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  #1  
Old June 17th 05, 12:10 PM
Neil Gould
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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


  #2  
Old June 17th 05, 02:21 PM
Jose
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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.
  #3  
Old June 17th 05, 03:39 PM
Neil Gould
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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



  #4  
Old June 17th 05, 03:55 PM
Dave Butler
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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
  #5  
Old June 17th 05, 04:34 PM
Neil Gould
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Recently, Dave Butler posted:

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.

If what you think is true, then it is possible to determine the AOA when
the aircraft is parked. Do so, and I'll revise my thinking. The wonderful
thing about this level of mathematics is that it is not ambiguous. If any
usage results in a violation of the definition, then the usage is wrong,
period.

Neil


  #6  
Old June 17th 05, 04:58 PM
Dave Butler
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Neil Gould wrote:

If what you think is true, then it is possible to determine the AOA when
the aircraft is parked. Do so, and I'll revise my thinking. The wonderful
thing about this level of mathematics is that it is not ambiguous. If any
usage results in a violation of the definition, then the usage is wrong,
period.


I don't give a flip whether you revise your thinking or not. Your loss.
  #7  
Old June 17th 05, 06:26 PM
Neil Gould
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Recently, Dave Butler posted:

Neil Gould wrote:

If what you think is true, then it is possible to determine the AOA
when the aircraft is parked. Do so, and I'll revise my thinking. The
wonderful thing about this level of mathematics is that it is not
ambiguous. If any usage results in a violation of the definition,
then the usage is wrong, period.


I don't give a flip whether you revise your thinking or not. Your
loss.

Not really.

Have a nice weekend, and fly safely.

Neil



  #8  
Old June 17th 05, 04:40 PM
Jose
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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.
  #9  
Old June 17th 05, 06:56 PM
Neil Gould
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Recently, Jose posted:
[...]
A vector has magnitude and direction. AOA has no direction in and of
itself.

[...]
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.

However, it is valid for a vector to have a magnitude of zero. It is NOT
valid for a scalar to have a directional component, and it is meaningless
to have an AOA with no directional component and magnitude (e.g. parked
aircraft have no AOA). Ergo, to have an AOA, you *must* also have velocity
(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).

The two aspects of the AOA is referenced to the wing chord and relative
wind, not the fuselage.

There is no "direction" to this angle (except perhaps an
algebraic sign). So it is not a vector.

I'd say that it is often "OK" to PRESUME the directional components and
IGNORE their value if they are unimportant to usages where only the angle
is needed. But, that's quite a different situation than calling AOA
something it can't be by definition.

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

We're not talking about generic "angles", but an "Angle Of Attack", i.e.,
a specific usage which is defined by and inseparable from the components
of motion (aka relative wind). Without those components, AOA doesn't
exist.

Neil


  #10  
Old June 18th 05, 12:35 AM
Jose
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it is valid for a vector to have a magnitude of zero.

Correct.

It is NOT
valid for a scalar to have a directional component


Correct.

and it is meaningless
to have an AOA with no directional component and magnitude


Incorrect. I can give you many examples of such AOAs. Can you give me
an example of an AOA that =itself= has a direction and magnitude? (Not
that it's derived from things that have direction and magnitude, but
that it, =itself= has such)

The two aspects of the AOA is referenced to the wing chord and relative
wind, not the fuselage.


The two aspects of the AOA are referenced to each other. I refereneced
them to the same other thing (fuselage) and then derived their relation
to each other.

I'd say that it is often "OK" to PRESUME the directional components and
IGNORE their value if they are unimportant to usages where only the angle
is needed.


It is not OK to presume anything in math. Things are what they are
defined to be.

You might be thinking of "unit vectors" in which case a magnitude of one
is used, but they are defined that way. Or you might be thinking of the
algebraic sign (which is part of a scalar quantity).

We're not talking about generic "angles", but an "Angle Of Attack"


An angle of attack =is= an angle. All angles are scalars. Therefore,
an angle of attack is a scalar. Which part of this do you disagree with?

i.e.,
a specific usage which is defined by and inseparable from the components
of motion (aka relative wind).


Defined by, yes. Inseperable from, no.

The price to earnings ratio (PE) of a stock is =defined by= the dollar
price of a stock, and the dollar earnings of the company divided by the
number of shares outstanding. Without those components, you don't have
a PE ratio. But the PE is a pure number. It is not a dollar amount.

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.
 




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