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#31
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Lancair crash at SnF
"WingFlaps" wrote in message ... On Apr 25, 3:12 am, Dylan Smith wrote: On 2008-04-24, Brian wrote: Depends on what you mean by "the impossible turn". If you mean turning back at 200 AGL, yeah, that one's pretty much impossible. If you mean 600 AGL, it's pretty much possible in the average aircraft. (Hell, that's pattern altitude at EFD!) The line lies somewhere in between. It is statements like this that get pilots killed. It's statements like 'never turn, always land straight ahead' that also gets pilots killed. There are plenty of airfields where going straight ahead is quite possibly the worst option, and the best survivability options are at least a 120 degree turn away from whatever point you're at when at 600' AGL. The only thing you can do is use the best judgement at the time. You get one chance - it may be wrong. Sometimes, trying to turn back might be wrong. Sometimes doing anything *other* than trying to turn back might be wrong. In gliders, every glider pilot is taught "the impossible turnback" from 200 feet (which, in the typical low performance training glider, is about equal to turning back at 600 feet in a C172). It's the L/D that makes it much harder in a typical powered plane. This means that all manouvers lose energy much faster. The turn back needs at least 2 turns as well as acceleration if there is any wind. You will note that nearly all the accidents are stall spins -a moments thought about the situation will make you realize why this is. The turns are made tight because there is not enough height/time for a lazy turn. Let's work some real numbers for a 172 at 500'. Say climb was a Vx 59 knots. The plane must first be accelerated to 65 for best glide. The pilot carries out some trouble checks say 10s. Calls on the radio =10 s and plans his return. Note that 20s have probably elapsed. The plane has already travelled ~0.4 miles and at a 10:1 glide ratio has lost 200' (assuming he did get it to best glide in the first place). Can he make 2 turns and land back -no way! Ah you say, I'm a much better pilot, I would loose not more than 10 seconds in starting my turn back., trimming etc. But how much does the turn back cost? Assuming you keep to 45 degrees of bank to stay _above_ stall (the stall is now damn close -better hope there's no significant wind) the turns are still going to cost you 35 seconds. 45 seconds lost = 450 feet! Now we add in the energy losses from having to accelerate with the wind and to glide speed. It's still an impossible turn. Try to tighten that turn more and you have to dive to accelerate to avoid the stall and what does that do to your energy management and turn radius? Now what safety margin is appropriate for you and you PAX? Say 100% in that case, unless you've climbed to 1000' don't even think about turning back but practice spotting good landing sites. I've also heard a lot of BS in this thread about not having good palces to put the plane. There is nearly always somewhere flat to put the plane within 90 degrees of the runway centerline -even a road. Malls have big parking lots! Put it down flat in landing config and you will probably survive, stall spin and you'll DIE along with your PAX. A good pilot looks at the airport environs in a strange airport and may ask about options at the runway end for this emergency. Cheers Why two turns? At 500' why not one turn and land with wind up derriere? And, at 500 ft I wouldn't be too worried about the radio. Brian |
#32
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Lancair crash at SnF
WingFlaps wrote in
: On Apr 25, 3:12*am, Dylan Smith wrote: On 2008-04-24, Brian wrote: Depends on what you mean by "the impossible turn". If you mean turning back at 200 AGL, yeah, that one's pretty much impossible. If you mean 600 AG L, it's pretty much possible in the average aircraft. (Hell, that's patter n altitude at EFD!) The line lies somewhere in between. It is statements like this that get pilots killed. It's statements like 'never turn, always land straight ahead' that also gets pilots killed. There are plenty of airfields where going straight ahead is quite possibly the worst option, and the best survivability options are at least a 120 degree turn away from whatever point you're at when at 600' AGL. The only thing you can do is use the best judgement at the time. You get one chance - it may be wrong. Sometimes, trying to turn back might be wrong. Sometimes doing anything *other* than trying to turn back might be wrong. In gliders, every glider pilot is taught "the impossible turnback" from 200 feet (which, in the typical low performance training glider, is about equal to turning back at 600 feet in a C172). It's the L/D that makes it much harder in a typical powered plane. This means that all manouvers lose energy much faster. The turn back needs at least 2 turns as well as acceleration if there is any wind. You will note that nearly all the accidents are stall spins -a moments thought about the situation will make you realize why this is. The turns are made tight because there is not enough height/time for a lazy turn. Let's work some real numbers for a 172 at 500'. Say climb was a Vx 59 knots. The plane must first be accelerated to 65 for best glide. The pilot carries out some trouble checks say 10s. Calls on the radio =10 s and plans his return. Note that 20s have probably elapsed. The plane has already travelled ~0.4 miles and at a 10:1 glide ratio has lost 200' (assuming he did get it to best glide in the first place). Can he make 2 turns and land back -no way! Ah you say, I'm a much better pilot, I would loose not more than 10 seconds in starting my turn back., trimming etc. But how much does the turn back cost? Assuming you keep to 45 degrees of bank to stay _above_ stall (the stall is now damn close -better hope there's no significant wind) the turns are still going to cost you 35 seconds. 45 seconds lost = 450 feet! Now we add in the energy losses from having to accelerate with the wind and to glide speed. It's still an impossible turn. Try to tighten that turn more and you have to dive to accelerate to avoid the stall and what does that do to your energy management and turn radius? Now what safety margin is appropriate for you and you PAX? Say 100% in that case, unless you've climbed to 1000' don't even think about turning back but practice spotting good landing sites. I've also heard a lot of BS in this thread about not having good palces to put the plane. There is nearly always somewhere flat to put the plane within 90 degrees of the runway centerline -even a road. Malls have big parking lots! Put it down flat in landing config and you will probably survive, stall spin and you'll DIE along with your PAX. A good pilot looks at the airport environs in a strange airport and may ask about options at the runway end for this emergency. Exactly. I've practiced turning back. It's difficult, at best, even when you know it's going to happen. It has to be planned before the departure and the pilot mjst be very sharp to get away with one. No way is comparable to a rope break in a glider at 200' unless it's something like a Luscombe. Even if I went out and practiced them continuously, I don't think I'd try one on the day unless the outlying terrain was really poor and the departure field was particularly well suited to me doing so. But doin git without proficiency is suicide. Better to het something at 55 under control than to spin from 100 feet. Bertie |
#33
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Lancair crash at SnF
Stefan wrote in news:9f9a3$481106d6$54487369
: WingFlaps schrieb: (the stall is now damn close -better hope there's no significant wind) ... Now we add in the energy losses from having to accelerate with the wind and to glide speed. Arrrgh! Not the old "turn into downwind" legend again! Better work out your understanding of physics before publicly reasoning about turns. There is nearly always somewhere flat to put the plane The operative word in this sentence is "nearly". so you've done this, have you? Bertie |
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Lancair crash at SnF
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#35
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Lancair crash at SnF
Why two turns? At 500' why not one turn and land with wind up derriere? And, at 500 ft I wouldn't be too worried about the radio. When you turn back for the runway, you will not be lined up with it! Still air, you have to turn in excess of 180 deg to get back on or near the RW centerline and then, you need to make on in the opposite direction to line up. Bertie |
#36
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Lancair crash at SnF
WingFlaps wrote:
A Lancair crashed just moments after takeoff here in Mesa, Arizona, today, too. Plane was headed for California. There was smoke trailing from the plane on takeoff and controllers cleared them to turn back around and land. They tried -- they made the left turn but crashed into the orange orchard. Three fatalities, all in their late 20s. Sympathies and prayers to the families. When will pilots learn to stop trying to do the impossible turn... and go for a straight ahead landing on soemthing horizontal? What a bummer. The Jeppeson Instrument/Commercial text has a great illustration of how turning to land after losing power on takeoff doesn't work. -c |
#37
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Lancair crash at SnF
Dylan Smith wrote:
We don't know it was an 'impossible turn'. We don't even know what altitude they were at, whether the engine was still developing power or not, or whether the plane caught fire, or ... there simply isn't enough information to Except we know they didn't make it. Assuming the pilot was reasonably proficient, that suggests the turn couldn't be made. Late '20s, flying a Lancair. Well, at least he seems to have lived well. -c |
#38
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Lancair crash at SnF
On Fri, 25 Apr 2008 00:48:45 +0200, Stefan
wrote in : WingFlaps schrieb: Try reading the statement again, here it is: "Now we add in the energy losses from having to accelerate with the wind and to glide speed." Now perhaps you would like to revise some physics and try to critcise it for us? It's the "having to accelerate with the wind" part which is complete BS unless I completely misunderstand what you are trying to say. Here's what a physics professor, pilot and author of Performance of Light Aircraft* had to say on the subject of downwind turns when he was active on Usenet: http://groups.google.com/group/rec.a...41feb8649bbe12 Nov 22 1996, 1:00 am The Kinetic Energy "Paradox" Several recent rec.aviation.xx threads have featured or involved the apparent paradox that an aircraft which turns downwind thereby undergoes a large gain of kinetic energy (KE) w.r.t. (with respect to) the earth but no gain of kinetic energy w.r.t. the air mass. Initial (incorrect) physical reasoning is often along these lines: there is only a constant velocity (that of the wind) difference between CSs (coordinate systems) fixed w.r.t. the earth and w.r.t. the air. Hence time rates of change of velocities (accelerations) are the same in the two systems (the derivative of a constant being zero); hence so are the forces the same in both (the mass never varies). Since it takes a difference in forces to produce a difference in kinetic energies, there SHOULD be no such KE difference. But there is. The error is in the seemingly innocuous statement that two identical forces F(t), t the time, produce the same KE difference. The solution is to see that even though the forces, as seen in the two CSs, are identical, the DISTANCES moved under those forces, and the parallelism between force and displacement, are NOT the same. Delta_KE = Int(F dot dr) = Int(F dot v)dt=F dot vdelta_t; Int' stands for integral, F and dr and v are vectors and dot' is that of the dot or scalar product, the product of the two flankers times the cosine of the angle between them. Angular brackets, as X , stand for the average of quantity X. If one keeps track of the displacements, and their orientation with respect to the forces, which is what vector analysis does, then the kinetic energy difference computes properly and, as always, there is NO PARADOX. The truth is that the forces ARE the same, but the kinetic energy changes w.r.t. the two systems are NOT the same. Here's a simple example. A jet is heading North, unit vector i, at speed V w.r.t. the earth. The stewardess, Linda, of mass m, is walking South at velocity -si w.r.t. the jet. V might be 600 mph, s might be 4 mph. She hands a copy of Mountain Pilot magazine to a passenger, reverses course, and walks North at velocity +si. W.r.t. the jet, Linda has undergone a momentum change delta_p = 2msi. And, also w.r.t. the jet, a KE change delta_KE = 0. W.r.t. the earth, Linda has undergone a momentum change delta_p = 2msi (the same as w.r.t. the jet). And, also w.r.t. the earth, a KE change delta_KE = (m/2)*(V+s)^2 - (m/2)*(V-s)^2 = 2mVs. Not at all the same as w.r.t. the jet. The force of the deck carpeting on Linda's pumps, F(t), is in the +i direction during her turnabout maneuver; some sort of Gaussian peak over a relatively small time interval delta_t. Ignoring the small walking speed difference, her speed w.r.t. the earth throughout that maneuver was essentially V. (A relatively tiny bit less before she got stopped, a relatively tiny bit more afterwards.) So F dot V = FV (they're in the same direction). Now her momentum change is Int(F dt) = F delta_t = 2msi. We can use this fact to evaluate Fdelta_t, plug it in the average of the integral formulation in the second paragraph, and find that indeed her delta_KE = 2msV, just as calculated by taking simple differences. The short time intervals, integrals approximated as averages, etc., are only incidental features which have no bearing on the essential argument. (Not any more than accelerations at the end and beginning of one twin's journey at 0.9c has any real bearing on the twin "paradox" in special relativity.) The only trouble with treating the original airplane case, which I've done, is that you get two and a half pages of fairly dense vector calculus and the Internet is not yet sophisticated enough to let us put that across uniformly in this kind of forum. The result: though the forces are the same in the two coordinate systems (air mass and earth), the greater distance travelled in the earth-based system, and the fact that those displacements in space are not perpendicular to the forces (as they always are in the air mass based system), means that indeed kinetic energy IS gained in a downwind turn w.r.t. the earth. Here's the final result: Delta_KE(w.r.t. earth) = -m*omega*R*Vw*(cos(omega*tf)-1), where omega is the angular speed (yaw rate) in radians/sec, R is the radius of the turn w.r.t. the air mass, Vw is the wind speed, and tf is the time of flight. When you do a 180-degree turn, tf = T/2, where T is the time needed to make a circle w.r.t. the air mass. Then delta_KE = 2m*omega*R*Vw. If you continue around to make a 360, you find, as intuition suggests, that delta_KE(w.r.t. earth) = 0. NO paradox. There never is. There's only (hopefully, temporary) CONFUSION. Physics is wonderful. ------------------------------------- http://groups.google.com/group/rec.a...c4f35133062386 Jun 25 1996, 12:00 am Here's a posting I came up with this afternoon which *SHOULD* excite some discussion. At the moment, however, it is also my genuine opinion. John You Do Lose Airspeed in a Downwind Turn John T. Lowry, PhD Flight Physics Billings, Montana (406) 248 2606 June 1996 I came across a side reference to the infamous "downwind turn controversy" in an Internet user group posting. Then I located a reference to that conundrum, in Joe Christy's "Good Takeoffs and Good Landings," 2nd edition. On page 11 Mr. Christy writes: ...there is no difference between downwind, upwind, or crosswind turns in flight. They are all the same to the airplane. It seems that beginning flight students think that they will need to add power or angle of attack when turning from upwind to downwind because of the loss of airspeed inherent in the reversal of direction. Later on they learn the more sophisti- cated "truth." As expressed by Joe Christy farther down the same page: Remember, when airborne, you are carried *with* a moving air mass independently of your movement *through* it. His italics shown with *XX*. So the conventional wisdom is that there is no difference, except for the name, between a turn from downwind to upwind and one from upwind to downwind. Let's leave it at that for the moment. And let's change the subject; to that of horizontal wind shear, or gusts. Say you're heading into the wind at 100 KTAS and at first (Wind A) the wind is 40 knots. Suddenly (Wind B) there is a slackening gust to only 10 knots. (All wind speeds are ground speeds, speeds with respect to the earth just below them.) What happens? Your airspeed suddenly drops 30 knots to 70 KTAS. (For the moment, your 60 knot ground speed stays the same.) Let's say you take no immediate pilot control action. If your airplane is properly trimmed, the lower airspeed results in the airplane both losing altitude and nosing over. Airspeed begins to recover. When the original 100 KTAS is regained, you're back into trim. Albeit at a lower altitude. The point is: your air speed really did diminish. In fact there's a caveat to the conventional wisdom in a figure caption on that same page of Mr. Christy's book: To an airplane in flight, there is no difference between turning downwind, crosswind, or upwind, *except when flying through a wind shear*. This time, my (**) italics. So at this point we have it that turning towards downwind does not result in loss of air speed but flying into a slackening gust does result in loss of airspeed. But here's the kicker. *Turning to downwind is the same thing as a slow wind shear*. Say you start a coordinated turn from upwind towards downwind at standard rate, 3ø per second. At the start, say you have a 30 knot headwind. A second later you have a 30*cos(3ø) = 30*0.9986 = 29.96 knot headwind. It's not much less, but it *is* less. After turning for 10 seconds you've turned 30ø and the headwind has slacked off to "only" 30*cos(30ø) = 30*0.8660 = 25.98 knots. That's getting to be at least marginally perceptible, but remember the effect has been continuously spread over ten seconds. After a quarter turn from headwind to crosswind, 90ø, which took 30 seconds, your relative headwind has changed by 30 knots. That 30 knots is almost precisely the size (50 ft/sec) of the FAA's standard gust. But the FAA takes it that the gust develops its full force over only one quarter second! There's the difference. The same wind change, spread over a time interval 30/0.25 = 120 times as long and at a time while you've already increased angle of attack to counteract your banked and therefore off-vertical lift vector, and possibly also increased power a little to counteract increased induced drag and increased trim drag in the turn, is essentially imperceptible. The time scales are vastly different. So you do lose airspeed in a downwind turn. Only, normally, not much. I can imagine that students and instructors of aviation have infused their hangar flying sessions on this subject with several additional scientific red herrings. The Internet posting I mentioned even got into general relativity. Here are the more likely possibilities I can think of. Possible Red Herrings in the Downwind Turn Controversy 1. The air mass in which the airplane is embedded is an inertial frame and so . . . . If the air mass is steady, that's so. But if you fly from Wind A into slower moving Wind B, you're talking about two *different* inertial frames. Your air speed is your speed through the air in which you're currently flying, not your speed relative to the air a mile back. An airplane can change air speed quite rapidly because the speeds of air molecules can vary considerably from one place to another place fairly near by. That is, gusts exist. 2. Air speed is air speed, so . . . . Air speed is speed through that air *in the direction the airplane is pointed*. Say I told you an airplane was progressing northward against a 30 knot wind out of the north and its ground speed was northward at 70 knots. You'd probably say its air speed was 100 knots. But what if I told you, in addition, that that airplane's nose was pointed directly to the *south*. (The ultimate skidded flat turn.) Tilt! Unfair! Now the air speed has suddenly become -100 knots. The concept of air speed takes a bit more refinement than most of us give it. 3. The earth isn't actually an inertial frame of reference anyway . . . . True. But the non-inertial terms (due to rotation about its axis and revolution about the sun) are known. And quite small. So for all practical intents and short term purposes, the earth is an inertial frame. Therefore, a force is necessary to change the speed of the airplane relative to the earth. Even if the airplane flew into a vacuum, it would *initially* still move at the same speed in the same direction, just as does a tennis ball you drop out the window of a speeding car. Then, as the changed and unbalanced forces took over, the airplane would accelerate. In summary, my answer to the downwind turn controversy is: You do lose air speed in a downwind turn, but in a "noisy" environment and only over a time span that renders that loss almost imperceptible. Still, I think those beginning students were right. COMMENTS? CONTRARY VIEWS? ------------------------------------- http://groups.google.com/group/sci.p...37f136c766e351 Feb 10 1999, 1:00 am There is a problem similar to the ball/train/kinetic energy problem in aviation. When turning from heading upwind to heading downwind, the airplane apparently has a large gain of kinetic energy (as measured with respect to the Earth underneath). Even though the forces acting on the airplane during the turn (aerodynamic + weight) are the same in both the Air Mass- and the Earth-based systems (uniform motion between them), the kinetic energies are not. It's because Delta_KE is the integral of Force dot dr and the space increments dr are NOT the same in the two systems. Rather than go into this with pilots I usually simplify to a flight attendant handing a passenger a magazine, then turning on her pumps and walking back the way she came from. Similar to the bouncing ball. John. John T. Lowry, PhD Flight Physics; Box 20919; Billings MT 59104 Voice: 406-248-2606 ---------------------- http://groups.google.com/group/aus.a...e03975757e040d Mar 16 1999, 1:00 am Gavan Cook asked me to address this issue. It's a repost of an older response on the same subject in rec.aviation.piloting. For me, the convincing reason there is NO DIFFERENCE between a downwind turn and an upwind turn is: **all the relevant FORCES in the turn (except gravity, which doesn't change) are aerodynamic.** With those forces being only between the airframe or control surfaces or propeller and the AIR, what possible effect could anything else have? It's only forces which make accelerations which are changes in velocity. This of course assumes the exact same air mass motion (no gusts or wind shear), the same control inputs, etc., throughout the two (to downwind or to upwind) turn maneuvers. Now the airplane DOES slow down (in air speed) in a turn, due to increased induced drag due to larger lift due to bank, but it doesn't slow down any MORE in a downwind turn than in an upwind one. Phil's insistence on a **coordinated** turn is very important in a practical sense. Say you turned by kicking hard rudder, with a real big rudder, so that you skidded around 180 degrees in (say) two seconds. (Strange to say, one airline pilot told me another airline pilot told him that this slewing-around skidding turn is just the thing to do!) THEN you've got a problem all right. You've then got (almost) your original 50 knots of ground speed, in the same direction (now backwards), plus a 20 knot tailwind. Seventy knots air speed, but a NEGATIVE 70 knots air speed with reference to the direction the airplane is pointed. You'd "fall out of the sky." However, there is no difference in this foolish maneuver whether it's a turn to downwind or upwind. In the other case, where you'd initially have 70 knots air speed and 90 knots ground speed, an instantaneous skidded reversal would leave you at 90 knots ground speed backwards and a 20 knot headwind, net result still the exact same negative 70 knots air speed with reference to the direction the airplane is pointed. So still a real bad idea to skid around, but the SAME bad idea whether to downwind or upwind. There's no AERODYNAMIC difference between turns to dowwind or to upwind. The only differences are between their final ground speeds, and in the perception of the pilot who unwittingly slows his air spseed when he notices, by looking at the earth, increased ground speed during and after his turn to downwind. Hope this clarifies. Sometime later on I'll address the question "Where does the additional kinetic energy (with respect to the earth, an (approximate) inertial frame of reference), after a downwind turn, come from?" It DOES come, even though all the relevant forces, as argued above, are the same. It's a subtler issue. John Lowry. -- John T. Lowry, PhD Flight Physics; Box 20919; Billings MT 59104 Voice: 406-248-2606 --------------------- * http://www.bookfinder.com/search/?ac...t%2520aircraft |
#39
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Lancair crash at SnF
Hold on while I try to correct some nits in what Bertie said and see
what happens Bertie the Bunyip wrote: Nope, the wind is going to help you in almost every way if you're turning back. [ many good points supporting this assertion, but... ] also, your best LD speed is going to occur at a lower airspeed, Well, technically, your best LD speed is related to angle of attack, and not the groundspeed, so that won't change. Your best glide speed certainly will be less... There is no inertia involved in making a downwind turn. None. Here's why I wonder about that. Let's suppose 65 KAS before and after a 180 turn from a 10 KT headwind. OK, before the turn, your groundspeed is 55KTS and after the turn your groundspeed is 75KTS. Your intertial frame of reference is tied to the g/speed, not the a/speed. So -- the kinetic energy of the aircraft and contents is about 33% higher (75/55)^2. That energy is only going to come from one place with no power -- trading in altitude (potential energy) for kinetic energy. |
#40
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Lancair crash at SnF
Buttman wrote:
On Thu, 24 Apr 2008 17:05:06 +0200, Stefan sayeth: Brian schrieb: Your right in that many aircraft it is possible. But the problem is it isn't possible for many pilots when the engine quits. It is not a maneuver that is routinly practiced. Now this problem could be solved. You're suggesting instructors practice engine failures with their students on takeoff? Oh boy, better hope Dudly doesn't see this... One way to practice this would be to establish a "runway altitude" at, say, 1000ft AGL, get the airplane into takeoff configuration on heading at that altitude over a road or something, simulate a failure at a specified altitude--say, 1,500 feet--and see what altitude you're at when you get back to your reciprocal heading. If it's above your starting altitude, you made it. Wind, density altitude and aircraft weight are significant variables. Of course, a proficient pilot will have considered all these variables as well as the terrain downrange before takeoff, so they already know what they will do if the engine quits at a specific altitude. On probably as many checkrides and flight reviews as not, the instructor has asked me what I will do if I lose power on takeoff so I already know where there transmission lines are, about how far it is to the lake, etc. -c |
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