Yeah, makes sense 🙂

..OK, I have nothing more to say, but.. this guy has! – By Gordon Penner, MCFI-Aerobatic, FAA Gold Seal Instructor – Very helpful stuff, which he introduces with “So let us now define the true, non-competition, coordinated, smooth, corkscrew Aileron Roll. I call it the Bob Hoover Roll.”

..http://www.iac34.com/media//DIR_24801/4dba26778a30c360ffff9630ffffe415.pdf – “A Roll is a Roll is a Roll – Or Is It?”

One sentance in particular caught my eye, bearing in mind my zero-G roll bugbear….

“The two main kinds of smooth Aileron Rolls are the 1G and the Zero G Aileron Rolls. First, after a good entry speed is
reached, the pilot pitches up with the wings level to between 20 to 45 degrees nose up. Bob Hoover uses about 60 to
70 degrees nose up with the Shrike Commander. The power should be full at this point and remain full throughout the
maneuver. Once the pitch up target is reached comes the pause, where the pilot pushes the elevator control forward to
establish either 1G or Zero G before rolling. The pilot must not change this G load as the roll input is initiated, and the G
load should stay the same throughout the roll. If more than 1G is felt while the roll is happening the pilot is pulling
enough to be moving over into Barrel Roll territory”.

With respect and deference to your vast experience, Douglasdriver, the ‘Zero G’ option does seem to chime with the manouver Henshaw described, over which you have doubts.

Anyway, a lot of clearly described and seemingly authoritative stuff here – from the point of view of the sport aerobatic sector of aviation only, of course, but nevertheless worth a read.

James, on the ‘ballistic roll’ thing – it seems that this phrase is universally used when describing the manouver sometimes taught to get out of unusual, awkward or confusing attitudes (what Henshaw said he was doing the first time he did it), as a more specific term than ‘aileron roll’. This is because of the complete removal of loads from the aircraft that the manouver is aiming at – called ‘unloading’ – in other words, zero G. To quote a post on another forum from another aerobatic pilot, Eric Sandifer:

“3) Ballistic rolls – this is basically pitching the airplane up a bit, unloading the elevator to the zero lift position (zero G) and rolling the airplane while maintaining zero wing lift. This roll requires no rudder, since there is no adverse yaw at zero AOA. You might throw a tiny bit of oil out the breather doing this one. Secure loose cockpit items too”.

Note how this matches (no rudder) the description by Gordon Penner (in the full version) of the second category of smooth aileron roll.

Thanks for the AP!

I went quiet because I had nothing useful to add 🙂

From http://www.awm.gov.au/catalogue/research_centre/pdf/rc09125z020_1.pdf: – Interesting stuff about his escape and evasion during the battle of France. Also note his remarkably high ‘score’ during such a short period of action – another lost ‘ace’ and a true hero.

It has fibreglass cowlings and wings which will be replaced with alloy and fabric.

Doing a fabric-winged job? Would you like some (new) drawings?

I’ll be in Taunton around New Year, visiting family.

Thanks Carl – very surprising, and interesting.

Of course, to be clear ‘lift’ is measured along the vertical axis of the aircraft, while ‘weight’ in the load factor equation is always vertically downward. In the first of the attached the aircraft is banking and not decending, but the principle is the same, with the resultant load measured in G which happily (locally) agrees with the ‘Newtonian’ definition.

Interesting point there, fanavion. If I understand what you are saying, then the point is that in aviation this equation doesn’t resolve forces into diametrically opposite components – it is ‘dimensionless’ in that it looks at and compares absolute forces as a ratio, though this is messy when the resultant force has a vector in real space. In fact, the actual angle between lift and its vertical vector component is directly related to G as a vector quantity*. In the second diagram I have adapted your own to show why the aircraft is decending in the case of your glider – notice that although lift and weight are of the same magnitude, the vertical component of the lift (which directly opposes the weight) is less than the weight.

The pilot will experience 1G, as described previously by Douglasdriver. That fact and the lift/weight ratio being ‘1’ both equate, as I say, to the wider definition of G.

DD, thanks again. If you could see my desktop at home (and might guess from some of my other posts), you’d know that I am painfully all-too-familiar with chord lines (as well as profile, incidence, washout, camber, thickness to chord, taper, sweep and dihedral)! It’s always nice to learn more about what actually happens in ‘real life’ 😀

*Edit – KISS method: In a balanced turn, bank angle is proportionate to G