One G rolls: a physicist's view

[AES]

As a follow-on to the discussion of "one G rolls" in this NG some time
back, an analysis of a simplified "physicist's version" of the one G
roll problem, with formulas and graphs, is available at

<http://www.stanford.edu/~siegman/one_g_roll.html>

I'm not sure why I was driven to do this -- maybe nothing better to do
with my time these days? -- but I was, so here it is.

I'll give a quick summary of results first, then a more detailed summary
of the assumptions behind them. Any comments on this will be welcome
(but on the NG, please; not by private email). And before anyone jumps
too hard on my approach or results, please look at the assumptions to
see what I'm claiming and what I'm not.

RESULTS:

Unless I've made some dumb mistake (always a possibility), if an
aircraft starts out in level flight and rolls through 360 degrees
clockwise about its forward axis in 10 seconds, while maintaining 1 G
force on the aircraft and the pilot, and a constant forward velocity, it
will perform a circular corkscrew "barrel roll" type of motion with a
radius of about 80 feet around a curved "guiding axis" (although the
actual orbit will not look much like a corkscrew), and will end the
maneuver displaced about 500 feet to the right, with an altitude loss of
about 1600 feet, and a screaming final downward velocity component of
300 feet/second.

If on the other hand it can enter the roll maneuver *inverted* and with
an initial upward velocity of 150 feet/second (i.e., an initial roll
angle of 180 degrees), and again do a 360 degree, one G roll starting
from and ending up back at that roll angle, it can end up with no final
elevation loss; only a 200 foot maximum vertical elevation rise and fall
during the maneuver; the same 500 foot displacement to the right; and
only half the downward final velocity of the first case.

Those are the theoretical predicitions -- I'll leave it to the pilots on
the group to do the experiments . . .

ASSUMPTIONS:

1) The basic assumption here is that some object up in the sky is going
to be acted on by a constant 1 G transverse force, due to lift and other
aerodynamic foces, gas jet thrustors, attached wires, whatever).

The direction of that force is then rotated around through 360 degrees
in a vertical plane called the transverse plane, with sideways and
vertical coordinate axes x and y; and we ask how the object moves in
those x and y directions as a result of this rotating applied force.

2) At the same time the object may also be moving forward in a "z"
direction perpendicular to that x, y plane -- in fact, if it's an
airplane it is surely doing so. A second assumption is that during the
one G roll maneuver the object's forward velocity in that z direction
(i.e.e, over the ground) remains unchanged.

Note that if this forward velocity were to change during the maneuver,
then that would mean that some additional net force must have been
applied on the object along the z direction, and this would then become
a more complicated problem, and maybe no longer a "one G problem".

3) Drag effects associated with the sideways and vertical motions are
assumed to be either negligible or absorbed in the applied 1 G force.
Angular momentum effects associated with the roll are ignored, because
they can be.

4) Finally, I'm not a pilot. I don't claim to know what pilots believe
is a one G roll; and I have no idea whether a real plane could move in
the way described in the analysis, or what control inputs would be
needed to make it do so. The analysis nonetheless seems to to match up
with what some of the people in the earlier discussion described as a
one G roll; and it gives at least a starting point for understanding
what sort of motions will be required to make that kind of one G roll.

[Denny]

Ain't no such thing as a 1 G roll... 1 G is is sitting motionless in
your chair with your arms on the arm rests... Any perturbation
(physicist jargon for shake rattle and roll) from that requires
accelleration of mass, which translates to G forces.. Just raising
your arms off the arm rests to reach your keyboard requires
accellerating tens of pounds upwards, which thrusts your mass downwards
into your seat bottom, into the chair, into the floor, to the center of
the earth, and you are no longer at 1 G, even if momentarily...
Now ignoring the sideways G loads from the rolling moment, just
beginning a roll in the aircraft requires swinging the pilots mass in
an arc... Taint 1 G anymore... Just ask the rock on the end of a
string...
It's a poor physicist that thinks it is a 1 G roll... I might call it
a Low Gee roll, perhaps, but not a 1 gee... Smoothly done, a roll does
not have abrupt changes in perceived G load, and so is thought of as a
1 G roll... Yet, they overlook the fact that when inverted they have
to have +1G simply to neutralize the -1G from being inverted in the
gravitational field, and another +1G to retain the sensation of normal
weight into the seat of their pants, so they are actually pulling 2G at
that moment...


denny ... who took physics long, long, ago in a world far, far, away,
when logarithms were looked up in a table, and the only 'calculator'
you had was your brain and a fancy pair of rulers ( slide rule..)

[Doug]

Vertical airspeed of 300 hundred feet per second? That is a whopping
18,000 feet per minute! Typical airplane decent is less than 2000 feet
per minute. Also 18,000 feet per second is 180 knots. I dunno..... I
can assure you that it's not normally done this way.

[AliR]

Is this why we have so many problems with the space shuttle? It's trying to
do 1G rolls ;)

AliR.

"Denny" > wrote in message
oups.com...
> Ain't no such thing as a 1 G roll... 1 G is is sitting motionless in
> your chair with your arms on the arm rests... Any perturbation
> (physicist jargon for shake rattle and roll) from that requires
> accelleration of mass, which translates to G forces.. Just raising
> your arms off the arm rests to reach your keyboard requires
> accellerating tens of pounds upwards, which thrusts your mass downwards
> into your seat bottom, into the chair, into the floor, to the center of
> the earth, and you are no longer at 1 G, even if momentarily...
> Now ignoring the sideways G loads from the rolling moment, just
> beginning a roll in the aircraft requires swinging the pilots mass in
> an arc... Taint 1 G anymore... Just ask the rock on the end of a
> string...
> It's a poor physicist that thinks it is a 1 G roll... I might call it
> a Low Gee roll, perhaps, but not a 1 gee... Smoothly done, a roll does
> not have abrupt changes in perceived G load, and so is thought of as a
> 1 G roll... Yet, they overlook the fact that when inverted they have
> to have +1G simply to neutralize the -1G from being inverted in the
> gravitational field, and another +1G to retain the sensation of normal
> weight into the seat of their pants, so they are actually pulling 2G at
> that moment...
>
>
> denny ... who took physics long, long, ago in a world far, far, away,
> when logarithms were looked up in a table, and the only 'calculator'
> you had was your brain and a fancy pair of rulers ( slide rule..)
>

[Tony]

You can maintain I G into the seat with a coordinated banking and
decending flight path. Someone else pointed out if the airplane were
allowed to free fall with gravity a coordinated roll pulling back hard
enough to maintain 1 G into the seat (local upward acceleration of 32
fps^2) does the trick. You have to agree if you bank an airplane and
keep the ball centered there's no left or right G force, it's all into
the seat. If you bank and decend, you can make that force 1 G.

I think the OP ran the numbers to demonstrate that.

[Kyle Boatright]

"AES" > wrote in message
...
> As a follow-on to the discussion of "one G rolls" in this NG some time
> back, an analysis of a simplified "physicist's version" of the one G
> roll problem, with formulas and graphs, is available at
>
> <http://www.stanford.edu/~siegman/one_g_roll.html>
>
> I'm not sure why I was driven to do this -- maybe nothing better to do
> with my time these days? -- but I was, so here it is.
>
> I'll give a quick summary of results first, then a more detailed summary
> of the assumptions behind them. Any comments on this will be welcome
> (but on the NG, please; not by private email). And before anyone jumps
> too hard on my approach or results, please look at the assumptions to
> see what I'm claiming and what I'm not.
>
> RESULTS:
>
> Unless I've made some dumb mistake (always a possibility), if an
> aircraft starts out in level flight and rolls through 360 degrees
> clockwise about its forward axis in 10 seconds, while maintaining 1 G
> force on the aircraft and the pilot, and a constant forward velocity, it
> will perform a circular corkscrew "barrel roll" type of motion with a
> radius of about 80 feet around a curved "guiding axis" (although the
> actual orbit will not look much like a corkscrew), and will end the
> maneuver displaced about 500 feet to the right, with an altitude loss of
> about 1600 feet, and a screaming final downward velocity component of
> 300 feet/second.

<<snip>>>

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.

Also, most of us who do rolls in aircraft with non-inverted systems begin
the roll with a pull-up to a 20 degree (or so) climb. In my case (RV-6), I
pull 1.5 G's to the 20 degree upline, roll at more or less 1 G, and pull out
at 1.5 G's when the roll is complete without any net altitude loss...

KB

[alexy]

"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.

[vincent p. norris]

> 1 G is is sitting motionless in your chair

One G does not require the absence of motion; a pilot flying straight
and level at constant speed in smooth air experiences 1 G.

vince norris

[Kyle Boatright]

"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

[alexy]

"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.

[AES]

In article >,
"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.

Thanks for reaction (I'm always a bit worried when I'm calculating
something outside my personal experience). Results for a 6 sec roll are
at:

<http://www.stanford.edu/~siegman/one_g_roll_6_sec.html>

As expected, a lot less altitude loss and downward velocity buildup.


> Also, most of us who do rolls in aircraft with non-inverted systems begin
> the roll with a pull-up to a 20 degree (or so) climb. In my case (RV-6), I
> pull 1.5 G's to the 20 degree upline, roll at more or less 1 G, and pull out
> at 1.5 G's when the roll is complete without any net altitude loss...
>
> KB

I'll run this when I get a few minutes to do it. Or, if anyone has
access to Mathematica (a truly great software program, but expensive
unless you can get an academic discount), I'll be glad to send the
original notebook.

[Tony]

I really liked your analysis, but offer the following.

If we fly coordinated -- that is, use the rudder to keep the ball
centered our degrees of freedom are ailerons to control rate of bank,
elevators to control pitch (pulling the nose up, or pushing it down)
and throttle/flaps to control acceleration/deceleration.

Someone pointed out if the airplane was allowed to accelerate downward
at 1 G, the 1 G roll could be accomplished by simply pulling back on
the yoke hard enough to cause 1 G acceleration in the pitch direction
while a coordinated roll was flown. This is different than a constant
velocity model, but also works.

Thanks for crunching the numbers!