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Lear 35 down approaching Teterboro

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  • LH-B744
    replied
    Originally posted by Gabriel View Post
    130 kts is 67 m/s.
    At 90 deg of bank you will not hold the altitude or attitude no matter how much you "pull up" (because "up" is 100% horizontal), but that doesn't prevent you from making the turn, and all the lift will contribute to the radial acceleration needed to bend the speed vector (i.e. turn).
    So let's say that you pull with a load factor of 2.5, that means that the lift will be 2.5 times the weight, and since the lift will be the only horizontal force with a radial component (neglecting an radial component that the thrust may have), we have

    Sum of F radial = 2.5 x weight = m x acc radial
    Since weight = m x g (g = acceleration of the gravity =~ 10 m/s2)
    2.5 x m x g = m x a
    So, a = 2.5 g = 25 m/s2

    Now, in a circular motion, a = V2/r, so r = V2/a = (67 m/s)2 / 25 m/s2 = 179 m = 0.1 NM

    How much with a 30 deg bank turn, assuming we hold the vertical speed constant?

    Well, the vertical component of the lift needs to be equal to the weight.
    The lift vector, tilted 30 degrees from the vertical, will need to measure weight / cos 30 deg.

    1/cos 30 = 1.15 will be the load factor (and, if you are interested, the stall speed will be sqrt(1.15) = 1.075 times, or 7.5% faster than, the 1G stall speed). But we are not interested in anything of this.

    The horizontal component of the lift, the one that will make the plane turn, will be lift x sin 30, and since lift was = weight / cos 30, we get that the horizontal component will be weight / cos 30 x sin 30 and that's weight x tan 30 = 0.58 times weight (and remember that weight = m x g).

    Horizontal force = horizontal component of the lift = 0.58 x m x g = m x a
    a = 0.58 g

    a = v2/r

    r = v2/a = (67 m/s)2 / 0.58x10 m/s2 = 773 m or 0.42 NM

    At 15 deg bank it would be 0.9 NM.
    Now I know where all of you "Senior members" got their 6,000 or 8,000 entries from... You always look when "the young boys" like me are not online, and then you ask your neighbor, who must be a mathematics professor, to write 300 or 900 entries...

    Gabriel, but you don't wanna tell me that you do all this inflight, with only 1 of 2 engines, or only 2 of 4 engines running.

    Flight Captain Chesley Sullenberger in his Airbus A320 is a hero in my eyes. Sullenberger deserves a golden star with a diamond because his decision is unique, until today. And why?
    Because he knew what his a/c is able to do, in a jet with ZERO engines running. And he still knows how to get the best results in an A320. Evidence? "Sully The Film" (2016), where Sullenberger as senior advisor and reason for the film appeared at the end.

    I don't think that he safely could've transformed his jet into a ship if he used 208 seconds for calculation. In German, it is called - it is late enough isn't it - Arschgefühl. You either have it for a special a/c type, or not.
    Last edited by LH-B744; 2017-06-04, 05:17. Reason: Thank God, I have never sat in a jet with zero engines.

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  • elaw
    replied
    ... thanks to a couple of well-timed feats of cowboy ejectmanship!

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  • Gabriel
    replied
    Originally posted by 3WE View Post
    An SU-35 at a similar bank and altitude might have done the same thing...
    An Su-35 I don't know, but this Su-27 sliced several heads (but the pilots survived).

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  • 3WE
    replied
    Originally posted by TeeVee View Post
    so basically the guy was trying to get a lear 35 at approach speed to act like an SU-35...and paid the ultimate price for his actions.
    An SU-35 at a similar bank and altitude might have done the same thing...

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  • Gabriel
    replied
    Originally posted by 3WE View Post
    Are you ardvark2zz?
    No. I don't have bunny ears.

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  • TeeVee
    replied
    so basically the guy was trying to get a lear 35 at approach speed to act like an SU-35...and paid the ultimate price for his actions.

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  • 3WE
    replied
    Originally posted by Gabriel View Post
    [ATTACH=CONFIG]7746[/ATTACH]

    and

    130 kts is 67 m/s. Sum of F radial = 2.5 x weight = m x acc radial Since weight = m x g (g = acceleration of the gravity =~ 10 m/s2) 2.5 x m x g = m x a So, a = 2.5 g = 25 m/s2, a = V2/r, so r = V2/a = (67 m/s)2 / 25 m/s2 = 179 m = 0.1 NM 1/Cos 30, stall speed will be sqrt(1.15) = 1.075 times, or 7.5% faster than, the 1G stall speed). Horizontal force = horizontal component of the lift = 0.58 x m x g = m x a a = 0.58 g a = v2/r r = v2/a = (67 m/s)2 / 0.58x10 m/s2 = 773 m or 0.42 NM At 15 deg bank 0.9 NM.
    Are you ardvark2zz?

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  • Gabriel
    replied
    Originally posted by 3WE View Post
    Actually, I agree in that with 10 miles visibility, you ought to see the airport, and make nice big fat dumb and happy turns to line up...

    If the winds are severe and/or you cramp yourself a bit and gotta push your bank all the way to 30 degrees as part of cowboy improvisation...well, that's why you get the big bucks and have all the training and take the time to hand fly...and I always thought landing in good gusty winds were about as fun as it got...
    Yep. Agreed and concur.

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  • Gabriel
    replied
    Originally posted by Evan View Post
    What's the climbing radius of a Lear 35 in a 90* bank? I guess its an ever tightening... spiral...
    Bank 90 degrees, pull (or not-pull) zero G, and the plane will describe a nice ballistic parabola (i.e. it will be basically a stone, or better a dart).

    Pull any G that you like other than zero and keep it constant, and that same parabola will be wrapped around a widening cone as the speed builds up.

    Curiosity/Quiz: Once the nose starts going down, if you are pulling up it requires continuous roll input to keep the bank angle at 90 deg, very significant roll input if the nose goes way down. Why?

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  • 3WE
    replied
    Originally posted by Gabriel View Post
    ...miles...miles....blah blah blah....miles...miles...mile...blah blah blah...
    Actually, I agree in that with 10 miles visibility, you ought to see the airport, and make nice big fat dumb and happy turns to line up...

    If the winds are severe and/or you cramp yourself a bit and gotta push your bank all the way to 30 degrees as part of cowboy improvisation...well, that's why you get the big bucks and have all the training and take the time to hand fly...and I always thought landing in good gusty winds were about as fun as it got...

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  • Gabriel
    replied
    Originally posted by Evan View Post
    Ok, so, one mile out... .42NM radius turn to the right... .42NM radius turn to the left... nothing a self-respecting cowboy in a hotrod 35 can't handle...
    Oh, and that's with a 30 deg bank. That's not cowboy worthy. With 90 deg and 2.5G it's 0.1 NM radius.

    Not only that, it doesn't work either. Look at the graph, 1 mile out of 06 means much less than 1 mile out of 01. Even if you could do a 0 radius razor sharp right turn to establish yourself in a 100 deg heading base leg, you would be barely a few feet yards out of 01 by when you intercept the extended centerline of 01 for another zero-radius left turn to final.

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  • Gabriel
    replied
    Originally posted by 3WE View Post
    LOL- it's beautiful making half-donkey-ass-hat posts after a very quick scan of the flying article, and having Gabe do the deep dive read. I owe you a beer- on my second scan of the article, I did not see ceiling and visibility.

    Seeing that ceiling / visibility is a non issue...I also see that I was right that 5 miles was a bit much since 4 miles is standard...

    OK, good...I get it...the 4 miles (to me) seems like a fat, dumb and happy figure. I am pretty sure I have seen a 3 mile-or-so, near-90-degree base-to-final turn while riding an MD-80, heading West and then South into KKCI...nice gentle bank...plenty of time to straighten it out on final...speed under control...solid airmanship. (Yeah, sure, my foffie side imagined them running a wingtip into the ground...but the reality was Fat, Dumb, Happy, Stabilized).

    So, the guy broke it off 1 mile out...I concur that is too tight. I do still stand by my 'gross deviation from really fundamental fundamentals'...So, maybe it's 1) Don't bank the hell out of a jet (especially down low), and 2) Fast planes will get ahead of you if you are behind. I can't totally discount the wind, but what...without the wind, they would have only needed an 85-degree bank?
    The problem is that there are miles and miles. If you are approaching 06 and start the right turn 4 miles out, the point where you START THE TURN will be barely 2 and a little miles out of 01. By when you finish the right turn you will be in a 2 miles-out base, then you need to turn 90 deg to the left and you'll line up at about 1.5 miles out with some 500 ft AGL.

    Start right turn the turn 1 mile out of 06 and, if you make it tight, you will be heading directly toward the numbers, but perpendicular to the runway

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  • Gabriel
    replied
    Originally posted by Evan View Post
    Ok, so, one mile out... .42NM radius turn to the right... .42NM radius turn to the left... nothing a self-respecting cowboy in a hotrod 35 can't handle...
    Cowboys knew their stuff and they were bold and daring but not suicidal. They may try to ride a brave bull in a rodeo, but they would not stay standing in front of a stampede of bulls.

    If this pilot intentionally attempted a say 60 deg bank turn at approach speed and 500 ft, "cowboy" is not the word I'd use. I may use it if he attempted an aileron roll after staying in a shallow climb after take off to gain enough airspeed.

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  • 3WE
    replied
    Originally posted by Evan View Post
    What's the turning radius of a Lear 35 at approach speed in a 90* bank?
    Depends upon how much you pull up relentlessly and how far above maneuvering speed you are.

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  • Evan
    replied
    Originally posted by Gabriel View Post

    r = v2/a = (67 m/s)2 / 0.58x10 m/s2 = 773 m or 0.42 NM
    Ok, so, one mile out... .42NM radius turn to the right... .42NM radius turn to the left... nothing a self-respecting cowboy in a hotrod 35 can't handle...

    Leave a comment:

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