One G rolls: a physicist's view

"Kyle Boatright" wrote:


Good analysis. One thing you might try is running the model again at 6 or 8
seconds to complete the roll? This would be would be representative for many
GA aircraft, which have roll rates of ~45 degrees/second? I imagine it
would make a huge difference in the vertical velocity at the end of the
roll.

That was my gut reaction, too, but it is proportional, 1 g downward
for the duration of the roll. So a 10 second roll would end up with a
final downward velocity of 32 ft/sec^2 x 10 sec = 320 ft/sec.

Of course, that is added to the upward velocity at the begriming. So
if you are on a 20=degree upslope at 150mph, and roll at 1 g for 6
seconds, the final downward velocity will be 192ft/sec - initial
upward velocity of 150 mph *sin(20) * 1.46 ft/sec / mph =
192-75=117ft/sec, still a pretty steep dive. But in your RV-6, you
roll faster than that, don't you? If 4 seconds and 130 mph, that would
convert your 20-degree climb into a 20-degree dive.
--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

"alexy" wrote in message
...
"Kyle Boatright" wrote:


Good analysis. One thing you might try is running the model again at 6 or
8
seconds to complete the roll? This would be would be representative for
many
GA aircraft, which have roll rates of ~45 degrees/second? I imagine it
would make a huge difference in the vertical velocity at the end of the
roll.

That was my gut reaction, too, but it is proportional, 1 g downward
for the duration of the roll. So a 10 second roll would end up with a
final downward velocity of 32 ft/sec^2 x 10 sec = 320 ft/sec.

I think the net velocity would be what you suggest, but wouldn't the
downward acceleration for the first and last quarters of the roll be less
than one G, because the aircraft is still generating lift in the "up"
direction? However, during the middle 1/2 of the roll, the aircraft's
acceleration is between 1 and 2 G's downward. At the 90 degree point in the
roll, the aircraft is, in essence, falling at 1 G. At the 180 degree mark,
the aircraft is falling at 1 G and its lift is generating another G in the
down direction, for a total of 2 G's.


Of course, that is added to the upward velocity at the begriming. So
if you are on a 20=degree upslope at 150mph, and roll at 1 g for 6
seconds, the final downward velocity will be 192ft/sec - initial
upward velocity of 150 mph *sin(20) * 1.46 ft/sec / mph =
192-75=117ft/sec, still a pretty steep dive. But in your RV-6, you
roll faster than that, don't you? If 4 seconds and 130 mph, that would
convert your 20-degree climb into a 20-degree dive.
--
Alex -- Replace "nospam" with "mail" to reply by email. Checked
infrequently.

I'd bet the 6 seconds is probably realistic unless I'm really trying for a
quick roll. Also, let's be honest... I probably cheat on the pull-out and
begin a slight pull once the bank angle is under 90 degrees.

KB

"Kyle Boatright" wrote:


"alexy" wrote in message

That was my gut reaction, too, but it is proportional, 1 g downward
for the duration of the roll. So a 10 second roll would end up with a
final downward velocity of 32 ft/sec^2 x 10 sec = 320 ft/sec.

I think the net velocity would be what you suggest, but wouldn't the
downward acceleration for the first and last quarters of the roll be less
than one G, because the aircraft is still generating lift in the "up"
direction? However, during the middle 1/2 of the roll, the aircraft's
acceleration is between 1 and 2 G's downward. At the 90 degree point in the
roll, the aircraft is, in essence, falling at 1 G. At the 180 degree mark,
the aircraft is falling at 1 G and its lift is generating another G in the
down direction, for a total of 2 G's.

Right. I certainly didn't mean a constant 1 g downward. But when
integrating, pair each point on the top half of the circle with the
point directly below it, and the downward components of acceleration
of each pair totals 2 g, so integrating with constant angular velocity
gives you 1 g times the time.

Same logic would also indicate that you will be at the same heading,
although to the side of the original flight path (in the direction of
the roll). Does that jibe with your experience?
--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.