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.

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, ρπd^{2}/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 ρπd^{2}/4 v^{2}dt (1).

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

dP/dt = K ρπd^{2}/4 v^{2} (2)

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 = Kρπd^{2}/4 v^{2} (4).

Rearranging (4) by Grade 10 algebra, gives

dv/v = Kρπd^{2}/(4m) vdt (5).

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Integration (first year calculus) gives

ln v = ln v_{0} + πKρ/(4m) d^{2} x (6),

where v_{0} 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(v^{2}) / (2v^{3}) = Kρπd^{2}/(4m) dt (7).

Rearranging again gives

d(mv^{2}/2) / (mv^{2}/2) = 2Kρπd^{2}/(4m) vdt. (8)

Because the energy of the projectile is just the kinetic energy, E = mv^{2}/2,

d(E) / (E) = 2Kρπ d^2/(4m) dx. (9)

The kinetic energy is an exponential decay function,

E = E_{0} exp(2Kρπd^{2}/(4m) x), (10)

where E_{0} is the initial energy.

v = v_{0} exp(Kρπd^{2}/(4m) x) (11)

and,

dx/dt = v_{0} exp(Kρπd^{2}/(4m) x)

exp(-Kρπd^{2}/(4m) x)/v_{0} dx = dt

t = (1 - exp(-Kρπd^{2}/(4m) x))/ (Kρπd^{2}/(4m))/v_{0} (12)

(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 = v_{0} sin α t+ 1/2 g t^{2} (13),

where α is the angle of elevation and g is the acceleration of gravity, -32.2 ft/s^{2}. 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/v_{0} (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 = v_{0}sin α t/2 + 1/2 g(t/2)^{2} (15)

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.

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.

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

Type of bullet | C |
---|---|

Light weight pointed | 0.2 |

Medium weight pointed | 0.3 |

Heavy weight pointed | 0.5 |

Heavy weight pointed boattail | 0.6 |

Light weight blunt | 0.1 |

Medium weight blunt | 0.2 |

Heavy weight blunt | 0.3 |

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.

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

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; var d,v,v0,t,x,y,range,k,C,H,alpha,E,E0,m,s:real; answer:char; 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; write('weight of bullet in grains>');readln(m); write('muzzle velocity in feet/second>');readln(v0); write('ballistic coefficient (0 if unknown)>');readln(C); if c < 0.0001 then (*estimate BC from sectional density if BC is unknown*) begin s:=m/d/d; go_back: write('is the bullet pointed? (y/n)');readln(answer); if upcase(answer)='Y' then begin if s < light then c:=0.2 else if s< medium then c:=0.3 else c:=0.5; write('is the bullet boattailed (premium)? (y/n)>');readln(answer); if upcase(answer)='Y' then c:=c*1.2 end else if upcase(answer)='N' then 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; write('zero range (yards)>');readln(range); (*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.

Sighting Your Rifle

There's no point having all this wonderful information if you do not use it. One of the best uses is to see how much to adjust sights when changing zero ranges. Another is to adjust sights so that one does not have to worry much about the range in the field, because it can be difficult to estimate. A useful concept is point-blank range. This is the maximum range at which you can hit a vital zone of a game animal without adjusting the point of aim. The trajectories of most high velocity pointed hunting bullets are incredibly flat at the highest point but plunge like rocks past the zero. To use this knowledge, I sight in at the 6 o'clock position of the vital zone at the maximum likely range of encounter and where the remaining energy is lethal and capable of expanding the bullet. With many rifles, the height of the trajectory is still in the vital zone under these conditions. If not, reduce the zero range until it is. This way a .308 Win with a 150 grain pointed bullet will be good to 350 yards, considering remaining energy and height of trajectory. A light magnum (7mm!) will be good to 400 yards. In hunting situations one does not want to have to shoot much farther than this because of accuracy. Now, why do most hunters zero for 200 yards and have to do mental arithmetic, check a chart or guess holdover when they meet a deer at 350? By the way, my small finger nail extended in front of me is apparently as wide as a deer is long at 100 yards. A deer half a fingernail long means 200 yards. A deer a third of a fingernail long means 300 yards and so on. Good luck!