1. Garrett Murphy ID: 102045928 Sec. 505
Tyler Perkins ID: 100781554 Sec. 505
12/16/13 Fall 2013
APPM 2350 Project 3: Airfoil Lab
Introduction: By parameterizing two different equations representing the leading and trailing
edges of an airfoil, we are able to use calculus 3 techniques to analyze certain fluid dynamic
aspects of the given airfoil and rotating cylinder of same surface area. We will first calculate the
mean camber line by finding the mean of the two airfoil-edge functions. We will then plot the
airfoil with its camber line. We will then compute the lift per unit span by integrating a
circulation equation and multiplying it with the air density and free stream velocity. By
linearizing a function of angle of attack and lift, we will then find the maximum lift by plotting
lift vs. angle of attack. By using arc length integrals, we will calculate the perimeter of the
airfoil, and then equate it to the circumference of a cylinder with same surface area and solve
for its radius. (Perimeter of airfoil=circumference of cylinder). Using a velocity potential
equation converted to Cartesian coordinates, we will use partial derivatives to relate the
equation to the coefficients of a given velocity field vector and plot the velocity field. Using
square closed path integrals, we will then calculate the circulation of the cylinder as a function
of its angular velocity. We will then compute the cylinder’s lift as a function of its angular
velocity. By using given bounds, we will Plot lift vs. angular velocity to find the maximum lift
of the cylinder. By comparing the maximum lift between both the airfoil and the cylinder, we
will prove which shape has a better performance.
Background:
Airfoil theory, a part of fluid dynamics, uses certain equations, variables, and constants to
analyze the performance of different shapes in a fluid medium. In this lab, our fluid medium is
air, and the geometrical shapes are an airfoil constructed from a leading and trailing edge, and a
cylinder with identical surface area. The leading edge and trailing edge are the points on an
airfoil where the two surfaces meet. The leading edge comes in contact with the fluid first, and
the trailing edge is where the fluid loses contact. A chord line is a straight line that connects the
leading and trailing edges. Similar to the chord line, the mean camber line is the curve that
contains the points halfway between the upper and lower surfaces, and is measured
perpendicular the mean camber line itself. The angle of attack α, is the angle between the chord
line and the freestream velocity represented by the symbol V∞. The freestream velocity
represents the velocity field of the fluid through which the airfoil or cylinder travels. An airfoil
with its mean camber line, chord line, angle of attack, and freestream velocity is shown below.

2. Garrett Murphy ID: 102045928 Sec. 505
Tyler Perkins ID: 100781554 Sec. 505
12/16/13 Fall 2013
The lift L, is the componentof the aerodynamic force on the airfoil or cylinder perpendicular to
the freestream velocity. The lift is usually measured in Newtons per meter.
Velocity Field: In our lab, we will be using a 2-dimensional velocity field represented by
V=ui+vj. The velocity potential ϕ is a way of representing a fluid flow around an airfoil. By
setting the velocity potential equal to constant, one can implicitly find a function of x and y.
The velocity potential is related to the velocity field with these equations:
The velocity potential representing a fluid flow around a cylinder of radius R and angular
velocity ω is shown below in polar coordinates:
ω is taken to be positive counter-clockwise, and the freestyle velocity is taken to be the initial
freestyle velocity given.
Circulation: The circulation, Γ, is defined as:
V is the velocity field around an airfoil and Ω is a closed path that encloses the airfoil. In some
cases, the velocity field around an airfoil may be unknown; however, the pressure on the airfoil
may be measured by finding the strength of the vortex sheet, γ, and compute the circulation
using the equation below:

3. Garrett Murphy ID: 102045928 Sec. 505
Tyler Perkins ID: 100781554 Sec. 505
12/16/13 Fall 2013
Calculations:
Mean camber line and plot of airfoil:
By using the two given functions
we were able to compute the mean camber line using the formula1
where f and g are the leading and trailing functions respectively. We then plotted the airfoil
along with its camber line2 and scaled the x and y axis so that the airfoil is scaled correctly with
correct visual orientation. The units of the axis are in meters.
Lift per unit span:
We then used the equation to calculate the lift per unit span in (N/m) of the airfoil
at zero angle of attack. ρ∞ is given to be 1.23kg/m3 and represents the density of air at sea level.
V∞ is given to be 132 m/s. Because the velocity field is unknown, we used the equation
to calculate the circulation using the given strength of the vortex sheet to be

4. Garrett Murphy ID: 102045928 Sec. 505
Tyler Perkins ID: 100781554 Sec. 505
12/16/13 Fall 2013
We used a parameter, t, from 0 to 1 because that is the domain of the airfoil function. We
parameterized the airfoil edge function, and then took its derivative. We found ds by dividing
the derivative by its magnitude. Finally, we used Mathematica’s built in NIntegrate to integrate
the strength of the vortex sheet equation multiplied by ds. We then multiplied the newly found
circulation by the air density and airstream velocity and calculated lift per unit span to be
1998.66 N/m.3
Lift as a function of Angle of attack:
We know that when α=-2 degrees, L’=0. We can assume that in the range -2< α<15 degrees, L’
is a linear function of α. We then computed L’ as a function of α by using a linear relation. We
calculated L’ earlier to be 1998.66 N/m when α was 0 degrees. By dividing L’ in two, we
knokw that L’ is 999.33 N/m when α is -1 degrees. Using this relation, we were able to compute
L’ as a linear function of α. L’(α)=999.33 α+1998. 66. 1998.66 is our L’ intercept because L’ is
zero at that α value, and 999.33 is the slope. We then plotted lift vs. angle of attack below:
Maximum Lift:
The domain of α goes from -2 to 15 degrees, so we know that when α= 15 degrees, the
maximum value of L’ must be L’(15) which is 16988.6 N/m.4
Perimeter of Airfoil and Radius of Cylinder:
We then calculated the perimeter of the airfoil by splitting the piecewise function for the leading
edge of the airfoil into two distinct functions. We kept the trailing edge the function the same,

5. Garrett Murphy ID: 102045928 Sec. 505
Tyler Perkins ID: 100781554 Sec. 505
12/16/13 Fall 2013
so then we were left with three different functions each with their own domain. We then took
the derivatives of each of the functions. We then used the arc length integral
For each function and added up the total arc length to be 2.03335 meters.5 We found the radius
of the cylinder by equating the perimeter of the air foil to the circumference of the cylinder and
solving for R. We found the radius to be 0.323618 m.
Velocity Field:
We then calculated the Velocity field around the cylinder by using the potential velocity
function around a cylinder
We converted the function into Cartesian coordinates using the transformation x2+y2=r2 ,
θ=artan-1(y/x), and x=rcosθ. We then used the transformation
and took the partial derivatives to find the coeffients of the Veloctiy field V=ui+vj6. We then
plotted the cylinder of radius 0.323618 m and the velocity field shown below:

6. Garrett Murphy ID: 102045928 Sec. 505
Tyler Perkins ID: 100781554 Sec. 505
12/16/13 Fall 2013
Circulation: We used the formula
to calculate the circulation on a closed path. We parameterized a square path around the
cylinder using 4 equations.
Using the equation above, we took the negative sum of the different line integrals of our
Velocity function, and dotted it with ds which we calculated to be the derivative of the path
function. We found the circulation to be a function of angular velocity to be Γ(ω)=-0.658029 ω7
Lift of cylinder as function of angular velocity:
The original lift equation, can be changed so that lift is a function of angular
velocity. We found previously circulation to be a function of angular velocity
Γ(ω)=-0.658029 ω. By plugging this equation into circulation, can find the lift function in terms
of angular velocity to be L’(ω)= 162.36(0.-0.658029 ω).
Maximum value of lift: By restraining the domain of ω to be between -2 V∞ /R and 0, we can
plot Lift as a function of angular velocity. 𝜔 is taken to be the positive counter clockwise
direction, so 𝜔 ranges from 0 to -2 V∞ /R on our plot below because only when a cylinder
rotates clockwise does it producelift, therefore having a negative angular velocity.

7. Garrett Murphy ID: 102045928 Sec. 505
Tyler Perkins ID: 100781554 Sec. 505
12/16/13 Fall 2013
We then do L’(-2 V∞ /R) to be 87155.6 N/m because it is the extreme point on the domain of w
where lift is the greatest.8
Comparison of Results: An airfoil creates a maximum lift when the angle of attack (α) is 15
degrees and the lift per unit spanis 16988.6 N/m. A cylinder of identical surface area creates a
lift of 87155.6 N/m when the angular velocity is -815.777 rad/s. According to our calculations,
the cylinder creates a larger lift compared with an airfoil of identical surface area, therefore it
has a better performance in terms of lift.
Let’s be real: While in this lab we found the cylinder to create more lift, it is not a realistic
situation. Airplanes do not use rotating cylinder for generating lift for a few reasons. There are
four forces in flight: lift, drag, mass, and weight (mass * acceleration of gravity). Although the
cylinder creates more lift than the airfoil with identical surface area, it needs to be spinning very
rapidly in order to do so. This spinning needs to be propelled by some force (thrust). This
magnus effect produces lift, however, is inefficient in terms of net force required for lift. An
airfoil is designed so that the upper edge is curved, and causes a lower air pressure flowing over
it, and a higher pressure underneath it, causing lift. There is no internal revolution in an airfoil,
and the lift can be increased or decreased by changing the angle of attack. A glider can achieve
lift with no thrust, solely based on angle of attack, and air thermals, providing an upward
pressure on the bottomof the airfoil. Airplanes do not use rotating cylinders becauseit requires
too much energy to producethe same lift as an airfoil. While cylinders only provide lift in one
direction, the thrust from a propeller or jet engine causes a plane to go forward, while the airfoil
wings can change the lift or decrease the lift, making the airplane much more efficient than a
cylinder.

8. Garrett Murphy ID: 102045928 Sec. 505
Tyler Perkins ID: 100781554 Sec. 505
12/16/13 Fall 2013
Appendix:
1.
2.

9. Garrett Murphy ID: 102045928 Sec. 505
Tyler Perkins ID: 100781554 Sec. 505
12/16/13 Fall 2013
3.
4.
5.
6.
7.

10. Garrett Murphy ID: 102045928 Sec. 505
Tyler Perkins ID: 100781554 Sec. 505
12/16/13 Fall 2013
8.

11. Garrett Murphy ID: 102045928 Sec. 505
Tyler Perkins ID: 100781554 Sec. 505
12/16/13 Fall 2013