Me438 Aerodynamics (week 1-2-3)

Aerodynamics
ME-438
Spring’16
ME@DSU
Dr. Bilal A. Siddiqui

Course Outline
• Introduction to aerodynamics
• aerodynamics of incompressible flow
• Compressible
• ideal fluid flow
• airfoils theory
• finite wing aerodynamics
• blade element theory and aircraft propellers
• Cascade aerodynamics
• jet propulsion
• intake and nozzle performance
• aircraft performance measurement.

Course Objectives
• To understand the concepts in incompressible airfoil theory, including
• symmetric and cambered airfoils using
• analytical and numerical approaches.
• To understand the incompressible wing theory including
• down wash
• lifting-line theory
• elliptic wings
• general twisted wings
• Application of fundamentals to the design of a wing to meet given
performance criteria.
• Recommended Books
• Fundamentals of Aerodynamics by John D. Anderson
• Aerodynamics by L.J. Clancy [Available in library]

So what is Aerodynamics?
• Aerodynamics is the study of
motion of air.
• It is a sub branch of Fluid
Mechanics and Gas Dynamics
• However, there is a significant
overlap between all these
fields
• Formal study of aerodynamics
began 300 years ago, so it is a
relatively young science!
Fluid Mechanics
Hydrodynamics Gas Dynamics
Aerodynamics
External
Aerodynamics
Internal
Aerodynamics

Difference between Aerodynamics
and CFD
• Think of mechanics of materials. It is a science which predicts
relationships between forces, stresses and deflections.
• FEA is the numerical solution of mechanics of materials. It is a TOOL.
• Machine Design is the application of mechanics of materials. It is the
PRODUCT.
• Similarly, aerodynamics is applied fluid mechanics.
• CFD is just a tool to find numerical solutions to analytically difficult
fluid mechanical problems.
• Aerodynamics is the key driver behind external design of many
vehicles in the air and on land….and surface boats.

What do we do in Aerodynamics?
• Aerodynamics is an applied science and practical engineering for:
• Prediction of forces / moments on, and heat transfer to aerodynamic bodies
• For example, generation of lift, drag, and moments on airfoils, wings, fuselages, engine
nacelles, and whole vehicle configurations.
• Estimation of wind force on buildings, ships, and other surface vehicles.
• Hydrodynamic forces on surface ships, submarines, and torpedoes.
• Aerodynamic heating of flight vehicles
• Determination of flows moving internally through ducts.
• Flow properties inside rocket and jet engines and to calculate the engine thrust.
• Flow conditions in the test section of a wind tunnel.
• Fluid flow through pipes under various conditions.
• Gas dynamic lasers are little different than high speed wind tunnels.
Internal Aerodynamics
External Aerodynamics

What to expect in first four weeks
Introductory Concepts Flow Quantities
Sources of Aerodynamic
Forces/Moments
Lift/Drag/Moment
Coefficients
Center of Pressure Types of Flow
Boundary layers

Historical Context – Motivating Examples
• In 1799, Sir George Cayley became the first person to identify the four
aerodynamic forces of flight
• In 1871, Francis Herbert Wenham constructed the first wind tunnel,
allowing precise measurements of aerodynamic forces.
• In 1889, Charles Renard became the first person to predict the power
needed for sustained flight.
• Otto Lilienthal was the first to propose thin, curved airfoils that would
produce high lift and low drag.
• However, interestingly the Wright brothers, mechanics – not engineers-
found most of the initial work flawed and did something else….a hundred
years after Cayley.

27th of Ramadhan….a
Friday
• On Dec 15, 1903 (corresponding to Hijri
date above), Wilbur and Orville Wright
made history after failing to achieve it for
3 years.
• Cycle mechanics, enthusiastic about flight,
they designed airplanes based on
aerodynamic data published by Leinthall
and Langley.
• All attempts were splendid failures, so
they began to doubt the crude theories of
their times.
• They built themselves a windtunnel and
started testing airfoils. To their surprise,
they obtained reliable results and the rest
is history.

Motivating Examples….DC-3
• The Douglas DC-3 is one of the most famous
aircraft of all time
• It is a low-speed subsonic transport designed
during the 1930s.
• Without a knowledge of low-speed
aerodynamics, this aircraft would have never
existed.
• Notice the dorsals, wing-fuse fairings and
nacelles.
• The fuse shape has not changed fundamentally
for this class of aircraft in a century

Motivating Examples…Boeing 707
• The Boeing 707 opened high-
speed subsonic flight to
millions of passengers
• Designed in the late 1950s.
• Without a knowledge of high-
speed subsonic aerodynamics,
this design would not have
taken off.
• Notice how many modern
aircraft retain many features
after 70 years
• Notice the swept wings

Motivating Examples…Bell X-1
• Bell X-1 became the first piloted
airplane to fly faster than sound
• Captain Chuck Yeager broke the
speed barrier in October 14,
1947..a couple of months after
we broke away from the British ;)
• Without a knowledge of
transonic aerodynamics (near the
speed of sound), the X-1 would
have crashed like all previous
attempts at breaking the Sound
Barrier.
• Notice the pointed nose, thin
wings and high tail.

Motivating Examples…the F-104 Starfighter
• Designed in the 1950s, it was the
first supersonic fighter to reach
Mach 2 in straight flight!
• For a long time, its speed was
unmatched, but it had poor flight
performance, as it was aptly
called “Rocket with a man in it”.
• Knowledge of Supersonic
Aerodynamics was fundamental
• The wings were so sharp
mechanics could cut hands
touching it!

Motivating Examples…F-22 Raptor
• Supersonic fighter with
much improved flight
performance.
• Notice blended wing/body
• Notice twin tails and all
movable tails
• The aerodynamic shape
has also to serve the
purpose of stealth from
radar

Ancient Egyptian Aerospace!
• https://en.wikipedia.org/wiki/Helicopter_hieroglyphs [1290 BC]

Forces and Moments in Flight
• Straight level flight means constant velocity and altitude.
• There are four main forces which govern straight level flight
• For level flight, Lift=Weight and Thrust=Drag
• In other words,
𝑇
𝑊
=
𝐿
𝐷
in level flight
• Except weight, all other variables depend
on Aerodynamics.
• Aerodynamics also causes moments in all three axes

Aerodynamic Moments
• Aerodynamics also causes moments in all three axes.
• Performance of aircraft depends on aerodynamics!

Source of all Aerodynamic Forces &Moments
• No matter how complex the body shape and flow, the aerodynamic
forces and moments on the body are due to only two basic sources:
a) Pressure distribution p over the body surface
b) Shear stress distribution τ over the body surface
• Pressure varies with velocity of air over the surface and acts normal
to it. For incompressible, inviscid flow, it follows Bernoulli principle.
• Shear stress is due to friction in the boundary layer and acts tangent
to the surface. Typically 𝜏 ≪ 𝑝

Net Effect of Pressure and Shear Distribution
• Each body shape and flow condition creates unique p & τ distribution
• The net effect of the p and τ distributions integrated over the
complete body surface is a resultant aerodynamic force R and
moment M on the body.
• Far ahead of the body, the flow is undisturbed and called free stream.
• V∞ = free stream velocity=flow velocity far ahead of the body.

Components of Aerodynamic Force R
• Let distance between leading and trailing edges be
“chord”=c
• “Angle of attack” is the angle between 𝑉∞ and c
• R can be resolved into two sets of components: either
wrt 𝑉∞ or c
•
𝐿 = 𝑙𝑖𝑓𝑡 = 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑉∞
𝐷 = 𝑑𝑟𝑎𝑔 = 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑉∞
•
𝑁 = 𝑛𝑜𝑟𝑚𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 = 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑐
𝐴 = 𝑎𝑥𝑖𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 = 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑐
• But, {L,D} and {N,A} are related through 𝛼
𝐿 = 𝑁 cos𝛼 − 𝐴 𝑠𝑖𝑛α
𝐷 = 𝑁 𝑠𝑖𝑛𝛼 + 𝐴 𝑐𝑜𝑠𝛼

Source of {N,A} and {L,D}
• Pressure p, shear τ , surface slope θ are
functions of path length s.
• For a unit span l=1, the forces are
𝑁 = −
𝐿𝐸
𝑇𝐸
𝑝 𝑢 𝑐𝑜𝑠𝜃 + 𝜏 𝑢 𝑠𝑖𝑛𝜃 𝑑𝑠 𝑢
+
𝐿𝐸
𝑇𝐸
𝑝𝑙 𝑐𝑜𝑠𝜃 − 𝜏𝑙 𝑠𝑖𝑛𝜃 𝑑𝑠𝑙
𝐴 =
𝐿𝐸
𝑇𝐸
−𝑝 𝑢 𝑠𝑖𝑛𝜃 + 𝜏 𝑢 𝑐𝑜𝑠𝜃 𝑑𝑠 𝑢
+
𝐿𝐸
𝑇𝐸
𝑝𝑙 𝑠𝑖𝑛𝜃 + 𝜏𝑙 𝑐𝑜𝑠𝜃 𝑑𝑠𝑙

Source of Aerodynamic Moment
• The aerodynamic moment exerted on the body depends on the point about
which moments are taken.
• Moments that tend to increase α (pitch up) are positive, and moments that
tend to decrease α (pitch down) are negative.
• Moment about the leading edge is simply the forces x moment arms.
𝑀𝐿𝐸 =
𝐿𝐸
𝑇𝐸
𝑝 𝑢 𝑐𝑜𝑠𝜃 + 𝜏 𝑢 𝑠𝑖𝑛𝜃 𝑥 − 𝑝 𝑢 𝑠𝑖𝑛𝜃 − 𝜏 𝑢 𝑐𝑜𝑠𝜃 𝑦 𝑑𝑠 𝑢
+
𝐿𝐸
𝑇𝐸
−𝑝𝑙 𝑐𝑜𝑠𝜃 + 𝜏𝑙 𝑠𝑖𝑛𝜃 𝑥 + 𝑝𝑙 𝑠𝑖𝑛𝜃 + 𝜏𝑙 𝑐𝑜𝑠𝜃 𝑦 𝑑𝑠𝑙
• In equations above, x, y and θ are known functions of s for given shape.
• A major goal of aerodynamics is to calculate p(s) and τ(s) for a given body
shape and freestream conditions (𝑉∞ and α) aerodynamic forces/moments

Getting rid of the Dimensions…convenience
• It will become clear later that it is of benefit to non-dimensionalize forces
and moments.
• Let 𝜌∞ be the free stream air density and S and l be reference area and
reference length respectively.
• Dynamic pressure, 𝑄∞ =
1
2
𝜌∞ 𝑉∞
2
• Lift coefficient, CL =
L
Q∞S
• Drag coefficient, CD =
D
Q∞S
• Lift coefficient, CL =
M
Q∞S𝑙
• Axial and normal force coefficients are similarly defined.
Coefficients makes the math
manageable. An aircraft with a 50m2
wing area and weight of 10,000kg at sea
level cruise will have a lift coefficient of
0.3 at a speed of 100m/s, rather than a
lift of 9.8x104 N.

Reference Area and Length
• In these coefficients, the reference area S
and reference length l are chosen to
pertain to the given geometric body shape
• E.g., for an airplane wing, S is the planform
area, and l is the mean chord length c.
• for a sphere, S is the cross-sectional area,
and l is the diameter
• Particular choice of reference area and
length is not critical
• But, when using force and moment
coefficient data, we must always know
what reference quantities the particular
data are based upon.

A note on coefficients
• When we talk about 2-D object (like airfoils) rather than 3-D (like
aircraft), it is customary to use small letters as the lift, drag etc are
reported on unit span (depth). i.e. cl, cd,cm instead of CL, CD,CM.
• Therefore, for 2-D airfoil, S=c(1)=c.
• Additionally, we also have the pressure and friction coefficients.
𝐶 𝑃 =
𝑝 − 𝑝∞
𝑄∞
𝐶 𝐹 =
𝜏
𝑄∞

Rewriting the Force/Moment Integrals
• More convenient to write integrals in terms of coefficients and
transforming length variable ds to more familiar dx and dy
𝑐 𝑛 =
1
𝑐 0
𝑐
𝑐 𝑝,𝑙 − 𝑐 𝑝,𝑢 𝑑𝑥 +
0
𝑐
𝑐𝑓,𝑢
𝑑𝑦𝑢
𝑑𝑥
+ 𝑐𝑓,𝑙
𝑑𝑦𝑙
𝑑𝑥
𝑑𝑥
𝑐 𝑎 =
1
𝑐 0
𝑐
𝑐 𝑝,𝑢
𝑑𝑦𝑢
𝑑𝑥
− 𝑐 𝑝,𝑙
𝑑𝑦𝑙
𝑑𝑥
𝑑𝑥 +
0
𝑐
𝑐𝑓,𝑢 + 𝑐𝑓,𝑙 𝑑𝑥
𝑐 𝑚 𝐿𝐸
=
1
𝑐2
0
𝑐
𝑐 𝑝,𝑢 − 𝑐 𝑝,𝑙 𝑥𝑑𝑥 −
0
𝑐
𝑐𝑓,𝑢
𝑑𝑦𝑢
𝑑𝑥
+ 𝑐𝑓,𝑙
𝑑𝑦𝑙
𝑑𝑥
𝑥𝑑𝑥
+
0
𝑐
𝑐 𝑝,𝑢
𝑑𝑦𝑢
𝑑𝑥
+ 𝑐𝑓,𝑢 𝑦𝑢 𝑑𝑥 +
0
𝑐
−𝑐 𝑝,𝑙
𝑑𝑦𝑙
𝑑𝑥
+ 𝑐𝑓,𝑙 𝑦𝑙 𝑑𝑥
𝑐𝑙 = 𝑐 𝑛 cos𝛼 − 𝑐 𝑎 𝑠𝑖𝑛α
𝑐 𝑑 = 𝑐 𝑛 𝑠𝑖𝑛𝛼 + 𝑐 𝑎 𝑐𝑜𝑠𝛼

So are there any other benefit from changing
everything into CP ??
• In wind tunnels and flight tests, we often do a pressure survey using
manometers and pressure taps!

Example 1.1
• Consider the supersonic flow over a 50 half-angle wedge at
zero angle of attack. The freestream Mach number ahead
of the wedge is 2.0, and the freestream pressure and
density are 1.01×105 N/m2 and 1.23 kg/m3, respectively
(this corresponds to standard sea level conditions).
• The pressures on the upper and lower surfaces of the
wedge are constant with distance s and equal to each
other, namely, pu = pl = 1.31×105 N/m2.
• The pressure exerted on the base of the wedge is equal to
p∞.
• Shear stress varies over both the upper and lower surfaces
as τw =431s−0.2. Chord length c of the wedge is 2 m.
• Calculate the drag, lift and moment coefficients for the
wedge.

Solution continued
• Since angle of attack is zero, there is no lift or moment!
• For symmetric bodies (symmetry about chord), zero α means zero lift
and moment but nonzero drag.

Solution 2
• We could also have solved using pressure coefficients in a simpler way

Variation of Forces with Angle of Attack
• The behavior of lift with α is like behavior of stress vs strain in metals.
• It is linear until a linear limit called “stall” .
• For some shapes there is lift CL0 even at zero angle of attack
• Like 𝜎 = 𝐸𝜖, we have 𝐶𝐿 − 𝐶𝐿0
= 𝐶𝐿 𝛼
(𝛼)
• Drag is quadratic with angle of attack.
• Therefore, Cm 𝛼
=
𝑑𝐶 𝑚
𝑑𝛼
=
𝑑𝐶 𝑚/𝛼
𝑑𝐶 𝐿/𝛼
=
𝑑𝐶 𝑚
𝑑𝐶 𝐿

Center of Pressure
• If the aerodynamic force on a body is specified in terms of a resultant
single force R, or its components such as N and A, where on the body
should this resultant be placed?
• The answer is that the resultant force should be located on the body
such that it produces the same effect as the distributed loads.
• This location is called the center of pressure CP=[xcp, ycp]
• If axial force passes through the chord, then only normal force
produces moment and yCP=0
𝑥 𝐶𝑃 =
𝑀𝐿𝐸
𝑁

Center of Pressure … continued
• Xcp is the location where the resultant of a distributed load effectively
acts on the body.
• If moments were taken about the center of pressure, the integrated
effect of the distributed loads would be zero.
• An alternate definition of the center of pressure is that point on the
body about which the aerodynamic moment is zero.
• For low angles of attack, 𝐿 = 𝑁cos𝛼 − 𝐴𝑠𝑖𝑛α ≈ 𝑁.
• Thus, for 𝛼 ≤ 15°,
𝑥 𝐶𝑃 ≅
𝑀𝐿𝐸
𝐿

From CP to Aerodynamic Center (AC)
• As L or N approach zero, the center of pressure moves to infinity (tail)!
• As L approaches a high value, xcp approaches zero (nose)
• Also, both MLE and L change with α, which means xCP changes with α
• Since moment about xcp is zero, we can translate R to any location
using moment laws.
• About the quarter-chord point c/4, we can write
𝑀𝑐/4 = 𝑀𝐿𝐸 +
𝑐
4
𝐿
𝑥 𝐶𝑃 ≅
𝑀𝐿𝐸
𝐿

Aerodynamic Center
• It turns out there is a point where the aerodynamic moment does
not change with 𝛼
• This point is called the Aerodynamic Center (AC)
• It is located near the
• Quarter chord c/4 for subsonic flows
• Half chord c/2 for supersonic flows
• Since we know the moment of the major lift producing element of
the aircraft (wing!) is fixed for all angles at AC, we would like to place
the wing’s AC near the aircraft’s CG….why?

Cm 𝛼
=
𝑑𝐶 𝑚
𝑑𝛼
=
𝑑𝐶 𝑚/𝛼
𝑑𝐶 𝐿/𝛼
=
𝑑𝐶 𝑚
𝑑𝐶 𝐿

Examples 1.3 and 1.4
• In low-speed, incompressible flow, the following experimental data
are obtained for an NACA 4412 airfoil section at an angle of attack of
4◦: 𝑐𝑙= 0.85 and 𝑐 𝑚 𝑐/4
= −0.09. Calculate the location of the center of
pressure.
• Consider the DC-3. Just outboard of the engine nacelle, the airfoil
chord length is 15.4 ft. At cruising velocity (188 mi/h) at sea level, the
moments per unit span at this airfoil location are M’c/4=−1071 ft lb/ft
and M’LE =−3213.9 ft lb/ft. Calculate the lift per unit span and the
location of the center of pressure on the airfoil.

Parameters which influence Aerodynamics
• By intuition, the resultant aerodynamic force R should depend on
• Freestream velocity 𝑉∞ (faster air  more force and moment)
• Freestream density 𝜌∞ (denser air  more force and moment)
• Freestream viscosity 𝜇∞ (viscosity  shear stress)
• Size of the body (more reference area and length  more force and moment)
• Compressibility of air (density changes if flow speed is comparable to speed of sound 𝑎∞)
• Angle of attack 𝛼
Therefore,
𝐿 = 𝑔𝑙(𝜌∞, 𝑉∞, 𝑆, 𝑐, 𝜇∞, 𝑎∞, 𝛼)
𝐷 = 𝑔 𝑑(𝜌∞, 𝑉∞, 𝑆, 𝑐, 𝜇∞, 𝑎∞, 𝛼)
M= 𝑔 𝑚(𝜌∞, 𝑉∞, 𝑆, 𝑐, 𝜇∞, 𝑎∞, 𝛼)
for some nonlinear functions 𝑔𝑙, 𝑔 𝑑 and 𝑔 𝑚.
• This is not useful as this means a huge combination of parameters needs to be
tested or simulated to find the relationships necessary for designing aerodynamic
vehicles and products.

Dimensional Analysis
• Fortunately, we can simplify the problem and considerably reduce our time and effort by
first employing the method of dimensional analysis.
• Dimensional analysis is based on the obvious fact that in an equation dealing with the
real physical world, each term must have the same dimensions.
• So it is equivalent to find relationships between dimensionless groups of parameters
rather than the parameters themselves.
• We can show that force and moment coefficients depend on the Reynold and Mach
numbers and flow angles only, i.e.
𝐶𝑙 = 𝑓𝑙(𝑀∞, 𝑅𝑒∞, 𝛼)
𝐶 𝑑 = 𝑓𝑑(𝑀∞, 𝑅𝑒∞, 𝛼)
𝐶 𝑚 = 𝑓𝑚(𝑀∞, 𝑅𝑒∞, 𝛼)
i.e. fl, fd and fm.
• Notice that all parameters are dimensionless!
• Notice that we have reduced the number of parameters from 7 to just 3!
• There may be other “similarity parameters” other than these 3, depending on problem.
𝑅𝑒∞ =
𝜌∞ 𝑉∞ 𝑐
𝜇∞
𝑀∞ =
𝑉∞
𝑎∞

Flow Similarity
• Consider two different flow fields over two different bodies. By definition,
different flows are dynamically similar if:
1. Streamline patterns are geometrically similar.
2. Distributions of
𝑉
𝑉∞
,
𝑝
𝑝∞
,
𝑇
𝑇∞
etc., throughout the flow field are the same when plotted
against non-dimensional coordinates.
3. Force coefficients are the same.
• If nondimensional pressure (CP) and shear stress distributions (
𝜏
𝑄∞
) over different
bodies are the same, then the force and moment coefficients will be the same.
• In other words, two flows will be dynamically similar if:
1. The bodies and any other solid boundaries are geometrically similar for
both flows.
2. The similarity parameters are the same for both flows.
• This is a key point in the validity of wind-tunnel testing.

Flow Similarities Application – Wind Tunnel
• If a scale model of a flight vehicle is tested in a wind tunnel, the
measured lift, drag, and moment coefficients will be the same as for
free flight as long as the 𝑀∞ and 𝑅𝑒∞ of the wind-tunnel test-section
flow are the same as for the free-flight case.
• Wind tunnels are large tubes with air moving inside, used to copy the
actions of an object in flight. Some wind tunnels are big enough to
hold full-size versions of vehicles, but most can hold only scale models
1/3 scale model of space shuttle in
NASA’s 40-foot-by-80-foot WT
Mercedes-Benz’s Aeroacoustic Wind Tunnel Educational Wind Tunnel

Example 1.5
Consider the flow over two circular
cylinders, one having four times the
diameter of the other. The flow over
the smaller cylinder has a freestream
density, velocity and temperature given
by ρ1, V1, and T1, respectively. The flow
over the larger cylinder has a
freestream density, velocity, and
temperature given by ρ2, V2, and T2,
respectively, where ρ2 = 0.25ρ1, V2 =
2V1, and T2 = 4T1. Assume that both μ
and a are proportional to T1/2. Show
that the two flows are dynamically
similar.

Example 1.6
Consider a Boeing 747 airliner cruising at a velocity of 550 mi/h at a
standard altitude of 38,000 ft, where the freestream pressure and
temperature are 432.6 lb/ft2 and 390◦R.
A one-fiftieth scale model of the 747 is tested in a wind tunnel where
the temperature is 430◦R. Calculate the required velocity and pressure
of the test airstream in the wind tunnel such that the lift and drag
coefficients measured for the wind-tunnel model are the same as for
free flight.
Qinetiq’s pressurised, 5m
(16.4ft)-diameter windtunnel
in Farnborough, UK.

A note on Example 1.6
• In Example 1.6, the wind-tunnel test stream must be pressurized
far above atmospheric pressure in order to simulate the free-flight
Re.
• However, most standard subsonic wind tunnels are not
pressurized, because of the large extra financial cost involved.
• This illustrates a common difficulty in wind-tunnel testing:
simulating both flight M and flight Re number simultaneously.
• In most cases we do not attempt to simulate all the parameters
simultaneously
• Mach number simulation is achieved in one wind tunnel, and Re in
another tunnel.
• Results from both tunnels are then analyzed and correlated to
obtain reasonable values for CL and CD appropriate for free flight.
The NACA variable density
tunnel (VDT). Authorized in
March of 1921, VDT was a large,
subsonic wind tunnel entirely
contained within
an 85-ton pressure shell,
capable of 20 atm. It was de-
commissioned in 1940s

Me438 Aerodynamics (week 1-2-3)

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Me438 Aerodynamics (week 1-2-3)