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Ballistics

Projectile paths without air resistance

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Acceleration | Differential Equations | Ballistics AppletVariations on an Equation | Free-Fall with Air Resistance

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Acceleration

In this section we will cover the dynamic motion of objects in free-fall, that is, in a state of motion where gravitational mass and inertial mass cancel, objects of different masses fall at the same rate, and acceleration is the predominant issue rather than force. This is not to say there are no forces — there always are — but that two forces acting on a small free-falling mass (gravitational and inertial) cancel, leaving only the gravitational attraction of a primary mass like a planet:

(1) $ \displaystyle a_g = \frac{GM}{r^2}$Where:Equation (1) is a formality — in most cases we will use little-g to accelerate our objects, because we know its value, because our results will then agree with those from other sources, and because most of the equations require that

- $a_g$ = gravitational acceleration
- G = universal gravitational constant
- M = mass of attracting body
- r = distance between the accelerated object and the center of mass of the attracting body
gbe a constant. On that issue, most of the methods to come require some simplifying assumptions:

- That the gravitation acceleration is constant (only really true for small altitude changes)
- The there is no air resistance (true on the moon but not Earth).
- That there is no gravitational field curvature (only apparently true on a small scale)

Differential Equations

In setting up the theoretical basis for what follows, it's best to use some ideas from the field of differential equations. Those who don't want to dwell on this aspect of ballistics can just skip forward — the ideas that follow won't suddenly become incomprehensible because you skipped this section.

We will construct a differential equation for the position of a mass with respect to time. The terms for the unknown function are:

- $p''(t) = -g$ (acceleration)
- $p'(0) = v_0$ (velocity)
- $p(0) = p_0$ (position)
Where:

- $t$ = Time, seconds
- $g$ = Little-g, m/s
^{2}- $v_0$ = Initial velocity, m/s
- $p_0$ = Initial position, meters
For those unfamiliar with differential equation nomenclature, the above describes an unknown function p(t) that is meant to define position with respect to time. The term p'(t) (note the asterisk) refers to the first derivative of position or the rate of change in position, i.e. velocity. The term p''(t) refers to the second derivative of position, or acceleration. In what follows, remember:

- Velocity is the rate of change in position with respect to time, or the time derivative of position.
- Acceleration is the rate of change in velocity with respect to time, or the time derivative of velocity.
Here is the solution for the above terms, the result of a mathematical process that provides the unknown function:

(2) $ \displaystyle p(t) = p_{0} + t v_{0} -\frac{g t^{2}}{2}$Equation (2) is a position function that produces a

y(vertical) coordinate for a free-falling object as a function of time. We will be using this equation as the basis for a number of derived equations and results below, so it might be wise to describe it in some detail:

- Variable p
_{0}is used to set an initial position. For example if we wanted to model an object falling from heighth, we would set p_{0}=h.- Variable v
_{0}is used to set an initial velocity, separate from gravitation. This might be used while simulating throwing a ball into the air — v_{0}would be the ball's initial upward velocity.- Variable g is the gravitational acceleration. In many cases this will be set to little-g from our earlier discussion, but it could be any gravitational acceleration — that of the the moon, Mars, or another planet.
- Variable t is time. The time variable is the only value not assumed to be a constant.
I have written Equation (2) in a very general way, with plenty of flexibility to handle different kinds of physical problems. For example:

- For the case of an object released from a height
h, we would set p_{0}toh, the initial height, and v_{0}to zero, since we're just releasing the object, not giving it an initial velocity.- For the case of a ball thrown into the air, we would set p
_{0}to zero (ground level) and v_{0}to the vertical part of the throw velocity. Given time arguments, the function would show the arc described by the object as it flew through the air to its landing spot.

Ballistics Applet

I include a Java Ballistics Applet on this page (Figure 3) so the reader can set up and run dynamic ballistic experiments. Here are some examples:

Zoom = mouse wheel, Translate = mouse dragFigure 3: Ballistics Applet

- Ball throw — the default settings. Press "Defaults" or:

- Set p
_{0}= 0 (always press Enter when making a change)- Set v
_{0}= 30- Verify that g = 9.80665
- If the image doesn't update, click your mouse in the applet window.
- The graph line should achieve a maximum height of 45.88 meters and cross zero at a time of 6.12 seconds.
- Ball drop from a height:

- Set p
_{0}= 40- Set v
_{0}= 0- Verify that g = 9.80665
- If the image doesn't update, click your mouse in the applet window.
- The graph line should cross zero at a time of 2.85 seconds.
- Ball drop from higher:

- Change only p
_{0}= 200- Point your mouse at the applet and use your mouse wheel to zoom out to see the drop point at 200 meters.
- Drag your mouse on the applet to change the viewing position.
- Now zoom in to the point where the line crosses zero.
- The graph line should cross zero at a time of 6.38 seconds.
Feel free to experiment with this applet — you can't break it, and you can always recover by pressing the "Defaults" button. Remember about the outcomes that they make some simplifying assumptions, like no air resistance — the drop from 200 meters wouldn't really work as shown if there was any air resistance.

Variations on an Equation

Equation (2) has many practical applications, as long as we remember that it is only accurate at velocities below that at which air resistance becomes significant. Here are some variations on equation (2) to solve specific problems:

When using these equations, remember these associated values:

- $t$ = Time, seconds
- $g$ = Little-g
- $v_0$ = Initial velocity
- $p_0$ = Initial position

Ballistic flight values as a function of time

t:

- Vertical velocity: (3) $ v = v_0 - g t$
- Height: (4) $ h = p_0 + t v_0 - \frac{g t^2}{2}$
Apex Values:

- Velocity: 0
- Time: (5) $ t = \frac{v_0}{g}$
- Height: (6) $ h = \frac{v_0^2}{2 g}$
Terminal Values:

- Velocity: (7) $ v = -v_0$
- Time: (8) $ t = \frac{2 v_0}{g}$
- Height: 0
Freefall:

- Fall distance
dfor timet: (9) $ d = \frac{g t^{2}}{2}$- Time
tfor fall distanced: (10) $ t = \sqrt{\frac{2d}{g}}$- Velocity
vfor timet: (11) $ v = gt$- Velocity
vfor fall distanced: (12) $ v = \sqrt{\frac{2dt}{g}}$- Kinetic energy
keof a massmmoving at velocityv: (13) $ ke = \frac{mv^2}{2}$- Kinetic energy
keof a falling massmfor timet: (14) $ ke = \frac{g^2 m t^2}{2}$

Free-Fall with Air Resistance

It is important to stress that a very accurate result for an object falling through air cannot be expressed in a closed-form equation — it must be modeled using numerical methods. But if we accept certain limitations, there are useful closed-form solutions. We can't assume that the result rigorously takes changes in altitude into account (air pressure varies with altitude, and gravitational acceleration does also), but it's useful as long as these limitations are kept in mind. Here are new differential terms for an unknown equation that takes air resistance into account:

$ \displaystyle p''(t) = -g + p'(t)^2\ k$:

This acceleration term incorporates the fact that air resistance varies as the square of velocity ($p'(t)^2$) times an empirical factorkthat depends on the object's shape and surface roughness. In free-fall, gravitation is a constant, but as air resistance builds up, it eventually equals the gravitational force, acceleration falls to zero, and velocity becomes constant (see Figure 2).- $ \displaystyle p'(0) = 0$: Initial velocity is zero
- $ \displaystyle p(0) = h$: Initial height
hAs before, a mathematical analysis process leads to this result:

Figure 4: Free-fall altitude profile(15) $ \displaystyle p(t) = \frac{h k - \log\left(\cosh\left(t \sqrt{g k}\right)\right)}{k}$Where:

- p(t) = position with respect to time
- t = Time, seconds
- h = Initial height
- k = An empirical factor that takes the object's air friction into account
- g = Little-g, described previously
Figure 4 shows the altitude profile for this function, where the chosen air friction coefficient leads to relatively fast convergence on an unchanging descent rate of 53 m/s or 190.8 kph, a typical skydiver terminal velocity.

There is more on this topic here.

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