
English: 
Hello, welcome to my talk on thin airfoil theory.
In this talk the thin airfoil theory will be introduced, in which I will show
how the problem is described and how the problem
is solved. This theory was developed by Munk in
1922. This theory is based on the vortex principle and
Kutta condition and it first theoretically show the importance
of the Kutta condition, if lift is generated
on the airfoil. This thin airfoil theory
can be regarded as the first theory to prove the basic features of airfoils,
such as, the slope of the lift coefficient

English: 
Hello, welcome to my talk on thin airfoil theory.
In this talk the thin airfoil theory will be introduced, in which I will show
how the problem is described and how the
problem
is served this theory
was developed by monk in
1922 this theory
is based on the vortex principle and the
cutter condition and it first
theoretically show the importance
of the cuta condition if lift is
generated
on the airfoil this thing aerofoil
theory
can be regarded as the fossil series
to prove the basic features of airfoils
such as the slope of the lift
coefficient

English: 
and the aerodynamic center. For simplifying the problem, in here
we solve the problem for the symmetrical airfoils,
and we will make some comparisons with the experiment data for different
airfoils. These simple cases could be extended
to more complicated case, for instance, the cambered thin airfoils.
An airfoil is a cross-section of a wing
of an airplane, or of a blade of a rotor, or
a propeller or a turbine and of the sail for a sailingboat,
see the practical airfoils: different airfoils for different

English: 
and the aerodynamic center
for simplifying the problem in here
we solve the problem for the symmetrical
airflows
and we will make some comparisons
with the experiment data for different
airflowers
these simple cases could be extended
to more complicated case for instance
the cambered seeing airfoils
and aerofoil is a close section of a
wing
of an airplane or of a
plate of a rotter or
a propeller or a turbine
and offer sale for a selling boat
see the practical airfoils
different airfoils for different

English: 
applications: low speed ultra-light machine;
this airfoil for the airliners; this airfoil for the propeller blades, and
this one for supersonic interceptors; and in here we can also find some
natural airfoils, like the blackbird,
dragonfly, dolphin etc. In studying the aerofoil, we need
to calculate the aerodynamic forces acting on the aerofoil, so these
include: the lift, L, which is perpendicular to the direction
of the incoming flow, so this is the
main topic in this talk; the drag, D, is parallel to the direction
of the incoming flow, see here. And these lift and drag

English: 
applications
low speed ultralight machine
this therefore for the air nanos this
airfoil for the propeller blades and
this one for symbol sonic interceptors
and in here we can also find some
natural
aerofoils like the blackboard
dragonfly dolphin et cetera
in studying the aerofoil we need
to calculate the aerodynamic forces
acting on the aerofoil so this
including the left air
which is perpendicular to the direction
of
the incoming flow so this is the
main topic in this talk the drug
d is parallel to the direction
of the incoming flow see here
and this lift and drag

English: 
could form a resultant force r
showing here and another important
aspect for the aerofoil is the
aerodynamic
center for most practical aerofoil
the aerodynamic center is at the quarter
of the chord
we will consider this in this talk as
well
and we also consider the aerodynamic
force as
the normal force which is the force
perpendicular to the chord of the
airfoil
and here and the extra force
a which is parallel to the chord
so in this case we can see this
extra force could push the airfoil
moving forward in the direction of
the airfoil cord and this is the force
making the wind turbine to rotate to
generate

English: 
could form a resultant force R, showing here. And another important
aspect for the aerofoil is the aerodynamic
center, for most practical aerofoils, the aerodynamic center is at the quarter
of the chord, we will consider this in this talk as well.
And we also consider the aerodynamic
force as the normal force, which is the force
perpendicular to the chord of the airfoil,
N, here, and the axial force, A, which is parallel to the chord.
So in this case, we can see this axial force could push the airfoil
moving forward in the direction of the airfoil chord, and this is the force
making the wind turbine to rotate to generate

English: 
power. However, we must say the airfoil will not move
forward in the direction of the air flow,  because the air drag is always in the same
direction of the airflow, see here.
For studying the airfoils and/or wings, we have many different
methods, some might be quite simple, while others may be very complicated.
Currently we have the methods: - the thin airfoil theory for
inviscid flows, this theory will examine the lift and
the moment, so this theory is valid for thin aerofoils in the small
angle of attack, so this is the topic for this talk.

English: 
power however we must say
the airfoil will not move
forward in the direction of the air flow
because the object is always in the same
direction of
the airflow see here
for studying the airflow and all
wings we have many different
methods some might be quite simple
while others may be very complicated
currently we have the method
the seeing error for your theory for
invasive flows
sincerely we are examined the lift and
the moment so this theory is
valid for seeing aerofoil in the small
angle of
attack so this is the topic for this

English: 
- the more complicated method would be
conformal transformation. It is the method working for inviscid flows,
so the lift and the moment can be calculated. so this method can be
varied for all airfoils in small angle of attack.
so this two methods were developed in 1920s and 1930s,
when the computer was not available.
So now we have the panel methods,
for inviscid flow. Panel methods are based on potential flow theory,
and in some cases the viscous modification
would be provided so this is good for the pressure and flow field:

English: 
talk
the more complicated method would be
conformal transformation it
is the method working for invasive flows
so the lift and the moment can be
calculated so this method can be
varied for all airfoils in small angle
of attack
so this two method was developed
in 1920s and 1930s
when the computer was not available
so now we have the panel method
for invasive flow panel methods
are based on potential flow theory
and in some cases the viscose
modification
would be provided so this is
good for the pressure and flow field

English: 
we have 2d panel method
for any air foils in reasonable angle of
attack
such as x foil the open source
of the panel method and this
panel method might include the air
compressibility 3d panel method
which can be used for tall aeroplane
with the lifting and the long lifting
bodies
and the air compressibility can be
included
in the analysis the most
complicated method would be the
navier-stock's equation
so this is the method for solving the
navier-stokes equation
for the viscose flow thus
the lift moment drag pressure
and the flow field would be calculated
this method should be valid for all
structures

English: 
we have 2D panel methods for any airfoils in reasonable angle of
attack, such as XFoil, the open source
of the panel method; and this panel method might include the air
compressibility;
-3D panel method, which can be used for whole airplane
with the lifting and the non-lifting bodies,
and the air compressibility can be included
in the analysis.
- the most complicated method would be the
Navier-Stokes equation, so this is the method for solving the
Navier-Stokes equation for the viscous flow, thus
the lift, moment, drag, pressure and the flow field would be calculated.
This method should be valid for all structures

English: 
all angles of attack in theory, but might be questionable in the flow
separations.
Thin airfoil theory is a simple theory of airfoil in incompressible
inviscid flows, it was developed by Max Munk in 1922
and further refined by Hermann Glauert and others.
In this thin airfoil theory, there are some essential
assumptions: for instance, - the aeordynamic forces including
the magnitude, direction and the line of action, is only depended
on the camberline, while the airfoil thickness is neglected;

English: 
all angles of attack in theory but
might be questionable in the flow
separations
seeing airfoil theory is a simple theory
of airfoil in incompressible
invasive flows it was developed
by max monk in 1922
and the father defined by hermann kloret
and others in this
thing aerofoil theory there are some
essential
assumptions for instance
the error dynamic forces including
the magnitude direction and
the line of action is only depend
on the chemboni while the
aerofoil thickness is neglected

English: 
- the ratio of maximum thickness to the chord
is considered to be very small; - and the theory is only working for the
small angle of attack, thus the aerodynamic features
would be all linear; - a Kutta condition must be satisfied
at the trailing edge of the thin airfoils.
In all, this theory idealizes
the flow around the airfoil, that is, the airfoil is represented by the
camber line only.
The thin airfoil theory was
particularly important because it
theoretically proved some important properties of the airfoil:
for instance, the theoretical lift coefficient

English: 
the ratio of maximum thickness to the
cord
is considered to be very small
and the theory is only working for the
small angle of
attack thus the aerodynamic features
would be all linear
a coda condition must be satisfied
at the cheating age of the zinc airfoils
in all this theory idealizes
the flow around the aerofoil that is
therefore is represented by the
camber 9 only
the thin air for your crd was
particularly
important because it is
theoretically prove some
important properties of the airflow
for instance the theoretical left
coefficient

English: 
for thin aerofoil would be given by this formula, here C_l0
would be zero for symmetrical airfoils; and it is proved the aerodynamic center
lies at the quarter of the chord behind the leading edge.
For studying the thin airfoil, vorticity is distributed on the
campanile, see here. The vorticity strength is given by
Gamma(s), and the circulation on the thin aerofoil due to the vorticity is
given by this (capital) GAMMA. And based on the Kutta-
Joukowski theorem, the lift of the airfoil would be calculated as this.
It is well known to generate the

English: 
for seeing aerofoil would be given
by this formula here cl0
would be zero for symmetrical airflows
and it is proved the aerodynamic center
lies at the quarter of the court
behind the leading age
for studying the scene air foil
verticity is distributed on the
campanile see here
the vorticity strength is given by
gamma s and the circulation on the
scene aerofoil due to the vorticity is
given by this
gamma and based on the kota
yukovsky cerium the lift
of the airfoil would be calculated as
this
it is well known to generate the

English: 
that event lived on their foil and
invasive flow coda condition must
be satisfied that is the vorticity
must be zero at the treating age
mathematically in a form as this
so generally for the cuda condition we
have
two different cheating age for the
aerofoil
the sharp angle of the chilling age in
this case
v1 and v2 must be 0
so that the flow at the cheating edge
can be
defined and for the cosmetic cheating
age
we have velocity v1 and v2
from the up surface and the low surface
the coda condition would ask v1
equals v2 so in these two cases

English: 
the relevant lift on the airfoil in inviscid flow, Kutta condition must
be satisfied, that is, the vorticity must be zero at the trailing edge,
mathematically in a form as this.
So generally for the Kutta condition, we
have two different trailing edges for the
aerofoil: the sharp angle of the trailing edge, in
this case, V1 and V2 must be 0.
so that the flow at the trailing edge can be
defined; and for the cusped trailing edge,
we have velocity V1 and V2 from the upper surface and lower surface,
the Kutta condition would ask V1 = V2, so in these two cases,

English: 
the coda condition must be given by
gamma c equaling to zero
so for the single fourier this is the
situation
we need to consider
in this slide we will calculate
the induced velocity due to the
vorticity distributed on the cambodi
for instance the verticity in the
segment ds we are create a velocity
dv at the point p
so the magnitude of the dv is calculated
by this formula here r
is the distance between the verticity
segment
and the target point p
however for the political purpose
we are more interested in the loma
velocity

English: 
the Kutta condition must be given by Gamma(c) = 0.
So for the single aerofoil, this is the situation
we need to consider.
In this slide, we will calculate the induced velocity due to the
vorticity distributed on the camberline,
for instance, the vorticity in the
segment ds will create a velocity dV at the point P,
so the magnitude of the dV is calculated by this formula, here r
is the distance between the vorticity segment
and the target point P, however for the practical purpose,
we are more interested in the normal velocity

English: 
at the point P, which would be given by this, here Delta_3 is the angle
of the induced velocity and the normal direction, and
the minus sign means the normal component of the induced velocity is
in the opposite direction of the normal vector n,
so consider the relation r, given by this, and ds given by this,
so we can formulate the resultant vortex induced normal velocity
Vn as this.
now we are trying to solve the vorticity strength
Gamma, the first thing we need to build the equation for

English: 
at the point p which would be given
by this here let us read the angle
of the induced velocity and
the normal direction and
the minus psi means the normal component
of the induced velocity is
in the opposite direction of the normal
vector n so consider the
relation r given by this
and d s given by this
so we can formulate the resultant
vertex induced normal velocity
v n as this
now we are trying to solve the verticity
strength
gamma the fourth machine we need to
build the equation for

English: 
this vorticity. So now we consider the normal component of the free stream
at the point P, which would be given
by this, here U0 is the velocity of the free stream and
Alpha is the angle of attack here,
and delta_p is the angle between the tangent of the camberline
and the horizontal line, and delta_p in this figure
is the negative gradient, because the tangent of the camberline is
below the horizontal line.
Therefore we put these two together, we
have the normal component of the free stream
given by this expression, and consider the normal component of

English: 
this vortecity so now we consider
the normal component of the free stream
at the
point p which would be given
by this here u0 is
the velocity of the free stream and
alpha
is the angle of attack here
and the delta p is the angle between the
tangent of the cambodi
and the horizontal line
and delta p in this figure
is the negative gradient because
the tangent of the temporary is
below the horizontal nine
therefore we put these two together we
have
the normal component of the free stream
given by this expression
and consider the normal component of

English: 
the induced velocity, so on the surface of the camberline, the
total normal velocity on the component would be 0 given by
this expression, put all this together
we could have the mathematical equation as this, here we can see
the mathematical equation is quite complicated.
To solve this equation, two assumptions must be made here:
- the thin airfoil has a small camber, in the figure the camber line is enlarged
in the z direction for the purpose of an illustration.
Therefore, we could have the small angles,
Delta_1, Delta_2 and Delta_3, as well as the tangent at the

English: 
the induced velocity
so on the surface of the camber 9 the
total normal velocity
on the component would be 0 given by
this
expression put all this together
we could have the mathematical equation
as this here we can see
the mathematical equation is quite
complicated
to solve this equation two assumptions
must be made here
the sea airfoil has a small camber
in the figure the component is enlarged
in the z
direction for the purpose of a
illustration
therefore we could have the small angles
data 1 data 2 and the data 3
as well as their tangent at the

English: 
campanile at
the point p so we have
the approximations given by
this expression and we
also consider the angle of attack is
small
therefore the expression can be
approximated by this
simple expression as such
the mathematical equation can be
simplified
as this for the verticity's chance
here in this equation in the integral
of the expression we can see there is a
singularity on the left-hand side
when k-z equals to x
this singularity might cause the problem
in mathematics but the integration
itself has the limit value as
shown by this equation so the
integration
would have a value as this if we

English: 
camberline at the point P, so we have
the approximations given by these expressions; - and we
also consider the angle of attack is small,
therefore, the expression can be approximated by this
simple expression. As such the mathematical equation can be
simplified as this for the vorticity strength,
Here in this equation in the integral of the expression, we can see there is a
singularity on the left-hand-side when Xi equals to x
this singularity might cause the problem in mathematics, but the integration
itself has the limit value, as shown by this equation. So the
integration would have a value as this.

English: 
if we imagine when approach to Xi = x, from both sides, the denominator
would have a positive value in one direction
but a negative value in other direction,
so the integration would cancel each other
around the singularity, even around the singularity,
the integrand would be very large. Therefore,
a limit value would be for the integration.
We will see this later, and our next step is on how to solve this equation.
For simplifying the solution, we first examine
a simple case: the symmetrical thin airfoil,
which means the camberline of the airfoil would
be on x-axis, see the plot here. hence the gradient of the comberline

English: 
imagine when approach to k z equals
2x from both sides the denominator
would have positive value in one
direction
but the negative value in other
direction
so the integration would cancel each
other
around the singularity even around the
singularity
in the ground that would be very large
therefore
a limit value would be for the
integration
we will see this later and our next
step is on how to solve this equation
for simplifying the solution we first
examine
a simple case the symmetric thing
airfoil
which means the campanile of the airfoil
would
be on x-axis see the plot here
hence the gradient of the component i

English: 
with regard to x would be zero, given by this, so
the mathematic equation for the vorticity
strength can be simplified as this.
However, if the equation is in this form,
it is still very difficult to solve. Munk in 1922
found a way to solve the equation by introducing a
coordinate transformation, given by this Xi, and this coordinate transformation
will lead the limit for the leading edge
Xi=0 will lead Theta = zero; and at
the trailing edge Xi = c  would lead to Theta = Pi.

English: 
with regard to
x would be zero given by this so
the mathematic equation for the
verticity
stress can be simplified as this
however if the equation is in this form
it is there very difficult to solve
monk in 1922
found a way to solve the equation by
introducing a
coordinate transform given by this
case z and this coordinate transform
will delete the limit for the leading
edge
c equaling to zero with the lead
theta equaling zero and at
the chilling age c equaling
to c would lead to theta equaling to pi

English: 
and we have the expression for d_Xi, expressed as this.
Kutta condition now is expressed as Gamma(Pi) equaling to zero.
For the specific point, P, on the thin airfoil,
where Xi = x, which would correspond to a specific
angle Theta_0, given by
the expression as this. so we can have x-Xi
expressed as this.
So after the coordinate transformation, the mathematical equation becomes as this,
and the corresponding Kutta condition, given by this.
so following the reference a solution to the vorticity strength

English: 
and we have the expression for d z
expressed as this
coda condition now is expressed as
gamma pi equaling to zero
for the specific point the p on the
scene error forward
we are c equaling to x
which would correspond to a specific
angle
theta 0 given by
the expression as this
so we can have x minus
c expressed as this
so after the coordinate transformation
the mathematic equation becomes as this
and the corresponding cuda condition
given by this
so following the reference a solution to
the verticity strands

English: 
can be given in this form (this should be obtained first by Munk
in 1922). If we substitute this vortex strength into the
left-hand-side of this equation, we can see
the value on the left-hand-side would be given as this, and we separate the
integration into two terms, we have this.
Now we consider the integration identity (if we like to see, you can find the
derivation for this integration identity from the reference here)
so if we take n = 0
we have the integration by this, and n = 1, we have this. Obviously,
this integration is the same as this, so it would be 0, and this integration
would be same as this, so it equals to Pi, therefore

English: 
can be given in this form this
should be obtained first by monk
in 1922 if we
substitute this vertex strength into the
left-hand side
of this equation we can see
the value on the left hand side would be
given as this and then we separate the
integration
into two terms we have this
now we consider the integration identity
if we like to see you can find the
derivation for this
integration identity from the reference
here
so if we take n equaling to 0
we have the integration by this and
a equals 1 we have this obviously
this integration is the same as this so
it would be 0 and this integration
would be same as this so it
equals to pi therefore

English: 
the value of the left-hand-side would be this, so this
is exactly same as this. Therefore we can confirm the vorticity strength
in such a form is correct.
Now here we will have a look at the Kutta condition.
so if we apply the Kutta condition at the
trailing edge, when Theta equals to
Pi, and we can calculate the vorticity Gamma, given by this,
and we will have an undefined expression, meaning zero
divided by zero. This difficulty can be solved if we use the
l'Hopital's rule, following the reference,

English: 
the value of the left-hand side
would be this so this
is exactly the same as this therefore
we can confirm the verticity strength
in such a form is correct
now here we will have a look
at the kuda condition so if
we apply the kuda condition at the
cheating age
when theta equals to
pi and we can calculate
the vorticity gamma given by this
and we will have an undefined
expression meaning zero
divided by zero this difficulty
can be solved if we use the
lord patella's law following the
reference

English: 
given by this: the limit of both functions, f(x) and g(x) (both
are 0 in the limit) can be replaced by the limits of the differentiations
given by this, as such we apply this l'Hopital's rule into this vorticity
at the trailing edge, we have the expression as this,
and this, so this will lead to an expression as this,
thus we can see the limit of the vorticity at the trailing edge is 0.
In here I will show another approach to obtain the limit of
the vorticity strength at the trailing edge,
which is directly based on the trigonometric relations as this,

English: 
given by this the limit of both function
f x and g x both
are 0 in the limit can be replaced
by the limit of the differentiations
given by this as such we apply this
l'hopital's law into this vorticity
at the cheating age we have the
expression this
and this so this
will lead to an expression as this
thus we can see the limit
of the vertex at the tuning age is
zero in here i will show
another approach to obtain the limit of
the vorticity strands at the cheating
age
which is directly based on the
trigonometric relations as this

English: 
and we will have the expression as this
here we can cancel out the part of
cosine theta divided by two
so we have the expression at this
and we can see when theta
approaches to pi this expression would
be zero therefore
the expression of the vorticity strength
satisfies the cuta condition
once we have the verticity distribution
we can calculate the circulation on the
sink error foil capital gamma
given by this equation and
we can substitute the verticity
expression into this integral and
we can have the result at this
and then we can calculate the

English: 
and we will have the expression as this, here we can cancel out the part of
cosine(Theta/2), so we have the expression as this,
and we can see when Theta approaches to Pi, this expression would be zero.
Therefore, the expression of the vorticity strength
satisfies the Kutta condition.
Once we have the vorticity distribution, we can calculate the circulation on the
thin airfoil, capital GAMMA, given by this equation, and
we can substitute the vorticity expression into this integral, and
we can have the result as this,
and then we can calculate the

English: 
lift on the thin aerofoil, based on the Kutta-Joukowski theorem, given by this,
so we have the lift, this is the lift per unit span.
If we plot the lift, it would be something like this. Here we must
notice the lift is perpendicular to the incoming flow, not to the chord of
the airfoil,
and the lift coefficient would be given as this.
and the result is in a very simple form: 2*Pi*Alpha, here
Alpha is the angle of attack in radians.
In this slide, the aerodynamic moment on the thin airfoil with regard to the

English: 
lift on the scene aerofoil based on the
kuta yokowski theorem given by this
so we have the lift this is
the lift per unit span
if we plot the lift it would be
something like this here we must
notice the lift is perpendicular to the
incoming flow not to the cord of
the error foil and the lift
coefficient would be given as this
and the result is in a very simple form
2 pi times alpha here
alpha is the angle of attack in
radians
in this slide the aerodynamic moment
on the scene airfoil with regard to the

English: 
leading age
would be calculated given as this
here m0 means the moment with regard
to the leading age x equaling to zero
and the negative psi means
the moment is loads down which
is in the opposite direction with the
y-axis
in this plot it is pointing into the
screen using the following relation
and the vertex expression
we could calculate m0
and we have the expression as this so
this
will lead to a result as this
here again the trigonometric relation
given as this is used
now we can calculate the pressure center
x cp using this equation

English: 
leading edge would be calculated, given as this
here m0 means the moment with regard to the leading edge, x = 0,
and the negative sign means the moment is nosedown, which
is in the opposite direction with the y-axis,
in this plot, it is pointing into the screen. Using the following relations
and the vortex expression, we could calculate m0
and we have the expression as this, so this
will lead to a result as this, here again the trigonometric relation
given as this is used.
Now we can calculate the pressure center, x_cp, using this equation:

English: 
the moment equals the lift timing the pressure center x_cp,
here negative sign means nosedown. So if we use the values for m0,
and l, we can obtain the pressure center x_cp would
be a quarter of the chord after the leading edge, and the moment
coefficient would be given as this, and then we can
see this would be minus quarter of the lift coefficient.
Generally for most practical airfoils,
the aerodynamic center is at x = c/4,
since at the aerodynamic center the moment
coefficient would be a constant, here it's zero. If we check this

English: 
the moment equals the lift
time the pressure center xcp
here negative psi means nodes down
so if we use the value for m 0
and l we can obtain
the pressure center x cp would
be a quarter of the chord
after the leading age and at the moment
the coefficient
would be given as this and then we can
see
this would be minus quarter of the lift
coefficient
generally for most practical airfoil
the aerodynamic center is at
x equals quarter of the chord
since at the aerodynamic center the
moment
coefficient would be a constant here
is zero if we check this

English: 
moment with regard to the quarter chord, this moment would be zero (if you are
interested in the calculation, you can do this yourself).
In this slide, the comparisons are made for different
airfoils. The first comparison is for the NACA 009
and NACA 0015, both are the normal NACA 4-digit
symmetric airfoils: they have very similar features in geometry,
the main difference between them is the thickness,
9% compared to 15%.
Generally, NACA 0009 can be considered as

English: 
moment with regard to the quarter chord
this moment would be zero if you are
interested in the calculation you can do
this
yourself
in this slide the comparisons are made
for different
airflows the first comparison
is for the naka009
and the naga zero zero one five
both are the normal locker four digit
symmetric airfoils they have very
similar features in geometry
the main difference between them is the
sickness
nine percent compared to 15 percent
generally naka0009 can be considered at

English: 
the thin airfoil, but NACA 0015 is a relatively thick airfoil,
since the most useful airfoil would have a thickness of 12%.
so the prediction use the thin airfoil theory
and the experiment data for these two airfoils,
we can see in the small angles of attack, say less than 10
degrees, both airfoils have very close lift coefficient
as the theoretical lift coefficient, and the large difference between these
two airfoils is the lift coefficient in the stall
condition, here and here,
and we can see from the comparison, NACA0015 has a
mild stall lift drop.
The second comparison is for the

English: 
the same airfoil but the naka0015
it's a relatively sick airfoil
since the most useful evolver would
have a thickness of 12 percent
so the prediction uses the sync airfoil
cld
and the experiment data for this true
airfoil
we can see in the small angles of
attack say less than 10
degrees also foils
have very close lift coefficient
as the theoretical lift coefficient
and the large difference between this
true airfoil
is the lift coefficient in the store
condition
here and here and
we can see from the comparison
naka0015 has a
mired store lifted job
the second comparison is for the

English: 
naka0012
and the naka0012 hyphen
64. both symmetric
airfoil have same thickness
but the latter is the modified version
of
the formal one with the main difference
of the maximum thickness occurs
at 40 percent of the cord
the former at 30 percent
see the profile comparison
if we look at the experiment data
and the protection from the things are
for your theory
we can see again for the left
coefficient
in the small angle of attack both
symmetric airfoils
are very similar and the slope of
the lift coefficient evolves very close

English: 
NACA 0012 and NACA 0012-64,
both symmetrical airfoils have same thickness,
but the latter is the modified version of
the formal one, with the main difference of the maximum thickness occurs
at 40% of the chord (the former at 30%),
see the profile comparison.
If we look at the experiment data
and prediction from the thin airfoil theory,
we can see, again, for the lift coefficient
in the small angle of attack, both symmetric airfoils
are very similar, and the slopes of the lift coefficient are very close

English: 
to that from the thin airfoil theory, and from the comparison, we can see
both airfoils have very similar stall characteristics, see here and here
In this slide, the moment coefficients
are compared for different airfoils. In the small angle of attack,
both NACA 0009 and NACA 0012 would have very small moment coefficient
regarding to the quarter chord, again from this comparison
the large deviation occurs at the large angle of attack,
where the stall happens.
Similarly for the NACA 0012
and NACA 0012-64, when the angle of attack is small, the

English: 
to that from the zinc aerofoil theory
and from the comparison we can see
both airfoils have very similar
store cracked elastics see here
and here
in this slide the moment the coefficient
are compared for different airfoils
in the small angle of attack
both naka0 and the naka002l
would have very small moment coefficient
regarding to the quarter chord
again from this comparison
the large deviation occurs at the large
angle of attack
where the store happens similarly for
the naka0012
and naka0012 164
when the angle of attack is small the

English: 
moment the coefficient
from the scene alpha theory is very
close
to the experiment data but
when the angle is more than 12 degrees
the large deviation would have happened
from the zinc elevation see here
you

English: 
moment coefficient from the thin airfoil theory is very
close to the experiment data,
but when the angle is more than 12 degrees (should be 15 degrees), the large deviation would have happened
from the thin airfoil theory, see here.
