Bernoully's principle
John Roncallo
Cankmaster wrote:
>
> Can anyone tell me the formula for bernoully's principle and the basic jist of
> it. I've got a paper due soon and i need some help. Anyway, thanks a lot.
>
> ~Stephen Shirk
I would say that if you have a paper due on it shortly, you would at
this point be able to tell us the basic jist of Bernoulli. I would also
recomend you start by spelling his name right.
--
/---------------------------------------/
/ John Roncallo /
/ ron...@ibm.net /
/---------------------------------------/
Jerry Bransford
Cankmaster wrote:
>
> >I would say that if you have a paper due on it shortly, you would at
> >this point be able to tell us the basic jist of Bernoulli. I would also
> >recomend you start by spelling his name right.
>
> on an aviation paper due in math. I'm sorry if I don't live up to your
> standards. My brother is the one who graduated in aviation sciences. All i
> was looking for was a little help and obviously you've got something stuck so
> far up your ass you can't realize it. F*** you (John Roncallo) a**hole.
It's apparent you not only have something 'stuck' up yours even further
than John's, it's also apparent you're lacking the maturity of even a
typical 10th grader with that very crude and immature response. When
you grow up and stop responding like you did, perhaps you'll get answers
from people that know all about Bernoulli's principle. Your response to
John stopped me cold from wanting to help with your question at all.
Jerry
--
Jerry Bransford
To send me email, remove the 'nospam' from my address.
PP-ASEL, C.A.P., KC6TAY
The Zen hotdog... make me one with everything!
Cankmaster
HLAviation
>Can anyone tell me the formula for bernoully's principle and the basic jist
>of
>it. I've got a paper due soon and i need some help. Anyway, thanks a lot.
>
>
The B. principle. It doesnt really exist. It is just an observation of
Newtonian Physics. Air is deflected down lifting the deflecting body causing
the change in pressure not the other way around. Any way that was what my CFI
instructot explained to me and he was a senior aerodynamics engineer for
Douglas Aircraft in Long Beach.
Cankmaster
>I would say that if you have a paper due on it shortly, you would at
>this point be able to tell us the basic jist of Bernoulli. I would also
>recomend you start by spelling his name right.
I'm in 10th grade and my brother told me that i might be able to find some info
on an aviation paper due in math. I'm sorry if I don't live up to your
standards. My brother is the one who graduated in aviation sciences. All i
was looking for was a little help and obviously you've got something stuck so
far up your ass you can't realize it. Fuck you (John Roncallo) asshole.
Ron Parsons
In article <199804100323...@ladder01.news.aol.com>,
hlavi...@aol.com (HLAviation) wrote:
A voice crying in the wilderness. I can't count the times I've tried to
pound through heads that Bernoulli is only a shortcut to get from one
observation to another and does not represent the real world.
Indeed, it's Newton that holds aircraft up and only Newton.
Ron
n55bz
Ron Parsons wrote in message <6gl2al$5id$1...@gte1.gte.net>...
Not so! It is money- just try to commit aviation without it.
rober...@osi.com
> The B. principle. It doesnt really exist. It is just an observation of
> Newtonian Physics. Air is deflected down lifting the deflecting body
causing
> the change in pressure not the other way around. Any way that was what my
CFI
> instructot explained to me and he was a senior aerodynamics engineer for
> Douglas Aircraft in Long Beach.
It is common for CFI's to say that, in fact I've seen some CFI literature
say to tell students that. The reason is *NOT* that the B principle is
not true, but that its confusing to students and thinking of only
Newton give you what you really need to understand which is critical
angle of attack.
The reason some people claim its best to deny B is that some will confuse
it and assume that it just requires airspeed to avoid stall, when in real
life its the angle of attack on the wing. B is affected by angle
of attack though.
If you read "Stick and Rudder" the first chapter says something like...
"B does exist but its not very useful for pilots to know, and Newton
is easier, so we'll pretend its all Newtow"
In real live though about 70% of lift is B and about 30% is Newton,
depending on the wing shape
-----== Posted via Deja News, The Leader in Internet Discussion ==-----
http://www.dejanews.com/ Now offering spam-free web-based newsreading
Paul Lee
Cankmaster wrote:
>
> Can anyone tell me the formula for bernoully's principle and the basic jist of
> it. I've got a paper due soon and i need some help. Anyway, thanks a lot.
>
P + (1/2)(v*2)(Dens) + gh(Dens) = constant.
(as derived in intro college physics books)
Where:
P is the pressure in the fluid,
v*2 is the sqare of the velocity of the fluid,
g is gravitaional accn (32ft/sec/sec or 9.8m/s/s depending on units),
h the vertical position above a reference point,
and Dens is the density of fluid.
In a carburetor or near a wing, for example, h is relatively constant
so in that case you can approximate
P + (1/2)(v*2)(Dens) = constant
This means if the velocity of a fluid increases the pressure
decreases (and vice versa). Hence there is a net lift on a wing
shape due to lower pressure on upper side and suction in a carburetor
due to lower pressure there.
Hope that helps.
Now can anybody tell me why flight instruction still includes
the mythical centrifugal force on an airplane making a turn?
The sideway forces on a turning airplane are NOT BALANCED.
-----------------------------------------------------------------
Paul Lee ........... Abri Technologies ......... http://abri.com/
benjamin j snyder
In article <352D98...@cts.com>,
Jerry Bransford <jer...@nospam.cts.com> wrote:
>Cankmaster wrote:
>> far up your ass you can't realize it. F*** you (John Roncallo) a**hole.
>
>It's apparent you not only have something 'stuck' up yours even further
>than John's, it's also apparent you're lacking the maturity of even a
>typical 10th grader with that very crude and immature response.
Actually I agree with his comment, but maybe it's just because you would think
I am an immature college student who has nothing better to do than drink beer
and make an ass out of myself whenever the opportunity arises. You made a
*VERY* similar stereotype when you said a 'typical 10th grader'...many people
have the beer drinking, party going, hell raising stereotype of college
students. I think most people will agree with me when I say that stereotypes
are a bad thing.
Perhaps the response should not have been posted in the ng, though. I do not
remember seeing the original response, but why was he attacked for a simple
misspelling?
IMO, the only true ignorant (stupid/dumb/moronic/whatever) person is someone
who attacks (in any way) the lack of knowledge of another. Just because he
misspelled Bernoulli he was attacked. Hell, four years of college loaded with
sciences and I don't know if I can spell it correctly. Before you slam another
person's spelling look at your own posts, how many times has 'the' been spelled
'teh'? How many times has someone used a word that a spell-checker will not
ketch (catch) because it is spelled correctly, but makes know (no) cents
(sense) in the sentence? Or rather then (than) accept your own downfalls you
must give others a herd (hard) time. There are countless examples of commonly
misspelled words, why aren't we all being attacked for our own ignorances?
I understand that many will make the argument that if he is doing a paper on
someone he should be able to spell the individual's name, but 'y' and 'i' have
one 1 key between them on the keyboard. Also, it could have been explained
to him that he had misspelled it without any sort of attack, a simple
explanation would do.
I know it sounds like I am a PC shrink (I don't think that I am), but for some
reason I feel sorry for the young man. If you think for a while, I am sure you
too can think of a time when someone attacked an ignorance of yours. Give
the kid a break.
Please note that all of the above was written by a beer drinking, party going,
college student who makes an ass out of himself (and acts immature) whenever
given the chance to do so.
Give a man a fish and he'll eat for a day. Teach him to fish and he'll eat for
a lifetime.
---
The thoughts and opinions here express those of the author, and do in no way
reflect the opinion of The Ohio State University, or the Computer and
Information Science Department of said University.
--
Ben Snyder
Student Programmer
Ohio Supercomputer Center
General Stark
Out of the mouths of babes.... wasn't there a time oh so long ago when
kids didn't talk like drunken sailors? *sigh* I'm dating myself I
guess...
Marty
PP-ASEL
7B2
David Kornreich
In article <352E4E...@spam.please>, n...@spam.please wrote:
>
> Now can anybody tell me why flight instruction still includes
> the mythical centrifugal force on an airplane making a turn?
> The sideway forces on a turning airplane are NOT BALANCED.
>
> -----------------------------------------------------------------
> Paul Lee ........... Abri Technologies ......... http://abri.com/
They certainly are if you think about the (accelerating) reference frame
of the airplane, which presumably is where you are sitting when you really
want to know about these things.
I suggest an experiment. Next time you go into a steep right turn, grease
down your seat, unbuckle your seat belt and open the door. When you are
shot out of the airplane by centrifugal force, then will you still be
claiming that it is "mythical?" How about if I put you on a space station
which produces artificial gravity by rotation? Will you stubbornly float
into the air because, as Bugs Bunny says, you never studied the laws of
physics?
Despite the unfortunate term "fictional force" often used to describe
centrifugal and coriolis forces, they are indeed very real in accelerated
frames, which is where you are when you're in a turning airplane. So why
not use CF to describe what's going on? It's what the you and the wing
experience every time you go into a turn. Eliminating it just because you
have a bias for inertial frames is confusing because it is non-intuitive,
and because you end up descibing the situation from the standpoint of a
hypothetical observer hovering in midair. Then you end up (in essence)
explaining CF anyway when your passenger asks why (s)he is pressed against
the left bulkhead during a right turn. Better to describe it from the
pilot's perspective in the first place.
dave ::)
--
David A. Kornreich, PP-ASEL -><- dk...@alderan.tn.cornell.edu
Cornell University Space Sciences
The Fraternal Order of the Eternal Employees of Floyd
** We Specialize in Circumstances Beyond Our Control **
John Lowry
Dear Paul and All:
On the question of forces in a turn, about which you remarked
Paul Lee wrote in message <352E4E...@spam.please>...
>
>Now can anybody tell me why flight instruction still includes
>the mythical centrifugal force on an airplane making a turn?
>The sideway forces on a turning airplane are NOT BALANCED.
>
The only reason that nonsensical idea persists is that the FAA is
intellectually challenged. And fails to respond to that challenge.
John.
John T. Lowry, PhD
Flight Physics; Box 20919; Billings MT 59104
Voice: 406-248-2606; E-mail: jlo...@mcn.net
sue***
General Stark wrote in message <352e685d...@news-server.amherst.edu
>Out of the mouths of babes.... wasn't there a time oh so long ago when
>kids didn't talk like drunken sailors? *sigh* I'm dating myself I
>guess...
>Marty
>PP-ASEL
>7B2
not so bad as not dating, Marty. and yes, it always amazes me when the
Very Unpleasant Child (6th Grader) down the street continually flips me the
bird when i tell him to clean out his mouth or take a different route. i
wonder if his mom would sue me if i got out the dishsoap and chased him
down.
course, then there's what i said the other day when i latched the truck door
onto my index finger....
sue**better wash my mouth out first***kramer
Paul Lee
David Kornreich wrote:
>
> In article <352E4E...@spam.please>, n...@spam.please wrote:
>
The issue is the motion of the plane as one body and not accelerated
frames INSIDE
the airplane. What accelerated reference frame factors can you include
for this calculation? The motion of the airplane body IS accelerated
toward the center
and that is produced by the unbalanced centripetal force (=mv*2/R). The
rate of
turn is directly related to the unbalanced centripetal force (the
DEFLECTING force
towards the center). A constant unbalanced centripetal force produces
a constant rate of turn. A body with balanced sideways forces would
continue in a straight line and not be deflected towards the center.
> .... as Bugs Bunny says, you never studied the laws of
> physics?
I got a PhD in theoretical physics and thought it for over 20 years.
Lets leave buggs bunny out of this and stick to physics.
Sorry to be so voluminous about this. Shortest answer is often
the best.
Don
rober...@osi.com wrote:
> It is common for CFI's to say that, in fact I've seen some CFI literature
> say to tell students that. The reason is *NOT* that the B principle is
> not true, but that its confusing to students and thinking of only
> Newton give you what you really need to understand which is critical
> angle of attack.
>
> The reason some people claim its best to deny B is that some will confuse
> it and assume that it just requires airspeed to avoid stall, when in real
> life its the angle of attack on the wing. B is affected by angle
> of attack though.
>
> If you read "Stick and Rudder" the first chapter says something like...
> "B does exist but its not very useful for pilots to know, and Newton
> is easier, so we'll pretend its all Newtow"
>
> In real live though about 70% of lift is B and about 30% is Newton,
> depending on the wing shape
>
>
Yep, you're right. In fact, B accounts for more than 70%. I've been watching
people on these groups say it's Newton forever, and couldn't understand how they
thought that. Nobody ever offered any proof. Now I know why they say that.
Thanks.
Ken McCarley
Paul basically gave you the right fomula.
Start from there and visualize air moving through a cylinder with a short
section that is tapered to form a neck of smaller diameter. The mass flow
rate through the cylinder is the same in the neck of the cylinder as it is
where the cylinder diameter is constant. Therefore, if the mass flow rate
is constant through the neck, the velocity the air must increase as it
moves through the portion of the cylinder with smaller diameter.
It's conservation of energy. Static pressure is exchanged for kinetic
energy (an increase in velocity). In short the pressure decreases inside
the neck of the tube as the velocity increases.
How does this apply to a wing? Draw a picture of a cylinder with a neck in
it the shape of an airfoil, and cut in half length wise (axially). Just one
half of the neck looks somewhat like a wing. And there you have it;
pressure decreases along the curved surface of the wing, on an airfoil
that's the top, and lift is created.
Look in some aerospace engineering books or check out these sites for some
decent illustrations and explanation.
http://ldaps.ivv.nasa.gov/Physics/bernoulli.html
http://ldaps.ivv.nasa.gov/Physics/lift.html
By the way, some of the folks that answered were wrong in saying that
Bernoulli's principles don't exist, but they were right in saying that lift
(and Bernoulli's principles) follow Newtonian mechanics.
If you're in tenth grade, are you taking flying lessons? It won't be long
before you can solo.
Ken
Paul Lee <n...@spam.please> wrote in article <352E4E...@spam.please>...
> Cankmaster wrote:
> >
> > Can anyone tell me the formula for bernoully's principle and the basic
jist of
> > it. I've got a paper due soon and i need some help. Anyway, thanks a
lot.
> >
> > ~Stephen Shirk
>
> P + (1/2)(v*2)(Dens) + gh(Dens) = constant.
> (as derived in intro college physics books)
>
> Where:
> P is the pressure in the fluid,
> v*2 is the sqare of the velocity of the fluid,
> g is gravitaional accn (32ft/sec/sec or 9.8m/s/s depending on units),
> h the vertical position above a reference point,
> and Dens is the density of fluid.
>
> In a carburetor or near a wing, for example, h is relatively constant
> so in that case you can approximate
>
> P + (1/2)(v*2)(Dens) = constant
>
> This means if the velocity of a fluid increases the pressure
> decreases (and vice versa). Hence there is a net lift on a wing
> shape due to lower pressure on upper side and suction in a carburetor
> due to lower pressure there.
>
> Hope that helps.
>
> Now can anybody tell me why flight instruction still includes
> the mythical centrifugal force on an airplane making a turn?
> The sideway forces on a turning airplane are NOT BALANCED.
>
Hilton Goldstein
Bernoulli accounts for 100% of the lift.
Newton accounts for 100% of the lift.
Bernoulli
=========
Examining the pressures around the aircraft (including airfoils) shows
that pressure differences account exactly for the forces exerted on the
aircraft.
Newton
======
The aircraft pushes on the air, and according to Isaac, the air pushes
on the aircraft.
Therefore an aircraft flies entirely because of three things (order
optional):
1. Bernoulli
2. Newton
3. Money
Hilton
--
Hilton Goldstein..............................hilton@sgi.com
650-933-5254 (phone)......................(fax) 650-390-6159
M/S 42M-945, 2011 N. Shoreline Blvd, Mountain View, CA 94043
http://reality.sgi.com/hilton
Love thy neighbor..., but don't get caught!
AJG
"Where the velocity is the greatest, the pressure is the least"
John Roncallo
Cankmaster wrote:
>
> I'm in 10th grade and my brother told me that i might be able to find some info
> on an aviation paper due in math. I'm sorry if I don't live up to your
> standards. My brother is the one who graduated in aviation sciences. All i
> was looking for was a little help and obviously you've got something stuck so
> far up your ass you can't realize it. Fuck you (John Roncallo) asshole.
Stephen
I would like to apologise for my rudeness. Perhaps it was a bit over the
top. But I would like to say that the reason for my response was not
because you dont know Bernoulli. It was because (and correct me if Im
wrong) the way your request was phrased it apeard to me you wanted us to
do your homework. A reasearch paper dosent start by asking people for
the basic jist. That you get on your own. The finner details are what
you ask for help with. The request you made looked more like something
someone would answer and you would print out and hand in. I'm sorry
that doesent wash with me but perhaps I'm jumping to conclusions and if
so I'm way out of line.
Gustavo Flores
In article <199804100339...@ladder01.news.aol.com> cankm...@aol.com (Cankmaster) writes:
>From: cankm...@aol.com (Cankmaster)
>Subject: Re: Bernoully's principle
>Date: 10 Apr 1998 03:39:23 GMT
>>I would say that if you have a paper due on it shortly, you would at
>>this point be able to tell us the basic jist of Bernoulli. I would also
>>recomend you start by spelling his name right.
>I'm in 10th grade and my brother told me that i might be able to find some info
>on an aviation paper due in math. I'm sorry if I don't live up to your
>standards. My brother is the one who graduated in aviation sciences. All i
>was looking for was a little help and obviously you've got something stuck so
>far up your ass you can't realize it. Fuck you (John Roncallo) asshole.
It is really not that difficult to look it up in a dictionary, if you can't
manage that, I think you have no business calling anybody an asshole.
Look yourself in the mirror first.
Gustavo Flores
TALLMAN
The one time when a flame-out can be a GOOD thing.
Gustavo Flores
In article <dkorn-10049...@alderan.tn.cornell.edu> dk...@alderan.tn.cornell.edu (David Kornreich) writes:
>From: dk...@alderan.tn.cornell.edu (David Kornreich)
>Subject: Re: Bernoully's principle
>In article <352E4E...@spam.please>, n...@spam.please wrote:
>>
>> Now can anybody tell me why flight instruction still includes
>> the mythical centrifugal force on an airplane making a turn?
>> The sideway forces on a turning airplane are NOT BALANCED.
>>
>> -----------------------------------------------------------------
>> Paul Lee ........... Abri Technologies ......... http://abri.com/
>They certainly are if you think about the (accelerating) reference frame
>of the airplane, which presumably is where you are sitting when you really
>want to know about these things.
>I suggest an experiment. Next time you go into a steep right turn, grease
>down your seat, unbuckle your seat belt and open the door. When you are
>shot out of the airplane by centrifugal force, then will you still be
>claiming that it is "mythical?" How about if I put you on a space station
>which produces artificial gravity by rotation? Will you stubbornly float
>into the air because, as Bugs Bunny says, you never studied the laws of
>physics?
>Despite the unfortunate term "fictional force" often used to describe
>centrifugal and coriolis forces, they are indeed very real in accelerated
>frames, which is where you are when you're in a turning airplane. So why
>not use CF to describe what's going on? It's what the you and the wing
>experience every time you go into a turn. Eliminating it just because you
>have a bias for inertial frames is confusing because it is non-intuitive,
>and because you end up descibing the situation from the standpoint of a
>hypothetical observer hovering in midair. Then you end up (in essence)
>explaining CF anyway when your passenger asks why (s)he is pressed against
>the left bulkhead during a right turn. Better to describe it from the
>pilot's perspective in the first place.
>dave ::)
>--
> David A. Kornreich, PP-ASEL -><- dk...@alderan.tn.cornell.edu
> Cornell University Space Sciences
> The Fraternal Order of the Eternal Employees of Floyd
> ** We Specialize in Circumstances Beyond Our Control **
I f you were to grease the seat of your airplane and slid out on a turn ( it
would aslo have to be a hell of an uncoordinated turn) , you would see that
your path is TANGENTIAL to the turn and NOT radial as the pseudo force called
centrifugal would suggest. This is consistent with Newton's first law.
There is no centrifugal force, what you are experiencing is the effect of
centripetal acceleration, the very same accelerattion that causes the body to
turn, otherwise it would continue to move in a straight line.
Gustavo Flores
" One who paid attention in freshman physics "
Jerry Bransford
True... such as in an automobile you'll experience being 'pressed
against the door' in a hard/fast turn. In a coordinated turn in an
airplane however, the force is felt down through the seat, not
sideways. The only time you'd feel the sideways force during a normal
turn inside an airplane would be in a heck of an uncoordinated turn!
Notice in commercial flights when the pilot is making very smooth turns
to keep the passengers happy, that no side-to-side movement or feeling
that you're being forced against the window is ever encountered.
David Kornreich
In article <352E7F...@spam.please>, n...@spam.please wrote:
> David Kornreich wrote:
> >
> > In article <352E4E...@spam.please>, n...@spam.please wrote:
> >
>
> The issue is the motion of the plane as one body and not accelerated
> frames INSIDE
> the airplane. What accelerated reference frame factors can you include
> for this calculation? The motion of the airplane body IS accelerated
> toward the center
> and that is produced by the unbalanced centripetal force (=mv*2/R). The
> rate of
> turn is directly related to the unbalanced centripetal force (the
> DEFLECTING force
> towards the center). A constant unbalanced centripetal force produces
> a constant rate of turn. A body with balanced sideways forces would
> continue in a straight line and not be deflected towards the center.
I know all this, but is tangential to my point. In the frame rotating with
the airplane, there is an additional force (centrifugal) which balances
the centripetal force of lift. In this frame, the airplane doesn't
accelerate at all. When we talk about "forces being in balance," we're
usually talking about what you need to do in the pilot's seat to keep the
airplane in a coordinated turn. The *easiest* way for most people to think
about it is that while you are flying along, in your turn, you want to be
at the correct bank angle such that you feel all accelerations in the
"downward" (through your seat) direction. This direction is changing wrt
any inertial co-ordinates. It is perfectly correct, semantically and
physically, to say that this direction is the direction of the resultant
of the centrifugal and gravitaional forces. And since this is the way most
non-physicists think about it, why not teach it that way?
> > .... as Bugs Bunny says, you never studied the laws of
> > physics?
>
> I got a PhD in theoretical physics and thought it for over 20 years.
> Lets leave buggs bunny out of this and stick to physics.
>
> Sorry to be so voluminous about this. Shortest answer is often
> the best.
Knowing the internet, I should have smiley faced that sentence. Sorry. It
was not meant as any kind of slur, but as a joke. Obviously you know
plenty of physics. But you didn't answer the question. If you're in a
rotating spaceship with no windows, is the centrifugal force which holds
you to the "ground" still unreal? Would you do all your calculations in an
inertial frame that had nothing really to do with the lab, and go though
the awful trouble of converting back and forth? Do you calculate weather
patterns in an inertial frame, or do you use coriolis forces to explain
atmospheric circulation? One is certainly much more difficult than the
other.
Sorry about all this, but derision of "ficticious" forces is a pet peeve
of mine. Poor centrigual and corilois effects. They can make everything so
much simpler, but everyone maligns them. It is all very sad.
dave.
David Kornreich
In article <pilot.507...@iamerica.net>, pi...@iamerica.net (Gustavo
Flores) wrote:
> In article <dkorn-10049...@alderan.tn.cornell.edu>
dk...@alderan.tn.cornell.edu (David Kornreich) writes:
> >From: dk...@alderan.tn.cornell.edu (David Kornreich)
> >Subject: Re: Bernoully's principle
> >Date: Fri, 10 Apr 1998 15:12:30 -0500
>
> >In article <352E4E...@spam.please>, n...@spam.please wrote:
>
> >>
> >> Now can anybody tell me why flight instruction still includes
> >> the mythical centrifugal force on an airplane making a turn?
> >> The sideway forces on a turning airplane are NOT BALANCED.
> >>
> >> -----------------------------------------------------------------
> >> Paul Lee ........... Abri Technologies ......... http://abri.com/
>
> >They certainly are if you think about the (accelerating) reference frame
> >of the airplane, which presumably is where you are sitting when you really
> >want to know about these things.
>
> >I suggest an experiment. Next time you go into a steep right turn, grease
> >down your seat, unbuckle your seat belt and open the door. When you are
> >shot out of the airplane by centrifugal force, then will you still be
> >claiming that it is "mythical?" How about if I put you on a space station
> >which produces artificial gravity by rotation? Will you stubbornly float
> >into the air because, as Bugs Bunny says, you never studied the laws of
> >physics?
>
> >Despite the unfortunate term "fictional force" often used to describe
> >centrifugal and coriolis forces, they are indeed very real in accelerated
> >frames, which is where you are when you're in a turning airplane. So why
> >not use CF to describe what's going on? It's what the you and the wing
> >experience every time you go into a turn. Eliminating it just because you
> >have a bias for inertial frames is confusing because it is non-intuitive,
> >and because you end up descibing the situation from the standpoint of a
> >hypothetical observer hovering in midair. Then you end up (in essence)
> >explaining CF anyway when your passenger asks why (s)he is pressed against
> >the left bulkhead during a right turn. Better to describe it from the
> >pilot's perspective in the first place.
>
> >dave ::)
>
> >--
> > David A. Kornreich, PP-ASEL -><- dk...@alderan.tn.cornell.edu
> > Cornell University Space Sciences
> > The Fraternal Order of the Eternal Employees of Floyd
> > ** We Specialize in Circumstances Beyond Our Control **
>
> I f you were to grease the seat of your airplane and slid out on a turn ( it
> would aslo have to be a hell of an uncoordinated turn) , you would see that
> your path is TANGENTIAL to the turn and NOT radial as the pseudo force
called
> centrifugal would suggest. This is consistent with Newton's first law.
> There is no centrifugal force, what you are experiencing is the effect of
> centripetal acceleration, the very same accelerattion that causes the body to
> turn, otherwise it would continue to move in a straight line.
You're right about the uncoordinated bit. I thought I had said that.
Apparently not. Think about that, or a turn in a car. Sorry. Anyway, the
rest of your observations are totally incorrect in the pilot's (driver's)
frame of reference. To an outside observer in an inertial frame, the
cessation of centripetal force causes the unfortunate driver to move
tangentially to the orbit. But if you had read my post critically instead
of giving the canned Phys 1 answer, you would know that is not what I'm
talking talking about. I'm talking about the pilot's (driver's) version of
things, and he is in the frame rotating with the airplane (car). If you
want me to spell it out for you, he feels, in addition to all the inertial
forces acting on him, an additional force equal to m*omega X (omega X r)
where X is a cross product and r is the radius of the turn. This force is
_directly_outward_, which is exactly where you will be thrown from a
turning car in the car's (and your) reference frame.
To prove it to yourself, roll down your car window and place a tennis ball
on the "sill" (or whatever you call it in a car). Go into a gradual turn
(which cuts down on coriolis forces). You will observe the tennis ball
initially fly directly outwards (and down, of course) away from the car.
Only your friend standing on the ground will see the ball fly tangentially
away from you.
(This is applicable also to a plane in a skid. In coordinated flight the
ball should stay put. This is the idea behind the, well, ball in a turn
coordinator. _That_ ball certainly moves inwards and outwards, not
tangentially, wouldn't you say?)
This is inconsistent with Newton's first law. But then, Newton's first law
is only applicable to inertial frames.
There _is_ centrifugal force. It exists. It is measurable. It is required
for a rotating observer to get the right answer without switching frames.
> Gustavo Flores
>
> " One who paid attention in freshman physics "
dave, champion of downtrodden noninertial forces.
" One who paid attention in graduate physics "
Gustavo Flores
>In article <pilot.507...@iamerica.net>, pi...@iamerica.net (Gustavo
>Flores) wrote:
>
>> In article <dkorn-10049...@alderan.tn.cornell.edu>
>dk...@alderan.tn.cornell.edu (David Kornreich) writes:
>> >From: dk...@alderan.tn.cornell.edu (David Kornreich)
>> >Subject: Re: Bernoully's principle
>> >Date: Fri, 10 Apr 1998 15:12:30 -0500
>>
>>
Ok Dave. champion of downtrodden noninertial forces.
First of all you have a sense of humor, that's good.
Second , you are correct that from the airplane's occupants'
perspective the force is real and measurable,
( g meter, pressing of ass against seat and all that)
However... I always have trouble when people include it when viewing
the whole phenomenon form the outside forcing the horizontal
component of lift balance the nonexistent centrifugal force.
They even draw neat, but inaccurate vector diagrams.
(Your friend on the ground will not see the cenrtrifugal force)
And yes, you in the car or airplane will only see the ball get away
from you in a straight line.
I see that you not only paid attention in graduate physics, but also
have not forgotten freshman physics.
There are many who have, if you don't believe me I refer you to Barry
Schiff's article on the downwind turn and all the "rebuttals"
( I have copies if you want them , most amusing)
One more thing I also call those devices in the lab centrifuges and
not " Newton's first law machines"
Gustavo Flores
******* NOTICE ************
Due to unforeseen circumstances, the clairvoyant society meeting
is cancelled tonight.
TALLMAN
Gustavo Flores wrote:
> On Sun, 12 Apr 1998 18:47:10 -0500, dk...@alderan.tn.cornell.edu
> (David Kornreich) wrote:
>
>
> perspective the force is real and measurable,
> ( g meter, pressing of ass against seat and all that)
>
> However... I always have trouble when people include it when viewing
> the whole phenomenon form the outside forcing the horizontal
> component of lift balance the nonexistent centrifugal force.
>
So what's it going to be? Centripetal force or not? Acceleration
or no acceleration?
> ******* NOTICE ************
> Due to unforeseen circumstances, the clairvoyant society meeting
> is cancelled tonight.
The meeting will be rescheduled on ... well, you know.
Julian Scarfe
Paul Lee wrote in message <352E4E...@spam.please>...
> >
> >the mythical centrifugal force on an airplane making a turn?
> >The sideway forces on a turning airplane are NOT BALANCED.
John Lowry wrote:
> The only reason that nonsensical idea persists is that the FAA is
> intellectually challenged. And fails to respond to that challenge.
I don't understand John. You were quite happily talking about "coriolis
force" in the discussion of two aircraft flying in opposite directions at the
equator. Why now do you seem to be rejecting its centrifugal neighbour?
--
Julian Scarfe
rober...@osi.com
In article <35337091...@scigen.co.uk>,
In the five years I spend getting my engineering degree, I took
a lot of physics classes. Centrifugal force was not a debated topic,
every prof. just said it isn't real. It make sense when explained,
and we never thought twice about it.
I guess if you are talking from the point of someone in the plane they
may feel something they call 'centrifugal' force, but its really
just their body's reacting to the plane's continuous change in direction
(like Newton predicted)
I guess I don't understand why, as a bunch of pilots, we are debating
something that physicists have since finished.
John Lowry
Hi, Julian:
You have a point in the sense that the FAA does not explicitly lay out
what coordinate system they are assuming. And so there is some wiggle room.
But in their Flight Training Handbook, to take one of two examples I have
handy, pp 283 ff, they IMPLICITLY use an inertial frame. For example: "...
an object at rest or moving in a straight line will remain at rest or
continue to move in a straight line until acted on by some other force... in
the SAME paragraph in which they say "Centrifugal force is the 'equal and
opposite reaction' of the airplane to the change in direction and acts equal
and opposite to the horizontal component of lift."
That sort of thing.
John.
P.S. I just finished the gliding chapter of my new book "Performance of
Light Aircraft" and came up with one little thing you'll be interested in.
Take a helical gliding flight path. The radius of curvature of the path,
rho, is greater than the radius r of the cylinder on which the path is
wound. So is V = r*omega or is V = rho*omega? Neither, it's V =
sqrt(r*rho)*omega. Not particularly of practical import, but interesting.
J.
Julian Scarfe wrote in message <35337091...@scigen.co.uk>...
>Paul Lee wrote in message <352E4E...@spam.please>...
>> >
>> >Now can anybody tell me why flight instruction still includes
>> >the mythical centrifugal force on an airplane making a turn?
>> >The sideway forces on a turning airplane are NOT BALANCED.
>
>John Lowry wrote:
>
>> The only reason that nonsensical idea persists is that the FAA is
>> intellectually challenged. And fails to respond to that challenge.
>
>I don't understand John. You were quite happily talking about "coriolis
>force" in the discussion of two aircraft flying in opposite directions at
the
>equator. Why now do you seem to be rejecting its centrifugal neighbour?
>
>
>Julian Scarfe
Gustavo Flores
In article <3532F8...@nr.infi.net> TALLMAN <tal...@nr.infi.net> writes:
>From: TALLMAN <tal...@nr.infi.net>
>Subject: Re: Bernoully's principle
>Gustavo Flores wrote:
>> On Sun, 12 Apr 1998 18:47:10 -0500, dk...@alderan.tn.cornell.edu
>> (David Kornreich) wrote:
>>
>> >This is inconsistent with Newton's first law. But then, Newton's first
>law is only applicable to inertial frames.
>>
>> Second , you are correct that from the airplane's occupants'
>> perspective the force is real and measurable,
>> ( g meter, pressing of ass against seat and all that)
>>
>> However... I always have trouble when people include it when viewing
>> the whole phenomenon form the outside forcing the horizontal
>> component of lift balance the nonexistent centrifugal force.
>>
>> Gustavo Flores
>
>So what's it going to be? Centripetal force or not? Acceleration
>or no acceleration?
>> ******* NOTICE ************
>> Due to unforeseen circumstances, the clairvoyant society meeting
>> is cancelled tonight.
> The meeting will be rescheduled on ... well, you know.
BOTH in physics as well as in mathematics there often is more than one
explanation for a phenomenon.
In this case it depends on your frame of reference. But you must be very
careful to be consistent.
Gustavo Flores
David Kornreich
In article <3532b1c7...@nntp.iamerica.net>, xpi...@iamerica.net
(Gustavo Flores) wrote:
>
> Ok Dave. champion of downtrodden noninertial forces.
>
> First of all you have a sense of humor, that's good.
>
> Second , you are correct that from the airplane's occupants'
> perspective the force is real and measurable,
> ( g meter, pressing of ass against seat and all that)
>
> However... I always have trouble when people include it when viewing
> the whole phenomenon form the outside forcing the horizontal
> component of lift balance the nonexistent centrifugal force.
> They even draw neat, but inaccurate vector diagrams.
I'm still not so sure of that. I know the vector diagrams you're speaking
of, but (and I don't know the asnwer to this offhand) do they explicitly
say that the vector digrams are drawn in the frame of a ground observer?
Most non-physicists understand what "centrifugal force" is, but few relate
to "centripetal force." That, I assume, is why the diagrams are like that.
> (Your friend on the ground will not see the cenrtrifugal force)
>
> And yes, you in the car or airplane will only see the ball get away
> from you in a straight line.
Yep, it's all just a matter of perspective. I just prefer the perspective
of the pilot, rather than the perspective of the observer on the ground,
when trying to explain the turn coordinator and "balance of forces" while
in the air.
> One more thing I also call those devices in the lab centrifuges and
> not " Newton's first law machines"
*Snicker* Cool!
> Gustavo Flores
Dave ::)
John Lowry
Dear Julian, Paul, Robert and All:
About the statement --
>In the five years I spend getting my engineering degree, I took
>a lot of physics classes. Centrifugal force was not a debated topic,
>every prof. just said it isn't real. It make sense when explained,
>and we never thought twice about it.
>
>I guess if you are talking from the point of someone in the plane they
>may feel something they call 'centrifugal' force, but its really
>just their body's reacting to the plane's continuous change in direction
>(like Newton predicted)
>
>I guess I don't understand why, as a bunch of pilots, we are debating
>something that physicists have since finished.
>
I'd say we're still debating because to internalize a concept takes a lot
more than having seen it "demonstrated" once on some blackboard. That's part
of the charm and frustration of the subject (physics and aviation).
Tobus
Actually, the problem you two (Dave and Gustavo) are facing is not new
-- the neophyte in physics vs the expert. Richard Feynman, in his
famous "Lectures on Physics" sums it up nicely, and I'll paraphrase:
If you teach physics correctly from the beginning, you'll confuse
everybody, because you'd have to teach all sorts of esoteric stuff
(quantum mechanics, relativity, etc.). Thus you compromise and teach
simple stuff (classical mechanics, conservation of mass, etc.) which are
only "true" in a very narrow set of circumstances, but in general, are
wrong. Eventually, you have to teach people that all that early stuff
has limited application, and then teach them the "right" stuff (with
apologies to Tom Wolfe and Chuck Yeager.)
So what you two guys are rehashing is the same argument that apparently
has been thrashed in physics departments for decades.
Nice to see that Feynman was right. Gustavo took a few semesters of
physics (as did I) and learned enough to understand the everyday
phenomena he needed to understand. Dave took six bzillion advanced
courses, and learned the limitations of the undergrad knowledge
(presumably, enabling him to do cool stuff). The system worked!
Julian Scarfe
> In article <35337091...@scigen.co.uk>, I wrote:
> I don't understand John. You were quite happily talking about "coriolis
> force" in the discussion of two aircraft flying in opposite directions at
> the equator. Why now do you seem to be rejecting its centrifugal neighbour?
rober...@osi.com wrote:
> In the five years I spend getting my engineering degree, I took
> a lot of physics classes. Centrifugal force was not a debated topic,
> every prof. just said it isn't real. It make sense when explained,
> and we never thought twice about it.
>
> I guess if you are talking from the point of someone in the plane they
> may feel something they call 'centrifugal' force, but its really
> just their body's reacting to the plane's continuous change in direction
> (like Newton predicted)
>
> I guess I don't understand why, as a bunch of pilots, we are debating
> something that physicists have since finished.
Well, this physicist hasn't "finished" it!
"Is centrifugal force real?" is a slightly different question from "Is
centrifugal force a useful concept in solving problems?". And what's useful
depends on the sort of problem that you're solving.
In engineering, it may be possible to solve every problem you come up against
in an inertial frame of reference, and avoid fictitious forces entirely. And
if you can, then all well and good -- you can dispense with the idea of
centrifugal force. Likewise, if you solve all your problems in mechanics using
Hamilton's equations, you can probably dispense with the idea of *force*
itself. Does that therefore cease to be "real" too?
If you try solving anything more than the most trivial problem involving
three-dimensional motion in naturally rotating frames, it gets tough if you
don't use fictitious forces. Yes, you *can* demonstrate why circulation
around a low is counterclockwise in the northern hemisphere by solving the
problem in an inertial frame in which the earth's surface is rotating, but
it's a lot easier if you introduce coriolis force, which is no more or less
real than centrifugal force, and use the rotating frame of the earth.
There are two aspects to modern physics: one is about problem solving, the
other about natural philosophy. The former demands useful concepts, and I
think centrifugal force, used appropriately and consistently, can be one of
those. (That's why we had to teach our physics students at Cambridge about
fictitious forces, sometimes asking them to "unlearn" what they learned about
its non-existence at high school.) The latter is more concerned with what is
real in physics, and when you start to learn about relativity, what is "real"
in mechanics starts to depend on who's asking, and what frame of reference
they use. Again, fictitious forces have their uses in a deeper understanding
of that.
--
Julian Scarfe
rober...@osi.com
In article <353491B9...@scigen.co.uk>,
I guess that I agree with you if using a non-existant force helps you
to get a correct solution in less time, than its useful.
A similar useful example would be a "spin force". Rather than explaining
all the things that are going on in a plane during a spin, you can
just tell the physics sthudent that there is a mysterious "spin force"
that spins the plane around. Then ask the student to solve a problem using
it. The student should be able to get the correct answer and understand his
part of the issue without spending a year learning what a spin really is
(from a dynamic forces point of view).
Of course, this is how real world engineering problems are solved.
Not by relearning and figuring out things that people spent lifetimes
learning hunderds of years ago, but rather take their ending as a starting
point.
So, if you are saying that centrifugal force is useful like this, I'd
agree with you.
When a two year old asks why a car moves, you can tell him because
the wheels turn. For him, its better to just assume that there is a
strange force that magicly turns the wheels. That gets him
the answers that he really needs. Then in 10 years you can have him
assist in rebuilding the engine. :)
John Lowry
Dear Julian, Robert, and All:
Fictitous forces (centrifugal and Coriolis, for us now) are only
fictitious in the sense that no external force (electromagnetic,
gravitation, thrust, drag, lift, etc.) underlies them. That does NOT mean
they don't have REAL effects (deflections of projected objects, Foucault's
pendulum's plane rotating, increased wear on right river banks in the
Northern hemisphere, etc.).
To add to the confusion, 'centrifugal' is used in two different senses:
(1) a fictitious force due to our referring motions to an accelerated
(rotating) coordinate system; and (2) the equal and opposite real force to
centripetal. For the latter, consider a stone being whirled in a roughly
circular path by your hand holding a string attached to the stone. There is
a centripetal force (of the string) on the stone; there is an equal and
opposite centrifugal force (of the string) on your hand.
But, in the airplane case, which is relatively simple, and depicted as
by an observer in an inertial frame momentarily behind the airplane (in the
usual picture), there is no need to invoke pseudo-forces. When the airplane
banks, part of lift (an aerodynamic force, of the air on the airframe)
pushed the airplane to the side. The airplane actually accelerates (moves in
a circular path) so, in this inertial frame, no centrifugal force, on the
airplane, is present. There is a centrifugal force, in sense #2 above, but
it is on the AIR, swooshed outwards.
None of this has anything to do with either special or general
relativity, only with ordinary Galilean relativity.
When I taught this stuff at Colorado Rocky Mountain School (and I
suppose in various colleges, though I don't remember) I used an accelerated
cart (via weight on string over pulley attached to table edge) with a
scaffold supporting a plumb bob. We could EITHER consider the inertial frame
of the laboratory and incorporate the weight making the cart accelerate, OR
(better, XOR) we could consider the non-inertial frame of the accelerated
cart and invoke a fictitious force to explain the rearwards deflection of
the plumb bob. Didn't matter, the REAL deflection was the same in either
case. Ordinary mechanics is invariant under Galilean transformations. The
rotating coordinate system is simply more of the same, though considerably
more complicated.
John.
P.S. Did you know that your inclinometer can be used (via the off-center
deflection of the ball) to assess your "excess" yaw rate in a slip or skid?
One of the least useful relations I've ever derived.
John T. Lowry, PhD
Flight Physics; Box 20919; Billings MT 59104
Voice: 406-248-2606; E-mail: jlo...@mcn.net
Julian Scarfe wrote in message <353491B9...@scigen.co.uk>...
>--
>
>Julian Scarfe
rober...@osi.com
In article <6h2kug$nu8$1...@news.mcn.net>,
"John Lowry" <jlo...@mcn.net> wrote:
>
> Dear Julian, Robert, and All:
> Fictitous forces (centrifugal and Coriolis, for us now) are only
> fictitious in the sense that no external force (electromagnetic,
> gravitation, thrust, drag, lift, etc.) underlies them. That does NOT mean
> they don't have REAL effects (deflections of projected objects, Foucault's
> pendulum's plane rotating, increased wear on right river banks in the
> Northern hemisphere, etc.).
>
Another example is gravity. Do we really want to tell people that
gravity is not a force (not like they probably think it is) but
rather the affect of the earth curving space inward, causing us to
be continuosly falling against the earth (thus giving the feeling of gravity).
Its much easier to just say 'gravity' and it allows people to get correct
answers to difficult problems.
Tallman
I'm just curious if the folks at the FAA have these same discussions
prior to implementing policy and procedure....
I attached a little joke that sort of reminded me of this thread:
=================================================================
> >How Hot Is It In Hell?
> >(A True Story)
> >
> >A thermodynamics professor had written a take-home exam for his
> >graduate
> >students. It had one question: "Is Hell exothermic (gives off heat)
> >or endothermic (absorbs heat)? Support your answer with a proof."
> >Most of the students wrote proofs of their beliefs using Boyle's Law
> >(gas
> >cools off when it expands and heats up when it is compressed) or some
> >variant.
> >
> >One student, however, wrote the following:
> >
> >First, we need to know how the mass of Hell is changing in time.
> >So,we need
> >to know the rate that souls are moving into Hell and the rate they
> >are
> >leaving. I think that we can safely assume that once a soul gets to
> >Hell, it
> >will not leave. Therefore, no souls are leaving. As for how many
> >souls are
> >entering Hell, let's look at the different religions that exist in
> >the world
> >today. Some of these religions state that if you are not a member of
> >their
> >religion, you will go to Hell. Since there are more than one of
> >these
> >religions and since people do not belong to more than one religion,
> >we can
> >project that all people and all souls go to Hell. With birth and death rates
> >as they are, we can expect the number of souls in Hell to increase
> >exponentially. Now,we look at the rate of change of the volume in Hell
> >because Boyle's Law states that in order for the temperature and
> >pressure in
> >Hell to stay the same, the volume of Hell has to expand as souls are
> >added.
> >This gives two possibilities:
> >
> >1) If Hell is expanding at a slower rate than the rate at which souls enter
> >Hell, then the temperature and pressure in Hell will increase until
> >all Hell breaks loose.
> >2) Of course, if Hell is expanding at a rate faster than the > >increase of
> >souls in Hell, then the temperature and pressure will drop until Hell
> >freezes over.
> >
> >So which is it? If we accept the postulate given to me by Ms.
> >Therese > Banyan
> >during my Freshman year that, "It will be a cold night in Hell before I go out
> >with you," and take into account the fact that I still have not
> >succeeded in
> >dating her, then #2 cannot be true, and so Hell is exothermic.
> >
> >The student got the only A.
===================================================================
Andrew M. Sarangan
> In the five years I spend getting my engineering degree, I took
> a lot of physics classes. Centrifugal force was not a debated topic,
> every prof. just said it isn't real. It make sense when explained,
> and we never thought twice about it.
>
> I guess if you are talking from the point of someone in the plane they
> may feel something they call 'centrifugal' force, but its really
> just their body's reacting to the plane's continuous change in direction
> (like Newton predicted)
>
> I guess I don't understand why, as a bunch of pilots, we are debating
> something that physicists have since finished.
>
Whether or not centrifugal force exists is a meaningless discussion.
A more important question is whether or not it is a useful quantity,
and whether it leads to results that agree with the observations we
make in the real world.
As a matter of fact, there is nothing "real" about any force. It is a
fictitious quantity. Newton defined force in terms of other fictitous
quantities.
An important principle in Engineering is that everything is a model.
Everything from Newton's laws to Quantum Physics to Relativity are just
models that seem to fit the observations of our environment. There is
nothing "true" or "deep" about them. We think of something as being real
if it intuitively makes sense to us. That is hardly a scientific
definition of "real".
--
Andrew Sarangan
PP-ASEL-IA http://lights.chtm.unm.edu/~sarangan/aviation/
Carsten Mueller
rober...@osi.com wrote:
: I guess that I agree with you if using a non-existant force helps you
: to get a correct solution in less time, than its useful.
You cannot call the centrifugal and coriolis forces non-existant. They are
only 'fictitious' as long as they don't interact with the opposite rotating
frame. As soon as they do, they become real mechanical forces (or in the
non-rotating reference frame, their counterparts - centripetal force and
(is there a name for the negative coriolis force?)). Which part of the couple
is considered real depends only on the reference frame you are using, and
if one restricts oneself to the non-rotating frame because it's an inertial
frame, then those opposite forces (centripetal..) are present which are real
and differ from the 'fictitios forces' only in sign.
Example: The river washing out one of the shores due to coriolis force.
Described in a non-rotating frame, it is the rotating earth that forces
the flow of water to change it's direction.
: A similar useful example would be a "spin force". Rather than explaining
: all the things that are going on in a plane during a spin, you can
: just tell the physics sthudent that there is a mysterious "spin force"
: that spins the plane around. Then ask the student to solve a problem using
: it. The student should be able to get the correct answer and understand his
: part of the issue without spending a year learning what a spin really is
: (from a dynamic forces point of view).
Nice example, but this "spin force" would be real too, because it is the sum
of all the aerodynamical forces acting on the plane. It's a crude simplifi-
cation but a real force.
John S. Denker
Hi Folks --
In this thread, several people have posted the right answer. Alas, several
people have vehemently advocated various wrong answers, and no clear consensus
has yet emerged, so I thought I would weigh in (so to speak :-).
Let me make an analogy:
The gravitational field at any given point is an acceleration. It acts on
objects, producing a force in proportion to the object's mass. Most people
are willing to believe that gravity exists.
The centrifugal field at any given point is also an acceleration. It
too acts on objects, producing a force in proportion to the object's mass.
In an inertial frame, neither gravity nor centrifugity exist. Both emerge as
consequences of working in a non-inertial frame.
For a ground-bound observer analyzing the flight of an airplane, it is often
convenient to use a frame where gravity exists and centrifugity does not.
However, for the pilot and passengers in an airplane, it is usually more
convenient to use a frame that includes both gravity and centrifugity. I can
feel the centrifugal force in my cheeks.
I must also refute the suggestion centrifugal effects are like magic and are
used as a substitute for understanding the situation or performing a real
analysis. The problem with magic is that it can explain anything, real or
unreal. Centrifugal and gravitational effects are the quantitatively
predictable ingredients of an accurate analysis in an accelerated frame.
Cheers --- jsd
To reply to me, change ".nospam" to ".com" in the address above.
Bill Leonard
In article <6h5o19$f9o$1...@news.monmouth.com>, j...@monmouth.nospam (John S. Denker) writes:
> Let me make an analogy:
>
> The gravitational field at any given point is an acceleration. It acts on
> objects, producing a force in proportion to the object's mass. Most people
> are willing to believe that gravity exists.
>
> The centrifugal field at any given point is also an acceleration. It
> too acts on objects, producing a force in proportion to the object's mass.
Er, no. Your analogy breaks down right here, because there is no such
thing as a "centrifugal field". Centrifugal force is only an *apparent*
force, apparent only to an observer in a non-inertial reference frame.
> In an inertial frame, neither gravity nor centrifugity exist. Both emerge as
> consequences of working in a non-inertial frame.
Wrong. Gravity most certainly does exist in an inertial frame of
reference. I think you are confusing two things: being in the presence of
a force (such as gravity), but not being accelerated, and being accelerated
by a force. The former is still an inertial frame of reference, while the
latter is not.
Thus, while standing on the ground and not moving with respect to the
earth, you are in an inertial frame of reference, yet you feel the effects
of gravity.
If you are inside a container that is rotating (such as inside a
centrifuge), though, you are *not* in an inertial frame. What you perceive
as centrifugal force is the result of your senses believing that you are
motionless when in fact you are being accelerated (i.e., your direction of
motion is constantly changing). What is actually happening is that a
*centripetal* force (one directed towards the center of rotation) is acting
on you and your surroundings to change your direction of motion. If you
release an object from your hand, you (in your non-inertial frame) think
that the object is acted on by centrifugal force because it (apparently
spontaneously) moves outward from the center of rotation. What is actually
going on is that the object was in motion before you released it, and after
being released continues to move in a straight line according to Newton's
Laws of Motion.
The G forces you feel, which many people describe as centrifugal force, in
circumstances such as a turning airplane is actually the airplane pushing
against you in order to change your direction of motion. Your senses tell
you that some force is trying to pull you through the bottom of the plane,
but in fact it is the plane pushing against you.
--
Bill Leonard
Concurrent Computer Corporation
2101 W. Cypress Creek Road
Fort Lauderdale, FL 33309
Bill.L...@mail.ccur.com
These opinions and statements are my own and do not necessarily reflect the
opinions or positions of Concurrent Computer Corporation.
------------------------------------------------------------------------------
It doesn't matter which way you are headed if you're not going anywhere.
------------------------------------------------------------------------------
Julian Scarfe
> In article <6h5o19$f9o$1...@news.monmouth.com>, j...@monmouth.nospam (John S. Denker) writes:
> > Let me make an analogy:
> >
> > The gravitational field at any given point is an acceleration. It acts on
> > objects, producing a force in proportion to the object's mass.
> > Most people are willing to believe that gravity exists.
> >
> > The centrifugal field at any given point is also an acceleration. It
> > too acts on objects, producing a force in proportion to the object's mass.
Bill Leonard wrote:
>
> Er, no. Your analogy breaks down right here, because there is no such
> thing as a "centrifugal field".
Sure there is. The centrifugal field is Omega x Omega x r, per unit mass,
where r is the position vector within the field and Omega is a constant (just
like G for the gravitational field). What's the difference?
> > In an inertial frame, neither gravity nor centrifugity exist. Both emerge
> > as consequences of working in a non-inertial frame.
>
> Wrong. Gravity most certainly does exist in an inertial frame of
> reference.
How can I tell if a frame is inertial or not? Surely only by launching a test
particle and seeing if Newton's First Law is obeyed!?
In a "rotating frame", the test particle accelerates without a visible force
acting on it. I'd like to say it deviated because of "centrifugal force", but
you won't let me ;-) So we disqualify the frame as being non-inertial.
In a frame where you are "standing on the ground and not moving with respect
to the earth", the test particle accelerates without a visible force acting on
it. You'd like to say it deviated because of "gravity", but I won't let you
;-) So we disqualify the frame as being non-inertial.
Where's the difference?
Don't get me wrong -- most people, including me most of the time, are
"framist" like Bill seems to be. But the physics of the universe, if done
properly, has to be consistent. It's somewhere in the Constitution.
[with apologies to John and Albert, both of whom could probably have put it
better!]
--
Julian Scarfe
Raul Alvarez
Gun One wrote:
>
> In article <352E4E...@spam.please>, Paul Lee <n...@spam.please> writes:
> |> Now can anybody tell me why flight instruction still includes
> |> the mythical centrifugal force on an airplane making a turn?
> |> The sideway forces on a turning airplane are NOT BALANCED.
Centrifugal force is the force pushing inward on a rotating object.
Banked wings have a horizontal component of lift pointing inwards which
caused the plane to turn; nothing mythical about this. I guess all your
physics professors lied to you.
George R Patterson
Julian Scarfe wrote:
>
> rober...@osi.com wrote:
>
> > In the five years I spend getting my engineering degree, I took
> > a lot of physics classes. Centrifugal force was not a debated topic,
> > every prof. just said it isn't real.
> > I guess I don't understand why, as a bunch of pilots, we are debating
> > something that physicists have since finished.
>
> Well, this physicist hasn't "finished" it!
>
> "Is centrifugal force real?" is a slightly different question from "Is
> centrifugal force a useful concept in solving problems?".
I would also bet that Robert has a consistent misunderstanding here.
I've
not heard a physicist say centrifugal force isn't real in the sense that
centrigugal force doesn't exist. What I've heard them (including my
father,
whose degree was in physics) say is that it isn't a real force, which is
an
entirely different kettle of fish.
My physics text calls it a "pseudo-force", and says it's quite real.
George Patterson, N3162Q.
PHAR...@post.omnitel.net
I would like to try submitting my explanation to the centrifugal force
question.
There are no unreal forces nor pseudo forces. A force does exist or
does not only with reference to a specific reference system.
A system can be "inertial" (I don't know the English term, but it
means that it is steady or animated by rectilinear uniform movement)
or not.
If we sit on the Earth surface (can be considered inertial) and look
at an aircraft making a turn, there is no centrifugal force at all. As
a matter of fact, an object not subject to any force would proceed
with uniform rectilinear movement. If it turns, it means there is a
force that makes it turn, and it is the centripetal force.
But, if we sit in the aircraft, we experience a lateral acceleration
(we are pushed against the side wall): it means that there is a force,
and that force is called centrifugal force.
Regards.
P. Staffini (Italy)
John S. Denker
Hi Folks –
I like to stick to standard terminology whenever possible. Imagine my
surprise when I discovered (in the course of preparing this note) that
physicists are grossly inconsistent in their use of the term "inertial frame".
1) It appears that when discussing classical physics (Newton’s laws and such)
it is common practice to apply the term "inertial frame" to a reference frame
attached to the surface of a planet (provided we can neglect the rotation of
the planet). In an inertial frame, free particles are supposed to move in
straight lines, but in such a frame there *cannot be* any free particles,
because the frame is permeated by a gravitational field.
A rotating frame would not be considered inertial.
2) When discussing modern physics (general relativity and such) the *only*
acceptable inertial frames are free-falling. My laboratory is not
free-falling because it is being accelerated skyward by a huge force produced
by contact with the soil beneath the building.
Einstein’s principle of equivalence states that (at any given point) the
gravitational field is indistinguishable from an acceleration of the reference
frame. This statement applies equally well to steady, straight-line
accelerations (such as those produced by the surface of a gravitating planet)
as to the ever-changing accelerations produced by a centrifuge (or an airplane
in a steep turn).
**********
I regret any confusion arising from my inadvertent use of a term that was open
to misunderstanding.
I stand by the essential points of my previous note, namely
A) There is nothing wrong with working in an accelerated reference frame,
provided the work is done carefully. When people think this can’t be done,
it is just because they don’t know how to do it.
B) Accounting for centrifugal effects is only slightly trickier than
accounting for gravitational effects. There is a profound analogy between the
two phenomena.
C) The term "gravitational force field" is a slight misnomer. The field is an
acceleration (i.e. force per unit mass) not a force per se. Similar remarks
apply to "centrifugal force" – please remember it is a force per unit mass.
**********
In article <6h7o21$1...@hawk-hcsc.hcsc.com>, Bill.L...@mail.ccur.com wrote:
>Gravity most certainly does exist in an inertial frame of
>reference.
You are evidently using the classical-physics definition of inertial frame
(which is of course reasonable and appropriate to this newsgroup).
> I think you are confusing two things: being in the presence of
>a force (such as gravity), but not being accelerated, and being accelerated
>by a force. The former is still an inertial frame of reference, while the
>latter is not.
I was quite intentionally making no distinction between a gravitational field
and an accelerated frame, according to the modern-physics definition of
inertial frame.
Dave Thomas
>Einstein’s principle of equivalence states that (at any given point) the
>gravitational field is indistinguishable from an acceleration of the
reference
>frame.
I've never understood that.
Under acceleration, the force acting across all points in the field is
parallel. Under gravitation, the forces converge at the center of mass. So,
for anything other than an infinitely small point, it _is_ possible to
distinguish the two. If the point is infitintely small, then all the rules
go out the window anyway, so the statement becomes meaningless.
Ignorant Dave
John S. Denker
In article
<7F4570EC6D574705.70E24AA5...@library-proxy.airnews.
Fear not. It's not as complicated as it might seem.
Sure, as we move from point to point the centrifugal field changes in
direction and magnitude, and (as you point out) so does the gravitational
field.
It's really not much different from say, the velocity field that describes the
wind. The wind blows one way here and blows another way there, but we can
deal with it. We can still speak of the wind velocity at a particular point.
The rules do NOT break down when we consider points (or very small regions).
To consider what happens to a larger, rigid object, we see what happens at
each point and add up the contributions. No big deal.
For multiple points that are not rigidly connected, we observe eddies in the
atmosphere and orbits in the gravitational field. Again, no big deal.
Cheers --- jsd
John S. Denker
Hi –
My previous post may have missed the main point of your note, so let me come
at it from another direction to see if this is any better.
As you point out, a typical straight-line acceleration affects a reference
frame "here" the same as a reference frame "there", so we can establish one
big reference frame and forget about it. This is equivalent to the uniform
gravitational field of a really big planet really far away, or to a centrifuge
with a really long arm.
More generally, the gravitational and centrifugal fields change from point to
point. We cannot have one big reference frame; there has to be a different
frame at different points. With such a profusion of references, a non-expert
has to worry about getting lost, but really it’s not all that bad.
Consider an analogy: the direction called "North" in Oshkosh is not the same
as the direction called "North" in Seattle. If you fly from one place to the
other, you need to reset your Directional Gyro (by a total of about 25
degrees) to take this into account. The relationship between the various
local reference frames is not very complicated. If you know about the
curvature of the earth, you can program a computer to keep your DG aligned
with the local reference frames during your trip. Conversely, you could use a
good DG to *measure* the curvature of the earth.
Similarly, a free-falling reference frame in Oshkosh falls in a different
direction from one in Seattle. Once again if you know about the curvature of
spacetime (which is determined by the distribution of mass in your vicinity)
you can 100% predict and account for what happens to your instruments if you
switch from one reference frame to another. And again conversely, during a
long trip you can use gyros and accelerometers to measure the curvature of
spacetime.
Cheers --- jsd
Bill Leonard
In article <6hfio4$mi3$1...@news.monmouth.com>, j...@monmouth.nospam (John S. Denker) writes:
> Einstein's principle of equivalence states that (at any given point) the
> gravitational field is indistinguishable from an acceleration of the
> straight-line accelerations (such as those produced by the surface of a
> gravitating planet) as to the ever-changing accelerations produced by a
> centrifuge (or an airplane in a steep turn).
Consider the following experiment. I am placed inside a closed room, with
no opportunity to observe outside references. I observe that objects
released from my hand fall to the floor. My task is to determine whether I
am in the presence of a gravitational field, or whether my room is being
rotated at the end of a centrifuge arm (with the floor oriented outwards
from the center of rotation).
I proceed by using very precise instruments to measure the force on objects
at different locations in the room, paying special attention to the
*direction* of the force.
If I am in the presence of a gravitational field, I should see that the
forces all converge at a central point located "below" the floor, that
central point corresponding to the center of mass of the object producing
the gravitational field. If, however, I am at the end of a centrifuge, I
should observe that the forces all emanate from a central point somewhere
"above" my ceiling, that central point corresponding to the center of
rotation.
From this hypothetical experiment, I conclude that a gravitational field is
NOT equivalent to a non-linearly accelerated reference frame, such as a
centrifuge. I believe Einstein's general theory of relativity says only
that a uniformly accelerated reference frame is indistinguishable from a
gravitational field.
Bob Myers
In article <6hikvm$l...@hawk-hcsc.hcsc.com>,
John Denker said "at any given point". He's perfectly correct within
that limitation.
You chose to ignore that parameter, and created a finite sized room.
So you are arguing a different situation.
The gravitational field you mentioned above, converging below, *is*
distinguishable given sufficiently accurate measurements and a finite
sized room from a uniformly accelerated reference frame - uniform
acceleration doesn't converge. So your example does not hold for
what you believe Einstein's General Theory says, either, for the same
reasons.
In any case, I strongly suspect it is possible (in theory) to create a
gravitational field indistinguishable from a centrifugal force field
within a finite room, by very careful mass distribution. You don't have
to have a spherical mass generating the gravitational field. How
do you tell them apart now? Field geometry is not the issue you
think it is.
Sriram Narayan
In article <6hikvm$l...@hawk-hcsc.hcsc.com>, bi...@amber.ssd.csd.harris.com
says...
IMHO, if the arm of the centrifuge arm is sufficiently long, the force field
felt within your hypothetical room could be made "parallel" enough such that it
is indistinguishable form a uniform force field or a uniform gravity field. I
think it comes to a matter of how good a job you do of making the force vectors
all parallel to each other such that it is beyond the instruments' capability
of resolving the convergence direction (up or down as you claim is possible).
In such a situation, you couldn't tell if it were a gravitational field or a
pseudo field created by rotation or any other form of a linear force field.
--
sriram....@technologist.com http://members.home.net/snarayan/homepage.html
pp-asel:san francisco bay area:vfr flight planner:av articles:photo:sw dxing
Greg Boston
Ken,
Yours is the best answer so far. That reduced neck you refer to is often
used as, and is referred to, a venturi. Drill a hole in that reduced
area and you'll feel suction as the air or fluid(doesn't matter) rushes
through the length of the tube. The waterbed drain that hooks to your
faucet is a venturi. Another great example of Bernoulli's principle is
to use a high pressure air gun to direct a stream of air down between
your fingers with your hand as flat as possible. Place hand over a piece
of paper and watch it get 'sucked up' against your hand. You are
essentially recreating the airfoil thing where you get a low pressure
area above the paper with high pressure below, due to the rapid velocity
of air in all directions away from the bottom of your hand. I work in
the semiconductor industry and we used to have a tool that lifted wafers
with such an apparatus. It was called a Bernoulli head.
regards,
-gb-
Ron Parsons
It's nice that the calculations work enough to be able to predict results,
but there is no magic force of nature which makes increased velocity cause
lower pressure other than the friction between adjacent molecules moving
at different velocities. Bernoulli's equations are mathematical shortcuts,
not statements of physics.
Ron
Mark Mallory
Ron Parsons (jr...@gte.net) wrote:
: Bernoulli's equations are mathematical shortcuts,
: not statements of physics.
Pure BS!
Leonard Wojcik
I'm afraid that Ron is correct. The Bernoulli equations are a pragmatic
mathmatical description of what is observed. You can calculate things
from them, but they don't explain why it happens.
Anybody want to have a go at an explanation at a molecular level of what
goes on in the Bernoulli effect ? I am guessing it has something to do
with molecular momentum and mean free path ?????
--
Leonard Wojcik Leonard.d...@wojciktech.com
Bob Myers
In article <353FA0...@wojciktech.com>,
Leonard Wojcik <Leonard.d...@wojciktech.com> wrote:
>Mark Mallory wrote:
>>
>> Ron Parsons (jr...@gte.net) wrote:
>> : Bernoulli's equations are mathematical shortcuts,
>> : not statements of physics.
>>
>> Pure BS!
>
>I'm afraid that Ron is correct. The Bernoulli equations are a pragmatic
>mathmatical description of what is observed. You can calculate things
>from them, but they don't explain why it happens.
I would dispute that physics is about "why" things happen. It contend
that it is primarily and most importantly a pragmatic mathematical
description of what happens.
Example: Newton's Theory of Gravity says nothing about why masses
experience mutual gravitational attraction; it just describes that
attraction.
The philosophical models of most sciences aren't nearly as important
or as long-lasting as the pragmatic, often mathematical, descriptions,
IMHO.
>Anybody want to have a go at an explanation at a molecular level of what
>goes on in the Bernoulli effect ? I am guessing it has something to do
>with molecular momentum and mean free path ?????
Basically you're converting static pressure to dynamic pressure. Air
molecules are still moving at the same speed, but now they're moving more
in the direction of the overall flow than they are perpendicular to the
flow - and so produce reduced pressure perpendicular to the flow.
Harvey
>Anybody want to have a go at an explanation at a molecular level of what
>goes on in the Bernoulli effect ? I am guessing it has something to do
>with molecular momentum and mean free path ?????
>
>Leonard Wojcik Leonard.d...@wojciktech.com
I am cross-posting an item from a different newsgroup because it was a
very good description of how the Bernoulli principle works.
<QUOTE>
Dave Hall wrote:
There are three terms that need to be understood.
Static pressure is what a barometer measures. It is the pressure
either measured in a non-moving mass of fluid or measured against
a surface exactly parallel to the flow in a moving fluid stream.
Kinetic or dynamic pressure is that pressure created by fluid flow
and is given by the equation: 1/2 * density * (velocity squared).
Total or stagnation pressure for a given flow is simply the addition
of the above two pressures. It is called stagnation pressure
because it is the pressure you would measure at a point where the
velocity of the airflow "stagnates" or goes to zero. If you look
at the leading edge of a body in an airflow, the stagnation point
is where the airflow splits to go to either side of the body. At
that point, one can visualize a single line of flow that has been
decellerated to zero and the pressure measured would equal
stagnation pressure.
Bernoulli states that in a given fluid flow, stagnation pressure
remains constant. It is a simple energy conservation principle.
Since dymanic pressure is a squared function of velocity, it is
clear that as velocity increases, the static pressure must decrease.
Since static pressure is the one actually acting on surfaces in
airflows, you can see how it is possible to create a partial
vacuum on a surface in an airflow merely by virtue of the increased
velocity of the airflow.
The thing to get out of this is that the pressure that matters is
that measured against a surface PARALLEL to the airflow. As such,
it isn't really counter-intuitive that it decreases with increased
velocity.
> The airflow divides as the wing cuts into it; that
> flowing over the top of the wing has to go further in the same time,
> because of the aerofoil shape, but it rejoins as it leaves the wing,
> so the upper airflow must go faster. As you say, the air pressure
> over the wing is less than below, giving 'lift'.
One of the misconceptions I keep seeing in this thread is that
"lift" can be completely explained by the Bernoulli theorem. As a
first order approximation, it can, but it is way more complex than
that. The Bernoulli theorem is only really accurate in an idealized
one-dimensional flow through a channel. It also only works with
incompresible fluids, but for the speeds we are talking about, we
can consider air incompressible. You get out into the three-D
real world and there is a lot more involved than just Bernoulli.
We can leave it at that and live with the approximation.
The other misconception I have seen is that two air molecules that
are side-by-side but get separated at the leading edge somehow
miraculously make it to the trailing edge of the body at the same
instant and that this explains the increased velocity. It simply
doesn't happen that way. Again, the real world is not that simple.
Of course, it doesn't really matter that these two molecules don't
meet again, the air velocity is still greater on top in the idealized
model.
> The same idea applies to a carb venturi, which is a narrower
> section, that speeds up the airflow through it, and (Bernoulli)
> creates lower pressure. This "sucks" the gas (petrol) from the carb
> bowl through the pipes and it mixes with the air stream, then into
> your cylinders...
Actually, a carb venturi is a much better example of the Bernoulli
effect than a wing is. It really is close to one-dimensional
channel flow.
> But just a moment - the air isn't speeding up over the wing at
> all; the wing is moving through the air! Is the answer here simply
> that the air molecules are spaced wider apart over the top of the
> wing, so exert less pressure for that reason?!
Kinda sorta. This actually helps with the explanation a little. As
far as a molecule is concerned, it is just hanging around and suddenly
it must move up and out of the way and then back down again as a wing
passes by. Clearly, its velocity has increased during this disruption
of its boring life. If its partner molecule ends up under the wing
and
doesn't have to move nearly as far, its velocity increase is smaller.
Again, it is more complex than that in reality, but for lift on nice
rounded bodies like bugs, this is a good first order explanation.
> I've only ever seen the Bernoulli explanation for wing lift in texts,
> with or without wind-tunnel illustrations. Can you reverse the fluid
> dynamics in this way, when the fluid is stationary and the solid object
> moves?
Yes. The equations apply to relative velocities only -- it does't
matter what your frame of reference is attached to -- the fluid or
the body.
> If this is too far off topic, please e-mail
Since when has that been a problem? ;^/
At least I didn't add a binary graphic...
--
Mike Kohlbrenner
<kohlbren (-a t-) an dot hp dot com> sorry!
</QUOTE>
Harvey
remove 'nospam' to send me e-mail.
Unsolicited commercial e-mail is unwelcome.
Leonard Wojcik
Bob Myers wrote:
>
>
> I would dispute that physics is about "why" things happen. It contend
> that it is primarily and most importantly a pragmatic mathematical
> description of what happens.
>
> Example: Newton's Theory of Gravity says nothing about why masses
> experience mutual gravitational attraction; it just describes that
> attraction.
I will admit that the why of gravity is on the leading edge of
theoretical physics. But if you get to the correct why, then everything
else just falls out as a predictable calculation. No stupid rules to
memorize, just turn the crank.
>
> The philosophical models of most sciences aren't nearly as important
> or as long-lasting as the pragmatic, often mathematical, descriptions,
> IMHO.
Not true. It is more than just philosophy. If you know what is
actually happening then you don't make mistakes. Otherwise you always
have to look for the exception to the rule.
Example chemistry: If I know how molecules react and why (energetics,
electrons, etc), I can predict exactly what products will be made under
a given set of conditions.
Example physics: If I know the mechanism if the interactions of
electrons in semiconductors, I can design new neat stuff, If I don't it
is all by guess and by golly.
Leonard Wojcik
Harvey wrote:
>
>
> I am cross-posting an item from a different newsgroup because it was a
> very good description of how the Bernoulli principle works.
>
> <QUOTE>
>
> Dave Hall wrote:
>
<nice hand waving explanation snipped here>
>
> Bernoulli states that in a given fluid flow, stagnation pressure
> remains constant. It is a simple energy conservation principle.
> Since dymanic pressure is a squared function of velocity, it is
> clear that as velocity increases, the static pressure must decrease.
> Since static pressure is the one actually acting on surfaces in
> airflows, you can see how it is possible to create a partial
> vacuum on a surface in an airflow merely by virtue of the increased
> velocity of the airflow.
May be, but it doesn't explain why. It says a set of equations must
balance. It doesn't explain what is happening on a molecular level.
Also there is really no such thing as a vacuum.... really just the
presense of molecules more or less.... You can't make a vacuum, a
vacuum is not something that exists. You can take away molecules...
what I am saying is that we are better off dealing in terms of pressure,
not vacuum.
>
> One of the misconceptions I keep seeing in this thread is that
> "lift" can be completely explained by the Bernoulli theorem.
It can't. What we call "lift" is really caused by higher pressure on
the bottom of the wing. Pressure differential causes "lift", not a
great "sucking in the sky".
--
Leonard Wojcik Leonard.d...@wojciktech.com
Pete
Ron,
Sorry if this has already been answered, but:
I intrigued...how does the "friction beween adjacent moelcues moving at different
velocities" cause lower pressure?
The concept of air over a wing seems quite straightforward (in my simple mind at
least): since the air over the top of the wing has to travel a longer distance in
the same amount of time as the air on the bottom, the air on top must then be less
dense (effectively being "stretched" to fill the space)....and less density means
lower pressure (less happy molecules bumping against each other).
Pete
Ron Parsons wrote:
> It's nice that the calculations work enough to be able to predict results,
> but there is no magic force of nature which makes increased velocity cause
> lower pressure other than the friction between adjacent molecules moving
> at different velocities. Bernoulli's equations are mathematical shortcuts,
> not statements of physics.
>
> Ron
Mark Mallory
Leonard Wojcik (Leonard.d...@wojciktech.com) wrote:
: > : Bernoulli's equations are mathematical shortcuts,
: > : not statements of physics.
: >
: > Pure BS!
: I'm afraid that Ron is correct. The Bernoulli equations are a pragmatic
: mathmatical description of what is observed. You can calculate things
: from them, but they don't explain why it happens.
Baloney. Bernoulli's principle is simply a statement of one of the most
basic ideas in Physics, the principle of Conservation of Energy. It
explains *precisely* why it happens.
: Anybody want to have a go at an explanation at a molecular level of what
: goes on in the Bernoulli effect ? I am guessing it has something to do
: with molecular momentum and mean free path ?????
No, this has nothing to do it. Again, the "Bernoulli effect" is just a
result of Conservation of Energy: if the Kinetic Energy of the fluid flow
*increases*, the Potential Energy *decreases*, and vice-versa.
new...@algonet.se
In article <35402d22...@news.telepath.com>,
nospam...@innova.nospam.net (Harvey) wrote:
>
> On Thu, 23 Apr 1998 14:11:45 -0600, Leonard Wojcik
> <Leonard.d...@wojciktech.com> wrote:
>
> >Anybody want to have a go at an explanation at a molecular level of what
> >goes on in the Bernoulli effect ? I am guessing it has something to do
> >with molecular momentum and mean free path ?????
> >
> >Leonard Wojcik Leonard.d...@wojciktech.com
In article <353F73...@My.Mailbox>,
Mike Kohlbrenner <No.Ju...@My.Mailbox> wrote:
>
> Dave Hall wrote:
> >
> > The odd thing about Bernoulli's theorem is that it seems to run
> > opposite to what is expected:- faster-moving air produces lower
> > pressure.
>
> It isn't really counter-intuitive if you look at the whole theorem.
>
> There are three terms that need to be understood.
>
> Static pressure is what a barometer measures. It is the pressure
> either measured in a non-moving mass of fluid or measured against
> a surface exactly parallel to the flow in a moving fluid stream.
>
> Kinetic or dynamic pressure is that pressure created by fluid flow
> and is given by the equation: 1/2 * density * (velocity squared).
>
> Total or stagnation pressure for a given flow is simply the addition
> of the above two pressures. It is called stagnation pressure
> because it is the pressure you would measure at a point where the
> velocity of the airflow "stagnates" or goes to zero. If you look
> at the leading edge of a body in an airflow, the stagnation point
> is where the airflow splits to go to either side of the body. At
> that point, one can visualize a single line of flow that has been
> decellerated to zero and the pressure measured would equal
> stagnation pressure.
>
> Bernoulli states that in a given fluid flow, stagnation pressure
> remains constant. It is a simple energy conservation principle.
> Since dymanic pressure is a squared function of velocity, it is
> clear that as velocity increases, the static pressure must decrease.
> Since static pressure is the one actually acting on surfaces in
> airflows, you can see how it is possible to create a partial
> vacuum on a surface in an airflow merely by virtue of the increased
> velocity of the airflow.
Physically only a change in pressure can cause a change in airflow velocity.
Never the other way. This was stated by Leonard Euler 1754 when he devoted the
so called "Bernoulli theorem" to Daniel Bernoulli.
Who invented this faulty world wide spread "longer airflow path/equal time to
TE" theory? Here you can see how airflowphaseshift look in windtunnels:
At this website you can look at syncronized smokepulses in windtunnels
showing the airflow phaseshift:
http://www.eskimo.com/~billb/wing/airgif.html
FIG 2. picture is made by late Aerodynamics Alexander Lippish, The
German Me 163 creator,at Collins Radio windtunnels,USA, in 1953.It was
published in his book.Alexander Lippish was an expert on smoke in
windtunnels and his smoketrails was world famous.
FIG 3. picture is made by Aerodynamist Martin-Ingelman Sundberg at KTH
windtunnels in 1992. Sundberg,which first saw syncronized smokepulses
when visiting a windtunnel maker in USA 1962, made this smokepulsvideo
to show how wrong ICAO pilot education was,explaining winglift with
"airflow longer path over wing".
FIG 5. picture is a Navier-Stokes 2D airflow calculation around a SAAB
340 wing made by Aerodynamist Krister Karling,SAAB Aerospace.
Already 1754 Leonard Euler told us what causes the pressure gradient:
"centrifugal forces on curved airflow", very simple and physically
correct. dp=rho*V*V*dr/r.
> > The same idea applies to a carb venturi, which is a narrower
> > section, that speeds up the airflow through it, and (Bernoulli)
> > creates lower pressure. This "sucks" the gas (petrol) from the carb
> > bowl through the pipes and it mixes with the air stream, then into
> > your cylinders...
>
> Actually, a carb venturi is a much better example of the Bernoulli
> effect than a wing is. It really is close to one-dimensional
> channel flow.
In the carburator only the divergent part is active in lowering the pressure
by "bending the airstream" (The Coanda effect causes the airflow to follow
the walls of the divergent streamtube. Some German aircraft uses speed
indicators with only the rear part of the venturi pipe. You dont need the
front converging part.
Many authors write that "wing is working like a half venturi pipe/spilt
venturi".
How an asymmetric, cambered, wing can fly inverted is then not explained!
Professor John D. Anderson, Maryland University, writes in "Fundamentals of
Aerodynamics",2e, page 501-502 about "one dimensional flow in a streamtube":
"Strictly speaking, a streamtube for such flow must be of constant area; i.e,
the one-dimensional flow discussed in Chap 8, is a constant-area flow,
A=const.
"In a venturipipe, strictly speaking, this flow is three-dimensional; the
flow-field variables are functions of x,y and z... ... However, if the area
variation is moderate, the components in the y and z directions are small in
comparison with the component in the x direction. In such a flow-field
variables can be assumed to vary only with x only; i.e, the flow can be
assumed to be uniform across any cross section at a given x station. Such a
flow is defined as "A quasi-one-dimensional flow".
Prof. Anderson writes about "perfect fluid" that it is to found nowhere in
nature, it a hypotetical, non existing, mathematical created fluid. All fluid
are compressible to some extent.
Even steel is compressible (Se Hooks law).
An SAAB 340 with a approachspeed of 120 knots, the local airflow velocity at
LE is some Mach 0.85. Thats typical for an aircraft without leading slats. So
the airspeed is locally much higher than max velocity for "perfect fluid" use
(100 meter/second).
> > But just a moment - the air isn't speeding up over the wing at
> > all; the wing is moving through the air! Is the answer here simply
> > that the air molecules are spaced wider apart over the top of the
> > wing, so exert less pressure for that reason?!
>
> Kinda sorta. This actually helps with the explanation a little. As
> far as a molecule is concerned, it is just hanging around and suddenly
> it must move up and out of the way and then back down again as a wing
> passes by. Clearly, its velocity has increased during this disruption
> of its boring life. If its partner molecule ends up under the wing and
> doesn't have to move nearly as far, its velocity increase is smaller.
>
> Again, it is more complex than that in reality, but for lift on nice
> rounded bodies like bugs, this is a good first order explanation.
>
> > I've only ever seen the Bernoulli explanation for wing lift in texts,
> > with or without wind-tunnel illustrations. Can you reverse the fluid
> > dynamics in this way, when the fluid is stationary and the solid object
> > moves?
Bernoulli equation can never give you the velocities you need to put into the
formula itself.
So the "creation of lift" depends on other physical effects ( Eulers
equation):
Already 1754 Leonard Euler told us what causes the pressure gradient:
"centrifugal forces on curved airflow", very simple and physically
correct.
dp=rho*V*V*dr/r.
> Yes. The equations apply to relative velocities only -- it does't
> matter what your frame of reference is attached to -- the fluid or
> the body.
>
> > If this is too far off topic, please e-mail
>
> Since when has that been a problem? ;^/
>
> At least I didn't add a binary graphic...
>
> --
> Mike Kohlbrenner
> <kohlbren (-a t-) an dot hp dot com> sorry!
Jan-Olov Newborg
Stockholm
-----== Posted via Deja News, The Leader in Internet Discussion ==-----
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John Lowry
Dear Leonard, Ron, Mark, and All:
Integral principles (conservation of energy, momentum, Lagrange's
equations, Hamilton's equations, etc.) are physics just as much as
"ordinary" mechanical explanations. Neither type explanation is logically
prior to the other. BUT one or the other (and this may depend on the
readers' orientation and background) MAY be heuristically more attractive.
Here is a repeat of a mechanical explanation of Bernoulli's principle I put
together a few months ago. Only trouble is: I don't know whether or not it's
correct! And I don't have time, right now, to work on it further. But, for
what it's worth...
Ballpark Approximation of Bernoulli's Law Using Elementary Notions
A. Background
Bernoulli's Law -- p1 + Rho*V1^2/2 = p2 + Rho*V2^2/2 -- applies to inviscid
incompressible irrotational fluids along a stream line. We take position 1
to the left with larger pipe area A1 and position 2 to the right with
smaller pipe area A2. In between is a smooth transition region TR.
Take the fluid at 1 to be (aeronautical) standard air: Rho = 0.002377
slugs/ft^3, p1 = 2116.2 psf. The ideal gas law is p = nkT, n the number
density. Taking structureless molecules of mass m, the equipartition theorem
says m<v^2>/2 = 3kT/2, v the molecular speed and k the Boltzmann constant. T
is absolute temperature (which we don't need, but it's 59F = 518.7R for our
case). Take V1 = 100 ft/sec (near the speed of a fully-loaded Cessna 172
about to stall). The dynamic pressure Rho*V1^2/2 = 11.9 psf (quite small
compared with ambient atmospheric pressure 2116.2 psf).
Since Rho = nm, p = Rho*<v^2>/3.
B. Physical Reasoning
The gas in TR speeds up because molecules to the right are moving faster
than ones to the left. In TR, the gas is as though in a container with an
expanding wall to the right. Hence there are fewer collisions with molecules
to the right than with molecules to the left, less momentum is transferred,
and the pressure is less. It's a fluid, so the pressure (at a single place)
is the same on surfaces of all orientations. The flow speed V2 > V1 because
of the dictates of continuity and incompressibility: Fluxes: Rho*A1*V1 =
Rho*A2*V2. Since here Rho is constant, V2 = (A1/A2)*V1.
My idea is to see to what extent the flacid collisions to the right, as
though the molecules impinging on the molecular "wall" there had reduced
(relative) speed (v - (V2 - V1)), can account for the pressure drop between
positions 1 and 2.
C. Example 1, A2 = A1/2, V2 = 200 ft/sec.
By Bernoulli, p2 = p1 - 3*Rho*V1^2/2 and the pressure drop is only 35.7 psf,
a 1.68% drop.
<v1^2> = 3*2116.2/0.002377 = 2.671E6, so v1RMS = 1634 ft/sec. (It's "well
known" that average molecular speeds in a gas are somewhere around the speed
of sound in the gas.) Now if we reduce that speed by 100 ft/sec, to 1534, we
get <v^2> = 2.353E6. The pressure ratio goes as this ratio, 2.353/2.671 =
88.1%, so the pressure has dropped 11.9%.
But wait (this is "wait" #1)! ONLY the "wall" to the right is "moving." And
it's only one of six walls. So the effect just calculated is six times too
large. So the pressure drop should really be about 11.9%/6 = 1.98%. Compared
to the Bernoulli Law result 1.68%, not bad.
But wait! (#2) Let's check some other "experimental" values.
D. Example 2, A2 = 9*A1/10, V2 = 10*V1/9 = 111.1 ft/sec.
This time Bernoulli gives p2 = 2113.4 psf, a mere 0.13% drop.
The speed increase to the right is only 11.1 ft/sec, and so the relative
speed decrease is this amount. The pressure ratio this time is 2.634/2.671 =
0.9861, a 1.35% drop which, when divided by 6 gives a 0.23% drop. Compared
with 0.13% from Bernoulli, not very close at all.
E. Example 3, A2 = A1/4, V2 = 4*V1 = 400 ft/sec.
This last time Bernoulli gives p2 = 1937.7 psf, an 8.43% drop in pressure.
The speed increase to the right is 300 ft/sec. The pressure ratio is
(1634-300)^2/1634^2 = 0.6665, a 33.35% drop which, when divided by 6 gives a
5.56% drop. Sort of close to Bernoulli's 8.43%. Sort of.
F. Discussion.
This is a ballpark explanation. And only that. The situation is one of
steady flow in that at any one place in the fluid nothing is changing with
time, but if you follow a fluid "particle" it is certainly accelerating
between positions 1 and 2, in the transition region TR. So in that sense
this is not a very static situation. Elementary kinetic theory is full of
slightly subtle finagle factors sqrt(2) and sqrt(3). This "derivation" has
even larger ones and they aren't constant as you move from situation to
situation. It's in the ball park, but probably slightly foul. Maybe it hit
the flag pole.
Now of course Bernoulli's hands aren't so clean either. Air is
incompressible? Not in reality. And not even in TR! (The "moving wall"
rarifies the gas a bit, temporarily, until it gets well into region 2, where
the numbers of collisions with gas molecules from the left and the right are
again the same.) And no viscosity? Constant pressure in each region? Then
what's making the fluid move from left to right? It's not just a puzzlement,
it's a mess.
Comments are welcomed.
John Lowry, Flight Physics, jlo...@mcn.net, 406-248-2606.
Mark Mallory wrote in message ...
>Leonard Wojcik (Leonard.d...@wojciktech.com) wrote:
>: > : Bernoulli's equations are mathematical shortcuts,
>: > : not statements of physics.
>: >
>: > Pure BS!
>
>: I'm afraid that Ron is correct. The Bernoulli equations are a pragmatic
>: mathmatical description of what is observed. You can calculate things
>: from them, but they don't explain why it happens.
>
>Baloney. Bernoulli's principle is simply a statement of one of the most
>basic ideas in Physics, the principle of Conservation of Energy. It
>explains *precisely* why it happens.
>
>: Anybody want to have a go at an explanation at a molecular level of what
>: goes on in the Bernoulli effect ? I am guessing it has something to do
>: with molecular momentum and mean free path ?????
>
Hilton Goldstein
John S. Denker wrote:
>
[zap]
> Similar remarks
> apply to "centrifugal force" – please remember it is a force per unit mass.
Why? How can you 'define' force as being force per unit mass
(acceleration)? Centrifugal force has absolutely nothing to do with the
mass upon which it is acting. Yes, in special cases it is, such as
swing a bucket of water around you, or planetary motion, but those are
special cases.
Hilton
--
Hilton Goldstein..............................hilton@sgi.com
650-933-5254 (phone)......................(fax) 650-933-0836
M/S 42M-945, 2011 N. Shoreline Blvd, Mountain View, CA 94043
http://reality.sgi.com/hilton
GetRect is my favorite Windows function.
Dr. Christian Kurz
Hilton Goldstein wrote in message <35456B...@sgi.com>...
>Why? How can you 'define' force as being force per unit mass
>(acceleration)? Centrifugal force has absolutely nothing to do with the
>mass upon which it is acting
Oh yes it does. The centrifugal force felt by a body going around in a
circle of radius 'r' with a circumferential velocity 'v' is given by: F =
mv^2/r, where 'm' is its mass. Just imagine what it would otherwise feel
like going around on a merry-go-round with a fly sitting on your head...
First and foremost, don't forget that centrifugal force is not the only
force acting on a body in a rotating frame of reference. There is also
Coriolis force. So if you drop an object in a rotating frame of reference,
it fill follow a curved path.
And for the statement that you can mimick a centrifugal force field with
gravitational masses: the radial dependences are different, therefore you
can't. Gravity drops off like 1/r^2, whereas the centrifugal force goes like
1/r. And that's even neglecting Coriolis force. No go, can't do.
-- Chris.
---------------------------
Christian Kurz, Ph.D.
(Physics)
Hilton Goldstein
Dr. Christian Kurz wrote:
>
> Hilton Goldstein wrote in message <35456B...@sgi.com>...
>
> >Why? How can you 'define' force as being force per unit mass
> >(acceleration)? Centrifugal force has absolutely nothing to do with the
> >mass upon which it is acting
>
> Oh yes it does. The centrifugal force felt by a body going around in a
> circle of radius 'r' with a circumferential velocity 'v' is given by: F =
> mv^2/r, where 'm' is its mass. Just imagine what it would otherwise feel
> like going around on a merry-go-round with a fly sitting on your head...
Exactly, special case, just like I mentioned in my post; like swinging a
bucket of water around you.
Firstly: John said that centrifugal force is a force per mass. How can
a force be an acceleration?
Secondly: Put an object in space with a rocket which always acts
perpendicular to its velocity (motion). That rocket exerts a
centrifugal force on the object which has absolutely nothing to do with
the mass of the object.
sha...@nospam.com
>Firstly: John said that centrifugal force is a force per mass. How can
>a force be an acceleration?
I think what he meant was the centrifugal force should be more correctly
called the centrifugal acceleration, since the magnitude of the force
depends on the mass of the object.
>Secondly: Put an object in space with a rocket which always acts
>perpendicular to its velocity (motion). That rocket exerts a
>centrifugal force on the object which has absolutely nothing to do with
>the mass of the object.
Or, put a moving electron in a magnetic field perpendicular to its
velocity. The force experienced by it is the cross product of the
field and the velocity and is independent of the mass of the electron
i.e. this is exactly your model above.
So what happens when you do that? The electron travels in a circle. The
force is in fact a *centripetal* force. (The radius of the circle
depends, of course, on the mass of the electron.)
You need a different analogy for what you're trying to say (and I don't
know what that is).
-s
Shamim Mohamed
spm at drones dot com
Hilton Goldstein
sha...@noSpam.com wrote:
>
> In article <354629...@sgi.com>, Hilton Goldstein <hil...@sgi.com> wrote:
>
[zap]
> >Secondly: Put an object in space with a rocket which always acts
> >perpendicular to its velocity (motion). That rocket exerts a
> >centrifugal force on the object which has absolutely nothing to do with
> >the mass of the object.
>
> Or, put a moving electron in a magnetic field perpendicular to its
> velocity. The force experienced by it is the cross product of the
> field and the velocity and is independent of the mass of the electron
> i.e. this is exactly your model above.
>
> So what happens when you do that? The electron travels in a circle. The
> force is in fact a *centripetal* force. (The radius of the circle
> depends, of course, on the mass of the electron.)
>
> You need a different analogy for what you're trying to say (and I don't
> know what that is).
No, you've just supported what I'm saying. Centripital/centrifugal
force has nothing to do with the mass of the object. Given this force
"F", and we obviously know the mass "m" of the electron, we can
calculate the acceleration "a" by F=ma. The acceleration and thus
radius are dependent upon the mass, but the force is not necessarily.
Thanks for that additional example, I'd forgotten about that one.
JT
>>Why? How can you 'define' force as being force per unit mass
>>(acceleration)?
>Oh yes it does. The centrifugal force felt by a body going around in a
>circle of radius 'r' with a circumferential velocity 'v' is given by: F =
>mv^2/r, where 'm' is its mass. Just imagine what it would otherwise feel
>like going around on a merry-go-round with a fly sitting on your head...
I thought all the physicists decided that there is no centrifugal force,
it's just inertia. Is this debate still raging?
-John
John S. Denker
Yo, Hilton!
In article <35464C...@sgi.com>, you <hil...@sgi.com> wrote:
>
> Centripital/centrifugal
>force has nothing to do with the mass of the object. Given this force
>"F", and we obviously know the mass "m" of the electron, we can
>calculate the acceleration "a" by F=ma. The acceleration and thus
>radius are dependent upon the mass, but the force is not necessarily.
We've had this discussion before.
Consider the analogy: it is *not* advantageous to think of the electric field
as a "force" field per se. It is best thought of producing, at each point, a
force proportional to the electric charge at that point and independent of
mass and other factors.
Similarly: it is *not* advantageous to think of a centrifuge as producing a
"force" field per se. Just because you've been calling it "centrifugal force"
since childhood doesn't settle the issue. The centrifuge creates a field
which, at each point, produces a force proportional to the mass at that point
and independent of electrical charge and other factors. A force per unit mass
is called an acceleration.
You do nobody any good when you persist in claiming that gravity and/or
centrifugity produces a force having "nothing to do with" mass.
--- jsd
Dr. Christian Kurz
Mark Mallory wrote in message ...
>Ron Parsons (jr...@gte.net) wrote:
>: Bernoulli's equations are mathematical shortcuts,
>: not statements of physics.
>
>Pure BS!
>
Unfortunately the answer to this question lies squarely in the realm of
philosophy. The important fact, however, is to realize that when one does
physics and describes nature, then there is always an act of philosphy
involved.
Bernoulli's equation is a mathematical statement, which summarizes a large
number of experimental results. As such it makes use of what I call a
"hierarchy of influences". It takes into account influences and
observations starting with the most important ones and draws the line at
some point, beyond which it neglects influences. For instance, Bernoulli's
equation does not include the effect of gravity, or of viscosity.
Bernoulli's equation also describes these experimental results on a
particular level. In this case it is the level of fluid dynamics. One
could, equally well, describe the same effect at the level of particle
kinetics. That doesn't make one better than the other. Each physical
equation is a tool in our toolbox to interpret the large variety of natural
pheomena we see happening around us. It is not the truth. It is a tool.
So, Bernoulli's equation is a mathematical shortcut in the same sense that
all physical formulae are mathematical shortcuts to our experimental
observations. But it also statement of physics for the same reason.
I guess that Ron was thinking about the individual gas molecules and a
kinetic model which would describe what happens at the particle level.
Well, if you do the kinetic calculations uding the same hierarchy of
influences that underlies the fluid dynamics approach, you wind up with the
same mathematical formula. The point is that one has to be knowledgable
about the particular hierarchy of influences and what the point is beyond
which they are neglected. That's the important part. Everything else after
that is pure algebra.
-- Chris.
Dr. Christian Kurz
John S. Denker wrote in message <6i5tkc$r75$1...@news.monmouth.com>...
>You do nobody any good when you persist in claiming that gravity and/or
>centrifugity produces a force having "nothing to do with" mass.
Agreed 110%.
Hilton gave an example a couple of posts earlier:
>Secondly: Put an object in space with a rocket which always acts
>perpendicular to its velocity (motion). That rocket exerts a
>centrifugal force on the object which has absolutely nothing to do with
>the mass of the object.
This is a flawed example. In order for the rocket to produce a force
exactly equal to the centrifugal force, it would have to be proportional to
the other object's mass, its velocity squared, and inversely proportional to
the radius of curvature of its trajectory.
In the same way that electrical charge makes you susceptible to
electro-magnetic effects, mass will make you susceptible to inertial
effects. No mass -- no inertial effect. No charge -- no electro-magnetig
effect.
-- Chris.
Dr. Christian Kurz
JT wrote in message <35469...@206.103.97.91>...
>I thought all the physicists decided that there is no centrifugal force,
>it's just inertia. Is this debate still raging?
The physicists know exactly what centrifugal force is or is not. The
problem lies with the non-physicist understanding only half of it and
extrapolating the other half.
Centrifugal force is a virtual force, since it is only experienced in
rotating frames of reference. It is not experienced in inertial frames of
reference, which move at a constant velocity in a straight line. So the
correct term for it is a 'virtual' force, but nobody bothers to call it
that, since all the physicists know that it is a virtual force. To
differentiate: if you kick me in the butt (I switched the roles so as not to
be offensive), it hurts just the same, no matter wether you kicked me while
we were standing still or flying in an airplain at Mach 2. This was a real
force. The centrifugal force is not real in the sense that it depends on
your state of motion and position how you experience it. But since it jurts
just as much when you get squeezed in a centrifuge, we make amends and still
call it a force, albeit a 'virtual' one.
-- Chris.
Hilton Goldstein
John S. Denker wrote:
>
> Yo, Hilton!
>
Yo, John!
> You do nobody any good when you persist in claiming that gravity and/or
> centrifugity produces a force having "nothing to do with" mass.
I have never claimed that gravity has nothing to do with mass, that
would be silly and incorrect, given that gravity is defined as: F =
(Gm1m2)/d^2. Obviously gravity is a force proportional to the objects'
masses.
Let's clarify a few things: I said centrifugal force does not
necessarily depend on the object's mass. Yes, in special cases it does,
such as swinging a bucket of water and other circular motion. However
there are numerous examples where it does not hold, such as my rocket
example and the electron example. John, show us how the force exerted
by the rocket or the force exerted on the electron has anything to do
with the mass of the object. It doesn't. See, an example where
centrifugal force causes circular motion has ***NOTHING*** to do with
the object's mass.
I once said: "Gravity is a force, period.", you replied: "Gravity is not
a force, period." I still cannot believe you think gravity is not a
force. Gravity is a force as per its definition, and it is *useful* to
consider the gravitational field as an acceleration field. But gravity
is a force, which is only useful to consider it being equivalent to an
acceleration because of some general purpose equation F=ma. As I've
said before, if this general purpose F=ma just happened to be different,
then would gravity suddenly be a different thing?
Please don't claim I have said things which I haven't. I have never
ever claimed that gravity has nothing to do with mass as I explained
above.
Dr. Christian Kurz
Hilton Goldstein wrote in message <35476F...@sgi.com>...
>Let's clarify a few things: I said centrifugal force does not
>necessarily depend on the object's mass. Yes, in special cases it does,
>such as swinging a bucket of water and other circular motion. However
>there are numerous examples where it does not hold, such as my rocket
>example and the electron example
I don't want to offend you, Hilton, but you are plain wrong. The example
with the circular path of an electron in a magnetic field is another
misconception. Actually, if I may borrow it, I can show nicely that the
centrifugal force depends on mass. When calculating the electron trajectory
you are balancing the electron's centrifugal force with the Lorentz force.
The Lorentz forece does not depend on mass, but the centrifugal one does.
That's why an object with the same charge as the electron but only heaver
will go around in a much wider circle (e.g. an antiproton). It is precisely
because the centrifugal force depends on mass and the Lorentz force doesn't.
I also pointed out before why your example with the rocket is flawed; no
need to repeat my argumen here.
I have no idea where you got this notion that centrifugal force is
independent of mass most of the time, but I sure hope that no students
reading this follow your believe. Can you name any physicist who shares
this belief with you? None of my colleagues do, and none of the ones I have
worked with so far.
-- Christian Kurz, Phd.
(Experimental Plasma Physicist, MIT)
Chris Kurz
Hilton Goldstein wrote in message <35490E...@sgi.com>...
>Please list the external force(s) on the electron and show how they are
>a function of mass. You have said that the Lorentz Force does not
>depend on mass. What other forces are acting on the electron? None.
There are two forces acting:
Lorentz force = ev * B (forget vectors and constants)
Centrifugal force = mv^2/r
The radius of the circle the electron follows has a special name, it is the
Larmor radius. You can find it by solving the above two equations for r, in
which case you get:
r = m*v / (e * B).
As you can see, the radius depends on the mass of the electron simply
because the centrifugal force depends on mass. This is not theory, this is
experimental practice as well. A lot of things would stop functioning if
this were any different.
What units do you measure centrifugal force in, Hilton? A force is
measured, by definition, in Newtons (SI units), which is kg * m / (s^2).
Regardless whether you measure mass in pounds, slugs, or whatever, the unit
for force is "mass * distance / (time ^2)". How do you get these units in
the case of centrifugal force, if it doesn't depend on mass?
-- Chris.
Chris Kurz
Mark Mallory wrote in message ...
>Actually, centrifugal force is proportional to r, not 1/r. ( F = m w^2 r )
That depends. If you write it in terms of "v", the circumferential
velocity, it is dependent on 1/r. If you write it in terms of "w" (omega,
the angular velocity), it is proportional to r. This is not a
contradiction. One is exactly identical to the other if you remember how v
and w relate to each other: v = w*r.
-- Chris.
Mark Mallory
Chris Kurz (ck...@bnet.at) wrote:
: Hilton Goldstein wrote in message <35490E...@sgi.com>...
: >Please list the external force(s) on the electron and show how they are
: >a function of mass. You have said that the Lorentz Force does not
: >depend on mass. What other forces are acting on the electron? None.
: There are two forces acting:
: Lorentz force = ev * B (forget vectors and constants)
: Centrifugal force = mv^2/r
WRONG. There is only *one* force acting on the electron in this case,
the Lorentz force ev * B. This force produces a *centripetal*
acceleration of the electron. And as Hilton correctly points out, (as is
obvious from the Lorentz force equation) this force is *independent* of mass.
Remember: Centrifugal force "is not experienced" in the inertial frame.
Hilton Goldstein
Mark,
Thanks for your post.
I have maintained that centrifugal force/centripital force is not
necessarily a function of mass. In some case it is, and in others it is
not. I was told I'm completely wrong and centrifugal is always a
function of mass.
Example 1: We're sitting in a gizmo which swings us around. Since
the angular velocity and radius are constants, the force
exerted on and by the object is a function of our mass.
This is obvious. Trying to make a tennis ball go in
circles takes less force than trying to make a 1 ton
boulder do the same.
Example 2: We have an object in space and we exert a known force of,
say, 10 newtons on it. Note, a constant force totally
independent of the object's mass. Given that we have the
mass of the object and the force, we can determine the
acceleration of the object and the path it describes.
This centrifugal/centripital force is independent of the
object's mass.
Our first example shows where the centrifugal/centripital force is a
function of the object's mass, and the second example shows a case where
the force is not a function of the object's mass. Therefore, as I've
said all along, centrifugal/centripital force is not neccessarily a
function of the object's mass. QED.
Thank you for listening,
Robert Loer
Your example number 2 is obviously flawed. Calculating the curved path will
be dependent upon the mass of the object. If not then all this hot air being
expended on our side of the planet must have at least started a wobble in
trajectory by now.
Robert Loer
John S. Denker
In article <354A39...@sgi.com>, Hilton Goldstein <hil...@sgi.com> wrote:
> I was told I'm completely wrong and centrifugal is always a
>function of mass.
The centrifugal force is indeed *always* a function of mass.
>Example 1: We're sitting in a gizmo which swings us around.
Yes, that's called a centrifuge.
>Since
> the angular velocity and radius are constants, the force
> exerted on and by the object is a function of our mass.
> This is obvious. Trying to make a tennis ball go in
> circles takes less force than trying to make a 1 ton
> boulder do the same.
Good. You have seen the light.
Alas, your note now plunges back into darkness:
>Example 2: We have an object in space and we exert a known force of,
> say, 10 newtons on it. Note, a constant force totally
> independent of the object's mass. Given that we have the
> mass of the object and the force, we can determine the
> acceleration of the object and the path it describes.
> This centrifugal/centripital force is independent of the
> object's mass.
OK, if we apply a constant force then we get a constant force. Duh. The
fallacy here is the use of the word "centrifugal", since the constant force
you describe is not what arises in a centrifuge.
If you wish to redefine the word "centrifuge" to apply to this situation, then
you are just playing silly word games. Don't expect anyone to understand what
you're saying.
=====
An airplane in a steady turn is an example of a centrifuge. It creates a
field which is profoundly analogous to gravity. At any point this field
produces a force in proportion to the mass at that point. A force per unit
mass is called an acceleration.
You aren't helping yourself or anyone else by saying otherwise.
--- jsd
Hilton Goldstein
Correct, the trajectory in example 2 is dependant on the mass. We
have: F=ma and since F is independant of the mass, clearly the
acceleration is dependant on the mass. That's what I said.
Hilton Goldstein
John S. Denker wrote:
>
> In article <354A39...@sgi.com>, Hilton Goldstein <hil...@sgi.com> wrote:
> > I was told I'm completely wrong and centrifugal is always a
> >function of mass.
>
> The centrifugal force is indeed *always* a function of mass.
When one disagrees with someone with a Ph.D., a bunch of papers, etc
etc, I think it is necessary for that person to be sure of their facts.
Hilton takes a deep breath and says: John, you are wrong. You say that
centrifugal force is always a function of mass. That is incorrect.
I'll explain below.
> >Example 1: We're sitting in a gizmo which swings us around.
>
> Yes, that's called a centrifuge.
Yip.
> >Since
> > the angular velocity and radius are constants, the force
> > exerted on and by the object is a function of our mass.
> > This is obvious. Trying to make a tennis ball go in
> > circles takes less force than trying to make a 1 ton
> > boulder do the same.
>
> Good. You have seen the light.
Thank you very much, but I always knew this.
> Alas, your note now plunges back into darkness:
>
> >Example 2: We have an object in space and we exert a known force of,
> > say, 10 newtons on it. Note, a constant force totally
> > independent of the object's mass. Given that we have the
> > mass of the object and the force, we can determine the
> > acceleration of the object and the path it describes.
> > This centrifugal/centripital force is independent of the
> > object's mass.
>
> OK, if we apply a constant force then we get a constant force. Duh.
That is not what I said, or at least not what I was implying. I said
that if we apply a constant force which is not a function of the mass,
by F=ma, the acceleration of the object is indeed a function of the
object's mass, and hence, so is its path.
> The
> fallacy here is the use of the word "centrifugal", since the constant force
> you describe is not what arises in a centrifuge.
>
> If you wish to redefine the word "centrifuge" to apply to this situation, then
> you are just playing silly word games. Don't expect anyone to understand what
> you're saying.
I feel a little strange saying that you don't understand what a
centrifugal force is, because obviously you do. However, to solve any
ambiguities, I shall quote from the Webster's Dictionary for centrifugal
force: "The component of a force on a body in curvilinear motion that is
directed away from the center of curvature or axis of rotation." Note
that this says *nothing* about this object being in a centrifuge.
Anytime you make an object go in a circle, you have
centrifugal/centriptal forces. A centrifuge is *not* required. A
centrifuge is just a machine which swings something around at a defined
radius and angular velocity. Clearly therefore, in this case the
centrifugal force is dependent on the mass (my example 1). That is a
special case. And no, I'm not playing silly little word games, unless
you are changing the physics definition of "centrifugal force" and then
saying that I'm the one playing the word games.
Having said that, your definition of "Centrifugal Force" is wrong.
Peter Gottlieb
Hilton Goldstein wrote in message <354A39...@sgi.com>...
>
>Example 1: We're sitting in a gizmo which swings us around. Since
> the angular velocity and radius are constants, the force
> exerted on and by the object is a function of our mass.
> This is obvious. Trying to make a tennis ball go in
> circles takes less force than trying to make a 1 ton
> boulder do the same.
Constant radius and angular velocity ==> force is a function of mass
>
>Example 2: We have an object in space and we exert a known force of,
> say, 10 newtons on it. Note, a constant force totally
> independent of the object's mass. Given that we have the
> mass of the object and the force, we can determine the
> acceleration of the object and the path it describes.
> This centrifugal/centripital force is independent of the
> object's mass.
Constant force ==> radius AND angular velocity cannot both be independant
of mass, as you allude to.
I think these two examples are simply variants of one another and do
not show anything in particular.
The concept of a "centrifugal" force always bothered me. Is there such a
force? I mean different from ordinary acceleration? Really, it's just a
specialized
application of some other well known force. Think of the orbit of a
satellite
(geosynchronous) ... it's always falling down (in freefall), with the force
vector
pointing towards the center of the earth. But since it is moving "sideways"
so fast, it follows the curvature of the earth, but keeps falling! Of
course you
could say that the force of gravity causes the satellite to deviate from a
straight
path through space, but where is the centrifugal force? The same could be
said
for a rock you are spinning around you on a rope. The rock wants to move in
a
straight line, but you are exerting a force on it to bend its path. Just a
"normal"
force...
For a given circular path, there is an implied acceleration vector necessary
to
maintain that path. The force necessary to cause that acceleration is, of
course,
directly related to the mass of the object in question.
The force vector that is responsible for a circular path of a 5Kg rock going
around
you at 5 radians/sec will not be sufficient to maintain a 50Kg rock in the
same
configuration (it would be a much gentler curve).
John S. Denker
Hi Folks --
In article <354E3C...@sgi.com>, Hilton Goldstein <hil...@sgi.com> wrote:
>
>HG: Example 2: We have an object in space and we exert a known force of,
>HG: say, 10 newtons on it. Note, a constant force totally
>HG: independent of the object's mass. Given that we have the
>HG: mass of the object and the force, we can determine the
>HG: acceleration of the object and the path it describes.
>HG: This centrifugal/centripital force is independent of the
>HG: object's mass.
>
>jsd: OK, if we apply a constant force then we get a constant force. Duh.
>
>That is not what I said, or at least not what I was implying.
But that IS what Hilton said, and the implications are clear. In the scenarios
he has conjured up, the only reason there is a constant force is because he
IMPOSED a constant force.
His constant-force scenarios do not arise naturally in centrifuges, nor in
airplanes.
>jsd: The
>jsd: fallacy here is the use of the word "centrifugal", since the constant
>jsd: force you describe is not what arises in a centrifuge.
>
>jsd: If you wish to redefine the word "centrifuge" to apply to this situation,
>jsd: then you are just playing silly word games. Don't expect anyone to
>jsd: understand what you're saying.
>
> I shall quote from the Webster's Dictionary for centrifugal force:
> "The component of a force on a body in curvilinear motion that is
>directed away from the center of curvature or axis of rotation." Note
>that this says *nothing* about this object being in a centrifuge.
>Anytime you make an object go in a circle, you have
>centrifugal/centriptal forces. A centrifuge is *not* required.
Hilton's reply reinforces my suspicion that he is more interested in word
games than airplanes, and more interested in arguing than learning.
Consider two different masses moving past the same point, on circular paths.
There are three options:
1) the objects have different speeds
2) their paths have different centers
3) they have different centrifugal forces
Under options 1 and 2, the masses are together only for an instant. These
scenarios are so bizarre that Mr. Webster didn't even consider them when
writing the layman's definition of "centrifugal force". Even so, option 2 is
pretty well excluded because the definition speaks of "the" (singular) center.
Certainly neither option 1 or 2 has anything to do with airplanes.
Option 3 is the one that applies to steady situations, such as in airplanes
and centrifuges. It is also consistent with the common-sense definition of
centrifugal force, used by physicists, engineers, and pilots everywhere,
namely: the force (per unit mass) that arises when we work in a rotating
reference frame.
Hilton can twist words however he likes. The rest of us will continue to use
words according to their ordinary meanings, in ways that helpfully describe
what really happens in airplanes.
--- jsd
Dr. Christian Kurz
I am getting a bit tired of this thread. If anybody wants to BELIEVE that
centrifugal force is independent of mass, then that's their business.
Religion is beyond debate. But if somebody wants to argue on physical basis
that same is true then they are plain wrong. I think I have established that
I know what I am talking about (I've got my PhD in physics from MIT and am
working as an experimental plasma physicist at MIT and UCSD). I have also
given you several bullet proof (to a scientist) arguments why centrifugal
force is dependent on the mass of the object -- always and at all times. I
can't change your believes, I wouldn't want to.
You don't have to take it from me if you don't want to. Go talk to any
physicist of your choice.
Maybe I was unable to explain my point. I probably wasn't being clear
enough. In that case I apologize. But that still doesn't make what you
said true.
-- Chris.
new...@algonet.se
In article <35473...@server02.bnet.at>#1/1,
I addition to your good posting we must remember that the so called Bernoulli
equation setup by Leonard Euler as a simplification of his 3 D fluid equation,
Bernoulli equation is valid for 1 dimensional flow along the same streamline
in a "perfect fluid" or in a point. Going above and below an airfoil is not
the same streamline and its not one dimemsional, so Bernoulli can not be used
that way as said by one earlier posting.
In a curved streamline we have 3 dimensional flow and Bernoulli is not valid.
(due to 3d turbulens effects).
Hilton Goldstein
Chris Kurz wrote:
>
> Hilton Goldstein wrote in message <35490E...@sgi.com>...
>
> >Please list the external force(s) on the electron and show how they are
> >a function of mass. You have said that the Lorentz Force does not
> >depend on mass. What other forces are acting on the electron? None.
>
> There are two forces acting:
> Lorentz force = ev * B (forget vectors and constants)
> Centrifugal force = mv^2/r
>
> Larmor radius. You can find it by solving the above two equations for r, in
> which case you get:
>
> r = m*v / (e * B).
>
> As you can see, the radius depends on the mass of the electron simply
> because the centrifugal force depends on mass. This is not theory, this is
> experimental practice as well. A lot of things would stop functioning if
> this were any different.
Yes, the radius depends on the mass, but by the equation above for the
force, F = evB, clearly the centripital force is not a function of
mass. Are you suggesting that the centripital force is not equal in
magnitude to the centrifugal force, or are you agreeing with me that
centrifugal force is not neccessarily a function of mass?
Thanks for helping me in my argument.
Hilton Goldstein
Dr. Christian Kurz wrote:
>
> I am getting a bit tired of this thread. If anybody wants to BELIEVE that
> centrifugal force is independent of mass, then that's their business.
> Religion is beyond debate. But if somebody wants to argue on physical basis
> that same is true then they are plain wrong. I think I have established that
> I know what I am talking about (I've got my PhD in physics from MIT and am
> working as an experimental plasma physicist at MIT and UCSD).
Chris,
Congratulations on your PhD, however I do not feel that a University
degree allows anyone to change established physics concepts.
> I have also
> given you several bullet proof (to a scientist) arguments why centrifugal
> force is dependent on the mass of the object -- always and at all times.
Actually, the one example you present clearly showes that the force is
completely independent of the mass. Actually you showed that the radius
in the electron/proton example varied according to mass. That's pretty
obvious. However, as you quoted, the force on the object is given by:
F = evB
which is not a function of mass.
Mark Mallory
Diane (as...@slip.net) wrote:
: The factor "e" *IS* mass you twit.
: Electron volts and mass are interchangeable units.
The factor "e" is *not* "electron volts", you twit. It's charge.
Perhaps next time you'll THINK before you post. (But if you've got a
"real degree" I suppose you don't need to bother with thinking :)
: Go back to reading your "Starlog" magazines and leave the real people with real degrees
: alone.
Sriram Narayan
Mark Mallory (mmal...@netcom.com) wrote:
Looks like "someone" cancelled the post, before I got around to replying
to it (and trashing it). Anyway "e" has dimensions of Q (charge) and
eV is an energy unit.
Regarding the original argument, I agree that centripetal force acting
on an electron is evB (magnitude). From an inertial reference point
of view there is no centrifugal force that one needs to consider as
the body is clearly accelerating. Now, if we were to be "sitting"
on the electron, clearly there is a virtual force that would tend to
keep the forces balanced. Is this one way to determine the acceleration
of the electron? If we equate ma = evB, then a = evB/m. All we
are saying is that the centrifugal force is equal to the electromagnetic
force. Since the electron has a finite mass, its acceleration is
dependent on its mass, even though the cause of its acceleration
has nothing to do with the mass of the electron. Why is there a problem
with this concept?
--
sriram....@technologist.com http://members.home.net/snarayan/homepage.html
pp-asel:san francisco bay area:vfr flight planner:av articles:photo:sw dxing
new...@algonet.se
Professor Mark Drela, MIT Aero and Astro once wrote on Bernoulli Myths:
In article <1991May7.1...@doc.ic.ac.uk>, dw...@doc.ic.ac.uk writes:
|> When you blow across the top of a sheet of paper, it rises.
|> The standard explanation for this is that the high velocity air on the top
of
|> the paper gives rise to a lower pressure on top that on the bottom (where
the
|> air is still) so the paper rises.
|>
The "standard explanation" you describe is fallacious. The jet of air,
before it gets to the paper, has the same pressure as the surrounding air.
Bernoulli's Law doesn't apply in this situation. The jet imparts lift to
the paper because as it is curved downward, the pressure at the bottom
edge of the jet (where it contacts the paper) is lower than at the top
edge which is at ambient pressure. This pressure difference is set up
to force the jet into the curved path. not because the jet has a higher
velocity. If you blow across a flat sheet of paper, not much will happen.
|> Why (in molecular terms) does the high velocity air cause a lower pressure?
Why
|> are the molecules not hitting the top of the paper just as much as they
were
|> before air was blown across?
|>
The molecules do not hit the curved paper surface as hard because the
air pressure and air density is slightly lower there. I know this seems
like circular reasoning, but it is misleading to think of Bernoulli's Law
(when it applies) as a cause-effect relationship between velocity and
pressure. In response to the common explanation of lift:
1) The higher velocity over the top of an airfoil causes the pressure to be
lower there (via Bernoulli's Law), thus generating lift.
I could just as well claim:
2) The lower pressure region over the top of the airfoil causes the air
to accelerate towards it, thus increasing the velocity there.
Both of these statements are correct, but don't really explain anything
in concrete physical terms.
The best explanation I can give is that the pressure on top of an airfoil
is lower so that the air can be forced into a curved path along the airfoil
surface. The air velocity there will consequently also increase as the air
is "drawn" into the lower pressure region, a la the second "explanation"
above.
Bernoulli's Law then simply tells you the quantitative relation between the
pressure and velocity changes.
This also neatly explains the loss of lift after stall, where the flow
no longer closely follows the surface because of boundary layer separation.
The flow path curvature over the airfoil is reduced, which then requires
a smaller pressure decrease over the airfoil, which reduces the lift.
In either case, if the airflow were to curve over the airfoil surface
without the lower pressure, then you would have centripetal acceleration
without an inward force, and Sir Isaac would turn over in his grave.
Hence the lower pressure MUST BE THERE, Bernoulli or no Bernoulli.
Mark Drela
MIT Aero & Astro
>From
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com!think.com!mintaka!bloom-picayune.mit.edu!athena.mit.edu!drela Mon May 13
18:54:00 EDT 1991
Article 1496 of sci.aeronautics:
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>From: dr...@athena.mit.edu (Mark Drela)
Newsgroups: sci.aeronautics
Subject: Re: Molecular Explanations of Pressure Drop
Message-ID: <1991May11.0...@athena.mit.edu>
Date: 11 May 91 02:55:37 GMT
References: <1991May7.1...@doc.ic.ac.uk>
<1991May8.0...@athena.mit.edu> <44...@bnr-rsc.UUCP>
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In article <44...@bnr-rsc.UUCP>, m...@bnr-rsc.UUCP (Max Feil) writes:
|> In article <1991May8.0...@athena.mit.edu> dr...@athena.mit.edu
(Mark Drela) writes:
|> >The "standard explanation" you describe is fallacious. The jet of air,
|> >before it gets to the paper, has the same pressure as the surrounding air.
|> >Bernoulli's Law doesn't apply in this situation. ....
|>
|> Hmm... I always thought that a faster flowing fluid, say in a constriction
of
|> a pipe, would extert less pressure. Also, I assumed that this is how a
perfume
|> atomizer worked. What gives?
|>
When comparing pressures at two different points by using Bernoulli's Law
(i.e. where the velocity is higher, the pressure is lower), it is essential
that the two points in question have the same _total pressure_.
The jet of air which you blow from your lips has a higher total pressure
than the surrounding air, so the jet and the ambient air are not related
by Bernoulli's Law as most people know it.
For any steady fluid flow, the total pressure is the same for all points
along any streamline, as long as viscous effects are not too significant.
Viscous shear gradually decreases total pressure along a streamline
(in most cases). Thus, you can use Bernoulli to relate the pressure in
the air jet and the pressure inside your mouth, as these points lie
along a streamline. Likewise, you can use it to compare the pressures
between two closely-spaced points in your constricted pipe example.
You cannot compare two points far apart, because the total pressure
in a pipe flow gradually decreases downstream by viscous action.
In general, all points in an external aerodynamic flow (outside of
boundary layers, blown jets, engine exhausts, etc.) are at the same
total pressure, so Bernoulli applies to any selected pair of points.
Mark Drela First Law of Aviation:
MIT Aero & Astro "Takeoff is optional, landing is compulsory"
Article 1454 of sci.aeronautics:
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From: dr...@athena.mit.edu (Mark Drela)
Newsgroups: sci.aeronautics
Subject: Re: How do airplanes fly?
Message-ID: <1991Dec6.0...@athena.mit.edu>
Date: 6 Dec 91 02:46:54 GMT
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In article <1991Dec5.0...@math.ucla.edu>, ba...@arnold.math.ucla.edu
(Barry Merriman) writes:
|>
|>
|> How do planes fly? This thread started on sci.physics, but physicists
|> don't seem to know how flight works. I'm hoping the aero engineers
|> can give a good intuitive explanation.
|>
|> More precisely, here's what I'd like: starting from the
|> wing at rest, show, using obvious forces, how the
|> lift develops, and why the corresponding flow is stable.
|>
|> The explanation should make it intuitively obvious whether
|> such things as surface curvature, angle of attack and
|> sharp trailing edge are necessary for lift.
Whew! That's a toughie. Here's my shot ...
First, let's dispel some myths.
MYTH #1
"The air over the top of the airfoil has to go farther, so it goes faster
to meet up with the air going under the airfoil at the same time".
This is what Encyclopedia Britannica says. It is also totally wrong.
In fact, a collection of fluid "globs" lined up in a vertical line
will be anything but vertical once they pass the airfoil. The
"before & after" picture is crudely indicated below.
before . . . . . > after
o o
o o
o o
o o
o _-----____ o
o c______________ o
o airfoil o
o o
o o
o o
The deformation in the line is NOT due to the boundary layer! It looks
like this in inviscid flow. In fact, if the leading edge is blunt, the
one glob starting out exactly on the stagnation streamline never gets to
the airfoil, let alone past it! (it's a simple calculus exercise to show
this). A continuous line of particles therefore never gets "cut" by the
airfoil, but gets stretched out indefinitely. Clearly, MYTH #1 makes no
sense in this context.
MYTH #2
"The lowered pressure over the top of the wing is _caused_ by
the higher velocity there, in accordance with Bernoulli's Law."
It is misleading to use Bernoulli's Law in a cause-and-effect argument.
I could just as well say that the velocity over the wing is higher because
the air accelerates towards the lower pressure there. The squabble here
is over semantics more than anything else. I like to think of Bernoulli's
Law in the following terms:
"In an irrotational, effectively inviscid flow, pressure and velocity
are uniquely related by... (we all know the actual formula)",
and avoid using it in any kind of physical explanation of flow phenomena.
MYTH #3
"The flow around an airfoil and the lift on it are non-unique"
This is only true in a mathematical oversimplification of reality -- namely
inviscid flow. In a real viscous fluid, there is only one flow which
satisifies conservation of mass, momentum, and energy everywhere. Typically,
this "physically correct" (PC ?) flow will satisfy the Kutta condition (smooth
flow-off) at a sharp trailing edge. If the flow does not come off smoothly,
such as shortly after the start of the wing motion, viscous forces acting at
the sharp trailing edge will cause a vortex to roll off on the upper side,
if the angle of attack is positive. The lowered pressure at the core of
this vortex will accelerate the upper flow towards it, setting up the
"lifting" flow pattern.
I should add that in some rare instances, several distinct flows may be
possible (e.g. stall hysteresis), but not infinitely many. Also, the
only possible flow may be oscillatory (e.g. vortex-shedding off a cylinder).
These are mainly curiosities, however.
So to answer your questions...
|> Here's a few ground rules for the discussion:
|>
|> (1) No "explanations" of the form:
|> air must (for some bogus reason) flow faster over the
|> top than the bottom, therefore, by conservation of energy,
|> (= bournoulli's law) there is low pressure on top and net lift.
|>
|> This is the physicist's argument. The main flaw is that they
|> _assume_ a lifting flow pattern (fast on top, slow on bottom),
|> and then invoke conservation of E to verify it is lifting. THis
|> argument would be reasonable if they could give a valid argument that
|> the flow pattern would arise at take off, and is also stable
|> against perturbations. After all, there are other, non-lifting, flows
|> past airplane wings. The argument would get even better if they would
|> also scrap bournoulli, and show exactly how the forces acting arise,
|> so that would could get an intuitive feel for what is going on.
There is one unique flow past a wing (see MYTH 3). The pressure field
associated with this flow exerts a lift on the wing which depends on the
wing shape, angle of attack, dynamic presure, etc. To "see" why the
pressure must be lower over the wing, think of the curvatures of the
streamlines above and below an airfoil at an angle of attack, and the
pressure gradients necessary to force the fluid globs in curved paths
(I'm sorry, but you have to rely on at least F = ma at some point).
. .
. .
. . - p . .
_____
-----_____
FLOW >>> + p
. .
. .
. . . .
Note that the streamlines are roughly parallel to the airfoil at the
sharp trailing edge, as required by viscosity. Clearly, the pressure
over the wing must be decreased, and the pressure under the wing increased
to force the streamlines into this pattern.
I hope this answers some of your questions.
Mark Drela First Law of Aviation:
MIT Aero & Astro "Takeoff is optional, landing is compulsory"
Julian Scarfe
new...@algonet.se wrote:
>Bernoulli equation is valid for 1 dimensional flow along the same streamline
> in a "perfect fluid" or in a point. Going above and below an airfoil is not
> the same streamline and its not one dimemsional, so Bernoulli can not be
> usedthat way as said by one earlier posting.
>
> In a curved streamline we have 3 dimensional flow and Bernoulli is not
> valid. (due to 3d turbulens effects).
That's a little misleading, Jan-Olov. Bernoulli's relationship between
velocity and pressure *is* valid throughout an inviscid, irrotational fluid (a
fluid without vorticity). While that clearly does not perfectly describe a
real fluid, no physical model is a perfect match to reality, and potential
flow calculations (with the assumption that the fluid is inviscid and
irrotational) ignoring turbulence do a pretty good job of predicting the lift
coefficients of aerofoils. To me that suggests that the essential physics of
lift *is* captured within that model.
--
Julian Scarfe
John S. Denker
In article <35518BF0...@scigen.co.uk>,
Julian Scarfe <ja...@scigen.co.uk> wrote:
> In article <6iqaoc$jrr$1...@nnrp1.dejanews.com>,
I completely agree with Julian. Bernoulli's principle is a useful
approximation in 2 dimensions and in 3 dimensions.
Jan-Olov, I know you hate Bernoulli's principle. I realize it can be misused.
Butby the same token, kitchen knives can be used for murder. Still they have
many legitimate uses. If you don't like knives, get them out of *your*
kitchen but please don't accuse my knife of crimes it didn't commit.
--- jsd
John S. Denker
In article <6irm7r$nme$1...@nnrp1.dejanews.com>, new...@algonet.se wrote,
quoting Mark Drela,
>When comparing pressures at two different points by using Bernoulli's Law
>(i.e. where the velocity is higher, the pressure is lower), it is essential
>that the two points in question have the same _total pressure_.
>The jet of air which you blow from your lips has a higher total pressure
>than the surrounding air, so the jet and the ambient air are not related
>by Bernoulli's Law as most people know it.
Excellent point. Many of the common demonstrations that purport to illustrate
Bernoulli's principle are fallacious. But you can't blame that on Dr.
Bernoulli.
>For any steady fluid flow, the total pressure is the same for all points
>along any streamline, as long as viscous effects are not too significant.
>Viscous shear gradually decreases total pressure along a streamline
>(in most cases).
Yes.
>In general, all points in an external aerodynamic flow (outside of
>boundary layers, blown jets, engine exhausts, etc.) are at the same
>total pressure, so Bernoulli applies to any selected pair of points.
Agreed. To be explicit: Bernoulli's principle DOES APPLY to the natural flow
of air over the wings of an airplane, to a very useful approximation.
>MYTH #1
>"The air over the top of the airfoil has to go farther, so it goes faster
>to meet up with the air going under the airfoil at the same time".
>
>This is what Encyclopedia Britannica says. It is also totally wrong.
I completely agree with Mark. The Encyclopedia is wrong. The FAA Flight
Training Handbook is wrong.
>MYTH #2
>"The lowered pressure over the top of the wing is _caused_ by
>the higher velocity there, in accordance with Bernoulli's Law."
>
>It is misleading to use Bernoulli's Law in a cause-and-effect argument.
>I could just as well say that the velocity over the wing is higher because
>the air accelerates towards the lower pressure there. The squabble here
>is over semantics more than anything else. I like to think of Bernoulli's
>Law in the following terms:
>
>"In an irrotational, effectively inviscid flow, pressure and velocity
>are uniquely related by... (we all know the actual formula)",
I can't disagree with any particular word or phrase here, yet this passage
seems overstated. In particular, I worry that Mark's words can be misused by
Bernoulli-haters.
By way of analogy, a Newton-hater might argue that F=ma is misleading because
we can't decide whether the force is "caused by" the accelerating mass or vice
versa. Seriously, it is better to say that F and ma are "uniquely related" by
the standard formula. The same goes with Bernoulli's principle, or
practically any other valid principle of physics.
Sriram Narayan
In article <6ir07h$ort$1...@news.ncal.verio.com>, nar...@shell.dsp.com says...
I should make a correction that the virtual force is m(v^2/r) for a rotating
object but that doesn't change the argument, i.e. acceleration determined by the
mass.