Aerodynamics Midterm Take Home Exam Problem

7/2/2007 MAE430 - Exam Problem David Clark 1 of 11

Mid-Term Examination Problem

Consider a NACA 1412 airfoil. Using Thin Airfoil Theory for a combined airfoil, estimate the

specified aerodynamic parameters specified below and compare them with experimental data in

order to validate the theory.

---------------------------------------- Part 1 -------------------------------------------

Obtain the expression for zf and za in terms of x and c only.

The parametric equations for the camber line of a NACA 4 digit airfoil (NACA mpnk) are...

(1)zf

c

m

p2

2 p⋅x

c

⋅x

c

2

−

⋅= ahead of the max camber point

(2)za

c

m

1 p−( )2

1 2 p⋅−( ) 2 p⋅x

c⋅+

x

c

2

−

= aft of the max camber point

Since we are using a 1412 airfoil, we substitute can define the parameters m and p.

m 0.01:= p 0.40:=

Substituting these values into equation 1 and 2...

(3) zf

c

0.01

0.402

2 0.40⋅x

c

⋅x

c

2

−

⋅=

(4)za

c

0.01

1 0.40−( )2

1 2 0.40⋅−( ) 2 0.40⋅x

c⋅+

x

c

2

−

⋅=

And multiplying both sides by c to find expressions for zf and za exclusively (also

simplifying a little)...

(5a, b) zf c 0.0625⋅ 0.8x

c⋅

x

c

2

−

⋅= zf x c, ( ) 0.050 x⋅0.0625 x

2⋅

c−:=

(6a, b) za c 0.02778⋅ 0.2 0.80x

c⋅+

x

c

2

−

⋅= za x c, ( ) 0.005556 c⋅ 0.02222 x⋅+0.02778 x

2⋅

c−:=

---------------------------------------- Part 2 -------------------------------------------

Calculate the following:dzf

dx

dza

dx

Taking the derivative of equation 5b yields...

(7)x

zf x c, ( )( )d

d0.050 0.125

x

c⋅−=


Taking the equation of equation 6b yields...

(8)x

za x c, ( )( )d

d0.02222 0.05556

x

c⋅−=

---------------------------------------- Part 3 -------------------------------------------

Obtain the expression fordzf

dxand

dza

dxin the form

dz

dxa cosθ⋅ b+=

Using a slightly modified version of book equation 4.21...

(9)x

c

1

21 cosθ−( )⋅=

Substituting this identity into equation 7 and 8 yields...

(10)x

zf x c, ( )d

d0.050 0.125

1

21 cosθ−( )⋅

⋅−=

(11)x

za x c, ( )d

d0.02222 0.05556

1

21 cosθ−( )⋅

⋅−=

Performing some simple expansion, the expressions simplify to the form desired...

(12)x

zf x c, ( )d

d0.0625 cosθ⋅ 0.0125−=

(13)x

za x c, ( )d

d0.02778 cosθ⋅ 0.005556−=

---------------------------------------- Part 4 -------------------------------------------

Calculate the angle θc that locates the max camber point.

Using the identity used for part 3 (equation 9), we realize the max camber occurs at 40% of c, as

defined by the NACA naming convention.

Plugging this value into equation 9 and solving for θ

(14) 0.41

21 cosθ−( )⋅=

Rearranging to solve for θ...

(15) 0.8 1 cosθ0−=


(16) 0.2 cos θ0( )=

(17a, b,c) θ0 acos 0.2( ):= θ0 1.3694 rad⋅= θ0 78.463 deg⋅=

---------------------------------------- Part 5 -------------------------------------------

Calculate A0.

As defined by book equation 4.50, and accomidating for the two regions of integration...

(18) A0 α1

π

0

θ0

θ0x

zf x c, ( )d

d

⌠⌡

d

θ0

π

θ0x

za x c, ( )d

d

⌠⌡

d+

⋅−=

Substituting equation 12 and 13 into 18..

(19) A0 α1

π0

θ0

θ0.0625 cosθ⋅ 0.0125−⌠⌡

d

θ0

π

θ0.02778 cosθ⋅ 0.005556−⌠⌡

d+

⋅−=

To maintain clarity, we will analyze each integral separately.

First, consider the first integral in equation 19...

(20)

0

θ0

θ0.0625 cos θ( )⋅ 0.0125−⌠⌡

d

Performing the integration process and simplifying yields...

(21) 0.0625 sin θ0( )⋅ 0.0125 θ0⋅−( ) 0.0625 sin 0( )⋅ 0.0125 0⋅−( )−

(22) 0.0625 sin 1.3694( )⋅ 0.0125 1.3694⋅−( ) 0.0625 sin 0( )⋅ 0.0125 0⋅−( )−

(23) 0.06124 0.01712−( ) 0− 0.04412=

Thus the result for the first integral in equation 19 is...

(24)

0

θ0

θ0.0625 cos θ( )⋅ 0.0125−⌠⌡

d 0.04412=

Now, consider the second integral in equation 19 ...

(25)

θ0

π

θ0.02778 cos θ( )⋅ 0.005556−⌠⌡

d



(26) 0.02778 sin π( )⋅ 0.005556 π⋅−( ) 0.02778 sin θ0( )⋅ 0.005556 θ0⋅−( )−

(27) 0.02778 sin π( )⋅ 0.005556 π⋅−( ) 0.02778 sin 1.3694( )⋅ 0.005556 1.3694⋅−( )−

(28) 0 0.01745−( ) 0.02722 0.007608−( )− 0.03706−=

Thus, the result for the second integral in equation 19 is...

(29)

θ0

π

θ0.02778 cos θ( )⋅ 0.005556−⌠⌡

d 0.03706−=

To determine the value of A0, we will revisit equation 19

(19) A0 α1

π0

θ0

θ0.0625 cosθ⋅ 0.0125−⌠⌡

d

θ0

π

θ0.02778 cosθ⋅ 0.005556−⌠⌡

d+

⋅−=

... and substitute the values for both integrals as found in equation 23 and 29.

(30) A0 α1

π0.04412( ) 0.03706−( )+[ ]⋅−=

(31) A0 α1

π0.006965( )⋅−=

(32) A0 α( ) α 0.002217−( ):=

---------------------------------------- Part 6 -------------------------------------------

Calculate A1.


(33) A12

π

0

θ0

θx

zfd

d

cosθ⋅⌠⌡

d

θ0

π

θx

zad

d

cosθ⋅

⌠⌡

d+

⋅=

Substituting equation 12 and 13 into 33...

(34) A12

π0

θ0

θ0.0625 cosθ⋅ 0.0125−( ) cosθ⋅⌠⌡

d

θ0

π

θ0.02778 cosθ⋅ 0.005556−( ) cosθ⋅⌠⌡

d+

=

Distributing terms...


(35) A12

π0

θ0

θ0.0625 cos θ( )2

⋅ 0.0125 cos θ( )⋅−( )⌠⌡

d

θ0

π

θ0.02778 cos θ( )2

⋅ 0.005556 cos θ( )⋅−( )⌠⌡

d+

⋅:=

Consider the following algebraic identity..

(36) cos θ( )2 1

2

cos 2 θ⋅( )

2+=

Substituting this into equation 35 and distributing terms...

(37) A12

π

0

θ0

θ0.0625

2

0.0625 cos 2θ( )⋅

2+

0.0125 cos θ( )⋅−⌠⌡

d

θ0

π

θ0.02778

2

0.02778 cos 2θ( )⋅

2+

0.005556 cos θ( )⋅−

⌠⌡

d+

⋅=

To maintain clarity, we will perform the integration one term at a time...

Consider the first integral in equation 37 ...

(38)

0

θ0

θ0.0625

2

0.0625 cos 2θ( )⋅

2+

0.0125 cos θ( )⋅−⌠⌡

d 0.036671226=


(39)0.0625 θ0⋅

2

0.0625 sin 2 θ0⋅( )⋅

4+ 0.0125 sin θ0( )⋅−

0.0625 0⋅

2

0.0625 sin 2 0⋅( )⋅

4+ 0.0125 sin 0( )⋅−

−

(40)0.0625 1.3694⋅

2

0.0625 sin 2 1.3694⋅( )⋅

4+ 0.0125 sin 1.3694( )⋅−

0.0625 0⋅

2

0.0625 sin 2 0⋅( )⋅

4+ 0.0125 sin 0( )⋅−

−

(41) 0.04280 0.006124+ 0.01225−( ) 0 0+ 0−( )− 0.03667=

Thus, the result for the first integral in equation 37 is...

(42)

0

θ0

θ0.0625

2

0.0625 cos 2θ( )⋅

2+

0.0125 cos θ( )⋅−⌠⌡

d 0.03667=

Consider the second integral in equation ...

θ0

π

θ0.02778

2

0.02778 cos 2θ( )⋅

2+

0.005556 cos θ( )⋅−

⌠⌡

d(43)



(44)0.02778 π⋅

2

0.02778 sin 2 π⋅( )⋅

4+ 0.005556 sin π( )⋅−

0.02778 θ0⋅

2

0.02778 sin 2 θ0⋅( )⋅

4+ 0.005556 sin θ0( )⋅−

−

(45)0.02778 π⋅

2

0.02778 sin 2 π⋅( )⋅

4+ 0.005556 sin π( )⋅−

0.02778 1.3694⋅

2

0.02778 sin 2 1.3694⋅( )⋅

4+ 0.005556 sin 1.3694( )⋅−

−

(46) 0.04364 0+ 0+( ) 0.01902 0.002722+ 0.005444−( )− 0.02734=

Thus, the result for the second integral in equation 37 is...

(47)

θ0

π

θ0.02778

2

0.02778 cos 2θ( )⋅

2+

0.005556 cos θ( )⋅−

⌠⌡

d 0.02734=

To determine the value of A1, we will revisit equation 37...

(37) A12

π

0

θ0

θ0.0625

2

0.0625 cos 2θ( )⋅

2+

0.0125 cos θ( )⋅−⌠⌡

d

θ0

π

θ0.02778

2

0.02778 cos 2θ( )⋅

2+

0.005556 cos θ( )⋅−

⌠⌡

d+

⋅=

and substitute the values found in equation 42 and 47.

(48) A12

π0.03692 0.02734+( )⋅=

A1 0.04075=(49)

---------------------------------------- Part 7 -------------------------------------------

Calculate A2.


(50)A2

2

π

0

θ0

θdzf

dxcos 2 θ⋅( )⋅

⌠⌡

d

θ0

π

θdza

dfcos 2 θ⋅( )⋅

⌠⌡

d+

⋅=

Substituting equation 12 and 13 into 33...

(51) A22

π0

θ0

θ0.0625 cos θ( )⋅ 0.0125−( ) cos 2 θ⋅( )⋅⌠⌡

d

θ0

π

θ0.02778 cos θ( )⋅ 0.005556−( ) cos 2 θ⋅( )⋅⌠⌡

d+

⋅:=


Distributing terms yields...

(52) A22

π0

θ0

θ0.0625 cos θ( )⋅ cos 2 θ⋅( )⋅ 0.0125 cos 2 θ⋅( )⋅−⌠⌡

d

θ0

π

θ0.02778 cos θ( )⋅ cos 2 θ⋅( )⋅ 0.005556 cos 2 θ⋅( )⋅−⌠⌡

d+

⋅=

Consider the following identity...

(53) cos 2 θ⋅( ) 1 2 sin θ( )2

⋅−=

To maintain clarity, we will evaluate each integral of equation independently

Consider the first integral in equation 52 ...

(54)

0

θ0

θ0.0625 cos θ( )⋅ cos 2 θ⋅( )⋅ 0.0125 cos 2 θ⋅( )⋅−⌠⌡

d

Using the identity (equation 53 ) to further evaluate the integral, equation 54 becomes...

(55)

0

θ0

θ0.0625 cos θ( )⋅ 1 2 sin θ( )2

⋅−( )⋅ 0.0125 1 2 sin θ( )2

⋅−( )⋅−⌠⌡

d

Further distributing terms..

(56)

0

θ0

θ0.0625 cos θ( )⋅ 0.0625 2⋅ cos θ( )⋅ sin θ( )2

⋅− 0.0125− 0.0125 2⋅ sin θ( )2

⋅+⌠⌡

d

Using mathcad to perform the integration of each part...

(57) 0.06124 0.03919− 0.017118− 0.01467+ 0.0196=

Thus, the result for the first integral in equation 52 is...

(58)

0

θ0

θ0.0625 cos θ( )⋅ cos 2 θ⋅( )⋅ 0.0125 cos 2 θ⋅( )⋅−⌠⌡

d 0.0196=

Consider the second integral of equation 52 ...

(59)

θ0

π

θ0.02778 cos θ( )⋅ cos 2 θ⋅( )⋅ 0.005556 cos 2 θ⋅( )⋅−⌠⌡

d


Using the identity (equation 53 ) to further evaulate the integral, equal 59 becomes...

(60)

θ0

π

θ0.02778 cos θ( )⋅ 1 2 sin θ( )2

⋅−( )⋅ 0.005556 1 2 sin θ( )2

⋅−( )⋅−⌠⌡

d

(61)

θ0

π

θ0.02778 cos θ( )⋅ 2 0.02778⋅ cos θ( )⋅ sin θ( )2

⋅− 0.005556− 0.005556 2⋅ sin θ( )2

⋅+⌠⌡

d

Using mathcad to perform the integration of each part...

(62) 0.02722− 0.01742+ 0.009846− 0.01094+ 0.00871−=

Thus, the value of the second integral in equation 52 is...

(62)

θ0

π

θ0.02778 cos θ( )⋅ cos 2 θ⋅( )⋅ 0.005556 cos 2 θ⋅( )⋅−⌠⌡

d 0.00871−=

Now that we have evaluated both integrals, we will revisit equation 52

(52) A22

π0

θ0

θ0.0625 cos θ( )⋅ cos 2 θ⋅( )⋅ 0.0125 cos 2 θ⋅( )⋅−⌠⌡

d

θ0

π

θ0.02778 cos θ( )⋅ cos 2 θ⋅( )⋅ 0.005556 cos 2 θ⋅( )⋅−⌠⌡

d+

⋅:=

and insert the values found in equation 58 and 62...

(63) A22

π0.0196 0.00871−( )+[ ]⋅=

(64) A2 6.9302 103−

×=

---------------------------------------- Part 8 -------------------------------------------

Calculate cl and calculate the zero-life angle of attack.

The lift coefficient, as defined by book equation is...

(65) cl 2π A0

A1

2+

⋅=

Inserting the results from part 5 and 6 (equation 32 and 49)

(66) cl 2π α 0.002217−0.04075

2+

=


Thus, the coefficient of lift, as a function of alpha, is...

cl α( ) 2π α 0.018158+( )⋅:=(67)

To find the angle at which cl goes to zero, we will set equation to zero and solve for α

(68) 0 2 π⋅ α0 0.018158+( )⋅=

(69) 0 α0 0.018158+=

(70a,b) α0 0.018158− rad:= α0 1.04− deg⋅=

0.1− 0 0.1 0.2 0.31−

0

1

2

3

Cl vs Angle of Attack

Angle (Radian)

Cl

cl α( )

α

(See solution for further solutions)


The following matrix will be used for plotting

the calculated data against the experimental

data provided.αplot

8− deg

4− deg

0deg

4deg

8deg

12deg

16deg

20deg

24deg

28deg

:= cl αplot( )

0

0

1

2

3

4

5

6

7

8

9

-0.763

-0.325

0.114

0.553

0.991

1.43

1.869

2.307

2.746

3.185

=

---------------------------------------- Part 9 -------------------------------------------

Calculate cl and calculate the zero-life angle of attack.

The slope of cl is, by definition...

αcl α( )

d

d(71)

Taking the derivative of the lift coefficient yields...

α2 π⋅ α 0.018158−( )⋅[ ]

d

d α2 π⋅ α⋅ 2 π⋅ 0.018158⋅−( )

d

d=

(72)

Thus...

(73)α

cl α( )d

d2 π⋅→

---------------------------------------- Part 10 -------------------------------------------

Calculate the moment about the quarter-chord

The moment coefficient about the quarter chord can be obtained using book equation 4.64

(74) cmc4π

4A2 A1−( )⋅:=

Substituting the values of A2 and A1, as found above in equation 54 and 49 respectively, yields...

(75)cmc4

π

46.9302 10

3−× 0.04075−( )⋅=

(76) cmc4 0.02656−=

The value for the quarter chord varies slightly from actual data provided on the graph. In the given NACA 1412 Wing

Section chart, the moment coefficient behaves linear during many low angles of attack, however whener large angles

are applied, either positive or negative, the moment coefficient sharply increases or decreases. This is different from

thin airfoil theory, which assigns a constant value for the moment coefficient.


---------------------------------------- Part 11 -------------------------------------------

A comment regarding the moment coefficient and the quarter chord...

On page 333 of Fundamentals of Aerodynamics, a comment is made regarding the quarter-chord acting as the

theoretical location of the aerodynamic center. From equation 66, we can determine the quarter chord is not the

actual center of pressure. If the quarter-chord was the center of pressure, we'd expect the net moment to be zero.

As shown further in book equation 4.66, the center of pressure is shown to be a function of the angle of attack, α.

Thus, the center of pressure can not be described as one single point on the foil since the point will change with the

angle. The center of pressure is needed, however, to further calculate aerodynamic properties for the foil. Performing

these calculations on a variable point is inconvienient.

Describing the quarter-chord as the theoretical center of pressure within the airfoil is an acceptable method of

overcoming this problem. Doing so allows the calculations on a force-and-moment system within predictable points

on the airfoil.

---------------------------------------- Part 12 -------------------------------------------

Determine the lift-to-drag ratio at minimum drag...

Using the supplied NACA 1412 Wing Section Chart, calculate the minimum coefficient of drag and corresponding angle

of attack

cdmin 0.006:= at clmin 0.2:=(77,78)

Thus, the ratio of lift-to-drag is...

(79)clmin

cdmin

33.333=

References:

Anderson, John D. Jr. Fundamentals of Aerodynamics. 4th ed. New York: McGraw-Hill, 2007.

Aerodynamics Midterm Take Home Exam Problem

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