On armour.

Introduction

Here we shall determine the ability of armour to withstand being subjected to non- or weakly-penetrating radiation.

This one was a larger problem than anticipated, and seemed to spend a lot of time in that "almost finished" state, hence coming not one but three weeks after the last instalment. Previously I've mainly been checking over and typing up work I originally did ages ago, but this wasn't a problem I'd looked at before. My first approach to the problem proved to be overcomplicated and incorporated severely flawed hidden assumptions that rendered its results inconsistent. The method presented here is simpler, yet still more convoluted than previous topics, and while it has its fair share of assumptions I feel it gives useful insight into the physics.

Said fair share of assumptions.

• The armour is an infinite flat plate of uniform composition. This should be more or less valid away from corners or tight curves.
• The armour is solid and conducts heat passively. This should be valid for simple armour.
• The armour is a grey body. This should be valid for most materials.
• The armour has a single melting point, as opposed to a range. This should be valid for simple chemical compositions like pure metals or binary ceramic compounds.
• Variation of material properties with temperature, other than melting, is ignored. This should be a reasonable approximation for most materials.
• One face is in vacuum and subjected to a constant and uniform heat flux. We are ignoring how the radiation initially interacts with the armour (a topic for another post), including ignoring the reflectivity if it is thermal radiation, and considering only the thermal behaviour after the energy is deposited.
• The interior, protected by the armour, is isothermal and of finite heat capacity per unit surface area. Said heat capacity is much larger than that of the armour itself. If there is both considerable structural mass and plenty of fluid (eg air) circulation this should be valid.
• If the interior is heated above a critical temperature it is considered 'cooked' and thus the armour has failed. This seems sensible enough.
• Armour that has been melted is immediately shed. If the vehicle it's on is jinking around this should be valid.
• If the armour is melted all the way through, the interior will be directly exposed to the radiation and thus the armour has failed. This seems patently obvious.

Basics and thresholds

The following diagram depicts the various heat transfer process at work at the outer surface of the armour.

Since the surface has no volume and thus no heat capacity of its own, the heat flux in must equal the heat flux out at all times.

q_i is the incoming heat flux.
q_r is the thermal radiation.
q_c is the heat conducted inwards.
q_a is the heat carried away by melting and ablating armour material.
q_i = q_r + q_c + q_a

q_r is given by the Stefan-Boltzmann law.

T₀ is the absolute temperature at the outer surface, ie the temperature measured in Kelvin or degrees Rankine. Not that I've ever come across anyone using Rankine.
σ is the Stefan-Boltzmann constant.
ε is the emissivity.
q_r = εσT₀⁴

This allows us to determine a minimum intensity required to cook the interior. Since the interior cannot become warmer than the outer surface of the armour, because that would require heat to flow from a cold to a warm region, in order to cook the interior the outer surface must be raised to at least the critical temperature. Therefore, the incoming heat flux must be at least the thermal radiation at that critical temperature.

T_c is the absolute critical temperature.
q_cook is the minimum intensity for the incoming heat flux to cook the interior.
q_cook = εσT_c⁴

q_c depends on the thermal conductivity of the armour and the thermal gradient across it. Once a (quasi-)steady state has been reached, this gradient will be linear through the armour (and flat in the interior), as shown in the following diagram.

D is the thickness of the armour.
T_D is the absolute temperature at the inner surface.
k is the thermal conductivity.
q_c = k(T₀-T_D)/D

Before the steady state is reached, the temperature profile must be concave and therefore the gradient at the outer surface will be steeper. This is because if the profile was convex then some part of the armour would actually be cooling down, nonsensical considering the armour started cold and is subject to a constant heat flux.

The steady state will not actually be truly steady, since the interior has finite heat capacity so will warm. However, the assumption is that this is slow enough for the temperature profile to never depart significantly from linear.

Knowing the amount of heat conducted inwards allows us to determine the minimum intensity required to begin melting and ablating the outer surface before the interior is cooked. This will be equal to the heat conducted inwards when the interior and inner face are at the critical temperature while the outer face is at the melting point, plus the thermal radiation when the outer face is at the melting point. Any further incoming heat flux cannot be removed by radiation and steady-state conduction, so must be used to melt the armour.

T_m is the absolute melting point of the armour.
q_melt is the minimum intensity for the incoming heat flux to begin melting the armour.
q_melt = εσT_m⁴ + k(T_m-T_c)/D

Of course, once some material is ablated, the remaining armour is thinned. With D lowered, q_melt is raised. Indeed, the above equation predicts that it is impossible to melt the full thickness of the armour without first cooking the interior, since as D tends to zero the second term on the right-hand-side tends to infinity. Infinities appearing in science equations tend to indicate that one of the assumptions has become invalid. In particular, if the assumption that the interior is isothermal ceases to hold then T_D can rise to T_m without requiring the whole interior rise to T_m.

q_a depends on the thermal properties of the armour material, and on the speed at which it is being ablated. The following diagram shows how a small thickness of armour that's ablated within some small time interval has volume per unit surface area equal to that thickness.

The heat required to ablate this volume can be found by considering its mass, its specific heat capacity, the temperature it needs to be raised by, and its latent heat of fusion.

dx is the small thickness.
ρ is the density.
c is the specific heat capacity.
T_i is the initial temperature
L_m is the latent heat of fusion.
dH is the small energy per unit area required to ablate the thickness dx.
dH = ρ[c(T_m-T_i)+L_m]dx

Dividing through by a small time interval gives us the relation between the heat flux and the speed of ablation.

dt is the small time interval.
v is the speed of ablation.
q_a = dH/dt
v = dx/dt
q_a = ρv[c(T_m-T_i)+L_m]

We can therefore write the full equation for the heat flux balance at the outer surface of the armour.

q_i = εσT₀⁴ + k(T₀-T_D)/D + ρv[c(T_m-T_i)+L_m]

Note that for the third term on the right-hand-side to be non-zero T₀ must equal T_m, since obviously there can't be any melting if the outer surface, the hottest part of the armour, isn't at least at the melting point, and it can't be above the melting point because any liquid material falls off.

Time to cause damage

The next task is then to determine the time required for the armour to fail, depending on its properties and on q_i. Rather than attempt to present a single equation, we will consider the different cases in turn.

No damage case
If q_i is less than q_cook, the armour will never fail.

Cooking case
If q_i is greater than q_cook but less than q_melt, the armour will fail by the interior being cooked. T₀ and T_D will both change over time, which would make the problem complicated. To simplify things, we shall consider that higher T_D implies higher T₀, which implies higher q_r, which implies lower q_c. Thus we shall determine the lower and upper bounds for q_c, and thus bounds for the time to fail.

The following equation should be solved for T₀ in the cases where T_D=T_i and T_D=T_c, the initial and critical temperatures being of course the minimum and maximum respectively. Being a quartic I believe the analytical solution will be complicated, so trial and improvement will solve it straightforwardly and more than adequately.

q_i = εσT₀⁴ + k(T₀-T_D)/D

Then q_c should be found for the two cases.

q_c = k(T₀-T_D)/D

With the heat capacity of the armour itself neglected, q_c must equal the rate of heat rise in the interior, by virtue of heat flux balance at the inner surface. Since we're considering q_c constant (but in two cases), this rate of warming will also be constant.

q_h is the rate of heat rise.
t_cook is the time required to cook the interior.
C is the heat capacity (total, not specific) of the interior per unit surface area of the armour faces. This will be the volume weighted average of the volumetric heat capacity (specific heat capacity times density) of the various materials in the interior, multiplied by the width.
q_h = C(T_c-T_i)/t_cook
q_c = q_h
k(T₀-T_D)/D = C(T_c-T_i)/t_cook

This can be rearranged to give t_cook, which should be calculated for the higher and lower values of T0 and TD to give the upper and lower bounds on the time. The actual time will lie somewhere between, though not I believe midway.

The energy requirement per unit surface area is then simply given by the heat flux times the time.

H_cook is the energy per unit surface area needed to cook the interior.
H_cook = q_it_cook

Melting and ablation case
If q_i is very much greater than q_melt, then q_r and q_c can be neglected as insignificant and q_i equated to q_a.

q_i = ρv[c(T_m-T_i)+L_m]

This can be rearranged to give the speed of ablation.

The time required to melt through the whole thickness is then given simply by the usual relation between speed, time, and distance.

t_melt is the time required to melt through the whole thickness of the armour.
t_melt = D/v
t_melt = Dρ[c(T_m-T_i)+L_m]/q_i

The energy requirement per unit surface area is again the heat flux times the time.

H_melt is the energy per unit surface area required to melt the armour.
H_melt = Dρ[c(T_m-T_i)+L_m]

Notice that q_i does not come into the formula for H_melt. This is not surprising, as q_r and q_c were neglected, so all the energy input goes into melting the armour, therefore how quickly it's input does not affect the end result.

Borderline case
If q_i is greater than q_melt, but not very much greater, then the situation is the most complicated. q_c can no longer be neglected, and rises as material is ablated, slowing further ablation. For certain values of q_i it will be the case that some armour, but not all, is ablated before the interior is cooked, and a full treatment would require considering just when the assumption of an isothermal interior becomes invalid. I feel the three cases already considered will adequately cover the majority of situations, and thus shall not expend a great deal of effort on what I feel will be an uncommon situation.

Worked Example

For a worked example I will consider a spacecraft with ten metres of titanium armour, protecting an interior one kilometre wide made up of 90% (by volume) air and 10% diamond. Material properties, by the way, are mainly obtained from the excellent website Matweb.

Armour properties:
D = 10 m.
ε = 0.63 (dimensionless).
k = 17 W m^-1 K^-1.
T_i = 293 K. A typical room temperature, so sensible for a manned spacecraft.
T_m = 1923 K.
ρ = 4500 kg m^-3.
c = 528 J kg^-1 K^-1.
L_m = 4.35 x 10⁵ J kg^-1.

Interior properties:
T_c = 373 K. Ie 100 °C. With low humidity this is tolerable by humans for short periods; some saunas get this hot for example.
Width = 1000 m.
Density of diamond = 3500 kg m^-3.
Specific heat capacity of diamond = 508 J kg^-1 K^-1.
Volumetric heat capacity of diamond = 1.78 x 10⁶ J m^-3.
Density of air = 1.2 kg m^-3.
Specific heat capacity of air = 1012 J kg^-1 K^-1.
Volumetric heat capacity of air = 1214 J m^-3.
Weighted average volumetric heat capacity = 1.79 x 10⁵ J m^-3.
C = 1.79 x 10⁸ J m^-2.

Constants:
σ = 5.67 x 10^-8 J s^-1 m^-2 K^-4.

Thresholds:
q_cook and q_melt can be found.

q_cook = 691 W m^-2.
q_melt = 4.91 x 10⁵ W m^-2.

q_cook is quite low. Indeed, even considering reflectivity, this spacecraft would apparently "cook" if it were at the same distance as Earth from the Sun. (We haven't considered the shaded side at all, so this may not be an accurate conclusion). For our example, though we shall place it much closer, at about 0.09 AU.

Cooking time:
q_i = 10⁵ W m^-2. This is after taking into account reflectivity; the actual solar radiation is more intense.

Low case:
T_D = T_i = 293 K.
T₀ = 1288.0 K. Found using Goal Seek in LibreOffice Calc, which I believe implements trial-and-improvement. It could of course be found manually if needed.
t_cook = 8.5 x 10⁶ s. Or nearly a hundred days.
H_cook = 8.5 x 10¹¹ J m^-2.

High case:
T_D = T_c = 373 K.
T₀ = 1288.4 K. t_cook = 9.2 x 10⁶ s. Or just over a hundred days.
H_cook = 9.2 x 10¹¹ J m^-2.

Ablation time:
If q_i is significantly higher then melting will occur. This is likely to be the case with exposure to weapons fire.

q_i = 10⁹ W m^-2. Contemporary laser cutters can easily reach such intensities, albeit in a very small spot.
t_melt = 58 s
H_melt = 5.8 x 10¹⁰ J m^-2

Conclusions

This has been a long piece, but one rewarded with some good conclusions. Firstly, the dependency of the time and energy needed to cause damage on armour thickness and radiation intensity. In the ablation case and provided that the intensity is easily enough to ablate the material, it is simple: doubling the thickness doubles the time and energy, while doubling the intensity halves the time but leaves the energy requirement constant. In the case of cooking things are more complex. The following graphs show how T₀ and t_cook vary with D and qi. Other than changing D or q_i, all values are the same as in the example case above.

For thin armour t_cook varies very little depending on D but T₀ rises quite sharply, while for thicker armour t_cook rises, tending to double with increasing thickness just as for melting, while T₀ asymptotically approaches a maximum value. The variation of t_cook with q_i, by contrast, is less obvious, but an inverse square root relation is about right, meaning the intensity would need to be increased fourfold to halve the time required.

In short, for all but the thinnest armour doubling the thickness doubles the performance, while for weapons capable of ablating the target doubling the intensity doubles the performance.

The second conclusion is that 'cooking' the interior is an inefficient way to do damage. This is despite the fact that in our example the heat rise (H_melt) involved in melting the armour is two orders of magnitude higher than that required to cook the interior (C). The long duration, imposed by the thermal insulation of the armour, results in a long time for losing heat by radiation, vs the short sharp pulse that will melt and ablate the armour with less losses. Therefore, the idea of a weapon that kills the crew of a target ship by cooking them while leaving the ship itself largely intact is a poor one against all but the smallest ships. 'Cooking' is I feel more relevant when considering how close a spaceship can get to a star, and there we see that a large ship can be parked in orbit quite close for quite some time without any special cooling systems and yet not overheat, provided it doesn't generate too much heat internally of course.

The third conclusion is that relatively modest, by science-fiction standards, intensities and energies are sufficient to do damage. On the order of 10¹³ Joules, or a couple of kilotons of TNT, will melt and ablate a 10x10x10 m cube of typical armour (with broadly realistic physical properties) for example. If our example ship was 1 km in all three dimensions, the most powerful real-world weapon built - the 50 megaton Tsar Bomba - would if detonated in contact with the armour be capable of devastating the ship. Monstrous yields in the gigatons or teratons TNT equivalent are not needed to defeat armour unless either the targets are exceptionally large, they're armoured with physically-unrealistic unobtanium, or there are some major inefficiences at play.

A comparison with the accelerators looked at previously is also instructive. The 1 megaton yield accelerator had a mass of around a million tonnes. That energy would be capable of melting a 300x300x10 m piece of our titanium armour weighing in at around 4 million tonnes. Considering the variety of materials that might be used, in general we can expect a spaceship gun to be capable of destroying something roughly the same size as itself. This I feel is a pleasing result, indicating offense and defense are fairly well balanced.

Fourthly, the relevance or otherwise of different thermal properties of armour is established. Against low intensities, such as from proximity to stars, thermal conductivity is extremely important with lower better. For reasonably dense solid materials it can range over five orders of magnitude, from below 0.1 W m^-1 K^-1 for some ceramics up to 2000 W m^-1 K^-1 for diamond. As we've seen, with thick enough armour moderately-conductive metals will be fine, but if a spaceship or space station is to have a thin outer skin then the material needs to be chosen wisely. Against high intensities, such as weapons fire, however thermal conductivity is largely irrelevant, indeed if anything high conductivity might be desired to raise qmelt. More important are specific heat capacity, melting point, and latent heat of fusion. The gains to be made by material choice there are more modest, but still present. The biggest gains might be if the armour could somehow function while liquid, so destroying it required vaporisation which takes much more energy than melting.

Of course, thermal performance is only one aspect of armour. We haven't considered what I call 'primary penetration' - how far a weapon shot can penetrate from its initial momentum, before it even deposits its energy into the target material. It may be that protection against such imposes conflicting demands on material properties, necessitating a compromise. For example heavy elements offer high absorption of gamma rays but light elements offer high specific heat capacity.

Finally, thermal behaviour is mathematically complicated, and armour is just complicated period.

Consequences of violating assumptions

If the assumption that "the armour is an infinite flat plate of uniform composition" is violated, the exact equations may change in some cases. However, the general conclusions should remain valid.

Violating the assumption that "the armour is solid and conducts heat passively", for example by having liquid cooling ducts through it, will considerably change the thermal behaviour in the low-intensity cooking case, and may raise q_melt. For sufficiently high intensities, that overwhelm the cooling system, however things will revert to the equation given for t_melt.

If the assumption that "the armour is a grey body" is violated, the dependency of q_r on temperature will become more complex. However in most cases this is a relatively unimportant factor, and the general conclusions will I think be unchanged.

If the fourth assumption is violated, and the armour undergoes partial melting over a range, as most rocks do for example, then I believe the method presented here will be usable if some temperature is chosen as the "melting point", most likely that temperature at which sufficient melting has occurred for the armour to lose structural integrity, which in turn may be when it goes from being a solid with pockets of melt in it to a melt with crystals of solid in it.

If the assumption that "variation of material properties with temperature, other than melting, is ignored" is unjustified, the detailed maths will become far more complicated. However the increasingly familiar refrain of "the general conclusions should remain valid" is repeated.

The assumption that "one face is in vacuum and subjected to a constant and uniform heat flux" is an important one. Firstly as mentioned previously not all types of radiation will just hit the surface, some will penetrate. Secondly if the outer surface is in a fluid, then convective heat transfer will come into play and the equations become quite different. Not generally an issue for a spaceship in deep space, but it means the results are not valid for air or marine vehicles.

The assumption that "the interior, protected by the armour, is isothermal" we already suggested will break down when the armour is thin. When it does, it throws off the calculations of t_cook and q_melt, but t_melt is unaffected.

The assumption that "if the interior is heated above a critical temperature it is considered 'cooked' and thus the armour has failed" feels like more of a definition. All realistic systems will have a maximum operating temperature. If for some reason it is broken, then 'cooking' becomes impossible.

If the assumption that "armour that has been melted is immediately shed" is violated, then the equation for t_melt will cease to hold, and the process of further melting will become quite complex. This may be the case if the incoming radiation is so intense it would melt through the material very quickly, so it can't fall off fast enough and acts to slow down continued melting.

The last starting assumption, that melting through the armour constitutes damage, scarcely seems worth covering what it would mean if this wasn't true, but I'll do it anyway: ablation becomes irrelevant and only 'cooking' exists as a damage mechanism.

Final note

I'll admit this might be a bit of a shaky piece. Feedback in the comments is welcome as ever, and I'll get back to the comments I've recieved shortly. Normal service may or may not be resumed next weekend; I do still have plenty of material, but I kind of want a break from the science now. Future updates will be announced in my Nationstates signature anyway.