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Take Off!!!

Rosny Daniel Daniel Pappalardo Nadeline Rabot. Take Off!!!. 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.

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Take Off!!!

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  1. Rosny Daniel Daniel Pappalardo Nadeline Rabot Take Off!!!

  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

  7. 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.

  8. 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)

  9. 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

  10. 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

  11. 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

  12. 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 Euler’s Method is another method of calculating the total displacement. 0∫5.9 46.412x dx = 807.8 ft

  13. Landing… Coming to a STOP!

  14. Arresting Cables Arresting Cable

  15. 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.

  16. 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

  17. 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.

  18. Riemann Sum for Velocity Function

  19. 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.

  20. 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.

  21. The End

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