Path of an airplane in a 1G roll

David, the issue for me was 1 g down, into the seat. In a steady state
climb one experiences one G, but if the nose is 5 degrees up that force
is 5 degrees aft of down. My understanding of the question (and it
could not be an accurate understanding) was, can one somehow roll an
airplane without having it experience anything other than 1 g "down". I
think it's been shown the airplane can be rotated 360 degrees on its
axis with the pilot always experiencing 1 g down into the seat. At that
point though the airplane is going downward pretty fast, and the weight
vector would shift forward of straight down.

On Thu, 30 Jun 2005 at 18:35:25 in message
.com, Tony
wrote:
David, the issue for me was 1 g down, into the seat. In a steady state
climb one experiences one G, but if the nose is 5 degrees up that force
is 5 degrees aft of down. My understanding of the question (and it
could not be an accurate understanding) was, can one somehow roll an
airplane without having it experience anything other than 1 g "down". I
think it's been shown the airplane can be rotated 360 degrees on its
axis with the pilot always experiencing 1 g down into the seat. At that
point though the airplane is going downward pretty fast, and the weight
vector would shift forward of straight down.



Tony,

I see where you are. I would say that if that is what you using as a
datum then the 'g' down axis will vary from a few degrees at high speed
in level flight to up to around 12 degrees or more at touchdown in level
flight without any aerobatics at all..

But in your definition it would be impossible to have a flight that
included take off and landing and a modest climb and descent at a strict
'1g' down.

I was assuming a steady one 'g' at right angles to the free stream
airflow!
--
David CL Francis

In article ,
David CL Francis wrote:

David, the issue for me was 1 g down, into the seat. In a steady state

Tony,

I see where you are.

But in your definition it would be impossible to have a flight that
included take off and landing and a modest climb and descent at a strict
'1g' down.

I'm still struggling to think this whole problem through from the
viewpoint of someone who likes to solve "simple" physics problems, but
is absolutely not a pilot.

Let's just take the part of the flight that involves climbing at a
constant upward rate and then leveling off. Seems as if you will never
be able to convert to level flight without reducing the upward velocity
vector, ergo some (negative) vertical acceleration has to occur.

But what if you roll the plane, slowly and gently, about a longitudinal
axis that passes through the bathroom scales, simultaneously applying
control forces so that the plane begins turning right.

If you can roll slowly enough so you neglect the rotational inertia of
the pilot about this axis and simultaneously turn right at the correct
rate, during this time the seat will push the pilot (who's a point mass,
of course) up with *less* vertical force than previously, while pushing
(and accelerating) the pilot to the right with a small horizontal
component of force. If you do this just right, you ought to be able to
keep the total force pushing from the seat into the pilot equal to the
pilot's weight.

Do this carefully enough, keep it up for a while, then roll back to
level, and you ought to be able to bleed the vertical velocity down to
zero and thus be leveled off -- though with a different compass heading
-- while keeping the bathroom scales reading a constant value equal to
the pilot's weight.

Does this make sense?

--"The other Tony"

P.S. -- Takeoff and landing is more easily solvable. You just need a
long enough taxiway that can be curved but eventually feeds straight
into a (potentially very short) runway, with both of these at the point
where they join having exactly the same upward slope as the slope that
you want to climb at after takeoff, and with the runway ending at the
edge of a cliff.

So, all you have to do is accelerate up to full flying speed while
you're still on the taxiway -- which of course doesn't count since
you're still only taxiing -- until you get on the runway part and just
keep going.

Landing is obviously the same thing reversed.

About taking off and landing: If you take as the goal 1 g downward --
that's an integer, not 1.0000 plus or minus a little bit, you can't do
it. There is acceleration. The same thing is true for a loop, there
just are not enough degrees of freedom to allow the pilot to see the
horizon drop down as he climbs, then reappear inverted at the top of
the loop, without experiencing some incremental (even if small) g
forces.

I think a roll adds the additional variables one might need. Consider,
for example, a plane about 45 degrees into a roll. At that moment, in
coordinated (pilot talk for keeping the pilot's weight centered on the
seat) level flight there's a g toward the center of the earth, and
another along the radius of the turn. The pilot experiences 1.414 gs
into his seat. If, however, the airplane is also pitched down 45
degrees accelerating, and coordinated, you could choose numbers that'll
resolve to 1 g into the seat.

Take now a bank of 90 degrees. If the airplane is pointed straight down
and accelerating at 1 G, that is, in free fall vertically, there'd be
no fore and aft weight component. The pilot would, however, have to be
pulling back on the yoke hard enough to accelerate in the nose up
direction at 1 G.

At inverted if level he'd be experiencing 1 g "up", so he'd have to
have the yoke back far enough to accelerate in the nose up direction 2
gs worth.

You can, at each point point in the "roll", calculate how the airplane
must be accelerating in the nose up direction and what direction the
nose must be pointing for the pilot to experience 1 g down.

I don't know if it's a realizable manouver -- it'll take some serious
elevator "authority" to provide the nose up accelerations that are
needed. Some insightful person in this thread made the observation that
if the airplane was during the roll just accelerating downward at 1 G
-- in free fall, if you will, one need not worry about the gravity
effects and the pilot would just have to pull back on the yoke hard
enough to keep the nose accelerating up at 1 g.

It's an interesting problem. I think the airplane, as seen from
outside, would look like it was in a death sprial. The pilot, however,
would see the horizon rotate through 360 degrees, so he'd say he rolled
the airplane. At the end of the roll the airplane would be in a serious
nose down attitude, and going pretty fast. I don't think you can get
from there to straight and level while keeping a local 1 g down weight
component.

"Tony" wrote:


At the end of the roll the airplane would be in a serious
nose down attitude, and going pretty fast. I don't think you can get
from there to straight and level while keeping a local 1 g down weight
component.
Of course not. I think the only reasonable interpretation of the
"aileron roll is a 1-g maneuver" claim is that the maneuver itself
will be at one g, while setup and recovery can be at g-loads normally
experienced in non-aerobatic flight.

E.g, enter a 15-degree nose-up climb (requiring more than 1-g), do the
roll at 1-g, then recover from the resulting 15-degree nose-down dive
(requiring more than one g).
--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

On Fri, 1 Jul 2005 at 18:35:50 in message
, AES
wrote:
In article ,
David CL Francis wrote:

David, the issue for me was 1 g down, into the seat. In a steady state

Tony,

I see where you are.

But in your definition it would be impossible to have a flight that
included take off and landing and a modest climb and descent at a strict
'1g' down.

I'm still struggling to think this whole problem through from the
viewpoint of someone who likes to solve "simple" physics problems, but
is absolutely not a pilot.

It would help if you give a picture of what you mean by simple physics:
e.g. are you comfortable with Newton's basic mass and force equations?

Do you have any knowledge of vectors as applied to forces? Are you able
to calculate the forces required for a level banked turn of a given
angle?

Do you have any knowledge of the simple calculations of drag and lift?

There could be others but it is difficult to answer without knowing
something more about your starting point.

You may not believe this, but I have been caught up in discussions with
people, trying to help them when their sole object was to stir things
up.

I am not a pilot either although I have, many years ago, flown solo. Now
I am an elderly ex-aerospace engineer whose powers have faded somewhat!

Let's just take the part of the flight that involves climbing at a
constant upward rate and then leveling off. Seems as if you will never
be able to convert to level flight without reducing the upward velocity
vector, ergo some (negative) vertical acceleration has to occur.

Correct, but in some cases it may be quite a small effect.

But what if you roll the plane, slowly and gently, about a longitudinal
axis that passes through the bathroom scales, simultaneously applying
control forces so that the plane begins turning right.

You lost me there!

If you can roll slowly enough so you neglect the rotational inertia of
the pilot about this axis and simultaneously turn right at the correct
rate, during this time the seat will push the pilot (who's a point mass,
of course) up with *less* vertical force than previously, while pushing
(and accelerating) the pilot to the right with a small horizontal
component of force. If you do this just right, you ought to be able to
keep the total force pushing from the seat into the pilot equal to the
pilot's weight.

If by 'correct rate', you mean a properly balanced turn then you are
wrong. In a balanced turn the pilot will always detect slightly more
'g'. Remember what the pilot feels is the vector sum of any
accelerations.

Do this carefully enough, keep it up for a while, then roll back to
level, and you ought to be able to bleed the vertical velocity down to
zero and thus be leveled off -- though with a different compass heading
-- while keeping the bathroom scales reading a constant value equal to
the pilot's weight.

Does this make sense?

Not to me. Are your bathroom scales fixed to the aircraft and under the
pilot's seat?
--"The other Tony"

P.S. -- Takeoff and landing is more easily solvable. You just need a
long enough taxiway that can be curved but eventually feeds straight
into a (potentially very short) runway, with both of these at the point
where they join having exactly the same upward slope as the slope that
you want to climb at after takeoff, and with the runway ending at the
edge of a cliff.


You have lost me again.

So, all you have to do is accelerate up to full flying speed while
you're still on the taxiway -- which of course doesn't count since
you're still only taxiing -- until you get on the runway part and just
keep going.

Only a vague idea what you might be trying to get at here. If you are
referring obliquely to a 'ski jump' then the only difference is that the
slight acceleration required is provided a forced rotation on a curved
slope than the result is the same except the force is in the original
case is provided by lift (and thrust) and in the second by the change of
momentum caused by being forced around a curve. You feel it just the
same.

Keep up the search for enlightenment!

--
David CL Francis

David, first of all I'm the Tony who thinks I can take my Mooney into a
coordinated bank and accelerate downward enough to keep 1 g into the
seat. Mooney ain't made for rolling, although one could argue if my
model is correct the damned thing would never know it rolled, although
the AH would complain, wouldn't it?

You mentioned you're retired out of aerospace. I'd appreciate talking
with you about a different matter but your email address is obscured.
Have you an available IM address or the like?

On Sun, 3 Jul 2005 at 09:01:15 in message
.com, Tony
wrote:
David, first of all I'm the Tony who thinks I can take my Mooney into a
coordinated bank and accelerate downward enough to keep 1 g into the
seat. Mooney ain't made for rolling, although one could argue if my
model is correct the damned thing would never know it rolled, although
the AH would complain, wouldn't it?

You mentioned you're retired out of aerospace. I'd appreciate talking
with you about a different matter but your email address is obscured.
Have you an available IM address or the like?

Tony,

Just reply to my message as mail and I should get it. A working reply
email address is in there!

David
--
David CL Francis