Traectory recent
- 1. 1 Trajectory of a Projectile by RK4 methods By: Kanhaiya Jha
- 2. Outline of Presentation 2 Basic idea about Ordinary Differential Equations RK Method to solve ODE Application Part • Trajectory of a Projectile (Unguided Missile) • Trajectory of a Self-propelled missiles (Guided Missile)
- 3. 3 Idea about Ordinary Differential Equations A Ordinary differential equation is an equation containing derivatives of a dependent variable with respect to one independent variables. i.e. its involving a function and its derivatives. Example: 𝑑𝑦 𝑑𝑥 = 6𝑥𝑦 1 𝑥 𝑑𝑦 𝑑𝑥 + 3𝑦 = 𝑦3 2 We are going to solve the Differential equation of the form 𝒅𝒚 𝒅𝒙 = 𝒇 𝒙, 𝒚 with some initial condition of 𝐲 𝟎 = 𝐲 𝒙 𝟎 . 𝒚𝒊+𝟏 = 𝒚𝒊 + 𝒇 𝒙, 𝒚 ∗ 𝒉 method to predict future Note: The numerical solution of the initial value problem (IVP) for ordinary differential equations (ODE) is fundamental to most trajectory optimization methods.
- 4. 4 A fundamental source of error in Euler’s method is that the derivative at the beginning of the interval is assumed to apply across the entire interval. Euler’s Method (First order RK method) Ralston’s Method 𝑦𝑖+1 = 𝑦𝑖 + ( 1 3 𝑘1 + 2 3 𝑘2)h 𝑘1 = 𝑓 𝑥𝑖 , 𝑦𝑖 𝑘2 = 𝑓(𝑥𝑖 + 3 4 ℎ , 𝑦𝑖 + 3 4 𝑘1ℎ) Heun Method 𝑦𝑖+1 = 𝑦𝑖 + ( 1 2 𝑘1 + 1 2 𝑘2)h 𝑘1 = 𝑓 𝑥𝑖 , 𝑦𝑖 𝑘2 = 𝑓(𝑥𝑖 + ℎ , 𝑦𝑖 + 𝑘1ℎ) 𝑦𝑖+1 = 𝑦𝑖 + 𝑓 𝑥, 𝑦 ∗ ℎ Where 𝑓 𝑥, 𝑦 is slope at initial point Commonly used higher order method
- 5. 5 Most popular Method of solving ODE Here idea is ‘Slope is not always constant’, rather than it also depend upon step size h. Runge-Kutta (4th Order) Method of Solving ODE We can divide the whole interval 𝑥𝑖+1 − 𝑥𝑖 in to no. of parts and calculate slope of each parts then Effective slope K= 𝟏 𝟔 (k1+ 2k2+ 2k3+ k4) Where Slope at initial value k1 = h ∗f (x , y) , k2 = h ∗ f(x+h/2, y(x)+k1/2) , k3 = h ∗ f(x+h/2, y(x)+k2/2) and Slope at the end k4 = h∗ f(x+dx, y(t)+k3) 𝒚𝒊+𝟏 = 𝒚𝒊 + K ∗ 𝒉
- 6. What is Missile ? 6 Any object thrown at a target with the aim of hitting the target is called Missile. A stone thrown at a bird is also a missile. The bird, by using its power of reasoning may evade the missile (the stone) by moving either to the Left, right, top or bottom with respect to the flight path (trajectory) of the missile.
- 7. What is Missile ? 7 Thus, the missile in this case has been ineffective in its objective of hitting the bird (the Target).
- 8. 8 e.g.: The Stone thrown at a bird They are attractive because they are inexpensive and they can cover larger areas in short reaction times. Dispersion is quite large typically being around 5% of their range which for a typical 20 km range can be more than 1 km. Note: This rather large dispersion renders these unguided missiles not useful for Precision applications e.g. : Controlled Self-Propelled Missiles Provide energy source in a missile for its movement (propulsion). Intelligence to go in the correct direction (guidance). Effective control over entire flight of the missile. Unguided Missiles Guided Missiles
- 9. 9 Projectile Motion of the Stone Using Runge-Kutta Methods (Trajectory of the Stone) m 𝒅𝑽 𝒅𝒕 = -mg - 1 2 Cw A v2ρ0e−h/h0 Negative sign implies here that both forces are opposite to the direction of initial velocity. In the drag free case , i.e. with only gravity the entire flight follows simple parabolic motion. All applicable forces applied to the Stone Where Fdrag = 1 2 Cw A v2ρ Cw Drag coefficient A Cross-sectional area of the object ρ Air density (Varies with altitude) v Velocity of the object Ftotal = Fgrav + Fdrag
- 10. 10 Simulated values of distance travelled as a function of initial launch angle for the No drag case and two drag cases at different altitudes NY (altitude=0 m) and Denver(altitude=1600 m) It was also determined theoretically that a 450 home run launch angle was optimal in vacuum, while a 400 home run launch angle was in all environments with drag. Trajectory in different Atmospheric situations. Mass of the stone = 0.145 kg Area of cross section= 0.0043 m2 Drag Coeff (Cw) = 0.35 Air Density (ρ0) = 1.22kg/m3 Typical HR Velocity = 49.0 m/s Scale Height (h0) = 7000 m Input values for program
- 11. 11 The addition of a drag force can half the horizontal range of the stone path. The trajectories of stone hit in NY, all with the same initial velocity 45o without Drag 40o with Drag(NY)
- 12. 12 Self-Propelled Missiles (SPMs) An SPM free body diagram During the burn phase, the mass of the entire SPM decreases linearly in time mr(t) = m0 − (mf ∗ 𝒕 𝒕 𝒃 ) , t ≤ tb m0 − mf , t > tb −𝑮𝒎 𝒆 𝒎 𝒓 𝒓 𝒕 𝟑 Three forces acting on the rocket during the burn phase Fgravity = r (t) Fdrag = - ρ(|r (t)|)Cd A V2 V(t) Fthrust = g ∗ p , t ≤ tb 0 , t > tb 𝟏 𝟐 −𝑮𝒎 𝒆 𝒎 𝒓 𝒓 𝒕 𝟑
- 13. 13 M dM Ve dV Conservation of Linear Momentum 𝐝𝐕 𝐭,𝐫 𝐭 𝐝𝐭 = 𝐦 𝐫 −𝟏 (𝐭)(𝐅𝐠𝐫𝐚𝐯𝐢𝐭𝐲+𝐅𝐝𝐫𝐚𝐠 + 𝐅𝐭𝐡𝐫𝐮𝐬𝐭) The trajectories are governed by Ordinary Differential Equations (ODEs) which give the time rate of change of each state variable h(t), V (t), m(t). Note: In the presence of drag equation cannot be integrated analytically We need to predetermined ρ(h) according to the distance between the rocket and the center of the earth. A linear interpolation was used to calculate ρ(h) at these intermediate altitudes. ρ(h) = ρ(hi ) + (ρ(hi+1) − ρ(hi )) h−hi / hi+1−hi Fdrag Rate of change of momentum is force Fthrust
- 14. 14 Continuous time traces approximated by discrete time traces. Discretization Technique The basic procedure is to discretize over some small but finite time interval before numerically integrating equation. We replace the continuous time variable t with a time index indicated by the subscript i, so that the state variables are defined only at discrete times t0, t1, t2 . . . ti .
- 15. 15 Midcourse Phase In this phase a guidance system hold the missile on course and at required altitude for a Iong time. Launch Phase(Boost Phase) During the launch phase aerodynamic requirements of the missile are different and due to weight reduction and a consequent shift in the Centre of gravity, the load, and the various parametric requirements of the missile are altered. Terminal Phase In this phase a guided system bring the missile accurately to the target and make up for possible deviations which might have crept in during the flight. Generally this decision has to be taken at very short intervals of time (1/50th of a second) during the flight of the missile.
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- 17. 17 Parameter Sym. Value Thrust p 26,050 kg Launch Mass m0 16,000 kg Initial Fuel Mass mf 12,912 kg Burn Time tb 110 sec Cross-Sectional Area A 1.37 m2 Drag Constant Cd 0.25 Input value in the simulations were done with characteristics of a certain North Korean missile called the ‘Nodong’. Simulations predicted the Nodong had an optimal range of 808km; whereas North Korea claims an optimal range of ∼1,425km. Optimal range is obtained at 55o Trajectories of rockets launched at different initial angles X value Yvalue
- 18. 18 Rockets have a mass distribution and cannot be accurately approximated by their centroids. Over short ranges, the Earth‘s surface is approximately flat; however, over longer ranges, Earth‘s curvature must be taken into consideration. 𝒇 𝒙 = 𝑹 𝒆 𝟐 − 𝑿 𝟐 - Re Note that in the code, the angle of Fthrust was fixed constant during the duration of the burn time. This is not true in real world application. Many real factors that were neglected by this program
- 20. 20 If India and Pakistan detonated 100 nuclear warheads (each equivalent to a 15-kiloton Hiroshima bomb) More than 21 million people (Half the death toll of World War II) will die immediately. Blast effects, burns and acute radiation Another two billion people worldwide would face risks of severe starvation due to the climatic effects of the nuclear-weapon. Half the world's ozone layer would be destroyed. "Nuclear winter" would cripple monsoons and agriculture worldwide. What happens if India and Pakistan both fire nuclear warheads at each other? The NewsMinute report (Thursday, September 29, 2016)