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