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#31
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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. |
#32
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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. |
#33
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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" |
#34
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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. |
#35
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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. |
#36
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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)." |
#37
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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. |
#38
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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 |
#39
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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 |
#40
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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 |
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