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Sorry Greg, "no induced drag in a wind tunnel" is simply not true.
Once again, you don't understand. If you put a 3-D wing in a wind tunnel, you will get induced drag. The article you posted even documents the effect: ----------snip-------------- This drag occurs because the flow near the wing tips is distorted spanwise as a result of the pressure difference from the top to the bottom of the wing. Swirling vortices are formed at the wing tips, and there is an energy associated with these vortices. ----------snip-------------- See that? Flow "near the wing tips is distorted"....vortices are formed at the wing tips 2-D flow is achieved when the wing tips are flush against the sides of the wind tunnel. No wing tips = no pressure leaking around the wing tips = no wingtip vortices = no induced drag. |
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How did you arrive at 1.5%?
L = Wcos(theta) Come on Greg, you're telling me that wings in wind tunnels have no induced drag? That's ridiculous. How about if the wind tunnel was 1000 miles long by 10 miles high - would the wings in that wind tunnel have induced drag? (I seem to remember this same argument a few months ago) Argument? No, Todd Pattist and I attempted to educate you on this subject, but apparently failed. The size of the wind tunnel is irrelevant. What matters is that the wing tips are flush against the walls of the wind tunnel. This produces 2-D airflow, rather than 3-D. In 2-D flow, there are no wing tip vortices and thus will have no induced drag. (If you would read something other than Aerodynamics for Naval Aviators, you'd understand this.) |
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Greg Esres wrote:
How did you arrive at 1.5%? L = Wcos(theta) *If* you assume level and climb airspeeds are the same. [zap: induced drag disagreement] (If you would read something other than Aerodynamics for Naval Aviators, you'd understand this.) Yeah, I also hate it when people use decades of research, hours of wind tunnel testing, and accepted aerodynamic principals in these newsgroups. ![]() Hilton |
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On Fri, 31 Dec 2004 at 02:06:08 in message
, Peter Duniho wrote: Peter, I am sorry that this reply is a bit delayed as we have had visitors over the last three days. I have gone back to basic equations and looked again at them! I now have to confess that you are, at least in part correct. I offer my apologies. "David CL Francis" wrote in message ... [...] So under some conditions if you just raise the nose a little you can find a new steady state where speed is slightly reduced but with the same thrust you can climb at the same AoA.. Your theory sounds wonderful, but I doubt it holds water in practice. My statement above was wrong as written. It is true in the sense that you can find a new steady state at the same AoA and a slightly reduced speed but only by applying more thrust. You apply more thrust and if the AoA stays the same the velocity will initially increase and the flight path will curve upward arriving at a new steady state where there is a climb, but at a slightly reduced speed. Theta is the angle of climb in degrees. If the L/D is fixed at 10 then here is a little table. I hope it comes out not too screwed up by different fonts and line lengths: weight Lift/ Theta Drag Thrust Lift Lift Velocity lb. drag deg lb. lb. lb. % % 10000 10 0 1000 1000 10000 100 100.00 10000 10 1 1000 1174 9998 100 99.99 10000 10 2 999 1348 9994 100 99.97 10000 10 5 996 1868 9962 100 99.81 10000 10 15 966 3554 9659 97 98.28 10000 10 30 866 5866 8660 87 93.06 10000 10 45 707 7778 7071 71 84.09 10000 10 60 500 9160 5000 50 70.71 10000 10 90 0 10000 0 0 0.00 In practice aircraft cannot be controlled at zero velocity and more thrust would be needed at the higher angles of climb to maintain a controllable airspeed. Note that lift is reduced to zero in a vertical climb as it must be. I freely admit that these equations give only a simplified demonstration. There are many refinements to be added but the basics are generally accepted I believe. For example at the same power setting thrust is not independent of velocity, the thrust is not always exactly opposite to drag and control deflections also have an effect on drag and total lift. Weight is the gravitational force on the aircraft and points to the centre of the earth. Thrust is defined, in this simplification as a force along the flight path and drag is a force in the opposite direction. Lift (you seem to have got this wrong further down in your post) is defined, in the usual way, as acting at right angles to the flight path. Drag acts along the flight path in the opposite general direction to thrust. It has two components. One is the standard parasitic drag and the other is induced drag which can be looked at as the drag component due to the tilting back of the force vector to allow for the deflection of the air stream by the generation of the lift. By making a few assumptions about the size of the airflow that is actually deflected this can be shown to be proportional to the square of the lift coefficient. Thrust does not reduce the required lift by much, especially not in the light planes we tend to fly. Steady-state pitch angles in climbs tend to be modest, meaning a tiny fraction of the thrust vector is the downward component. Just 17% of the total thrust, for a 10 degree pitch angle. Given how little thrust a light plane has in the first place (only a relatively small fraction of the total weight of the airplane in the first place, less than 10% in at least some cases, perhaps most cases), even taking 20% (or even 34%) of that and applying it to lift just isn't going to help that much. The thrust must be equal to the drag or the aircraft will not fly at all! If the thrust is smaller than 10% of the weight then that implies that the lift/drag ratio is better than 10 for it to even fly at best lift/drag ratio. A motor glider requires little thrust to fly but may not climb very well. Note that the pitch angle, depending on how it is defined, is not normally the same as the angle of climb Even assuming an airplane with a thrust-to-weight ratio of 1.0 (a rare occurrance, but they do exist...some F-16s, for example), I'm not sure your theory holds up very well. You might think that you could simply increase thrust as you slow the airplane in order to allow a smaller AOA to suffice to provide the remaining necessary lift. But there's a problem with that idea. See the table above. I think you are using what you call 'my theory' in a strange way in the above. If an aircraft is capable of climbing vertically and steadily then thrust must always be greater than weight so that a velocity can be maintained sufficient to maintain control Under those circumstances the situation requires that lift is zero. If the AOA is kept small, then the increased thrust will prevent the airplane from decellerating. To have a vertical component high enough to support the airplane will require a horizontal component so high that the airplane won't slow. If the AOA is allowed to increase, then what the increased thrust is actually allowing is for the wing to stall without the airplane being pulled downward by gravity (the wing WILL stall at the appropriate AOA, regardless of airspeed). It's not demonstrating anything about some "new steady state". I do not understand the above and in particular your remark that the 'aircraft will not slow'. If AoA is not changed then increased thrust will increase speed and increased speed will increase lift. That, unless checked, will lead to the flight path curving upward and a climb beginning until a new state is reached. The gravity component then also plays its part in speed reduction. The above thought experiment should also illustrate another problem with your theory: it ignores the change in the portion of lift actually contributing to counteracting gravity that occurs due to pitch changes. See below for more commentary on that. Again I am not sure what you mean there. If you mean it ignores the effect of a climbing flight path then it does not. I have no special theory - just an attempt to explain basics in the simplest reasonable way using the simplest possible static equations. AFAIK there are an infinite number of different possible steady states in the flight of an aircraft. Of course, to me the biggest problem intuitively with your theory is that I am sure that aerodynamics involves only continuous functions. Given that, if you assume more than one steady state, you are claiming that there are multiple local minima/maxima between which are apparently "lower efficiency" areas. And of course, if there's more than one, I see no reason to believe that there are only two. This would then imply that the flight envelope has numerous of these local minima/maxima points. I certainly agree that aerodynamics is a series of continuous functions with changes of those functions at certain values. I am not sure about the rest of your paragraph above. An aircraft can fly at any speed and angle within its capabilities. Change any of the basic 4 forces and a new situation that can be maintained may or may not be possible. Given that in more than 100 years of study, this concept has never shown up as a noted element of the relationship between speed, drag, and lift, I'm inclined to believe that it's just not true (just as the idea of "cruising on the step" is not true). What concept are you talking about? I am just using simple balance of forces equations - apart from when I get it wrong of course! :-( I admit that I have not brought out the equations and proved my point irrefutably. Someone like Julian Scarfe or Todd Pattist would probably do a better job discussing this, since they seem to be more "math oriented" (that is, they don't mind crunching some equations now and then ![]() feel reasonably confident that there's no secondary "new steady state" one can achieve by increasing AOA and taking advantage of thrust. I would welcome Todd's participation, he has corrected me a few times but we have often agreed. The extra work comes ONLY from a net surplus of power. I agree with that. However that net surplus can come from either more engine power or reduced drag. Because even in a modest climb the engine thrust vector plus the lift vector combine to match the weight. I assume by "the engine thrust vector" you really mean "the vertical component of the engine thrust vector". Assuming that, I agree with your statement, but I don't find it informative. The engine thrust vector has a non-zero vertical component even during level cruise flight, and yes it does contribute to counteracting gravity, reducing the lift required. Fair enough and I have made a part of that point above somewhere. It is a part of the simplification involved. Put another way if you are flying below the maximum lift drag ratio and you increase the AoA to the optimum whilst keeping the same power the aircraft should climb. This is self evident if you are flying level at high speed and at a climbing power setting. Bring the nose up increasing the AoA and your aircraft will definitely climb. Agreed? I never said anything to the contrary. If you are at an angle of attack lower than the best L/D AOA (and thus at an airspeed higher than the L/Dmax airspeed), increasing pitch angle without a change in power will result in a climb, yes. I am glad we agree on that! But that has nothing to do with what happens at an airpseed near stall, which occurs below the L/Dmax airspeed, and well above the best L/D AOA. Agreed. You cannot decrease drag by increasing AoA near the stall. This is part of where we were at cross purposes. I was not thinking specifically about near the stall. This seems like a good point at which to mention something else you've left out and which I alluded to earlier... Lift is always generated perpendicular to the wing's chord. Some people like to call just the vertical component of this force "lift", but the amount of force acting through the wing is the force perpendicular to the chord. That is wrong. Lift is always defined as perpendicular to the free stream velocity of the aircraft. In the same way lift is not defined relative to gravity either except that in steady level flight it just happens to be exactly opposite and balancing gravity. However you label things, you cannot avoid the fact that when you change the angle of the lift vector, the portion of the force created by the wing used to counteract gravity is also changed. In particular, in the low-airspeed, power-on example, even as thrust is helping support the airplane, you are using your lift less efficiently, which means that the wing needs to generate more total lift just to provide the necessary vertical component. Insofar as I understand it, I disagree still with that statement. If thrust is providing a vector in the lift direction then in steady flight less lift is required as my table does show. This is similar to the required increase in lift while in a turn, but due to redirecting the lift vector in a different way. No, it is different to that because the increased force required in a turn is needed to produce an acceleration which is felt as 'g'. That does not apply in steady level, climbing or gliding flight when just the normal 1g is felt. Since the lift vector points slightly aft in level flight, even at high airspeeds when the angle of attack is low, it's easier to see how this negates at least some of whatever contribution thrust might make as the angle of attack is increased. However, in gliding flight, the vector is pointed forward, helping counteract the contribution drag makes to lift. The lift vector points at right angles to the line of flight no matter what. Lift and drag coefficients are measured like that unless things have changed a lot since my long ago student experiments in a wind tunnel.. The only time you consider a backward shift of the 'lift' vector is in calculating induced drag using a simplified method. By including a separate term for induced drag proportional to Cl^2 that effect is allowed for. Lift and drag coefficients for an airfoil are determined for infinite aspect ratio. This is getting too long to deal with like this! It has got to the point where we would need to discuss this face to face to achieve much more and come to a common understanding of terms and simple equations. I wrote "at least some" up there, but it should be apparent from the disparate magnitudes of the lift and thrust vectors that you get a much more significant change in lift than you do in thrust. Not at small angles of climb as my correction and table indicates. Small angles of climb mean a small change of the lift vector component in the truly vertical direction vector. The bottom line he even though thrust does contribute at least a little to counteracting gravity, it does not do so in a way significant enough to change the fact that, as you slow the airplane from any airspeed at or below L/Dmax airspeed, you experience increased drag. That is true and I do agree with that and apologise for the confusion. Only if the L/D increases with AoA do you get a benefit of reduced drag. Although the L/D of an aircraft may be 10 or more at maximum. at full speed in level flight it will be much lower, perhaps as low as 2 or less. At the zero lift AoA the Lift/drag ratio is also zero but you cannot then maintain level flight!!!. I hope that helps to clear some of the misunderstandings between us. My only excuse is my great age! I repeat my apology for my error. I always enjoy your posts even if I don't quite agree! -- David CL Francis |
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David CL Francis
David, I think you mentioned that you were retired aeronautical engineer of some sort. Can you relate something about your background? I'm just curious |
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On Wed, 5 Jan 2005 at 01:56:52 in message
, Greg Esres wrote: David CL Francis David, I think you mentioned that you were retired aeronautical engineer of some sort. Can you relate something about your background? I'm just curious You remember correctly Greg! Who is this guy eh? I took a degree in Aeronautically engineering many years ago and got involved mainly in missile design at Filton (It was Bristol Aeroplane Company then). I did a lot of structural work and then moved into Engineering management. As the years passed I moved into more and more other management type jobs but I was delighted that for the last two years of my career I was able to move back into engineering management. I have now been retired some years and a lot of my earlier detail knowledge has drifted away from me. At my age that is not surprising! I did start learning to fly, went solo but then I gave it up, partly I was not sure I had the abilities that I thought I should have and also because I felt I would never be any good at radio work! That tells you something about how long ago it was because the aircraft I flew had no radios despite sharing the airfield with a modest number of commercial flights! One of the highlights of my instruction was a brief spell of spin recovery training in a Tiger Moth (which does it for you unless you are quick). I did a lot of model work later with radio controlled models but eventually dropped that and found flight simulators. My first ever flight was with my father in WW2, would you believe, when he smuggled me up in a DH Dragonfly! It was being used as a communication aircraft. My father was a full timer in the RAF and flew in Hendon Air Displays in 1928-30. My Uncle was also a full time RAF pilot who ended his flying career as an Instructor in Canada in WW2. More recent highlights were a successfully take off, circuit and landing at Hong Kong in the REAL Concorde Simulator at Filton (with just a little help from the 'instructor') and a ride in the jump seat of a 747-400 flying into Kennedy on the 6th Sept 2001. I'll never get the chance to do that again! We drove away from New York to Philadelphia on the 10th Sept 2001 and found all about it in there in company with my friend who is a regular and respected contributor to this newsgroup. My wife and I stood on top of the South Tower on Sept 8th 2001. -- David CL Francis |
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Who is this guy eh?
Very interesting, thank you! Surprising that you could spend your life in this field, yet never get the pilot certificate. I'm not sure I would be interested in the subject if I were not able to fly. |
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On Fri, 7 Jan 2005 at 05:02:30 in message
, Greg Esres wrote: Who is this guy eh? Very interesting, thank you! Surprising that you could spend your life in this field, yet never get the pilot certificate. I'm not sure I would be interested in the subject if I were not able to fly. Just me I guess Greg. We are all different. My son was working in South Africa some years ago and he won his Private Licence out there. In 1989 my wife and I were out there and he took us for a flight over Johannesburg. Later back here he hired a Warrior and after a check ride from Filton we made a tour of the district. Before he left South Africa he checked out on a Cherokee 6 300 and with three friends (none of them were pilots) went on a touring cruise. They flew from Johannesburg to the mouth of the Orange River then north across the Namibian desert to Swakopmund. Then to the airport of Windhoek to Maun and then across Botswana to Francistown before heading back to Rand Airport In Johannesburg. Two desert crossings (Namibian and the Kalahari) 22.4 hours flying and 1012 litres of fuel. He no longer flies - it is very expensive in the UK and he has his own business to run. Anyway where in the UK can you fly for 3 hours without being able to even contact anyone on the radio? So my son, I am glad to say, has done many things that his father has not! If you wish to go to email my reply address should work OK. -- David CL Francis |
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