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
Get the gifts you want &
give the gifts they want
http://thewishzone.com

[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

[David CL Francis]

On Wed, 22 Jun 2005 at 20:09:25 in message
>, Bob Moore
> wrote:
>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.

That must of course be correct. To describe an actual horizontal helix
would require a lot of smooth changes in g and in necessary control
deflections, otherwise the helix would not be circular in cross section.

However perhaps there could be a manoeuvre that combines a roll with a
constant 1 g pressure on the crew?

My guess is that it would turn into rather odd sort of spiral dive. The
nose would certainly fall below a level flight path!
--
David CL Francis

[Gig 601XL Builder]

Just to add to all of this about if a barrel roll is a 1G maneuver, which I
don't think it is, I was watch a Military Channel show on Boeing and
the -80/707 and Tex Johnston was talking about the barrel roll he did and
said it was a 1G maneuver.




"David CL Francis" > wrote in message
...
> On Wed, 22 Jun 2005 at 20:09:25 in message
> >, Bob Moore
> > wrote:
>>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.
>
> That must of course be correct. To describe an actual horizontal helix
> would require a lot of smooth changes in g and in necessary control
> deflections, otherwise the helix would not be circular in cross section.
>
> However perhaps there could be a manoeuvre that combines a roll with a
> constant 1 g pressure on the crew?
>
> My guess is that it would turn into rather odd sort of spiral dive. The
> nose would certainly fall below a level flight path!
> --
> David CL Francis

[Corky Scott]

On Fri, 24 Jun 2005 13:08:09 -0500, "Gig 601XL Builder"
<wr.giacona@coxDOTnet> wrote:

>Just to add to all of this about if a barrel roll is a 1G maneuver, which I
>don't think it is, I was watch a Military Channel show on Boeing and
>the -80/707 and Tex Johnston was talking about the barrel roll he did and
>said it was a 1G maneuver.

Think about it, sitting at the keyboard typing, and flying along in
level flight, you are at 1G. Pull back on the yoke or stick and you
are no longer at 1G, you are at 1G plus whatever it takes to climb.
It's impossible to gain altitude without experiencing more than 1G.

Could it be that the person describing the 707 barrel roll meant that
the maneuver was 1G in excess of 1G? In other words 2G's? Could be,
but probably the announcer or "talent" speaking for the clip had no
idea what he was talking about.

Corky Scott

[Chris Colohan]

Corky Scott > writes:
> Could it be that the person describing the 707 barrel roll meant that
> the maneuver was 1G in excess of 1G? In other words 2G's? Could be,
> but probably the announcer or "talent" speaking for the clip had no
> idea what he was talking about.

Actually, I think Tex says it. You can watch the video clip yourself
and make up your mind, Jay has it archived on this page:

http://www.alexisparkinn.com/aviation_videos.htm

Search for "Tex" and you'll find it.

Chris
--
Chris Colohan Email: PGP: finger
Web: www.colohan.com Phone: (412)268-4751

[Bob Moore]

"Gig 601XL Builder" <wr.giacona@coxDOTnet> wrote

> Just to add to all of this about if a barrel roll is a 1G maneuver,
> which I don't think it is, I was watch a Military Channel show on
> Boeing and the -80/707 and Tex Johnston was talking about the barrel
> roll he did and said it was a 1G maneuver.


By the definition of a barrel roll used by the International Aerobatic
Club and the US Navy where I learned to fly, Tex was of course wrong
on two counts. First as pointed out by others, anything other than
"straight and level" must incur more than 1g, and secondly, Tex's roll
was not a barrel roll.

Bob Moore

[Gig 601XL Builder]

"Corky Scott" > wrote in message
...
> On Fri, 24 Jun 2005 13:08:09 -0500, "Gig 601XL Builder"
> <wr.giacona@coxDOTnet> wrote:
>
>>Just to add to all of this about if a barrel roll is a 1G maneuver, which
>>I
>>don't think it is, I was watch a Military Channel show on Boeing and
>>the -80/707 and Tex Johnston was talking about the barrel roll he did and
>>said it was a 1G maneuver.
>
> Think about it, sitting at the keyboard typing, and flying along in
> level flight, you are at 1G. Pull back on the yoke or stick and you
> are no longer at 1G, you are at 1G plus whatever it takes to climb.
> It's impossible to gain altitude without experiencing more than 1G.
>
> Could it be that the person describing the 707 barrel roll meant that
> the maneuver was 1G in excess of 1G? In other words 2G's? Could be,
> but probably the announcer or "talent" speaking for the clip had no
> idea what he was talking about.
>
> Corky Scott

Well Corkster, don't tell me tell Tex Johnston. He did it and he said it. As
I said in my post I don't think it is a 1G maneuver and for that mater it
wasn't a very good barrel roll except when you consider it had never been
done in that model of airplane and even the co-pilot didn't know he was
going to do it until right before he did it.

Gig G

[Tony]

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.

Think about it, sitting at the keyboard typing, and flying along in
level flight, you are at 1G. Pull back on the yoke or stick and you
are no longer at 1G, you are at 1G plus whatever it takes to climb.
It's impossible to gain altitude without experiencing more than 1G.

Could it be that the person describing the 707 barrel roll meant that
the maneuver was 1G in excess of 1G? In other words 2G's? Could be,
but probably the announcer or "talent" speaking for the clip had no
idea what he was talking about.

[Jose]

> 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.

.... and the string will break.

Or at least will be under more stress. There's more than the position
of the plumb bob to the total net force.

Jose
--
You may not get what you pay for, but you sure as hell pay for what you get.
for Email, make the obvious change in the address.

[Chris Colohan]

Bob Moore > writes:
> "Gig 601XL Builder" <wr.giacona@coxDOTnet> wrote
>
> > Just to add to all of this about if a barrel roll is a 1G maneuver,
> > which I don't think it is, I was watch a Military Channel show on
> > Boeing and the -80/707 and Tex Johnston was talking about the barrel
> > roll he did and said it was a 1G maneuver.
>
>
> By the definition of a barrel roll used by the International Aerobatic
> Club and the US Navy where I learned to fly, Tex was of course wrong
> on two counts. First as pointed out by others, anything other than
> "straight and level" must incur more than 1g, and secondly, Tex's roll
> was not a barrel roll.

Too much misinformation in this thread. Here is exactly what he said
(any transcription errors are my fault):

"...and I knew the prototype, and there is one maneuver you can do
with no hazard whatsoever. I decided that I would do a roll to
impress the people. So I came accross and did a chandelle..."

"...I was called into Mr. Allen's office on Monday morning, and
Mr. Allen asked me what I thought I was doing, and I explained
that I was selling airplanes. And I explained that it was a 1G
maneuver, and it was absolutely non-hazardous, and it was very
impressive."

Chris
--
Chris Colohan Email: PGP: finger
Web: www.colohan.com Phone: (412)268-4751

[Bob Moore]

Chris Colohan > wrote
> "...and I knew the prototype, and there is one maneuver you can do
> with no hazard whatsoever. I decided that I would do a roll to
> impress the people. So I came accross and did a chandelle..."

Well....It certainly was NOT a chandelle, which is a climbing 180
degree change of direction.

I have in my possesion, two videos of the Dash 80 roll and in one of
them, Tex does make the statement that you attribute to him. In the
other, a one hour TV program, his quote is:

"I came back over the course, pulled the nose up, put in full aileron,
and did a nice 1g roll."

Sure looks asn sounds like an aileron roll to me.

John Steiner, a retired Boeing Vice-President, was the narrator
in the video that dubbed it a "barrel roll".

I was disappointed somewhat later in the video when Tex stated that:

"I had perfected aerobatics years and years ago and as I said, it
was a 1g maneuver."

This from the Chief Test Pilot on what was perhaps the most important
aircraft project in Boeing's history. ??????

Bob Moore
17 years in the Grand Old 707. 1967-1985

[pullinggs]

Bob Moore wrote:
> Chris Colohan > wrote
> > "...So I came accross and did a chandelle..."
>
> Well....It certainly was NOT a chandelle, which is a climbing 180
> degree change of direction.

No arguments there.

> I was disappointed somewhat later in the video when Tex stated that:
>
> "I had perfected aerobatics years and years ago and as I said, it
> was a 1g maneuver."


It is probably dangerous and stupid to try to guess the meaning of
another's words taken out of context and years after the fact. OTOH,
this is usenet, so... :-)

It would seem to me that the forces acting on the airplane (and pilot)
during the part of the roll where the airplane is inverted are what
most people wouldn't understand. Tex Johnson obviously knew that one
can't fly an aileron roll or a barrel roll without pulling a little g
on the way in and on the way out. It's not hard, however, to maintain
1 positive g while the airplane is upside-down. Maybe that's what he
was talking about?

Assuming the maneuver starts with a bag full of speed, a judicious
pitch rate, an airplane capable of a reasonable roll rate, and proper
placement of the nose before starting the roll, the whole thing can be
done easily with 2-ish g's and little-to-no seat puckering. (Make a
mistake in timing or control, however, and all bets are off.) Tex knew
what he was doing, and the tape proves it.

-Dave Russell
N2S-3

[Corky Scott]

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

[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

[Tony]

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.

[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.

[AES]

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"

[Jose]

> There is a path that, if followed by a vehicle, produces a
> loop at exactly 1G. It can be visualized as a combination
> of a path that accelerates downward to produce 0 G and a
> path that makes a perfect circle at exactly 1 G in the 0 G
> background field.

This is true as far as it goes. However, in order to do this in still
air, using wings, the aircraft has to assume certain attitudes that
preclude the 1G from being pointed in the same direction relative to the
cabin throughout the maneuver. To illustrate, consider the aircraft at
the bottom of a loop done in this manner, after completion. It would be
essentially horizontal (since it's the bottom of the loop and the
centripetal forces are pushing outward (downward w.r.t the cabin).
However, it is descending at whatever rate a freefall would be after
however long it takes to complete the maneuver. I bet that's pretty
fast - probably much faster than the forward speed of the airplane.

Consider the relative wind against the wings - I bet you'd see much more
than a 1G load on them were they to actually get into this configuration.

Now, it would work if the surrounding air were also freefalling.
However, the circumstance that would lead to freefalling air with an
airplane inside of it is not a circumstance in which I want to be the
pilot. :)

Jose
r.a.student snipped - I don't follow that group
--
You may not get what you pay for, but you sure as hell pay for what you get.
for Email, make the obvious change in the address.

[Tony]

Tod, your comment about free fall is on the money and it makes the
analysis easy. Great insight. If we define a roll as the airplane
rolling on its axis I think it's realizable. Think about flying a roll.
It's probably complete in say 5 seconds. 5 seconds of 1 G acceleration
is about 160 fps, or about 100 kts. That might be a dive something less
than 45 degrees, and I agree a straight pull back to level would induce
some Gs.



You'll need enough elevator authority and air speed to maintain the 1
G.

[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!



--

PERCUSSIVE MAINTENANCE:
The fine art of whacking the cr*p out of an electronic device to get it to
work again.

[David Odum]

T o d d P a t t i s t > wrote:

>There is a path that, if followed by a vehicle, produces a
>loop at exactly 1G. It can be visualized as a combination
>of a path that accelerates downward to produce 0 G and a
>path that makes a perfect circle at exactly 1 G in the 0 G
>background field. At the end of the maneuver, the object
>following that path would be hurtling towards the ground at
>high speed (the speed the object would have reached if it
>had fallen towards the ground during the entire time it took
>to fly the 1G loop.) I doubt any vehicle I'm familiar with
>could actually traverse such a path through air.

I agree with all of the above, Todd, except the part about calling it
a "loop" <g>. The following is the vertical profile of such a "loop",
flown in the hypothetical 0 G free fall that you describe, and at a
constant "loop" speed of 100 kts.

http://www.airplanezone.com/PubDir/FauxLoop.php (9 KB gif file)

Note that the Y scale of the plot is about 10 times the X scale so the
path is exaggerated in the x direction. Also note that the resulting
path has no resemblance to a loop. Perhaps you already visualized the
path, but I am sure many others didn't (including me). Just trying to
add to the knowledge.

David Odum -- email: David at AirplaneZone dot com

[David Odum]

T o d d P a t t i s t > wrote:

>Cool! Thanks for doing the math (what did you use?)

Well, I could have used Mathematica, my zillion dollar workhorse (and
toy), or I could have written a program in Delphi or one of the
several C variants at my disposal, but, alas, I used what I usually
use for quick and dirty analysis such as this, I used Excel. :)

>Have you looked closely enough at this to figure out whether
>you could get a more practical result with a non-constant
>speed loop, or by using a different speed?

I can plug any constant loop speed I wish into the spreadsheet that I
made. Indeed, loop speed is the only input variable. Alas, however,
it doesn't matter what speed is input, the general shape of the
vertical profile remains the same, only the scale changes. As for
modeling a non-constant loop speed, I doubt whether it would make much
of a difference. Perhaps one day I'll do that too, but not now
because my new Canon D20 digital SLR has just arrived and I want to
play. :)

BTW, Todd, those "1 G free fall barrel rolls" proposed elsewhere in
this thread have about as much resemblance to a barrel as your "1 G
free fall loop" does to a loop, as I'm sure you can now well imagine.

Greetz,

David Odum - email: David at AirplaneZone dot com

[David CL Francis]

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

[Tony]

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.

[David CL Francis]

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

[AES]

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.

[Tony]

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.

[alexy]

"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.

[David CL Francis]

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

[Tony]

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?

[David CL Francis]

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