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# BALLISTICS

By Robert Pogson

This article will give the reader some of the important factors affecting the slowing of a projectile by resistance to passage by the air.

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Consider a projectile of diameter, d, passing through air with a speed, v. The projectile will collide with the mass of air in a cylinder of the same diameter and length, vdt, in a time of flight, dt. The mass of air in that cylinder will be the product, ρπd2/4 vdt, where ρ is the density of air, and π is the ratio of the circumference of a circle to its diameter, 3.1415926.... The air struck by the projectile will carry off momentum (product of mass and speed) in proportion to the speed of the projectile, assuming the air is at rest. The air does not fly off in the same direction as the projectile is moving and the speed of the air after collision will depend on the shape of the projectile and where the air collided with it. Thus, we must include a geometric fudge factor, K, to account for these unknown effects. We can supply a value for K by matching the change of speed of the theoretical projectile with an actual projectile. The momentum lost by the projectile and given to the air in a time, t, may be written

dP = K ρπd2/4 v2dt (1).

Dividing both sides of (1) by dt gives the rate of change of momentum of the projectile with time,

dP/dt = K ρπd2/4 v2 (2)

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which is just the force slowing the projectile, f, which is equal to the product of the mass and acceleration of the projectile. The acceleration is the rate of change of the velocity of the projectile with time, dv/dt.

f = ma = m dv/dt = dP/dt (3)

Thus,

m dv/dt = Kρπd2/4 v2 (4).

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Rearranging (4) by Grade 10 algebra, gives

dv/v = Kρπd2/(4m) vdt (5).

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Integration (first year calculus) gives

ln v = ln v0 + πKρ/(4m) d2 x (6),

where v0 is the initial velocity, x is the distance travelled and t is the time of flight.

We can rewrite (5) using calculus to obtain

d(v2) / (2v3) = Kρπd2/(4m) dt (7).

Rearranging again gives

d(mv2/2) / (mv2/2) = 2Kρπd2/(4m) vdt. (8)

Because the energy of the projectile is just the kinetic energy, E = mv2/2,

d(E) / (E) = 2Kρπ d^2/(4m) dx. (9)

The kinetic energy is an exponential decay function,

E = E0 exp(2Kρπd2/(4m) x), (10)

where E0 is the initial energy.

v = v0 exp(Kρπd2/(4m) x) (11)

and,

dx/dt = v0 exp(Kρπd2/(4m) x)

exp(-Kρπd2/(4m) x)/v0 dx = dt

t = (1 - exp(-Kρπd2/(4m) x))/ (Kρπd2/(4m))/v0 (12)

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(11) and (12) give us the speed of the projectile as a function of time and the time required to slow to a given speed from an initial speed. From this we can calculate the height of trajectory, H, energy at a given range, and the angle of elevation for the barrel to reach the initial height at a given range.

For low angles of elevation used on level ground and short ranges, the vertical motion is approximately independent of the horizontal motion. We can use the high school equations for motion under constant acceleration for the vertical part:

y = v0 sin α t+ 1/2 g t2 (13),

where α is the angle of elevation and g is the acceleration of gravity, -32.2 ft/s2. We can solve (13) for α to find the angle of elevation to a given range by setting y to 0 and substituting (12) for t:

sin α ≈ α = -1/2 gt/v0 (this will be in radians , 180/π degrees). (14)

Then the height of trajectory will be the value of y for one half of the value of (12).

H = v0sin α t/2 + 1/2 g(t/2)2 (15)

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These equations will permit writing computer programmes to calculate trajectories for a given range PROVIDED that the approximations involved in the derivations remain true. For practical hunting purposes, these approximations are good enough if the speed of the bullet does not fall below 70% of the muzzle velocity (about 1/2 the muzzle energy). This limits pointed bullets to 400 - 600 yards and blunt bullets to about 200 yards, near the limits for expansion and quick kills with these bullets.

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To put these equations to use, a value for K must be chosen. K has two functions. K accounts for the shape of the bullet and how air behaves when struck by a bullet. Air is complex. Its density and temperature and molecular speed distributions can vary. The drag on a bullet varies with the speed of the bullet varying with respect to the speeds of the molecules. These equations assume the molecules are at rest, which is only good for velocities much higher than the speed of sound (average molecular speed). In reality, molecules will have a variety of speeds and directions, but the drag will not change much for a small change in speed of the projectile, hence the 70% limit. On the other hand, one may well calculate with bullets of different shapes, so a factor representing properties of the bullet should be included. A number called the ballistic coefficient, C, is published for bullets from several manufacturers of bullets for reloaders. A long pointed bullet will have a much higher value of C than a short blunt bullet of the same material.

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Very roughly (errors like 20%) values for C will be near these:

Type of bulletC
Light weight pointed0.2
Medium weight pointed0.3
Heavy weight pointed0.5
Heavy weight pointed boattail0.6
Light weight blunt0.1
Medium weight blunt0.2
Heavy weight blunt0.3

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From (1), we can see that less momentum loss requires K to be smaller so that

K = B/C

Where B is a fixed constant to describe air and the units being used and C is a constant we get from published results or by testing particular bullets.

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This theory was first described by Newton, one of the inventors of calculus. He was working with cannon balls and rocks.

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A trivial PASCAL programme to demonstrate these equations follows:

```Program ballistics;
(*Copyright 2010 Robert Pogson - This is Free Software. You may run, examine, modify and distribute this software
under licence GPL v3 or later. See http://www.gnu.org/ *)
label go_back,start;
const light = 130.0/0.308/0.308; medium= 180.0/0.308/0.308;
(*these constants help choose BC in comparison to sectional densities of .308 bullets*)
pi=3.1415926;
b=-0.00017; (*b this number is empirical. It is chosen to approximate other published data*)
g=-32.2;(*acceleration of gravity in ft/s/s*)
begin
start:
Writeln(space(30),'POGSON''S TRIVIAL APPROXIMATE BALLISTICS PROGRAMME');
writeln('give 0 diameter to exit');
write('diameter of bullet in inches>');readln(d);if d<=0.001 then exit;
if c < 0.0001 then (*estimate BC from sectional density if BC is unknown*)
begin
s:=m/d/d;
go_back:
begin
if s < light then c:=0.2
else
if s< medium then c:=0.3 else c:=0.5;
end
else
begin
if s < light then c:=0.1 else if s < medium then c:=0.2 else c:=0.3
end
else goto go_back;
writeln('value of C is estimated to be ',C:5:3)
end;
(*convert units*)
range:=range*3.0;
m:=-m/7000.0/g;
d:=d/12.0;
k:=b/c;
x:=range;
e0:=0.5*m*v0*v0;
writeln('energy at the muzzle =',e0:5:0,' ft-lb');
e:=e0*exp(2.0*K*pi*d*d/(4.0*m)*x);
v := V0* exp(K*pi*d*d/(4.0*m)*x);
t := (1.0 - exp(-K*pi*d*d/(4.0*m)*x))/(k*pi*d*d/(4.0*m))/v0;
alpha:=-g*t/v0/2.0;
y := V0*alpha*t +0.5*g*t*t;
H := V0*alpha*t/2.0 +0.5*g*t*t/4.0;
writeln;
writeln('for the given range:'); writeln;
writeln('time of flight = ',t:5:2,' second');writeln;
writeln('remaining energy is ',e:5:0,' ft-lb'); writeln;
writeln('remaining speed = ',v:5:0, ' ft/s');
if v/v0 < 0.7 then write('  ***WARNING*** remaining spped is only ',v/v0*100.0:3:2,' % of muzzle velocity!');
writeln;
writeln('height of trajectory above horizontal for zero is ',h*12.0:4:1,' inches');
writeln;
writeln('angle of elevation for zero = ',alpha*60.0*180.0/pi:5:1,' minutes');
writeln;
writeln;
goto start
end.
```

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