Take Off!!!

1
Take Off!!!

• Rosny Daniel
• Daniel Pappalardo
• Nadeline Rabot

2
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.

3
The Four Aerodynamic Forces That Act Upon an
Airplane in Flight
Lift
Thrust
Drag
Weight
4
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

5
Take Off
6
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

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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.

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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)

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Stall Velocity(continued)

• We need enough lift to counteract the takeoff
  weight. So to solve for the stall velocity we
  get
• VS (2W/(p SRef CL ))1/2
• Weight 40,000 lbs
• P .00237 slugs/ ft3
• SRef 605 ft2
• CL 1.6 for supersonic jets
• Vs ((244,000)/(.002376051.6))1/2
• Vs 186.7 MPH

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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 Newtons 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

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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

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Thrust, Acceleration, Distance (continued)
By taking the integral of this function we find
total displacement of the aircraft during
takeoff. Calculating the area underneath the
graph or Eulers Method is another method of
calculating the total displacement. 0?5.9
46.412x dx 807.8 ft
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Landing

• Coming to a STOP!

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Arresting Cables
Arresting Cable
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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.

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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.
• Newtons 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
• -.59.8 -4.9 m/s2 ax

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Landing the F-15
The area under this curve is the total
displacement of the F-15 after touchdown, thus
the necessary length of the runway is at least
737.375 meters, or 2418.6 feet.
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Riemann Sum for Velocity Function
Number of steps Starting point Ending point Total area under curve
1 0 17.35 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
8 0 17.35 737.375
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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.

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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.

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The End