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