Is there a simple mathematical expression for the stopping power of a given thickness of armor, given the thickness of armor plate, the radius of a cannon ball, the density of the cannonball and the armor, the tensile strength and/or toughness of the armor, and the speed of the cannonball? For simplicity assume the cannonball is a solid metal sphere and that the armor plate is homogeneous. I realize that in modern warfare the projectiles are pointed and armor plate isn't a homogeneous slab, but I want to understand the simple case. (My question is inspired in part from reading about Civil War ironclads, but I also saw the question about chain mail and thought that if that question was legitimate this one should be more so.)

In case my question isn't clear, what I'm asking is something like the following. Suppose it took 10 cm of iron armor to stop a 20 cm diameter cannonball moving at 300 meters/sec. How thick would the armor have to be to stop a 40 cm cannonball moving at the same speed? Or what if you doubled the speed? Or what if you doubled the tensile strength of the armor? Etc...

  • $\begingroup$ a simple equation can be put like, $$x = E_0.e^{-kh}$$, but then all the pain would be in finding the proper value of $k$. Which would again be dependent on the initial energy, $E_0$. $\endgroup$ – Vineet Menon Nov 11 '11 at 5:55
  • $\begingroup$ I expect that the fine points and really interesting details are classified, as the U.S. military would NOT want everyone in the world knowing how to penetrate the armor of their best tanks. $\endgroup$ – David White Jun 24 '16 at 2:04

People spend their whole lives answering this question. The best I can do is give you a method for making a very rough estimate of the stopping power of a material. I'm going to assume a uniform slab of armor and a cubic bullet that will not deform at all. The relevant material property is the toughness. Toughness is defined as the amount of energy per unit volume a material can absorb before fracturing. It depends heavily on the rate at which the material is deformed and is typically determined empirically.

The reason I've chosen a cubic projectile is to have a uniform, time independent stress being applied to the armor. Let's also assume that the projectile applies force only to the region of the armor directly in its path. Under these many simplifications, the energy required to penetrate the armor can be estimated as $T\times t_{\text{armor}} \times A_{\text{bullet}}$, in which $T$ is the toughness of the armor, $t$ is the thickness of the armor, and $A$ is the area of the bullet. This is essentially the energy required to rip all the armor blocking the projectile out of the way.

This paper gives the high-strain rate (i.e., impact) toughness of one particular steel as roughly 2000 MPa, which translates to 2000 J/cm$^3$. So for 1 cm thick armor, a 50 gram bullet with a 1 cm cross section would need to have 2000 J of kinetic energy to make it through. Solving

$$2000 = \frac{1}{2}mv^2$$

gives $v = 282\text{m/s}$ to just penetrate.

I'm not a ballistics expert, but there are bullets that can make it through a cm of steel and bullets generally have velocities measured in the 100s of m/s so this all seems pretty reasonable.

  • $\begingroup$ You might also assume that C17 steel or C19 iron (depends whose civil war you are talking about!) is 1/2 or 1/4 as strong as modern pipe steel and so 1/4 to 1/16 the speed of projectile $\endgroup$ – Martin Beckett Nov 11 '11 at 5:28
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    $\begingroup$ Usual army rifle bullets have velocities of 800 to 900 m/s , but never will penetrate one cm of steel. If those bullets were made from tungsten, they might penetrate such armour. In practice such questions are business for engieers and are solved experimentally. Maybe very sophisticated fined element calculations can be used. An important thing in such processes is that the velocity of the impactor is not far from sound velocity in the target material. $\endgroup$ – Georg Nov 11 '11 at 10:38
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    $\begingroup$ I am not sure that the example sevenofdiamonds gave is correct, unless I have read it wrong. For example, following you maths I will solve for an old WW2 German Anti-Tank Gun, the PaK 35. Calibre = 37mm so Area in cross section is 1075mm2 Bullet Weight= roughly 690g (not including gunpowder and such of course) Velocity = 760m/s Bullet delivers 199272 Joules of energy on impact which means then (199272/1075/20*) should give the maximum thickness of armour that can be penetrated by this bullet (*about the toughness of steel) =9mm I know for a hard fact that these bullets can pierce 30mm armour q $\endgroup$ – user73539 Feb 19 '15 at 14:01

You may find this link interesting. 19th century research indicates (for iron/steel armor and iron balls) a formula along the lines of $$ T = .00005D\sqrt{\frac{W\times V^2}{D^3}}$$

where T is armor thickness, V is velocity, W is weight and D is diameter. This assumes pretty much perpendicular impact on the armor face.


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