Fatal crash Arizona

On 5/7/2014 11:18 AM, Waveguru wrote:
Can you really get 8kts of sink at 200ft? Where is the air going, into the ground?

Dunno, but one time I was checking out a new club member in a Blanik. We
were on short final, rwy 16 at 3B3, when I said something like "this is
perfect, the right position at the right speed". About 5 seconds later I
couldn't see the rwy 'cause the tree tops were in the way. The thing
that saved us was the extra speed in anticipation of the wind gradient.

Tony "6N"

We have a
rule here in the UK, launch failure on aerotow below 300ft a landing should
be made ahead, or slightly to one side. No attempt should be made to turn
back below this height.

As a UK FI(S) and FIC, previously a Full Cat instructor, this is news to me. Where is it expressed?

I thought the rule was to exercise judgement. I always address 'where would you go now' issues with students and on check flights. In most conditions and situations I would turn back from lower than 300 foot.

On Wednesday, May 7, 2014 11:49:46 PM UTC+12, James Metcalfe wrote:
At 06:55 07 May 2014, Bruce Hoult wrote:
[1] handy formula: X knots of kinetic energy is worth (X/5)^2 feet of
gravitational potential energy. e.g. 70 knots = (70/5)^2 = 14^2 = 196 ft.
50 knots = (50/5)^2 = 10^2 = 100 ft. Less drag loses of course.
You'll never turn speed into quite that much height, and you'll need more
height than that to get speed. But the differences are large in a high
performance glider at moderate speeds.

..or you could use my rule of thumb: a change of speed of 10 knots IAS
gives
you (or costs you) the number of feet in height of the speed you arrive at.

For example:

70 knots to 60 knots : plus 60 feet
60 knots to 70 knots: minus 60 feet
70 knots to 50 knots (i.e. 70 to 60, then 60 to 50) : (60+50) = plus 110
feet

It's nearly the same formula. If you made it "every 12.5 knots IAS" instead of 10 then it'd be nearly exact.

The derivative of my formula (X/5)^2 i.e. X^2/25 is X/12.5. So gaining or losing X feet at X knots takes roughly 12.5 knots of speed change.

Doing it using the speed you're changing *to* is clever, as it makes some allowance for drag losses. However your height estimates are systematically biased 20% too large.

If you said 70 to 50 is (60+50)-20% = 110-22 = a gain of 88 ft then that would be good.
And 50 to 70 is (60+70)-20% = 130-26 = a loss of 104 ft, which is pretty good too.

If you're doing something like "how high can I zoom to after a 140 knot low pass and still have 60 knots for the circuit?" then I'm not convinced that adding up eight numbers is easier than calculating two squares and subtracting them :-)

By me: 28^2 - 12^2 = ~800 - ~150 = 650 ft (784-144 = 640 if you do it exact)
By you: 130+120+110+100+90+80+70+60 = 760 ft
By you with my -20% correction: 760 - 152 = 608

(remember folks, that's 140 knots at the END of the low pass, not the start!)

Someone wrote:
"Turns do not cause stall/spin accidents. In fact, the steeper the turn, the greater the stall margin. That's because elevator authority is progressively used up making the glider turn as the bank gets steeper until the angle of attack can't be raised above stall. Try it. You'll find many gliders will run out of up elevator before they can be stalled at bank angles over 45 degrees."

Any comments on the above statement. Steep turns offer a greater stall margin--really?

On Thursday, May 8, 2014 1:20:33 AM UTC+12, Vaughn wrote:
The comparison isn't quite as simple as just looking at L/D. Turn
radius also has a lot to do with your chances of making it back to the
field, and turn radius is proportional to the SQUARE of airspeed.

Compare your example (40 knot stall) with a (horrors) 2-33. The highest
stall listed for a 2-33 is around 30 knots. If you do the math, you
will find that your DG1000 has nearly double the turn radius of the
slower glider.

I've done the math :-) With a 30 knot stall instead of 40, you'll want 41 knots for that 45º banked turn instead of 55 knots. And, yes, the radius will be about 56% as large, or about 65m instead of 115.

Whether that 100m difference laterally after the turn makes a difference is highly airfield dependent. On a wide field you can land straight in in either case. Somewhere else, it might put you on the wrong side of a row of trees.

In a typical place, I'd much rather be twice as far off the center line with twice the L/D :-)

On 5/7/2014 6:00 PM, 150flivver wrote:
Someone wrote: "Turns do not cause stall/spin accidents. In fact, the
steeper the turn, the greater the stall margin. That's because elevator
authority is progressively used up making the glider turn as the bank gets
steeper until the angle of attack can't be raised above stall. Try it.
You'll find many gliders will run out of up elevator before they can be
stalled at bank angles over 45 degrees."

Any comments on the above statement. Steep turns offer a greater stall
margin--really?


We're really drifting away from this thread's initial topic, but if some folks
learn a useful thing or two from so doing, then maybe thread drift ain't
always so bad!

I think I can relate to the (presumed on my part) bemusement underlying the
question...it took some cogitation on my part when first exposed to the
concept, too.

My short form answer to the question is: Yes.

Assuming a still atmosphere, consider what constitutes a wing stall, what
control is used to drive the wing to stalling angle of attack, and what that
control has to be doing to get/sustain the glider into a constant speed bank,
and the light bulb may begin to glimmer.

And if you're a glider pilot with access to a glider capable of stalling the
wing in steady, 1G flight, go try it out at a safe height (e.g. steep
thermalling)...and ponder some more. Remember, it's not the speed, it's the
angle (of attack)...glider pilots are just used to using speed in the pattern
as a proxy for AoA.

Bob W.

P.S. I am not a lawyer nor have I ever played one on TV, but do keep in mind
two things: 1) all free instruction is worth exactly what you paid for it; 2)
the reader assumes all risks involved in testing any advice they may presume I
may be offering... :-)

On Thursday, May 8, 2014 3:34:48 AM UTC+12, Bill D wrote:
You're exaggerating the stall speed differences. Regardless of what the 2-33 "manual" says, no 2-33 ever got as slow as 30 knots. 35 knots is a practical minimum speed. The flight test stall speed for a DG 1000 is 37 knots.

I knew 40 knots was conservative with the DG1000 stall speed, but that difference is less than I'd have expected. I've never flown a 2-33, but I've had Blaniks under 35 knots :-)


However, no one should consider a turn back at stall speed. Virtually all gliders will be at 50 - 55 knots so the turn radius will be essentially the same.

I've put together a spreadsheet for the calculations and In fact it turns out that for minimum loss of height in a 180º turn -- and also much smaller turn radius -- you should fly a bit faster and bank a bit steeper.

https://docs.google.com/spreadsheets...FXBnfP8tauGK1k

No matter what your stall speed or L/D, it turns out the optimum to minimise loss of height in a turn is to bank at 54.7 degrees.

This is the bank angle at which the total G loading is 1.732 (sqrt(3)) and the G available to turn you is 1.414 (sqrt(2)).

In that glass ship with 120 fpm min sink at 45 knots you're looking at 67m turn radius at 59 knots, with 24 feet loss of height in a 180º turn.

In a Blanik with 160 fpm at 42 knots you'll get a 58m turn radius at 55 knots, with 30 feet loss of height.

In a 2-33 with 168 fpm at 35 knots you'll get a 40.5m turn radius at 46 knots, with 26 feet loss of height.

Feel free to play.

On Thursday, May 8, 2014 12:00:45 PM UTC+12, 150flivver wrote:
Someone wrote:

"Turns do not cause stall/spin accidents. In fact, the steeper the turn, the greater the stall margin. That's because elevator authority is progressively used up making the glider turn as the bank gets steeper until the angle of attack can't be raised above stall. Try it. You'll find many gliders will run out of up elevator before they can be stalled at bank angles over 45 degrees."



Any comments on the above statement. Steep turns offer a greater stall margin--really?

Sure.

If you have a glider that stalls at 35 knots then in a 60º banked turn (2 Gs) at 49.5 knots your turn radius is about 38m.

If your glider has 6m (20ft) between the wing and the tailplane then they are 9 degrees around the circle from each other and will see 9º different angle of attack to each other compared to straight and level flight. That 9º comes off your available back elevator movement.

In a corresponding 5G turn the numbers are 78.46º of bank, 78.3 knots, 33..8m radius and over 10 degrees of difference between wing and tailplane airflow. Plus you're going to need a lot more back elevator to pull those Gs.

Can you stall your glider in straight and level flight if you don't use the last 9 or 10 degrees of elevator deflection?

Another factor is that in a high G steep turn the two wingtips are seeing more more nearly the same airspeed and same angle of attack. So even if you do manage to stall, the spin tendency is much less and the merest easing of back pressure will stop it instantly.

On Wednesday, May 7, 2014 2:40:09 PM UTC-4, Don Johnstone wrote:
Emergency procedures do not have to be formulated for experienced thinking pilots.

Why do so many experienced thinking pilots enter spins from low altitude turns?

I want to know because I'm becoming more experienced and thoughtful with every flight.

Why do so many experienced thinking pilots enter spins from low altitude turns?

I want to know because I'm becoming more experienced and thoughtful with every flight.

My sincere sympathies to the friends and family of this pilot. John did a great job explaining the the answer to this question: Why do experienced pilots spin in low. Here is a link to his thoughts:

https://groups.google.com/forum/?fromgroups#!searchin/rec.aviation.soaring/john$20cochrane|sort:date/rec.aviation.soaring/ibhUAkQ6Z1s/8WoOYyZIlqEJ

Stay safe,
Bruno - B4