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barton [ Centurion ]
Co to jest paradoks d'Alamberta
Jak wiesz to jestes kozak (kozaczka)
electra [ Konsul ]
czyzby cos zwiazanego z fizyka ??
Septi [ Starszy Generał ]
Też mi sie tak wydaje,bodajże z Równaniem falowym d’Alamberta.
electra [ Konsul ]
eee to niezle :) cos tam sie jeszcze pamieta :)
barton [ Centurion ]
Niestety. To nie ma nic wspolnego z rownaniem falowym d'Alamberta.
donzoolo [ Senator ]
google nie podpowiedzial wam tym razem dobrze :)
electra [ Konsul ]
donzoolo --->> skad wiedziales ?? hihihi
Didier z Rivii [ life 4 sound ]
d'Alambert? czyzby szeregi geometryczne......?
Verminus [ Prefekt ]
Paradoks polaga na tym, że wiesz co masz liczyć a nie wiesz po co :)
settogo [ Vote 4 TUSK ]
PARADOKS KŁAMCY :) O tym jest teraz głośno :)
barton [ Centurion ]
Eeee... Myslalem, ze jestescie kozaki. A tu jakies Sofixy tylko :)
alan09 [ Konsul ]
Czy to chodzi o to, że idąc i tak stoimy w miejscu??
agentzel [ Pretorianin ]
jak zes taki kozak to sam napisz co to jest!!
gereg [ zajebisty stopień ]
Spring 2003
22 Towards airplane flight In the limit where the flow is irrotational we just need to find solutions to Laplace's equation to find solutions to the Euler equations. Let's write down a couple of these to gain some intuition: our aim being to acquire techniques to begin to think about airplane flight.
22.1 Point source. We know from electrostatics that a good solution of Laplace's equation is just
OE = \Gamma
c
4ssr
; (17)
where c would be the charge in an electrostatic problem. What does this solution correspond to for us ? The velocity field
u =
c
4ssr
2
r (18)
is a radial source or sink of fluid.
22.2 Uniform flow. Another trivial solution is simply uniform flow
OE = U \Delta r; (19) which works for any velocity vector r
22.3 Vortex Solutions We can also guess solutions of the form OE = f (`)g(r), where ` and r denote cylindrical coordinates. Laplace's equation in cylindrical coordinates is
1
r
@ @r
`
r
@OE
@r
'
+
1
r
2
@
2
OE
@`
2
= 0: (20)
Plugging this ansatz into the equation gives that
r g
d dr
`
r
dg
dr
'
+
1
f
d
2
f
d`
2
= 0: (21)
Each term in must be a constant, i. e.
d
2
f
d`
2
= f k
2
; (22)
65
r
d
dr
`
r
dg
dr
'
= gk
2
: (23)
For k 6= 0 f = A sin(k`) + B cos(k`), with k being an integer. However, if k = 0 then f = A + B`. Turning to the radial part we guess that g = r
ff
. The radial equation then
requires that ff = \Sigma k, giving g(r) = Ar
k
+ Br
\Gamma k
. If k = 0 then g(r) = A + Blnr. So the
most general solution is
OE(r; `) = (A
0
+ B
0
lnr)(C
0
+ D
0
`) +
k=1
X
k=1
(C
k
sin k` + D
k
cos k`)(A
k
r
k
+ B
k
r
\Gamma k
) (24)
The corresponding velocity field is
v = rOE =
@OE
@r
^r +
1
r
@OE
@`
^ `: (25)
Putting in our general solution with k = 0 we get
v =
B
0
(C
0
+ D
0
`)
r
^r +
D
0
r
(A
0
+ B
0
lnr)
^ ` (26)
Setting B
0
= 0 we get a flow with no radial component, and angular component
v
`
=
D
r
; (27)
which is irrotational by virtue of it's construction. This is called a point vortex solution. If we consider the circulation about a loop containing the origin
v \Delta dl =
Z
2ss
0
v
`
rd` = 2ssD = \Gamma (28)
Thus D = \Gamma =2ss, where \Gamma is the circulation about the point vortex.
22.4 Flow around a cylindrical wing. Okay, so this isn't the true shape of an airplane wing, but it's a good place to start. Let's see if we can calculate the lift and drag on a wing of radius R moving with velocity u
0
. In
the frame of reference of the wing the boundary conditions are
OE ! u
0
x as r ! 1;
@OE
@r
= 0 at r = R:
Using the general solution we found above the first boundary condition requires that
OE =
`
u
0
r +
D
1
B
1
r
'
cos ` + A
0
(C
0
+ D
0
`): (29)
The second boundary condition gives us
OE = D
0
` + u
0
cos `
r +
R
2
r
!
; (30)
where we have set C
0
= 0 since rOE is all that matters. Physically we can see that D
0
=
\Gamma =2ss, where \Gamma is the circulation about the wing (check this by integrating around a circular loop containing the wing).
66
22.5 Forces on the circular wing The lift and drag forces on the wing are respectively given by
F
L
=
Z
2ss
0
p sin `rd`; (31)
F
D
=
Z
2ss
0
p cos `rd`: (32)
. We can determine the pressure distribution from Bernoulli's Law
p = p
0
\Gamma
ae
2
(rOE)
2
R
= p
0
\Gamma
ae
2
`
2u
0
sin ` \Gamma
\Gamma
2ssR
'
2
: (33)
Putting this into the above relations we find that
F
D
= 0: (34)
This result is known as D'Alambert's paradox. Furthermore, the lift on the wing is linearly proportional to the circulation about the wing;
F
L
= \Gamma aeu
0
: (35)
Thus the lift on the wing is zero (unless it is spinning !) so there really is a problem. What could be wrong ? Firstly, airplane wings are not circular and maybe if consider an alternative shape we would find lift. This will be our next avenue of investigation. We could also be worried about the fact that our problem is 2D. However, given that the aspect ratio of a wing is roughly 10:1 it is acceptable to consider 2D flow, and it is unlikely that this changing this will generate the lift and drag we are missing. Indeed, we hope that this is not the case otherwise theoretical aerodynamics would become even more complex. Lastly, we have neglected viscosity, which we considered to be small enough to be neglected. This last point turns out to be the source of our troubles, but let's first convince ourselves that altering the shape of the wing doesn't change things. To do so will require conformal mapping.
Szwaroc [ ]
/\ /\ cokolwiek to znaczy ;-)
gereg [ zajebisty stopień ]
zainteresowanym polece tez to
Gasto [ Pretorianin ]
gereg --> a teraz badz tak mily i nam "sofiksom" wszystko wytlumacz ;)
Ezrael [ Very Impotent Person ]
AAAAA!!! Co to było?! :)
gereg [ zajebisty stopień ]
to chyba nam objasni barton, ktory pokazaczyl, a jak wszystko przedstawilem zniknal jak kamfora ;-)
nie moja dzialka - ja o tym wiem, ale sie na tym nie znam ;-)
Bich [ Pretorianin ]
gereg --> twój post był chyba najdłuższy w historii foum...
gereg [ zajebisty stopień ]
Bich - widac, zes niedawno tu zawital ;-)
zapewniam Cie, ze daleko mi do najlepszych ;-)))
PROSZATAN [ Apokalipsa ]
no to ja nie jestem Kozakiem :) bo nie wiem
AK [ Senator ]
Taki ze mnie fizyk, jak z kangura helikopter (jesli chodzi o dynamikę płynów), lecz czy nie chodziło tu o stwierdzenie że strumień cieczy idealnej, opływający idealnie cylindryczny obiekt, nie jest w stanie go poruszyć, gdyż siły parcia działające od przodu są wyrównywane przez ssącą siłę cieczy, która działa od tyłu?