Under the hood

The physics of interior ballistics

A projectile is initially accelerated (interior ballistics), passes through some intermediate space which may be filled with a fluid such as air (exterior ballistics), and then interacts with its target (terminal ballistics). This document will be concerned primarily with the inital aceleration of bullets fired from gunpowder-based firearms. The basic results should be easily adaptable to any slugthrower that uses the expansion of a fixed quantity of expanding gas to launch its projectile (such as air guns or steam guns).

The physical model starts with a projectile of mass mb initially at rest in a tube of area A which is closed at one end. The space between the projectile and the closed end is filled with gunpowder with mass mp. Upon ignition, the gunpowder becomes a hot, high pressure gas which expands and pushes the projectile down the tube. We will assume that the projectile fills the tube so that expanding gas cannot jet out around it, making this essentially the same problem as an expanding piston or syringe plunger. The key to solving this problem is to assume that the explusion of the projectile is so fast and the insulating properties of gas is good enough that there is negligible heat conducted from the gas to the tube or the projectile. Systems which obey this approximation are called adiabatic, and the physics of adiabatic gases have been worked out in detail.

When a gas expands adiabatically from an initial pressure P0 and intitial volume V0 while doing work against an external load (the projectile, in this case), the pressure P and volume V are related thereafter by
P Vγ = P0 V0γ.
The parameter γ depends on various properties of the gas. You can say that it is the ratio of the constant pressure and constant temperature specific heats, for example, or relate it to the number of quadratic degrees of freedom among which the thermal energy can be distributed. For our purposes, all we need to know is that the gunpowder combustion gasses will have a γ of about 1.2. We also need to know what the initial pressure is for combusting gunpowder, which can be found from the specific energy of black powder of 2.86 MJkg and the mass density of black powder ρp of about 1 gcm³. This tells us that the energy density of black powder is about 2.86 GJm³. The energy density of a gas is related to the pressure by
P = (γ-1) EV.
It immediately follows that once the gunpowder has combusted but has not had time to expand its pressure is P0 = 0.572 GPa. The initial volume of the powder is
V0 = mpρp
and the length of the barrel taken up by powder is
L0 = V0A.
As the combustion products expand, pushing the projectile down the barrel, the pressure varies as
P(L) = P0 (L0L)γ.
The energy that the internal motions of the gas molecules give to the kinetic energy of the projectile after moving a small distance dL is found by the relation
dEk = P(L) A dL.
This is easily integrated to find
Ek = A L0 P0 (γ-1) (1 - L0L)γ-1 = E0 (1 - L0L)γ-1.
This energy is distributed between the projectile and the now fast-moving jet of gas behind it that is pushing it. If we just want the energy of the projectile, we can approximate
Ep = Ek     mb     .
mb + mp
The projectile speed is now given by
v = √(2 Ep ⁄ mb).