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 Armyduring World War II in order to produce a maximum impact on targets. We will demonstratethrough calculations of differential equations how the speed of the plane, linear distance theplane is from the target, and altitude of the plane determines the time that the bombs mustbe deployed.

Intro

During World War II the role of the bombardier was critical in the accuracy of placing the freefalling 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 airspeedwhich gives the bomb its initial forward velocity and in turn affects the trail (horizontal distancethat the bomb is behind the plane upon impact). The bomb ballistics such as the size, shape, anddensity determine the air resistance the bomb will encounter as it falls. The actual time of fall isthe length of time the bomb is sustained in the air from the instant of release to the instant ofimpact. The actual time of fall is affected by the altitude and bomb ballistics. The ground speed isthe 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 bombwill travel downwind. With the help of a mechanical analog computer, the then top secret Nordenbomb 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 sufficientfor 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 bombleaves 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 dimensionalmotion: one along the x − axis and the other along the y-axis. We will initially study the onedimensional cases separately, without considering air resistance.

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Figure 1: Vertical Motion

1.1 Vertical Motion

We now develop a mathematical model of the bomb’s trajectory. For convenience, we will think ofour bomb as an object moving in the xy plane and make 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 nearthe Earth’s surface that the acceleration of gravity is equal to the differential change in velocity ina given time interval.

dvy

dt= a = −g

We know that gravity g is 9.8,m/s2. In our coordinate system, with y upward, this acceleration is

a = −gy = −9.8 m/s2 (1)

This is the acceleration along the y-axis. The velocity in the y direction can now be determinedby taking 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 initialvertical velocity vy = 0.

dy

dt= −

∫gdt

vyf = −gt+ c,

where c is a constant of integration. By letting c = vyi and vyi = 0 then it follows that

vyf = vyi − gyt

vyf = 0− gtvyf = −gt

(2)

Equation (2) can be used to solve for the velocity along the y-axis at any instant t. By the sameprocess, we can determine the displacement along the y-axis at any instant t. Integrating thevelocity with respect to time gives

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y =∫

(vyi − gt)dt

y = vyit−12gyt

2

y = 0− 12gyt

2

y = −12gt2.

(3)

We have now described all of the components of the bomb’s vertical drop. We can now calculatethe 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 finaldisplacement by taking the anti-derivative of each equation. Determining the velocity and displace-ment of the bomb in the horizontal direction can be calculated in the same manner. First, we willneed to establish an initial condition. Since the bomb is attached to a plane which is flying in ahorizontal direction, it will have the same initial velocity as the plane. However, due to air resis-tance and gravitation, the bomb will fall in an arc like path towards the ground. In other words,there is no horizontal acceleration and the plane maintains its velocity from the time of release toimpact. Therefore, the initial conditions are ax = 0 and vxi = vx. We will need to use ax to solvefor our final displacement. Now the component of initial velocity along the x-axis can be writtenby taking the integral of the acceleration component

dx

dt=∫axdt

vxf = axt+ c

vxf = 0 + c

vxf = c.

(4)

The c represents an arbitrary constant which we can solve for by plugging in our initial values ofvxi = vxf and ax = 0

vxi = axt+ c

vxi = 0 + c

vxi = c.

Plugging this c value into equation (4) with the initial conditions, will give us a first order differentialequation that describes the horizontal speed of our bomb.

vxf = axt+ c

vxf = axt+ vxi

vxf = 0 + vxi

vxf = vxi.

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Figure 2: Horizontal Motion

This is the component of the velocity along the x-axis at any instant t. This means that thehorizontal component of velocity does not change throughout the projectile’s motion. Let

vxf = vxi = vx (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.

xf =∫

(axt+ vx)dt

xf =12axt

2 + vxt+ c.

We will need to solve for c to determine the final displacement equation. Plugging in the initialvalues and letting t = 0 we will solve for the initial horizontal position

xi =12axt

2 + vxt+ c

xi = 0 + 0 + c

xi = c.

With this c value the displacement along the x-axis at any instant t can be described by

xf = axt2 + vx + c

xf = 0 + vxt+ xi

xf = vxt+ xi.

(6)

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 inFigure 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 byneglecting 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 themotion of an object in a fluid. This fluid can be anything from liquid to gas. As our bomb leaves

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Figure 3: Horizontal and Vertical trajectory.

the plane, the air that it moves through will provide this resistance. Drag R depends on the densityof the air ρ, the velocity v, the air’s viscosity and compressibility D, and the reference area A.

R =12DρAv2. (7)

To derive the drag equation is a tedious task that would take hours of computation. So for the sakeof 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 bombwill now have a drag force. Using Newtons law of force F = ma we will sum the forces in the ydirection to derive our acceleration formula.

F = Fg − Fd.

Knowing that F = ma, Fg = mg, and Fd = 12DρAv, the equation looks like

ma = mg − 12DρAv

a = g − DρAv

2m.

(8)

Equation (8)includes all of the forces that are acting on a bomb as it falls to the ground. Beforewe attempt to integrate this equation, we will simplify the drag by letting k = DρA/2m. This willhelp keep our values straight. Therefore, with the new substitution, equation (8) becomes

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a = g − kv

a =dv

dt= g − kv.

(9)

Equation (9) is the acceleration due to the velocity and drag. By integrating this function, we willget a formula for the velocity with respect to time. In order for us to differentiate equation (9), wewill need to separate the variables as follows

dv

dt= g − kv

dv = (g − kv) dtdv

g − kv= dt

Now that the function is set up, we will integrate∫dv

g − kv=∫dt

ln (g − kv)k

= t+ C

ln (g − kv) = −k (t+ C)

eln(g−kv) = e−kt+kC

g − kv = e−ktekC

Through the integration, we are left with an arbitrary constant ekc. To simplify this, let b = ekc.With the substitution we can now solve for the velocity as a function of time.

g − kv = be−kt

kv = g − be−kt

v =g

k− b

ke−kt.

However, this equation is not yet correct. We will need to substitute our initial values to solve forthe constant b. Setting v = 0 at t = 0 produces

v =g

k− b

ke−kt

0 =g

k− b

ke−k(0)

0 =g

k− b

kg = b.

Our calculations reveal that b is equal to g. This will make the velocity equation a little bit easierto look at.

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v =g

k− g

ke−kt

v =g

k

(1− e−kt

)v =

g

k

(1− e−kt

).

(10)

This formula tells us that at any time t we can evaluate the velocity of the bomb. We would liketo note that as t→∞, v → g/k, called the terminal velocity.

Up to this point, we have formulated equations for the acceleration and velocity of the bombin the vertical direction. Neither of these equations, however, will tell us where the bomb will landwhen a given height and time are known. We will integrate once more to find the displacement ofthe bomb

v =g

k

(1− e−kt

)dy

dt=g

k

(1− e−kt

)∫dy =

g

k

∫ (1− e−kt

)dt

y =g

k

(t+

e−kt

k

)+ c.

Yet again, we are left with a constant of integration c. Providing some initial conditions willeliminate the constant. With the initial condition y = y0 at t = 0 gives

y0 =g

k

(0 +

e0

k

)+ c

y0 =g

k

(1k

)+ c

y0 =g

k2+ c

c = y0 −g

k2.

Plugging in the c value will give the final equation for displacement

y =g

k

(t+

e−kt

k

)+ y0 −

g

k2

y =g

kt+

g

k2e−kt + y0 −

g

k2

y = y0 +g

kt− g

k2

(1− e−kt

).

(11)

This final equation represents the general solution to our system in the vertical direction. We canplot 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 trajectoryin the vertical direction. We will then follow up with the derivation for the equations with drag inthe horizontal direction.

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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. Inthe same way, we can derive the general solution to the horizontal dislpacement. By taking thesame steps of integration we can go through this process without too much explanation. Startingwith the initial forces in the x direction

F = −Fd. (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 k =DρA/2m for our drag force.

a =dv

dt= −kv

Seperate variables then integrate to solve for the final velocity equation

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dv

dt= −kv

dv

v= −kdt∫

dv

v=∫−kdt∫

dv

v= −k

∫dt

lnv = −kt+ c

v = e−kt+c

v = c1e−kt.

(13)

The integration constant c1 can be evaluated with the initial conditions t = 0 and v = v0.

v = c1e−kt

v0 = c1e0

v0 = c1.

Then plugging this value back into the equation (13) will give

v = v0e−kt (14)

This is the final eqaution for the velocity with drag. We will then integrate, just as we did with thevertical equation, to solve for the displacement.

v = v0e−kt

dx

dt= v0e

−kt∫dx =

∫v0e

−ktdt

x = v0

∫e−ktdt

x = v0e−kt

−k+ c.

Yet again we are left with the integral constant that can be solved with the initial conditions t = 0and x = 0

x = v0e−kt

−k+ c

0 = v0e0

−k+ c

0 =v0−k

+ c

v0k

= c

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Figure 5: Horizontal Motion with Drag

Therefore

x = v0e−kt

−k+v0k

x =v0k

(1− e−kt

) (15)

Equation (15) is our final formula needed to solve for the discplacement of the bomb on thex-axis. Setting an altitude and initial speed of the plane, we can plot this trajectory as shown infigure 5.

3 Conclusion

With equations (11) and (15),

y = y0 +g

kt− g

k2

(1− e−kt

)x =

v0k

(1− e−kt

)we can plot the trajectory of the bomb in an xy coordinate system. These equations can be usedto solve the position of the bomb when dropped from any height and distance. Setting an altitudeand velocity of the plane, we have plotted these equations as demonstrated in figure 6.

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Figure 6: Horizontal and Vertical Motion with Drag

References

[1] Probable Bomb Impact Areas (PJR) I04FL040501

[2] Research and Education Association. 1996 Differential Equations.

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