Who does flight plans?

"Neil Gould" wrote in message
...
[...] So, in context, how is your theoretically available lift relevant?

It is relevant only to your false claim that a stalled wing provides no
lift. Had you not made that false claim, I would have had no reason to
bring up that element of the discussion.

[...]
To be a scalar, it would have to lack motion, ergo no "attack".

Wrong. "To be a scalar" it needs to be a single value. And it is.
Angle-of-attack is just an angle. A single value.

Every angle requires two reference lines in order to define that angle.
That doesn't change the fact that the angle itself is a single value,
without any direction component. Likewise, the fact that two reference
lines (one defined by a direction of travel) are used to define
angle-of-attack DOES NOT MAKE ANGLE-OF-ATTACK ITSELF A VECTOR. It's still
just an angle.

[...]
I responded to that. In the context of landing, if one flies slowly
enough to stall, one can stall "flat" relative to the ground because
the decrease in forward "relative wind" increases the AOA. That is
what my remark addresses.

Your claim is incorrect. As long as the airplane is flying just
above the ground, the relative wind is parallel to the ground. No
change in the angle-of-attack will occur from any decrease in speed,
not directly.

My claim is that if the aircraft is flying parallel to the ground just
before touch-down, it isn't stalled.

That's a new claim. Your previous claim (quoted above) was that you COULD
stall while flying parallel to the ground. That is, one could "stall
'flat'".

In any case, other than the issue with the geometry of the airplane, there
is absolutely no justification in claiming that flight parallel to the
ground precludes a stall.

[...]
It is simply impossible to do what you suggest one might do. If one
"flies slowly enough to stall", the angle-of-attack is at the stalling
angle-of-attack, period.

And all I'm saying is that this is independent of the pitch angle relative
to the ground.

It is NOT independent of the pitch angle relative to the ground if the
airplane is being flown in a flight path parallel to the ground.

[...]
What WILL happen is that as the aircraft slows, the pitch angle of the
aircraft will need to be increased, so as to continually increase the
angle-of-attack of the wing.

We are describing the same phenomena from two perspectives.

I am fairly certain we're not.

In the context
of my usage, if one maintains the pitch angle as the aircraft slows, the
AOA will continually increase (normally, the pitch angle changes as the
aircraft slows).

You cannot "maintain the pitch angle as the aircraft slows" without touching
the runway. If the aircraft slows and the pitch angle is not changed, lift
is reduced and the airplane will descend onto the runway.

I'll say it again: the scenario you propose is an impossibility.

Pete

Recently, Peter Duniho posted:

"Neil Gould" wrote in message
...
[...] So, in context, how is your theoretically available lift
relevant?

It is relevant only to your false claim that a stalled wing provides
no lift. Had you not made that false claim, I would have had no
reason to bring up that element of the discussion.

The only falsity here is your claim that I made such a statement. So,
apparently you have no reason to bring up that element in the discussion.

What I *did* claim is:
"You're holding the aircraft at just above the stall speed. When you're
stalled, you're falling, not flying."

This makes *no* reference to the amount of lift that a stalled wing
provides beyond it being inadequate to support flight, which should be
obvious to one who pays so much attention to details as you.

[...]
To be a scalar, it would have to lack motion, ergo no "attack".

Wrong. "To be a scalar" it needs to be a single value. And it is.
Angle-of-attack is just an angle. A single value.

Wrong. A scalar can not contain elements of direction by definition. Ergo,
AOA has no meaning as a scalar.

Every angle requires two reference lines in order to define that
angle. That doesn't change the fact that the angle itself is a single
value, without any direction component. Likewise, the fact that two
reference lines (one defined by a direction of travel) are used to
define angle-of-attack DOES NOT MAKE ANGLE-OF-ATTACK ITSELF A VECTOR.
It's still just an angle.

Wrong, it's AOA only has meaning when referenced by a direction and
motion. Without that direction and motion, THERE IS NO ANGLE OF ATTACK. If
you still don't understand this, what is the AOA when the aircraft is
parked in the hangar (with the doors closed and no fans running, if you
insist on picking nits)? ;-)

Finally, when you apply a direction to a scalar, what does it become?
;-)

[...]
I responded to that. In the context of landing, if one flies slowly
enough to stall, one can stall "flat" relative to the ground
because the decrease in forward "relative wind" increases the AOA.
That is what my remark addresses.

Your claim is incorrect. As long as the airplane is flying just
above the ground, the relative wind is parallel to the ground. No
change in the angle-of-attack will occur from any decrease in speed,
not directly.

My claim is that if the aircraft is flying parallel to the ground
just before touch-down, it isn't stalled.

That's a new claim.

Wrong, that's my original claim:

Jose: " Then what am I doing when I practice stalls at altitude, holding
the aircraft at the stall buffet?"

I wrote: "You're holding the aircraft at just above the stall speed."

Your previous claim (quoted above) was that you
COULD stall while flying parallel to the ground. That is, one could
"stall 'flat'".

Again, I made no such statement. My above claim is actually a
clarification of my original statement for your benefit.

My "stall flat" comment refers to pitch angle relative to the ground (as
was stated in the same paragraph). I made no claim about travelling
parallel to the runway. Indeed, one would *not* be going parallel to the
ground in a flat stall.

In any case, other than the issue with the geometry of the airplane,
there is absolutely no justification in claiming that flight parallel
to the ground precludes a stall.

I made no such claim as this, either.

[...]
It is simply impossible to do what you suggest one might do. If one
"flies slowly enough to stall", the angle-of-attack is at the
stalling angle-of-attack, period.

And all I'm saying is that this is independent of the pitch angle
relative to the ground.

It is NOT independent of the pitch angle relative to the ground if the
airplane is being flown in a flight path parallel to the ground.

Again, "a flight path parallel to the ground" is *your* assumption, and
not a claim of mine. A flight path parallel to the ground, but with a high
AOA is Todd's assumption, and again not a claim of mine. I have no problem
with these scenarios, they just have nothing to do with what I've written.

[...]
What WILL happen is that as the aircraft slows, the pitch angle of
the aircraft will need to be increased, so as to continually
increase the angle-of-attack of the wing.

We are describing the same phenomena from two perspectives.

I am fairly certain we're not.

I am fairly certain that you are confused about this.

In the context
of my usage, if one maintains the pitch angle as the aircraft slows,
the AOA will continually increase (normally, the pitch angle changes
as the aircraft slows).

You cannot "maintain the pitch angle as the aircraft slows" without
touching the runway. If the aircraft slows and the pitch angle is
not changed, lift is reduced and the airplane will descend onto the
runway.

That is NOT incongrous with what I've written. The aircraft "descends"
because the AOA increases as the aircraft slows and the pitch angle is
maintained. Do you *really* disagree with this?

I'll say it again: the scenario you propose is an impossibility.

And, I'll say it again, you are confused about what I've written. You've
repeatedly attributed claims to me that I never made, and set up straw men
that have nothing to do with the issue at hand.

Neil

"Neil Gould" wrote in message
.. .
This makes *no* reference to the amount of lift that a stalled wing
provides beyond it being inadequate to support flight,

Interestingly enough, a stall occurs at the critical angle of attack, which
is the AOA at which the coefficient of lift is the *maximum possible*. Just
past the critical angle of attack (that is, further into the stall), the
lift coefficient is no longer maximal, but is still well above what it is in
ordinary cruise flight.

What *does* happen just past the critical AOA--that is, just into the
stall--is *not* that there's insufficient lift to support the plane's
weight, but rather that there's a loss of *vertical damping*. John Denker (a
physicist and a pilot) has a nice explanation he
http://www.av8n.com/how/htm/vdamp.ht...rtical-damping.

Wrong. "To be a scalar" it needs to be a single value. And it is.
Angle-of-attack is just an angle. A single value.

Wrong. A scalar can not contain elements of direction by definition. Ergo,
AOA has no meaning as a scalar.

No, an angle is unquestionably a scalar, not a vector. Check any
introductory math text. If an angle were a vector, then a symbol
representing an angle would be set in boldface; but it is not.

You're right that an angle is defined by reference to vectors, but so is
(for example) the *dot product* of two vectors (yet the dot product is a
scalar); or so is the *magnitude* of a vector (but the magnitude is a
scalar). So being defined by reference to vectors does not preclude a
quantity from being scalar.

--Gary

Recently, Gary Drescher posted:

"Neil Gould" wrote in message
.. .
This makes *no* reference to the amount of lift that a stalled wing
provides beyond it being inadequate to support flight,

Interestingly enough, a stall occurs at the critical angle of attack,
which is the AOA at which the coefficient of lift is the *maximum
possible*. Just past the critical angle of attack (that is, further
into the stall), the lift coefficient is no longer maximal, but is
still well above what it is in ordinary cruise flight.

What *does* happen just past the critical AOA--that is, just into the
stall--is *not* that there's insufficient lift to support the plane's
weight, but rather that there's a loss of *vertical damping*. John
Denker (a physicist and a pilot) has a nice explanation he
http://www.av8n.com/how/htm/vdamp.ht...rtical-damping.

Thanks for that! I'm not sure that resolves the issue of trying to land
with these parameters.

Wrong. "To be a scalar" it needs to be a single value. And it is.
Angle-of-attack is just an angle. A single value.

Wrong. A scalar can not contain elements of direction by definition.
Ergo, AOA has no meaning as a scalar.

No, an angle is unquestionably a scalar, not a vector. Check any
introductory math text. If an angle were a vector, then a symbol
representing an angle would be set in boldface; but it is not.

Mathematically, an "angle" by itself *is* a scalar, and I'm not arguing
otherwise. I'm saying that "Angle Of Attack" requires direction to have
meaning. Without direction, there is no AOA.

You're right that an angle is defined by reference to vectors, but so
is (for example) the *dot product* of two vectors (yet the dot
product is a scalar); or so is the *magnitude* of a vector (but the
magnitude is a scalar). So being defined by reference to vectors does
not preclude a quantity from being scalar.

Well, OK. Then, how do you determine the AOA when the aircraft is parked?
If the component of direction is inseparable from the definition of AOA,
how can it be a scalar?

Neil

Mathematically, an "angle" by itself *is* a scalar, and I'm not arguing
otherwise. I'm saying that "Angle Of Attack" requires direction to have
meaning. Without direction, there is no AOA.
[...]

Well, OK. Then, how do you determine the AOA when the aircraft is parked?
If the component of direction is inseparable from the definition of AOA,
how can it be a scalar?

How long is it between 11 AM and 1:30 PM? The answer is 150 minutes.
Now, is that 150 minutes AM or 150 minutes PM?

In order to form the difference, you NEED to know whether the original
times are AM or PM, but the result is a pure number of minutes. 11 AM
is a =time of day=, 1:30 pm is a =time of day= but the difference is
=not= a time of day, it is just a number (of minutes).

In a similar vein, in order to form an angle (a pure number), you need
to have not just one direction, but two. You need TWO quantites that
have direction (they don't even have to have magnitude!). However, the
result (the angle between them) has no direction (beyond the algebraic
sign).

When you say:

I'm saying that "Angle Of Attack" requires direction to have
meaning.

all you're really saying is that you don't have an angle of attack if
you don't have the requisite components (a relative wind, and a chord).
But don't confuse the components with the result. Area is made up of
length and width, but area is not in itself one dimesional. Cakes are
made with raw eggs and flour, but I'm not likely to confuse the two any
time soon.

Jose
--
"Never trust anything that can think for itself, if you can't see where
it keeps its brain."
(chapter 10 of book 3 - Harry Potter).
for Email, make the obvious change in the address.

Recently, Jose posted:

Mathematically, an "angle" by itself *is* a scalar, and I'm not
arguing otherwise. I'm saying that "Angle Of Attack" requires
direction to have meaning. Without direction, there is no AOA.
[...]

Well, OK. Then, how do you determine the AOA when the aircraft is
parked? If the component of direction is inseparable from the
definition of AOA, how can it be a scalar?

[...]

When you say:

I'm saying that "Angle Of Attack" requires direction to have
meaning.

all you're really saying is that you don't have an angle of attack if
you don't have the requisite components (a relative wind, and a
chord).

I'm not sure that I follow your analogies, here, Jose. But, it may be a
good idea for you to look up the definition of "scalar". It *can not*
include a directional component. Conversely, AOA can not exist without
one.

Neil

Neil Gould wrote:

But, it may be a
good idea for you to look up the definition of "scalar". It *can not*
include a directional component. Conversely, AOA can not exist without
one.

Angle of attack does not "include a directional component". It is just an angle,
which is a scalar quantity.

You have evidently looked up the dictionary definition of scalar, and you read
it, but you didn't understand it.

Dave

I'm not sure that I follow your analogies, here, Jose.

The analogy is merely that you can use one kind of quantity to derive
another kind of quantity. You can use eggs to derive cake, you can use
"time of day" to derive "time", you can use length to derive area, and
you can use vectors to derive scalars.

The simplest example, I suppose, is a ratio. Fifteen kilograms is THREE
times as much as five kilograms. Fifteen inches is THREE times as much
as five inches. The "three" in both cases is the same - it is a pure
scalar quantity. It is the same "three" as the number of fingers on my
hand that are surrounded by other fingers and the number of days in a
long weekend.

Fifteen kilograms is =not= three times as much as five inches. The
units are important when =deriving= the result, but once the result is
correctly derived, it has its own units (or lack of them).

Similarly, two vectors can intersect at an angle. The angle is not a
vector, it is a scalar. As an aside, two vectors (of the same units)
can also define an area; that area is not a vector, it is a scalar (with
units of square fubars, where "fubars" are the unit both vectors are
measured in).

A vector has magnitude and direction. AOA has no direction in and of
itself. To see this, imagine a wing chord which is inclined three
degrees (the leading edge higher) from some reference plane (say, the
fuselage), and a relative wind which is blowing up from ahead and
underneath at an angle of eighteen degrees to that same fuselage, at
seventy knots. This is typical of an approach in a light aircraft.

What is the angle of attack? To be a scalar, it would have just
magnitude (which could include an algebraic sign). To be a vector, it
would have to have magnitude AND direction.

In this case, the angle of attack is twenty-one degrees. It is the
difference between the two angles given (with reference to the same
fuselage). There is no "direction" to this angle (except perhaps an
algebraic sign). So it is not a vector.

One source of confusion arises because in other contexts angles are also
used to define direction, for example wind velocity is a vector whose
angle is a direction component, not a magnigude component. For example,
"zero three zero at ten knots" is a vector, where the magnitude part is
ten knots, and the direction part is 30 degrees East of North. However,
if you put a weight on an old fashioned butcher scale, the pointer moves
through some angle. That angle does =not= represent a direction, it is
a magnitude only, and thus a scalar (related to the weight of the meat
put in the pan). And if you weigh two cuts of meat, note the angles of
the pointer for each weighing, and subtract those angles, the result is
also an angle - a magnitude with no direction component. This is a scalar.

So, depending on context, angles can be magnitudes =or= directions, but
not both at once.

Jose
--
"Never trust anything that can think for itself, if you can't see where
it keeps its brain."
(chapter 10 of book 3 - Harry Potter).
for Email, make the obvious change in the address.

"Neil Gould" wrote
Mathematically, an "angle" by itself *is* a scalar, and I'm not arguing
otherwise. I'm saying that "Angle Of Attack" requires direction to have
meaning. Without direction, there is no AOA.

Neil, give us an example of AOA having a "direction".

Well, OK. Then, how do you determine the AOA when the aircraft is parked?

When parked with no wind, there is no relative wind and therefore NO AOA.

Bob Moore

Recently, Bob Moore posted:

"Neil Gould" wrote
Mathematically, an "angle" by itself *is* a scalar, and I'm not
arguing otherwise. I'm saying that "Angle Of Attack" requires
direction to have meaning. Without direction, there is no AOA.

Neil, give us an example of AOA having a "direction".

Well, OK. Then, how do you determine the AOA when the aircraft is
parked?

When parked with no wind, there is no relative wind and therefore NO
AOA.

Stated another way, AOA doesn't exist *without* a directional component.

Neil