Download Take Off!

Take Off!!!
Rosny Daniel
Daniel Pappalardo
Nadeline Rabot
Introduction
In this project we will discuss the
advantages of arresting cables and
steam catapults on aircraft carriers by
comparing the distance needed with and
without these factors for take off and
landing.
The Four Aerodynamic Forces That
Act Upon an Airplane in Flight
Lift
Thrust
Drag
Weight
F-15 Eagle: Specifications
Weight: 40,000 lbs loaded
Powerplant: Engine
Pratt and Whitney F100-229
Afterburning Turbo Fans 29,000
lbf
Wing Area: 608 ft2
Max speed: 1,875 MPH
Cruising Speed: 570 MPH
Armament:
Guns: 1 x M61 Vulcan 20 mm
Gatling Gun w/ 940 rounds
Missles: Combination of AIM-7F
Sparrows, AIM-120
AMRAAMS and AIM-9
Sidewinders
Take Off…
Steam Catapult
• Device used to launch the aircraft from aircraft
•
•
•
carriers
Generally a track built into the flight deck
Below is a piston that attaches up through the
track to the nose of the aircraft
At launch, a release bar holds the aircraft in
place as the steam pressure builds up and then
it releases the aircraft at a high speed
Steam Catapult (continued)
A catapult can accelerate an aircraft from 0 to 182
mph in 2 seconds. This allows an aircraft to take
off safely on a 300ft deck.
Stall Velocity
Stall Velocity is the minimal velocity necessary for the
aircraft to take flight or remain airborne
To solve for the stall velocity we need the Lift equation:
L = (1/2)p * V2 * SRef * CL
L = Lift force
P = air density (.00237 slug/ ft)
V = aircraft velocity
SRef = reference area (surface area of wings)
CL = Coefficient of lift (1.6)
Stall Velocity
(continued)
We need enough lift to counteract the
takeoff weight. So to solve for the stall
velocity we get:
VS = (2*W/(p* SRef * CL ))1/2
Weight = 40,000 lbs
P = .00237 slugs/ ft3
SRef = 605 ft2
CL = 1.6 for supersonic jets
Vs = ((2*44,000)/(.00237*605*1.6))1/2
Vs = 186.7 MPH
Thrust, Acceleration, Distance
Thrust:
Pratt and Whitney F100-229 Afterburning Turbo Fan Engine =
29,000 lbs of Thrust/ engine
1 pound of force = 4.44822162 Newtons
2 Engines = 257,996.85 Newtons
Thrust is a force, therefore Newton’s Second Law applies.
F = MA
257,996.85 = mass of F15 * acceleration
257,996.85 = 18,181.8 kg * acceleration
acceleration = 14.2 m/s2
acceleration = 46.576 ft/s2
Thrust, Acceleration, Distance
(continued)
If we anti-differentiate the acceleration we will end up with the velocity
equation. From here, we can solve for the amount of time it will take
the plane to reach the stall velocity and then the amount of distance
it will take the aircraft to take off.
A = 46.576 ft/s2
V = 46.576 ft/s * time + V0
186.7 MPH = 273.83 ft/s
V0 = 0
273.83 = 46.576 * t
t = 5.9 seconds
Thrust, Acceleration, Distance
(continued)
Velocity Vs. Time of an F15 During Take Off
300
250
Velocity (ft/s)
200
Velocity vs. Time
Linear (Velocity vs. Time)
150
100
y = 46.412x
50
0
0
1
2
3
4
5
6
7
Time (seconds)
By taking the integral of this function we find total
displacement of the aircraft during takeoff. Calculating the
area underneath the graph or Euler’s Method is another
method of calculating the total displacement.
0∫
5.9 46.412x
dx = 807.8 ft
Landing…
Coming to a STOP!
Arresting Cables
Arresting
Cable
Arresting Cables
• Thick steal cables
• Fitted at the end of the flight deck on an aircraft
•
•
•
carrier
Planes have a tail hook that catches onto the
cable
Cable takes up the slack by a hydraulic
mechanism that rapidly decelerates the aircraft
Arresting cables stop F-15s within two seconds
of engaging the cable, and within 320 ft of touch
down.
Kinetic Friction
When the aircraft touches down, there are THREE main forces acting upon it: Normal Force from the
runway, the Force of Gravity, and the Kinetic Friction Force.
The Normal force and the Force from gravity act upon the plane in the vertical axis. The net force in the
vertical direction equals zero because there is no acceleration of the plane in the vertical direction. The
only force acting upon the plane in the horizontal direction is the force due to kinetic friction.
Newton’s 2nd Law:
F = MA
Fg + FN + FF = MA; Fg = mg, g = 9.8 m/s
FF = -μ Fn ,, μ = .5
Horizontal axis : -μ FN = Max
Vertical axis: -mg + FN = 0
FN = Mg
-μ Mg = Max
-μg = ax
-.5*9.8 = -4.9 m/s2 = ax
Landing the F-15
Velocity vs. Time
90
80
velocity (meters/second)
70
60
50
Series1
40
30
20
10
0
0
2
4
6
8
10
12
14
16
18
20
time (seconds)
The area under this curve is the total displacement of the F15 after touchdown, thus the necessary length of the runway
is at least 737.375 meters, or 2418.6 feet.
Riemann Sum for Velocity Function
Number of
steps
Starting
point
Ending
point
1
0
17.35
Total area
under
curve
0
10
0
17.35
663.49
50
0
17.35
722.5
100
0
17.35
729.87
250
0
17.35
734.29
∞
0
17.35
737.375
Conclusions
To take off with the steam catapult, an F-15
fighter jet only needs 300ft of runway. We
found that without the catapult, it will take
at least 808ft to take off.
To land with arresting cables, an F-15 only
needs 320ft of runway. We found that
without the arresting cables, it will take at
least 2419ft to land safely.
Conclusions (continued)
Through applying and using
our extensive knowledge
of calculus, and what it is
related to, we have
decided that the
engineers in the United
States Government and
at N.A.S.A. are pretty
smart and efficient. The
inventions of the steam
catapult and arresting
cables are great.
The End