Thinking about stalls

On Mar 14, 11:58*am, Bertie the Bunyip wrote:
wrote :



On Mar 14, 11:37*am, Bertie the Bunyip wrote:
TakeFlight wrote in
news:935d6394-8224-482e-9428-
:

Put me in the "not enough info" column.

Plane #2 could be in fact _in_ a stall (or spin), "descending fast
with 50% power" or _more_. *Think Delta Flight 191, for example.

That was something else entirely. That was a microburst. The rules
pretty much go out the window with one of those.
not to say the laws of physics are suspended, but it's a scenario
that is so different from what we learn as pilots that drastic
retraining was * introduced right across the board after it. Flight
guidance systems were modified to account for the new methods, so
it's not really relevant.

Just to give you some idea of what I mean, I'll give you a scenario.
You'v
e
just aken off and yoou're climibing away at best rate. Suddenly, your
airspeed increases by a fairly large lump. 15-20 knots, say. you
increase your pitcha bit to absorb it and your speed bleeds back a
tad. Still plent
y
in hand, though. all the sudden the pitch you have is dragging your
speed back and it's beginning to decrease as the wind that delivered
that extra speed vanishes. You're still OK and back to your orignal
pitch and have a couple of knots more than you had at the beginning.
All the sudden, the bottom falls out of your airplane. Your climb
stops and then a second late
r *
you begin to sink, and fast. another second or two and your speed
washes off even further and now you're sinkng and your stall warning
is starting to squeak.

you gotta do something and right now. you still have some altitude,
say 40
0
feet. what do you do?

Bertie

Alt-Ctl-Del

No, wait, change my underwear.

Yoke forward, nose down and max power?

That's what the Delta guys did. And that 727 in New Orleans. A different
approach was needed and what they came up with was full power. and in a
jet that means firewall and overboost to your little heart's contenet,
and nose up as much as you can. The stall warnign should be ringing off
the wall ( we have stick shakers, but same thing) and you keep this up
til you fly out the other side of the mess. It goes against everything
we've learned but that's what they tell us to do. There's generally some
guidance form the flight director as well. On some it's a set of yellow
"antlers" that give you best pitch and on others the pitch bar on the
flight director just gives you all the pitch info you need ( you just
put the airplane wings on a magenta bar, no brains required)
note this is for a sustained and powerful microburst and not for
recovery form a tiny bit of wind shear in a 20 knot wind.

Bertie

That's what I couldn't remember; I recalled the DFW incident (drove by
a couple of weeks after- messy) and remembered reading about the sim
duplication. Yep. Pull back would take some sim work I'd think to
ingrain something that seems so counter-intuitive.

So, pull back, firewall, recover, enjoy the shaking hands/knees,
change underwear. Gotcha. Oh, land, have beer.

wrote in
:

On Mar 14, 11:58*am, Bertie the Bunyip wrote:
wrote
innews:b887a328-823a-4aa9-8af0-75e24eadf0f2@p25
g2000hsf.googlegroups.com:



On Mar 14, 11:37*am, Bertie the Bunyip wrote:
TakeFlight wrote in
news:935d6394-8224-482e-9428-
:

Put me in the "not enough info" column.

Plane #2 could be in fact _in_ a stall (or spin), "descending
fast with 50% power" or _more_. *Think Delta Flight 191, for
example.

That was something else entirely. That was a microburst. The rules
pretty much go out the window with one of those.
not to say the laws of physics are suspended, but it's a scenario
that is so different from what we learn as pilots that drastic
retraining was * introduced right across the board after it.
Flight guidance systems were modified to account for the new
methods, so it's not really relevant.

Just to give you some idea of what I mean, I'll give you a
scenario. You'v
e
just aken off and yoou're climibing away at best rate. Suddenly,
your airspeed increases by a fairly large lump. 15-20 knots, say.
you increase your pitcha bit to absorb it and your speed bleeds
back a tad. Still plent
y
in hand, though. all the sudden the pitch you have is dragging
your speed back and it's beginning to decrease as the wind that
delivered that extra speed vanishes. You're still OK and back to
your orignal pitch and have a couple of knots more than you had at
the beginning. All the sudden, the bottom falls out of your
airplane. Your climb stops and then a second late
r *
you begin to sink, and fast. another second or two and your speed
washes off even further and now you're sinkng and your stall
warning is starting to squeak.

you gotta do something and right now. you still have some
altitude, say 40
0
feet. what do you do?

Bertie

Alt-Ctl-Del

No, wait, change my underwear.

Yoke forward, nose down and max power?

That's what the Delta guys did. And that 727 in New Orleans. A
different approach was needed and what they came up with was full
power. and in a jet that means firewall and overboost to your little
heart's contenet, and nose up as much as you can. The stall warnign
should be ringing off the wall ( we have stick shakers, but same
thing) and you keep this up til you fly out the other side of the
mess. It goes against everything we've learned but that's what they
tell us to do. There's generally some guidance form the flight
director as well. On some it's a set of yellow "antlers" that give
you best pitch and on others the pitch bar on the flight director
just gives you all the pitch info you need ( you just put the
airplane wings on a magenta bar, no brains required) note this is for
a sustained and powerful microburst and not for recovery form a tiny
bit of wind shear in a 20 knot wind.

Bertie

That's what I couldn't remember; I recalled the DFW incident (drove by
a couple of weeks after- messy) and remembered reading about the sim
duplication. Yep. Pull back would take some sim work I'd think to
ingrain something that seems so counter-intuitive.


Exactly.

So, pull back, firewall, recover, enjoy the shaking hands/knees,
change underwear. Gotcha. Oh, land, have beer.

Absolutely!

Best to stay well away form them anyway!
Oh yeah. I almost forgot, we have a computer that detects windshear and
we get an aural warning and an annunciator. All a legacy of that
accident..


Bertie

On Mar 14, 12:08*pm, Bertie the Bunyip wrote:
wrote :



On Mar 14, 11:58*am, Bertie the Bunyip wrote:
wrote
innews:b887a328-823a-4aa9-8af0-75e24eadf0f2@p25
g2000hsf.googlegroups.com:

On Mar 14, 11:37*am, Bertie the Bunyip wrote:
TakeFlight wrote in
news:935d6394-8224-482e-9428-
:

Put me in the "not enough info" column.

Plane #2 could be in fact _in_ a stall (or spin), "descending
fast with 50% power" or _more_. *Think Delta Flight 191, for
example.

That was something else entirely. That was a microburst. The rules
pretty much go out the window with one of those.
not to say the laws of physics are suspended, but it's a scenario
that is so different from what we learn as pilots that drastic
retraining was * introduced right across the board after it.
Flight guidance systems were modified to account for the new
methods, so it's not really relevant.

Just to give you some idea of what I mean, I'll give you a
scenario. You'v
e
just aken off and yoou're climibing away at best rate. Suddenly,
your airspeed increases by a fairly large lump. 15-20 knots, say.
you increase your pitcha bit to absorb it and your speed bleeds
back a tad. Still plent
y
in hand, though. all the sudden the pitch you have is dragging
your speed back and it's beginning to decrease as the wind that
delivered that extra speed vanishes. You're still OK and back to
your orignal pitch and have a couple of knots more than you had at
the beginning. All the sudden, the bottom falls out of your
airplane. Your climb stops and then a second late
r *
you begin to sink, and fast. another second or two and your speed
washes off even further and now you're sinkng and your stall
warning is starting to squeak.

you gotta do something and right now. you still have some
altitude, say 40
0
feet. what do you do?

Bertie

Alt-Ctl-Del

No, wait, change my underwear.

Yoke forward, nose down and max power?

That's what the Delta guys did. And that 727 in New Orleans. A
different approach was needed and what they came up with was full
power. and in a jet that means firewall and overboost to your little
heart's contenet, and nose up as much as you can. The stall warnign
should be ringing off the wall ( we have stick shakers, but same
thing) and you keep this up til you fly out the other side of the
mess. It goes against everything we've learned but that's what they
tell us to do. There's generally some guidance form the flight
director as well. On some it's a set of yellow "antlers" that give
you best pitch and on others the pitch bar on the flight director
just gives you all the pitch info you need ( you just put the
airplane wings on a magenta bar, no brains required) note this is for
a sustained and powerful microburst and not for recovery form a tiny
bit of wind shear in a 20 knot wind.

Bertie

That's what I couldn't remember; I recalled the DFW incident (drove by
a couple of weeks after- messy) and remembered reading about the sim
duplication. *Yep. *Pull back would take some sim work I'd think to
ingrain something that seems so counter-intuitive.

Exactly.



So, pull back, firewall, recover, enjoy the shaking hands/knees,
change underwear. *Gotcha. Oh, land, have beer.

Absolutely!

Best to stay well away form them anyway!
Oh yeah. I almost forgot, we have a computer that detects windshear and
we get an aural warning and an annunciator. All a legacy of that
accident..

Bertie




Got me thinking about beer:30 after jumping. Sigh. Now I miss the
planes, the jumping, the women, the beer. Hanging under a ram air on
high altitude (12K) jump and floating for 20 minutes in the cool
air...unlimited visibility. Just you, the harness and 200 sqft of
nylon. Think I'll have a beer. But I do miss the flying.

On Mar 14, 9:27 am, WingFlaps wrote:
On Mar 15, 12:08 am, Dan wrote:

On Mar 13, 9:57 pm, WingFlaps wrote:

On Mar 14, 2:46 pm, Dan wrote:

On Mar 13, 8:50 pm, WingFlaps wrote:

On Mar 14, 1:00 pm, wrote:

On Mar 13, 1:37 pm, WingFlaps wrote:

Nope, if the airspeed is constant, the lift from the two wings is not
the same. This is thought provoking discussion I was hoping to start!
Can you see why lift does not equal weight in both cases?

Common misconception: that a climbing wing is generating more
lift than a descending wing. If the flight paths are both straight
lines, whether climbing or descending, the lift is the same in both
cases. As Jim said, only a change in the direction of flight will
change the lift/weight ratio. A G-meter (such as in our Citabrias)
will prove it.
If the airspeeds are the same and the flight paths are both
straight, the AOAs are both the same, too. But change the speeds while
leaving the flight paths alone, and the AOA will change. It's why the
airplane has a nose-high attitude in level slow flight as opposed to a
lower nose attitude in level cruise.

Hi Dan see my rely to Jim. In fact, lift is reduced in a steady climb.

Cheers

The trust vector is added to the lift vector in a climb, as the drag
is added to weight.

Are you saying that wing lift does not change with attitude in a non-
accelerating frame?

Cheers

Of course it does.

However -- In a climb thrust acts contrary to drag some component of
weight (depending on the angle of climb). Thus the angle of attack is
not *necessarily* equal to the angle of climb.

I'm sorry but I'm having trouble understanding where you are coming
from. In my equations above I wrote that D=Tcos(alpha) and this is
based on the idea that W,D and L are 3 orthogonal forces. Of course
you can rewrite them non orthogonally if you please but my expression
makes good sense (to me anyway). This is why: Imagine a jet in a
steady vertical climb (alpha=90 degrees). My equation says D=0 so how
can that be? The answer is that this simplified (wing + engine) model
is really only considering induced drag and that the thrust line is
close to 0 AOA (not bad approximations IMHO). For induced drag to be
zero, wing LIFT _must_ be zero and so we see that Weight =Tsin(90) + L
- W=T -exactly as it should be for a vertical climb! The pilot has
reduced AOA to zero, the wing produces NO lift and the plane climbs
vertically. Again, I say L should be dropped to zero for a true steady
vertical climb and L=W only in straight and level flight or if the
plane is gliding (T=0). In all steady climbs LW and all descents LW.

I was a bit surprised when I realized that to be in a steady climb
the pilot must be operating the plane in a condition where wing lift
and AOA are actually lower than in straight and level -all thanks to
a component of thrust adding to lift !!! Although the effect may not
be large for low power planes (TW) , if what I'm saying makes sense,
one may see some advantages in this approach -e.g. why increasing
power leads to a nice steady climb or cutting power causes the nose to
drop and a descent to begin...

Cheers

Consider -- if you are straight and level at 150 knots with a 3 degree
AOA and you increase the angle of attack to 8 degrees, with no change
in power, what happens?

On Mar 15, 10:32*am, Dan wrote:
On Mar 14, 9:27 am, WingFlaps wrote:





On Mar 15, 12:08 am, Dan wrote:

On Mar 13, 9:57 pm, WingFlaps wrote:

On Mar 14, 2:46 pm, Dan wrote:

On Mar 13, 8:50 pm, WingFlaps wrote:

On Mar 14, 1:00 pm, wrote:

On Mar 13, 1:37 pm, WingFlaps wrote:

Nope, if the airspeed is constant, the lift from the two wings is not
the same. This is thought provoking discussion I was hoping to start!
Can you see why lift does not equal weight in both cases?

* * * Common misconception: that a climbing wing is generating more
lift than a descending wing. If the flight paths are both straight
lines, whether *climbing or descending, the lift is the same in both
cases. As Jim said, only a change in the direction of flight will
change the lift/weight ratio. A G-meter (such as in our Citabrias)
will prove it.
* * * If the airspeeds are the same and the flight paths are both
straight, the AOAs are both the same, too. But change the speeds while
leaving the flight paths alone, and the AOA will change. It's why the
airplane has a nose-high attitude in level slow flight as opposed to a
lower nose attitude in level cruise.

Hi Dan see my rely to Jim. In fact, lift is reduced in a steady climb.

Cheers

The trust vector is added to the lift vector in a climb, as the drag
is added to weight.

Are you saying that wing lift does not change with attitude in a non-
accelerating frame?

Cheers

Of course it does.

However -- In a climb thrust acts contrary to drag some component of
weight (depending on the angle of climb). Thus the angle of attack is
not *necessarily* equal to the angle of climb.

I'm sorry but I'm having trouble understanding where you are coming
from. In my equations above I wrote that D=Tcos(alpha) and this is
based on the idea that W,D and L are 3 orthogonal forces. Of course
you can rewrite them non orthogonally if you please but my expression
makes good sense (to me anyway). This is why: Imagine a jet in a
steady vertical climb (alpha=90 degrees). My equation says D=0 so how
can that be? The answer is that this simplified (wing + engine) model
is really only considering induced drag and that the thrust line is
close to 0 AOA (not bad approximations IMHO). For induced drag to be
zero, wing LIFT _must_ be zero and so we see that Weight =Tsin(90) + L
- *W=T -exactly as it should be for a vertical climb! The pilot has
reduced AOA to zero, the wing produces NO lift and the plane climbs
vertically. Again, I say L should be dropped to zero for a true steady
vertical climb and L=W only in straight and level flight or if the
plane is gliding (T=0). In all steady climbs LW and all descents LW.

I was a bit surprised when I realized that *to be in a steady climb
the pilot must be operating the plane in a condition where wing lift
and AOA *are actually lower than in straight and level -all thanks to
a component of thrust adding to lift !!! Although the effect may not
be large for low power planes (TW) , if what I'm saying makes sense,
one may see some advantages in this approach -e.g. why increasing
power leads to a nice steady climb or cutting power causes the nose to
drop and a descent to begin...

Cheers

Consider -- if you are straight and level at 150 knots with a 3 degree
AOA and you increase the angle of attack to 8 degrees, with no change
in power, what happens?- Hide quoted text -


You initially increase lift so the plane starts to acclerate in the
vertical plane and the increased drag starts to slow you down -so your
airspeed also comes back. As the airspeed deacys lift starts to drop
(with V^2) and, if you have enough power on to establish a steady
climb at 8 degrees AOA, you achieve a new state where you have a lower
airspeed and the vertical component of lift from the wing is slightly
less than before you started the climb -the lost lift is replaced by
the engine thrust component. If you don't get it, keep thinking about
the steady vertical climb -what is the force that opposes weight? It's
thrust pure and simple right? That force is a sine function of
angle... OK?

Cheers

On Mar 14, 5:59 pm, WingFlaps wrote:
On Mar 15, 10:32 am, Dan wrote:



On Mar 14, 9:27 am, WingFlaps wrote:

On Mar 15, 12:08 am, Dan wrote:

On Mar 13, 9:57 pm, WingFlaps wrote:

On Mar 14, 2:46 pm, Dan wrote:

On Mar 13, 8:50 pm, WingFlaps wrote:

On Mar 14, 1:00 pm, wrote:

On Mar 13, 1:37 pm, WingFlaps wrote:

Nope, if the airspeed is constant, the lift from the two wings is not
the same. This is thought provoking discussion I was hoping to start!
Can you see why lift does not equal weight in both cases?

Common misconception: that a climbing wing is generating more
lift than a descending wing. If the flight paths are both straight
lines, whether climbing or descending, the lift is the same in both
cases. As Jim said, only a change in the direction of flight will
change the lift/weight ratio. A G-meter (such as in our Citabrias)
will prove it.
If the airspeeds are the same and the flight paths are both
straight, the AOAs are both the same, too. But change the speeds while
leaving the flight paths alone, and the AOA will change. It's why the
airplane has a nose-high attitude in level slow flight as opposed to a
lower nose attitude in level cruise.

Hi Dan see my rely to Jim. In fact, lift is reduced in a steady climb.

Cheers

The trust vector is added to the lift vector in a climb, as the drag
is added to weight.

Are you saying that wing lift does not change with attitude in a non-
accelerating frame?

Cheers

Of course it does.

However -- In a climb thrust acts contrary to drag some component of
weight (depending on the angle of climb). Thus the angle of attack is
not *necessarily* equal to the angle of climb.

I'm sorry but I'm having trouble understanding where you are coming
from. In my equations above I wrote that D=Tcos(alpha) and this is
based on the idea that W,D and L are 3 orthogonal forces. Of course
you can rewrite them non orthogonally if you please but my expression
makes good sense (to me anyway). This is why: Imagine a jet in a
steady vertical climb (alpha=90 degrees). My equation says D=0 so how
can that be? The answer is that this simplified (wing + engine) model
is really only considering induced drag and that the thrust line is
close to 0 AOA (not bad approximations IMHO). For induced drag to be
zero, wing LIFT _must_ be zero and so we see that Weight =Tsin(90) + L
- W=T -exactly as it should be for a vertical climb! The pilot has
reduced AOA to zero, the wing produces NO lift and the plane climbs
vertically. Again, I say L should be dropped to zero for a true steady
vertical climb and L=W only in straight and level flight or if the
plane is gliding (T=0). In all steady climbs LW and all descents LW.

I was a bit surprised when I realized that to be in a steady climb
the pilot must be operating the plane in a condition where wing lift
and AOA are actually lower than in straight and level -all thanks to
a component of thrust adding to lift !!! Although the effect may not
be large for low power planes (TW) , if what I'm saying makes sense,
one may see some advantages in this approach -e.g. why increasing
power leads to a nice steady climb or cutting power causes the nose to
drop and a descent to begin...

Cheers

Consider -- if you are straight and level at 150 knots with a 3 degree
AOA and you increase the angle of attack to 8 degrees, with no change
in power, what happens?- Hide quoted text -

You initially increase lift so the plane starts to acclerate in the
vertical plane and the increased drag starts to slow you down -so your
airspeed also comes back. As the airspeed deacys lift starts to drop
(with V^2) and, if you have enough power on to establish a steady
climb at 8 degrees AOA, you achieve a new state where you have a lower
airspeed and the vertical component of lift from the wing is slightly
less than before you started the climb -the lost lift is replaced by
the engine thrust component. If you don't get it, keep thinking about
the steady vertical climb -what is the force that opposes weight? It's
thrust pure and simple right? That force is a sine function of
angle... OK?

Cheers

I know the answer..wasn't sure if you were thinking beyond the slide
rule.

On Mar 15, 11:08*am, Dan wrote:
On Mar 14, 5:59 pm, WingFlaps wrote:





On Mar 15, 10:32 am, Dan wrote:

On Mar 14, 9:27 am, WingFlaps wrote:

On Mar 15, 12:08 am, Dan wrote:

On Mar 13, 9:57 pm, WingFlaps wrote:

On Mar 14, 2:46 pm, Dan wrote:

On Mar 13, 8:50 pm, WingFlaps wrote:

On Mar 14, 1:00 pm, wrote:

On Mar 13, 1:37 pm, WingFlaps wrote:

Nope, if the airspeed is constant, the lift from the two wings is not
the same. This is thought provoking discussion I was hoping to start!
Can you see why lift does not equal weight in both cases?

* * * Common misconception: that a climbing wing is generating more
lift than a descending wing. If the flight paths are both straight
lines, whether *climbing or descending, the lift is the same in both
cases. As Jim said, only a change in the direction of flight will
change the lift/weight ratio. A G-meter (such as in our Citabrias)
will prove it.
* * * If the airspeeds are the same and the flight paths are both
straight, the AOAs are both the same, too. But change the speeds while
leaving the flight paths alone, and the AOA will change. It's why the
airplane has a nose-high attitude in level slow flight as opposed to a
lower nose attitude in level cruise.

Hi Dan see my rely to Jim. In fact, lift is reduced in a steady climb.

Cheers

The trust vector is added to the lift vector in a climb, as the drag
is added to weight.

Are you saying that wing lift does not change with attitude in a non-
accelerating frame?

Cheers

Of course it does.

However -- In a climb thrust acts contrary to drag some component of
weight (depending on the angle of climb). Thus the angle of attack is
not *necessarily* equal to the angle of climb.

I'm sorry but I'm having trouble understanding where you are coming
from. In my equations above I wrote that D=Tcos(alpha) and this is
based on the idea that W,D and L are 3 orthogonal forces. Of course
you can rewrite them non orthogonally if you please but my expression
makes good sense (to me anyway). This is why: Imagine a jet in a
steady vertical climb (alpha=90 degrees). My equation says D=0 so how
can that be? The answer is that this simplified (wing + engine) model
is really only considering induced drag and that the thrust line is
close to 0 AOA (not bad approximations IMHO). For induced drag to be
zero, wing LIFT _must_ be zero and so we see that Weight =Tsin(90) + L
- *W=T -exactly as it should be for a vertical climb! The pilot has
reduced AOA to zero, the wing produces NO lift and the plane climbs
vertically. Again, I say L should be dropped to zero for a true steady
vertical climb and L=W only in straight and level flight or if the
plane is gliding (T=0). In all steady climbs LW and all descents LW.

I was a bit surprised when I realized that *to be in a steady climb
the pilot must be operating the plane in a condition where wing lift
and AOA *are actually lower than in straight and level -all thanks to
a component of thrust adding to lift !!! Although the effect may not
be large for low power planes (TW) , if what I'm saying makes sense,
one may see some advantages in this approach -e.g. why increasing
power leads to a nice steady climb or cutting power causes the nose to
drop and a descent to begin...

Cheers

Consider -- if you are straight and level at 150 knots with a 3 degree
AOA and you increase the angle of attack to 8 degrees, with no change
in power, what happens?- Hide quoted text -

You initially increase lift so the plane starts to acclerate in the
vertical plane and the increased drag starts to slow you down -so your
airspeed also comes back. As the airspeed deacys lift starts to drop
(with V^2) and, if you have enough power on to establish a steady
climb at 8 degrees AOA, you achieve a new state where you have a lower
airspeed and the vertical component of lift from the wing is slightly
less than before you started the climb -the lost lift is replaced by
the engine thrust component. If you don't get it, keep thinking about
the steady vertical climb -what is the force that opposes weight? It's
thrust pure and simple right? That force is a sine function of
angle... OK?

Cheers

I know the answer..wasn't sure if you were thinking beyond the slide
rule.- Hide quoted text -


You used slide rules too? Good on ya! So, you agree that vertical lift
is lower in a steady climb -if so is the wing closer to a stall when
climbing or descending at the same speed -is this as thought provoking
as I hoped?

Cheers

On Mar 14, 6:42 pm, WingFlaps wrote:
On Mar 15, 11:08 am, Dan wrote:



On Mar 14, 5:59 pm, WingFlaps wrote:

On Mar 15, 10:32 am, Dan wrote:

On Mar 14, 9:27 am, WingFlaps wrote:

On Mar 15, 12:08 am, Dan wrote:

On Mar 13, 9:57 pm, WingFlaps wrote:

On Mar 14, 2:46 pm, Dan wrote:

On Mar 13, 8:50 pm, WingFlaps wrote:

On Mar 14, 1:00 pm, wrote:

On Mar 13, 1:37 pm, WingFlaps wrote:

Nope, if the airspeed is constant, the lift from the two wings is not
the same. This is thought provoking discussion I was hoping to start!
Can you see why lift does not equal weight in both cases?

Common misconception: that a climbing wing is generating more
lift than a descending wing. If the flight paths are both straight
lines, whether climbing or descending, the lift is the same in both
cases. As Jim said, only a change in the direction of flight will
change the lift/weight ratio. A G-meter (such as in our Citabrias)
will prove it.
If the airspeeds are the same and the flight paths are both
straight, the AOAs are both the same, too. But change the speeds while
leaving the flight paths alone, and the AOA will change. It's why the
airplane has a nose-high attitude in level slow flight as opposed to a
lower nose attitude in level cruise.

Hi Dan see my rely to Jim. In fact, lift is reduced in a steady climb.

Cheers

The trust vector is added to the lift vector in a climb, as the drag
is added to weight.

Are you saying that wing lift does not change with attitude in a non-
accelerating frame?

Cheers

Of course it does.

However -- In a climb thrust acts contrary to drag some component of
weight (depending on the angle of climb). Thus the angle of attack is
not *necessarily* equal to the angle of climb.

I'm sorry but I'm having trouble understanding where you are coming
from. In my equations above I wrote that D=Tcos(alpha) and this is
based on the idea that W,D and L are 3 orthogonal forces. Of course
you can rewrite them non orthogonally if you please but my expression
makes good sense (to me anyway). This is why: Imagine a jet in a
steady vertical climb (alpha=90 degrees). My equation says D=0 so how
can that be? The answer is that this simplified (wing + engine) model
is really only considering induced drag and that the thrust line is
close to 0 AOA (not bad approximations IMHO). For induced drag to be
zero, wing LIFT _must_ be zero and so we see that Weight =Tsin(90) + L
- W=T -exactly as it should be for a vertical climb! The pilot has
reduced AOA to zero, the wing produces NO lift and the plane climbs
vertically. Again, I say L should be dropped to zero for a true steady
vertical climb and L=W only in straight and level flight or if the
plane is gliding (T=0). In all steady climbs LW and all descents LW.

I was a bit surprised when I realized that to be in a steady climb
the pilot must be operating the plane in a condition where wing lift
and AOA are actually lower than in straight and level -all thanks to
a component of thrust adding to lift !!! Although the effect may not
be large for low power planes (TW) , if what I'm saying makes sense,
one may see some advantages in this approach -e.g. why increasing
power leads to a nice steady climb or cutting power causes the nose to
drop and a descent to begin...

Cheers

Consider -- if you are straight and level at 150 knots with a 3 degree
AOA and you increase the angle of attack to 8 degrees, with no change
in power, what happens?- Hide quoted text -

You initially increase lift so the plane starts to acclerate in the
vertical plane and the increased drag starts to slow you down -so your
airspeed also comes back. As the airspeed deacys lift starts to drop
(with V^2) and, if you have enough power on to establish a steady
climb at 8 degrees AOA, you achieve a new state where you have a lower
airspeed and the vertical component of lift from the wing is slightly
less than before you started the climb -the lost lift is replaced by
the engine thrust component. If you don't get it, keep thinking about
the steady vertical climb -what is the force that opposes weight? It's
thrust pure and simple right? That force is a sine function of
angle... OK?

Cheers

I know the answer..wasn't sure if you were thinking beyond the slide
rule.- Hide quoted text -

You used slide rules too? Good on ya! So, you agree that vertical lift
is lower in a steady climb -if so is the wing closer to a stall when
climbing or descending at the same speed -is this as thought provoking
as I hoped?

Cheers

Slide rules -- oh so long ago! Though I can still spin a whiz wheel...

Vertical lift may be equal to lift in a descent, or in straight and
level.

N'est pas?

Dan Mcc

On Mar 15, 1:31*pm, Dan wrote:
On Mar 14, 6:42 pm, WingFlaps wrote:





On Mar 15, 11:08 am, Dan wrote:

On Mar 14, 5:59 pm, WingFlaps wrote:

On Mar 15, 10:32 am, Dan wrote:

On Mar 14, 9:27 am, WingFlaps wrote:

On Mar 15, 12:08 am, Dan wrote:

On Mar 13, 9:57 pm, WingFlaps wrote:

On Mar 14, 2:46 pm, Dan wrote:

On Mar 13, 8:50 pm, WingFlaps wrote:

On Mar 14, 1:00 pm, wrote:

On Mar 13, 1:37 pm, WingFlaps wrote:

Nope, if the airspeed is constant, the lift from the two wings is not
the same. This is thought provoking discussion I was hoping to start!
Can you see why lift does not equal weight in both cases?

* * * Common misconception: that a climbing wing is generating more
lift than a descending wing. If the flight paths are both straight
lines, whether *climbing or descending, the lift is the same in both
cases. As Jim said, only a change in the direction of flight will
change the lift/weight ratio. A G-meter (such as in our Citabrias)
will prove it.
* * * If the airspeeds are the same and the flight paths are both
straight, the AOAs are both the same, too. But change the speeds while
leaving the flight paths alone, and the AOA will change. It's why the
airplane has a nose-high attitude in level slow flight as opposed to a
lower nose attitude in level cruise.

Hi Dan see my rely to Jim. In fact, lift is reduced in a steady climb.

Cheers

The trust vector is added to the lift vector in a climb, as the drag
is added to weight.

Are you saying that wing lift does not change with attitude in a non-
accelerating frame?

Cheers

Of course it does.

However -- In a climb thrust acts contrary to drag some component of
weight (depending on the angle of climb). Thus the angle of attack is
not *necessarily* equal to the angle of climb.

I'm sorry but I'm having trouble understanding where you are coming
from. In my equations above I wrote that D=Tcos(alpha) and this is
based on the idea that W,D and L are 3 orthogonal forces. Of course
you can rewrite them non orthogonally if you please but my expression
makes good sense (to me anyway). This is why: Imagine a jet in a
steady vertical climb (alpha=90 degrees). My equation says D=0 so how
can that be? The answer is that this simplified (wing + engine) model
is really only considering induced drag and that the thrust line is
close to 0 AOA (not bad approximations IMHO). For induced drag to be
zero, wing LIFT _must_ be zero and so we see that Weight =Tsin(90) + L
- *W=T -exactly as it should be for a vertical climb! The pilot has
reduced AOA to zero, the wing produces NO lift and the plane climbs
vertically. Again, I say L should be dropped to zero for a true steady
vertical climb and L=W only in straight and level flight or if the
plane is gliding (T=0). In all steady climbs LW and all descents LW.

I was a bit surprised when I realized that *to be in a steady climb
the pilot must be operating the plane in a condition where wing lift
and AOA *are actually lower than in straight and level -all thanks to
a component of thrust adding to lift !!! Although the effect may not
be large for low power planes (TW) , if what I'm saying makes sense,
one may see some advantages in this approach -e.g. why increasing
power leads to a nice steady climb or cutting power causes the nose to
drop and a descent to begin...

Cheers

Consider -- if you are straight and level at 150 knots with a 3 degree
AOA and you increase the angle of attack to 8 degrees, with no change
in power, what happens?- Hide quoted text -

You initially increase lift so the plane starts to acclerate in the
vertical plane and the increased drag starts to slow you down -so your
airspeed also comes back. As the airspeed deacys lift starts to drop
(with V^2) and, if you have enough power on to establish a steady
climb at 8 degrees AOA, you achieve a new state where you have a lower
airspeed and the vertical component of lift from the wing is slightly
less than before you started the climb -the lost lift is replaced by
the engine thrust component. If you don't get it, keep thinking about
the steady vertical climb -what is the force that opposes weight? It's
thrust pure and simple right? That force is a sine function of
angle... OK?

Cheers

I know the answer..wasn't sure if you were thinking beyond the slide
rule.- Hide quoted text -

You used slide rules too? Good on ya! So, you agree that vertical lift
is lower in a steady climb -if so is the wing closer to a stall when
climbing or descending at the same speed -is this as thought provoking
as I hoped?

Cheers

Slide rules -- oh so long ago! Though I can still spin a whiz wheel...

Vertical lift may be equal to lift in a descent, or in straight and
level.

N'est pas?


Yes, but not equal to weight unless thrust is purely horizontal. Not
sure about the French tho' is that slang for n'est-ce pas?

Cheers

writes:
That's what I couldn't remember; I recalled the DFW incident (drove by
a couple of weeks after- messy) and remembered reading about the sim
duplication. Yep. Pull back would take some sim work I'd think to
ingrain something that seems so counter-intuitive.

So, pull back, firewall, recover, enjoy the shaking hands/knees,
change underwear. Gotcha. Oh, land, have beer.

One beer? Now that's nerves of steel.

--
James Carlson, Solaris Networking
Sun Microsystems / 35 Network Drive 71.232W Vox +1 781 442 2084
MS UBUR02-212 / Burlington MA 01803-2757 42.496N Fax +1 781 442 1677