Path of an airplane in a 1G roll

[Chris W]

Do we have any who is a math whiz here? I want to find a formula to
calculate the position of an airplane throughout a 1G roll. The reason
I'm doing this is so I can build a "roll track" for a remote control car
so the car will alway have a positive g force on it to keep it on the
track. Anyone have any ideas? So far my attempts have have all come up
short. They don't pass what my college calculus instructor called the
"warm and fuzzy" test. I think it has been too long since I took those
classes.

--
Chris W

Gift Giving Made Easy
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give the gifts they want
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[Luke Scharf]

Chris W wrote:
> Do we have any who is a math whiz here? I want to find a formula to
> calculate the position of an airplane throughout a 1G roll. The reason
> I'm doing this is so I can build a "roll track" for a remote control car
> so the car will alway have a positive g force on it to keep it on the
> track. Anyone have any ideas? So far my attempts have have all come up
> short. They don't pass what my college calculus instructor called the
> "warm and fuzzy" test. I think it has been too long since I took those
> classes.

Most folks who do this do it with a system of differential equations.
It's not a simple thing to do, if you want to model the whole thing:
http://www.aoe.vt.edu/~durham/AOE5214/
But, then again, I'm the sysadmin for aoe.vt.edu (and not an aero
engineer) so I may not have looked at the simple solutions. A general
(in the mathematical sense) answer to your question is in "Chapter 7:
Equations of Motion". The previous 6 chapters are background knowledge.

I think I'd just try it in a flight simulator -- maybe you can use an
simulated-aircraft that has a G-meter.

One way to approach the calculation might be to model the aircraft as if
it were weightless. Then, have the aircraft accelerate with 1g worth of
lift (pitch-up, slam you into your seat). The model you've developed
should show the airplane looping in one way or another. Then, add a
roll at the maximum roll-rate of the aircraft into the model. And,
after that, wrap the resulting shape around a parabola. Some calculus
and a lot of vectors should do it. Or, you could just do a lot of
vector summing in a program like Mathematica or a program of your own
devising should do it.

But, please take what I have to say with a spoonful of salt since I am
merely an IT guy who gets really excited around airplanes.

-Luke

[zaphod]

On Sun, 19 Jun 2005 11:04:39 -0500, Chris W wrote:

> Do we have any who is a math whiz here? I want to find a formula to
> calculate the position of an airplane throughout a 1G roll. The reason
> I'm doing this is so I can build a "roll track" for a remote control car
> so the car will alway have a positive g force on it to keep it on the
> track. Anyone have any ideas? So far my attempts have have all come up
> short. They don't pass what my college calculus instructor called the
> "warm and fuzzy" test. I think it has been too long since I took those
> classes.


IIRC, There is a computer sim game for building roller coasters with some
realistic physics. While not exactly what you were looking for, maybe you
could build the coaster with the forces you want and then immitate the
shape produced for your track without going thru all the math? :)

peace,
chris

[Byron Covey]

You can't do a roll and retain 1 G positive throughout the roll.


BJC

"zaphod" > wrote in message
et...
> On Sun, 19 Jun 2005 11:04:39 -0500, Chris W wrote:
>
>> Do we have any who is a math whiz here? I want to find a formula to
>> calculate the position of an airplane throughout a 1G roll. The reason
>> I'm doing this is so I can build a "roll track" for a remote control car
>> so the car will alway have a positive g force on it to keep it on the
>> track. Anyone have any ideas? So far my attempts have have all come up
>> short. They don't pass what my college calculus instructor called the
>> "warm and fuzzy" test. I think it has been too long since I took those
>> classes.
>
>
> IIRC, There is a computer sim game for building roller coasters with some
> realistic physics. While not exactly what you were looking for, maybe you
> could build the coaster with the forces you want and then immitate the
> shape produced for your track without going thru all the math? :)
>
> peace,
> chris

[Roy Smith]

"Byron Covey" > wrote:
> You can't do a roll and retain 1 G positive throughout the roll.

Actually, you can't do ANY maneuver and maintain exactly 1G. The G's you
feel are the sum of the Earth's gravity and your acceleration. Since the
Earth's gravity is always 1G, if your total G force is always 1G, then your
acceleration must be zero, and you can not change your flight path.

You can certainly maintain positive G's through maneuvers (even inverted),
and you can certainly maintain something close to 1G though maneuvers, but
you cannot maintain exactly 1G through the whole thing.

[Tony]

Not quite true. Start a coordinated turn, decending at the same time
and you can keep the bathroom scale you're sitting on reading your
weight. At 45 degrees of back I think you'll find the airplane has to
be accelerating downward too, so the .707 horizontal G and the .707
vertical G combine to provide 1 G into the pilot's seat. At inverted,
you'll have to pull back pretty hard on the yoke to provide a relative
to the pilot upward acceleration of 64.4 f/sec*2 to keep pasted into
the seat at 1 g.

[Bob Fry]

>>>>> "BC" == Byron Covey > writes:

BC> You can't do a roll and retain 1 G positive throughout the
BC> roll. BJC

There's supposed to be a video of the great Bob Hoover doing a barrel
roll with a glass of water on the panel...not a drop spilled. If
anybody knows where a copy of the video is (or if it even exists) that
would be a worth addition to Jay Honeck's collection.

[David O]

Chris W > wrote:

>Do we have any who is a math whiz here? I want to find a formula to
>calculate the position of an airplane throughout a 1G roll. The reason
>I'm doing this is so I can build a "roll track" for a remote control car
>so the car will alway have a positive g force on it to keep it on the
>track. Anyone have any ideas? So far my attempts have have all come up
>short. They don't pass what my college calculus instructor called the
>"warm and fuzzy" test. I think it has been too long since I took those
>classes.

Chris,

I suggest that you forget about trying to model the path of an
airplane in a 1 G roll and, instead, make your car track a simple
helix. With a simple helix you should be able to keep your car's
front wheels straight as the car goes through the helix. Now for the
details...

Envision a helix laid out on the inside surface of a cylinder. The
cylinder will have a radius and a length. Let's assume for this
discussion that the helix makes one revolution in that length. Now
all we have to do is find a radius and a length for the cylinder that,
for a given car speed, will keep your car on the track throughout.

For your car to remain on the track at as it goes inverted, the
centripetal acceleration due to the car's rotation about the cylinder
axis will have to exceed the acceleration of gravity. We'll specify
the target centripetal acceleration by defining a multiplicative
factor which we will call the "G factor". A G factor of 1.5, for
example, would mean that the target centripetal acceleration is 1.5 G,
where G is the acceleration of gravity. With a G factor of 1.5, then,
at the top of the helix the net acceleration would be the centripetal
acceleration minus the acceleration of gravity or 1.5 G -1 G, or 0.5
G. The force of the car pushing on the track at that point would be
0.5mG where m is the mass of the car. We don't have to use 1.5 G for
the G factor. We could use, for example, 1.2 or 2.0. At the end of
this post, I'll give you a link to a couple of spreadsheets. In those
spreadsheets, "G factor" will be one of the user inputs.

A cylindrical helix is nothing but a straight line on a cylindrical
surface. "Unroll" that cylinder onto a plane surface and the helix
becomes a straight line. Knowing this, it becomes quite
straightforward to relate the path length of the helix to the cylinder
radius and cylinder length. Remember that we're talking about just
one turn of the helix for the cylinder length. Once the relationship
of helix path length to cylinder radius and cylinder length is
formulated, it is again straightforward to split the car speed (which
we shall assume is known), into two components, one along the cylinder
axis and the other around the cylinder circumference. With the
velocity around the cylinder circumference now formulated, and
specifying the cylinder radius as a known, the car speed as a known,
the acceleration of gravity as a known, and choosing a G factor, we
have all that is necessary to compute the cylinder length necessary to
achieve the target centripetal acceleration.

I'll not write the formula here because it would be too cumbersome to
write in text form. Instead, I will give you a link to an Excel (5.0)
spreadsheet in which you can inspect the formula if you wish.

http://www.airplanezone.com/PubDir/Helix01.xls

In the spreadsheet, I used 9.8 meters per second squared for G, the
acceleration of gravity, so all distances are in meters and all
velocities are in meters per second. If you changed G to 32.2 then
all distances would be in feet and all velocities would be in ft/sec.
The numbers in green are user inputs and the numbers in burnt red are
the calculated results. Note that I've not locked any cells.

Of course, you are free to alter the user inputs as you wish but let's
talk about the spreadsheet with numbers that I put in. Note that I
specified a car speed of 4 m/s and a G factor of 1.5. The results
table shows the cylinder radii and the resulting cylinder lengths to
achieve the specified G factor. Note the interesting result that
there are clearly two usable radii for most of the cylinder lengths
within the solution range. The smaller radius results in a long
narrow corkscrew while the larger radius result in a short wide
corkscrew. Also note that at the extreme, with a cylinder radius of
1.0884 (for the inputs I used), the cylinder length becomes quite
small. At this extreme, the solution is quickly approaching a loop
instead of a corkscrew. :)

As an aside, it should be noted that the formula I used in the
spreadsheet was not derived to solve for a loop (i.e. for a cylinder
length of zero) and it is ill suited for that purpose. In the argot
of numerical analysts, the calculation is "ill conditioned" for that
purpose. For completeness, then, for the data given, the radius for a
loop is 1.088435374... .

Of course, the spreadsheet results will change as you change Vcar or G
factor, or whatever. You will also note as you play around that some
speeds just won't work. I don't know your model scale or your model
speeds so you will have to play with the data yourself to find a good
solution for your needs.

Once you choose a cylinder radius and cylinder length for your helix,
you can use the following spreadsheet to see how the centripetal
acceleration varies with your model car speed. Of course, you'll not
want to let your car's centripetal acceleration fall below 1 G at the
inverted point. :)

http://www.airplanezone.com/PubDir/Helix02.xls

Now let's talk about the approach and exit from the helix. Let's call
the tracks leading to and away from the helix the "approach tracks".
You'll probably not want to have to turn your car as you enter the
helix so the approach tracks should be straight for some distance
before reaching the helix and should be tangent to the helix at the
helix entry and exit points. This means that the helix cylinder axis
will be at an angle to the approach tracks and that the approach
tracks will be parallel to each other but will be offset laterally.

Lastly, I'm fairly sure of my physics and math but I'll leave it to
others to vet. Good thing you posted your query on a Sunday. :)

Cheers,

David O -- (David at AirplaneZone dot com)

[Peter Duniho]

"Tony" > wrote in message
oups.com...
> Not quite true. Start a coordinated turn, decending at the same time
> and you can keep the bathroom scale you're sitting on reading your
> weight.

Only if that descent involves a vertical acceleration. That is, it's not a
constant rate descent.

A constant rate descent would require 1G of *vertical* lift, which means
greater than 1G of actual lift from the wing (where I blatantly misuse "1G"
as a way of describing the amount of lift equal to the weight of the
airplane :) ). Using your 45 degree bank angle example that comes to about
1.41G.

Alternatively, maintaining 1G of lift would mean that the descent rate would
be increasing throughout the turn. Depending on the bank angle, this could
turn into a pretty dramatic descent rate in short order.

Pete

[Byron Covey]

I've seen it. It was years ago. I borrowed the 8 mm tape from EAA for a
chapter program. Not only was the glass sitting there, Bob poured water
into it during the roll.


BJC

"Bob Fry" > wrote in message
...
>>>>>> "BC" == Byron Covey > writes:
>
> BC> You can't do a roll and retain 1 G positive throughout the
> BC> roll. BJC
>
> There's supposed to be a video of the great Bob Hoover doing a barrel
> roll with a glass of water on the panel...not a drop spilled. If
> anybody knows where a copy of the video is (or if it even exists) that
> would be a worth addition to Jay Honeck's collection.

[Roy Smith]

In article >, Bob Fry >
wrote:

> >>>>> "BC" == Byron Covey > writes:
>
> BC> You can't do a roll and retain 1 G positive throughout the
> BC> roll. BJC
>
> There's supposed to be a video of the great Bob Hoover doing a barrel
> roll with a glass of water on the panel...not a drop spilled. If
> anybody knows where a copy of the video is (or if it even exists) that
> would be a worth addition to Jay Honeck's collection.

All that shows is that he maintained positive G's and coordination.

[Chris W]

David O wrote:

>Chris W > wrote:
>
>
>
>>Do we have any who is a math whiz here? I want to find a formula to
>>calculate the position of an airplane throughout a 1G roll. The reason
>>I'm doing this is so I can build a "roll track" for a remote control car
>>so the car will alway have a positive g force on it to keep it on the
>>track. Anyone have any ideas? So far my attempts have have all come up
>>short. They don't pass what my college calculus instructor called the
>>"warm and fuzzy" test. I think it has been too long since I took those
>>classes.
>>
>>
>
>Chris,
>
>I suggest that you forget about trying to model the path of an
>airplane in a 1 G roll and, instead, make your car track a simple
>helix. With a simple helix you should be able to keep your car's
>front wheels straight as the car goes through the helix. Now for the
>details...
>
>
Why didn't I think of that. That is a much simpler solution. I can
even do those calculations but thanks for doing them for me. If I get a
3d model going I will send you an image.

--
Chris W

Gift Giving Made Easy
Get the gifts you want &
give the gifts they want
http://thewishzone.com

[David O]

Chris W > wrote:

>If I get a
>3d model going I will send you an image.

Yes, please do.

David O -- email: David at AirplaneZone dot com

[CB]

A properly performed barrel roll is a 1G manuever. The aircraft's
flight path describes a helix, as David described below. An aileron
roll is a variable-G operation, since you feel -1G while inverted.

[Ron Natalie]

CB wrote:
> A properly performed barrel roll is a 1G manuever.

Nope. It's a small amount of positive G's but it's not a constant
1G. Did you actually read David's post?

[Bob Moore]

"CB" > wrote
> A properly performed barrel roll is a 1G manuever. The aircraft's
> flight path describes a helix, as David described below. An aileron
> roll is a variable-G operation, since you feel -1G while inverted.

Check the following web sites, they all contain the same paragraph.
Care to give us your references for the definition of a barrel roll.

http://www.iac.org/begin/figures.html#Barrel%20Rolls
http://acro.harvard.edu
http://web.winco.net/~efildes/slowroll/barlroll.html
The Barrel Roll is a not competition maneuver. The barrel roll is a
combination between a loop and a roll. You complete one loop while
completing one roll at the same time. The flight path during a barrel roll
has the shape of a horizontal cork screw. Imagine a big barrel, with the
airplanes wheels rolling along the inside of the barrel in a cork screw
path. During a barrel roll, the pilot experiences always positive G's. The
maximum is about 2.5 to 3 G, the minimum about 0.5 G.

Bob Moore

[Happy Dog]

"CB" > wrote in message

>A properly performed barrel roll is a 1G manuever. The aircraft's
> flight path describes a helix, as David described below.

No.

> An aileron
> roll is a variable-G operation, since you feel -1G while inverted.

No.

Ever done one?

moo

[CB]

OK, folks, brace yourselves, you don't often see this on RAH....


I was wrong.


That's what I get for relying on an aging memory rather than looking it
up.
CB

[Roger]

On Sun, 19 Jun 2005 11:04:39 -0500, Chris W > wrote:

4 Groups?

>Do we have any who is a math whiz here? I want to find a formula to
>calculate the position of an airplane throughout a 1G roll. The reason

At any rate, do you want to maintain 1G, or just positive G? it's a
big difference.

You can do a barrel roll and maintain positive G all the way around.
It's a very simple maneuver and very easy to do. It's also probably
one of the easiest to screw up.

>I'm doing this is so I can build a "roll track" for a remote control car

Remember that in straight and level flight you are pulling 1G. If you
start a roll you will have to start adding nose up stick to maintain
1G to the point of 1G when inverted. "
"Theoretically" as you rolled past inverted you would start reducing
back pressure until you were back wings level.

At this point it takes some one much more versed in aeronautic theory
(and practice) than I, but... A barrel roll comes the closest to what
you are asking. It, however starts out at more than 1G. Typically 2Gs
and it can be more. With a 2G pull up at the start, you will be
pulling 1G when passing inverted.

Remember you started out in a nose high attitude to get to this point.
So in the theoretically description you would most likely be way nose
low at the 180 degree inverted position and I think you will probably
get well past 2 Gs getting back to the wings level position.

>so the car will alway have a positive g force on it to keep it on the

But, if it's just the positive Gs you need, shape the track like the
path a plane would take through a barrel roll. It would go up and
curve to the right forming a corkscrew shape with the end right back
at the same level as the beginning. You can add turns as well "as
long as the car is changing direction in relation to *its" own
vertical axis. For example if the car is on its right side the track
needs to be curving right, if on its left then the track needs to be
curving left. If the car is inverted the track needs to be curving
down.

Remember too that the car has to be going fast enough to maintain the
desired G forces and traction. Slow down and it'll just fall off.

Roger Halstead (K8RI & ARRL life member)
(N833R, S# CD-2 Worlds oldest Debonair)
www.rogerhalstead.com
>track. Anyone have any ideas? So far my attempts have have all come up
>short. They don't pass what my college calculus instructor called the
>"warm and fuzzy" test. I think it has been too long since I took those
>classes.

[Ernest Christley]

My take on this is that an airplane in a 1G roll would follow the same
path as any other object.

Imagine your in space. A 1G roll would be a perfect circle with a
constant 1G acceleration.

Now bring that path into the Earth's gravity well. Now the 1G roll is
all messed up by the Earth's 1G. How can we fix that? Just like the
Vomit Comet does, by accelerating down at 9.8m/s^2. Superimpose a roll
on top of a parabolic descent and you have the path of a theoretical
airplane in a 1G roll.

I don't think there is a plane that could actually perform this maneuver
in reality.

--
This is by far the hardest lesson about freedom. It goes against
instinct, and morality, to just sit back and watch people make
mistakes. We want to help them, which means control them and their
decisions, but in doing so we actually hurt them (and ourselves)."

[B S D Chapman]

On Wed, 29 Jun 2005 04:01:56 +0100, Ernest Christley
> wrote:

> My take on this is that an airplane in a 1G roll would follow the same
> path as any other object.
>
> Imagine your in space. A 1G roll would be a perfect circle with a
> constant 1G acceleration.
>
> Now bring that path into the Earth's gravity well. Now the 1G roll is
> all messed up by the Earth's 1G. How can we fix that? Just like the
> Vomit Comet does, by accelerating down at 9.8m/s^2. Superimpose a roll
> on top of a parabolic descent and you have the path of a theoretical
> airplane in a 1G roll.
>
> I don't think there is a plane that could actually perform this maneuver
> in reality.

Obviously, the quicker you can roll, the easier it would be, but
essentially it would be impossible to complete a constant 1G roll back to
S&L. You would have to end up in a nose-down attitude in order to
maintain 1G while inverted. The greater your roll rate, the less time
you'll need to maintain positive G while inverted, and hence the nose
won't need to drop as far. The closest you're going to get to this is a
simple aileron roll where you start nose high... but then you've pulled
more than 1G to get the nose in to that position!



--

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