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Thread: Attn: Gabriel

  1. #21
    Senior Member 3WE's Avatar
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    Quote Originally Posted by Evan View Post
    This does throw a rock into your theory that an airplane can be stalled at any airspeed and attitude...
    Wow...a near ultimate demonstration that you do are so enthralled with procedure to the detriment of one of the most mega important fundamentals...

    Not sure what FCOM you should check for type specific information on the relationship of attitude and AOA on that particular model...Boeings might be slightly better, but not sure.
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  2. #22
    Senior Member 3WE's Avatar
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    Quote Originally Posted by Gabriel View Post
    Only if your reference 0 deg AoA line is the zero lift AoA.
    Oh shit and lol...I send a ban-worthy flurry of personal attacks and expletives in your general direction...pedant!
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  3. #23
    Senior Member Gabriel's Avatar
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    Quote Originally Posted by LH-B744 View Post
    the AoA is +45°, during climb, and, -45° (mathematics..) during descent (!).
    No. That is pitch. The AoA is the angle between the wing airfoil and the relative wind and AF-447 is one of the few airliners that ever achieved 45 deg of AoA (and with a reasonably normal pitch).

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  4. #24
    Senior Member Gabriel's Avatar
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    Quote Originally Posted by BoeingBobby View Post
    Short duration and there are positive pressure boost pumps.
    And I guess continuous ignition on, just in case....

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  5. #25
    Senior Member Gabriel's Avatar
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    Quote Originally Posted by 3WE View Post
    "Only" is your word..."Roughly" is mine.

    I know trig functions are not linear and I assumed not too much airspeed change for the few seconds at push over

    BUT

    at low angles (like less than 16 degrees), angles and 'rise' and lift are ROUGHLY 1:1...yes?

    I anticipate 'the' equation...

    Oh, and I would ALSO guess that the AOA just before push over was slightly less than 16 degrees...
    It is not a matter of any trig function. CL (lift coefficient) vs AoA is pretty linear if you are far enough from the stall.
    But linear does not mean proportional. I.e. in y=mx+b b cab be other than zero.

    There are conventions for what is the zero degrees reference to measure AoA. When you are dealing with an airfoil in a wind tunnels (like in all those "famous" NACA and NASA papers where they were studying one airfoil after the other), then the typical convention is to take the "chord line" as reference. While this is pretty intuitive, it has its own ambiguities. What is the chord line? It is the line that joins the leading edge with the trailing edge. What is the leading and trailing edge? The leading edge is typically a curve of ample radius, what point on it do we take? The trailing edge is more clear... most of the times. Many airfoils (especially the not very old ones) have a trailing edge that is not the vertex of an acute angle but a small radius or even a a vertical cutoff, since the NASA engineers eventually realized that those razor sharp trailing edges where not practical for the real life industrial manufacturing methods and for the health of the wing once it is in service. One definition of the chord line is the line that joins the 2 points in the airfoil that are the farthest away one from the other. Even that definition has its problems. For example, what happens when you test the same airfoil with and without flaps? It is a convention that, once you define the chord line for the base airfoil, you will keep that line fixed so when you extend flaps or slats the chord line (end even the chord length) will still be the original one, not joining the "new" leading and trailing edges anymore. Also, when study a particular family of airfoils... that is when you take a camber line yc=m*f(x) and a thickness distribution line yt=n*g(x), where the airfoil shape is y = m*f(x) +/- n*g(x), and you keep f and g fixed and start to play with m and n, sometimes the trailing edge doesn't remain on the X axis after playing with m and n. In those cases, the chord line is preserved as the X axis, because if not it would complicate the comparison between airfoils of the same family.

    Now, regardless of all of the above, the lift at zero AoA (with whatever of the above references you want) will NOT be zero. Cambered (i.e. non-symmetrical) airfoils produce lift at zero AoA. You need to pitch the airfoil a bit down (could be 0.5 to 4 degrees, depending on how non-symmetric i.e. how much camber).

    That makes the lift equation (for the airfoil) being something L = k * (dCL/da * a + CLo), where dCL/da is the slope of the CL vs AoA straight line and CLo is the CL(0) i.e. the CL when the AoA is zero.

    Sometimes, we engineers take the zero-lift AoA as our zero AoA reference, because this simplifies the above formula to L = k * dCL/da * a.

    And that just the airfoil. Now let's move to the wing, where you have a spanwise distribution of chords (which together with the sweep gives you the planar wing shape) plus spanwise distribution of airfoils (aerodynamic twist) plus spanwise distribution of incidence angles (geometric twist). What on Earth do we take now as zero reference for the AoA???? The typical (and very arbitrary) convention is to take the AoA of the MAC (Mean Aerodynamic Chord, which I won't explain here, you are welcome). Again, the lift will not be zero when the AoA is zero here.

    And then we have the wing mounted on an airplane, Not only we have wing-to-fuselage and wing-to-tail interactions here, but we also have the incidence, which is the angle between the chord of the wing (however it is defined) and the deck one of the fuselage (however it is defined). Most of the times, the wing is mounted with a positive incidence, so when the airplane attitude is level (picth = zero), the wing will still have a positive AoA.

    Many times, especially for pilot's applications, in the whole airplane the AoA is defined as the angle between the deck line and the relative wind. That convention is nice because it makes the AoA = pitch - slope. So for example if you are in a 3-deg slope ILS and you are holding a pitch of 2 deg nose up, the AoA is 2 - (-3) = 5 egress (if no wind of course).

    Most times the wing is mounted with a positive incidence, meaning the when the plane pitch is level the wing is still at a positive AoA (by the typical wing AoA definition), so you need to pitch maybe 1 or 2 degrees down to have the wing at zero AoA (by the typical wing definition) and, still the wing as a postie lift at zero AoA, pitch down 1 or 2 egress more to have the needed negative AoA that will give you zero lift and let you fly at zero G.

    So, again, the AoA t 1.8G will be the 1.8 times the AoA at 1G ONLY IF YOU TAKE THE THE ZERO-LIFT AoA AS THE ZERO AoA REFERENCE.

    It is not a matter of trigonometry, it is not a matter of anything strange or complicated like my explanation above.

    It is a matter of the difference between y=mx and y=mx+b, and the fact that, in most conventions for what is the AoA, b is not zero.

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  6. #26
    Senior Member Gabriel's Avatar
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    Quote Originally Posted by 3WE View Post
    https://en.wikipedia.org/wiki/Angle_of_attack

    Errrrr straight enough that I stand behind my statement....curved enough that Gabeee will stand behind his...
    No. I am being as straight as you are (oh my gosh, this can be totally misleading since I don't know how straight you are )



    1) Say that, at a given weight and indicated airspeed, you need a CL of 0.5 to fly at 1G.
    2) See what AoA is needed for a CL of 0.5
    3) Now go and tell me that you need 1.8 times that AoA to fly at 1,8G
    4) Realize that you need a CL of 1.8 times 0.5 to fly at 1.8G.
    5) Realize that 3 and 4 are totally different.
    6) Perform the facepalm maneuver.

    Get it?

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  7. #27
    Senior Member Gabriel's Avatar
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    Quote Originally Posted by Evan View Post
    This does throw a rock into your theory that an airplane can be stalled at any airspeed and attitude...

    https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2598414/
    Absolutely not, this absolutely CONFIRMS my (my?) theory (theory?).

    An airplane CAN (vs will) be stalled at any airspeed and attitude, which (as I said many many times before) that also means that an airplane CAN (vs will) be not-stalled at any airspeed and altitude. And that includes ridiculously slow airspeed like, I don't know... 1 knot? (I am avoiding zero knots because there the AoA is as well defined as division by zero) and ridiculously extreme pitches like + or - 90 degrees (like an airplane doing a loop and not stalling in the process). Actually, and any COMBINATION of airspeed and pitch the airplane can be not-stalled (at least momentarily).

    One normally takes the accelerated stall saying that when pulling more than 1G the airplane will stall at a faster speed than the "official" stall speed (defined for 1G), but this also implies that flying at less than 1G the airplane will stall at a speed slower than the "official" stall speed. That's why pushing down can recover you from the stall faster that it takes for your speed to get back to normal flying speeds, and why you need to be so cautious when pulling back up after the stall horn stopped to avoid a secondary stall, and why the stall is considered recovered not when the stall warning stops bu when the plane is again flying level at a healthy airspeed.

    One important distinction, though, is that you can sustain a more-than=1G flight (you can turn for as long as you want). You cannot sustain a less-than-1G flight (unless you go as far as lass than negative 1G, aka inverted flight). Flight regimes between 1 and -1 G are not sustainable. Flights of 2G can be sustained for as long as the fuel last and then some more while you glide. Just keep turning with a bank of 60 degrees. And remember your stall speed will be 1.41 times the one indicated in the airspeed indicator.

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  8. #28
    Senior Member Gabriel's Avatar
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    Quote Originally Posted by 3WE View Post
    Oh shit and lol...I send a ban-worthy flurry of personal attacks and expletives in your general direction...pedant!
    Why do you reply thrice to the same post, before I even counter-reply even once? The realization process is slow nowadays?

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  9. #29
    Senior Member 3WE's Avatar
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    Quote Originally Posted by Gabriel View Post
    Why do you reply thrice to the same post, before I even counter-reply even once? The realization process is slow nowadays?
    Yes.

    I went ahead and made a second reply- flaming you for making me realize one thing...and to allow you to follow the process.

    The process:

    1. 3BS discovers graph that one could argue 'contains a moderately linear phase'...take that Gabriel, I am right and moderate flame!

    2. 3BS considers Gabby's comment that "AOA is a slightly vague term and that slight negative or 0 AOA's can result in positive lift...doubling negative numbers...seemed problematic to my 1:1 analogy...(Sarcastic, extreme flame for that nitpicky thing that it doesn't go through zero-zero...pedant!!!!!! )

    3. Well, since 1 and 2 have elapsed, 3BS has SINCE paused and looked at the Wiki AOA-Lift graph...the slope is nowhere near 1:1, more like 2:1.5...

    Argh...now what...

    3BS is a mechanistic thinker...to hell with your mathematical equations....tell me why in aggie English!

    If I have long board on a 2-degree slope and am rolling a wheel barrow up it, I get dang near almost twice the "lift" if I raise the board to a 4-degree slope. (minor trig adjustments ignored).

    Mechanistically, it seems to me that (with lower AOA's of course) if I double the AOA, I ought to roughly double the lift....

    I look at the example and instead of a 1:1 slope, it's more of a 2:1.5 slope...

    Ok, I'll go read your stuff- but scared it does not have a big fat clear summary statement to explain a 2:1.5 relationship...
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  10. #30
    Senior Member 3WE's Avatar
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    Quote Originally Posted by Gabriel View Post
    Blah blah blah-dissertation-blah blah blah with this bottom line:

    It is a matter of the difference between y=mx and y=mx+b, and the fact that, in most conventions for what is the AoA, b is not zero.
    Ummm errrrrr....

    Ok phase 2 (referenced just above) was my realization of your other-than-zero-intercept comment.

    Sure, doubling a -1 AOA (or 0 AOA) does what? to lift????

    But that still isn't it....I could force that line through 0-0 with the simplest transformation- but even then, it seems the relationship- although linear is not 1:1.

    You said AOA of 0 was something of a judgement call as to exactly how to define it...

    OK, I DEFINE 0 AOA as the AOA of zero lift...even then...my 2-degree vs 4-degree AOA does not give me 2X (times some sin, cosine, tangent slight correction figure of 0.98312) of lift...

    Something else is afoot...the slope of the line is 2:1.5....

    Total friggen accident, or some natural rule of airfoils shoving air downward?????

    There...maybe I can make a THIRD reply before you come back with more...
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  11. #31
    Senior Member Gabriel's Avatar
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    Quote Originally Posted by 3WE View Post
    Yes.

    I went ahead and made a second reply- flaming you for making me realize one thing...and to allow you to follow the process.

    The process:

    1. 3BS discovers graph that one could argue 'contains a moderately linear phase'...take that Gabriel, I am right and moderate flame!

    2. 3BS considers Gabby's comment that "AOA is a slightly vague term and that slight negative or 0 AOA's can result in positive lift...doubling negative numbers...seemed problematic to my 1:1 analogy...(Sarcastic, extreme flame for that nitpicky thing that it doesn't go through zero-zero...pedant...

    3. Well, since 1 and 2 have elapsed, 3BS has SINCE paused and looked at the Wiki AOA-Lift graph...the slope is nowhere near 1:1, more like 2:1.5...

    Argh...now what...

    3BS is a mechanistic thinker...to hell with your mathematical equations....tell me why in aggie English!

    If I have long board on a 2-degree slope and am rolling a wheel barrow up it, I get dang near almost twice the "lift" if I raise the board to a 2-degree slope. (minor trig adjustments ignored).

    Mechanistically, it seems to me that (with lower AOA's of course) if I double the AOA, I ought to roughly double the lift....

    I look at the example and instead of a 1:1 slope, it's more of a 2:1.5 slope...

    Ok, I'll go read your stuff- but scared it does not have a big fat clear summary statement to explain a 2:1.5 relationship...
    The problem why you fail to understand that you are wrong is that you are right, almost.

    a) Take a wing and fly it at an angle where it will make zero lift.
    b) Increase the angle (keeping the speed) until you produce lift = weight (i.e. 1G).
    c) Double that angle and you doubled the lift and the Gs.

    But, for example, a) is -2.5 deg, b) is 2.5 deg (and increase of 5 deg from a) and c) is 7.5 deg (another 5 deg from b, or total 10 deg from a).
    But obviously 7.5 is not 2*2.5.

    Now, if you measure AoA taking zero as AoA as the point of zero lift, then a) is zero, b) is 5 and c) is 10. Exactly as you expected.

    That's what I said, what you say is true if you put your zero AoA reference at the point of zero lift.

    And it doesn't matter if the ratio is 1:1 or 1:100.
    If it is 1:100, for 1 you have 100, and for 2 you have 200. It still hold that when you double one you doubled the other.
    But, you need the straight lie to cross the x and y axes in the origin, otherwise you have a linear function but NOT a proportional function.

    Again, it is the difference between y=mx and y=mx+b. An don't tell me that that is an engineers thing not an aggie one. My daughter is studying exactly that now in 7th grade and my other daughter is still studying that in 9th grade. And I bet that you studied that in the Advance Manuring class in the Aggie School.

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  12. #32
    Senior Member Gabriel's Avatar
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    Quote Originally Posted by 3WE View Post
    You said AOA of 0 was something of a judgement call as to exactly how to define it...
    Let me be less polite and say that it is totally arbitrary and that you can set it wherever you want.

    OK, I DEFINE 0 AOA as the AOA of zero lift...even then...my 2-degree vs 4-degree AOA does not give me 2X (times some sin, cosine, tangent slight correction figure of 0.98312) of lift...
    Then it works perfectly right. You are doing exactly what I said in my first post that you quoted and replied to thrice.

    Here it goes again, just in case:
    Only if your reference 0 deg AoA line is the zero lift AoA.

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  13. #33
    Senior Member 3WE's Avatar
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    Quote Originally Posted by Gabriel View Post
    No. I am being as straight as you are (oh my gosh, this can be totally misleading since I don't know how straight you are )
    ...
    6) Perform the plamface maneuver.

    Get it?
    No...I'm not that much of a dumbass, even though I am a dumbass...

    I understand what you said in the truncated quote above...but you are not addressing WHY there is an other than 1:1 relationship of angle and lift.

    The non-zero intercept is just a math game.

    If you read the two other replies (before you have had a chance to reply), I kick scream and yell that there must be an explainable, natural 'law of fiziks or something' to explain why it's not 1:1...even if I declared AOA of 0 is the AOA of zero lift...
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  14. #34
    Senior Member 3WE's Avatar
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    Ok, we are posting on top of each other...

    I will pause and do a safety check...currently, I do not believe that moving the line to 0-0 will cause it to yield ~2X lift for 2 degree vs. 4 degree AOA...

    Y = mx + b is not a foreign concept (other than it being Evan cryptic acronyms for slope and intercept)...
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  15. #35
    Senior Member 3WE's Avatar
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    Quote Originally Posted by 3WE View Post
    Oh shit and lol...I send a ban-worthy flurry of personal attacks and expletives in your general direction...pedant!
    @!#%@%#[email protected]#%@%@!#@$^!#@!$#^!#@%@#%!%!#%!%[email protected]^%@!#$

    Everyone please PM Brian to tell him I laid multiple ban-worthy, expletive laden attacks on Gabriel, his family and his pets...

    Here's the deal...I DID figure it out on the second reply to Gabriel.

    When I move the damn line to zero zero...and double a small, lift-producing AOA, I get about double the lift...

    I KNEW IT WAS SORT OF DAMN NEAR A 1:1 RELATIONSHIP!!!!!!!!!!!!!!!! (Told you I think conceptually!)

    I do wanna say that the AOA to produce 1.8 G is not all that super impressive (since any number of maneuvers can produce 1.8 g)...maybe it ain't normal airliner operations...but I THINK 60 degree banks are considered well within an airliner's structural abilities.

    Yeah...sure, I'll face palm myself...I sort of got it all along, until Mr. Mathematics had to correct me on an arithmetic quirk.
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  16. #36
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    Quote Originally Posted by 3WE View Post
    PS: Article by "Daddy Rich", former head of Obscure Aviation Typists, ATL crew's top unofficial mentor...
    Not surprisingly, the point of the article seems to be to once again, for at least the eleven thousandth time, show what a guru Ol' Dick is...

  17. #37
    Senior Member Gabriel's Avatar
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    And a bonus track... you will love this:

    Since you were talking about the actual value of the slope of the CL vs AoA quasi-straight line, and wondering why it is not 1:1. I don't know why would you expect it to be 1:1, there is no reason for that, especially since the CL is dimensionless and the degrees that we use to measure are totally arbitrary, just a circle split in 360 pizza slices. God knows why some old primitive mathematician decided to split the circle in 360 slices, and not 200, or 100, or 10, or 12 or 60 that were more used numbers back then.

    In theory (and it is a relatively simple theory called thin airfoil, purely mathematical, where you assume very little other than the chord is much bigger than the thickness, the flow is non-viscous, and the flow will separate at the trailing edge), the slope of CL vs AoA is 2*Pi. Yes, that's tight, Pi like the ratio of the circumference to the diameter. 3.1516....
    And not only that, but in practice it is very very close to 2*Pi.

    I know, you will say "Come on, an airfoil 1 Deg above the point of zero lift will not make a CL of 2*3.14=more than 6!!! Just look at the chart, 10 deg above the zero lift point you have a CL of only about 1.1".
    And you would be right (although I doubt that you would have the presence of mind to say that, because that is an engineering thing rather than an aggie one)

    That's because I forgot to mention that the theory says that the CL will increase 2*Pi per RADIAN of increase in AoA. Not per degree.

    Now, how much is 10 degrees in radians? It is 0.175 radians. Multiply by 2*Pi and you get 1.1. Surprise!!!

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  18. #38
    Senior Member brianw999's Avatar
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    Everyone please PM Brian to tell him I laid multiple ban-worthy, expletive laden attacks on Gabriel, his family and his pets...
    Don’t you bloody dare !!! If the slagging off was badly translated from German I might consider it but otherwise I’m keeping well out of this one !
    If it 'ain't broken........ Don't try to mend it !


  19. #39
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    Co-incidentally I just received the February Aerospace America magazine that contains an article on microgravity experiments using a 727 aircraft (G-Force 1).
    The article contains the following flight information:
    After reaching a flight altitude of 24,000 ft the pilot increases the plane’s ascent angle to about 45 degrees to 32.000 ft. The aircraft eases over to the top of the parabola and rapidly descends, 20 degrees nose down, creating microgravity for 20 to 30 seconds.

    The prior aircraft used for these type flights was a converted KC-135 fuel tanker dubbed “the vomit comet”.


    The aircraft had several modifications, made to gain FAA approval, including a modified hydraulic system to prevent dangerous cavitation, or bubbles, from forming.

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