Arial Bombing Techniques

Transcription

1 Arial Bombing Techniques By Crystal Pepper and Chris Wilson March 30, 2009 Abstract In this article we will explore the bombing techniques used by the United States Army during World War II in order to produce a maximum impact on targets. We will demonstrate through calculations of differential equations how the speed of the plane, linear distance the plane is from the target, and altitude of the plane determines the time that the bombs must be deployed. Intro During World War II the role of the bombardier was critical in the accuracy of placing the free falling bombs on target. The conditions needed were: altitude, true airspeed, bomb ballistics, trail, actual time of fall, ground speed, and drift. The altitude, which is controlled by the pilot, determines the amount of time the bomb is free falling. The pilot also controls the true airspeed which gives the bomb its initial forward velocity and in turn affects the trail (horizontal distance that the bomb is behind the plane upon impact). The bomb ballistics such as the size, shape, and density determine the air resistance the bomb will encounter as it falls. The actual time of fall is the length of time the bomb is sustained in the air from the instant of release to the instant of impact. The actual time of fall is affected by the altitude and bomb ballistics. The ground speed is the speed of the plane in relation to the Earth s surface. This affects the range of the bomb. Lastly, the drift is determined by the direction and velocity of the wind and affects the distance the bomb will travel downwind. With the help of a mechanical analog computer, the then top secret Norden bomb sight, the bombardier was able to input the parameters altitude, airspeed, bomb ballistics, and air density. With these parameters, the accuracy of the placement of the bombs was sufficient for the technology available at the time. 1 Projectile Motion When the bomb leaves the airplane it will not fall in a straight line. From the moment the bomb leaves the plane its trajectory will vary in curvature. Projectile motion is two-dimensional motion. However, any case of two dimensional motion can be resolved into two cases of one dimensional motion: one along the x axis and the other along the y-axis. We will initially study the one dimensional cases separately, without considering air resistance. 1

2 Figure 1: Vertical Motion 1.1 Vertical Motion We now develop a mathematical model of the bomb s trajectory. For convenience, we will thin of our bomb as an object moving in the xy plane and mae use of this idea in our graphical analyses. Figure 1 shows the force on an object with a vertical motion. It is always true for an object near the Earth s surface that the acceleration of gravity is equal to the differential change in velocity in a given time interval. dv y dt = a = g We now that gravity g is 9.8, m/s 2. In our coordinate system, with y upward, this acceleration is a = g y = 9.8 m/s 2 (1) This is the acceleration along the y-axis. The velocity in the y direction can now be determined by taing the anti-derivative of g with respect to time. In order to set up our velocity equation, two initial conditions will be required. Our parameters are the initial time t = 0, and the initial vertical velocity v y = 0. dy dt = gdt v yf = gt + c, where c is a constant of integration. By letting c = v yi and v yi = 0 then it follows that v yf = v yi g y t v yf = 0 gt v yf = gt Equation (2) can be used to solve for the velocity along the y-axis at any instant t. By the same process, we can determine the displacement along the y-axis at any instant t. Integrating the velocity with respect to time gives (2) 2

3 y = (v yi gt)dt y = v yi t 1 2 g yt 2 y = g yt 2 (3) y = 1 2 gt2. We have now described all of the components of the bomb s vertical drop. We can now calculate the acceleration at any time, the velocity at any time, and the position at any time. 1.2 Horizontal Motion In the previous section, we described the motion of the bomb from its acceleration to its final displacement by taing the anti-derivative of each equation. Determining the velocity and displacement of the bomb in the horizontal direction can be calculated in the same manner. First, we will need to establish an initial condition. Since the bomb is attached to a plane which is flying in a horizontal direction, it will have the same initial velocity as the plane. However, due to air resistance and gravitation, the bomb will fall in an arc lie path towards the ground. In other words, there is no horizontal acceleration and the plane maintains its velocity from the time of release to impact. Therefore, the initial conditions are a x = 0 and v xi = v x. We will need to use a x to solve for our final displacement. Now the component of initial velocity along the x-axis can be written by taing the integral of the acceleration component dx dt = a x dt v xf = a x t + c (4) v xf = 0 + c v xf = c. The c represents an arbitrary constant which we can solve for by plugging in our initial values of v xi = v xf and a x = 0 v xi = a x t + c v xi = 0 + c v xi = c. Plugging this c value into equation (4) with the initial conditions, will give us a first order differential equation that describes the horizontal speed of our bomb. v xf = a x t + c v xf = a x t + v xi v xf = 0 + v xi v xf = v xi. 3

4 Figure 2: Horizontal Motion This is the component of the velocity along the x-axis at any instant t. This means that the horizontal component of velocity does not change throughout the projectile s motion. Let v xf = v xi = v x (5) represent our final velocity in the horizontal direction. By incorporating a horizontal acceleration, and integrating once more, we will be able to determine the displacement of the bombs trajectory. x f = (a x t + v x )dt x f = 1 2 a xt 2 + v x t + c. We will need to solve for c to determine the final displacement equation. Plugging in the initial values and letting t = 0 we will solve for the initial horizontal position x i = 1 2 a xt 2 + v x t + c x i = c x i = c. With this c value the displacement along the x-axis at any instant t can be described by x f = a x t 2 + v x + c x f = 0 + v x t + x i x f = v x t + x i. We can now determine the velocity and displacement at any time during the flight of the bomb. With equation (6), we can plot the bomb s trajectory on an xy coordinate system as shown in Figure 3. 2 The Real Forces of Projectile Motion 2.1 Drag and the Vertical Motion In sections 1.1 and 1.2 we showed you the idealized formulas for the motion of a projectile by neglecting the force of drag. However, in order to establish a more real world basis for our model, we will need to determine and incorporate this force. Drag is defined as the force that opposes the motion of an object in a fluid. This fluid can be anything from liquid to gas. As our bomb leaves (6) 4

5 Figure 3: Horizontal and Vertical trajectory. the plane, the air that it moves through will provide this resistance. Drag R depends on the density of the air ρ, the velocity v, the air s viscosity and compressibility D, and the reference area A. R = 1 2 DρAv2. (7) To derive the drag equation is a tedious tas that would tae hours of computation. So for the sae of time we state that this equation is just the drag force without any further explanation. Since air resistance is present in all real applications, the motion of the vertical fall of the bomb will now have a drag force. Using Newtons law of force F = ma we will sum the forces in the y direction to derive our acceleration formula. F = F g F d. Knowing that F = ma, F g = mg, and F d = 1 2DρAv, the equation loos lie ma = mg 1 2 DρAv a = g DρAv 2m. (8) Equation (8)includes all of the forces that are acting on a bomb as it falls to the ground. Before we attempt to integrate this equation, we will simplify the drag by letting = DρA/2m. This will help eep our values straight. Therefore, with the new substitution, equation (8) becomes 5

6 a = g v a = dv dt = g v. (9) Equation (9) is the acceleration due to the velocity and drag. By integrating this function, we will get a formula for the velocity with respect to time. In order for us to differentiate equation (9), we will need to separate the variables as follows dv dt = g v dv = (g v) dt dv g v = dt Now that the function is set up, we will integrate dv g v = dt ln (g v) = t + C ln (g v) = (t + C) e ln(g v) = e t+c g v = e t e C Through the integration, we are left with an arbitrary constant e c. To simplify this, let b = e c. With the substitution we can now solve for the velocity as a function of time. g v = be t v = g be t v = g b e t. However, this equation is not yet correct. We will need to substitute our initial values to solve for the constant b. Setting v = 0 at t = 0 produces v = g b e t 0 = g b e (0) 0 = g b g = b. Our calculations reveal that b is equal to g. This will mae the velocity equation a little bit easier to loo at. 6

7 v = g g e t v = g ( t ) (10) v = g ( t ). This formula tells us that at any time t we can evaluate the velocity of the bomb. We would lie to note that as t, v g/, called the terminal velocity. Up to this point, we have formulated equations for the acceleration and velocity of the bomb in the vertical direction. Neither of these equations, however, will tell us where the bomb will land when a given height and time are nown. We will integrate once more to find the displacement of the bomb v = g ( t ) dy dt = g ( t ) dy = g (1 e t ) dt y = g ) (t + e t + c. Yet again, we are left with a constant of integration c. Providing some initial conditions will eliminate the constant. With the initial condition y = y 0 at t = 0 gives y 0 = g ) (0 + e0 + c y 0 = g ( ) 1 + c y 0 = g 2 + c c = y 0 g 2. Plugging in the c value will give the final equation for displacement y = g ) (t + e t + y 0 g 2 y = g t + g 2 e t + y 0 g 2 (11) y = y 0 + g t g 2 ( t ). This final equation represents the general solution to our system in the vertical direction. We can plot this, as shown in figure 4, given any initial height of the plane. We have now established formulas for the acceleration, velocity, and position for our trajectory in the vertical direction. We will then follow up with the derivation for the equations with drag in the horizontal direction. 7

8 Figure 4: Vertical Motion with Drag 2.2 Drag and the Horizontal Motion In section 2.1, we derived the the general solution to the vertical displacement by integration. In the same way, we can derive the general solution to the horizontal dislpacement. By taing the same steps of integration we can go through this process without too much explanation. Starting with the initial forces in the x direction F = F d. (12) We can see from equation (12) that the only force in the horizontal direction is the force of drag. Solving this equation will provide a formula for the velocity. For convienance, we will let = DρA/2m for our drag force. a = dv dt = v Seperate variables then integrate to solve for the final velocity equation 8

9 dv dt = v dv v = dt dv v = dt dv v = dt lnv = t + c v = e t+c v = c 1 e t. The integration constant c 1 can be evaluated with the initial conditions t = 0 and v = v 0. (13) v = c 1 e t v 0 = c 1 e 0 v 0 = c 1. Then plugging this value bac into the equation (13) will give v = v 0 e t (14) This is the final eqaution for the velocity with drag. We will then integrate, just as we did with the vertical equation, to solve for the displacement. v = v 0 e t dx dt = v 0e t dx = v 0 e t dt x = v 0 e t dt x = v 0 e t + c. Yet again we are left with the integral constant that can be solved with the initial conditions t = 0 and x = 0 x = v 0 e t + c 0 = v 0 e 0 + c 0 = v 0 + c v 0 = c 9

10 Figure 5: Horizontal Motion with Drag Therefore e t x = v 0 + v 0 x = v 0 ( t ) (15) Equation (15) is our final formula needed to solve for the discplacement of the bomb on the x-axis. Setting an altitude and initial speed of the plane, we can plot this trajectory as shown in figure 5. 3 Conclusion With equations (11) and (15), y = y 0 + g t g 2 ( t ) x = v 0 ( t ) we can plot the trajectory of the bomb in an xy coordinate system. These equations can be used to solve the position of the bomb when dropped from any height and distance. Setting an altitude and velocity of the plane, we have plotted these equations as demonstrated in figure 6. 10

11 Figure 6: Horizontal and Vertical Motion with Drag References [1] Probable Bomb Impact Areas (PJR) I04FL [2] Research and Education Association Differential Equations. 11