Wednesday, 2 August 2017

Liquid Droplet Radiator Inter-Reflection

We follow up 'All the Radiators' with two questions on Liquid Droplet Radiators: How do we take into account inter-reflection in a droplet sheet? How does it affect the radiative efficiency?

Introduction to the task

Currently available calculators for the performance of liquid droplet radiators (LDRs) do not take into account inter-reflection between droplets.

They assume that the droplet sheet is only one droplet thick. They do not take into account the fact that some of the energy radiated by one droplet will be absorbed by an adjacent droplet. 

We'll make several assumptions here:
-The droplets are perfectly spherical and can be perfectly formed down to 1um diameters.
-Droplets are nearly black (emissivity about 0.8).
-The droplet sheet have zero vertical variance and droplets are perfectly spaced.
-External effects are ignored.

With these assumptions, we can focus on the efficiency losses in droplet sheets.

It is known that regular radiators that radiate away waste heat, placed in close proximity to each other or facing each other, fully or partially, are less than 100% efficient. This is because some of the radiated energy is intercepted by the other radiator, an effect known as inter-reflection.

An emissivity rating of 0.8 also means an absorptivity rating of 0.8: Kirchhoff's Law of thermal radiation. This means a radiator that effectively radiates waste heat at a specific wavelength is equally effective at absorbing that wavelength. 

All of these apply to droplets. When one droplet radiates away energy, it does so in all directions. Some of these directions are occupied by other droplets; they will absorb the energy and lead to a loss of overall efficiency.

We will now try to quantify this loss of efficiency. 

A tale of two droplets

We will start off by considering two droplets in isolation. 

The droplets will have the following properties:
-Diameter D and radius R
-An inter-droplet separation S
-An emissivity E
-A temperature T

Since we are dealing with small, hot droplets, usually of liquid metal, the values D, R and S will be expressed in micrometers (um, 10e-6 meters). E is between 8.0 and 0.99. T is in Kelvin.

From those properties, we obtain:
-Droplet surface area A
-Projected surface area P
-Radiative power emitted W

A and P are in um^2. The projected area is the surface area of a sphere, centered on a droplet, but with a radius equal to the separation S. W, in watts, is evenly divided across the droplet's surface area A. The projected sphere is how the droplet's heat is spread if we consider it to be radiating point source.

So, what do we do with all of this information?

Picture looking up at the night sky. How much of the sky does the Moon occupy? We are trying to answer a similar question with the two droplets. How much of one droplet's 'sky' is occupied by the other droplet? 
Droplets A and B. In grey, the droplet sheet's plane. In light red, droplet A's projected sphere.
To work this out, we must first find out the surface area of the projected sphere at the second droplet's distance. Using the diagram above, generated by Geogebra's graphical calculator, we determine that droplet B's cross section, divided by the proected sphere's surface area, has the same ratio as the energy re-absorbed by droplet B divided by the total energy emitted by droplet A.

Droplets A and B have a diameter of 1um. 

Both have a surface area of 3.14um^2 and a cross section of 0.79m^2.
The radius of droplet A's projected sphere is the distance A<-->B.
A<-->B = 0.5 + 10 + 0.5 = 11um
The surface area of the projected sphere is 1520um^2.
Droplet B occupies 0.79/1520 = 0.052% of droplet B's 'sky'.

So, we calculate that the inter-reflection between two droplets of 1um separated by 10um is so small as to be negligible. 

Let's calculate the performance loss. We must take into account emissivity.

0.05% of each droplet's radiations are intercepted by another droplet.

80% of the inter-reflected radiations are re-absorbed. 
Only 100 - 0.05*0.8 = 99.96% of the waste heat reaches empty space. 

If we repeat the same calculations, but assume that the separation S is only 5 or 1um, then we get the following list of values:

Separation: 10um

Efficiency: 99.96%
Separation: 5um
Efficiency: 99.86%
Separation: 1um
Efficiency: 98.75%

What if we included even more droplets?

Droplets A, B and C are aligned and in the same sheet
If we align a third droplet C, behind A and B, it will not absorb any radiations from A. It is in droplet B's 'shadow'. Droplet C has no effect on droplet A's efficiency, and we can suppose the same applies for any droplet in C's shadow and so on.

Using all this data, we can verify that the loss of efficiency caused by inter-reflection by adjacent droplets follows this equation:

  • Loss of efficiency = [R^2 / (4*(D + S)^2)] * E * 100
Again, R, D and S are in micrometers.
R is the droplet radius, D is the diameter. S is the separation between droplets. If we input R=0.5, D=1, S=10 and E=0.8, we obtain 0.04%. 


A droplet sheet contains hundreds of thousands or even millions of droplets. We have assumed above that they are evenly spaced.

An even spacing of 10um requires that each droplet is separated from other droplets by at least that distance. Achieving this arrangement requires a specific 'tiling' of the surface area available: geometric tessellation can describe this tiling. 

Regular, equilateral triangles with droplets at the vertices.
The most compact arrangement is triangular tessellation. 
The simplest to create is squares, but it is less efficient with its use of space. Tessellations with other geometric shapes are not able to pack as many droplets per square meter as the triangle or square. 

To calculate the efficiency of a droplet sheet with each arrangement, we must consider each droplet within it's 'tile'. 

Here is a triangular tile: 

Droplet arrangement in triangle tessellation.
The droplets are labelled A to S. Droplet A is radiating waste heat, represented by the red circle. The first ring of droplets to intercept these radiations are droplets B to G. They are separated from A by a distance S. A further ring of droplets should be considered, labelled here H to S. They are separated by S * 2.

Continuing our example from the previous section, each droplet is 1 micrometer wide and the separation distance is 10 micrometers, with an emissivity of 0.8. 

The equation which determines the loss of efficiency between droplets A and B is the same as in the previous section. However, we must adapt it to take into account the fact that there are six droplets within droplet A's projected sphere. 

[(6 * 0.5^2) / (4 * (1 + 10)^2)] * 0.8 * 100 = 0.25%

The loss of efficiency is six times greater at 0.25%.

Let's add the inter-reflection of the droplets H to S. We must note that droplets H, J, L, N, P and R are directly behind the first ring of droplets and therefore do not affect droplet A's efficiency. They are also twice as distant.

[(6 * 0.5^2) / (4 * (1 + 20)^2)] * 0.8 * 100 = 0.07%

We assume that third, fourth and further rings of droplets will have a totally negligible effect on efficiency. In a triangular tesselation, 1um droplets separated by 10um have an efficiency of 99.68%.

Here is a table of values for separations of 10, 5 and 1um:

Separation: 10um

Efficiency: 99.68%
Separation: 5um
Efficiency: 98.98%
Separation: 1um
Efficiency: 91.3%

For a triangular tesselation, the loss of efficiency equation for the first two rings is:

  • Loss = [6*R^2/(4*(D + S)^2)+6*R^2 /(4*(D+2*S)^2)] * E * 100
Here is a square tile:
Square tessellation droplets.
The droplets are labelled A to I. Droplet A is radiating waste heat. The first droplets to intercept radiations from A are droplets B to E. They are separated from A by a distance S. Droplets F to I are separated from A by S*1.412.

Using the same equations as for the triangle tessellation, adapted for squares.

  • Loss = [R^2/((D + S)^2)+ R^2 /((D+1.412*S)^2)] * E * 100
Using R=0.5, D=1, S=10 and E=0.8, we calculate a loss of efficiency equal to 0.25%.

Neglecting the influence of further droplets is justified as they only contribute less than 0.1% to efficiency losses.

Here is a table of values for separations of 10, 5 and 1um:

Separation: 10um
Efficiency: 99.74%
Separation: 5um
Efficiency: 99.13%
Separation: 1um
Efficiency: 91.6%

As for comparing between the two tessellations, you can pack twice as many triangles as squares into the same area. As the triangle has one less vertex than a square, this translates into 50% more droplets in the same area. However, creating the triangular arrangement requires a third nozzle in the droplet sprinklers.

200um droplets spaced by 1000um would suffer an efficiency loss of 2.86%, in a square configuration. This indicates that for small droplets with reasonable spacing, inter-reflection is a small influence on waste heat removal capacity.  

Two sheets

What if we stacked two sheets of droplets on top of each other? This would double the number of droplets radiating. If the inter-sheet separation if equal to the inter-droplet spacing, then we could obtain a massive increase in waste heat removal capacity with only a small increase in the volume and equipment required.
Two square tiles in two sheets stacked above each other.
Side view of the two droplet sheets as parallel planes.
In a two-sheet arrangement, we must consider the inter-reflection between droplets both adjacent and above the radiating droplet. 

Let us calculate the efficiency loss for droplet A. We already have equations for the inter-reflection between droplets in the same sheet. 

To continue with the previous examples, the droplets are 1 micrometer wide and separated by 10 micrometers. 

Droplet J is directly above droplet A and is therefore separated by 10um. Droplets K, L, M and N are diagonal to A and the separation is 1.412 times greater; 14.12um. The remaining droplets O, P, Q and R are 15.53um distant from A.

Putting all these together, we calculate that K, L, M and N reduce efficiency by R^2 /((D+1.412*S)^2) * E * 100 : 0.087% and O, P, Q, and R contribute to a loss of R^2 /((D+1.553*S)^2) * E * 100 : 0.073%. Together with the losses due to same-sheet droplets, we obtain a total of 0.41%.

An efficiency drop from 99.74% to 99.58% is insignificant compared to a doubled output from having two droplet sheets instead of one. 

Several sheets deep

So far, the efficiency losses have been nearly negligible, always less than 1%. What about droplets in the middle of a droplet cloud that is is dozens of sheets thick?
Most illustrations of liquid droplet radiators depict them as rather thick and not one sheet thin. 
One might assume that the far greater number of droplets present in the cloud would create an obfuscating wall around the innermost droplets. Could it be that only the droplets at the edge of the cloud radiates effectively?

This is not the case!

Let's consider a droplet inside an a very thick cloud of droplets - perhaps 100 sheets deep or more. 

Droplet shadows is a significant factor in this case - only the first droplet in a line of droplets absorbs radiations, all those in its shadow do not contribute to efficiency losses. 

Imagine the expanding spheres as the radiating waste heat, with the horizontal lines representing droplet sheets.
The most important configuration is an octahedron. The closest six droplets absorb more radiation than any number of droplets further away and are the biggest contributors to efficiency losses. 
The droplets intercepted at one separation. The projected sphere is three sheets tall.
If we increase the number of sheets and droplets intercepted, we obtain significantly lower contributions to efficiency despite the increasing number of absorbing droplets. In the octahedron above, there are six droplets. In the cube below, 27 droplets are intercepting radiations but the innermost six droplets contribute 
Here, we have included the diagonals, which form a cube around the previous octahedron.
There are eight diagonal droplets per sheet intercepted, four at 1.412*S separation and four at 1.553*S separation for a total of sixteen. 

A section of the droplet cloud.
It could be surmised that the loss of efficiency in a droplet radiator due to inter-reflection will depend on the number of droplet sheets and how many intercepted droplets they represent. However, a few calculations represent a simpler picture.

The projected sphere encounters more and more droplets as we increase the number of layers. Let's call the number of layers N. N is an always an odd number because we count the number of sheets above and below a central sheet. Therefore, N=5 corresponds to a cube of dimensions 4*S x 4*S x 4*S. 

4x4x4 cube in isometric view. 
The number of droplets intercepted is 26 at N=3, 124 at N=5, 342 at N=7 and 728 at N=9. This indicates that the intercepted droplets follows a patterns of [(N)^3 - 1]. 

The number of new droplets is we encounter as N rises is actually smaller. It should follow an [(N^3 - (N-2)^3)] pattern; basically, the difference between the larger and smaller cubes from 2-sheet steps.
At N=5, there should be 98 new, at N=7 there'll be 218. This is a much slower progression.

Finally, we must find out which of the new droplets are shadowed behind the previous droplets. We can find this simply by deducing the number of new droplets by the number of new droplets from the previous step.

For example, if I wanted to calculate the new non-shadowed droplets in a droplet cloud 7 sheets deep (N=7) compared to a cloud five sheets deep (N=5), I'd first need to know the number of droplets creating shadows in N=3. 

This creates a series. It is best to use an Excel sheet to handle series, as they cannot be properly represented by an equation:
N=3 gives 26 shadows. N=5 has 98 new droplets. 98-26: 72. So, N=5 creates 72 new shadows. For N=7, we calculate 218 new droplets. We deduct 26 shadows from N=3 and 72 shadows from N=5 for a total of 120 new shadows. 
Seems to be 11 sheets thick.
How do we use the number of shadows?

Well, we can estimate the efficiency of a droplet radiator as the number of sheets increases. In a cubic tessellated volume, in other words a grid, the distance between a radiating droplet and an inter-reflecting droplet in a sheet N is at a minimum equal to S*N, at a maximum at 1.553*S. S*N represents a droplet directly above the radiating droplet. 1.553*N represents a droplet absorbing radiations from the corner of a cube. 

Any other position occupied by any droplet is further away than S*N or closer than 1.553*N*S.

Using these distances, we can restrict the range of efficiency losses caused by inter-reflected at any number of sheets.

Inter-reflection efficiency loss estimates 

At a distance S*N, a droplet covers a percentage of the radiating droplet's sky equal to (R^2)/(4*(D + S*N)^2)*E*100.

We multiply the percentage by the number of new shadows to obtain the maximum losses caused by the droplets in the sheet. 

At a distance S*N, a droplet covers a percentage of the radiating droplet's sky equal to (R^2)/(4*(D + 1.553S*N)^2)*E*100.

We multiply the percentage by the number of new shadows to obtain the minimum losses caused by the droplets in the sheet. 

Here is a table of values for the percentage of efficiency losses caused by droplets of 1um diameter with a separation S=10um and an emissivity of E=0.8.

We find that the biggest contributors to efficiency losses are the first few sheets. Despite the increasing numbers of droplets, sheets further away contribute less and less to inter-reflection losses. Even at 100 sheets, losses are at 6%.

Here is a table of values for the percentage of efficiency losses caused by the same droplets above separated instead by 5um. 
We see that the losses are quite a bit more significant. The increases in inter-reflection start petering out after 50 sheets, with the next 50 sheets contributing only 3% to the accumulated average losses. Losses are offset by the droplets being packed 8 times more densely. 

Here is a table of values for the percentage of efficiency losses caused by the same droplets above but separated now by 1um. 
An accumulated average of over 100% means that all of the radiating droplet's radiations are blocked and absorbed by the surrounding droplets. At 9 sheets deep, we can confidently say that the central sheet contributes nothing towards the droplet cloud's performance. 

Do note that these tables are calculating the efficiency of the innermost sheet only. The average losses once sheets closer to the edges of the droplet cloud are considered is lower than what the accumulated average suggests. 


For properly spaced droplets, where the separation is much greater than the droplet diameter, the efficiency losses caused by inter-reflection is quite negligible.

However, for densely packed droplet clouds, inter-reflection becomes a significant issue that greatly reduces output despite the greater number of droplets.


  1. This is actually interesting. A radiator shaped like a cube or cylinder is much more compact compared to a standard "wing", and the shape of the ship can be more compact as well. This may not save much on mass, but a compact ship should be much easier to manoeuvre without large booms or radiators.

    And of course colonies and other structures may benefit from this as well.

    1. Quite true. In the 10um spaced radiators, you can get clouds of several hundreds of sheets thickness and suffer only about 7% inter-reflection losses. It would still be less than a centimeter thick and yet it would pack so many more droplets.

      The best part is that you'd only have to use a single spraying and collecting arm.

  2. I think there is one factor that you overlooked: you take into account the energy that is being asorbed by other droplets, but I do not see anywhere where you take into account the increased emission by those other droplets.
    Remember that emission is dependent upon temperature. When a droplet absorbs energy from neighboring droplets, it effectively increases the temperature of that droplet... and the emission increases at a factor of T^4 with temperature.
    Of course, factoring this in would probably require factoring in specific material used for the droplets. This is because the actual temperature increase would be dependent upon the heat capacity of the material.

    Something to keep in mind: the situation we are describing is essentially exactly what happens when dealing with a solid or liquid material. Note that the blackbody radiation of a material is independent of its thickness. The situation would be different if there were a hot core layer that is trying to emit through cooler outer layers (a situation equivalent to wrapping a blackbody source with an insulator), but as is, we are dealing with an essentially homogenous heat source.

    What is actually critical is that the given available radiative area will be a constant. To a certain extent, the density of the droplets will be low enough that the system will act AS IF it were a monofilament, regardless of how thick it actually is. However, there is a limit (specifically, the number of droplets you would actually be able to fit into a real monofilament sheet) beyond which you are no longer creating greater emissive area by adding more droplets. This would then be the equivalent of just making a thicker blackbody: same emission per surface area exposed, but more mass.

    1. Quite right, more steps would be required to work out the 'effective' efficiency of a droplet radiator when we consider that 'inefficiency' can be translated into droplets staying hotter for longer.

      If we take the case of a 93% efficient radiator, then each droplet eventually re-absorbs 7% of the energy it emitted (after bouncing back and forth between not-perfect-blackbody objects).

      A simple fix would be reduce the emissivity.

      A single droplet might have an emissivity of 0.8, but inside the cloud the effective emissivity would become 0.73: this would account for efficiency losses while allowing us to use a straightforward drop in temperature.

  3. Looking at the entire radiator as a single object, we can look at it from any direction, choose any single point too look at and see if there is a droplet radiating toward us there.
    The two extreme examples are an empty radiator with no droplet, where no point radiates, and the full radiator, who is so dense that any point radiates. The second one is equivalent to a solid radiator.
    As such, it should be possible to describe any droplet radiator configuration (droplet size, distance, positioning) and assign it a solid radiator equivalent percentage.

    1. This only works for perfectly distributed droplet clouds where all droplets maintain a distance S from each other. This allows us to scale up and down their density of distribution seamlessly.

      In a random distribution, some droplets are closer, others father, and the distances average out. However, droplets closer to each other are disproportionately inefficient compared to how much more efficient droplets farther away from each other are: the efficiency losses do not average out like distances do.

      Nonetheless, it is a nice idea that might be worked into a calculator.

  4. I am not necessarily saying that there would not be ANY loss of efficiency due to droplet absorbtion. However, I suspect that any such loss would be much lower than your initial projections.

    Again, once you have more droplets than the number that you could actually fit into an actual monofilament sheet, there WILL be a definite loss of efficiency, equivalent to having a thicker blackbody block.

  5. A design idea I ran across on Worldbuilding Stack Exchange, which covers many of the ideas talked about on this blog, but with some interesting twists:

    "I think that a space fighter would in a way act more like a submarine than a modern warplane. Paint the thing black and it'll be hard to see. It'd be shaped to scatter radar reflections in all directions, since aerodynamics would be unimportant. Think of an anechoic chamber turned inside out. And anything emitting radar would be making itself much more visible than the target (of course, the radar emitter could be on a drone, but then the fighter could be a drone too).

    It'd want to keep its heat signature down, because that's always going to be broadcasting its location to passive sensors. I'd imagine a liquid droplet radatior with deployable arms, and using a railgun with explosive projectiles as its armament. The explosives wouldn't be to make a bigger hole in the target, but to turn the projectile into shrapnel for a better chance at a hit.

    Also, presenting more mass toward the likely location of the enemy would serve two purposes -- one, as armor of a sort, and, two, it'd be nice if that could be used to shadow the radiators from enemy heat sensors. We might as well keep the propellant there, as water or hydrogen slush.

    So, I'd say a spiky mushroom with a long stem, periodically with what appear to be wings sprouting up behind the head when it seems safest to deploy the radiator."

    Using multi layered droplet curtains allows for smaller radiators behind the "mushroom cap", but the "stem" might have to be shielded by dielectric mirrors to reflect the IR radiation away from the ship.

    Somewhat related is the newest entry in Atomic rockets: In-A.22 Nikto-ega class interceptor.

    A pulse fusion ship designed to carry a brace of (presumably) NEFP missiles, and having a secondary particle beam weapon as a CIWS looks a lot like the "Firestar" concept. While I am somewhat dubious of the author's projected employment of the ship, tactics are dependent on your assumptions. I would think of a Firestar built along these lines as being part of the outer layer of a constellation or moving in independent battlegroups, but YMMV.

    1. I think the importance of detection and defense are reversed in the quoted comment.

      The space fighter being described is no Hydrogen Steamer. It will have surfaces bright and hot enough to ensure detection through passive emissions alone, something an 'anechoic chamber turned inside out' won't help with.

      Droplet radiators are also very hard to hide from any direction. You'd have to use something like a membrane covering all the droplets, half opaque and half transparent... but you'd run into the problem of cooling down the opaque half. This is ignoring the fact that it is very simple to place sensor behind your spaceship and get a strong signal from the exposed side.

      Since detection is so hard to avoid unless we rely on very specialized designs or plot-dependent situations, it is better to rely on defense. This means concentrating armor at your opponent. A spinning, sharp cone has the highest defensive capability per kilogram against lasers. A layer of reactive armor plates is most efficient against high velocity projectiles.

      If we place the radiators edge-on behind the widest point of the cone, we'll end up with something like a flechette shape. Bonus points for looking cool.

  6. From re reading the comment, the author is obviously looking at the idea of making the ship difficult to "see" in the microwave/radar spectrum, and painting it black should make it difficult to see in the visible light spectrum.

    Of course, making it invisible in the infra red spectrum is not being considered at all in this design. The only thing that I could think of to make it more difficult to target is placing a metamaterial "shield" in front of it which refracts IR radiation. An umbrella shaped shield held in front of the craft and designed to refract IR radiation in a large open "cone" (think of the cones used around an animal to keep them from scratching themselves) would provide a rather strange"view" to any IR target seekers coming from the frontal arc. Like most things, this is not a 100% solution, and in the sorts of space battles where the sky is fun of target seeking drones and SCoDs (not to mention laserstars) this would not provide a great deal of protection at all.

    1. Refracting the IR emissions still means that some percent of the emissions end up straight in the enemy's CCD sensors. A statistical analysis of the concentration and distribution of 'hits' on the sensor's pixels will give a 'soft detect' of any target, as described in Stealth in Space is Possible IV.

      What the metamaterials could help with is avoiding a long-range 'target lock' using passive sensors only. A laser beam would have to be focused on your ship, forcing your enemy to in turn reveal themselves.

    2. In modern stealth tech, you actually don't want to scatter radar in all directions, as this allows detectors in any irection to pick up the radar energy. No matter how weak that energy is, there are means to detect that energy if it comes in your direction. Of course, there are both benefits and detractors with any given tactic for detection, so picking the wrong one could easily cause the energy to be missed... but this is straying from my point.
      Modern stealth, does not shape forms to radiate energy in all directions, but to radiate in specific directions away from the source. The first step is to use mostly flat forms in such a way that incident angles naturally reflect the energy away from the source, except when that source is emitting orthagonally to the reflecting plane... wherever possible, the planes are then angled so that any such orthagonal points are in places where interception (if detection does occur) would be difficult. From here, the forms are further shaped to bounce the reflected signal in such a way that it is virtually impossible to get an orthagonal angle to a surface. For example, the SR71 used concave triangular sections to direct radar away from a source in controlled directions.
      There are, of course, stealth techniques designed to scatter radar. However, these are used in radar absorbant materials that scatter the signal INTERNALLY. This spreads out the radar return signal over time, preventing accurate ranging. This is also combined with shaping so that when the signal energy IS emitted, the emission can be directed, in a controlled manner, away from the source signal.

    3. The problem with trying to find a neat-cut solution to how to respond to incoming energy and how to handle outgoing emissions is that it crashes head-on with the obstacle that is author choice.

      For example, in a setting where the two opponents start far away from each other and everything behind the space forces is 'safe', we can imagine an artificial horizon where radiating in front of you is bad and radiating behind you is safe.

      That sort of thing lends itself to directional radiator and hulls shaped to deflect RADAR and LIDAR in one direction instead of another. It is the situation we live in, where there are multiple horizons, both natural and artificial.

      However, an author can just as easily create a setting where sensors are all over the place, or everyone has allies and neutrals reporting the position of the space forces. There is no horizon, the enemy can be looking from anywhere and attack from a wide range of directions.

      This time, directional tactics would not be very useful, as you cannot know where to point your emissions if you don't know where the enemy is, and you cannot know where the enemy is without revealing your own position. So, it is best to simply reduce your emissions to gain some level of surprise on strategic scales (where did your mid-course burn move you?) even if you cannot hide you on tactical scales (evading a missile lock).

    4. It appears that might original reply to this got lost, so here it goes again.

      First: stealth never relies on absolutes. There is ALWAYS a non-zero chance of detection. Thus, even though there is no guaranty that there will be no one in the direction of reflection, those who employ stealth will still take the risk.
      There are actually a number of factors that are taken into consideration. These include probable placements, known or projected sensitivities/signal strengths, etc. radiating in all directions might be prefered if you KNOW that enemy platforms have a given sensitivity that you KNOW is above the reflection amplitude at a given intended range of operation. However, far more often, planners will opt to take chances on minimalised windows of peak reflection, in exchange for virtually no reflection in other directions.

      Second: you are quite correct that it is not possible to be certain where all the detection platforms are going to be. But that works in both scenarios. You never know how close a platform is going to be, either, and detectability is a function of distance (among other factors, of course). The one thing you CAN be certain of: if you reflect at all angles, you WILL be reflecting toward at least one detector platform, including the source... and that can be very bad.

      Third: you ABSOLUTELY do NOT want to reflect toward a source. If the source can detect the reflected signal, the enemy will have presence, direction, AND range information... possibly velocity as well. However, if a non-source detector DOES receive the reflected signal, all it will have is presence, and a more or less accurate estimate of direction. It will have zero range info.


    5. (cont)

      Fourth: "all over the place" is relative. In terms of resources, it is highly unlikely that it would be possible to establish a network of platforms of the size you suggest. It is PHYSICALLY impossible to establish a network that would be able to detect an asset regardless of reflection angle, with no gaps.
      Let's assume that an asset has an unreasonable reflection scatter of 1°^2 (that is, the peak, or near peak, energy levels will be as strong 1/2° off the norm as it is along the norm itself). You would already need over 40 000 platforms surrounding the asset, WITHIN detection range, for a given location in space (these platforms will have to cover every square degree projected from the asset). The problem is, as the asset drifts from the geometric centre of this array, there will be gaps to exploit... to say nothing about assets that are outside this array. The asset will be detectable only so long as there is a detector in EVERY 1°^2 window.


    6. I will answer your points. I must apologize for not always responding to the comments you have sometimes made on Google+ or on this blog.

      I must note that reflecting a signal from an enemy's active sensor will only be a small part of the detection game. Only two likely scenarios involve handling an incoming signal: a hard target lock for weapons fire after a strong confirmation of your location and identity has been obtained through passive sensors, or if the enemy if blindly waving a laser around like a flashlight without fear of being attacked by any number of 'silent' passively-sensing opponents.

      In the first case, throwing off the active sensor's signals will only lead to a worse targeting solution. In the second case, we would be dealing with a very assymetrical conflict where being detected is the least of your worries!

      The scatter numbers you have selected for the example in your second comment seem unrealistic. Active sensors will use RADAR of long wavelengths or LIDAR of short wavelengths.

      A signal that reaches your spaceship will continue expanding at an equal rate after you bounce it off in another direction. This means it could cover a very large area (spot size) after a certain distance. The critical point I'd like to made is that after a certain distance, the reflected signal will cover the entire hemisphere of some sensor, even if that distance is several AU.

      But, even at that distance and after the signal has been spread out, it will likely remain detectable by sensors of the level of sensitivity needed to detect passive IR emissions.

      A quick calculation tells me that a 10MW LIDAR beam is detectable by 10e-17W/m^2 sensors even after it has been spread over an area 1.12 billion km in diameter. What distance this corresponds to depends on the sensor's LIDAR array, but even for a small 1m wide mirror and 400nm wavelength beam set-up it is 1.15e15 meters... that's over 7800 AU and well outside the solar system. Can you ensure that there isn't a single sensor within such a massive cone?

    7. I had intended to continue my previous message, but I will elay that for a bit in order to address your recent response.

      First, bringing the discussion back into context, in this particular thread I was pointing out that the favourable strategy would be to risk projecting a higher peak reflection in (a) limited direction(s), rather than to risk diffusing the signal over all directions (absorbing the signal and diffusing it over time is not being considered, as the stategy can be applied to either technique). Actually, given your values for a 10MW LIDAR, you actually support my argument somewhat, as it would suggest that (for reasonable ranges, at least), any return at that source strength, no matter how deffuse over area, would easily be detected... including by the source platform.

      That said:

      The second thing I would like to mention is that you are not looking at the appropriate figures. Yes, the area of the window you describe is quite large (although, as I will explain shortly, not QUITE as large as you indicate). However, you are failing to take into consideration exactly how astronomical the scales we are dealing with actually are.

      I ran the numbers over in my head, so feel free to double check. However, the following are some important contexts, assuming your own calculations are accurate (I have no reason to doubt them):

      *1 billion+ km IS a rather sizeable swathe of space. However, this window is actually only on the order of 1/10^7 of the projected sphere of 7800 AU. A little more accurately, you would need upwards of 130 or 140 MILLION platforms to cover each of the windows of this size, rather than the 40 thousand I was originally estimating. I think this works out to an actual angular window of a bit under 1/40°^2, rather than my assumed 1°^2.
      Please note, when I did these mental calculations, I was using the (mis-remembered) distance figure of 7600 AU... at 7800 AU, you might need a few thousand more platforms.

      *1 billion km is indeed the likely order of magnitude for the original source beam. However, at this spread, the beam will already be at 112 km diameter by the time you reach a distance equivalent to half the distance between the Earth and the moon. Even with a target with a cross section as large as 1 km (roughly, IIRC, the size of an Executer class super star destroyer, illuminated full flank), the reflected energy will only be approximately 1/12 000th of this value. The energy density will progress unchanged, but this means the actual reflected cone will only be on the order of 1% the radius you calculate. The vast majority of the actual cone projected will be the unhinderedsignal from the source itself. BTW, this means that there is a fraction of a % chance that a reflected signal will be indistinguishable from the source signal, unless the receiver knows EXACTLY where the source was, and the exact direction of transmission, when the signal was sent.

    8. I see. You're going for a high-risk either I'm detected or not approach to detection.

      In the best case scenario, your opponent pings you with a LIDAR beam and you bounce it off into an empty corner of space. Because it is so hard to cover all the areas, the reflected signal is never intercepted.

      The calculation I did was [(10e7/10e-17)/3.14]^0.5*2. It gives the diameter.

      There are multiple problems with this scenario. I'm not saying it will never happen or that it is even unrealistic to assume that it is actually the likely and expected outcome. These are just ideas a military planner would suggest to prevent this situation from benefiting opponents.

      The first is the ability to bounce a LIDAR signal within a very very confined cone. By definition, the incident radiation is reflected at the same angle and spread: if you are pinged with a narrow, focused beam, the reflection can be pinpointed in another direction. If however the incoming beam is a broad RADAR wavelength, how can you be sure that the signals you are bouncing won't spread out so much that they'll be intercepted on the way back? This is especially problematic when 100MW lasers implies a 100MW+ output from RADARS.

      There's also the problem with geometry and materials.

      Is there even a material that can interact equally well both RADAR and LIDAR and all the frequencies in between? There are existing materials that can handle radio, microwave, infrared ect... but can you put them all together in the same reflecting face without interfering with each other?

      Let's assume such a material exists. Let's also assume that you need to bounce the incoming signals a full 45 degrees of more to the side to prevent a broad beam from spreading out on the way back to the opponent.

      The geometry becomes problematic. You will able to handle signals coming in from the intended direction. What about signals coming in from other directions? What about two directions at once? How well can you handle a pencil-wide beam that can catch an exposed corner on the hull or a sensor shroud, lighting you up like a Christmas tree, and at the same time deflect RADARs with wavelengths longer than the entire spaceship?

      If we do a calculation for a sacrificial 'sonar buoy'-type sensor platform that pulses 100MW LF-band radio emitted by a 10m wide phased array, we find that a 10e17W/m^s cheap sensor (more expensive sensors will have 10e-19 or better) will detect a signal after it has been spread out over a disk 3.56 billion km in diameter. The LF radio signal reaches this size after travelling 29.2 million km, so at a 14.6 million km distance from the target. You'll notice that the separation between the you and the platform is much smaller than the diameter of the disk - it is unescapable unless you somehow manage to deflect the signals by more than 180 degrees!

      Finally, the problem with distinguishing between your own signal and others can be simply solved by scanning your output over a specific range of wavelengths or pulsing it in a specific patterns or using more advanced modulating signals from AESAs.

    9. I might have to break up this response.

      The calculations you were using are self-evident. I believe I already said that I had at least confirmed the order of magnitudes when I calculated the figures in my head. However, I think one of us might be miscalculating for the 400nm LIDAR divergeance figure. On the one hand, if my calculations are correct, the range for LIDAR would actually be MUCH greater. On the other, the divergeance angle would be much smaller.
      I am using the formula Total Divergeance (radians) = 2 * wavelength / (pi*waist). Multiply this figure by 57 to get the value in degrees.

      Sorry, gotta go. TBC.

    10. I got that number from the Laser Weapons Calculator, but I might have made a mistake.

      I'll await the rest of your response.

    11. (CONT)

      So... to estimate the total divergence, assuming a waist of 1m, multiply the wavelength by 38 (2*57/3). For a 400 nm LIDAR, this comes out to about 1.5*10^-5°, or 0.000015°. This is quite a bit less then the 1/20° that I originally estimated from your figures, although that might be the result of miscalculation on my part.

      Moving on, however, you are quite correct that there is much greater risk from divergence at RF frequencies. Notably, at around 26 mm RF, the divergence from the proposed 1m transmitter works out to the 1° figure that I originally proposed. At 0.05m, the window of reflection increases from 1/40 000th of the full sphere to 1/10 000. At 0.5m, chances of detection by a single receiver increases to 1%. So, yes, there is definitely a risk with RF frequencies.
      I will get back to that shortly.
      This, however, assumes a waist of 1m. The problem is, once the emitted beam is large enough to cover the entire cross section, the area of the reflecting surface becomes the new waist. This means that you need togo to a 500m wavelength in order to have that 1% chance of detection by a single window.
      That said, it is highly unlikely that we will be dealing with "stealth" assets that large. Therefor, we will assume the significant risk factor when getting into the range of millimetric RF.

      I am glad that you mention the issue with material.
      There are three basic ways that photons interact with matter. Reflection, absorbtion, and transmittance (translucense/transparency).
      As far as reflection is concerned, the difference in material does not change the property of divergence. Divergence will depend only upon wavelength and waist. Now, there IS the issue of reflective diffusion. However, this is not so much a matter of material used as it is a matter of how that matterial is configured. If you have a very grainy surface, the sizes of the grains essentially replace the waist values of the surface area. This means that it would be desirable to carefully polish the surfaces of the stealth asset.
      The reflectivity of the material DOES become important, however. As it happens, as RF photons approach millimetric wavelengths, and especially as they approach increasingly longer wavelengths, most materials lose their ability to reflect the photons. So long as you avoid metals, the vessels will essentially become transparent to these longer wavelengths. Thus, significantly less energy available for detection. There is aso the possibility that some of the RF energy will be absorbed, rather than transmitted, so we are not necessarily dealing with transparency... but the end result is that there will not be sufficient reflected signal at these wavelengths.
      SO... you increase the source strength.
      Bad idea. Probably.
      As it happens, amplitude of the signal is ALSO important. It is possible to chart out the reflectivity of a material according to the amplitude of the signal. In general, as the amplitude increases, reflectivity decreases. For a while, this graph will be almost flat, with a negligible decrease in reflectivity. However, all materials have a region, refered to as the plasma barrier, where the reflectivity drops off sharply (in aluminum, for example, this occurs at about 18mV, IIRC). This is usually followed by anothehr region, known as the band gap (IIRC), where the energy is absorbed rather than transmitted. Interestingly, this does NOT mean that you have a plateau in reflected energy. It means that if you trnsmit too much energy (you cross the threshold into the plasma barrier), instead of a metal (for example, being one of the rare materials that will usually reflect RF regardless of wavelength) reflecting almost all of the energy, that same material will suddenly reflect virtually none of the energy. Note that this will apply to frequencies that SHOULD reflect off a given material.


    12. (CONT)

      Regarding multiple signals: multiple sources (sufficiently spaced) and multiple receivers will, of course, increase the risks of being detected. This is true regardless of the physical configuration of the stealth asset.

      Once again, Stealth is NOT guaranteed. Ever. There are ALWAYS conditions under which an asset might be detected. Military planners considering stealth options consider numerous factors, depending upon specific scenarios, and the specific intelligene gathered concerning any party that might become involved.
      OTOH, there are configuration issues that are as problematic for the would-be hunters. No one has, or will ever have, unlimited resources. Building, deploying, maintaining, and operating detection assets draws from these resources. Resources applied to these functions are not available for other functions... like farm machinery, transportation infrastructures, weapons and deffence developement, communications infrastructures, housing, etc.

      IIRC, the 10^-17 W/m^2 limit figure was specifically presented in terms of IR detectors. This is a function of the theoretical limit imposed by the number of actual photons emitted at a specific frequency, and by the amount of "noise". The absolute limit is the point where there is only a single photon available to strike the detector, assuming that a single photon can produce motion in a single electron. This is combined with the noise limit. The limit is usually expressed in terms of electrons. Thus, theoretical limit is the point where there is activation of a single electron. Current noise limits are at about 5 to 10 electrons (readings less than this are unreliable). This means that for the specific IR wavelength in question, 10^-18 W/m^2 would be the theoretical limit. UV detectors are much less sensitive, because much more energy is bound up in the single photons. However, existing, rather cheap, RF detectors would already be much more sensitive than this. Additionally, this figure made other assumptions, such as a 1s exposure window.
      I bring this up because, again, the thread I have been addressing here has never been about whether or not it would be possible to detect an asset with a sufficiently powerful RADAR (or LIDAR, or other equivalent method), but about which configuration is preferable for stealth applications: sharp reflection or diffuse reflection.
      Yes, of course, the reflected signal will be uch less for diffuse reflection. My point, however, was that diffuse reflection will effectively guaranty that photons WILL arrive at a detector, regardless of position. Notably, the reflection will arrive at the location of the source.
      Hunters will make decisions based upon sets of conditions such as optimal ranges. They will also take into effect all of the possible stealth considerations. You can enhance probability of detection by increasing the amount of energy. You can determine the likely ranges where transparency due to plasma barrier will become a problem for likely materials, and scan at intensities approaching this limit. You can also increase signal duration: with a floodlight approach, you can transmit constantly; or, with a ranging approach, you can transmit for as long as it takes a photon to reach the inner defined range. Either way, this multiplies the probability of detection. You can also increase the size of the detectors, and/or increase the exposure time frames. You might have to expend incredible amounts of energy, and you might have to wait for hours or days... but if there is a signal being projected in your direction, you WILL receive it.
      But you can't see what is not there. Directed reflection is no guaranty that you will avoid detection, but it is the only way to avoid a guaranty that you WILL be detected by an appropriately configured platform.


    13. Regarding my point about signal separation: I was refering to original and reflected signal separation, NOT resolution between different sources.

      This does NOT apply to all conditions.

      Coding won't help because the reflected signal will retain the same coding.
      The principle problem addressed here is for a target that is in near alignment beteen the source and the receiver. Ideally, the target will be in the signal cone, but will project toward a receiver that is outside of that signal cone, with the receiver registering only the reflection off the target. However, there is a possibility that the reflection might be directed toward a receiver that is in the original transmission cone.
      Under most conditions, this would not matter, because the receiver would take in both signals, and would be able to eliminate the signal from the original source. However, angular resolution is an issue. Longer wavelengths have progressively poorer resolution, as a physical limit. Thus, it is entirely possible that the reflected signal will appear from the same direction as the source, and it will be impossible to resolve the two.

      This also happens if the original signal has a long enough wavelength that it diffracts around the target, rather than being reflected by the target. It also happens if the target is transparent to the wavelength (non-metals, long RF wavelengths).

    14. Sorry, I think I need to correct a statement. The plasma barrier (this might be "plasma LIMIT", I forget the exact phrase) for Aluminum should be written as 18 eV, NOT 18 mV.

    15. Okay then.

      Let's see, laser spot size equation is 1.2 * Distance * Wavelength / Lens Diameter. We want a diameter of 1.12 billion km and have wavelength as 400nm and lens diameter as 1m, we need to work out distance. I get distance: 1 * 1.12 * 10^12 / 1.2 * 400 * 10^-9 : 2.33e18m or 2333 trillion km.
      My number was off by a factor 2000.

      To work out the angle, I'd use this calculator (: and get an angle of 0.00138 degrees. I think that translates to 5.98e-6 degrees^2.

      I agree with your assessment that you would not want to try diffusing the LIDAR signal in all directions even with a very low reflectivity surface, because even a factor 1000 reduction in return signal would just lower the detection range from trillions to billions of kilometers.

      On the other hand... let us imagine that the LIDAR target has a perfectly hemispherical face and is coated with Vantablack that absorbs 99.965% of the incoming wavelength. The Vantablack yields a signal reduction factor of 2857. The signal power being bounced back by a hemispherical depends on the size of the sphere and the distance, so let's fix the distance and find the signal reduction factor. At 1 billion km, the 400nm laser beam has spread to a diameter of 480km. If the target is 480km wide, it would project the bounced signal onto a sphere with a radius 1 billion km, so the signal reduction factor is 1000000000^2/480^2: 4340277777778. The return signal strength of a 10MW beam would be 8.06e-10W/m^2. A sensor sensitive to 10e17W/m^2 of radiation would be able to pick up this signal after it has been spread out over 80600000m^2. This is a disk of diameter 101.3km. I'm not sure which 'virtual laser' parameters I need to use to find the distance required for the 400nm beam to spread out to that spot size. Do I use a 1m focusing lens? If so, the distance becomes 211 million km.

      That's still a lot, but it's not across-the-solar-system distances. If we factor in a more natural spaceship size, like 10m diameter instead of 480km, it would only intercept a factor 10^2/480000^2 of the incoming signal. This would further reduce the detection distance by a factor 2.3 billion.

      If I re-done all calculations with the LIDAR being emitted from 1 million km instead of 1 billion km, the spot size at the target becomes 480 meters, the signal is reflected into a sphere 1 million km in radius and the 10m target intercepts 10^2/480^2 of the incoming signal. Add in the Vantablack, and a we get a signal reduction factor of 2857*(480^2/10^2)*(1000000000^2/480^2): 28570000000000000000. A 10MW signal gets reduced to 3.5e-13W/m^2. A sensor able to pick up 10e-17W/m^2 can detect this after it has spread to a disk 105km is diameter... which corresponds to a distance of 218 million km?

      I could be entirely wrong about how I calculate the separation between target and sensor, but I am right about the signal strength reduction factor. It doesn't seem too bad.

      I'll get to the rest later, but I just wanted to make the point for now.

  7. @ Mikkel Haaheim
    Under normal circumstances (like here on Earth) a network of 40,000 sensors would be difficult to establish, but in the space environment something like that might become trivial.

    I suggest you look up the "Server Sky" concept of Keith Lofstrom (yes, that Keith Lofstrom), where small computer chips backed by equally small solar cells are floating in free space to establish what is essentially a massively distributed computer network ( Individual chips are cranked off the assembly line relatively cheaply these days, and it can be assumed that a space civilization will have at least that level of ability as well.

    Planets. moons and asteroids will likely have a Server Sky network in orbit to deal with the large scale computational needs of the polity or planet, and other Server Sky clouds can be expected in stable orbits in places like the Lagrange points in various orbits. Whoever controls the Server Sky platform or has access to it could conceivably have millions of sensors built into these chips and be able to generate finely grained 3D views of the surrounding space. If they have access to multiple server sky clouds (for example Earth might have one in GEO, plus ones at L1, L2 and L4 and L5). Other polities might not have the same density of coverage, but it will still be considerable.

    1. I don't know why you would want to expose your computing hardware to the cosmic rays, thermal variations and general harshness of the space environment, when you could instead use the distributed satellite network as a simple communications network while keeping the computing hardware on the ground...

      As for the 40000 sensors number, using longer wavelengths that diffract faster and compensating for the loss of power by using the 100MW+ power supplies meant for laser weapons can drastically reduce this requirement.

    2. @Thucydides

      I will address this when I have the opportunity.

    3. Actually, it is MUCH easier to deploy such platforms on Earth. First, because you wouldn't have to worry about most of them moving out of position. Second, because you wouldn't have to expend extreme amounts of energy to deploy them, nor to access the for maintenance. Third, because they would be better protected against the majority of natural threats. Nor is there any reason why you would not be able to deploy the same wafer tech that you would use in space.
      The only real limitations are that planetside arrays can not cover the vast amount of space required to detect objects light-minutes away, and that the same atmosphere that protects the platforms from harm will also tend to block out certain wavelengths of photons, and other sources of detectable energy.
      The problem with the Server Sky approach is that it does not really address the actual problem. Yes, you need great numbers of platforms/arrays... but these can not just be put anywhere. You need a large number of platforms at the right locations.
      Server Sky might make an exceptionally competent array that would serve far better than a single massive platform. However, the reality is that you would then need 40 000 of these ARRAYS, and these arrays would only be able to assure coverage over a specific point. The further removed you are from this "ideal" point, the more you suffer from overlap and gaps in coverage, requiring even more arrays to cover those gaps.

      From a tech standpoint, the arrays would be much better performing than single massive platforms. However, there are certain liabilities to take into consideration.
      First, all of the units will need to communicate with one another, or (at least) with a centralised control relay. Now, this is also true of microchip components; however, at the distances involved, you no longer have the option of fixed pathway guides that define which units are communicating with which other units (requiring info to be location tagged, which complicates processing requirements cinsiderably). You also have an increased amount of signal activity that could result in network, processing, and sensory interference. Furthermore, although photons travel much faster than electrons, this advantage is largely dampened by the distances involved, as well as the (current) need to encode and decode the photon signals into electron signals that the unit processors can perform operations with.
      Second, although the site addresses the most common processing problems associated with radiation exposure, it does not currently provide fixes for all of these problems. More importantly, it does not address the primary damage that radiation can cause. Here, I am refering to transmutation, neutron embrittlement, physical radiation "burns", and other sources of physicl damage caused by alpha partical, beta partical, neutron, and osmic radiations that could easily render such thin strata microcomponents unusable.
      Third, although the site addresses the issue of avoiding physical damage due to collison hazards from other assets and debris, it does not address the issue of micrometeoroid damage. IIRC, analyses conducted on the micrometeoroid protection layers of the apollo space suits revealed that micrometeoroid impacts can be as frequent as dozens to hundreds of impacts per second per m^2. These impacts often penetrated as far as a few mm into the protective layer. An unprotected chip measured in microns are unlikely to survive such an environment for any meaningful length of time. Furthermore, the avoidance parameters against debris collision suggested by the site would not be available in some of the locations where coverage would be required.

  8. @Matter Beam

    The 40 000 sensor figure addresses the number of 1°^2 windows that would need to be covered (for a total reflection divergeance of 1°). Amplitude is irrelevant to this base figure, which essentially assumes infinite range (or, in other words, assumes that the platforms are already in the arbitrary range limit).
    This is a MINIMUM platform requirement for the given condition. The requirement is increased dramatically for Gh or greater frequencies (millimetric or lesser wavelengths). The requirement would, of course, increase for more limited ranges. It also increases for targets that move outside of an optimal position, or for platforms that move out of position due to differing orbits.

    Using longer RF wavelengths DOES decrease the number of required platforms. However, there are other considerations that yield such RF frequencies less suitable (RF transparency, for example).