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#1




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 
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#2




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 simulatedaircraft that has a Gmeter. 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 (pitchup, 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 rollrate 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




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




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




"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




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




"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




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




"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




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. 
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