Glide Distance for Cessna 172N
Wwplummer
writes:
>The handbook for a 1985 Cessna 172 N contains a chart which estimates
>glide distance in the event of an engine failure. It is based on no
wind,
>flaps retracted, and a 65 KIAS. I'm trying to estimate the effect of
wind
>and flaps on the distance.
>
>At 4,500 AGL the no wind distance is approx. 6.75 N.M based on the chart
>and interpolation. The way I figure it, that works out to a 1200 fpm
>altitude loss at 65K KIAS or 00:03:45 of time. Now, all I need is to
>figure ground speed to see how wind affects distance.
>
>Am I on track here? Also, if flaps are added how would I calculate the
>effect on distance?
That sounds about right -- a glide ratio of 8.2. Gliders have ratios of
20 to 50 and motor gliders are 90. Check out "the Joy of Soaring" and
read about the polar curve. Every airplane and glider in every
configuration (flaps, speed brakes, etc) has such a curve that relates
vertical sink rate to horizontal speed. It is a parabolic shape, usually
plotted so the open part is down. The peak identifies the speed to fly to
get the minimum sink or max. time in the air. The tangent which goes
through the origin identifies the maximum glide speed or max distance
covered while in the air. Max glide is not the same as min sink speed.
Up and down drafts just slide the y-axis, while head winds and tail winds
effectively move the x-axis. --Bill
_________________________________________________________________________
Wm W. Plummer, 7 Country Club D., Chelmsford MA 01824 508-256-9570
PP-ASEL,G
Roy Smith
> At 4,500 AGL the no wind distance is approx. 6.75 N.M based on the chart
> and interpolation. The way I figure it, that works out to a 1200 fpm
> altitude loss at 65K KIAS or 00:03:45 of time. Now, all I need is to
> figure ground speed to see how wind affects distance.
>
> Am I on track here?
Yup, that's exactly it. But, more interestingly, the way I see it, from
any altitude, you have a "withing gliding range" circle. The diameter of
the circle only depends on your altitude and the gliding characteristics
of your airplane. With no wind, the center of the circle is your current
position. Any wind simply shifts the center downwind, but doesn't change
the shape or size of the area that's within gliding range. Thus, if want
to be sure of being able to glide to land while crossing a body of water,
you only have to make sure you do it at enough altitude to have the
diameter of your glide circle equal or exceed the width of the body of
water. Any wind will not change that altitude, it will only change where
you change your engine-out strategy from "turn back" to "keep going to the
other shore".
> Also, if flaps are added how would I calculate the effect on distance?
I don't know of any quantitative formula for this, but it's a pretty good
bet flaps will decrease your glide distance. There are some complex
transient effects of raising flaps, so it's not clear what the right
strategy is if you have the flaps down at low altitude and the engine
dies. But, at altitude, with flaps 0, I'd say leave them that way until
your landing site is made.
> As a midwesterner who has relocated to Long Island, NY.; I'm not very
> comfortable flying across LI Sound.
I've done all my flying around here, and I agree. I once got into a fight
with ATC over this. I filed IFR White Plains to somewhere out on Long
Island. I filed a route and altitude that would keep me at gliding
distance from land over the sound. My clearance came back taking me over
the widest point of the sound at 3000 feet, or something like that. I
turned it down, "unable that route at that altitude", and explained why.
The response was essentially, "This is what we're giving you, you can take
it or not, but you're not getting anything else". Fortunately, the
weather was such that I could go VFR, so I went the way I wanted, at the
altitude I wanted.
The next day, I made a bunch of phone calls to the tower supervisor, and
to the procedures people at tracon. Everybody was very nice and polite on
the phone, but stuck to the party line that you get what you get, and
nobody's making you fly it. I even tried pointing out to them that if I
was flying part-135, such a routing (out of glide distance) would be
illegal. This did not impress them. I guess that's just life in
congested airspace.
--
Roy Smith <r...@nyu.edu>
Hippocrates Project, Department of Microbiology, Coles 202
NYU School of Medicine, 550 First Avenue, New York, NY 10016
"This never happened to Bart Simpson."
Roy Smith
I wrote:
> With no wind, the center of the circle is your current
> position. Any wind simply shifts the center downwind, but doesn't change
> the shape or size of the area that's within gliding range.
d...@opal.usna.navy.mil (PROF D. Rogers {EAS FAC}) replied:
> This statement is technically incorrect. Yes, the `center' shifts downwind
> BUT the shape is no longer a circle.
I'm having trouble seeing why not. Once you go into glider mode, you have
a certain amount of time in the air, determined by your initial altitude
and sink rate. During that time, you're going to move forward (in
whatever direction you choose) a fixed amount, determined by your forward
velocity and time aloft. During that time, you'll also move downwind a
fixed amount, which is simply your time aloft times the wind speed. The
circle is just uniformly shifted downwind. I don't see how it changes
shape to anything other than a circle.
PROF D. Rogers {EAS FAC}
!t...@systec.com (Tim Bachert) wrote:
!! At 4,500 AGL the no wind distance is approx. 6.75 N.M based on the chart
!! and interpolation. The way I figure it, that works out to a 1200 fpm
!! altitude loss at 65K KIAS or 00:03:45 of time. Now, all I need is to
!! figure ground speed to see how wind affects distance.
!!
!! Am I on track here?
!
!Yup, that's exactly it. But, more interestingly, the way I see it, from
!any altitude, you have a "withing gliding range" circle. The diameter of
!the circle only depends on your altitude and the gliding characteristics
!of your airplane. With no wind, the center of the circle is your current
!position. Any wind simply shifts the center downwind, but doesn't change
!the shape or size of the area that's within gliding range. Thus, if want
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
This statement is technically incorrect. Yes, the `center' shifts downwind
BUT the shape is no longer a circle.
!to be sure of being able to glide to land while crossing a body of water,
!you only have to make sure you do it at enough altitude to have the
!diameter of your glide circle equal or exceed the width of the body of
!water. Any wind will not change that altitude, it will only change where
!you change your engine-out strategy from "turn back" to "keep going to the
!other shore".
!
!! Also, if flaps are added how would I calculate the effect on distance?
!
!I don't know of any quantitative formula for this, but it's a pretty good
!bet flaps will decrease your glide distance. There are some complex
!transient effects of raising flaps, so it's not clear what the right
!strategy is if you have the flaps down at low altitude and the engine
!dies. But, at altitude, with flaps 0, I'd say leave them that way until
!your landing site is made.
If you know certain things about the effect of flaps on parasite
drag the decrease in glide distance is easy to calculate.
Dave Rogers
Anton Verhulst
>glide distance in the event of an engine failure. It is based on no wind,
>flaps retracted, and a 65 KIAS. I'm trying to estimate the effect of wind
>and flaps on the distance.
>
This is not right. At 65kts you are flying at slightly more that 1nm per
minute so it should take around 6 minutes to fly 6.5nm - not 3:45.
The figures are:
4500 AGL and 6.75nm yields a glide ratio of 9:1.
time to fly 6.75nm at 65kts is 6 min 14 sec.
sink rate is 722 fpm.
details:
(nm flown) * (feet per nm) / (feet AGL) = glide ratio
6.75 * 6072 / 4500 = 9.1
((distance in nm) / (speed in kts)) * (min per hour) = time aloft
(6.75 / 65) * 60 = 6.23 = 6:14
(AGL) / (min aloft) = sink rate
4500 / 6.23 = 722 fpm
Tony V. CFI-Glider
Ron Blaylock
exactly how much time you will get regardless of wind conditions. You
are flying relative to the wind for that length of time.
If that equates to 6.75 miles, no wind, through the air, it is also
6.75 miles in the wind through the air. What counts is your ground
track.
Assuming your ground speed and true airspeed are the same at 65 mph,
you will travel forward 95.3 feet per second for 405 seconds or
7.3 statute miles. If you subtract a dead ahead headwind of 5 mph, you
will only travel 88 feet per second or 6.75 statute miles.
Further if you have a 15 mph headwind you would be traveling at 73.3
feet per second for 405 seconds or 5.63 statute miles.
Of course I'm using 5,280 feet per mile for the calculations.
You can do the same math using 6,000 feet per mile to get nautical miles
traveled.
65 knots times 6,000 feet = 390,000 feet per hour. Divide 390,000
by 3600 (number of seconds in an hour) to get feet per second of 65.
Multiply this by your flying time of 405 seconds and you will get 26,325
feet. Divide this by 6,000 feet in a nautical mile and you will get
4.39 nautical miles.
I assume you could do the same using actual minute calculations but I
always break it down into feet per second. I know that at 60 statute
miles per hour you are traveling 88 feet per second. At 30 statute
miles per hour you are traveling at 44 feet per second. It makes some
estimates easy.
Also at 60 mph you are traveling one mile every minute. This be the
same for nautical miles. If you are traveling 60 knots, you will cover
one nautical mile every minute.
I use these for rough estimates before calculating flying time. A true
airspeed of 120 knots equals two nautical miles a minute. 50 miles to
fly equals 25 minutes. If your true airspeed is 110, subtract about
8% of the time.
Anyway, you are absolutely on the right track. Calculate your flying
time to ground zero, then subtract or add the winds as appropriate.
Ron B.
--
Ron Blaylock - < rg...@aimnet.com >
Living in beautiful Santa Clara, California
"You are but one flight physical away from an ultralight.", Steve
Larghi.
* Disclaimer: The view(s) presented are my own. I take full responsibility
* for them. They do not represent my employer or employment.
Stefan Plazier
and the reason it is a circle..
in nil wind it is a circle, no debate.
When a wind is applied it displaces all points on the diameter by an
equal amount (same applied velocity), hence it is still a circle but it
has moved.
okay i'll crawl back in my hole now.
Stefan..
Stefan Plazier
SNIP
>So what you have is an ellipse with the main axis running parallel to the
>wind with an attainable region 13.5 nautical miles long (this length will
>never change, more maths), this main axis will shift further downwind as
>the wind strength increases. The ellipse will be fatter on the downwind
>side an narrower at the upwind end, the stronger the winds the narrower
>the top.
>
SNIP
I agree with everything that this person said except for the ellipse bit,
I think if he went away and made some pretty little drawings he'd find it
is in fact a CIRCLE, yes that's right a circle. A perfectly round circle
which is displaced by the wind.
Oh the shame,
Stefan
Ron Natalie
: > BUT the shape is no longer a circle.
: I'm having trouble seeing why not. Once you go into glider mode, you have
: a certain amount of time in the air, determined by your initial altitude
: and sink rate.
If you keep your airspeed constant, then I agree. However, you may go
farther by altering your glide speed. Your "certain amount of time in
the air" presumes a constant glide airspeed. The best glide speed however
isn't the minimum sink rate, it's that balances time aloft with forward
speed, and is predecated on zero wind. You will may actually go farther
downwind by decreasing your speed towards the minimum sink speed or upwind
by nosing it over for some forward speed and less time aloft.
-Ron
John Rourke
Smith) writes:
>
>I wrote:
>> With no wind, the center of the circle is your current
>
>
>
>I'm having trouble seeing why not. Once you go into glider mode, you have
>a certain amount of time in the air, determined by your initial altitude
>whatever direction you choose) a fixed amount, determined by your forward
>velocity and time aloft.
PLUS the velocity of the air mass you are in, if with the wind, or MINUS the
velocity of the air mass, if you're trying to glide against the wind.
Look at it this way: For simplicity sake, let's say you're 1 mile up, and your
glide ratio is 10:1 - if you have no wind, you have a perfect circle of 10 mile
diameter to glide to.
Let's say your best glide speed is 80 mph. Let's also say the velocity of the
air mass you're in is also at 80 mph (Yikes...
sowhaddayadoiftheenginequitsinahurricane?) If you try to glide against
the wind, you will land directly underneath you, because you have a
groundspeed of zero (Hey, that was *easy*!), OTOH, if you glide in the
same direction of the airmass, you will have a groundspeed of 160 mph -
and you will be 20 miles downrange, because while you were descending 1
mile, you moved 10 miles through the air mass (10:1 glide), AND the air
mass *also* moved 10 miles. If you tried to glide perpendicular to the
wind direction, you would find you travel somehat less than 10 miles,
but that involves tangents and stuff, and I have so far successfully
avoided using any math (OK, you would be able to go 7.07 miles, if you
want to get techy).
During that time, you'll also move downwind a
>fixed amount, which is simply your time aloft times the wind speed.
The
>circle is just uniformly shifted downwind. I don't see how it changes
>shape to anything other than a circle.
As described above it would be an ellipse. Only in the no-wind
scenario do you get a perfect circle. OK, one more gedankenexperiment:
let's say you're going 60 mph in your glide, but the airmass is going
1000 mph. Intuitively, you'll probably realize your landing area
*cannot* be a circle. In this particular universe, anyway.
>
>--
>Roy Smith <r...@nyu.edu>
>Hippocrates Project, Department of Microbiology, Coles 202
>NYU School of Medicine, 550 First Avenue, New York, NY 10016
>"This never happened to Bart Simpson."
Happy gliding,
-jpr
Rod Cadanau
>
>
>and the reason it is a circle..
>
>in nil wind it is a circle, no debate.
>When a wind is applied it displaces all points on the diameter by an
>equal amount (same applied velocity), hence it is still a circle but it
>has moved.
>
Actually, you're right the first time, but for a different reason. Remember that
the 65 kts glide speed is a trade-off between sink rate and forward speed to get you
the maximum distance. If there is a wind, this trade-off point is different. You
want to balance ground speed with sink rate to get the maximum distance. Gliding
with a strong tail-wind you might want to slow down to more like 60 kts get the
minimum sink rate. In a strong headwind, increase your airspeed. The example of a
65 kt headwind is good: Airspeed of 65 kt gets you nowhere. Airspeed of 80 knots
won't keep you up as long, but at least you'll make some forward progress.
Rod
Jgs3mstr
writes:
>You can do the same math using 6,000 feet per mile to get nautical miles
>traveled.
Isn't 1nm=6080 feet instead of 6000 - maybe i'm wrong
Fred Black
You're right, but when you are in a situation where you have to be calculating
glide distance, you probably don't have the time to be that precise.
--
Fred G. Black, P. Eng. E-mail: bl...@bnr.ca Nortel North America
PP-ASEL,G Phone: (613)-763-7452 (W) P.O. Box 3511 StationC
(613)-823-6017 (H) Ottawa, Ontario
My opinions only Canada K1Y 4H7
Thomas K. Dibenedetto
: >
: >This statement is technically incorrect. Yes, the `center' shifts downwind
: >BUT the shape is no longer a circle.
If you have a headwind of a mile a minute, and your glide speed is a mile
a minute, and you glide into the wind (lets say North), you will glide
forward a mile a minute, but will also be pushed back a mile a minute
and so will land immediatly below where you started right? If you turned
180 (South), you would glide a mile a minute plus being pushed a mile a
minute = 2 miles a minute, right? If you turn 90, i.e. directly
crosswind, you will glide east a mile a minute, but be pushed south a
mile a minute right? Picture a circle. Picture yourself on a point on the
circle (not inside it). Situation 1, you go straight down. Situation 2,
you glide to the opposite side of the circle. Situation 3, you glide an
equal distance east and south,(two perpendicular radii),, would put you on
the circle again, wouldn't it?
With no wind, you start out in the center of the circle. With wind=glide
speed, you start out on the circle. Anything inbetween, you start out
somewhere between the middle and the edge. but it is still a circle...
-tom
Michael Masterov
>: >
>: >This statement is technically incorrect. Yes, the `center' shifts downwind
>: >BUT the shape is no longer a circle.
And then a bunch of people attempt to prove this statement wrong. The
problem is in the assumptions. If you fly at best-glide speed regardless
of wind, then the area you can land in is a circle, with only the center
affected by wind - the center is downwind of you.
But if there are winds, the best speed to fly IS NOT best glide - it
depends on the winds. If headed into a strong headwind, you would want
to fly faster than best glide. If you have a strong tailwind, you would
want to fly slower - closer to minimum sink.
The possible landing area is then not a circle. But the center does
move downwind anyway.
Every pilot who routinely flies gliders of any type knows this, but for
some reason I fail to comprehend power-only pilots sometimes don't.
Perhaps poor training from 300 hr CFI's... :)
Michael Masterov, PP-ASEL, AGI
Mike Ciholas
mast...@atom.ecn.purdue.edu (Michael Masterov) writes:
>But if there are winds, the best speed to fly IS NOT best glide -
>it depends on the winds. If headed into a strong headwind, you
>would want to fly faster than best glide. If you have a strong
>tailwind, you would want to fly slower - closer to minimum sink.
>
>The possible landing area is then not a circle. But the center
>does move downwind anyway.
For a thought problem, I tried to reason whether the wind made
the possible landing area larger or smaller if you operated the
aircraft in the best possible manner.
I think the answers is that wind *increases* the available
landing area.
Here is a way to reason about it:
Assume a wind equal to your best glide. If you fly at best glide
in all directions, your area would be a circle. When flying
directly into the head wind, your ground speed would be zero, so
your point of touchdown would be right on the edge of the circle.
Directly downwind flight would take you across the diameter of the
circle, you'd be doing a ground speed of twice your best glide
speed.
But, as we know, when flying into such a headwind, you need to
pick up your speed. If you do that, then you can make some
forward progress into the wind, and thus the area will be
extended in that direction. Likewise, going downwind, you can
slow down to give the tailwind more time to push you further
along. Thus you can go further than the original circle
perimeter. For a flight directly perpendicular to the wind
(heading, not course!), you will still fly at best glide and
still make the same distance over the ground.
So, the ellipse that results is the same "width" (minor axis
perpendicular to the wind) but has a larger "length" (major axis
parallel to the wind). So the area has increased.
The effect at wind less than best glide speed will be less, but
in the same direction. The faster the wind, the more area
available.
>Every pilot who routinely flies gliders of any type knows this,
>but for some reason I fail to comprehend power-only pilots
>sometimes don't. Perhaps poor training from 300 hr CFI's... :)
That was a tasteless joke. I don't think stereotyping power-only
pilots as ignorant due to poor training from low time CFI's is
very endearing. I happen to think part of general aviation's
problem is the lack of comfort and support we give to beginners.
Ignorance is not a sin, neither is inexperience. You were both
at one time.
Mike Ciholas (812) 858-1355 voice
CEDAR Technologies (812) 858-1360 fax
5855 Fiesta Drive mi...@flownet.com
Newburgh, IN 47630 mi...@lcs.mit.edu
PROF D. Rogers {EAS FAC}
!!: !
!!: !This statement is technically incorrect. Yes, the `center' shifts downwind
!!: !BUT the shape is no longer a circle.
!
!And then a bunch of people attempt to prove this statement wrong. The
!problem is in the assumptions. If you fly at best-glide speed regardless
!of wind, then the area you can land in is a circle, with only the center
!affected by wind - the center is downwind of you.
You are absolutely correct, the problem is in the assumptions.
Unfortunately, the large majority of people discussing this statement
are considering an INCOMPLETE problem. Let me restate the problem
so we are ALL discussing the SAME problem.
You are in cruise flight at a heading theta with velocity Vc at
an altitude h. Assume the air mass (wind) is moving at a
velocity Vw from a direction beta. The engine quits instantly.
From the original altitude h determine the shape and size of
the possible landing footprint (area of possible landing).
Hint: it will NOT be circular.
!But if there are winds, the best speed to fly IS NOT best glide - it
!depends on the winds. If headed into a strong headwind, you would want
!to fly faster than best glide. If you have a strong tailwind, you would
!want to fly slower - closer to minimum sink.
!The possible landing area is then not a circle. But the center does
!move downwind anyway.
The wind is NOT the only effect that you must consider.
!Every pilot who routinely flies gliders of any type knows this, but for
!some reason I fail to comprehend power-only pilots sometimes don't.
!Perhaps poor training from 300 hr CFI's... :)
Glider pilots are generally much better at this sort of question
than power-only pilots.
Dave Rogers
Michael Masterov
>The wind is NOT the only effect that you must consider.
OK, I'll bite. Besides wind, and obviously the possible landing areas
and glide performance, what other factors would you consider? The
only gliders I fly regularly are of the ram-air variety, so it's
entirely possible my education, too, has been very incomplete. And
I'm sure I'm not the only one here who would like to read what you
have to say on the topic...
Michael Masterov, PP-ASEL, AGI
John Rourke
writes:
>>>: >
>>: >This statement is technically incorrect. Yes, the `center' shifts downwind
>>: >BUT the shape is no longer a circle.
>
>And then a bunch of people attempt to prove this statement wrong. The
>
OK, I'm one of those low-time power-pilots, so let's see if I get what
your getting at: If headed into a strong headwind, you abandon
'best-glide' speed because it's not going to get you very far, correct?
And if you have a good tailwind, you may opt for min sink rate,
because the tailwind will help you cover distance, but the min sink
rate will keep you aloft the longest, which in turn keeps you in that
wonderful tailwind.
So, it would seem to me that maybe a good engine-out plan would be to
(in the absence of a known alternate airport) orient yourself with the
wind, and go for min sink rate - this maximizes your time aloft, gets
the wind to work for you (covering more possible landing areas) and you
can even think of it as a real long 'downwind' for whatever spot you do
find, allowing you to pass over it before committing to land.
Comments?
>
>The possible landing area is then not a circle. But the center does
>move downwind anyway.
>
>Every pilot who routinely flies gliders of any type knows this, but
for
>some reason I fail to comprehend power-only pilots sometimes don't.
>Perhaps poor training from 300 hr CFI's... :)
>
>Michael Masterov, PP-ASEL, AGI
>
-jpr
Tom Turton
>
> Isn't 1nm=6080 feet instead of 6000 - maybe i'm wrong
1 nautical mile = 6076.115 feet, but
who`s picking nits?
--Tom T.
Mike Ciholas
j...@ix.netcom.com(John Rourke ) writes:
>
>So, it would seem to me that maybe a good engine-out plan would
>be to (in the absence of a known alternate airport) orient
>yourself with the wind, and go for min sink rate - this maximizes
>your time aloft, gets the wind to work for you (covering more
>possible landing areas) and you can even think of it as a real
>long 'downwind' for whatever spot you do find, allowing you to
>pass over it before committing to land.
Sounds like a great plan to me, except for one observation:
Heading downind, you're moving in the direction that will
maximize impact forces, so you only want to do this plan when
favorable terrain is sparse (you need to find better) and you
have altitude (time to search). You'll want to do a 180 for
landing into the wind, and it will be embarassing to fly past
your field, then not make it back into the wind, so plan
carefully!
It is probably better to land into the wind (especially if
strong) on less than ideal terrain, than to land downwind on
better terrain. Of course, this "depends" :-). So at night,
you'd maybe opt for a headwind.
If I'm at 18,000 feet with a (fairly typical) 40 knot wind, I can
cover about 60nm glide distance to sea level (Piper Comanche).
That's a *lot* of terrain to choose from. At a 1000 feet, it is
going to be slim pickings.
PROF D. Rogers {EAS FAC}
!In article <4g96mp$7...@reader2.ix.netcom.com!
! j...@ix.netcom.com(John Rourke ) writes:
!!
!!So, it would seem to me that maybe a good engine-out plan would
!!be to (in the absence of a known alternate airport) orient
!!yourself with the wind, and go for min sink rate - this maximizes
!!your time aloft, gets the wind to work for you (covering more
!!possible landing areas) and you can even think of it as a real
!!long 'downwind' for whatever spot you do find, allowing you to
!!pass over it before committing to land.
!
!Sounds like a great plan to me, except for one observation:
Actually it is a terrible plan if your optimization criteria
is maximum range. For a tailwind equal to 1/3 of your flight
speed you want to decrease your flight velocity by approximately
5-6%, less for lower wind fractions while for a headwind equal
to 1/3 of your flight speed you want to increase you flight
velocity by approximately 10% and less for lower wind fractions.
Reference: Hale, J., "Introduction to Aircraft Performance,
Selection and Design", Wiley, 1984 p 230 fig. 10-3
Dave Rogers
Wwplummer
Rourke ) writes:
>So, it would seem to me that maybe a good engine-out plan would be to
>(in the absence of a known alternate airport) orient yourself with the
>wind, and go for min sink rate - this maximizes your time aloft, gets
No. You really do want the best glide speed. The idea is to maximize the
distance you can travel, not the time in the air. These are _not_ the
same. --Bill
John Rourke
Thanks for the reference, I'll look for it. But in the meantime, how
do you figure your conclusion? Given a 60-knot flight velocity,
20-knot head- or tail- wind, 6000-ft altitude (1 n.m. approx.) and a
glide ratio of 10:1 (somewhat optimistic), here's the figures for a
worst-case for the above plan (starting off directly into the
headwind), using your figures:
(I don't know how increasing flight velocity 10% into a headwind will
affect the glide ration, but it probably isn't much, and would likely
be worse than what you lose with the 5-6% reduction in a tailwind)
"KEEP PUSHIN' ON" "TURN TAIL AND RUN"
Time aloft: 10 min. 10 min.
Time lost to turns: 0 min. 2 min. (std rate)
Time enroute: 10 min. 8 min.
Flt Velocity (Your Ref) 66 kts 57 kts
Ground Speed 46 kts 77 kts
Distance to landing: 7.66 n.m. 10.26 n.m.
Now, as someone mentioned, once you pick a spot to land, you'd better
plan your approach *carefully*, and making turns while low, slow and
with a dead engine is not an ideal situation. OTOH, speaking as a
low-time pilot myself, setting up for a standard approach into a
hayfield as if it were a grass strip may actually be more
're-assuring'.
Also, I was taught to *always* cross over the proposed landing site,
especially off-airport scenarios where there may be ditches, bogs,
animals etc. Of course, one really should not *choose* a landing spot
at the end of one's glide ability, so if you spot one earlier you're
going to spiral over the approach end anyway, whether or not you
started from a head- or tail- wind bearing.
Lastly, this was the worst case for the 'turn tail' plan - if you start
off with a tailwind, or from 12,000 ft, the advantage will be
multiplied.
I'm not saying it is a good plan (I don't know, that's why I'm asking)
- I just don't understand what you think is so terrible about it.
-jpr
Adviser
>In <4g4v67$7...@mozo.cc.purdue.edu> mast...@atom.ecn.purdue.edu (Michael
>Masterov)
>writes:
>>
>>>: d...@opal.usna.navy.mil (PROF D. Rogers {EAS FAC}) wrote:
>>>: >
>>>: >This statement is technically incorrect. Yes, the `center' shifts
>downwind
>>>: >BUT the shape is no longer a circle.
>>
>>And then a bunch of people attempt to prove this statement wrong. The
>>problem is in the assumptions. If you fly at best-glide speed
regardless
>>of wind, then the area you can land in is a circle, with only the center
>>affected by wind - the center is downwind of you.
>>
>>But if there are winds, the best speed to fly IS NOT best glide - it
>>depends on the winds. If headed into a strong headwind, you would want
>>to fly faster than best glide. If you have a strong tailwind, you would
>>want to fly slower - closer to minimum sink.
>
>OK, I'm one of those low-time power-pilots, so let's see if I get what
>your getting at: If headed into a strong headwind, you abandon
>'best-glide' speed because it's not going to get you very far, correct?
> And if you have a good tailwind, you may opt for min sink rate,
>because the tailwind will help you cover distance, but the min sink
>rate will keep you aloft the longest, which in turn keeps you in that
>wonderful tailwind.
>
>So, it would seem to me that maybe a good engine-out plan would be to
>(in the absence of a known alternate airport) orient yourself with the
>wind, and go for min sink rate - this maximizes your time aloft, gets
>the wind to work for you (covering more possible landing areas) and you
>can even think of it as a real long 'downwind' for whatever spot you do
>find, allowing you to pass over it before committing to land.
>
>Comments?
>
>
Unless, of course a suitable forced landing site is in front, or to either
side of you when the engine quit. Did the previous author mean forced
landing site when they said "alternate airport"?
My guess would be that someone that tries to figure out which way the air
is moving (at altitude)(and it will change direction as your altitude
decreases, will it not?), and turn to go with the wind, and establish min.
sink rate, THEN ?? look for a place to land, is probably not going to
have a good outcome.
Whatever happened to establish best glide, look for forced landing site,
front, or to the side--NEVER behind unless you specifically remember just
flying over one (the altutude loss to do a 180 is tremendous), do your
emergency checklist(s), then land.
Just some thoughts-- H.T.
Common sense doesn't seem to be so common anymore---Dexter Yager
John Rourke
>
>In article <4g96mp$7...@reader2.ix.netcom.com>, j...@ix.netcom.com(John
>Rourke ) writes:
>
<snipping my original post>
>>
>>
>Unless, of course a suitable forced landing site is in front, or to either
>side of you when the engine quit.
of course...
>Did the previous author mean forced
>landing site when they said "alternate airport"?
>
I interjected the parenthetical "in the absence of a known alternate airport",
because when I fly cross-country, I always choose a flight path that will take
me within 10-20 n.m. of some kind of airport at all times (if not *too* far out
of the way). I also use this as a way to check on my navigating and observation
skills, to predict where the next airport is (along with scanning for traffic,
of course)
> My guess would be that someone that tries to figure out which way the air
>is moving (at altitude)(and it will change direction as your altitude
>decreases, will it not?), and turn to go with the wind, and establish min.
>sink rate, THEN ?? look for a place to land, is probably not going to
>have a good outcome.
I'm always checking wind direction (it's part of good navigation) although you
have a good point about wind changing at different altitudes.
I don't even know what my min. sink rate speed is on my 172, and I realize now
that that is inexcusable. Once I figure out what speed that is (for my
particular plane) I hope I don't forget it.
Like anything else, the plan I was positing is a possible tool. If using it
precludes looking for a place to land, trying engine restart, etc., then I
wouldn't use it.
>
>Whatever happened to establish best glide, look for forced landing site,
>front, or to the side--NEVER behind unless you specifically remember just
>flying over one (the altutude loss to do a 180 is tremendous), do your
>emergency checklist(s), then land.
>
I NEVER use the words 'always' or 'never', because there's ALWAYS an
exception. ;)
What *is* the altitude loss in a 30-degree banked s.r. turn? Or maybe
20 or 15 would be better? It may take longer to make the turn, but if
you've got lots of altitude, it doesn't matter what direction you go or
how long it takes you to turn to look for a place to put down - you're
covering ground (and possible landing sites) the whole time. It still
seems to me that covering more ground in your search (as long as you
don't fixate on 'looking for a little better spot') is a good thing.
If you are at a low altitude to begin with (like right after take-off)
when the engine fails, I absolutely wholeheartedly agree: look for a
landing site straight ahead or +/- 20 degrees or so to your flight
path. But at 9,500 or 10,500 (my usual cruise altitude, which will
probably increase when I get my Velocity flying) why not use any
advantage you can find?
>Just some thoughts-- H.T.
>
>Common sense doesn't seem to be so common anymore---Dexter Yager
Admittedly, adding a couple more things to remember in an emergency
situation is not always beneficial; that's why I am presisting in
beating this thread to death - once I come to an *intuitive*
understanding of the forces at work and the options they provide, I
won't have to 'remember' anything - it will seem perfectly natural.
Just some thoughts. And thanks to all the good points I've read from
everyone.
-jpr
Jgs3mstr
Turton) writes:
wow !!! been flying over 10 yrs and sailing for 15 yrs and never saw
6076.115 ft/nm. I truely appreciate you setting me straight on this one -
i don't know how i've been able to get by for so long and be wrong all
this time.
jgs3mstr
md88-fo
Anton Verhulst
>your getting at: If headed into a strong headwind, you abandon
>'best-glide' speed because it's not going to get you very far, correct?
> And if you have a good tailwind, you may opt for min sink rate,
>because the tailwind will help you cover distance, but the min sink
>rate will keep you aloft the longest, which in turn keeps you in that
>wonderful tailwind.
Yes, you have it right. The rule of thumb is that, if you're penetrating a
headwind, you add half the speed of the wind to your best glide speed.
So, if your best glide speed is 65kts, you should select 85kts if gliding
into a 40kt headwind. This will give you the best distance over the ground.
Note that the above is a rule of thumb and an approximation (a pretty good
one, IMHO). In case of an engine failure, an approximation is the best you can
hope for because precise calculations are non-trivial. Racing glider pilots
have panel mounted glide computers to help them answer the questions of
how far and how fast they should go. More specific, they want to know how
high they need to thermal at point A to get to point B at altitude C. Extra
time spent climbing reduces your average speed and loses contests.
Tony V. CFI-G
Barry H. Silverman
> > who`s picking nits?
Thought someone out there might like to know that the definition of the
nautical mile is 1852 meters exactly. Easier to remember than
6076.115... The fact that it's close to 6000 feet (along with
having 60 minutes in an hour, and almost 60 degrees in a radian) makes
possible some quick calculations and rules of thumb.
Barry
PROF D. Rogers {EAS FAC}
!In <4gcdbj$k...@newsbf02.news.aol.com! adv...@aol.com (Adviser) writes:
!!
!!In article <4g96mp$7...@reader2.ix.netcom.com!, j...@ix.netcom.com(John
!!Rourke ) writes:
!!
!
!<snipping lots
!I don't even know what my min. sink rate speed is on my 172, and I realize now
!that that is inexcusable. Once I figure out what speed that is (for my
!particular plane) I hope I don't forget it.
Ah... you probably will NOT find it in the POH.
!What *is* the altitude loss in a 30-degree banked s.r. turn? Or maybe
!20 or 15 would be better? It may take longer to make the turn, but if
!you've got lots of altitude, it doesn't matter what direction you go or
!how long it takes you to turn to look for a place to put down - you're
!covering ground (and possible landing sites) the whole time. It still
!seems to me that covering more ground in your search (as long as you
!don't fixate on 'looking for a little better spot') is a good thing.
Minimum altitude loss in a 180 degree turn occurs with a 45 degree
bank at stall velocity.
Dave Rogers
William Joseph Levenson x38205
Wwplummer <wwpl...@aol.com> wrote:
>
>No. You really do want the best glide speed. The idea is to maximize the
>distance you can travel, not the time in the air. These are _not_ the
>same. --Bill
>
If you have a big tail wind, they are virtually the same.
Bill Levenson
PP-ASEL-IA
Terra Corp.
: > > 1 nautical mile = 6076.115 feet, but
: > > who`s picking nits?
Let's confuse it even more. My engineering reference (Eshbach, 1975)
defines a nautical mile as: The length of a minute of longitude of the
earth at the equator at sea level.
Since the geodetic types keep updating the earth measurements, I have a
reference that shows 1 nm to be 6080.27 ft ('75) and another that says it
is 6076.10 ft ('87). I believe the second may correlate to WGS 84.
And then there are the practical matters. Eshbach notes that "The
British Admiralty uses the round figure of 6080 feet."
Maybe the ISO got tired of all this and set it equal to 1852 meters as
Barry said. But then 1852 meters is equal to 6076.41 feet...
I need my morning cup of coffee. 8^)
Gerry
--
Gerry Caron "Opinions are mine, not my employer's."
gca...@rt66.com PH: 800-328-1995 or 505-884-2321
Terra Corp. ABQ FAX: 505-884-2384
Thibaud Taudin-Chabot
"Barry H. Silverman" <bsilv...@admin.tc.faa.gov> wrote:
>> > 1 nautical mile = 6076.115 feet, but
>> > who`s picking nits?
>Thought someone out there might like to know that the definition of the
>nautical mile is 1852 meters exactly. Easier to remember than
>6076.115... The fact that it's close to 6000 feet (along with
>having 60 minutes in an hour, and almost 60 degrees in a radian) makes
>possible some quick calculations and rules of thumb.
>Barry
Alisdair Gurney
Of course, that would make it more than 1852.0 m at altitude, wouldn't it?
Barry H. Silverman
>
> Maybe the ISO got tired of all this and set it equal to 1852 meters as
> Barry said. But then 1852 meters is equal to 6076.41 feet...
You've got to be careful not to use conversion factors that have already
been rounded off (though it doesn't matter for practical purposes, of
course). Using one inch = 2.54 cm (EXACTLY) gives one foot = 0.3048
meters (EXACTLY), and so 1852 m = 6076.11548556... feet (NOT exact).
Barry
Terra Corp.
: In article <4ge359$l...@ixnews3.ix.netcom.com> j...@ix.netcom.com(John
: !Rourke ) writes:
: !I don't even know what my min. sink rate speed is on my 172, and I
: !realize now
: !that that is inexcusable. Once I figure out what speed that is (for my
: !particular plane) I hope I don't forget it.
: Ah... you probably will NOT find it in the POH.
No, but as a practical matter, it will be somewhere near Vx. It will also
vary with gross weight. For a 172, I used 60 knots as a reasonable estimate.
Course you could do a little flight test to be sure. Pick a safe
altitude, say 4000-5000 feet. Go up to 5500 and pull the power to idle.
Trim for 65 kts and start a stopwatch as you go thru 5000 feet. Stop the
watch at 4000 feet. Climb back up and record the time.
Repeat at various speeds. When you are done, calculate your sink rates
and plot them versus a/s. This is your polar.
: Minimum altitude loss in a 180 degree turn occurs with a 45 degree
: bank at stall velocity.
Don't think I'd want to try this at low alt. One good gust and you'd be
another stall-spin stat.
PROF D. Rogers {EAS FAC}
!PROF D. Rogers {EAS FAC} (d...@asgard.usna.navy.mil) wrote:
!: In article <4ge359$l...@ixnews3.ix.netcom.com! j...@ix.netcom.com(John
!: !Rourke ) writes:
!: !I don't even know what my min. sink rate speed is on my 172, and I
!: !realize now
!: !that that is inexcusable. Once I figure out what speed that is (for my
!: !particular plane) I hope I don't forget it.
!
!: Ah... you probably will NOT find it in the POH.
!
!No, but as a practical matter, it will be somewhere near Vx. It will also
!vary with gross weight. For a 172, I used 60 knots as a reasonable estimate.
!
!Course you could do a little flight test to be sure. Pick a safe
!altitude, say 4000-5000 feet. Go up to 5500 and pull the power to idle.
!Trim for 65 kts and start a stopwatch as you go thru 5000 feet. Stop the
!watch at 4000 feet. Climb back up and record the time.
!
!Repeat at various speeds. When you are done, calculate your sink rates
!and plot them versus a/s. This is your polar.
Yes, that will work if done carefully. We've done it many times
with students in a flight test course.
!: Minimum altitude loss in a 180 degree turn occurs with a 45 degree
!: bank at stall velocity.
!Don't think I'd want to try this at low alt. One good gust and you'd be
!another stall-spin stat.
Still does not change the engineering results. Now that you know
the right answer add your own `safety' factor but with the knowledge
you are giving up something.
Dave Rogers
Jeffrey M. Matthews
>
>Whatever happened to establish best glide, look for forced landing site,
>front, or to the side--NEVER behind unless you specifically remember just
>flying over one (the altutude loss to do a 180 is tremendous), do your
>emergency checklist(s), then land.
>
Ahh, but the territory I know best is behind me. I've had a chance to
look at it from varying perspectives, so I may have seen flashes of
water reflections, fences, surface condition, and many other features
of interest.
The altitude lost in a 180 degree turn isn't really that bad, using Prof.
Dave's 45 degree banked turn. From a cruise speed you should even be able
to gain a bit in slowing to gliding speed. You've got to be willing and
able to lay the airplane over a bit, and you've got to know what to do
with the nose. To give a back-of-the-envelope idea, at 65 mph in a 45
degree bank, the turning radius is 282 ft. The flight path is 887 ft.
for 180 degrees, and will take a bit under ten seconds. If the L/D is
10:1, the sink rate in a 45 degree bank will be about 800 fpm and we'll
lose about 135 feet in the turn. Add a bit for rolling in and out, and
subtract a bit for converting excess speed if you want, but we'll be in
the ballpark of less than a couple hundred feet.
Many pilots have never experienced a true power-off glide, and have only
a sketchy idea of how far their plane will go--not in a textbook sense,
but in a "can I make it to yonder field?" sense. Most of these pilots
will overestimate their gliding range somewhat. Even an idling engine
produces some thrust--depending on one's reference frame that can be con-
sidered thrust or decreased drag, but either way, it's there. I've flown
with some instructors who even _added_ a bit of power to get to "zero
thrust"--a technique valid only to duplicate a feathered prop, not to
simulate a windmilling engine.
Given the choice, I'd have to be quite low, or quite sure of what lay
ahead of me to continue on course into the wind with a dead engine.
As for the danger of stalling or spinning out of a 45 degree banked turn
trying to get lined up, either downwind initially, or into the wind for
landing, I think that's down some on the list of things to worry about.
A pilot should be able to make such a turn in fairly turbulent air with-
out stalling. And if a pilot can't induce a stall banked 45 degrees and
recover without losing altitude or control, he is either a marginal pilot
or flying a very feisty airplane.
Jeff Matthews
Terra Corp.
: >
: > Maybe the ISO got tired of all this and set it equal to 1852 meters as
: > Barry said. But then 1852 meters is equal to 6076.41 feet...
: You've got to be careful not to use conversion factors that have already
: been rounded off (though it doesn't matter for practical purposes, of
: course). Using one inch = 2.54 cm (EXACTLY) gives one foot = 0.3048
: meters (EXACTLY), and so 1852 m = 6076.11548556... feet (NOT exact).
Got me. Yes, I used the meters to feet factor which does have a rounding
error in it.
Remember: Measure with a micrometer, mark with chalk, cut with an axe.
Guess I'll go back to the TLAR method. 8^)
Benjamin Setnick
a minute of longitude at the equator.
Ben Setnick
Tom Turton
> 1852.0 m exactly it is.
> "Barry H. Silverman" <bsilv...@admin.tc.faa.gov> wrote:
>
> TT>> > who`s picking nits?
>
> >Thought someone out there might like to know that the definition of the
> >nautical mile is 1852 meters exactly. Easier to remember than
> >6076.115... The fact that it's close to 6000 feet (along with
> >having 60 minutes in an hour, and almost 60 degrees in a radian) makes
> >possible some quick calculations and rules of thumb.
>
Man, you guys just can't let a thread die (and look who is calling
the kettle black!).
Someone asked if naut. mi to feet conversion was 6080 or what. I
posted the 6076.115 for all the techno-weenies (such as myself) who
LIKE precision, while realizing the extra precision isn't always needed.
Now we have people posting 1852 meters per nautical mile (as if THAT'S going
to do anyone any good who wants to go between ft and n.mi!).
For purposes of day-to-day navigating, yes, I agree that using 6000 ft per
naut mile is perfectly acceptable (on the order of 1% accuracy).
---Tom (just havin' fun with numbers, guys - let's give it a rest) Turton
Roy Smith
> And then there are the practical matters.
As a practical matter, I generally don't worry about measuring my position
to a precision smaller than the vehicle I'm in :-)
--
Roy Smith <r...@nyu.edu>
Hippocrates Project, Department of Microbiology, Coles 202
NYU School of Medicine, 550 First Avenue, New York, NY 10016
"This never happened to Bart Simpson."
Pete Brown
TC>Repeat at various speeds. When you are done, calculate your sink rates
TC>and plot them versus a/s. This is your polar.
TC>: Minimum altitude loss in a 180 degree turn occurs with a 45 degree
TC>: bank at stall velocity.
Gerry:
I agree that min altitude loss in a turn will occur at approximately a
45 degree bank..but I am not sure that the airspeed for such a minimum
altitude loss turn would be at stall velocity, regardless of how smart
it would or would not be.
Minimum loss of alitude in a given period occurs at the speed for
minimum sink. In level flight, this will be somewhat above stall
velocity, probably near 1.2x stall velocity. (Glider pilots would know
this from the polar of their wing, 172 drivers would have to calculate
it as you suggest.)
What would be different in turning flight except that stall and m/s
would increase with load factor? In any case, they would not be the
same. A glider thermalling in lift is confronted with a similar problem,
ie to fly at minimum sink for a given angle of bank. For gliders the
margin between min sink and stall can be so close that they are almost
but not quite the same.
Some gliders are flown in lift slowly until you feel the burble and
then just a bit of speed is added get to min sink (and whatever
additional wife and kids factor the pilot throws in). Others are flown
right at the burble if the pilot believes that the trade off for the
higher sink rate at the stall is compensated by the ability to make a
smaller radius turn to keep the glider in the core of the stronger
lift....but I digress.... this would be irrelevant to the power pilot
making a 180.
He should make a 45 degree turn at what probably will be about 1.2X min
stall at a 45 degree bank, whatever that would be.. That would be a lot
to calculate in a hurry.
The short, quick response is to execute a quick, coordinated (you power
pilots do know what a rudder's for don't you?) 45 degree turn with the
nose well below the horizon. The 45 degrees gets you turned with minimum
altitude loss and the nose below the horizon will keep your speed up to
avoid the stall/spin.
Pete Brown
---
* OLX 2.1 TD * Al Gore: Roadkill on the Information Highway
CEP
>>
>> Maybe the ISO got tired of all this and set it equal to 1852 meters as
>> Barry said. But then 1852 meters is equal to 6076.41 feet...
>
>You've got to be careful not to use conversion factors that have already
>been rounded off (though it doesn't matter for practical purposes, of
>course). Using one inch = 2.54 cm (EXACTLY) gives one foot = 0.3048
>meters (EXACTLY), and so 1852 m = 6076.11548556... feet (NOT exact).
>
>
I though a nautical mile was derived as the distance
travelled equal to one minute of longitude.
Broke, But No Longer Renting!
N4566F
Louis A. Ramsay
(Benjamin Setnick) writes:
>
>I thought it was one minute of lattitude. I guess that's the same as
>a minute of longitude at the equator.
>
That's the ONLY place it will be true.
Julian Scarfe
<bsilv...@admin.tc.faa.gov> wrote:
> Thought someone out there might like to know that the definition of the
> nautical mile is 1852 meters exactly. Easier to remember than
> 6076.115... The fact that it's close to 6000 feet (along with
> having 60 minutes in an hour, and almost 60 degrees in a radian) makes
> possible some quick calculations and rules of thumb.
There are a number of different definitions of the nautical mile,
depending on which authority you consult. Becuase the Earth is not a
prefect sphere, the length of a 1 minute arc depends on where you measure
it. I think the British admiralty used to use 1 minute at 52N, the
"tropical nautical mile" is measured at the equator, someone else measures
it at 45N... They're all slightly different.
Is the 1852 metres the ICAO definition then (I never managed to find that)?
Julian Scarfe
ja...@cus.cam.ac.uk
Julian Scarfe
wrote:
> I agree that min altitude loss in a turn will occur at approximately a
> 45 degree bank..but I am not sure that the airspeed for such a minimum
> altitude loss turn would be at stall velocity, regardless of how smart
> it would or would not be.
>
...
>
> What would be different in turning flight except that stall and m/s
> would increase with load factor? In any case, they would not be the
> same. A glider thermalling in lift is confronted with a similar problem,
> ie to fly at minimum sink for a given angle of bank. For gliders the
> margin between min sink and stall can be so close that they are almost
> but not quite the same.
>
You're missing something in the model, I think. For minimum altitude loss
in a 180 degree turn, you need the minimum altitude loss per unit *angle
turned* not per unit *time*. So if you know the rate of descent as a
function of speed, you must divide by the rate of turn as a function of
speed before you do the minimization. The rate of turn depends on 1/v^2,
so it turns out that there's no minimum with respect to airspeed, you just
fly as slowly as you can to get round 180 degrees (or any other angle)
with least altitude loss.
Hope that helps,
Julian Scarfe
ja...@cus.cam.ac.uk
Rhea Wood
>> > 1 nautical mile = 6076.115 feet, but
>> > who`s picking nits?
>Thought someone out there might like to know that the definition of the
>nautical mile is 1852 meters exactly. Easier to remember than
>6076.115... The fact that it's close to 6000 feet (along with
>having 60 minutes in an hour, and almost 60 degrees in a radian) makes
>possible some quick calculations and rules of thumb.
>Barry
Reading all of this nautical mile stuff led me to dig out my old copy
of Nathaniel Bowditch's "Piloting", (in the nautical sense,) and HE
says:
The "nautical mile" or the "international nautical mile" has been
established at "1852 meters exactly." This was done in 1929 by the
International Hydrographic Bureau and has since been adopted by most
maritime nations. Then, on July 1, 1959, the United States adopted
the exact relationship of 1 yard = 0.9144 meter. The length of the
international nautical mile is consequently equal to 6,076.11549 U. S.
feet (approximately.) (Damn! there’s that word approximately again!)
As for the other points in this thread here's some more Bowditch
wisdom:
For most navigational purposes the nautical mile is considered the
length of one minute of latitude, or of any great circle of the earth,
regardless of location. On the Clarke shpheroid of 1866, used for
mapping North America, the length of one minute of latitude varies
from about 6,046 feet at the equator to approximately 6,108 at the
poles. The length of one minute of a great circle of a sphere having
an area equal to that of the earth, as represented by this spheroid,
is 6,080.2 U. S. feet. This was the standard value of the nautical
mile prior to adoption of the international value. A "geographical
mile" is the length of one minute of the equator, or about 6.087 feet.
Any Flat Earth believers out there? I don't always believe what I
read but Bowditch is considered a reliable reference by those that ply
the sea in ships. Besides, I saw the earth at 45,000 feet in a Lear
25 and it IS round!
Rhea Wood
N3489Y C-185
Alaska-Based Floatplane
Rhea Wood
<clip>
>I though a nautical mile was derived as the distance
>travelled equal to one minute of longitude.
Don't you mean one minute of Latitude ALONG the great circle known as
a line of Longitude?
>Broke, But No Longer Renting!
>N4566F
PROF D. Rogers {EAS FAC}
!TC!Repeat at various speeds. When you are done, calculate your sink rates
!TC!and plot them versus a/s. This is your polar.
!
!dfr!: Minimum altitude loss in a 180 degree turn occurs with a 45 degree
!dfr!: bank at stall velocity.
This is my statement.
Reference: Rogers, David F. , "The Possible `Impossible' Turn,
AIAA (American Institue of Aeronautics and Astronautics),
Vol. 32, No. 1, pp 392-397, March-April 1995
!Gerry:
!
!I agree that min altitude loss in a turn will occur at approximately a
!45 degree bank..but I am not sure that the airspeed for such a minimum
!altitude loss turn would be at stall velocity, regardless of how smart
!it would or would not be.
The minimum altitude loss occurs at stall velocity. See the above
reference. What you are failing to consider is the TIME it takes
you to make the turn.
!Minimum loss of alitude in a given period occurs at the speed for
!minimum sink. In level flight, this will be somewhat above stall
!velocity, probably near 1.2x stall velocity.
The velocity for minimum sink is easily determined from the aircraft
parameters. In general, it has little to do with the stall velocity.
!(Glider pilots would know
!this from the polar of their wing, 172 drivers would have to calculate
!it as you suggest.)
!What would be different in turning flight except that stall and m/s
!would increase with load factor? In any case, they would not be the
!same. A glider thermalling in lift is confronted with a similar problem,
!ie to fly at minimum sink for a given angle of bank. For gliders the
!margin between min sink and stall can be so close that they are almost
!but not quite the same.
!Some gliders are flown in lift slowly until you feel the burble and
!then just a bit of speed is added get to min sink (and whatever
!additional wife and kids factor the pilot throws in). Others are flown
!right at the burble if the pilot believes that the trade off for the
!higher sink rate at the stall is compensated by the ability to make a
!smaller radius turn to keep the glider in the core of the stronger
!lift....but I digress.... this would be irrelevant to the power pilot
!making a 180.
!He should make a 45 degree turn at what probably will be about 1.2X min
!stall at a 45 degree bank, whatever that would be.. That would be a lot
!to calculate in a hurry.
Nope, as close to stall velocity as you dare.
!The short, quick response is to execute a quick, coordinated (you power
!pilots do know what a rudder's for don't you?) 45 degree turn with the
!nose well below the horizon. The 45 degrees gets you turned with minimum
!altitude loss and the nose below the horizon will keep your speed up to
!avoid the stall/spin.
Keeping the speed up will significantly INCREASE your altitude loss
even in a 45 degree bank. See the reference.
When I do these things I increase the angle of attack until the
stall warning just goes on then relax the back pressure until
it goes off, increase the back pressure until it comes on
again, etc. For most aircraft at gross weight this gives you
a 5-7 mph margin over actual stall. If you are under gross weight,
then margin is higher. In this case the additional altitude loss
in executing the turn with a 45 degree bank is minimal.
Dave Rogers
PROF D. Rogers {EAS FAC}
>In article <96022317...@proom.com>, pete....@proom.com (Pete Brown)
!wrote:
!
!! I agree that min altitude loss in a turn will occur at approximately a
!! 45 degree bank..but I am not sure that the airspeed for such a minimum
!! altitude loss turn would be at stall velocity, regardless of how smart
!! it would or would not be.
!!
!...
!!
!! What would be different in turning flight except that stall and m/s
!! would increase with load factor? In any case, they would not be the
!! same. A glider thermalling in lift is confronted with a similar problem,
!! ie to fly at minimum sink for a given angle of bank. For gliders the
!! margin between min sink and stall can be so close that they are almost
!! but not quite the same.
!!
!
!You're missing something in the model, I think. For minimum altitude loss
!in a 180 degree turn, you need the minimum altitude loss per unit *angle
!turned* not per unit *time*. So if you know the rate of descent as a
!function of speed, you must divide by the rate of turn as a function of
!speed before you do the minimization. The rate of turn depends on 1/v^2,
!so it turns out that there's no minimum with respect to airspeed, you just
!fly as slowly as you can to get round 180 degrees (or any other angle)
!with least altitude loss.
Yes, but in fact it is a bit more complex than that. In effect, what
you want is the minimum altitude loss per unit angle turned per unit time.
However, the time gets folded into the analysis.
In fact, the equation is
8*(drag coeff)*(weight)
d(altit)/d(angle) = --------------------------------------------------
((lift coeff)^2)*density*wing area*g*sin(2*bankangle)
Dave Rogers
John Rourke
Matthews) writes:
>
>In article <4gcdbj$k...@newsbf02.news.aol.com>, Adviser <adv...@aol.com>
wrote:
>>
>>Whatever happened to establish best glide, look for forced landing site,
>>front, or to the side--NEVER behind unless you specifically remember just
>>flying over one (the altutude loss to do a 180 is tremendous), do your
>>emergency checklist(s), then land.
>>
>
>Ahh, but the territory I know best is behind me. I've had a chance to
>look at it from varying perspectives, so I may have seen flashes of
>water reflections, fences, surface condition, and many other features
>of interest.
My point, as well.
>
>The altitude lost in a 180 degree turn isn't really that bad, using Prof.
>Dave's 45 degree banked turn. From a cruise speed you should even be able
>to gain a bit in slowing to gliding speed. You've got to be willing and
>able to lay the airplane over a bit, and you've got to know what to do
>with the nose. To give a back-of-the-envelope idea, at 65 mph in a 45
>degree bank, the turning radius is 282 ft. The flight path is 887 ft.
>for 180 degrees, and will take a bit under ten seconds. If the L/D is
>10:1, the sink rate in a 45 degree bank will be about 800 fpm and we'll
>lose about 135 feet in the turn. Add a bit for rolling in and out, and
>subtract a bit for converting excess speed if you want, but we'll be in
>the ballpark of less than a couple hundred feet.
At first I was going to thank you for providing actual data, but I started
running a few figures myself (actually on the back of an envelope, in fact!).
Let me start off by claiming little expertise in this area (I have been known
to use faulty logic/math before <g>) - but working backwards from a
standard-rate turn of 1 minute/180 degrees at a bank angle of 30 degrees, at 65
mph that would be 5720 fpm for a turn of 1820 ft radius, correct? Will
tightening up the turn from 30 degrees to 45 degrees really change it from a
1,820-ft radius to the 282 ft you described?
In any event, even if you use a s.r. turn, taking 1 minute to do a 30-degree
turn at a load factor of 1.5 (I believe), I'm guessing your sink rate would
change from 572 fpm to 858 fpm, or an excess of 286 ft lost for the minute you
were at the higher load factor. Making a 45-degree bank will increase your
load factor by another 20.7%, or 114 fpm (again, just a guess), but I suspect
the increased drag at that "weight" will take away even more altitude than is
saved by making the turn shorter duration. Why not just 15 or 20 degree bank?
Your turn will be a lot wider and take a lot longer, but the extra load factor
is practically nil, so you waste very little altitude, and you'll still roll
out within a couple miles of your original flight path (assuming you were
headed directly into the wind to begin with, and you make a complete 180).
In any event, I think you're right in stating the extra altitude lost due to
the turn is not "tremendous" and should be considered as another tool you can
use in an engine-out situation.
>
>Many pilots have never experienced a true power-off glide, and have only
>a sketchy idea of how far their plane will go--not in a textbook sense,
>but in a "can I make it to yonder field?" sense.
BTW, does anyone recommend true power-off practice? For instance, on a 172,
will the engine keep windmilling? Can that flood the engine, create backfiring
on restart, etc? I've done it in a motorglider, but there it's a
common practice (by definition!)
> Most of these pilots
>will overestimate their gliding range somewhat. Even an idling engine
>produces some thrust--depending on one's reference frame that can be
con-
>sidered thrust or decreased drag, but either way, it's there. I've
flown
>with some instructors who even _added_ a bit of power to get to "zero
>thrust"--a technique valid only to duplicate a feathered prop, not to
>simulate a windmilling engine.
>
>Given the choice, I'd have to be quite low, or quite sure of what lay
>ahead of me to continue on course into the wind with a dead engine.
Good point.
>
>As for the danger of stalling or spinning out of a 45 degree banked
turn
>trying to get lined up, either downwind initially, or into the wind
for
>landing, I think that's down some on the list of things to worry
about.
>A pilot should be able to make such a turn in fairly turbulent air
with-
>out stalling. And if a pilot can't induce a stall banked 45 degrees
and
>recover without losing altitude or control, he is either a marginal
pilot
>or flying a very feisty airplane.
>
Sounds reasonable to me.
-
Wwplummer
Joseph Levenson x38205) writes:
No. The speed to fly for min sink and for max glide are characteristics
of the airframe itself; both can easily be picked off the polar for the
airframe in question. Tom Knauff's _Transition_ book has a good
discussion of this. [Bill Levinson: send me your fax number if you want a
page or two on this -- Tom is much better at explaining it!]
--Bill
_________________________________________________________________________
Wm W. Plummer, 7 Country Club D., Chelmsford MA 01824 508-256-9570
PP-ASEL,G
Louis A. Ramsay
>
>
>Any Flat Earth believers out there? I don't always believe what I
>read but Bowditch is considered a reliable reference by those that ply
>the sea in ships. Besides, I saw the earth at 45,000 feet in a Lear
>25 and it IS round!
>
>
considering he left grammar school at quite a young age. He even found
a few errors in Newton's theories but was too awed by the man to put
forward his own conclusions.
And yes, the majority of his formulae work for aviation as well as
nautical applications.
VincentLee
>>>front, or to the side--NEVER behind unless you specifically remember just
>>>flying over one (the altutude loss to do a 180 is tremendous), do your
>>>emergency checklist(s), then land.
>>>
>>Ahh, but the territory I know best is behind me. I've had a chance to
>>look at it from varying perspectives, so I may have seen flashes of
>>water reflections, fences, surface condition, and many other features
>>of interest.
Yeah, there is no point going for distance unless you are trying to
reach a pre-selected location. What the pilot really needs is TIME to
spot the best field, which is just as likely to be right underneath or
behind than ahead.
On my Private Pilot flight test, when the Examiner did the simulated
engine out, I frantically scanned the far horizons. I passed the
test, but he said something like, "well I think you could have got in
there.... but that ag strip with the windsock that was right below us
would have been a better choice".
My s.o.p. now would be to calm down and circle in an engine out,
looking in all directions and right below. I believe if I continued
straight I would tend to look straight ahead, plus could not see well
to the right and down. Plus if I went for a far ahead field, I would
only get one look at it; spiralling down at least I get to see where I
am going the whole way. (Of course I'd straighten out at the end.)
Just my $.02
-Vince
Neil Fraser
>>Thought someone out there might like to know that the definition of the
>>nautical mile is 1852 meters exactly. Easier to remember than
>>6076.115... The fact that it's close to 6000 feet (along with
>>having 60 minutes in an hour, and almost 60 degrees in a radian) makes
>>possible some quick calculations and rules of thumb.
FWIW, here's a note I made some time ago about the difference between the US and UK
definitions of the NM:
The US nautical mile = 6076.1 feet; the UK nautical mile = 6080 ft.
Because the UK nautical mile is longer than the US one, the
conversion to kilometres and statute miles will be different,
depending upon which nautical mile you decide to use.
There are supposed to be 60 NM between each line of latitude going
N/S or between each line of longitude going E/W along the equator.
In other words, the circumference of the earth is 60 * 360 = 21600
NM. In reality, the earth is slightly squashed, so that the distance
round the equator > 21600 NM, and round the poles < 21600 NM. The
average is 40,006 km which makes the US NM slightly nearer the mark.
Neil A Fraser
TEL: 27 (11) 468 2892
FAX: 27 (11) 468 2895
** There's more fun in getting there
than being there **
Rhea Wood
>
>Oh dear, and I thought a nautical mile was one minute of arc (at the
equator).
>Of course, that would make it more than 1852.0 m at altitude, wouldn't it?
>
>
>alis...@agurney.demon.co.uk
Yup it would.....cept you don't count miles up there, you count them as you
pass over the ground...........damn! that means KIAs needs to be calibrated
based on the distance above the Clarke spheriod of 1866!!! I wonder will my
new GPS have that built in?
Rhea Wood
Jeffrey M. Matthews
Roy Smith <r...@popmail.med.nyu.edu> wrote:
>gca...@mack.rt66.com (Terra Corp.) wrote:
>> And then there are the practical matters.
>
>As a practical matter, I generally don't worry about measuring my position
>to a precision smaller than the vehicle I'm in :-)
>
Bet you're not very popular in parking lots.
Jeff Matthews
who has found that three significant digits are sufficient for most
everyting except balancing his checkbook, which sometimes takes four.
John Rourke
{EAS FAC}) writes:
>In article <4g4v67$7...@mozo.cc.purdue.edu> mast...@atom.ecn.purdue.edu
(Michael Masterov) writes:
>!!: d...@opal.usna.navy.mil (PROF D. Rogers {EAS FAC}) wrote:
>!!: !
>!!: !This statement is technically incorrect. Yes, the `center' shifts downwind
>!!: !BUT the shape is no longer a circle.
>!
>!And then a bunch of people attempt to prove this statement wrong. The
>!problem is in the assumptions. If you fly at best-glide speed regardless
>!of wind, then the area you can land in is a circle, with only the center
>!affected by wind - the center is downwind of you.
>
>You are absolutely correct, the problem is in the assumptions.
>Unfortunately, the large majority of people discussing this statement
>are considering an INCOMPLETE problem. Let me restate the problem
>so we are ALL discussing the SAME problem.
>
>You are in cruise flight at a heading theta with velocity Vc at
>an altitude h. Assume the air mass (wind) is moving at a
>velocity Vw from a direction beta. The engine quits instantly.
>From the original altitude h determine the shape and size of
>the possible landing footprint (area of possible landing).
>Hint: it will NOT be circular.
>
I intuitively thought it would not be circular if you have any wind,
using best-glide speed only, and I am now convinced I was wrong. - It
is a circle in *any* wind, if you only use best-glide speed. Does your
hint assume flight speed modification as per the following
recommendations? If not, and there is some other factor, please
elaborate - I'm one of those power-pilots that needs more than a hint!
(Although I am looking forward to getting into gliders!)
>!But if there are winds, the best speed to fly IS NOT best glide - it
>!depends on the winds. If headed into a strong headwind, you would
want
>!to fly faster than best glide. If you have a strong tailwind, you
would
>!want to fly slower - closer to minimum sink.
>
>!The possible landing area is then not a circle. But the center does
>!move downwind anyway.
>
>The wind is NOT the only effect that you must consider.
For this post to be valuable to me, you have to say what I *do* need to
consider, not what I *don't*. Or are you simply agreeing with the
statement you are commenting on?
-jpr
(In other words, I thought I had it pretty straight from Michael
Masterov - or are you saying there's more?)
William Joseph Levenson x38205
Wwplummer <wwpl...@aol.com> wrote:
>In article <4gi8d3$a...@tel.den.mmc.com>, ss8...@hp26.den.mmc.com (William
>Joseph Levenson x38205) writes:
>
>>In article <4gbsmg$h...@newsbf02.news.aol.com>,
>>Wwplummer <wwpl...@aol.com> wrote:
>>>No. You really do want the best glide speed. The idea is to maximize
>the
>>>distance you can travel, not the time in the air. These are _not_ the
>>>same. --Bill
>
>>If you have a big tail wind, they are virtually the same.
>
>No. The speed to fly for min sink and for max glide are characteristics
>of the airframe itself; both can easily be picked off the polar for the
>airframe in question. Tom Knauff's _Transition_ book has a good
>discussion of this. [Bill Levinson: send me your fax number if you want a
>page or two on this -- Tom is much better at explaining it!]
>--Bill
C'mon, Bill (Plummer, that is ;-)
Think! Yes, both airspeeds can easily be determined from the polar (I assume
you mean a plot of vertical speed vs. airspeed ). But this method only applies
to a no wind situation. What is needed to consider wind is to plot
vertical speed vs. *ground*speed*. If you have a tail wind, this will
push the entire curve to the right. You need to plot vs. ground speed
because you are interested in glide distance relative to the ground.
When you plot this up, the min sink speed will be the highest point on
the curve. The max glide will the the min glide angle which is determined
graphically by drawing a tangent line from the origin to the curve. The
point of tangency is the speed you want to fly for best glide.
If you start plotting with zero wind, as I said, you will get the max glide
speed that you are refering to. As you pick up a tail wind, the curve
shifts right and the point of tangency of the line from the origin moves
closer and closer to the min sink speed.
The argument that max glide is dependent only on airframe can be seen
if you consider the following scenario: A plane w/ a max glide speed
of 70 knots is heading into a 70 knot headwind. If the pilot maintains
70 knots he will glide exactly *zero* feet! But if he increases his
airspeed, he will descend faster but will glide farther.
The bottom line is if you have a tailwind helping you, it benefits you
to sacrifice some distance through the air for time in the air because
the more time you spend in the air, the more the tailwind helps you relative
to the ground.
Bill Levenson
[BTW, Bill Plummer, you don't need to fax me...I am familiar with most
of the texts on the subject...it's sorta my job ;-) ]
Wwplummer
Joseph Levenson x38205) writes:
>When you plot this up, the min sink speed will be the highest point on
>the curve. The max glide will the the min glide angle which is
determined
>graphically by drawing a tangent line from the origin to the curve. The
>point of tangency is the speed you want to fly for best glide.
Agree.
>If you start plotting with zero wind, as I said, you will get the max
glide
>speed that you are refering to. As you pick up a tail wind, the curve
>shifts right and the point of tangency of the line from the origin moves
>closer and closer to the min sink speed.
Agree.
>The argument that max glide is dependent only on airframe can be seen
>if you consider the following scenario: A plane w/ a max glide speed
>of 70 knots is heading into a 70 knot headwind. If the pilot maintains
>70 knots he will glide exactly *zero* feet! But if he increases his
>airspeed, he will descend faster but will glide farther.
Agree.
>The bottom line is if you have a tailwind helping you, it benefits you
>to sacrifice some distance through the air for time in the air because
>the more time you spend in the air, the more the tailwind helps you
relative
>to the ground.
Agree.
But it is still true that the polar is peculiar to a specific airframe. I
agree that the effect of wind is to move it left or right; updrafts and
downdrafts move it up and down. BUT,
the min sink is the speed at the peak and max glide is the speed where the
tangent through 0,0 strikes. True, wind can make them become numerically
the same.
>[BTW, Bill Plummer, you don't need to fax me...I am familiar with most
>of the texts on the subject...it's sorta my job ;-) ]
You're a luck man, Bill Levenson. I have to work for a living! --Bill
Jeffrey M. Matthews
John Rourke <j...@ix.netcom.com> wrote:
>In <1996Feb22....@apgea.army.mil> jmma...@apgea.army.mil (Jeffrey M.
>Matthews) writes:
>
>>
>>The altitude lost in a 180 degree turn isn't really that bad, using Prof.
>>Dave's 45 degree banked turn. From a cruise speed you should even be able
>>to gain a bit in slowing to gliding speed. You've got to be willing and
>>able to lay the airplane over a bit, and you've got to know what to do
>>with the nose. To give a back-of-the-envelope idea, at 65 mph in a 45
>>degree bank, the turning radius is 282 ft. The flight path is 887 ft.
>>for 180 degrees, and will take a bit under ten seconds. If the L/D is
>>10:1, the sink rate in a 45 degree bank will be about 800 fpm and we'll
>>lose about 135 feet in the turn. Add a bit for rolling in and out, and
>>subtract a bit for converting excess speed if you want, but we'll be in
>>the ballpark of less than a couple hundred feet.
>
>At first I was going to thank you for providing actual data, but I started
>running a few figures myself (actually on the back of an envelope, in fact!).
>Let me start off by claiming little expertise in this area (I have been known
>to use faulty logic/math before <g>) - but working backwards from a
>standard-rate turn of 1 minute/180 degrees at a bank angle of 30 degrees, at 65
>mph that would be 5720 fpm for a turn of 1820 ft radius, correct? Will
>tightening up the turn from 30 degrees to 45 degrees really change it from a
>1,820-ft radius to the 282 ft you described?
No, a 30 degree banked turn at 65 mph will have a radius of about 489 feet.
A standard rate turn at 65 mph will have a bank angle of a bit less than
9 degrees. The quick and dirty way to these numbers is\
speed squared
R = ----------------
g tan (bank angle)
where g is the acceleration due to gravity (32.2 ft/sec^2), and speed is in
feet per second. Radius will be in feet.
To go from a standard rate turn, multiply the speed, still in ft/sec, by
120 seconds, then divide by 2pi, to get a radius of 1821 ft, then put that
value in the equation above and solve for tan(bank angle).
>
>In any event, even if you use a s.r. turn, taking 1 minute to do a 30-degree
>turn at a load factor of 1.5 (I believe), I'm guessing your sink rate would
>change from 572 fpm to 858 fpm, or an excess of 286 ft lost for the minute you
>were at the higher load factor. Making a 45-degree bank will increase your
>load factor by another 20.7%, or 114 fpm (again, just a guess), but I suspect
>the increased drag at that "weight" will take away even more altitude than is
>saved by making the turn shorter duration. Why not just 15 or 20 degree bank?
>Your turn will be a lot wider and take a lot longer, but the extra load factor
>is practically nil, so you waste very little altitude, and you'll still roll
>out within a couple miles of your original flight path (assuming you were
>headed directly into the wind to begin with, and you make a complete 180).
Not quite. Your 30 degree bank will result in a load factor of 1.15, and
a 45 degree bank will give 1.41. The load factor, usually denoted "n",
can be quickly found by:
1
n = -----------------
cos (bank angle)
Perhaps Prof Rogers will post you his article on the possible impossible
turn to demonstrate mathematically that a 45 degree bank gives the least
altitude loss. The formula, should you want to play with it, is
h = 2 x pi x r sin(L/D)cos(L/D)
L/D and turn radius are interrelated, so the solution requires a bit of
calculus or many trial and error calculations.
I've done any number of practice rope breaks with glider students, and
one for real on takeoff, and find that a student ready to solo, in a
training glider, can easily handle a 180 from 150-200 feet, but it works
best right near a 45 degree bank. That's a comfortable turn, not one with
a wingtip brushing the trees. There's plenty of height after rolling out
to set up a normal landing.
Jeff Matthews
William Joseph Levenson x38205
John Rourke <j...@ix.netcom.com> wrote:
>In <4g8b9o$6...@asgard.usna.navy.mil> d...@asgard.usna.navy.mil (PROF D. Rogers
>{EAS FAC}) writes:
>>
>>!The possible landing area is then not a circle. But the center does
>>!move downwind anyway.
>>
>>The wind is NOT the only effect that you must consider.
>
>For this post to be valuable to me, you have to say what I *do* need to
>consider, not what I *don't*. Or are you simply agreeing with the
>statement you are commenting on?
>
>-jpr
>
>(In other words, I thought I had it pretty straight from Michael
>Masterov - or are you saying there's more?)
I'd say that you also must consider the fact that you can't glide as far
if you were to head to a point behind you as if you head for a point
ahead of you. This is because, in order to glide to a point behind you,
you must first turn around, thus losing some altitude. So, even with
no wind, the shape of the area reachable after the engine quits is still
not a circle.
Bill Levenson
PP-ASEL-IA
patterson,george r
>Rourke ) writes:
>
>>So, it would seem to me that maybe a good engine-out plan would be to
>>(in the absence of a known alternate airport) orient yourself with the
>>wind, and go for min sink rate - this maximizes your time aloft, gets
>
>No. You really do want the best glide speed. The idea is to maximize the
>distance you can travel, not the time in the air. These are _not_ the
>same. --Bill
Well, they're not the same, but you do not necessarily want best glide
speed. Fly the plane. If the best landing spot is close by, set up for
the maximum time in the air. This gives you more time for things like
Mayday calls or restart attempts. You should only use best glide speed
if you actually need that distance endurance.
-----------------------------------------------------------------------
|
George Patterson - | In order to get a loan, you must first prove that
| you don't need it.
|
-----------------------------------------------------------------------
John Rourke
x38205) writes:
>
>In article <4gtanf$5...@reader2.ix.netcom.com>,
>John Rourke <j...@ix.netcom.com> wrote:
<snipping some other stuff about the landing area not being a` circle>
>>
>>(In other words, I thought I had it pretty straight from Michael
>>Masterov - or are you saying there's more?)
>
>I'd say that you also must consider the fact that you can't glide as far
>if you were to head to a point behind you as if you head for a point
>ahead of you. This is because, in order to glide to a point behind you,
>you must first turn around, thus losing some altitude. So, even with
>no wind, the shape of the area reachable after the engine quits is still
>not a circle.
>
>Bill Levenson
>PP-ASEL-IA
>
While that is true, the fact is you are still overflying possible
landing areas, even in the turn, as large or small as that might be.
Plus, since my questions were mostly related to 'what to do when
starting from a headwind', and since it is inadvisable to land straight
ahead into unknown and unseen territory (unless you are already real
low, like on take-off), if you continue flying into the headwind and
find what looks like a possible landing area, you would be overflying
it on upwind, do a 180 to downwind then another 180 to approach anyway.
In terms of procedures, it is probably safest to do your 180 to
downwind immediately, then all you have to do is one more 180 (or a
spiral if you are way too high!) to approach and land. This, of
course, is all a moot point if for some reason (low altitutde, fire,
heart attack, etc.) you must land *now*.
-jpr
John R. Johnson
> In article <4h989i$e...@newsbf02.news.aol.com>, adv...@aol.com (Adviser)
> writes:
> > I'd like to see the 45 degree bank at 65 mph. Without looking at the POH
> > I think one wing will stall, the plane will spin, and...(numbers
> > might look good on paper)
>
> Just a simple question: Assuming you are flying a coordinated turn, why would
> only one wing stall?
>
Because, believe it or not, the inside wing is not going as fast as the
outside wing. If it were you wouldn't be turning! In a 45 degree bank
the stall speed is increased slightly by the load factor increase in the
turn. Stall speed is about 1.2 Vso. Approach speed is about 1.3 Vso.
This gives you a fairly small margin, especially in a trainer with a
reasonable low Vso. A gust or sudden cessation of the wind can cause
a momentary increase of AOA that will stall one or both wings. The
inside wing is likely to develop deeper into the stall than the outboard
wing. If there is a loss of lift, the inside wing will drop. When the
pilot then tries to pick up this falling wing with aileron, the aileron
increases the camber of the inside wing, effectively increasing its
angle of attack and driving it deeper into the stall. The resulting
maneuver looks to bystanders like a sudden nosedive into the ground.
It is actually a developing spin out the bottom of a turn. If it
happened to be the turn from base to final, you probably do not have a
lot of altitude to recover. However, you have more altitude than you
want to dive into the ground from. The resulting mess is often fatal.
It has happened several times at Oshkosh when a pilot trying to fit into
a crowded pattern with a hot homebuilt lets it get a little slow while
watching traffic and falls out the bottom.
That is the reason most flight instructors suggest shallow turns in the
pattern. The increase in load factor and resulting stall speed increase
in a 30 degree bank is minimal. A 45 degree bank is noticeable but not
real bad. A 60 degree bank increases the load factor to 2, which
increases the stall speed to about 1.4 Vso. If you are holding a 1.3
Vso approach speed, you are in trouble. :^)
John "sorry for the rant" Johnson
Julian Scarfe
>
> Just a simple question: Assuming you are flying a coordinated turn, why
would
> only one wing stall?
In article <Pine.SOL.3.91.960306163911.14028A-100000@reliant>, "John R.
Johnson" <jo...@siu.edu> wrote:
> Because, believe it or not, the inside wing is not going as fast as the
> outside wing.
In article <4hnart$6...@hacgate2.hac.com>, ttu...@samson.hac.com wrote:
> If the center of the aircraft is in a turn and flying at stall speed, then
> the wing on the outside is having to move faster to keep up, and the wing
> on the inside is likewise moving slower. And if it is moving slower than
> stall speed, guess what happens.
Hang on a minute guys, doesn't a wing stall at a critical *angle of
attack* (rather than critical airspeed)?
Julian Scarfe
ja...@cus.cam.ac.uk
Roy Smith
>Hang on a minute guys, doesn't a wing stall at a critical *angle of
>attack* (rather than critical airspeed)?
That's what I was taught. On the other hand, I've been drawing pictures and
free body diagrams of airplanes in coordinated level turns for quite some
time, and I still havn't been able to convince myself that the AOA is
uniform along the wingspan. I'm not convinced, either, that the wing
loading is uniform along the span, nor am I even sure if that matters!
If AOA is spanwise uniform (assumption), and airspeed varies spanwise
(simple geometry), than lift must vary spanwise too. But if lift varies
spanwise, why isn't the airplane still rolling? Once it reaches a constant
roll angle, doesn't the lift from both wings have to be equal? If so, than
the AOA can't be constant.
I keep going around and around, and all I manage to convince myself is that
it's a lot easier to fly an airplane than to design one.
--
Roy Smith <r...@nyu.edu>
Hippocrates Project, Department of Microbiology, Coles 202
NYU School of Medicine, 550 First Avenue, New York, NY 10016
"This never happened to Bart Simpson."
Julian Scarfe
> > But if lift varies
> > spanwise, why isn't the airplane still rolling? Once it reaches a constant
> > roll angle, doesn't the lift from both wings have to be equal? If so, than
> > the AOA can't be constant.
In article <jas12-10039...@atroute1.pem.cam.ac.uk>, I wrote:
> All other things being equal it would be...
That was misleading. I meant "All other things being equal the airplane
would still be rolling". Sorry.
Julian Scarfe
ja...@cus.cam.ac.uk
Dean Tresner
> [deletia]
> Yes, the stall is only dependent on the angle of attack. The angle of attack,
> however, depends on both the vertical and horizontal components of the
> airspeed. Since the inside wing is travelling slower than the outside wing,
> the horizontal component of the airspeed is lower while the vertical component
> is the same, resulting in a very slightly higher angle of attack on the inside
> wing. This would result in a tendency for the inside wing to stall slightly
> before the outside wing.
> Here's my theory:
AOA variations are only a function of angular velocities.
In a level turn pitch, roll and yaw rates are all equal.
Pitch has no spanewise effect on AOA.
Yaw causes AOA to decrease spanewise from the inside wingtip.
Roll causes AOA to increase spanewise from the inside wingtip.
The two effects cancel. (offered w/o proof. maybe someone will do the math.)
A climbing turn reduces the yaw rate causing an outside first stall and visa
versa.
-Dean
--
'inkster. nonce-wd.
[f. ink v. or n.1 + -ster.]
A scribbler, an inferior writer.
1860 Reade Eighth Commandm. 343 These inksters are the enemies not only of
the country but of the human race.