"Robert11" wrote in message
...
Hello:

I'm not a pilot.

Saw a show on the first 707 on the WINGS TV channel regarding the barrel
roll
the test pilot did (unexpectedly) on the first test flight of the dash-

They said that he was able to maintain a constant 1g during the maneuver.

Here's where i'ma bit confused.

Seems to me that at the top of the roll, he would have had to be rolling
at
a rate
sufficient to have centripital force equal to 2 g; such that when you
subtract the normal downward
1 g, there's a resultant 1 g left acting in the conventional direction
(tending to pull the engines away from the wings-the same
as if in a normal level flight)

If so. what happens to the 2g when the plane is at the 3 and 9 o'clock
positions?
The resultant there would be over 1g, wouldn't it ?

What am I misssing, or mis-interpreting ?

If someone could walk me thru the maneuver, and the resultant g's at the
various
positions, would be most appreciative.

Also, could exactly the same result(s) be obtained via an aileron roll ?

Thanks,
Bob



Several answers and a correction...

The roll wasn't on the first test flight. It occurred sometime later in the
program.

If you listen to acro folks, what the -80 did wasn't a barrel roll. A
barrel roll is more or less a skewed loop. The -80 did a modified aileron
roll, which began with the aircraft in a climb, so the nose didn't drop too
far by the time the roll was complete. The pilot simply pitched up, held
the elevator in it's normal (cruise) position, and held the yoke over to
command a roll.

The airplane flew at 1 g or so except for the pitch up and the (probable)
pull-out at the end of the roll when the nose was probably slightly below
the horizon.

As to your 1 G vs 2 G at the top of the roll question, aerodynamically, the
airplane was at more or less 1 g the whole time. However, coming over the
top, it accelerated downwards at roughly 2 g's - one from gravitational
forces and one from aerodynamic forces.

Hope this makes things clear...


KB