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'[TECH] How planes fly'
2011\01\26@135955 by Vitaliy

face
flavicon
face
Olin Lathrop wrote:
> Planes already fly by "using the air".  As the plane passes thru some air,
> it pushes it down, which in turn pushes the plane up to counter gravity.

So you're in the Newton camp? :)  I'd love to hear you debate my Bernoulli friend.

Vitaliy

2011\01\26@153154 by Olin Lathrop

face picon face
Vitaliy wrote:
>> Planes already fly by "using the air".  As the plane passes thru
>> some air, it pushes it down, which in turn pushes the plane up to
>> counter gravity.
>
> So you're in the Newton camp? :)

You make that sound like a bad thing somehow.  Since what we're talking
about here is all much much less than the speed of light, Newton's laws
apply.

> I'd love to hear you debate my
> Bernoulli friend.

Bernoulli may explain how a wing causes the pressures around it and how it
causes the air to move.  However, none of the "how" can change the basic
force ballance.  To be clear, we're talking about a plane in steady level
flight thru (originally) still air.

Since there is no net accelleration, all the external forces must sum to 0.
Taking the vertical component of that, there is gravity pulling the plane
down.  There must therefore be a equal and opposite force pushing it up
since it's not falling out of the sky.  The only thing the plane has to push
on is the air it flys thru.  It follows that it is therefore pushing this
air down.  If you know the weight of the plane, you can go further and
compute the downward momentum the plane is imparting on the air per unit
time.  That means you can figure out how many (Kg)(m/s) it must cause air to
move dowwards every second.

Note that Newton only says what the (Kg)(m/s) product must be, not any
specific Kg and m/s combination.  In fact, this is a one of the tradeoffs
the airplane designer can make.  While Newton doesn't care about the Kg
versus m/s tradeoff, whoever is paying for the fuel does.  The energy
required to get this momentum is proportional to (m/s)**2.  Ideally
therefore you'd like to move lots of mass at low velocity.  This basic
physics explains why designs where power efficiency is important have wide
wings.  The wide wing pushes more Kg down, thereby not needing to push them
down with as many m/s.

Think of this another way.  Planes must impart momentum on the air to stay
up.  The velocity component of that momentum however represents energy left
in the air as a byproduct of producing the necessary momentum.  Nothing in
the basic free body analisys says a plane has to expend energy to stay up,
just that we don't know of a way to avoid it in realizable airplanes.

It is also interesting to look at that energy per unit time imparted on the
air.  Energy per time is power, and that is a minimum power that must be
expended to keep the plane flying.  Since power in airplanes is generally
applied as a force in the direction of flight, this power can be converted
to a force given the plane's speed.  This force is opposite the direction of
motion, and therefore looks just like drag.  In other words, planes with
finite wings have a minimum drag at each speed regardless of any actual
friction with the air.  Even if you could eliminate all lateral friction
between the plane's surface and the air, this apparent drag would still be
there.

Anyway, back to your comment about Bernoulli.  None of this says *how* the
plane causes air to be pushed down, only that it must somehow do so.
Bernoulli may be useful in understanding how a wing achieves this.  In the
end however, everything must happen within the basic force and momentum
ballance according to Newton.  Often it is useful to look at the same thing
from different ways, but you still can't violate the basic principles I
described above.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000

2011\01\26@160335 by Ing. Marcelo Fornaso

picon face
Actually, planes with wings with aerodynamic profiles are pulled up by vacuum (in a way to speak), more than pushing air down.

Regards,

Marcelo Fornaso



________________________________
From: Olin Lathrop <spam_OUTolin_piclistTakeThisOuTspamembedinc.com>
To: Microcontroller discussion list - Public. <.....piclistKILLspamspam@spam@mit.edu>
Sent: Wed, January 26, 2011 5:32:46 PM
Subject: Re: [TECH] How planes fly

Vitaliy wrote:
>> Planes already fly by "using the air".  As the plane passes thru
>> some air, it pushes it down, which in turn pushes the plane up to
>> counter gravity.
>
> So you're in the Newton camp? :)

You make that sound like a bad thing somehow.  Since what we're talking
about here is all much much less than the speed of light, Newton's laws
apply.

> I'd love to hear you debate my
> Bernoulli friend.

Bernoulli may explain how a wing causes the pressures around it and how it
causes the air to move.  However, none of the "how" can change the basic
force ballance.  To be clear, we're talking about a plane in steady level
flight thru (originally) still air.

Since there is no net accelleration, all the external forces must sum to 0.
Taking the vertical component of that, there is gravity pulling the plane
down.  There must therefore be a equal and opposite force pushing it up
since it's not falling out of the sky.  The only thing the plane has to push
on is the air it flys thru.  It follows that it is therefore pushing this
air down.  If you know the weight of the plane, you can go further and
compute the downward momentum the plane is imparting on the air per unit
time.  That means you can figure out how many (Kg)(m/s) it must cause air to
move dowwards every second.

Note that Newton only says what the (Kg)(m/s) product must be, not any
specific Kg and m/s combination.  In fact, this is a one of the tradeoffs
the airplane designer can make.  While Newton doesn't care about the Kg
versus m/s tradeoff, whoever is paying for the fuel does.  The energy
required to get this momentum is proportional to (m/s)**2.  Ideally
therefore you'd like to move lots of mass at low velocity.  This basic
physics explains why designs where power efficiency is important have wide
wings.  The wide wing pushes more Kg down, thereby not needing to push them
down with as many m/s.

Think of this another way.  Planes must impart momentum on the air to stay
up.  The velocity component of that momentum however represents energy left
in the air as a byproduct of producing the necessary momentum.  Nothing in
the basic free body analisys says a plane has to expend energy to stay up,
just that we don't know of a way to avoid it in realizable airplanes.

It is also interesting to look at that energy per unit time imparted on the
air.  Energy per time is power, and that is a minimum power that must be
expended to keep the plane flying.  Since power in airplanes is generally
applied as a force in the direction of flight, this power can be converted
to a force given the plane's speed.  This force is opposite the direction of
motion, and therefore looks just like drag.  In other words, planes with
finite wings have a minimum drag at each speed regardless of any actual
friction with the air.  Even if you could eliminate all lateral friction
between the plane's surface and the air, this apparent drag would still be
there.

Anyway, back to your comment about Bernoulli.  None of this says *how* the
plane causes air to be pushed down, only that it must somehow do so.
Bernoulli may be useful in understanding how a wing achieves this.  In the
end however, everything must happen within the basic force and momentum
ballance according to Newton.  Often it is useful to look at the same thing
from different ways, but you still can't violate the basic principles I
described above.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000

2011\01\26@161345 by N. T.

picon face
Marcelo Fornaso wrote:
> Actually, planes with wings with aerodynamic profiles are pulled up by vacuum
> (in a way to speak), more than pushing air down.

That vacuum is pushing air down. The vacuum is produced by moving
wings. Therefore "planes with wings with aerodynamic profiles are
pulled up by ... pushing air down :-) Elementary ..

2011\01\26@161508 by Mark Rages
face picon face
On Wed, Jan 26, 2011 at 3:03 PM, Ing. Marcelo Fornaso
<mfornasospamKILLspamyahoo.com> wrote:
> Actually, planes with wings with aerodynamic profiles are pulled up by vacuum
> (in a way to speak), more than pushing air down.
>
> Regards,
>
> Marcelo Fornaso
>

That sounds like the popular "equal transit time" fallacy:

http://en.wikipedia.org/wiki/Aerodynamic_lift#Other_alternative_explanations_for_the_generation_of_lift


Regards,
Mark
markrages@gmail
-- Mark Rages, Engineer
Midwest Telecine LLC
.....markragesKILLspamspam.....midwesttelecine.co

2011\01\26@164047 by mikecreid

picon face
Bernoulli's Principle!!
Sent from my Verizon Wireless BlackBerry

-----Original Message-----
From: "Ing. Marcelo Fornaso" <EraseMEmfornasospam_OUTspamTakeThisOuTyahoo.com>
Sender: piclist-bouncesspamspam_OUTmit.edu
Date: Wed, 26 Jan 2011 13:03:33 To: Microcontroller discussion list - Public.<@spam@piclistKILLspamspammit.edu>
Reply-To: "Microcontroller discussion list - Public." <KILLspampiclistKILLspamspammit.edu>
Subject: Re: [TECH] How planes fly

Actually, planes with wings with aerodynamic profiles are pulled up by vacuum (in a way to speak), more than pushing air down.

Regards,

Marcelo Fornaso



________________________________
From: Olin Lathrop <RemoveMEolin_piclistTakeThisOuTspamembedinc.com>
To: Microcontroller discussion list - Public. <spamBeGonepiclistspamBeGonespammit.edu>
Sent: Wed, January 26, 2011 5:32:46 PM
Subject: Re: [TECH] How planes fly

Vitaliy wrote:
>> Planes already fly by "using the air".  As the plane passes thru
>> some air, it pushes it down, which in turn pushes the plane up to
>> counter gravity.
>
> So you're in the Newton camp? :)

You make that sound like a bad thing somehow.  Since what we're talking
about here is all much much less than the speed of light, Newton's laws
apply.

> I'd love to hear you debate my
> Bernoulli friend.

Bernoulli may explain how a wing causes the pressures around it and how it
causes the air to move.  However, none of the "how" can change the basic
force ballance.  To be clear, we're talking about a plane in steady level
flight thru (originally) still air.

Since there is no net accelleration, all the external forces must sum to 0.
Taking the vertical component of that, there is gravity pulling the plane
down.  There must therefore be a equal and opposite force pushing it up
since it's not falling out of the sky.  The only thing the plane has to push
on is the air it flys thru.  It follows that it is therefore pushing this
air down.  If you know the weight of the plane, you can go further and
compute the downward momentum the plane is imparting on the air per unit
time.  That means you can figure out how many (Kg)(m/s) it must cause air to
move dowwards every second.

Note that Newton only says what the (Kg)(m/s) product must be, not any
specific Kg and m/s combination.  In fact, this is a one of the tradeoffs
the airplane designer can make.  While Newton doesn't care about the Kg
versus m/s tradeoff, whoever is paying for the fuel does.  The energy
required to get this momentum is proportional to (m/s)**2.  Ideally
therefore you'd like to move lots of mass at low velocity.  This basic
physics explains why designs where power efficiency is important have wide
wings.  The wide wing pushes more Kg down, thereby not needing to push them
down with as many m/s.

Think of this another way.  Planes must impart momentum on the air to stay
up.  The velocity component of that momentum however represents energy left
in the air as a byproduct of producing the necessary momentum.  Nothing in
the basic free body analisys says a plane has to expend energy to stay up,
just that we don't know of a way to avoid it in realizable airplanes.

It is also interesting to look at that energy per unit time imparted on the
air.  Energy per time is power, and that is a minimum power that must be
expended to keep the plane flying.  Since power in airplanes is generally
applied as a force in the direction of flight, this power can be converted
to a force given the plane's speed.  This force is opposite the direction of
motion, and therefore looks just like drag.  In other words, planes with
finite wings have a minimum drag at each speed regardless of any actual
friction with the air.  Even if you could eliminate all lateral friction
between the plane's surface and the air, this apparent drag would still be
there.

Anyway, back to your comment about Bernoulli.  None of this says *how* the
plane causes air to be pushed down, only that it must somehow do so.
Bernoulli may be useful in understanding how a wing achieves this.  In the
end however, everything must happen within the basic force and momentum
ballance according to Newton.  Often it is useful to look at the same thing
from different ways, but you still can't violate the basic principles I
described above.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000

2011\01\26@170308 by Olin Lathrop

face picon face
Ing. Marcelo Fornaso wrote:
> Actually, planes with wings with aerodynamic profiles are pulled up
> by vacuum (in a way to speak), more than pushing air down.

Vacuum versus pressure may describe how the forces are transferred to the
wing surfaces, but ultimately a plane has to push down air to keep itself
up.  You don't have to know anything about how wings work to show this has
to be true.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000

2011\01\26@171213 by N. T.

picon face
I wrote:
> Marcelo Fornaso wrote:
>> Actually, planes with wings with aerodynamic profiles are pulled up by vacuum
>> (in a way to speak), more than pushing air down.
>
> That vacuum is pushing air down. The vacuum is produced by moving
> wings. Therefore "planes with wings with aerodynamic profiles are
> pulled up by ... pushing air down :-) Elementary ...
>
Correction: replace "pushing" with "pulling", please

2011\01\26@171613 by Carl Denk

flavicon
face
The lift is provided by lower pressure on the top surface of the wing. The top surface is longer than the bottom surface parallel to the direction of travel. The air on the top surface is stretched since it has a longer route to travel. This stretching reduces the air density and the air molecules are further apart and there are less collisions with the wing top surface than the wing bottom surface. Then there are more collisions in the upward direction than the gravity direction, and this creates the lift. This of course is assuming the plane is in equilibrium flight in a normal upright mode.

On 1/26/2011 5:04 PM, Olin Lathrop wrote:
> Ing. Marcelo Fornaso wrote:
>    
>> Actually, planes with wings with aerodynamic profiles are pulled up
>> by vacuum (in a way to speak), more than pushing air down.
>>      
> Vacuum versus pressure may describe how the forces are transferred to the
> wing surfaces, but ultimately a plane has to push down air to keep itself
> up.  You don't have to know anything about how wings work to show this has
> to be true.
>
>
> ********************************************************************
> Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
> (978) 742-9014.  Gold level PIC consultants since 2000.
>

2011\01\26@172643 by Chris McSweeny

picon face
On Wed, Jan 26, 2011 at 9:03 PM, Ing. Marcelo Fornaso
<TakeThisOuTmfornasoEraseMEspamspam_OUTyahoo.com> wrote:
> Actually, planes with wings with aerodynamic profiles are pulled up by vacuum
> (in a way to speak), more than pushing air down.

So you're suggesting they fly without pushing any air down?

Chri

2011\01\26@174013 by Michael Watterson

face picon face
On 26/01/2011 22:12, N. T. wrote:
> I wrote:
>> Marcelo Fornaso wrote:
>>> Actually, planes with wings with aerodynamic profiles are pulled up by vacuum
>>> (in a way to speak), more than pushing air down.
>> That vacuum is pushing air down. The vacuum is produced by moving
>> wings. Therefore "planes with wings with aerodynamic profiles are
>> pulled up by ... pushing air down :-) Elementary ...
>>
> Correction: replace "pushing" with "pulling", please.

I know little of this, But I thought in ordinary flight it's Mr Newton the opposite and equal force caused by deflecting the airflow. Even a flat wafer will work (poorly). Nice Aerodynamic stuff reduces drag, turbulence and such I thought? Same as Sail advances on boats?

But an ekranoplane, Ground effect Aircraft may be using Bernoulli effects. Like a giant bloated disk head. :-) Certainly not working like a hovercraft, hydrofoil or regular plane.

http://www.airforce.ru/en/ekranoplanes/ekranoplanes.htm

Military value may be close to zero as a Transport if it's spotted

2011\01\26@174812 by Olin Lathrop

face picon face
Carl Denk wrote:
>> Vacuum versus pressure may describe how the forces are transferred
>> to the wing surfaces, but ultimately a plane has to push down air to
>> keep itself up.  You don't have to know anything about how wings
>> work to show this has to be true.
>
> The lift is provided by lower pressure on the top surface of the wing.

And to a lesser extent higher than ambient pressure on the bottom surface.

> The top surface is longer than the bottom surface parallel to the
> direction of travel. The air on the top surface is stretched since it
> has a longer route to travel.

This is a commonly quoted misconception.

> This stretching reduces the air density
> and the air molecules are further apart and there are less collisions
> with the wing top surface than the wing bottom surface. Then there are
> more collisions in the upward direction than the gravity direction,
> and this creates the lift. This of course is assuming the plane is in
> equilibrium flight in a normal upright mode.

I'm not sure what you're overall point is.  You seem to be trying to
describe how the lift forces on the wing are generated, but how ever they
are generated it doesn't change the fact that ultimately a plane stays up by
pushing down on air.  There are all kinds of nuances in how a wind actually
accomplishes this, but none of them can negate the basic physics of force
and momentum ballance.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000

2011\01\26@175432 by Olin Lathrop

face picon face
Michael Watterson wrote:
> But an ekranoplane, Ground effect Aircraft may be using Bernoulli
> effects. Like a giant bloated disk head. :-) Certainly not working
> like a hovercraft, hydrofoil or regular plane.

That's a bit different.  In EE terms it's sortof relying on "near field"
effects.  What I was talking about was when the plane was flying well above
the ground.

In your case the ground is so close that the plane pushes on the air, the
ground pushes back on the air, and this effect gets back to the plane in
time for it to "feel" it before it goes past.  Basically the ground stops
the air going down causing its pressure to increase, which in turn pushes up
more on the plane than just the momentum of pushing the air down would if
the plane were higher.  This "ground effect" gets rather complicated.


********************************************************************
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(978) 742-9014.  Gold level PIC consultants since 2000

2011\01\26@180056 by N. T.

picon face
Olin Lathrop wrote:
> I'm not sure what you're overall point is.  You seem to be trying to
> describe how the lift forces on the wing are generated, but how ever they
> are generated it doesn't change the fact that ultimately a plane stays up by
> pushing down on air.

I think he is trying to state that a plane hangs on by pulling the air
down, not "stays up", by "pushing down on air". Probably you should
have used "moving air down", not "pushing air down". But that's
absolutely not important in the context as for me.

2011\01\26@181447 by Michael Watterson

face picon face
On 26/01/2011 22:55, Olin Lathrop wrote:
> Michael Watterson wrote:
>> >  But an ekranoplane, Ground effect Aircraft may be using Bernoulli
>> >  effects. Like a giant bloated disk head.:-)  Certainly not working
>> >  like a hovercraft, hydrofoil or regular plane.
> That's a bit different.  In EE terms it's sortof relying on "near field"
> effects.  What I was talking about was when the plane was flying well above
> the ground.
>
> In your case the ground is so close that the plane pushes on the air, the
> ground pushes back on the air, and this effect gets back to the plane in
> time for it to "feel" it before it goes past.  Basically the ground stops
> the air going down causing its pressure to increase, which in turn pushes up
> more on the plane than just the momentum of pushing the air down would if
> the plane were higher.  This "ground effect" gets rather complicated.
>
Agreed

The "ordinary" plane work as you describe, the wing simply deflects the airflow. Since it's changed direction there must be an opposite and equal force vector, which is why there is lift. this is why the flaps must be changed with speed and to stop rising.

I only mentioned the "ekranoplane" as an example of something that looks very like a plane, but is actually not working the same way.

I believe in days of Flying boats (from Foynes down the road here on the Shannon Estuary) across the Atlantic that it the weather was good they did fly close to sea to get "ground effect" and save fuel, but not to the extreme of the Lun. Probably a bad idea in an A380 or 747!

Lun has 8 scary engines just behind pilot

http://www.schneiderism.com/tag/lun-ekranoplan/

http://igor113.livejournal.com/51213.html

2011\01\26@181703 by Oli Glaser

flavicon
face
On 26/01/2011 23:00, N. T. wrote:
> Olin Lathrop wrote:
>> I'm not sure what you're overall point is.  You seem to be trying to
>> describe how the lift forces on the wing are generated, but how ever they
>> are generated it doesn't change the fact that ultimately a plane stays up by
>> pushing down on air.
> I think he is trying to state that a plane hangs on by pulling the air
> down, not "stays up", by "pushing down on air". Probably you should
> have used "moving air down", not "pushing air down". But that's
> absolutely not important in the context as for me.
>

I think we are beginning to split hairs here - if one thing goes one way, then something generally goes the other way too. Out of interest - what do you see is the distinction between "pushing air down" and "moving air down"?

2011\01\26@182746 by Carl Denk

flavicon
face
Every action has a reaction. It's all relative (what you measure pressure or force) with respect to some other point. The sum of the vertical forces must equal zero, assuming stable (no acceleration)  flight. Once one gets out of the local effects of the air movement, all the air pressure is equal with no air movement. Draw a free body diagram (a physics exercise).Therefore the forces of holding the plane up  must be contained within that local area since there is no air movement. The air does get disrupted by the passing plane. With a plane (could be at least as big as a DC-9) passing under a large plane, you can pass at right angle directly under that plane by 100' with no issues, but the wake turbulence (wing tip vortices) settle earthward at approximately 300 - 1000 feet per minute (I forget the close number). A DC-9 got into that area 3 miles behind a larger aircraft, and did a un-commanded 180 degree roll.

The ground effect is effective for the wing being closer to the ground than 1/2 the wing span and yeas then the air is getting squeezed between the air and the ground. This starts to define the local effects I discussed above, and I'm not to try to define any further, get into the mechanics of the wig tip vortices.

On 1/26/2011 5:26 PM, Chris McSweeny wrote:
> On Wed, Jan 26, 2011 at 9:03 PM, Ing. Marcelo Fornaso
> <RemoveMEmfornasospamTakeThisOuTyahoo.com>  wrote:
>    
>> Actually, planes with wings with aerodynamic profiles are pulled up by vacuum
>> (in a way to speak), more than pushing air down.
>>      
> So you're suggesting they fly without pushing any air down?
>
> Chris
>

2011\01\26@183941 by RussellMc

face picon face
part 1 890 bytes content-type:text/plain; charset="iso-8859-1" (decoded quoted-printable)

> Actually, planes with wings with aerodynamic profiles are pulled up by vacuum
> (in a way to speak), more than pushing air down.

Planes with wings *without* aerodynamic profiles would be interesting
to see :-).

For some values of "wings" and "planes" these people make planes with
wings without aerodynamic profiles.
And, they are indeed interesting to see:

      http://www.armadilloaerospace.com/n.x/Armadillo/Home

                    http://media.armadilloaerospace.com/2004_05_30/flight_3.jpg

http://media.armadilloaerospace.com/2004_08_08/2004_08_07_e.jpg

http://media.armadilloaerospace.com/2005_09_24/2005_09_24_e.jpg


Attached - not your average factory carpark.


FWIW a rectangular cross section plate (say a plank of "12 x 1") can
be used to generate lift.



               Russell


part 2 15335 bytes content-type:image/jpeg; name="RROC_Armadillo_X_testing_J70.jpg" (decode)


part 3 181 bytes content-type:text/plain; name="ATT00001.txt"
(decoded base64)

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2011\01\26@185607 by Carl Denk

flavicon
face


On 1/26/2011 6:39 PM, RussellMc wrote:
>> Actually, planes with wings with aerodynamic profiles are pulled up by vacuum
>> (in a way to speak), more than pushing air down.
>>      
> Planes with wings *without* aerodynamic profiles would be interesting
> to see :-).
>
>    More typically aerobatic (planes designed to do stunts including flying upside down) have symmetric wings, with same top and bottom profiles. The difference of length of the top and bottom  surfaces is created when the wing impact point on the leading edge is moved down by increasing the angle off attack (move the front up relative to the trailing edge)

2011\01\26@190829 by Chris McSweeny

picon face
On Wed, Jan 26, 2011 at 11:56 PM, Carl Denk <cdenkEraseMEspam.....windstream.net> wrote:
>
>
> On 1/26/2011 6:39 PM, RussellMc wrote:
>>> Actually, planes with wings with aerodynamic profiles are pulled up by vacuum
>>> (in a way to speak), more than pushing air down.
>>>
>> Planes with wings *without* aerodynamic profiles would be interesting
>> to see :-).
>>
>>
> More typically aerobatic (planes designed to do stunts including flying
> upside down) have symmetric wings, with same top and bottom profiles.
> The difference of length of the top and bottom  surfaces is created when
> the wing impact point on the leading edge is moved down by increasing
> the angle off attack (move the front up relative to the trailing edge).

Aerodynamic and symmetric aren't mutually exclusive.

Chris

2011\01\26@192400 by John Gardner

picon face
> I believe in days of Flying boats (from Foynes down the road here on the
Shannon Estuary) across the Atlantic that it the weather was good they
did fly close to sea to get "ground effect" and save fuel...

That's interesting. Any idea where you ran across this?

 Jac

2011\01\26@193811 by RussellMc

face picon face
>>> Planes with wings *without* aerodynamic profiles would be interesting
>>> to see :-).

>> More typically aerobatic (planes designed to do stunts including flying
>> upside down) have symmetric wings, with same top and bottom profiles.
>> The difference of length of the top and bottom  surfaces is created when
>> the wing impact point on the leading edge is moved down by increasing
>> the angle off attack (move the front up relative to the trailing edge).

> Aerodynamic and symmetric aren't mutually exclusive.

Indeed.
Examination of he many NACA profiles is informative - they include
things that look disturbingly  close to rounded fence posts. And a lot
of other interesting shapes as well.

I have an experimental "wind turbine" (to give it a glorified name)
with a 1 metre+ prop with a fully rounded rectangular blade section.
More similar due sometime soon.

Various NACA and related  examples:

http://www.google.co.nz/images?q=naca+wing+profiles&hl=en&prmd=ivns&source=lnms&tbs=isch:1&ei=xbtATbeWJpCusAPfxN3jCg&sa=X&oi=mode_link&ct=mode&cd=2&sqi=2&ved=0CAoQ_AUoAQ&biw=1066&bih=754

Roll your own flying fencepost::

Set this to eg 2 0 33 400 1 0 400 0 0 for an example
More fencpostish is easy but the screen isn't disposed to dealing with it well.

                        http://www.ppart.de/aerodynamics/profiles/NACA4.html

Also:

http://en.wikipedia.org/wiki/NACA_airfoil
http://www.pdas.com/avd.html
www.desktop.aero/appliedaero/airfoils1/airfoilgeometry.html
http://www.allstar.fiu.edu/aero/wing31.htm

4 parameter  NACA profile formula:

           upload.wikimedia.org/math/4/0/a/40aaa61a06109d113af58b7e571285a9.png
Part of http://en.wikipedia.org/wiki/NACA_airfoil

2011\01\26@193931 by RussellMc

face picon face
>> I believe in days of Flying boats (from Foynes down the road here on the
> Shannon Estuary) across the Atlantic that it the weather was good they
> did fly close to sea to get "ground effect" and save fuel...

> That's interesting. Any idea where you ran across this?

Probably about 3 metres ASL.


2011\01\27@002704 by Ing. Marcelo Fornaso

picon face
Yes. Lots of different shapes can be used to generate lift. I'm sure you understood I was referring to efficient airfoils.
And what I wanted to stress in may bad english is that the effects produced in the top (upper camber?) are far more important than those produced in the lower camber.
Of course all the effects (even at the top of the wings) result in air moving down.

Thank you for your comments.

Marcelo Fornaso



________________________________
From:RussellMc <EraseMEapptechnzspamgmail.com>
To:Microcontroller discussion list - Public. <RemoveMEpiclistEraseMEspamEraseMEmit.edu>
Sent:Wed, January 26, 2011 8:39:00 PM
Subject:Re: [TECH] How planes fly

> Actually, planes with wings with aerodynamic profiles are pulled up by vacuum
> (in a way to speak), more than pushing air down.

Planes with wings *without* aerodynamic profiles would be interesting
to see :-).

For some values of "wings" and "planes" these people make planes with
wings without aerodynamic profiles.
And, they are indeed interesting to see:

     http://www.armadilloaerospace.com/n.x/Armadillo/Home

                   http://media.armadilloaerospace.com/2004_05_30/flight_3..jpg

http://media.armadilloaerospace.com/2004_08_08/2004_08_07_e.jpg

http://media.armadilloaerospace.com/2005_09_24/2005_09_24_e.jpg


Attached - not your average factory carpark.


FWIW a rectangular cross section plate (say a plank of "12 x 1") can
be used to generate lift.



               Russell



2011\01\27@005533 by N. T.

picon face
Oli Glaser wrote:
> On 26/01/2011 23:00, N. T. wrote:
>> Olin Lathrop wrote:
>>> I'm not sure what you're overall point is.  You seem to be trying to
>>> describe how the lift forces on the wing are generated, but how ever they
>>> are generated it doesn't change the fact that ultimately a plane stays up by
>>> pushing down on air.
>> I think he is trying to state that a plane hangs on by pulling the air
>> down, not "stays up", by "pushing down on air". Probably you should
>> have used "moving air down", not "pushing air down". But that's
>> absolutely not important in the context as for me.
>>
>
> I think we are beginning to split hairs here - if one thing goes one
> way, then something generally goes the other way too.

Let's leave the statement to philosophers :-)


> Out of interest - what do you see is the distinction between "pushing
> air down" and "moving air down"?

A wing can develop lift even if its bottom surface is at zero angle of
attack. This way the lift develops by upper surface of the wing. It's
only natural to say the upper surface pulls the air downwards, as the
low pressure region would make air to move towards the wing. That's
why I'd say "a wing moves air down" to describe both cases when air is
either pushed or pulled.

Normally, the language nuances don't make much sense, but as the
discussion goes as far as into all those lower/higher pressure things,
I'd use words that would better correspond to the physics, just to
avoid misunderstandings in the discussion.

2011\01\27@021848 by Oli Glaser

flavicon
face
On 27/01/2011 05:55, N. T. wrote:
> Oli Glaser wrote:
>> On 26/01/2011 23:00, N. T. wrote:
>>> Olin Lathrop wrote:
>>>> I'm not sure what you're overall point is.  You seem to be trying to
>>>> describe how the lift forces on the wing are generated, but how ever they
>>>> are generated it doesn't change the fact that ultimately a plane stays up by
>>>> pushing down on air.
>>> I think he is trying to state that a plane hangs on by pulling the air
>>> down, not "stays up", by "pushing down on air". Probably you should
>>> have used "moving air down", not "pushing air down". But that's
>>> absolutely not important in the context as for me.
>>>
>> I think we are beginning to split hairs here - if one thing goes one
>> way, then something generally goes the other way too.
> Let's leave the statement to philosophers :-)
>
>

I think that's a very good idea.. :-)

{Quote hidden}

Ah, I see what you meant now (should have seen it before if I'd taken the context into account - read too quickly)
I thought you were referring to the air on the underside of the wing rather than the air above the wing - somehow I "missed" the first line of your reply...
Certainly easy to get lost in the nuances of it all, with many ways to describe what is essentially the same thing.

2011\01\27@034055 by Michael Watterson

face picon face
On 27/01/2011 00:23, John Gardner wrote:
>> I believe in days of Flying boats (from Foynes down the road here on the
> Shannon Estuary) across the Atlantic that it the weather was good they
> did fly close to sea to get "ground effect" and save fuel...
>
> That's interesting. Any idea where you ran across this?
>
>    Jack
I don't remember where I read it.

Maybe in the Museum at Foynes

http://www.se-technology.com/wig/html/main.php?open=aero

Boeing 314 was the Clipper

http://www.flyingboatmuseum.com/b314.html

2011\01\27@034701 by Michael Watterson

face picon face
On 27/01/2011 05:27, Ing. Marcelo Fornaso wrote:
> I'm sure you understood I was referring to efficient airfoils.
> And what I wanted to stress in may bad english is that the effects produced in
> the top (upper camber?) are far more important than those produced in the lower
> camber.
> Of course all the effects (even at the top of the wings) result in air moving
> down.
>
you don't need an aerofoil. A flat plate will work. Just not efficiently or very stabl

2011\01\27@041445 by Tamas Rudnai

face picon face
On Thu, Jan 27, 2011 at 5:27 AM, Ing. Marcelo Fornaso <RemoveMEmfornasospam_OUTspamKILLspamyahoo.com>wrote:

> Yes. Lots of different shapes can be used to generate lift.
>

Exactly! Even a golf ball is generating enough lift force when it has plenty
of back spin (Magnus effect). This phenomenal was used even in a weird plane
design which called FanWing: http://www.fanwing.com/index.htm.

Tamas




{Quote hidden}

>

2011\01\27@044128 by Pete Restall

flavicon
face
On Thu, 27 Jan 2011 12:39:00 +1300, RussellMc wrote:

> For some values of "wings" and "planes" these people make planes with
> wings without aerodynamic profiles.
> And, they are indeed interesting to see:
>
>       http://www.armadilloaerospace.com/n.x/Armadillo/Home

Ah yes, I thought Armadillo sounded familiar...  Many happy teenage
hours spent using John Carmack's more well-known offerings*, it has to
be said.  Makes me want to fire up that old DOS 6.22 box I have
upstairs...

Regards,

Pete Restall

* http://en.wikipedia.org/wiki/DOOM

2011\01\27@083827 by Olin Lathrop

face picon face
Carl Denk wrote:
> The sum of the
> vertical forces must equal zero, assuming stable (no acceleration)
> flight.

Yes.

> Once one gets out of the local effects of the air movement,
> all the air pressure is equal with no air movement.

I'm not sure the point you're trying to make, but what I think you mean is
wrong.

> Draw a free body diagram (a physics exercise).

Yes, please do.

> Therefore the forces of holding the plane up  must
> be contained within that local area since there is no air movement.

No.  A passing plane (again let's stick to steady level flight at constant
speed) imparts downward momentum on the air.  It leaves a net downdraft
behind it.  If you put the free body diagram around the earth instead of the
plane, you will see that this downdraft hitting the ground is the
counter-acting force on the earth to the gravitational attraction to the
plane.

Think back to the classic physics thought experiment of a bunch of pigeons
in a large box.  From the outside, does the box weigh less when all the
pigeons are in flight?  No, since the pigeons push down on the air, which
eventually pushes down on the box.  There can be short term variations as
the pigeons move air before it hits the box, but in the long term the
average weight is exactly the same as the weight when everything in the box
is at rest.  Mythbusters actually did the experiment, and confirmed what
everyone that paid attention in physics class already knew.

> The
> air does get disrupted by the passing plane. With a plane (could be at
> least as big as a DC-9) passing under a large plane, you can pass at
> right angle directly under that plane by 100' with no issues, but the
> wake turbulence (wing tip vortices) settle earthward at approximately
> 300 - 1000 feet per minute (I forget the close number).  A DC-9 got
> into  that area 3 miles behind a larger aircraft, and did a
> un-commanded 180 degree roll.

You are confusing anecdotes with physics.  The physics says that a plane
must leave a *net* downdraft behind it.  There will be vortexes produced at
the wing tips and other irregular structures of the plane.  Some molecules
will be moving up or sideways.  Some of these vortexes can be strong enough
to be a danger to other planes.  However, the net motion is still down, and
none of the anecdotes about the resulting turbulence disprove that.  The net
momemtum exactly counters the plane's weight.  You can see that from your
free body diagram of the plane.


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2011\01\27@084343 by Olin Lathrop

face picon face
RussellMc wrote:
> For some values of "wings" and "planes" these people make planes with
> wings without aerodynamic profiles.

Those aren't planes, they are rockets.  They use a different mechanism to
counteract the force of gravity than something that is generally understood
as a "airplaine" does.


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2011\01\27@085635 by RussellMc

face picon face
> You are confusing anecdotes with physics.

Alas, flight 587, ran into excessive levels of anecdote and didn't survive.
The US's second most lethal air crash.

Interestingly, while the crash WAS precipitated by the effects of wake
turbulence much as Carl (I think it was) described, the overall cause
as reported by the NTSB was excessive rudder control force in response
to the turbulence. They concluded that if the pilot had left things
alone instead of fighting it as he did then the craft would have
survived.

                   http://en.wikipedia.org/wiki/American_Airlines_Flight_587


 Russel

2011\01\27@090045 by William \Chops\ Westfield

face picon face

On Jan 27, 2011, at 5:39 AM, Olin Lathrop wrote:

> The physics says that a plane must leave a *net* downdraft behind it.

So does a strict Newtonist explain the action of airfoils as something  like 'redirecting air that passes over the top surface in a downward  direction, due to surface effects' or something like that, rather than  the "longer path = reduced pressure = lift" ?

BillW

2011\01\27@090826 by Olin Lathrop

face picon face
Ing. Marcelo Fornaso wrote:
> Yes. Lots of different shapes can be used to generate lift.
> I'm sure you understood I was referring to efficient airfoils.
> And what I wanted to stress in may bad english is that the effects
> produced in the top (upper camber?) are far more important than those
> produced in the lower camber.
> Of course all the effects (even at the top of the wings) result in
> air moving down.

OK, I can agree with all that.  For most airfoils actually in use, the
descriptions I've seen do show more upwards force on the wing from the top
surface than from the bottom surface.  So a wing is "sucked" up more from
the top than it is "pushed" up from the bottom.  However, as you say, all
this is only a implementation detail and doesn't change the fact that the
plane must produce a net downward push on the air as it goes by.

I've occasionaly mused about a different mechanism than a airfoil to keep a
plane up.  Think of how a pitcher attempts to manipulate the path of a
baseball by spinning it.  The classic curve ball rises more than a purely
ballistic path would.  A baseball certainly isn't a airfoil.  It generates
this lift by changing the angle of its wake due to its spinning.

Let's say you're viewing a pitch in crossection with the pitcher on the
right throwing towards home plate on the left.  In this view, a curve ball
is rotating clockwise.  The wake behind the ball has a downward angle.
Since the ball is imparting downward momentum on the air, it is receives a
net upwards force.

Now imagine a plane that instead of wings has long rotating cylinders, with
the rotation axis along the length of these cylenders, which is left to
right with respect to the plane.  Small lateral grooves in the cylinders
would probably help, just like a baseball pitcher gets different effects
depending on how he orients the spin axis relative to the laces (laces are
roughness and more laces around the equator makes a stronger lift effect).

I'm not saying any of this is a good idea or that it would be useful, but it
is fun to think about.  Maybe even fun enough to build a prototype merely to
prove the point.


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2011\01\27@091559 by RussellMc

face picon face
>> For some values of "wings" and "planes" these people make planes with
>> wings without aerodynamic profiles.

> Those aren't planes, they are rockets.  They use a different mechanism to
> counteract the force of gravity than something that is generally understood
> as a "airplaine" does.

You noticed! :-)

"... generally understood ..." nicely covers the ante-locus of the
range I had in mind when using the term "for some values of "

In fact, the big tall greenish (AFAIR) one doing a Flash Gordon
takeoff, (it's the Mixed Monoprop Lander AFAIR)(or was, that was it's
final flight - spectacular demise) would have substantial forces
acting on its aeroshell at any sort of air speed (which is why they
call it an aeroshell) and these would absolutely have to be taken
account of in flight control - as John C will be very very very aware
and as he of course handles.

A significant part of getting rockets to go the  places and ways
intended is understanding and dealing with  interestingly variable
lift to drag ratios as the machines eg transit sound-speed or progress
through various stages of hypersonic flight with simultaneously
changing atmospheric denisty and related properties.

So, while "for some values of" largely doesn't cover "things that look
like wings"  there is a very large amount of aerodynamic stuff
happening and having to be accounted for.

The frozen in (some of) our memories "Go on throttle up!" was marker
for the point of maximum dynamic pressure, termed Qbar, where
increasing pressure related to velocity cubed is overcome by the
reduction in pressure due to logarithmically reducing air density.  If
you don't rteat your system as aerodynamic (and not just draggy) then
somewhere about there is where things often come apart, usually
literally.

Note that modern (last 50 years :-) ) satellite launchers may have no
fins at all. Note that Saturn V had teensy weensy little ones that
were patently totally useless given the rockets sie. Except, they
weren't. System survival depended in part on their being there.
It may look like a cross between a flying skyscraper and a super
caber, but there's a lot of aerodynamics there as well. (If thrust had
been lost at one point in ascent it would have been touch and go
whether the airframe would have remained intact due to the sudden
change in aerodynamic forces )(so Henry asserts, and Henry is always
right*)


          Russell

* 3 to 4 9's.

2011\01\27@091901 by Olin Lathrop

face picon face
Michael Watterson wrote:
> you don't need an aerofoil. A flat plate will work.

A flat plate is a airfoil.

Actually the lift of a airfoil comes mostly from how it looks like a flat
plate.  The fancy shapes are mostly to get better stall characteristics than
a flat plate, keep the flow laminar over some regions, etc.  A flat plate
has a very abrupt stall, when looking at lift as a function of angle of
attack.

Many toy airplanes use flat plate wings and work just fine.


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2011\01\27@092051 by RussellMc

face picon face
> A baseball certainly isn't a airfoil.  I

No.
But

> It generates
> this lift by changing the angle of its wake due to its spinning.

Mayhaps         http://en.wikipedia.org/wiki/Coand%C4%83_effect


              R

2011\01\27@092958 by Djula Djarmati

flavicon
On 27-Jan-11 15:09, Olin Lathrop wrote:
{Quote hidden}

If you have a great idea, someone has already tried it, at least this is true for most of my ideas:

www.pilotfriend.com/photo_albums/potty/2.htm
http://en.wikipedia.org/wiki/Flettner_airplane

Djul

2011\01\27@093516 by alan.b.pearce

face picon face
> Now imagine a plane that instead of wings has long rotating cylinders, with
> the rotation axis along the length of these cylenders, which is left to
> right with respect to the plane.  Small lateral grooves in the cylinders
> would probably help, just like a baseball pitcher gets different effects
> depending on how he orients the spin axis relative to the laces (laces are
> roughness and more laces around the equator makes a stronger lift effect)..

Or dimples on a golf ball - another example of how a modification to something makes it fly further.

IIRC the dimples on a golf ball were another of Barnes-Wallis' ideas (him of the Dam Busters bouncing bomb).
-- Scanned by iCritical.

2011\01\27@094604 by Olin Lathrop

face picon face
'William Chops" Westfield ' <RemoveMEwestfwKILLspamspammac.com wrote:
>> The physics says that a plane must leave a *net* downdraft behind it.
>
> So does a strict Newtonist explain the action of airfoils as something
> like 'redirecting air that passes over the top surface in a downward
> direction, due to surface effects' or something like that, rather than
> the "longer path = reduced pressure = lift" ?

When looking at the airplane as a whole, the physics says nothing about how
wings or any other structure accomplishes the pushing down of air, only that
there must be a certain net downward momentum per time imparted on the air.
Trying to look at how airfoils work obscures rather than illuminates this
big picture view.

Airfoils is a complicated and legitimate subject on its own, but irrelevant
to the original point.


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2011\01\27@094943 by Olin Lathrop

face picon face
Olin Lathrop wrote:
> Let's say you're viewing a pitch in crossection with the pitcher on
> the right throwing towards home plate on the left.  In this view, a
> curve ball is rotating clockwise.  The wake behind the ball has a
> downward angle. Since the ball is imparting downward momentum on the
> air, it is receives a net upwards force.

Oops.  I re-reading this I realized I flipped curve ball and rising
fastball.  A curve ball drops more than a ballistic path, and spins
counter-clockwise in the above illustration.  A rising fastball rises more
than a ballistic path and spins clockwise.  Sorry for the confusion.

2011\01\28@125550 by Gerhard Fiedler

picon face
Olin Lathrop wrote:

> 'William Chops" Westfield ' <westfwSTOPspamspamspam_OUTmac.com wrote:
>>> The physics says that a plane must leave a *net* downdraft behind it.
>>
>> So does a strict Newtonist explain the action of airfoils as
>> something like 'redirecting air that passes over the top surface in
>> a downward direction, due to surface effects' or something like
>> that, rather than the "longer path = reduced pressure = lift" ?
>
> When looking at the airplane as a whole, the physics says nothing
> about how wings or any other structure accomplishes the pushing down
> of air, only that there must be a certain net downward momentum per
> time imparted on the air.
"Momentum per time" is what exactly, physically speaking? I don't think
you wrote this with your physicist's hat on.

Which one of Newton's rules says that air must move down to keep a plane
in steady height? And please, don't put it in fuzzy and easy to
misunderstand (both in the writer's and in the reader's mind) words --
put it in clear formulae, using proper terms used in physics.

Gerhar

2011\01\28@142641 by Olin Lathrop

face picon face
Gerhard Fiedler wrote:
>> When looking at the airplane as a whole, the physics says nothing
>> about how wings or any other structure accomplishes the pushing down
>> of air, only that there must be a certain net downward momentum per
>> time imparted on the air.
>
> "Momentum per time" is what exactly, physically speaking? I don't
> think you wrote this with your physicist's hat on.

My hat is intact, but you might want to check where your duncecap is.

Momentum per time is both correct and unambiguous.  Momentum is mass times
velocity.  Momentum per time is therefore mass times velocity per time, or
mass times accelleration, or force.  Remember, F=ma.

So "momentum per time" is a fancy way to say "force", but I chose to say it
that way to help illuminate where that force is coming from.

> Which one of Newton's rules says that air must move down to keep a
> plane in steady height?

I think he mumbled something about equal and opposite forces.

> And please, don't put it in fuzzy and easy to
> misunderstand (both in the writer's and in the reader's mind) words --
> put it in clear formulae, using proper terms used in physics.

I think I've been plenty clear enough, especially if you read back thru this
thread.  Your problem seems to be with basic physics.  Surely there is a lot
of accessible stuff out there that explains forces, momentum, velocity,
accelleration, and Newton's laws.

However, let's do a example.  Suppose a plane weighs 100 Newtons.  Let's
further stipulate we're talking about a normal airplane that uses power to
push itself forward.  In other words, the force applied by the propeller,
jet engine, or whatever, is in the forward direction, which is also the
direction of motion.  As said before, the plane is in steady level flight
flying into still air.

Gravity is pulling down on the plane with 100N.  Something else must be
pushing up on the plane with 100N since it's not accellerating.  According
to Newton's law (the equal and opposite one), this in turn means the plane
is pushing down on something.  That something is the air around it.

So the plane is pushing down on the air with a force of 100N.

 Force = 100N = 100 Kg m / s**2 = (Kg m/s) / s

Note that (Kg m/s) is momentum.  So every second, the plane has to impart a
momentum of 100 Kg m/s on the air around it, and in the downward direction
since the resulting force on the plane is up.

One of the original points many posts ago is that there is a tradeoff here
between Kg and m/s.  The plane could push 100 Kg of air downward at 1 m/s,
or 10 Kg at 10 m/s, or 1 Kg at 100 m/s, etc, all resulting in the same net
force.  At a fixed speed and weight of the plane, the tradeoff actually
chosen is dictated largely by the wing span.  A wider wingspan gives access
to more air to push over the same time.

The point I made about this tradeoff earlier is that while the force is the
same, the energy is not.  The energy it takes to move the air is
proportional to the square of the speed.  To impart a momentum of 100 Kg m/s
to 100 Kg of air is:

 (1/2) 100Kg (1m/s)**2 = 50 Joules

Since this is what happens each second, the power is 50 Watts.  Now consider
when only 10 Kg are moved.  To keep the same momentum, those 10 Kg must be
moved at 10 m/s.  The energy required to do that is:

 (1/2) 10Kg (10m/s)**2 = 500 Joules

which means a power of 500 Watts is required to keep doing this.  This basic
physics explains why wide wingspans are more efficient.  They move more air,
which therefore needs to be moved with a lower speed, which therefore
requires less power to sustain the flight.


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2011\01\28@151346 by MICHAEL REID

picon face

This is from a children's science website!

Bernouilli's Principle

>From the side view of your plane, you notice that the top surface is slightly curve upwards, while the bottom is flat. So when you fly your plane through the air, the air flow speed on top has to travel faster to cover a longer distance. The air flow below the wing is slower.
Remember to what Mr. Bernoulli said, “a steady flow of fluid (gas or liquid), the pressure of the fluid decrease when the velocity(speed) if the fluid increase”
Ah…forget it!!! In the case of your plane, the air flow on top of the wing is faster, therefore creating a lower pressure region on top of the wing. While the slower air flow under the wing creates an higher pressure region. Thus, the higher pressure helps the plane to take off or flying upwards.                                          

2011\01\28@161253 by N. T.

picon face
MICHAEL REID wrote:
> In the case of your plane, the air flow on top of the wing is faster,
> therefore creating a lower pressure region on top of the wing.
> While the slower air flow under the wing creates an higher pressure
> region. Thus, the higher pressure helps the plane to take off or flying upwards.
> --

Does "Zonda C12" take off at 200 mph? According to your logic, it
should :-)  Instead, it develops the most down force at the speed, I'd
suggest

2011\01\28@170253 by mikecreid

picon face
This isn't my logic. I noted my source. When an airliner slows to land they deploy flaps to increase the surface area for more lift. Bernoulli's principle states that the higher speed airflow on top creates a low pressure area and the wing moves upward. The comments here have been interesting but the science is well defined. The size of the wings is determined by the lift needed. Lift off speed is a ratio of weight vs. surface area to create the lift. Sent from my Verizon Wireless BlackBerry

{Original Message removed}

2011\01\28@171556 by Carl Denk

flavicon
face


On 1/28/2011 5:04 PM, spamBeGonemikecreidSTOPspamspamEraseMEmsn.com wrote:
> This isn't my logic. I noted my source. When an airliner slows to land they deploy flaps to increase the surface area for more lift. Bernoulli's principle states that the higher speed airflow on top creates a low pressure area and the wing moves upward. The comments here have been interesting but the science is well defined. The size of the wings is determined by the lift needed. Lift off speed is a ratio of weight vs. surface area to create the lift.
>
Add to the list, profile of the lifting surface, angle of attack, elevation (higher, the air is less dense), air temperature, runway surface (hard smooth, grass, water, snow, tall grass), and we are assuming a constant weight. Our plane at sea level would need 2000', at 7000' elevation and 70 degrees F, required 5200'

2011\01\28@173402 by Olin Lathrop

face picon face
mikecreid@msn.com wrote:
> This isn't my logic. I noted my source. When an airliner slows to
> land they deploy flaps to increase the surface area for more lift.

That's a bit misleading.  Yes, flaps tend to increase the surface area, but
the main point is that flaps give a wing more lift.  This allows the plane
to fly slower and still stay in the air.  It also increases drag, which is
why you don't want the flaps during normal flight when you're going fast
enough.


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2011\01\28@174802 by Robert Csaba Molnar

picon face
I have just one problem with this: where does the force generated go? One thing pops in my mind, communicating vessels. As soon as the additional pressure appears it would just get equalized, so I don't see how that could generate lift. Well... that much about my physics skills :-)

Now, pushing down on the air, should increase air pressure under the plane. So when planes fly close above our head there should be an increase of atmospheric pressure for a short time, no?  And if a 747 flies 20 meters above my head from the pressure probably my eyes should pop out or skull get crushed? Well I bet that's not gonna happen... maybe I'll go deaf because of the engine noise, otherwise will be fine.



--- On Fri, 1/28/11, Olin Lathrop <KILLspamolin_piclistspamBeGonespamembedinc.com> wrote:

From: Olin Lathrop <EraseMEolin_piclistspamEraseMEembedinc.com>
Subject: Re: [TECH] How planes fly
To: "Microcontroller discussion list - Public." <@spam@piclist@spam@spamspam_OUTmit.edu>
Date: Friday, January 28, 2011, 9:27 PM

Gerhard Fiedler wrote:
>> When looking at the airplane as a whole, the physics says nothing
>> about how wings or any other structure accomplishes the pushing down
>> of air, only that there must be a certain net downward momentum per
>> time imparted on the air.
>
> "Momentum per time" is what exactly, physically speaking? I don't
> think you wrote this with your physicist's hat on.

My hat is intact, but you might want to check where your duncecap is.

Momentum per time is both correct and unambiguous.  Momentum is mass times
velocity.  Momentum per time is therefore mass times velocity per time, or
mass times accelleration, or force.  Remember, F=ma.

So "momentum per time" is a fancy way to say "force", but I chose to say it
that way to help illuminate where that force is coming from.

> Which one of Newton's rules says that air must move down to keep a
> plane in steady height?

I think he mumbled something about equal and opposite forces.

> And please, don't put it in fuzzy and easy to
> misunderstand (both in the writer's and in the reader's mind) words --
> put it in clear formulae, using proper terms used in physics.

I think I've been plenty clear enough, especially if you read back thru this
thread.  Your problem seems to be with basic physics.  Surely there is a lot
of accessible stuff out there that explains forces, momentum, velocity,
accelleration, and Newton's laws.

However, let's do a example.  Suppose a plane weighs 100 Newtons.  Let's
further stipulate we're talking about a normal airplane that uses power to
push itself forward.  In other words, the force applied by the propeller,
jet engine, or whatever, is in the forward direction, which is also the
direction of motion.  As said before, the plane is in steady level flight
flying into still air.

Gravity is pulling down on the plane with 100N.  Something else must be
pushing up on the plane with 100N since it's not accellerating.  According
to Newton's law (the equal and opposite one), this in turn means the plane
is pushing down on something.  That something is the air around it.

So the plane is pushing down on the air with a force of 100N.

  Force = 100N = 100 Kg m / s**2 = (Kg m/s) / s

Note that (Kg m/s) is momentum.  So every second, the plane has to impart a
momentum of 100 Kg m/s on the air around it, and in the downward direction
since the resulting force on the plane is up.

One of the original points many posts ago is that there is a tradeoff here
between Kg and m/s.  The plane could push 100 Kg of air downward at 1 m/s,
or 10 Kg at 10 m/s, or 1 Kg at 100 m/s, etc, all resulting in the same net
force.  At a fixed speed and weight of the plane, the tradeoff actually
chosen is dictated largely by the wing span.  A wider wingspan gives access
to more air to push over the same time.

The point I made about this tradeoff earlier is that while the force is the
same, the energy is not.  The energy it takes to move the air is
proportional to the square of the speed.  To impart a momentum of 100 Kg m/s
to 100 Kg of air is:

  (1/2) 100Kg (1m/s)**2 = 50 Joules

Since this is what happens each second, the power is 50 Watts.  Now consider
when only 10 Kg are moved.  To keep the same momentum, those 10 Kg must be
moved at 10 m/s.  The energy required to do that is:

  (1/2) 10Kg (10m/s)**2 = 500 Joules

which means a power of 500 Watts is required to keep doing this.  This basic
physics explains why wide wingspans are more efficient.  They move more air,
which therefore needs to be moved with a lower speed, which therefore
requires less power to sustain the flight.


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2011\01\28@181226 by Olin Lathrop

face picon face
Robert Csaba Molnar wrote:
> I have just one problem with this: where does the force generated go?

Forces don't "go" anywhere.  This is a nonsensical question.

> Now, pushing down on the air, should increase air pressure under the
> plane.

Only a little bit for a little time since the air tends to get out of the
way.  What's left is not so much significant pressure above ambient, but a
net downward motion of the air.

> So when planes fly close above our head there should be an increase of
> atmospheric pressure for a short time, no?

Think downdraft more than pressure.  Of course when that downdraft hits the
ground, the pressure increases slightly.  That's how the ground ultimately
bears the weight of the plane in the air above it.  Remember the pigeons in
the box.

> And if a 747 flies 20 meters above my head from the pressure
> probably my eyes should pop out or skull get crushed?

Depends how thick your skull is.  Seriously though, the pressure increase
would not be much, but you definitely would notice the downdraft shortly
after the plane passed over.

If you see a picture of a low flying plane (just a few meters up), look for
how it seriously blows things around a short distance behind it on the
ground.  I remember seeing this effect shown very will in a movie once.  It
might have been Independence Day.  A military plane is flying a few meters
above the flat desert floor and you can see it kicking up a large plume of
dirt behind it.  It's a movie, so you don't know what's real.  However, the
plume had nothing to do with the story and it looked real enough.

You can probably find clips of low flying planes with a little searching.


I wonder if Gerhard has adjusted his headgear appropriately yet.


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2011\01\29@080552 by Gerhard Fiedler

picon face
Olin Lathrop wrote:

> Gerhard Fiedler wrote:
>>> When looking at the airplane as a whole, the physics says nothing
>>> about how wings or any other structure accomplishes the pushing
>>> down of air, only that there must be a certain net downward
>>> momentum per time imparted on the air.
>>
>> "Momentum per time" is what exactly, physically speaking? I don't
>> think you wrote this with your physicist's hat on.
>
> My hat is intact, but you might want to check where your duncecap is.
>
> Momentum per time is both correct and unambiguous.  Momentum is mass
> times velocity.  Momentum per time is therefore mass times velocity
> per time, or mass times accelleration, or force.  Remember, F=ma.

You're of course right here -- that was a brain fart.

However, that wasn't the only one here :)

>> Which one of Newton's rules says that air must move down to keep a
>> plane in steady height?
>
> I think he mumbled something about equal and opposite forces.

Yes, he did. Because he did say other things too, I needed to know to
which one you referred.
> I think I've been plenty clear enough, especially if you read back
> thru this thread.  
While completely besides this point, I couldn't resist in commenting
here that whenever you're told something like this, you seem to find it
almost an insult. "Real men don't read threads, and don't keep history"
or something like that :)


> Your problem seems to be with basic physics.  
It may seem so, on occasion, but isn't that the same with all of us? :)

Anyway, I found the "pigeon in a box" experiment interesting. Let's put
our physicists' hats on and dive into it for a while.

At first I'll put a fan in a box. In a box like a long hallway. No
doors. Just a fan blowing down the hallway. Then I take an elastic foil,
like a thin plastic foil, and cover the whole profile of the hallway,
perpendicular to the air stream, a few meters away from the fan. We
probably all agree that the foil will be stretched, due to the force. We
explain this by thinking of air as tiny, elastic particles, and of the
wind generated by the fan as an average movement of those particles --
momentum, power and all that.

But the inquisitive mind then puts the foil a few hundred meters down
the hallway. (Remember, this is a thought experiment, and I don't have a
problem imagining an imaginary hallway a few hundred meters long. All
flat, no doors, no change in profile.) I hope that we all can agree that
the foil will remain flat. (If a few hundred meters is not enough for
you to imagine that the foil will remain flat, imagine a few
kilometers.) I'm not sure it was Newton, but his rules suffice to
explain that, too. Since the wind is only an average movement, created
by the particles bumping into each other more in one direction than in
others, and this bumping isn't fully elastic, there's some kinetic
energy transformed into other energy (heat mostly). That slows the
average movement down. There's also statistics. Since the wind is the
effect of many "white noise" bumps, they go into all directions, and
spread out, and create those localized "momentum compensation loops".
That is, the wind generated by the fan doesn't stream all the way down
to the foil several hundred meters away, bumps into it, then moves
laterally back, like maybe students in a cafeteria line would, they just
say "what the heck" and turn back before they reach the end of the
hallway. All with Newton and the idea of air as almost elastic little
particles that move around arbitrarily to create pressure. (I'm not sure
it was Newton who first came up with this idea, but it is necessary
here.)

Now let's put the hallway vertically, with the fan on top blowing
downwards, extend it in one direction and replace the fan with an
airplane with wings that have a symmetric profile -- which is
essentially the same, just that blowing downwards directly, it uses the
fan (propeller) to move forwards, then uses the wings to move air
downwards.
To keep the plane in the air, there is a balancing force needed, which
we assume stems from the (almost elastic) collisions with the air
particles on the lower side of the wing. That generates a downwind,
which makes perfect sense. However, considering our first experiment
with the foil, if the plane flies in a box that is high enough, say a
few kilometers, it's also very likely that nothing of that average
momentum of the air particles reaches the floor -- the power is
partially transformed into heat, and the rest of the momentum is used to
create those local "flow backs". Without resulting average momentum at
the bottom of the box, there is no force, and hence, I'm convinced that
if you make the box big enough, what you'll have is a box that has the
same weight with and without a plane flying in it, Newton and everything
considered.


Now with all this in mind, and out of the way, back to the original
question...

> Gravity is pulling down on the plane with 100N.  Something else must be
> pushing up on the plane with 100N since it's not accellerating.  According
> to Newton's law (the equal and opposite one), this in turn means the plane
> is pushing down on something.  That something is the air around it.
>
> So the plane is pushing down on the air with a force of 100N.
>
>   Force = 100N = 100 Kg m / s**2 = (Kg m/s) / s

Oh man, am I ever glad that you used SI units here :) Whoever doubts
that they are most easy to work with should re-do Olin's calculations
using "customary" units...

I snipped the rest, because it follows quite nicely and
straightforwardly from this, and I don't have much to add or remove.
However, I think the mechanical "net effect" depends a lot on a number
of other effects and assumptions (for example, over what volume you are
"netting"), as exemplified above with the "pigeons in a box" thought
experiment (even though I didn't use pigeons... :). Also, while I agree
with the basic assumption that the force stems from (almost) elastic
collisions with stochastically moving air particles, it seems to me that
the conditions around a wing are too complex to be able to tell with
such simple "napkin calculations" what net effect you get (down, up,
none) with what size of box around a wing -- but the one thing I'm quite
certain of: if you make the box large enough, there is no mechanical net
effect (force) reaching the border of the box. There is probably some
radiated heat, though, that reaches the border.

Also, Newton didn't exactly say that any air particle needs to go
downwards to keep a plane in the air. As you correctly said, he mumbled
that forces need to be in balance, so /something/ needs to balance the
gravitational force -- but it doesn't necessarily have to be air
particles going downwards. We assume that air pressure is created by air
particles moving stochastically around and bouncing off surfaces. So if
one would design a device that works the particles above a (flat)
surface so that the pressure above it is lower than the pressure below
it, we have a force that may balance the gravitational force. I'm not
sure it is possible to design such a device without a net downwards
movement of particles, but it seems possible. For example, by giving the
particles a horizontal speed, the resulting pressure is lower in other
directions (including vertical), it may be possible to create vertical
forces without an actual vertical air movement. I'm not claiming that
this is what happens in a typical plane, but that maybe Newton's rules
don't require that air flows downwards to keep something up in the air.
The reason is that the air particles don't move straight, they move in a
stochastic way, and something like wind (or air movement around a wing)
is not properly modeled by just looking at air like tiny elastic spheres
that stand still when there is no wind. This stochastic movement adds
another degree of complexity, and the possibility of vertical forces
with horizontal movement (average movement of air particles). And it's
this that is the basis of Bernoulli's view of pressures and resulting
forces.

Gerhar

2011\01\29@104259 by Olin Lathrop

face picon face
Gerhard Fiedler wrote:
> At first I'll put a fan in a box. In a box like a long hallway. No
> doors. Just a fan blowing down the hallway. Then I take an elastic
> foil, like a thin plastic foil, and cover the whole profile of the
> hallway, perpendicular to the air stream, a few meters away from the
> fan. We probably all agree that the foil will be stretched, due to
> the force.

Yes.

> We explain this by thinking of air as tiny, elastic
> particles, and of the
> wind generated by the fan as an average movement of those particles --
> momentum, power and all that.

Not so fast.  Keep momentum and power separate.  The lack of distinction is
getting you into trouble later.

> But the inquisitive mind then puts the foil a few hundred meters down
> the hallway. (Remember, this is a thought experiment, and I don't
> have a problem imagining an imaginary hallway a few hundred meters
> long. All
> flat, no doors, no change in profile.) I hope that we all can agree
> that the foil will remain flat.

No, I don't.  Think of the limiting case where the crossection of the
hallway is exactly the size of the fan.  Instead of a hallway, think of a
pipe with the fan blades just barely not touching the inside of the pipe as
they spin around.  In that case I hope you can agree that if both ends of
the pipe are sealed, the fan will cause a pressure difference between the
air on both sides of it.  The entire pipe on the outflow side of the fan is
at higher pressure than the entire pipe on the inflow side of the fan.  The
pressure on each side is constant, extending to the full length of that side
no matter how long it is.

Let's say we hold this pipe fixed on the outside but can measure the lateral
force it exerts on the fixture.  Hopefully you agree that it exerts no net
force once equillibrium is reached regardless of how long the pipe is.  The
fan is exerting a force (thrust) in one direction on the pipe.  That force
is exactly countered by the force of the pressure difference on the ends of
the pipe.

In your case I think you intended the fan not to be sealed against the sides
of the hallway.  This is the same as making the pipe diameter larger than
the fan.  Now the fan actually moves air to the outflow side, which then
eventually returns to the inflow side thru the gap between the fan and the
inside of the pipe.  This does make the force ballance a bit more
complicated, but it still has to ballance.  As you make the pipe diameter
larger, the pressure difference between the two ends goes down due to the
high pressure air "leaking" back to the low pressure side around the outside
of the fan.  Note though that the area of each end of the pipe is increased
too.  That means a smaller pressure difference is needed to ballance the
thrust from the fan.

I maintain that as before, the closed pipe will not exert any net force on
the outside world once the airflow inside has reached equillibrium.  Since
there are no net forces, the sum of forces must be equal along any
direction.  To keep things simple, let's only look at the force ballance in
the lateral direction (down the long axis of the pipe, also the direction
the fan is pointing).

There are only three forces that stuff on the inside of the pipe can exert
on it in the lateral direction.  There is the fan, the pressure difference
at the ends times their area, and friction of air moving in the lateral
direction by the inside of the pipe.  All three of these have to sum to zero
else the pipe would exert a force on its surroundings.

Do you agree it doesn't exert any such force?

In this case there is a force on the fan not so much due to a pressure
difference but because it is imparting momentum per time (= force) on the
air flowing thru it.  For the same fan speed, this is actually a greater
force than just resulting from pressure as in the previous case when the
pipe was sealed around the fan.  You can prove this to yourself by thinking
what happens when you hold your hand over a vacuum cleaner hose.  The motor
speeds up, not slows down.  This shows it's "harder" to maintain air flow
than it is to maintain pressure.  There is more backward push on the fan
blades due to imparting momentum on the air passing thru it than there is
due to any pressure difference the fan can cause.

Anyway, the fan is still pushing the pipe in one direction from the inside.
In the steady state case, this push is exactly countered by the total of the
pressure difference on each end of the pipe times the pipe area, plus the
friction forces of the air sliding against the inside of the pipe.  There
will still be a pressure difference but I think the problem is that it's a
little less obvious to visualize.  Note also that the required pressure goes
down with the square of the pipe diameter.

A short distance from the fan, your foil will perceive a push because it is
stopping the momentum of the moving jet of air coming from the fan.  This is
a force due to momentum per time just like the fan's, except in the opposite
direction.  As you go further down the pipe, the jet spreads out and
eventually stalls.  This stalling is a momentum change, and exerts a force
on the air further down the pipe, which causes a pressure increase in the
air the remaining distance down the pipe.  The pressure increase persists to
the end of the pipe where it exerts a force on the pipe.

> Since the wind is only an average movement, created
> by the particles bumping into each other more in one direction than in
> others, and this bumping isn't fully elastic, there's some kinetic
> energy transformed into other energy (heat mostly). That slows the
> average movement down.

You are mixing energy and momentum arguments.  The two can't just be
exchanged.  Both momentum and energy have to ballance separately.  In this
case, trying to figure out the energy ballance is a lot harder.
Fortunately, the momentum ballance is easier to consider in this case.
Let's stick to that.  In any case, you can't use loss of kinetic energy as a
argument for momentum change.  The units don't match, for one.

> There's also statistics. Since the wind is the
> effect of many "white noise" bumps, they go into all directions, and
> spread out, and create those localized "momentum compensation loops".

This is voodoo physics.

> That is, the wind generated by the fan doesn't stream all the way down
> to the foil several hundred meters away, bumps into it, then moves
> laterally back,

I agree.

> All with Newton and the idea of air as almost elastic little
> particles that move around arbitrarily to create pressure. (I'm not
> sure it was Newton who first came up with this idea, but it is
> necessary here.)

You are supposing a micro-mechanism to explain a macro effect.  Momentum
still has to ballance regardless of what micro effect you think impedes
motion, converts kinetic energy into heat, etc.

> To keep the plane in the air, there is a balancing force needed, which
> we assume stems from the (almost elastic) collisions with the air
> particles on the lower side of the wing. That generates a downwind,
> which makes perfect sense.

I agree about the downwind part.  Trying to assume the micro effect causing
it will only confuse and possibly mislead.

> However, considering our first experiment
> with the foil, if the plane flies in a box that is high enough, say a
> few kilometers, it's also very likely that nothing of that average
> momentum of the air particles reaches the floor

Oh yes it does.  This is exactly the pigeon in the box problem.

> the power is
> partially transformed into heat, and the rest of the momentum is used
> to create those local "flow backs".

Again, you're mixing energy and momentum ballance.  You can't convert one to
the other to show ballance.  Each has to ballance separately.  In this case,
let's stick to just showing the momemtum (and thereby the force) ballance.

> Without resulting average momentum at the bottom of the box,
> there is no force, and hence, I'm convinced that
> if you make the box big enough, what you'll have is a box that has the
> same weight with and without a plane flying in it, Newton and
> everything considered.

No!  Absolutely not!  This is the crux of the matter.  Instead of a plane
passing thru, it might be easier to envision a hovering helicopter.  The
weight of that helicopter is ultimately borne by the ground somehow.  That
somehow is in the form of slightly elevated pressure over a area.  The
higher the helicopter, the smaller the pressure and larger the area.

Perhaps it helps to draw the free body diagram around the earth instead of
the helicopter.  Earth with helicopter sitting on the ground is the obvious
case.  The earth is gravitaionally attracted to the helicopter, which pushes
back on the earth thru the pressure it excerts on the ground.  When the
helicopter is hovering, the earth is still gravitionally attracted to the
helicopter.  Unless you think the earth is now accellerating towards the
helicopter, some force must be pushing back on the earth to counter the
gravitational attraction.  That force is the atmosphere pressure increase
over some area underneath the helicopter.

Note that depending on how high the helicopter is, this force that is
ultimately pressure on the ground comes from a mix of two causes.  One is
that the air pressure under the helicopter is slightly higher than ambient.
This effect decreases as the helicopter is higher.  The other is the force
due to the momentum change of the downdraft as it hits the ground.  This
will be the dominant effect when the helicopter is "high".  It will then
also be spread over such a large area and the velocity change so little that
a human standing in the area below the helicopter would not notice with
ordinary human senses.

To put this in terms of my previous post, the helicopter imparts a momentum
of some (Kg m/s) on the air.  That same momentum is ultimately imparted on
the ground, but the Kg to m/s tradeoff is very different.  The ground sees
many many more Kg, which are therefore moving with much much less m/s.  Add
to that the fact that this effect is spread over a much larger area on the
ground than the helicopter's rotors sweep, and the helicopter doesn't have
to be too high for a unaided human observer to not notice the effect.

> Also,
> while I agree with the basic assumption that the force stems from
> (almost) elastic collisions with stochastically moving air particles,
> it seems to me that the conditions around a wing are too complex to
> be able to tell with
> such simple "napkin calculations" what net effect you get (down, up,
> none) with what size of box around a wing

Again, you are supposing micro effects to explain a macro effect.  Your
explanation may or not be right, but that doesn't change the fact that the
macro effect is there, in fact must be there.  The napkin calculations I
made were only about the macro effect.  I specifically avoided talking about
the micro effects that explain it since these confuse more than illuminate.

For the plane to stay up, it has to push down on air.  This remains true
regardless of whether you can imagine the correct micro mechanism for this
or not.  In fact if you can't, then your perceptions of the micro mechanisms
are wrong.

> -- but the one thing I'm
> quite certain of: if you make the box large enough, there is no
> mechanical net effect (force) reaching the border of the box.

Again, no!  This would mean that the box with a pigeon in it would weigh
less when the pigeon is in flight.

> There
> is probably some radiated heat, though, that reaches the border.

Again, heat is a energy argument, which does not apply to a momentum or
force argument.

> Also, Newton didn't exactly say that any air particle needs to go
> downwards to keep a plane in the air. As you correctly said, he
> mumbled that forces need to be in balance, so /something/ needs to
> balance the gravitational force -- but it doesn't necessarily have to
> be air particles going downwards.

What else is there?  If you claim the plane is not pushing down on the air,
you have to show something else it is pushing down on.  Newton says it has
to push down on something.

> We assume that air pressure is created by
> air particles moving stochastically around and bouncing off surfaces.

For this argument, there is no need to assume anything about the micro
structure of air.  It only serves to obfuscate in this case.

> So if
> one would design a device that works the particles above a (flat)
> surface so that the pressure above it is lower than the pressure below
> it, we have a force that may balance the gravitational force.

That is in fact what a wing does.

> I'm not
> sure it is possible to design such a device without a net downwards
> movement of particles, but it seems possible.

It's not, at least not if the thing you are pushing on can move (like the
air can).  Pushing something with a force over time IS imparting momentum on
it.

 Force = mass x accelleration    (see Newton)
 = mass x velocity / time        (see definition of accelleration)
 = momentum / time               (see definition of momentum)


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2011\01\30@083033 by Gerhard Fiedler

picon face
Olin Lathrop wrote:

{Quote hidden}

Agreed. (FWIW, that's not the situation I was talking about, as you said
later.)

> Let's say we hold this pipe fixed on the outside but can measure the
> lateral force it exerts on the fixture.  Hopefully you agree that it
> exerts no net force once equillibrium is reached regardless of how
> long the pipe is.  The fan is exerting a force (thrust) in one
> direction on the pipe.  That force is exactly countered by the force
> of the pressure difference on the ends of the pipe.

Assuming both ends sealed, I agree.

> In your case I think you intended the fan not to be sealed against the
> sides of the hallway.  This is the same as making the pipe diameter
> larger than the fan.  
This is correct.

> Now the fan actually moves air to the outflow side, which then
> eventually returns to the inflow side thru the gap between the fan
> and the inside of the pipe.  This does make the force ballance a bit
> more complicated, but it still has to ballance.  As you make the pipe
> diameter larger, the pressure difference between the two ends goes
> down due to the high pressure air "leaking" back to the low pressure
> side around the outside of the fan.  Note though that the area of
> each end of the pipe is increased too.  That means a smaller pressure
> difference is needed to ballance the thrust from the fan.

I think I agree with all this.

{Quote hidden}

Yes.

{Quote hidden}

I follow you so far. Do we agree that there is no net air movement in
either direction, in the steady state situation?

{Quote hidden}

Let's look at a plane from a macro perspective. Assume still air without
a resulting momentum (not considering that the air particles move).
Let's also only look at the vertical component of the momentum. Since
momentum is a vector, each directional component of it must be
preserved, so the vertical component alone must also be preserved. Now
there comes a plane in steady flight, also without a vertical momentum
(no vertical movement). Since there is no vertical momentum anywhere,
and we postulate that the plane creates a downward wind to stay at the
same height, the downward wind would have a downward momentum, which
would create an imbalance in the vertical momentum balance. So it seems
that this downward wind must be counterbalanced by an upward movement to
maintain the macro balance, no? And that's not the plane, because it is
assumed to be in steady flight, same height, no vertical velocity and
therefore no vertical momentum. The only other mass around is the air.

So it seems that the rule of the balance of forces says that there must
be a downward air movement to keep the plane at steady height, but the
rule of the preservation of (vertical) momentum says that this downward
wind must be balanced by an upward wind elsewhere. (See the fan in a
pipe above.)


{Quote hidden}

Again, the question of the preservation of (vertical) momentum. If there
is a downward momentum imparted on (part of) the air, and no vertical
momentum on the plane, there must be an upward momentum on something --
and the only other thing around that could have an upward momentum seems
to be (another part of) the air. Which seems to say that there is no net
downward momentum (movement) of the air.

Gerhar

2011\01\30@093608 by Olin Lathrop

face picon face
Gerhard Fiedler wrote:
>> A short distance from the fan, your foil will perceive a push
>> because it is stopping the momentum of the moving jet of air coming
>> from the fan.  This is a force due to momentum per time just like
>> the fan's, except in the opposite direction.  As you go further down
>> the pipe, the jet spreads out and eventually stalls.  This stalling
>> is a momentum change, and exerts a force on the air further down the
>> pipe, which causes a pressure increase in the air the remaining
>> distance down the pipe.  The pressure increase persists to the end
>> of the pipe where it exerts a force on the pipe.
>
> I follow you so far. Do we agree that there is no net air movement in
> either direction, in the steady state situation?

I agree with what I think you mean.  No net air movement means that at any
point along the pipe there may be air movement, but there will be equal mass
of air moving in each direction.  In steady state, the sum of all air
movement in the pipe is zero.

> Again, the question of the preservation of (vertical) momentum. If
> there
> is a downward momentum imparted on (part of) the air, and no vertical
> momentum on the plane, there must be an upward momentum on something

Yes.

> and the only other thing around that could have an upward momentum
> seems
> to be (another part of) the air. Which seems to say that there is no
> net downward momentum (movement) of the air.

No.

If you're going to do a momentum ballance, you have to look at the whole
system, unlike a force ballance where you can draw your imaginary
encapsulating surface anywhere you like and call it a free body diagram.
Free body diagrams are a force ballancing visualization technique, not a
momentum ballancing one.

The big thing you forgot is the earth.  The plane pulls upwards on the earth
with the same force (its weight) that the earth pulls downward on it.  This
is where your missing momemtum is.  The long term net momentum imparted on
the earth is zero since eventually the downward movement of the air caused
by the plane hits the ground and imparts a momentum on the earth to exactly
cancell the one imparted by the plane earlier.

The plane, atmosphere, and earth are much like the pigeon in the box,
although the box is rather bigger.


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2011\01\30@154429 by Gerhard Fiedler

picon face
Olin Lathrop wrote:

>>> A short distance from the fan, your foil will perceive a push
>>> because it is stopping the momentum of the moving jet of air coming
>>> from the fan.  This is a force due to momentum per time just like
>>> the fan's, except in the opposite direction.  As you go further
>>> down the pipe, the jet spreads out and eventually stalls.  This
>>> stalling is a momentum change, and exerts a force on the air
>>> further down the pipe, which causes a pressure increase in the air
>>> the remaining distance down the pipe.  The pressure increase
>>> persists to the end of the pipe where it exerts a force on the
>>> pipe.
>>
>> I follow you so far. Do we agree that there is no net air movement
>> in either direction, in the steady state situation?
>
> I agree with what I think you mean.  No net air movement means that
> at any point along the pipe there may be air movement, but there will
> be equal mass of air moving in each direction.  In steady state, the
> sum of all air movement in the pipe is zero.
----

{Quote hidden}

I don't understand how you get from force to momentum. In this case,
neither the earth nor the plane have a vertical momentum. (Imagine the
earth as large enough so that the non-flat shape of its surface doesn't
matter.) So neither goes into the momentum preservation calculation.

FWIW, I'm using the earth as reference frame for the momentum. (Momentum
needs to be preserved in each reference frame.)

> The long term net momentum imparted on the earth is zero since
> eventually the downward movement of the air caused by the plane hits
> the ground and imparts a momentum on the earth to exactly cancell the
> one imparted by the plane earlier.
Momentum needs to be preserved by each action, at the time and place
when and where the action happens. Whether the downstream of the air
later hits the earth is irrelevant for the momentum balance at the
moment it hits the plane. Each single process needs to preserve
momentum.

I don't see a way to preserve (vertical) momentum when the plane hits
the (standing) air than to have both air flowing downwards (to provide
the uplifting force to the wing) and air flowing upwards at the same
time (to keep the momentum balance). With "at the same time" I mean the
time/moment when the wing hits the still air.

In order for it to work, the downwards pushing effect needs to be
different from the upwards pushing effect, and more specifically it
needs to create a different (higher) vertical force on the wing.
How do you explain a wing moving through (previously still) air with
vertical momentum preserved -- /and/ a net downward stream?
> The plane, atmosphere, and earth are much like the pigeon in the box,
> although the box is rather bigger.
Sure, but that's irrelevant for the momentum situation of the flying
plane. The air hitting the earth later is a different process that has
nothing to do with the plane hitting the air earlier, in terms of
preservation of (vertical) momentum. You can't do a "delayed momentum
compensation" -- at least not according to Newton. Each collision has to
preserve momentum by itself.

Gerhar

2011\01\30@160908 by Olin Lathrop

face picon face
Gerhard Fiedler wrote:
>> The big thing you forgot is the earth.  The plane pulls upwards on
>> the earth with the same force (its weight) that the earth pulls
>> downward on it.  This is where your missing momemtum is.
>
> I don't understand how you get from force to momentum. In this case,
> neither the earth nor the plane have a vertical momentum.

Not true.  The earth does have vertical momentum because the plane is
pulling on it.

> (Imagine the
> earth as large enough so that the non-flat shape of its surface
> doesn't matter.)

That's fine.  The fact the the earth curves is not important.

> So neither goes into the momentum preservation
> calculation.

But you can't leave it out of the momentum equation just because it's not
highly curved.

> FWIW, I'm using the earth as reference frame for the momentum.
> (Momentum needs to be preserved in each reference frame.)

You can't do that, as the earth is not a inertial reference frame in this
case.  It is being accellerated towards the plane.  You have to look at the
whole system from outside.

> Momentum needs to be preserved by each action, at the time and place
> when and where the action happens. Whether the downstream of the air
> later hits the earth is irrelevant for the momentum balance at the
> moment it hits the plane. Each single process needs to preserve
> momentum.

You are right.  I waved my hands around a bit too much here as I thought it
was already clear.  Let me be more clear:

Step way back and consider the three players in this momentum game: The
plane, the air, and the earth.  When the plane is flying thru still air the
immediate momentum ballance is the plane's stays the same, the atmosphere
goes down, and the earth goes up.  Later as the downdraft hits the ground,
the atmosphere stops going down and the earth stops going up.  All the
momentums ballance at every instance.


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2011\01\31@053421 by Gerhard Fiedler

picon face
(This is not really top-posting; I just want to get a few things out of
the way before getting to the actual post.)

Before being able to apply Newtonian physics, we need to select a
reference system. I thought we had an agreement on the Earth as
reference system, with the assumptions that it is big enough so that we
can consider its surface flat and the gravitational force vectors all
parallel and of same size. I don't think that these simplifications
affect the effects we're talking about.

We assumed a plane that flies with constant speed at a constant height
(both relative to our reference system, which is the Earth).

Momentum is mass multiplied by speed, no matter the forces. In each
single collision of two masses, momentum is preserved. There is no
"delayed compensation" of momentum imbalance: the momentum is preserved
at all times, by all individual collisions. Since momentum is a vector
(speed is a vector), each directional component of it is preserved. The
speed vectors (and therefore the momentum vectors) are dependent on the
reference system; that's why it is important to choose a reference
system before starting to think about speeds and momentums.

The existence of forces doesn't mean there is momentum; momentum is
"created" by speed alone. (Speed change of a mass requires a net force
on that mass; this is the connection between momentum and force.) Also,
the reference system by definition has no speed and therefore no
momentum.


With all this out of the way (and hopefully agreed upon), I go now into
the individual answers.

Olin Lathrop wrote:

> Gerhard Fiedler wrote:
>>> The big thing you forgot is the earth.  The plane pulls upwards on
>>> the earth with the same force (its weight) that the earth pulls
>>> downward on it.  This is where your missing momemtum is.
>>
>> I don't understand how you get from force to momentum. In this case,
>> neither the earth nor the plane have a vertical momentum.
>
> Not true.  The earth does have vertical momentum because the plane is
> pulling on it.
I think you're mixing forces and momentum. Since the Earth is the
reference system, it has no momentum. Even if you chose the plane as
reference system, the Earth wouldn't have a vertical momentum because we
assumed previously that the plane flies at constant height, so the
distance between plane and Earth is constant, hence no vertical speed,
hence no vertical momentum. Even if you chose any other reference system
that has a constant speed relative to the Earth and the plane, the
component of their momentums in "vertical" direction (if we take
"vertical" to mean "the direction from the plane towards its projection
on the Earth")  would be zero.

Momentum is /not/ about forces, it is about speed (and mass).
>> (Imagine the earth as large enough so that the non-flat shape of its
>> surface doesn't matter.)
>
> That's fine.  The fact the the earth curves is not important.

Ok, at least this we agree upon :)

>> So neither goes into the momentum preservation calculation.
>
> But you can't leave it out of the momentum equation just because it's
> not highly curved.

I didn't mean this; I meant to say that since the vertical component of
the speed of the plane relative to the Earth is zero, there is no
vertical component of the momentum of the plane (if you take the Earth
as reference system) or of the Earth (if you take the plane as reference
system).

>> FWIW, I'm using the earth as reference frame for the momentum.
>> (Momentum needs to be preserved in each reference frame.)
>
> You can't do that, as the earth is not a inertial reference frame in
> this case.  It is being accellerated towards the plane.  You have to
> look at the whole system from outside.

This is wrong. While there are forces, we assumed a plane with a
constant speed at constant height (that is, at constant distance from
the Earth). "Constant distance" means "constant vertical speed of zero"
which means "no vertical acceleration" and also "no vertical momentum".

If you don't like the Earth as reference frame, please suggest another
one. To stay within Newtonian physics and make things simple enough, it
needs to be well-defined and move at a constant speed relative to the
events we're looking at. To be well-defined, you need to suggest a fixed
point as origin, and our base vectors for a Cartesian coordinate system.
I suggest staying with the Earth as reference system, assuming a flat
surface and a homogeneous gravitational field.

{Quote hidden}

No. This violates the assumption of constant height; since we said that
the plane does not go down (using the Earth as reference system), this
also means that the Earth doesn't go up (choosing the plane as reference
system). Also, you really need to choose a reference system and clearly
state it if you want to apply Newtonian physics.

As I said, all my observations were made with the Earth as reference
system. You said this is not possible, but I think you are wrong. In any
case, please state clearly your reference system for each observation (I
suggest we use the Earth for all of them; it makes things much easier)
-- and please consider the assumption of a plane that flies with
constant speed at constant height ("height" here is meant to mean
"distance to the Earth").

Considering all this, please re-read my previous message and answer it
again. (Or, of course, if you disagree with any of these "ground rules",
let's get to an agreement on them first.)

Thanks,
Gerhar

2011\01\31@062417 by Chris McSweeny

picon face
On Sun, Jan 30, 2011 at 8:44 PM, Gerhard Fiedler
<spamBeGonelistsspamKILLspamconnectionbrazil.com> wrote:
> Momentum needs to be preserved by each action, at the time and place
> when and where the action happens. Whether the downstream of the air
> later hits the earth is irrelevant for the momentum balance at the
> moment it hits the plane. Each single process needs to preserve
> momentum.
>
> I don't see a way to preserve (vertical) momentum when the plane hits
> the (standing) air than to have both air flowing downwards (to provide
> the uplifting force to the wing) and air flowing upwards at the same
> time (to keep the momentum balance). With "at the same time" I mean the
> time/moment when the wing hits the still air.

OK - if I understand your argument correctly, this is where you're
going wrong. You're saying that there must be conservation of momentum
in a system, and that the plane and the air is such a system. You're
wrong - it's not a closed system (which is required for conservation
of momentum) as there is an external force acting on it, gravity. You
only get a closed system if you consider the complete system all the
way to the ground. You can only say that momentum has to be preserved
at each time and place if you also include external forces as part of
your "momentum balance" (in which case it's not really conservation of
momentum any more).

Chri

2011\01\31@093058 by Olin Lathrop

face picon face
Gerhard Fiedler wrote:
> because
> we assumed previously that the plane flies at constant height, so the
> distance between plane and Earth is constant, hence no vertical speed,
> hence no vertical momentum.

We loosly said "constant height".  That was not strictly true.  It is true
for any ordinary human observation and for any practical use of flying a
airplane, but it is not strictly true.  However, you're trying to split
hairs so the definition of "constant height" becomes important.

Let's say the atmosphere around the earth is uniform and perfectly still
before the plane flies thru it.  The plane is flying at a steady speed, and
such that it is neither climbing or falling from it's point of view of
flying thru the air.  This is ever so slightly different from not climbing
of falling from the point of view with respect to the ground.  The
difference would be probably not be measureable even with today's most
advanced instruments, yet it is there.

For the plane to fly "level" thru the air (perpendicular to the gravity
vector), it is actually getting a tiny amount closer to the earth.  For it
to fly at level altitude with respect to the earth, it is actually going
down a tiny amount with respect to the air (slightly down from perpendicular
to the gravity vector).

So in the end we're left with the three players, the plane, the earth, and
the air.  When the plane is flying perfectly perpendicular to the gravity
vector, it is imparting downward momemtum to the air near it and upward
momentum to the earth.  Since the earth is so massive, the speed from the
momentum imparted by the plane is very very small.

I think we're going in circles and not making any headway in this argument.
I've got things to do and I think I've explained things as well as I can.
Either someone else can jump in here that may have a different way to
illuminate the issue, or you can look up some aerodynamics texts if you want
to be convinced that a plane leaves a net downdraft.  I have run out of ways
to explain it, and patience to argue it.


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2011\01\31@102022 by N. T.

picon face
Gerhard Fiedler  wrote:
>
> I don't understand how you get from force to momentum. In this case,
> neither the earth nor the plane have a vertical momentum. (Imagine the
> earth as large enough so that the non-flat shape of its surface doesn't
> matter.) So neither goes into the momentum preservation calculation.
>

That's wrong statement "neither goes into the momentum preservation
calculation" as the assumption "neither the earth nor the plane have a
vertical momentum" is wrong. They have variable momentums strictly
within momentum preservation low.

Imagine a plane is a molecule, air is molecule and Earth is a molecule
too. Plane and Earth will be playing ping-pong, a ball being an air
molecule. Clearly Plane and Earth will be experiencing variable
momentums. You can glue more molecules into each object, this will
affect numbers, but physics would remain the same - variable
momentums

2011\01\31@113600 by Oli Glaser

flavicon
face
On 29/01/2011 13:05, Gerhard Fiedler wrote:
> Without resulting average momentum at
> the bottom of the box, there is no force, and hence, I'm convinced that
> if you make the box big enough, what you'll have is a box that has the
> same weight with and without a plane flying in it, Newton and everything
> considered.

This is incorrect.
All very interesting though, I enjoyed the exchange here...

Anyway, I went back to see roughly where it started and the above looked about right. Thought I'd have a quick go at changing Gerhards mind..
I don't think it matters too much about the exact mechanism of flight, as this makes no difference to the bigger picture anyway (the "box" feels the same force either way - essentially the atmosphere will get heavier when the plane leaves the ground)
With the above statement - essentially what you are saying is that it is possible to place an object (plane, heavier than air) into a closed system (box, filled with air) and have it make no difference to the weight. In this case, if the box is initially weighed with the plane before take off - does it get lighter after the plane takes off? Where does the weight go?
Will this be the same with a box filled with water and a heavier than water craft keeping itself afloat?

If you take all the air inside the box, and add the plane - regardless of how big the box is, or what the plane is doing, ask this question: Is the total weight more than before the plane was added?
Is it possible to change a closed (and steady state) systems weight?  (assuming no change in gravity, nor observing from speeds close to light)




2011\01\31@114720 by Michael Watterson

face picon face
On 31/01/2011 14:31, Olin Lathrop wrote:
> Let's say the atmosphere around the earth is uniform and perfectly still
> before the plane flies thru it.  The plane is flying at a steady speed, and
> such that it is neither climbing or falling from it's point of view of
> flying thru the air.  This is ever so slightly different from not climbing
> of falling from the point of view with respect to the ground.  The
> difference would be probably not be measureable even with today's most
> advanced instruments, yet it is there.

Or, it might be easier to consider a fixed wind and a tethered kite. Put a scale on the cord to measure the lift force

A plane in still air is a slightly more complicated version

2011\01\31@121440 by Gerhard Fiedler

picon face
Chris McSweeny wrote:

{Quote hidden}

I included the Earth here, under the assumptions listed before: The
Earth is the reference system, so by definition its momentum is zero.
Momentum is speed times mass. For "speed" to be of any use, we need a
reference system. My chosen reference system is the Earth, and by
definition its speed is zero in this system, so its momentum is also
zero. Gravity is not an "external force" in this system, it's part of
the system. But gravity is not a momentum, it is a force.
Gerhar

2011\01\31@121942 by Gerhard Fiedler

picon face
N. T. wrote:

> Gerhard Fiedler  wrote:
>>
>> I don't understand how you get from force to momentum. In this case,
>> neither the earth nor the plane have a vertical momentum. (Imagine
>> the earth as large enough so that the non-flat shape of its surface
>> doesn't matter.) So neither goes into the momentum preservation
>> calculation.
>
> That's wrong statement "neither goes into the momentum preservation
> calculation" as the assumption "neither the earth nor the plane have a
> vertical momentum" is wrong. They have variable momentums strictly
> within momentum preservation low.

(I assume you meant "momentum preservation law".)
You seem to have forgotten to consider that our assumption is that the
plane flies with constant speed at constant height ("height" meaning
"distance from Earth surface").

> Imagine a plane is a molecule, air is molecule and Earth is a molecule
> too. Plane and Earth will be playing ping-pong, a ball being an air
> molecule. Clearly Plane and Earth will be experiencing variable
> momentums. You can glue more molecules into each object, this will
> affect numbers, but physics would remain the same - variable
> momentums.

We assumed a plane flying at constant height. How it got there is
another question for another day, and what it does once fuel goes out
(or shortly before :) is also another question for another day -- maybe.
But for now, I was only looking at the time where the plane flies with
constant speed at constant height.
The second issue to consider is the reference system. In order to talk
about speed, you need to say "speed relative to ...", where "..." is
called the "reference system". I chose the Earth as the reference
system. Being the reference system, its speed and therefore its momentum
in this system is by definition zero. Since I assume a plane flying at
constant height, its vertical speed and therefore its vertical momentum
is also zero -- in this reference system.

Gerhar

2011\01\31@124355 by Gerhard Fiedler

picon face
Olin Lathrop wrote:

> Gerhard Fiedler wrote:
>> because we assumed previously that the plane flies at constant
>> height, so the distance between plane and Earth is constant, hence
>> no vertical speed, hence no vertical momentum.
>
> We loosly said "constant height".  
"We" did not "loosly" say "constant height" -- when I wrote "constant
height", I actually meant "constant height". This apparently wasn't true
for you... (Back to the issue about messages that lead to
misinterpretations? Maybe another time.)

> That was not strictly true.  It is true for any ordinary human
> observation and for any practical use of flying a airplane, but it is
> not strictly true.  However, you're trying to split hairs so the
> definition of "constant height" becomes important.
>
> Let's say the atmosphere around the earth is uniform and perfectly
> still before the plane flies thru it.  
Ok.

> The plane is flying at a steady speed, and such that it is neither
> climbing or falling from it's point of view of flying thru the air.
This is not what I was saying... I meant "constant height" as in
"constant distance from Earth", as I wrote a few times.

> This is ever so slightly different from not climbing of falling from
> the point of view with respect to the ground.  The difference would
> be probably not be measureable even with today's most advanced
> instruments, yet it is there.
I explained very clearly before what I mean with "constant height", and
I think my definition of "constant height" is in alignment with the
typical understanding of it:

>> ... we assumed a plane with a constant speed at constant height (that
>> is, at constant distance from the Earth). "Constant distance" means
>> "constant vertical speed of zero" which means "no vertical
>> acceleration" and also "no vertical momentum".

To simplify things, I assumed so. Your idea was to look at this from a
"macro" point of view. (I assume you don't have the thread's history in
front of you, and I don't know how good your memory is, but I can quote
you if you don't remember and are unable to look up the thread's
history.) So I followed your way to look at this.

In a "macro" point of view, there are no individual collisions with
individual air molecules, there is just an amorphous mass of air that
may move or stand still, and a plane wing that moves /at constant speed
and height/ through this mass.


> For the plane to fly "level" thru the air (perpendicular to the
> gravity vector), it is actually getting a tiny amount closer to the
> earth.  For it to fly at level altitude with respect to the earth, it
> is actually going down a tiny amount with respect to the air
> (slightly down from perpendicular to the gravity vector).
When I say "at constant height", I didn't mean to imply an angle of
attack or anything, just "at constant height". It was assumed that the
AOA is adjusted so that at the given speed and with the given wing
profile and air density and ..., the plane flies at constant height.
What's so difficult to understand with "constant height"?

> So in the end we're left with the three players, the plane, the earth,
> and the air.  When the plane is flying perfectly perpendicular to the
> gravity vector, it is imparting downward momemtum to the air near it
> and upward momentum to the earth.  Since the earth is so massive, the
> speed from the momentum imparted by the plane is very very small.
You again have not defined your reference system. How can you even think
of applying Newton's rules without defining a reference system? How do
you measure speed if you don't define a reference system? You seem to
look at this with a pre-Newtonian understanding that assumed an absolute
frame of reference, sort of an "ether". I don't want to get into
discussions around this, so let's stick with Newton -- but in order to
talk in Newton's terms, you /have/ to define a reference system. Which
you never do, so pretty much everything you say is not valid, or at best
unintelligible, from a Newtonian point of view. There is no (Newtonian)
meaning of "speed" or "momentum" without a reference system.
The normal reference system for things happening on Earth is the Earth,
often with a few assumptions (we talked about this). You rejected the
Earth as reference system, for reasons that are not quite clear, but you
never defined what else you're using as reference system when you talk
about "speed" and "momentum". They need a reference system to make any
sense at all. If you just started to do clean physics, and began with
defining your reference system, you'd realize that some things are not
exactly the way you tried to put them while avoiding to define a
reference system.

> I think we're going in circles and not making any headway in this
> argument. I've got things to do and I think I've explained things as
> well as I can. Either someone else can jump in here that may have a
> different way to illuminate the issue, or you can look up some
> aerodynamics texts if you want to be convinced that a plane leaves a
> net downdraft.  I have run out of ways to explain it, and patience to
> argue it.
I understand... :)  But I thought you could, too. I may be wrong...

Gerhar

2011\01\31@133935 by Olin Lathrop

face picon face
Gerhard Fiedler wrote:
> I understand... :)  But I thought you could, too. I may be wrong...

I am not sure what that is supposed to mean.  I have spent considerable time
and effort trying to explain the phenomenon of a airplane leaving a net
downdraft as it passes thru air.  I'm sorry I couldn't come up with a
explanation you can believe, but I did try.  I'm out of ideas how to explain
it differently and have spent too much time on it already, so I'm giving up..

Maybe someone else can explain it from a fresh perspective.  In any case,
there must be many references to the basic mechanics of flight out there
that you should be able to find.  I mentioned a book on the subject I
thought was a great introduction maybe a year ago.  I don't have the book in
front of me now nor remember the title or authors off the top of my head.
If I see a picture of the front cover, I'll be able to say yes or no that
it's the book I had in mind.  There should be a record in the archives, but
it may be difficult to find.


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(978) 742-9014.  Gold level PIC consultants since 2000

2011\01\31@135331 by Jim Franklin

flavicon
face
These guys used to do a demo version of their software (flight modelling,
trim etc and their website has some info on...


{Original Message removed}

2011\01\31@143440 by N. T.

picon face
Ok, You can be convincing. I agree, I was wrong with the post. What
else am I expected to say?


{Quote hidden}

>


'[TECH] How planes fly'
2011\02\01@071004 by Gerhard Fiedler
picon face
Olin Lathrop wrote:

> Gerhard Fiedler wrote:
>> I understand... :)  But I thought you could, too. I may be wrong...
>
> I am not sure what that is supposed to mean.  I have spent
> considerable time and effort trying to explain the phenomenon of a
> airplane leaving a net downdraft as it passes thru air.  I'm sorry I
> couldn't come up with a explanation you can believe, but I did try.
> I'm out of ideas how to explain it differently and have spent too
> much time on it already, so I'm giving up.

You never explained how the momentum is preserved, staying in a single
reference system. You never even defined which reference system you're
using. You rejected the Earth as reference system, but I'm pretty sure
if you use Alpha Centauri as reference system things don't get easier.

I'm not claiming to have the answer, and I agree that it is very likely
that there is a downdraft in the air to provide the upward force on the
wing that balances the force of gravity (to keep the plane at constant
height), but I just don't see the conservation of momentum here --
without something else.

To recap:

- Reference system is the Earth. (This doesn't mean that the Earth
itself is excluded; it just means that we measure all speeds relative to
the Earth.) Whenever we do Newtonian physics (actually, all physics
since Newton at least, excluding some theories that postulate an
absolute reference) we need to define a reference system. Speed and
momentum only make sense within a reference system, and are only
conserved within a reference system. It is not "allowed" to change
reference systems while talking about conservation of momentum.

- The Earth is considered "big" relative to our experiment. Surface is
flat, gravity vectors are parallel and same size. I think we already
have agreed that the curvature of the Earth and its gravitational field
is not relevant for our problem, and it makes things easier.

- I'm looking only at the vertical component of momentum (and speed).
Since momentum is a vector, each directional component has to be
preserved in order for the whole to be preserved.

- Air is standing still before the wing moves through. (Remember, this
is "still" WRT the Earth.) No (net) movement whatsoever, also no
vertical movement. Which means no vertical momentum either.
- Wing moves at constant speed and constant height through the air.
"Height" is meant to be "distance from Earth", not something relative to
wind or other odd references.
- Movement at constant height means there is no vertical component of
the wing's speed, hence it has no vertical momentum.

- The Earth, as reference frame, has no momentum whatsoever, by
definition.

- A downward wind, air moving downwards, does have a vertical component
of its speed, and since it has a mass, there is a vertical component of
the momentum.


Olin suggested a "macro" view; that is, not considering that the air is
made up of individual molecules that move around stochastically. We
never established what exactly this means, but from what I understood
Olin, he meant a simple before/after comparison of involved momentums.
This is exactly what I'm trying to do here.

Olin's postulate is that there is a net downdraft after the wing went
through the air.
My doubt is:
- Before the wing went through a piece of air there is no vertical
momentum, neither in the air nor in the wing.

- If there is a net downdraft after the wing passed through a given
piece of air, there must be a net downward momentum.
- How can this be?


My postulate is that there can't be a downdraft without an updraft that
balances the vertical momentum. I don't know where it is, and I don't
know how this all exactly works, but conservation of momentum means that
if there was no vertical momentum before, and there is a downdraft
afterwards, there must be something that moves upwards to balance the
downward momentum of the downdraft. (It's like if after a collision of
billiard balls you see a ball move left compared to the initial
movement, you know there's another one moving right, too -- even if you
don't see it.)


My only question is: how is the initial vertical momentum of zero
preserved? And Olin, however hard he tried, wasn't able to stay within
one single reference system and do a momentum balance that would result
in a downdraft, parting from still air and a wing in horizontal
movement. (Neither was I, but I at least acknowledged as much and
introduced the idea of a balancing updraft to keep the momentum
preserved.)

Gerhar

2011\02\01@071429 by Gerhard Fiedler

picon face
Oli Glaser wrote:

> On 29/01/2011 13:05, Gerhard Fiedler wrote:
>> Without resulting average momentum at the bottom of the box, there is
>> no force, and hence, I'm convinced that if you make the box big
>> enough, what you'll have is a box that has the same weight with and
>> without a plane flying in it, Newton and everything considered.
>
> This is incorrect.

I came to this conclusion, too. But I don't yet fully understand the
mechanisms that make this work, especially around the momentum
preservation (see also the other part of the question in an email in
this thread I just sent).
Gerhar

2011\02\01@081734 by Olin Lathrop

face picon face
Gerhard Fiedler wrote:
> I agree that it is very
> likely that there is a downdraft in the air to provide the upward
> force on the wing ...

What, so the remaining unlikely part is magic?

> And Olin, however hard he tried, wasn't able to stay within
> one single reference system and do a momentum balance that would
> result in a downdraft, parting from still air and a wing in horizontal
> movement.

Actually I mentioned it several times.  The earth goes towards the plane a
little bit to ballance the momentum of the air going down.


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(978) 742-9014.  Gold level PIC consultants since 2000

2011\02\01@083211 by RussellMc

face picon face
> Maybe someone else can explain it from a fresh perspective.  In any case,
> there must be many references to the basic mechanics of flight out there
> that you should be able to find.  I mentioned a book on the subject I
> thought was a great introduction maybe a year ago.  I don't have the book in
> front of me now nor remember the title or authors off the top of my head.
> If I see a picture of the front cover, I'll be able to say yes or no that
> it's the book I had in mind.  There should be a record in the archives, but
> it may be difficult to find.


<http://www.abebooks.com/servlet/SearchResults?sts=t&tn=understanding+flight&x=0&y=0>

>From $US4.10 used.

Book is perhaps (as author unspecified)(always helpful, if known,
which it may not have been in this case).

Understanding Flight (ISBN: 0071363777 / 0-07-136377-7)
Anderson, David F.;Eberhardt, Scott
McGraw-Hill, 2000.
Book Condition: Good. Light shelf wear and minimal interior marks. Buy
all your books used. Help the earth. Bookseller Inventory #

Synopsis: The simplest, most intuitive book on the toughest lessons of
flight--addresses the science of flying in terms, explanations, and
illustrations that make sense to those who most need to understand:
those who fly. Debunks long-rooted misconceptions and offers a clear,
minimal-math presentation that starts with how airplanes fly and goes
on to clarify a diverse range of topics, such as design, propulsion,
performance, high-speed flight, and flight testing. Not-to-be missed
insights for pilots, instructors, flight students, aeronautical
engineering students, and flight enthusiasts.

_________

Aha- second edition from $19.53
Different cover art.

http://www.abebooks.com/servlet/SearchResults?kn=second+edition&tn=understanding+flight&x=0&y=0

Understanding Flight, Second Edition
(ISBN: 0071626964 / 0-07-162696-4 )
David Anderson, Scott Eberhardt

Discover how planes get--and stay--airborne

Now you can truly master an understanding of the phenomenon of flight.
This practical guide is the most intuitive introduction to basic
flight mechanics available. Understanding Flight, Second Edition,
explains the principles of aeronautics in terms, descriptions, and
illustrations that make sense--without complicated mathematics.
Updated to include helicopter flight fundamentals and aircraft
structures, this aviation classic is required reading for new pilots,
students, engineers, and anyone fascinated with flight.

Understanding Flight, Second Edition, covers:

Physics of flight
Wing design and configuration
Stability and control
Propulsion
High-speed flight
Performance and safety
Aerodynamic testing
Helicopters and autogyros
Aircraft structures and materials

2011\02\01@103408 by RussellMc

face picon face
> <http://www.abebooks.com/servlet/SearchResults?sts=t&tn=understanding+flight&x=0&y=0>
>
> From $US4.10 used.
>
> Book is perhaps (as author unspecified)(always helpful, if known,
> which it may not have been in this case).
>
> Understanding Flight (ISBN: 0071363777 / 0-07-136377-7)
> Anderson, David F.;Eberhardt, Scott
>  McGraw-Hill, 2000.

Useful subset of book here

                 http://bit.ly/UnderstandingFlight


            RM

2011\02\01@110122 by Gerhard Fiedler

picon face
Olin Lathrop wrote:

> Gerhard Fiedler wrote:
>> I agree that it is very likely that there is a downdraft in the air
>> to provide the upward force on the wing ...
>
> What, so the remaining unlikely part is magic?

I don't know about you, but science is about probabilities, not about
certainties. It is not possible to reach certainty with science; if you
want certainty, you need to resort to religion. Sometimes it sounds as
if you take stuff "religiously", though... :)

>> And Olin, however hard he tried, wasn't able to stay within one
>> single reference system and do a momentum balance that would result
>> in a downdraft, parting from still air and a wing in horizontal
>> movement.
>
> Actually I mentioned it several times.  The earth goes towards the
> plane a little bit to ballance the momentum of the air going down.

No, this is at best inconsistent, and I'm sure it is way below what you
could achieve if you really thought about what I wrote -- however
difficult this may be for you.


I don't know what your problem is, but it seems to have to do with
either textual understanding or maybe simply not reading what others
write because you think it's not worth it. In any case, our
communication is severely limited here, by you not having captured or
having forgotten things I wrote several times before.


A few of the things I mentioned repeatedly before (including in the
email to which you just responded):

1- You need to choose a reference system, define it so that we (and you)
know what you are talking about, /and stay within it/ if you want to
talk about preservation of momentum. "The Earth [I'm assuming you're
talking about the Earth, and not some earth] goes towards the plane"
means "the distance between the Earth and the plane becomes smaller",
which, in the reference system of the Earth means that now not only has
the air a downward momentum, but also the plane.
2- From the get-go we have assumed a plane flying at constant speed and
constant height (that is, constant distance from the Earth). Not on a
slight down slope, but at constant height. According to your
explanation, the plane would not be able to keep height, because it
always goes down a little bit. If it keeps height, it needs to go up a
little bit after going down a little bit, to keep the same height (on
average), but even /if/ the going down of the plane would explain the
preservation of the momentum, where's the preservation when the plane
goes up a little bit, to keep its average constant height?


We assumed a plane flying at constant height, in a "macro" vision, as
you called it, and you said how this all works is easily explained by
Newtonian physics, especially the preservation of momentum. I'm just
trying to /really/ apply physics here. For this you need to define a
reference system within which we then can apply the rule of preservation
of momentum. You started with this, so I assume you already have a
reference system in your mind -- don't you? So why don't you define the
reference system, and then stay with it and explain how the momentum is
balanced?

If you really want to understand the problem, re-read the last message
and try to understand it. Most of it is pretty clearly spelled out
there.


To summarize the most important parts in short, simple sentences for the
attention span challenged:

1- We need to define a reference system.

2- We need to stay within the reference system.

3- The speed of the reference system within itself is zero, and so is
its momentum -- at all times.

4- Momentum is preserved only within a reference system.

5- We look at a "macro" view of things, because that's what was
suggested by Olin as enough to explain what happens, with basic
Newtonian physics.

6- Item 5 means that a plane flying at constant height doesn't go micro
up and micro down to achieve a macro constant height; it simply flies at
constant height.
(If you really think the micro down and micro up helps you explain
preservation of momentum, by all means use it -- but explain how the
momentum is preserved in both phases, not only in one.)

Gerhar

2011\02\01@121415 by Olin Lathrop

face picon face
RussellMc wrote:
{Quote hidden}

Thanks, Russell.  That's the book I was thinking of.  I really liked the way
they explained things.  I kept thinking to myself "Oh, yeah, that makes
sense" as I was reading it.  Other descriptions of flight I had run accross
relied too much on voodoo explanations that didn't seem to be well grounded
in physics.  This is not a aeronautical engineering text, but rather a
physics-based introduction to the principles.

I definitely recommend this book if you have a basic physics background and
want to understand how flight works, and the tradeoffs and restrictions
faced by aircraft designers.


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(978) 742-9014.  Gold level PIC consultants since 2000

2011\02\01@141128 by Denny Esterline

picon face
At the risk of adding fuel to this fire....

Consider a helicopter hovering in otherwise still air. We can all
agree there is a net downdraft, yes?

I don't see it as too difficult to consider the wing of a plane to be
a big linear fan blade doing essentially the same thing.

-Denny
<ducking for cover :-)

2011\02\01@143422 by Olin Lathrop

face picon face
Denny Esterline wrote:
> Consider a helicopter hovering in otherwise still air. We can all
> agree there is a net downdraft, yes?

Those that paid attention in physics class or ever stood underneath a
hovering helicopter certainly should.

> I don't see it as too difficult to consider the wing of a plane to be
> a big linear fan blade doing essentially the same thing.

Not too difficult because it is.  Usually helicopter blades are described as
rotating wings, but the reverse works too.


********************************************************************
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(978) 742-9014.  Gold level PIC consultants since 2000

2011\02\01@235505 by RussellMc

face picon face
   bcc: Ken wants to look at the first reference. John may

> Consider a helicopter hovering in otherwise still air. We can all
> agree there is a net downdraft, yes?
Those that paid attention in physics class or ever stood underneath a
hovering helicopter certainly should.
> I don't see it as too difficult to consider the wing of a plane to be
> a big linear fan blade doing essentially the same thing.
Not too difficult because it is.  Usually helicopter blades are described as
rotating wings, but the reverse works too.

________

The book or web version of the book has a nice photo of an aeroplane (Cessna
Jet) heading towards the camera & flying slightly above a cloud base. It has
"blown" a large slot into the cloud from its downdraft  and there are nice
swirly bits along the edge from the wingtip vortices.

Have a look at the web subset I posted that provides  a good summary of what
has been being talked about. Taught me things about wind turbine blades in
just skimming it :-).

Useful subset of book here

                http://bit.ly/UnderstandingFlight

I found that standing as close as they would let me (= surprisingly close
considering) under a slowly ascending harrier was a much less windy
experience than I had expected. And vastly noisier.




           Russell


> <
www.abebooks.com/servlet/SearchResults?sts=t&tn=understanding+flight&x=0&y=0

2011\02\02@072157 by N. T.

picon face
Oli Glaser wrote:
>>>  - if one thing goes one
>>> way, then something generally goes the other way too.
>> Let's leave the statement to philosophers :-)
>>
>
> I think that's a very good idea.. :-)
>

Yeah, good idea, cuz If I were a  philosopher, I would stumble on that
statement, does that mean "every action will have a symmetric
counter-action", or, probably it would mean "air resistance is
futile", or both?
:-

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