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)