A aviation & planes forum. AviationBanter

If this is your first visit, be sure to check out the FAQ by clicking the link above. You may have to register before you can post: click the register link above to proceed. To start viewing messages, select the forum that you want to visit from the selection below.

Go Back   Home » AviationBanter forum » rec.aviation newsgroups » Aerobatics
Site Map Home Register Authors List Search Today's Posts Mark Forums Read Web Partners

Path of an airplane in a 1G roll



 
 
Thread Tools Display Modes
  #1  
Old June 19th 05, 05:04 PM
Chris W
external usenet poster
 
Posts: n/a
Default Path of an airplane in a 1G roll

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
Ads
  #2  
Old June 19th 05, 07:20 PM
Luke Scharf
external usenet poster
 
Posts: n/a
Default

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
  #3  
Old June 19th 05, 07:51 PM
zaphod
external usenet poster
 
Posts: n/a
Default

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
  #4  
Old June 20th 05, 12:52 AM
Byron Covey
external usenet poster
 
Posts: n/a
Default

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


BJC

"zaphod" wrote in message
news
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



  #5  
Old June 20th 05, 12:58 AM
Roy Smith
external usenet poster
 
Posts: n/a
Default

"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.
  #6  
Old June 20th 05, 01:36 AM
Tony
external usenet poster
 
Posts: n/a
Default

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.

  #7  
Old June 20th 05, 04:40 AM
Bob Fry
external usenet poster
 
Posts: n/a
Default

"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.
  #8  
Old June 20th 05, 04:54 AM
David O
external usenet poster
 
Posts: n/a
Default

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)


  #9  
Old June 20th 05, 09:13 AM
Peter Duniho
external usenet poster
 
Posts: n/a
Default

"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


  #10  
Old June 20th 05, 10:46 AM
Byron Covey
external usenet poster
 
Posts: n/a
Default

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.



 




Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

vB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Forum Jump

Similar Threads
Thread Thread Starter Forum Replies Last Post
Parachute fails to save SR-22 Capt.Doug Piloting 72 February 10th 05 06:14 AM
"I Want To FLY!"-(Youth) My store to raise funds for flying lessons Curtl33 General Aviation 7 January 10th 04 12:35 AM
Rolling a 172 - or not Scott Lowrey Piloting 55 November 16th 03 01:15 AM
rec.aviation.aerobatics FAQ Dr. Guenther Eichhorn Aerobatics 0 September 1st 03 07:27 AM
rec.aviation.aerobatics FAQ Dr. Guenther Eichhorn Aerobatics 0 August 1st 03 07:27 AM


All times are GMT +1. The time now is 04:37 PM.


Powered by vBulletin® Version 3.6.4
Copyright ©2000 - 2019, Jelsoft Enterprises Ltd.
Copyright 2004-2019 AviationBanter.
The comments are property of their posters.