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.

On circular accelerators.

Here we shall determine the required size for a circular accelerator (whether of particles or massive objects) to deliver a chosen energy per shot.

Back to normal service now. I still don't understand circular motion (and in particular I don't understand angular momentum), but I've managed to muddle my way through the equations.

To business. Essentially a sci-fi circular accelerator is like the LHC on steroids. Projectile goes round and round the loop, and then it's let to fly off at a tangent towards its target. The assumptions are much the same as those for the linear case covered last weektwo weeks ago.

• The accelerator fires discrete shots.
• The yield is limited by material strength.
• The accelerator structure is a rectangular torus. This is a sensible shape that will make the maths easier than for a circular torus. The major radius is also significantly greater than the minor width, again for mathematical simplicity.
• The projectile is fired at ultrarelativistic speeds. Again this lets us simplify the equations.
• Recoil can be neglected.

Again we'll call whatever's being fired a projectile, even if it's a bunch of particles. And like before, I'll always be working in the accelerator frame.

Unlike in the linear case, a circular accelerator does not have to bring its projectile up to speed within a limited distance; rather the projectile can "spin up" over many laps. The accelerator structure is thus required to exert a centripetal force needed to keep the projectile going round in circles, as well as a linear accelerating force which can be arbitrarily small and shall thus be neglected. The equation for centripetal force in Newtonian mechanics, F = mv²/r, is of course not valid in special relativity, so we will derive the ultrarelativistic equation.

Those who last did vector calculus more recently than several years ago will have to forgive my doubtless slightly sloppy notation.

The diagram above shows the projectile with an initial momentum p. After a short time interval dt, its momentum is now p'. During that time interval it travels through an arc with length vdt and angle dθ. Its momentum keeps the same magnitude - we're considering uniform circular motion - but rotates through that same angle.

The above diagram shows two similar isosceles triangles: on the left the change in position of the projectile, on the right the change in momentum. As such, we can write

s/r = dp/p

If dθ is small, then s ≈ vdt, ie the arc and chord are nearly the same length (and tend towards the same length as dθ shrinks to zero). As such

vdt/r = dp/p

Rearranging, the equation for force can be found.

F = dp/dt = pv/r

The above equation is valid equally in Newtonian mechanics and special relativity, since it's just derived from geometry. The expression for momentum, however, varies. In the ultrarelativistic limit,

v = c
E = pc
F = E/r

Thus we have the equation linking centripetal force, energy, and radius of curvature for an ultrarelativistic projectile.

In the linear case, we were able to consider the forces as being borne simply in compression. For a circular accelerator, however, the stresses are more complex. A key simplifying assumption I feel justified in making is that the force is spread essentially equally about the ring, because the projectile and thus the centripetal force are moving much faster than the speed of sound in the accelerator structure. With that assumption, the situation can be considered as analogous to a pressurised cylinder.

The above shows a cross-section through a cylinder with an outward force acting on it. In our accelerator, the outwards force is the reaction to the centripetal force on the projectile. In the analogous pressurised cylinder, it's fluid pressure on the inner walls. Either way, the outwards force results in a circumferential, or 'hoop', tension in the accelerator structure.

Assuming the projectile travels centrally around the accelerator, the radius of curvature r corresponds to the mean radius of the cylinder. If the cylinder then has length l and wall thickness w, and is thin-walled (wall thickness small compared to radius), the pressure on the inside walls is given by the force divided by the surface area of the inside

σ_p = F/2πlr

The hoop stress is then give by the following equation, which can be derived by considering half of the cylinder as a free body

σ_θ = σ_pr/w

In both the above equations, the radius should correctly be the inside radius of the cylinder. However using the mean radius does not in practice result in a large error for the peak hoop stress; even for w = 0.2r, normally regarded as thick-walled, the value is just 11% too low.

As in the linear case, this stress is limited by material strength. The hoop stress cannot be greater than the ultimate tensile strength σ_t of the material of the accelerator. Equating σ_θ and σ_t, and substituting in expressions for σ_p, we can get an equation relating projectile energy, material strength, and accelerator geometry.

σ_t = F/2πlw
σ_t = E/2πlwr

The volume of the accelerator structure can be approximated as

V = 2πrlw

And thus the equation for energy simplifies down to

E = σ_tV

Apart from being limited by tensile not compressive strength, it's exactly the same as for the linear accelerator! As a result, I shan't be doing a worked example this time round.

Power we cannot determine from the geometry for a circular accelerator in the way we did for the linear case. Ignoring losses, it could be computed simply from the desired firing rate and could be arbitrarily low. In practice, however, losses may exist, such as synchrotron radiation.

The key consequence is thus that a circular accelerator design will have mass similar to a linear one, the exact ratio depending on the available materials' performances in compression compared to in tension. The circular design though will require less power, probably orders of magnitude less, to run. Within a certain range, the yield depends just on the overall mass of the accelerator structure, and not on the exact shape.

If the first assumption is violated, it actually won't make a difference, due to the assumption made during the working out that the centripetal force is essentially equally spread around the accelerator anyway.

If the second assumption is violated then an entirely different approach will be required to determine the yield.

If the third assumption is violated, a more detailed mathematical treatment will be required. In particular, the equations given here will tend to underestimate the stresses, and thus overestimate the yield, for 'thick walled' designs, especially for the extreme case where the structure is a solid disc which might be practical for a compact accelerator. That said, I believe the error does not increase without limit, so the calculations here will still be within the right order of magnitude.

If the fourth assumption is violated, likewise a more detailed treatment will be required. The mass and velocity of the projectile will have to be considered separately, instead of just its total energy. Unlike the linear case though, we won't have to deal with a varying speed.

The violation of the fifth assumption has two implications. Firstly, while the projectile is circulating, the whole accelerator structure ought to also circle due to conservation of momentum. This could simply be mitigated by having multiple smaller projectiles to balance things though. Secondly there is the linear recoil upon releasing the projectiles, which like in the linear case will sap some energy and so manifest as an inefficiency.

I plan on in future returning to issues of accelerator design, for both the linear and circular cases, and considering arguments based on magnetic field strength for when that is the specific accelerating force. For now, I feel confident in the basic conclusion that circular designs are better than linear ones - a conclusion backed up by most real-world particle colliders using circular designs.

On not understanding circular motion.

I never did really understand circular motion, even in Newtonian Mechanics. I never got the more complex aspects of the mathematics, nor how the equations are derived. As a result I never understood simple harmonic motion, which is mathematically similar, and because I never understood that I never understood waves either.

And now, I'm stuck on the question I was going to tackle last weekend, of which ultrarelativistic and relativistic circular motion is a big part.

If you have any good resources on the matter please post them in the comments. Just a good explanation of how things are derived in Newtonian mechanics would help, then I might be able to follow the same derivation but relativistically.

For now, one of the best sources I've come across is a series of video lectures from the 80s called The Mechanical Universe. Unfortunately they're prohibitively expensive to buy because they're not sold in editions for personal use, only for showing in schools and stuff. They can, however, be readily obtained by copyright-infringing means.

Normal service will be resumed once I've sussed the physics.

On linear accelerators.

Here we shall determine the required power and size for a linear accelerator (whether of particles or massive objects) to deliver a chosen energy per shot.

As ever, a few assumptions to focus the problem and make life simpler:

• The accelerator fires discrete shots, not a continuous beam. If it's firing a massive object this will always be the case, while particle accelerators vary.
• The accelerating force is limited by compressive strength of the accelerator structure. This is similar to how I handled spacecraft acceleration limits previously. Tensile strength is usually lower than compressive strength so support in compression seems reasonable.
• The accelerator structure is a prism. Obvious choice assumption.
• The projectile is fired at ultrarelativistic speeds, by which I mean in excess of 0.95c. For a particle accelerator this is natural. For a projectile accelerator I've explained previously that I feel this is necessary for it to be worthwhile as a weapon.
• Recoil can be neglected. This will be true if the accelerator's physically connected to something with enough mass.

I'm going to always call what the accelerator fires a projectile, regardless of whether it's a solid object or a bunch of detached particles. At the speeds involved it actually won't make any difference to the terminal ballistics, as to quote xkcd's what if, "the bonds holding the sphere together are completely irrelevant, it’s just a collection of carbon atoms".

I'm also going to always be working in the reference frame of the accelerator's structure.

Considering the force to be exerted on the projectile by some component (an electromagnet, perhaps) that is physically supported by the accelerator structure being loaded in compression, then

F = σ_cA

Where σ_c is of course the compressive strength.

The general definition of force, valid in special relativity as well as Newtonian mechanics, is the time derivative of momentum. With constant force, then for a projectile with no initial momentum and final momentum p, accelerated over time t,

F = p/t

In the ultrarelativistic limit, energy E is given by

E = pc

The time available for the acceleration depends of course on the length of the accelerator. Since we are taking the ultrarelativistic limit the projectile velocity is approximately c, and thus for an accelerator of length d,

t = d/c

Putting all the above equations together, we can express the projectile momentum and energy as depending on the volume and compressive strength of the accelerator

V = Ad
p = Ft = σ_cAd/c

E = σ_cV

Power is the time derivative of energy. Since energy is directly proportional to momentum, the constant time derivative of momentum (constant force) implies the constant time derivative of energy, ie constant power. This is thus given simply by

P = E/t = Ec/d

For a worked example, I'll consider a desired yield of 1 megaton of TNT, with a diamond accelerator structure or cylindrical shape and a length of 1 kilometre.

E is 4.2 x 10¹⁵ Joules.
σ_c is 1.2 x 10¹⁰ Pascals.
d is 1000 metres.

V = E/σ_c = 350000 m³
P = 1.3 x 10²¹ W

A = 350 m²
r = 10.6 m

The accelerator radius is quite modest, indeed positively svelte, though the overall size and thus mass are considerable, the latter being over a million tonnes. The power required, however, is extreme, many orders of magnitude above anything humanity has yet created.

The consequences here are relatively simple. For fixed material properties, the energy of the projectile depends only on the accelerator's volume, or equivalently its mass. The power drawn, by contrast, is lower for a long slender accelerator than for a short and bulky one. Therefore, assuming that lower power draw is desired, linear accelerators should be made as long as possible, and even then systems need to be capable of extreme bursts of power. Of course if the weapon is desired to fire in the direction the spacecraft accelerates, the previously established limitations on overall craft length come into play.

If the first assumption is violated, a more sophisticated treatment will be required. A continuous beam, or at least one long compared to the accelerator, would I expect bring the power requirements down drastically while still being able to deliver considerable energy to the target.

If the second assumption is violated, the accelerator might be made more slender, but the power requirements will still be the same since they depend only on accelerator length and not on the details of its construction.

If the third assumption is violated, I suppose you'd have to use calculus to handle the varying forces. I'm not sure why a non-prismatic shape would be used though.

If the fourth assumption is violated, then again calculus will probably be required to handle the varying velocity, unless the speeds are low enough for a Newtonian constant acceleration treatment. For an accelerator of fixed length, reducing projectile speed while increasing mass to compensate and give the same final energy will increase the time for acceleration and thus lower the power requirements.

If the fifth assumption being violated, it will manifest as an inefficiency since some of the energy input goes into accelerating the accelerator itself rather than the projectile.

Next week I will consider the case of a circular accelerator, which will have advantages and drawbacks compared to the linear variety.

On superconducting cables.

Here we shall determine the power transmission capacity of a superconducting cable, as a function of the cable size and material properties.

Running late again. I always was bad with leaving stuff until the last moment then finding it took more work than I expected. Hey, somewhere in the world it is still Sunday night!

Even though superconducting cables have zero electrical resistance, they still have a critical current. Put too much current through and it will stop superconducting, and if it's not properly controlled this can result in damage. Along with the operating voltage, this will determine the maximum power that can be put through the cable.

Starting assumptions this time.

• Circular cross section to the cable. It's the obvious shape.
• The operating voltage is limited only by the insulation of the cable.
• The superconductor is Type-II. Type-II superconductors can carry current through their whole bulk, and as such have a well-defined critical current density in Amps per unit area. Type-I superconductors confine current to their surface and thus the critical current depends on a wire's circumference not its cross-sectional area. They also have much smaller critical currents.
• The superconductor has zero electrical resistance. This isn't a tautology; a phenomenon called flux creep allows Type-II superconductors to have low but non-zero resistance while still superconducting!

We will consider a cable with overall radius r, a fraction f of that being the superconductor, the remainder of the cable's radius being the insulator.

The critical current depends on the cross-sectional area of the superconductor and its critical current density J_c.

I = π(fr)²J_c = πf²r²J_c

The maximum voltage is limited by the dielectric strength, d, of the insulator. This is the electric field strength, measured in volts per metre, above which the insulator will stop insulating and current will discharge across it in a phenomenon called dielectric breakdown, permanently damaging the insulator if it is solid. The dielectric breakdown of air is familiar to us all as lightning.

The electric field strength across the insulator is simply given by the operating voltage of the superconductor inside divided by the width of the insulator. Maximum operating voltage is when this equals the insulator's dielectric strength.

d = V / (1-f)r
V = d(1-f)r = dr - dfr

The maximum power that can be put through the wire, which I'll call the power capacity, is then simply given by voltage times current. Multiplying the equations for current and voltage together, we get

P = πf²r³J_cd - πf³r³J_cd
P = πr³J_cd(f² - f³)

If f is zero, there is no superconductor and obviously no current can flow. If f is one, there is no insulator so the superconductor can't be at any voltage and no power can be transmitted. For some f between 0 and 1, f² - f³ will be at a maximum, and thus the power capacity will be a maximum. This will tell us how much of the cable's overall radius should be superconductor, and how much insulator, for best results. To find this maximum, we differentiate P with respect to f.

DP/Df = 2πr³J_cdf - 3πr³J_cdf²
DP/Df = πr³J_cd(2f - 3f²)

(Capital Ds have been used for differentiation to avoid confusion with the lowercase d used for the dielectric constant)

When DP/Df = 0, P is at a maximum, minimum, or point of inflection.

0 = πr³Jcd(2f - 3f²)
0 = -3f² + 2f)
0 = 3f² - 2f

Using the quadratic formula

The solution at 0 is of no interest, P is 0 there. The remaining solution, at f = 2/3, must therefore be where P is at its maximum. There,

V = dr/3
I = 4πr²J_c/9
P = 4πr³J_cd/27

For a worked example, we shall determine the necessary radius of a cable capable of carrying 15 terawatts, the average power consumption of all of humanity, using contemporary to near-future materials.

Jc = 10⁹ Am^-2. Or 10⁵ Acm^-2. Various groups have reported values of this order of magnitude for different superconductors.
d = 10⁸ Vm^-1. The value for Teflon, among the best insulators.
P = 1.5 x 10¹³ W, as mentioned.

r = 0.069 m

A wire 14 centimetres across could carry the entire world's power consumption; testament to the capabilities of superconductors.

As ever, the equations have consequences, which are in fact what we're mainly interested in. The key factor is that the power capacity of the wire scale with the radius cubed, not squared, because a wider cable not only carries more current but also operates at higher voltage. A corollary of this is that it's better to use one big cable than lots of small ones with the same combined mass.

A second notable consequence is that there are gains to be had not just by improving superconductor technology but also by making better insulators.

A third consequence comes from a bit of knowledge about superconductors. Jc is temperature dependent; at the superconductor's critical temperature, the warmest it can superconduct, it approaches zero, while at lower temperatures it is higher. Thus, even given a much-vaunted room-temperature superconductor, for maximum power transmission it would still require cryogenic cooling.

If the first assumption is broken, you need more insulation to carry a given amount of power. It becomes rather a case of why would you do that?

If the second assumption is violated, and other factors limit the voltage, then for large cables the two-thirds superconductor, one-third insulation ratio will not be held, the insulation can be thinned due to the reduced voltage. Consequently the power capacity will then scale approximately with r squared, not r cubed. The drawback to using multiple smaller cables will be lessened, though not wholly eliminated as they will require more insulation than a single large one.

If the third assumption is violated the situation would considerably change. For this to happen though would require a novel Type-I superconductor with an unprecedently high critical current, or else something entirely unanticipated.

If the fourth assumption is violated we may have to consider power losses in the wire that would create inefficiency, as well as the implications for cooling system of heating in the wire. The exact impact of these factors I am unsure of, though qualitatively I expect larger cables would be required to carry the same current.

For some background information, the Open University has a free e-text on superconductivity. It focuses primarily on Type-I superconductors, which were historically discovered earlier and are better understood but have fewer practical applications than Type-II superconductors.

On effective range.

Here we shall determine the effective range, meaning the range at which you can hit what you are shooting at, of a weapon in space.

I'm planning on posting updates to the blog weekly, every Sunday night, for as long as I have material to work with. This one is a bit late. If you have any questions you'd like me to tackle, post them in the comments or telegram me on Nationstates.

To business. Our starting assumptions:

• The weapon's intrinsic range is much greater than its effective range. This is valid for solid projectiles, though may not be for realistic lasers or particle beams which diverge.
• The weapon shoots one bullet at a time. We aren't considering shotgun-like approaches.
• The bullets travel in straight lines and don't have any active manoeuvring systems. We aren't considering guided munitions.
• The weapon is intrinsically precise, it won't miss a non-moving target.
• The defender can change their acceleration without delay.
• The attacker and defender both have effectively-instant FTL scanners. This is not uncommon in science-fiction.

The scenario is thus simple. The attacker fires, aiming the bullet so it would strike the middle of the defending ship. The defender immediately detects this and takes evasive action by accelerating. If the defender can completely vacate the space it was occupying when the shot was fired, the shot misses, while if the defender is still in that place then it is a hit. We shall use the reference frame of the defender, prior to their taking the evasive action. The below diagram depicts a miss, the defender having just managed to move half its own length for the shot to pass harmlessly behind it.

With the projectile travelling at velocity v and having to cover distance r, the defender has a time to evade

t = r/v

To determine how far the defender can move within that time, we can use one of the SUVAT equations for uniform acceleration situations to find that distance

s = ut + at²/2

u, the initial velocity, is zero since we chose our reference frame so that would be the case. Therefore, substituting the first equation into the second, we get

s = ar²/2v²

If

s < l/2

Then the projectile will hit. Substituting and rearranging, we get

This I refer to as the "Range Equation". If the target is closer than r, it will be hit. If not, it will be missed.

For a worked example, consider the 1 km long ship capable of 100 g of acceleration from the previous post, being targeted by a projectile moving at 0.95c. Converting to SI units

v = 2.85 x 10⁸ ms^-1
l = 1000 m
a = 981 ms^-2

r = 2.88 x 10⁸ m

So the target can be reliably hit if it's as far as 288 thousand kilometres away. For context, the average Earth-Moon distance is 385 thousand km.

The range equation has various consequences. Most obviously, smaller targets can get closer to an attacker safely than larger ones. Critically, if two ships are in one-on-one combat, both using weapons that fire their projectiles at the same speed, the smaller one can sit at a range where it will always hit its enemy while never being hit itself. Compactness is thus advantageous, something only emphasised by the previous posts's result that shows a more compact ship will also accelerate harder.

Also, the faster the projectile can be fired the better. Laser shots will of course go at light-speed, and particle beams won't be far off, so massive projectiles need to be doing relativistic speeds or have other advantages in order to compete.

Increasing acceleration while holding size the same, for example by advancing technology, has a limited benefit compared to reducing size and thus benefiting from the natural consequent acceleration increase.

Finally we can tell that the intrinsic precision needs to be pretty precise, to well under a second of arc. The modern Hubble Space Telescope is capable of pointing with a precision of 0.01 arc seconds, so this is not an especially demanding requirement.

If the first assumption is violated, and range is actually limited by the intrinsic behaviour of the weapon, the argument becomes moot. A consequence of such a situation is that large ships can now get just as close to their attacker as smaller ones. It's also possible for the range equation to apply to small targets but not to large ones, creating a two-regime situation. This would likely result in the 'borderline' area being vacated as ships would either be larger for more general capability, or smaller for being hard to hit.

If the second assumption is violated, then the geometry needs to be considered in more detail. The overall conclusion that smaller ships are harder to hit - which is the intuitive result after all - will probably remain intact though.

If the third assumption is violated the argument becomes moot. Guided munitions are likely to have much longer effective ranges. However they may have other drawbacks.

If the fourth assumption is violated then the imprecise pointing will result in probabilistic concerns, but the overall effect is likely to be similar to limited intrinsic range.

If the fifth assumption is violated, then a 'reaction time' delay must be incorporated into the equations, making them more complicated.

If the sixth assumption is violated, then the defender of course cannot actively dodge the incoming fire if it's travelling at or near light-speed. However, since the attacker's information will always be out of date, the defender can make random manoeuvres. If they are within a range half that given by the above equation, they will surely be hit (the derivation of this will come in a future post). Beyond that, the situation becomes probabilistic instead of a certain miss.

On maximum acceleration.

Here we shall determine the maximum possible acceleration of a science-fiction spacecraft.

A few starting assumptions shall be made.

• The spacecraft is accelerated by reaction thrusters at its stern.
• The reaction thrusters can be built to produce as much thrust as desired. We shall not worry about exactly how they work, just treat them as black boxes.
• The spacecraft can be approximated as a simple prism of material with a certain compressive strength.

Since the thrusters can be as forceful as we like, the limit on spacecraft acceleration comes from its structure. If the thrusters push too hard, they will crush the spacecraft they are supposed to be pushing.

A spacecraft in deep space subject to a force giving it an acceleration a is an analogous situation to a spacecraft sitting bow-skywards on a planet with surface gravity g=a.

As such, the maximum height of a column you could build on that planet is equal to the maximum length of a spacecraft you could build to withstand that acceleration. A taller column will fail by crushing. This height can be derived as follows.

σ_c is the material compressive strength
ρ is the material density
a is acceleration
h is column height
A is column cross-sectional area

The mass of the column is given by

m = Ahρ

And the pressure at its base by

σ = ma/A = hρa

The cross-sectional area unsurprisingly cancelling out.

Obviously the pressure at the base of the column is greater than at any higher point. If this pressure is less than the material compressive strength, the column - or the spacecraft - will not fail by crushing. It may still fail by buckling, but buckling requires lateral deflection. As such, I feel it can be prevented by an actively-controlled restoring force to counter the deflection before it reaches failure, or by just not making the spacecraft too slender.

Rearranging to give acceleration as a function of the other variables,

a < σ_c/ρh

This, then, is the maximum possible acceleration of a science-fiction spacecraft.

A real spacecraft will not be a homogenous block of material. To treat it as such, we can calculate its average compressive strength and density as follows.

f is the volume fraction of the spacecraft that is structure
ρ_s is the density of the structure material
σ_cs is the compressive strength of the structure material
ρ_f is the mean density of the functional parts of the spacecraft, ie everything but its structure
The compressive strength of the functional parts is assumed to be negligible

The average compressive strength and density of the whole spacecraft are given then by

σ_c = fσ_cs
ρ = fρ_s + (1-f)ρ_f

For a worked example, consider a spacecraft with the following properties

f = 0.1, ie 10% of the craft is its structure
ρ_s = 3500 kg m-3, ie diamond
σ_cs = 12 GPa, again diamond
ρ_f = 1000 kgm-3, same as water, just feels like a good value
h = 1 km, feels like a good size for a fairly large spacecraft

Then
σ_c = 1.2 GPa
ρ = 1250
a < 960 ms-1

So the maximum acceleration is about a hundred Earth gravities.

If the shape is a pyramid or cone rather than a prism, the exact formula may change, I am uncertain on what to, but the general form will remain the same.

This limit implies various consequences. Most obviously, smaller spacecraft are capable of greater accelerations than larger ones, as most would intuitively expect. Also, to obtain maximum acceleration with a given spacecraft volume, a relatively flat spacecraft, for example a classic flying saucer, will be superior to the slender designs often seen in science fiction.

If the first assumption is violated, the entire argument can cease to hold. There  is unlikely to be much benefit from trying to place engines along the flanks of the spacecraft, material failure will still occur but in shear rather than compression. However reactionless drives that create a field acting on the entire bulk of the spacecraft, or on the space it sits in, will completely nullify the equations here. With such reactionless drives, small and large craft might accelerate equally, or large ones could even accelerate harder. Shape may not factor into acceleration performance, allowing it to be determined by other considerations.

If the second assumption is violated, the specific equations become invalid, but the generic square-cube law still indicates smaller spacecraft can probably accelerate harder than larger ones, and a flat craft still has more space on its surface for engines.

If the third assumption is violated, the entire argument again may cease to hold. One way to accomplish this is for the spacecraft to not rely solely on a physical structure for its strength, but to use dynamic support methods such as some sort of forcefield to transfer the thrust from the engines forward to the bows. Such technology may reverse the situation and make long slender spacecraft actually advantageous