Flight Investigations
TEXT BOOK CHAPTER 16 PAGE 362 - 377
Flight – the beginning
The Wright Brothers
https://www.youtube.com/watch?v=VcqxI-OJ1mk
Failures / Early Attempts
https://www.youtube.com/watch?v=fw_C_sbfyx8
Newton’s laws of motion
https://www.youtube.com/watch?v=iH48Lc7wq0U
Summary:
First Law The velocity of an object can change only if there is a non-zero net force
acting on it.
Second Law The relationship between the acceleration of an object, the net force
acting on it, and the object’s mass can be expressed as 𝐹 = 𝑚𝑎
Third Law When an object applies a force (action) to a second object, the second
object applies an equal and opposite force (reaction) to the first object
Forces acting on an aircraft
The main forces acting on an
aircraft in level flight can be
identified as vertical and
horizontal pairs.
Vertical Pair – Lift and Weight
Horizontal Pair – Thrust and Drag
Forces acting on an aircraft
Lift is the upwardacting force created by
a wing moving through
the air
Drag is the rearwardacting force that resists
the motion of an
aircraft through the air
Thrust is the forward
force that drives an
aircraft through the air
Weight is the force
applied to an object
due to gravity
Lift Force
• Acts upwards, at right angles to the airflow direction.
• Lift force is generated over the entire wing, although it is usually thought of as acting at
one position along the wing.
• This position is known as the centre of lift or the centre of pressure (CP).
Weight Force
• The weight force is considered to act through the centre of gravity (COG) - This is the point
where the mass of the aircraft is considered to be concentrated and is the point of balance.
• If an aircraft were hung from a cable attached to its centre of gravity, it would hang level
and perfectly balanced.
• The location of an aircrafts COG depends on the load it carries (fuel, cargo, passengers etc)
In level flight the Lift Force and Weight Force are equal in size and opposite in direction.
Drag Force
• As an aircraft moves through the air in flight, it experiences air friction or drag.
• The faster the aircraft moves, the greater the resultant drag force.
• There are several different types of drag forces that act on different parts of the aircraft
when in flight. The arrow used to represent the drag, refers to the resultant of all the drag
forces that act on every part of the aircraft.
Thrust Force
• Causes an aircraft to move through the air.
• Propeller blades or jet engines push the air backwards (an action).
• The air pushed backwards therefore pushes the plane forwards with a force of equal
magnitude (a reaction).
• The magnitude of the thrust depends on the total Power delivered by engines of the
aircraft.
• The Power output, P, when a Force, F, is applied to an object causing the object to move
with a speed, v, is given by the equation:
𝑃 = 𝐹𝑣
Which for an aircraft can be related to the mechanical power output of the engines:
𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑃𝑜𝑤𝑒𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 = 𝑡ℎ𝑟𝑢𝑠𝑡 × 𝑠𝑝𝑒𝑒𝑑
Forces acting on an aircraft
𝑃 = 𝐹𝑣
→
Power (Watts, W)
𝑀𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑃𝑜𝑤𝑒𝑟 𝑂𝑢𝑡𝑝𝑢𝑡 = 𝑡ℎ𝑟𝑢𝑠𝑡 × 𝑠𝑝𝑒𝑒𝑑
Force/Thrust (Newtons, N)
Speed/Velocity (ms-1)
A jet travels at a constant speed of 150𝑚𝑠 −1 while it’s engine provides a total thrust of 10kN.
What is the Power output of the engines?
𝑃 = 10 × 10 3 𝑁 × 150 𝑚𝑠 −1
= 1500000 𝑊
= 1.5 𝑀𝑊
If in level flight, what is the total drag on the jet?
In level flight, Thrust = Drag
∴ Drag = 10,000 N
Forces acting on an aircraft
If the net force acting on an aircraft in flight is zero, it maintains constant velocity.
If the net force acting on an aircraft is not zero, the magnitude and direction of the
net force determines the magnitude and direction of the acceleration of the aircraft.
Describe the direction of the resulting motion (acceleration) if an aircraft has the
following forces acting upon it in flight:
a) Lift = 5000 N, Drag = 800 N, Weight = 5000 N, Thrust = 1200 N
Net Force 400 Forward – Plane accelerates Forward
b) Lift = 6000 N, Drag = 900 N, Weight = 5000 N, Thrust = 900 N
Net Force 1000 Upward – Plane accelerates Upward
Forces acting on an aircraft
The Aerodynamics of Flight
https://www.youtube.com/watch?v=5ltjFEei3AI&feature=related
Now Do
Text Book - Chapter 16
Questions 1 – 3 Pg 376
( Applying Newton’s laws to Aircraft)
Moving through fluids
Aeronautics is concerned with the motion of aircraft through gases – in particular, air.
To understand how lift in an aircraft occurs, we need to understand a little about
movement through fluids.
All liquids and gases are fluids. Fluids, like solids, are composed of small particles.
Particles are packed less tightly in fluids than in solids, allowing movement of particles
more freely.
Moving through fluids
In the 1700’s Daniel Bernoulli developed the equation of continuity which stated,
“ All material that enters a pipe will leave the pipe” which can be expressed as:
𝑄 = 𝑣1 𝐴1 = 𝑣2 𝐴2
Q = flow rate ( measured in cubic metres per second 𝑚3 𝑠 −1 )
v = fluid speed ( measured in meters per second 𝑚 𝑠 −1 )
A = cross-sectional area of the pipe ( measured in square metres 𝑚2 )
Moving through fluids
𝑄 = 𝑣1 𝐴1 = 𝑣2 𝐴2
Q = flow rate ( measured in cubic metres per second 𝑚3 𝑠 −1 )
v = fluid speed ( measured in meters per second 𝑚 𝑠 −1 )
A = cross-sectional area of the pipe ( measured in square metres 𝑚2 )
eg1. In the diagram pictured, A1 is larger than A2.
Using the equation of continuity, explain the
difference in V1 and V2 in this scenario.
V1 < V2
Wider pipe, slower speed
Narrower pipe, faster speed
Moving through fluids
𝑄 = 𝑣1 𝐴1 = 𝑣2 𝐴2
Q = flow rate ( measured in cubic metres per second 𝑚3 𝑠 −1 )
v = fluid speed ( measured in meters per second 𝑚 𝑠 −1 )
A = cross-sectional area of the pipe ( measured in square metres 𝑚2 )
eg2. Air flows through a pipe with a cross sectional area
of 0.5 𝑚2 at a speed of 20 𝑚𝑠 −1 . If the air exits the
pipe at a speed 100 𝑚𝑠 −1 , what is the cross sectional
area of the end of the pipe?
𝑣1 𝐴1 = 𝑣2 𝐴2
20 × 0.5 = 100 × 𝐴2
𝐴2 =
10
100
= 0.1 𝑚2
𝑄 = 𝑣1 𝐴1 = 𝑣2 𝐴2
Q = flow rate ( measured in cubic metres per second 𝑚3 𝑠 −1 )
v = fluid speed ( measured in meters per second 𝑚 𝑠 −1 )
A = cross-sectional area of the pipe ( measured in square metres 𝑚2 )
eg3. Air flows through a pipe with radius of 0.3 𝑚 at a speed of 4 𝑚𝑠 −1 .
The end of the pipe has a radius of 0.8 𝑚. What is the velocity of the air
travelling at the end of the pipe?
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑎 𝑐𝑖𝑟𝑐𝑙𝑒 = 𝜋𝑟 2
2
𝑣1 𝐴1 = 𝑣2 𝐴2
𝐴1 = 𝜋 × (0.3) = 0.09𝜋 𝑚
𝐴2 = 𝜋 ×
(0.6)2
𝑣1 = 4 𝑚𝑠 −1
= 0.36𝜋
2
𝑚2
4 × 0.09𝜋 = 𝑣2 × 0.36𝜋
𝑣2 =
0.36𝜋
0.36𝜋
= 1 𝑚𝑠 −1
Text Book - Chapter 16
Now Do
Questions 4, 5, 6, 7a Pg 376
( Moving through fluids and Bernoulli’s Equation)
𝑄 = 𝑣1 𝐴1 = 𝑣2 𝐴2
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑎 𝑐𝑖𝑟𝑐𝑙𝑒 = 𝜋𝑟 2
𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
𝑟𝑎𝑑𝑖𝑢𝑠 =
2
Fluid Speed and Pressure – Bernoulli’s Principle
Applying the Law of Conservation of Energy to fluid flow, Bernoulli found that the total
energy of fluid is constant throughout the flow.
Regardless of the pressure/speed of the fluid in the system, the total energy stays the same.
Using this principle, he derived an equation which states:
1
𝜌𝑣 2
2
+ 𝜌𝑔ℎ + 𝑃 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
(total pressure)
𝜌 = fluid density ( 𝑘𝑔 𝑚−3 )
𝑣 = speed of the fluid ( 𝑚 𝑠 −1 )
𝑔 = acceleration due to gravity ( 𝑚 𝑠 −2 )
ℎ = vertical displacement of the fluid ( 𝑚 )
𝑃 = static pressure of the fluid ( 𝑁 𝑚−2 𝑜𝑟 𝑃𝑎 )
When dealing with aircraft travelling at high speeds, exceptions begin to occur due to
friction on the wing of the aircraft, which causes heat. For aircraft, we refer to another
version of this formula, developed by Leonhard Euler.
Fluid Speed and Pressure – Bernoulli’s Principle
Euler found that an increase in speed, created when fluid flows through a narrow section
of a pipe, produced a decrease in static pressure. He named this principle,
the Bernoulli Principle which says, the pressure of fluid decreases as its velocity increases.
For fluid at a constant height:
𝜌 = fluid density ( 𝑘𝑔 𝑚−3 )
𝑣 = speed of the fluid ( 𝑚 𝑠 −1 )
1
𝜌𝑣 2 + 𝑃 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (total pressure) ℎ = vertical displacement of the fluid ( 𝑚 )
2
𝑃 = static pressure of the fluid ( 𝑁 𝑚−2 𝑜𝑟 𝑃𝑎 )
Dynamic Pressure
(pressure
associated with
movement)
Fluid Speed and Pressure – Bernoulli’s Principle
We can apply these principles to the wing of the aircraft.
An aircraft’s wing has a curved top and a flat bottom – this shape is known as aerofoil.
The wing is shaped this way so that air travelling over the it’s top surface speeds up.
This in turn reduces the air pressure above the wing to below normal pressure, resulting
in an upward force known as the lift force.
Newtonian Lift
Newtonian Lift only accounts
for about 15% of the lift
required for a cruising aircraft.
…..is the name given to the effect of the action reaction pair on the wing of an aircraft.
Newtonian Lift is the lift created in addition to the effects described by the Bernoulli
principle. As Aerofoil passes through the air :
• It pushes air below it downward (an action)
• Consequently, the air beneath the wing pushes upward (a reaction)
• The greater the amount of air that is deflected downwards by the wing, the greater
the upward force, or the greater the Newtonian Lift.
Newtonian Lift
• The downward flow of air caused by Newtonian Lift is called downwash
• The amount of downwash depends on the shape of the wing, the and
the speed that the aircraft is travelling at.
• An effective wing design may involve trying to produce as
much downwash as possible, without causing turbulence.
• The amount of downwash can change, due to the angle
of the wing, called the angle of attack.
Angle of Attack
• The position that the wing is angled at is called the angle of attack.
• Increasing the angle of attack increases the lift
• BUT, as the airflow experiences a greater disturbance, drag is increased, so more thrust
must be produced by the engines to counteract the drag
• If the angle of attack is increased significantly, the air is disturbed too much, increasing
drag to the point of turbulence.
As the angle of attack increases, the planes lift increases, until the angle reaches the
critical angle. At this point, lift drops, drag increases and the plane stalls.
Text Book - Chapter 16
Now Do
Questions 12, 13, 14 Pg 376
( Moving through fluids and Bernoulli’s Equation)
Then
Continue on with your report
More on Thrust & Drag &
Drag Ratio
HOMEWORK
• Read text book pages 368 – 370 ( up until the turning effect of a force section )
• Answer questions 15 – 20
• You should include of some the questions and your responses in your report, to support
the concept / calculations
Glide Ratio
𝑔𝑙𝑖𝑑𝑒 𝑟𝑎𝑡𝑖𝑜 =
𝑔𝑙𝑖𝑑𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑙𝑜𝑠𝑠 𝑜𝑓 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒
The glide ratio of an aircraft is the ratio of the horizontal distance travelled (glide distance)
to the loss of altitude while gliding (no power to the plane).
Its value is equal to the ratio of lift to drag.
Applying the equation:
eg1. The engines of an aircraft fail while flying at an altitude of 500m , leaving the aircraft
gliding to a safe landing 2 kms from the point of failure.
a) The glide ratio =
𝑔𝑙𝑖𝑑𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑙𝑜𝑠𝑠 𝑜𝑓 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒
=
2000
500
4
1
= = 4: 1
b) The Lift to Drag Ratio = the glide ratio = 4: 1
Glide Ratio
𝑔𝑙𝑖𝑑𝑒 𝑟𝑎𝑡𝑖𝑜 =
𝑔𝑙𝑖𝑑𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑙𝑜𝑠𝑠 𝑜𝑓 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒
eg2. The engines of an aircraft fail while flying at an altitude of 800m.
If the glide ratio is 6:1, calculate the horizontal distance travelled by the aircraft after the failure.
𝑔𝑙𝑖𝑑𝑒 𝑟𝑎𝑡𝑖𝑜 =
6
1
=
𝑔𝑙𝑖𝑑𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑙𝑜𝑠𝑠 𝑜𝑓 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒
𝑥
800
𝑥 = 800 ×
6
1
𝑥 = 4800𝑚
Using your paper planes from last
lesson, estimate the glide ratio of
your plane and compare with
others in the class.
Amend your plane to try and
achieve a greater glide ratio.
Turning Effect of a force
So we know that for an aircraft to maintain steady, level flight, the vertical pair and
horizontal pair of forces must each be balanced.
Lift Force = Weight Force = Constant Altitude
Thrust Force = Drag Force = Constant Velocity
Turning Effect of a force
But if we look closer, it’s not only the size of the forces that we need to take into
account - we need to consider the distance between the line of action of the force and
the centre of gravity of the aircraft.
If lift is generated far back from the centre of gravity and weight far forward, the
aircraft will not be balanced and will rotate due to the turning effect of the forces.
This turning effect is called torque.
Torque = Force x distance
𝜏 =𝐹×𝑑
(Nm)
(N)
(m)
Turning Effect of a force
𝜏 =𝐹×𝑑
In an aircraft, this may be considered in the example below.
A plane has 4 engines each outputting a thrust force of 20,000N.
The inner engines are 10 m from the centre, the outer engines are 5 m from the centre.
The outer right engine cuts out. To keep the plane in a
straight line, what force must the right engine output to
compensate for this loss?
𝜏 = 𝐹𝑑 →
𝑭𝑳𝑬𝑭𝑻 × 𝒅𝑳𝑬𝑭𝑻 = 𝑭𝑹𝑰𝑮𝑯𝑻 × 𝒅𝑹𝑰𝑮𝑯𝑻
𝑭𝟏 × 𝒅𝟏 + (𝑭𝟐 × 𝒅𝟐 ) = (𝑭𝟑 × 𝒅𝟑 ) + (𝑭𝟒 × 𝒅𝟒 )
𝟐𝟎, 𝟎𝟎𝟎 × 𝟏𝟎 + 𝟐𝟎, 𝟎𝟎𝟎 × 𝟓 = 𝑭𝟑 × 𝟓
𝟐𝟎𝟎, 𝟎𝟎𝟎 + 𝟏𝟎𝟎, 𝟎𝟎𝟎 = 𝟓𝑭𝟑
𝟑𝟎𝟎, 𝟎𝟎𝟎
𝑭𝟑 =
= 𝟔𝟎, 𝟎𝟎𝟎 𝑵
𝟓
Eg2. A plane was loaded ready for flight, until 200kg of excess baggage was loaded
onto the plane, located 5 m behind the centre of gravity of the plane.
To ensure the plane remains in equilibrium, this weight must be balanced.
This is done by adding fuel to a reserve tank 2 m in front of the centre of gravity of the
plane. How much fuel must be added to maintain equilibrium?
𝜏 =𝐹×𝑑
𝜏𝐹𝑢𝑒𝑙 = 𝜏𝐵𝑎𝑔𝑔𝑎𝑔𝑒
𝐹𝐹𝑢𝑒𝑙 × 𝑑𝐹𝑢𝑒𝑙 = 𝐹𝐵𝑎𝑔𝑔𝑎𝑔𝑒 × 𝑑𝐵𝑎𝑔𝑔𝑎𝑔𝑒
𝑚𝐹𝑢𝑒𝑙 𝑎 × 𝑑𝐹𝑢𝑒𝑙 = (𝑚𝐵𝑎𝑔𝑔𝑎𝑔𝑒 𝑎) × 𝑑𝐵𝑎𝑔𝑔𝑎𝑔𝑒
10𝑚𝐹 × 2 = (200 × 10) × 5
20𝑚𝐹 = 10,000
𝑚𝐹 =
10,000
20
= 500 𝑘𝑔
Now Do
Text Book - Chapter 16
Questions 21, 23, 24 Pg 377
( Torque & Equilibrium)