Definition of Dambusters Simulation - Case Study Example

Dambuster simulation Dambuster simulation Student’s Name Professor’s Name Course Title Date Table of Contents 2 Introduction 3 Method 3 Theory 5 Elastic and inelastic collisions 6 Results 8 Conclusion 12 References 13 University Of Brighton 2014, Dambusters Simulation- Diagrams of the Dam. Available on online at < https://sites.google.com/site/universityofbrighton/dambusters>[ Accessed on 28/4/2014] 13 Introduction In dambuster simulation various variable plays an important role, among them the speed of the aircraft releasing the bomb, the mass of the bomb, gravitational force, the height at which the aircraft is flying and air drag or force. This will determine where the bomb will fall and the height at which it will bounce. The assumption is that the bomb is elastic and bounces before reaching edge of the dam. The aim of this paper is to conclude the work of dambuster, to investigate the sensitivity of the distance at which jet is at from the ground, speed plane, number of times the bomb will bounce and the height of bouncing. Utilizing simulations can increase the accuracy of estimating the point at which a bomb can be released, whereas testing on various parameters may take considerably longer. The tradeoff between selecting simulation over video means that a situation will not be totally based in reality and controls that make work during the simulated testing phase could fail in a real world application. Method In simulation of the dambuster the code for simulation will be edited to include bouncing. The code is transferred to Java or Matlab for simulation. The two programs are suitable for this kind of simulation, since they use command line codes are used in running of simulations. The command enables the change parameters and immediately see what happens, for "what if" exploration. The simulation results can be put in the workspace for post-processing and visualization. The code is shown below Theory Any projectile released from the jetliner or aircraft relies on speed of aircraft, mass of the object and force of air resistance. The drag force exerted on an object moving through air increases dramatically with speed; it is proportional to the square of the speed: Fd = bv2 Where b is a constant that depends on the size and shape of the object that is being dropped. For a given shape, b is proportional to the cross-sectional area of the object. The direction of the drag force is opposite to the direction of motion. Since the drag force increases as the speed increases, a falling object approaches an equilibrium situation in which the drag force is equal in magnitude to the weight but opposite in direction. The velocity at which this equilibrium occurs is called the object’s terminal velocity. The direction of the terminal velocity is always downward if there are no forces other than air resistance and gravity. The magnitude of the terminal velocity is called the terminal speed. As the velocity approaches the terminal velocity, the acceleration gets smaller and smaller. The acceleration is zero when the object falls at its terminal velocity. When the projectile is falling, the displacement of it in equal time intervals increases linearly, giving its acceleration is constant. The net force on an object falling at its terminal velocity is zero, so the drag force is equal in magnitude to the weight. Therefore, Fd = mg = bvt2, where vt is the terminal speed, and Vt = . The magnitude of the drag force at any speed v is Fd = mg . The direction is opposite to the object’s velocity. When the projectile is falling at half its terminal speed, the drag force is significant, but it is smaller than the weight, the net force is down and, therefore, the acceleration is still downward, though with a smaller magnitude. The magnitude of the drag force at any speed is given by Fd = mg . The drag force acts in the direction opposite to the velocity. The projectile is falling straight down, so the drag force is upward. The net vertical force is = - mg= mg ( - 1) Apply Newton’s second law: = may Solving for the acceleration yields Ay= g( - 1) At a time when the speed is half the terminal speed, V=vt and = ay=g( – 1) = g So that the acceleration of the projectile is = Where and both point downward. Elastic and inelastic collisions Collisions are often classified based on what happens to the kinetic energy of the colliding objects. A ball dropped drop a height h does not rebound to the same height. The kinetic energy of the ball just after the collision with the floor or ground is less than it was just before the collision; the amount of the kinetic energy decrease depends on the makeup of the ball and the ground. A racquetball dropped onto a hard wooden floor may rebound nearly to its original height, but a watermelon rebounds very little or not at all. Imagine a racquetball colliding with the floor. The bottom of the ball is flattened. The forces holding the ball together are like springs; the kinetic energy of the ball has been transformed largely into potential energy stored in these springs. When the ball bounces back up, this energy is transformed back into kinetic energy. The watermelon, too, is deformed when it collides with the floor, but this deformation is not reversible. The kinetic energy of the watermelon is changed mostly into thermal energy rather than into potential energy. A collision in which the total kinetic energy is the same before and after is called elastic. When the final kinetic energy is less than the initial kinetic energy, the collision is said to be inelastic. Collisions between macroscopic objects are generally inelastic to some degree, but sometimes the change in kinetic energy is so small that we treat them as elastic. When a collision results in two objects sticking together, the collision is as large as possible. Now that we have defined elastic and inelastic collisions, we can put together a problem-solving strategy for collision problems. There is no conservation law for kinetic energy by itself. Total energy is always conserved, but that does not preclude some kinetic energy being transformed into another type of energy. The elastic collision is just a special kind of collision in which no kinetic energy is changed into other forms of energy. Momentum is conserved regardless of whether a collision is elastic or inelastic. It can be proved that for any elastic collision between two objects, the relative speed is the same before and after the collision. Since the relative velocity is in the opposite direction after a one-dimensional collision first the objects move together, then they move apart we can write: V2ix-vlix = - (v2fx –vifx) Assuming the objects move along the x-axis. For a one-dimensional elastic collision, a useful alternative to setting the final kinetic energy equal to the initial kinetic energy. By measuring the length of the marks from the stop sign and estimating the coefficient of friction, the accident investigator can determine that the car was pushed onto the highway at a speed of 20.0m/s. witnesses confirm that the car was stopped before the collision using conservation of momentum and, finding that it exceeds the speed limit, adds speeding to the charges against the driver of the pickup. Results The aeroplane is moving at the speed of 100m/s. the speed of the aeroplane horizontally is xp=xp+vxp*dt while movement vertically yp=yp+vyp*dt; The bomb when is not released will have horizontal speed of xbomb=xp+100 and the vertical speed will be ybomb =yp+40. The velocity of the bomb will be vxbomb=vxp and the vertical velocity will be vybomb=vyp. The bomb will begin to move at initial speed to the ground that y-direction at v0y=0 m/s while the v0x=100 m/s. When the bomb hits on the ground the distance covered becomes 400m or displacement is y=–400 m. The speed is accelerated as using gravitational force ay=–9.81 m/s2. Therefore Time = = 9seconds However, the bomb does not fall directly because the jetliner is moving at a speed of 100m/s and it will have a horizontal velocity of 100m/s with air drag and gravitation pull to it. This gives a initial velocity of = = 100.48m/s Using the equation we will find Vybomb = 0m/s +(9.81m/s2)(9s) = -88.29m/s Thus the speed before landing on dam is = =133.4m/s The angle of the bomb at which it falls will be as at is Cos = cos-1 () = cos-1 () = 45.83o When the bomb is released by the aircraft it hits the surface of the dam and bounces back at an angle of 45.83o. When bomb hit the dam at first time momentum is conserved. There is no need to compare these two values. The sum of the magnitudes of the momenta before the collision is not equal is not equal to the sum of the magnitudes of the momenta after the collision. Conservation of energy is perhaps easier to understand intuitively since energy is a scalar quantity. Converting kinetic energy to potential energy is analogous to moving money from a checking account to a savings account; the total amount of money is the same before and after. This sort of analogy does not work with momenta VIF = = 23.8m/s After collision with water the moves to up with a speed of 23.8m/s The height of the first bounce is h= = =28.85m Coefficient restitution = 0.06 The simulate graph above show that the bomb fell 400meters from the point of release horizontally. The bomb is released in position that does not hit the target. The results indicate that the first bounce goes up to 28.9m and the second bounce; results show a significant variation in the functional graphical deduction. The velocity of the second bounce is before landing on dam is Time = = 2.4seconds Using the equation we will find Vybomb = 0m/s +(9.81m/s2)(2.4s) = -23.8m/s Thus the speed before landing on dam is = =4.9m/s The angle of the bomb at which it falls will be as at is Cos = cos-1 () = cos-1 () = 66.85o When the bomb is released by the aircraft it hits the surface of the dam and bounces back at an angle of 66.85o. When bomb hit the dam at second time momentum is conserved.. The sum of the magnitudes of the momenta before the collision is not equal is not equal to the sum of the magnitudes of the momenta after the collision. VIF = = 2.9m/s After collision with water second time the moves to up with a speed of 2.9m/s The height of the third time bounce is h= = =1.22m Conclusion From the case study it can be noted that relationship between the coefficient restitution and the velocity of the bomb is determined by the following formula. The simulation involved the division of the investigation into the jetliner velocity, altitude at which the bomb will be released, landing speed, point at which it lands, and a bouncing height and a sinking point. From the simulation results it can be noted the bomb was released from the plane at 250m that was moving at the constant velocity of 100m/s thus giving it an initial speed of 100m/s horizontally. The bomb will fall on the dam in a trajectory form. The results showed that it landed on the point at 400m from the point at which it was released horizontally. References University Of Brighton 2014, Dambusters Simulation- Diagrams of the Dam. Available on online at < https://sites.google.com/site/universityofbrighton/dambusters>[ Accessed on 28/4/2014] Read More