Take Off Rosny Daniel Pappalardo Nadeline Rabot Introduction

Take Off!!! Rosny 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 F 100 -229 Afterburning Turbo Fans 29, 000 lbf Wing Area: 608 ft 2 Max speed: 1, 875 MPH Cruising Speed: 570 MPH Armament: Guns: 1 x M 61 Vulcan 20 mm Gatling Gun w/ 940 rounds Missles: Combination of AIM-7 F 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 300 ft deck.

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 * V 2 * 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/ ft 3 SRef = 605 ft 2 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 F 100 -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 F 15 * acceleration 257, 996. 85 = 18, 181. 8 kg * acceleration = 14. 2 m/s 2 acceleration = 46. 576 ft/s 2

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/s 2 V = 46. 576 ft/s * time + V 0 186. 7 MPH = 273. 83 ft/s V 0 = 0 273. 83 = 46. 576 * t t = 5. 9 seconds

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. 412 x 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-15 s 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 2 nd 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 -μ g = a x -. 5*9. 8 = -4. 9 m/s 2 = ax

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.

Riemann Sum for Velocity Function Number of steps 1 Starting point 0 Ending point 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 300 ft of runway. We found that without the catapult, it will take at least 808 ft to take off. To land with arresting cables, an F-15 only needs 320 ft of runway. We found that without the arresting cables, it will take at least 2419 ft 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