Path of an airplane in a 1G roll

On 24 Jun 2005 16:30:32 -0700, "Tony" wrote:

Not to put too fine a point on this, Corky, but you can in fact enter a
climb and still maintain 1 G downward relative to the axis of the
airplane, but you have to decelerate while starting the climb. I agree
that you can't climb at a steady rate and not experience a shift in g
forces aft, but as your bring the nose up you have to slow down as
well, to add a little negative acceleration to offset the aft shifting
gravity vector. Think of it this way. If you hang a plumb bob in the
airplane, and start a climb, the bob will shift aft. If you're in level
flight and slow down, it'll shift forward. If you combine the two
correctly, it'll stay pointing at the same spot on the floor.

If you agree with this reasoning, you'll also agree that with a fast
enough entry speed you could pull through an entire loop -- it wouldn't
be round! -- and keep the plumb bob centered over the same spot.

Tony, it doesn't matter where the plumb bob hangs, going up means
adding some force in excees of 1G to do it. No matter how gently you
do it, a sensitive enough G meter will detect the additional force
that is required.

It's kind of like trying to fake out a bathroom scale. No matter how
gently you step onto it, it will eventually read your weight.

Climbing is like pushing against an inverted scale.

Corky Scott

Corky, you're right. Earlier the discussion was about a roll: I think I
can present a model for experiencing 1 g throughout what might be
called a roll, but that argument fails for a straight loop.

In article ,
Corky Scott wrote:


Tony, it doesn't matter where the plumb bob hangs, going up means
adding some force in excees of 1G to do it. No matter how gently you
do it, a sensitive enough G meter will detect the additional force
that is required.

It's kind of like trying to fake out a bathroom scale. No matter how
gently you step onto it, it will eventually read your weight.

Climbing is like pushing against an inverted scale.

Corky Scott

Corky,

"Net force = mass times acceleration" (Isaac Newton, ~1620)

"Acceleration = rate of **change** of velocity"

"Work done = force times distance"

If you're standing on a bathroom scale in a stationary elevator at the
ground floor, the scale will read your weight. Gravity (the invisible
gravitational field) is pulling down on you with a force mg (your mass
times the acceleration of gravity); the scale is pushing back up on you
with the same force mg (and that force is what its dial reads). The
total force on you (the sum of gravity pulling down on you plus the
scale pushing up on you) is thus zero, but it's zero not because you're
stationary, that is, remaining at a constant level; it has to be zero
because you're at a constant vertical speed, namely zero; you're not
*accelerating* (changing speed) -- not at this point anyway

The doors close, the elevator starts upward. For a brief initial
period, until the elevator reaches its steady-state upward speed, the
scales read more than mg. That's because the elevator (or whoever is
pulling on the elevator cable) is both pushing you upward with a force
equal to your weight (mg), which does work and adds to your potential
energy in the gravitational field; but there's also a slight excess
force at the start which is needed to accelerate you to the steady-state
speed of the elevator (and which therefore gives you a little kinetic
energy as well as your steadily increasing potential energy).

Once the elevator reaches its steady-state speed (around the 3rd floor,
let's say) from there up to the 104th floor you're traveling at a
*constant* vertical velocity: you're *not accelerating*. Therefore
there can't be any net upward force on you; the scales read mg.

[This does leave out the fact that the value of g changes very (very!
very!) slightly between the ground and the 104th floors, but this change
is so minute in going from the ground floor to the 104th floor that it's
just not in the discussion here.]

The rest of the way up, from the 3rd to 101st floors, the scales are
pushing up on you with a constant force mg; and that force eventually
acts through a distance h, the height from the ground floor to the 104th
floor. Total work done by the elevator on you is therefore force times
distance, or mgh. That work has gone into giving you an added potential
energy equal to mgh in the Earth's gravitational field.

[As you neared the 104th floor and the elevator slowed to a stop, you
also lost the kinetic energy you had gained between the ground and third
floors. The scales showed a small reduction below mg between 101 and
104 just as they showed a small increase between first and 3rd floors.]

You now own that extra potential energy of mgh; it's yours, as you step
off the elevator on the 104th floor. Want to get it back? Step outside
on the viewing deck and pop over the railing. Leaving aside questions
of air resistance, by the time you get back down to, say, the first
floor level you'll have all that potential energy converted into kinetic
energy -- a lot of kinetic energy! Unfortunately, one floor later all
that energy energy will pretty much have been converted into heat,
slightly warming up a small patch of sidewalk and some gunk on it.

Same deal in a plane: hang a seat from a butcher's scales hung from the
ceiling of a plane, and sit in it. Have someone fly the plane in a
constant forward speed, constant upward speed, straight line climb.
Except for a slight initial transient period when they begin the climb,
the scales will read your weight, mg, no more, no less.

--"Another Tony"

On Mon, 27 Jun 2005 at 12:30:45 in message
, Corky Scott
wrote:

Tony, it doesn't matter where the plumb bob hangs, going up means
adding some force in excees of 1G to do it. No matter how gently you
do it, a sensitive enough G meter will detect the additional force
that is required.

Although it is true that a climb cannot be initiated without exceeding
1g in the transition from level flight to climb, a climb at a steady
speed and climb rate does not mean that more than 1g is felt in the
aircraft.

Why should it? If there is no acceleration there is no g other than the
gravitational one. Of course this does ignore the fact that the earth's
gravity is reduced as the aircraft moves further from the surface of the
earth!
--
David CL Francis

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