Bunk.

In such a situation, the pilot may choose to push (not "pull") -2G, but it's definitely not necessary.  A simple zero-G
pushover will do nicely.  Why?  Because by definition the aircraft's apparent weight goes to zero, therefore the lift
required will also be zero.  For conventional asymmetrical airfoils, this implies a slightly negative AOA, and the wing
won't be stalled.

In a zero G regime, you can literally fly the aircraft in an unstalled condition with just a few knots of forward
motion.  The wing won't even stall during a hammerhead turnaround, in which the aircraft literally stops atop a vertical
upline, because the wing is not supporting any weight and the AOA is essentially zero.

It's also worth noting that pushing negative G's at low airspeeds can actually result in a "negative stall", in which
the airflow detaches from the underside of the wing.  The results can be *very* exciting, and if you want to experience
it, contact someone in your local IAC chapter -- but don't try it on your own.

Regards,
DJ Molny
Extra 300/L, N133DF
Rocky Mountain Aerobatic Club

Lancair? Maybe some day




-----Original Message-----
From: Lancair Mailing List [mailto:lml@lancaironline.net]On Behalf Of
Bruce Gray
Sent: Friday, June 28, 2002 20:27
To: Lancair Mailing List
Subject: [LML] Re: Legacy Spin Testing?



If you want to have some eye opening fun, next time you're flying at a
safe altitude (5k agl or above), put the airplane at best angle of climb
airspeed with full power, wait a couple of seconds, and pull the power
to simulate an engine failure. Now, wait 2-3 seconds before you do
anything to allow for reaction time and then try to get the nose down
before the airplane stalls. The chances are you'll pull at least 2
negative g's doing it.

Part of the factory checkout in Glasair III's was an example of this
maneuver. Just their way of trying to convince you to climb out at a
high airspeed to give you time to control the airplane if you lost the
engine.

Bruce
Glasair III
www.glasair.org