I enjoy designing games, and in one that I'm working on the ships have shields, similar to the shields in Star Trek. They are created via some kind of force field projection mechanism and are spheroidical (that is, they take the shape of a spheroid). I would expect that those shields have strength relative to:

  • The power used (more is better)
  • The distance from the emitter (less is better)
  • The curvature of the shield (more is better - which implies smaller)

Now, on a small fighter or shuttlecraft, the second and third criteria imply that the shield would be stronger. I would naïvely expect both power and distance to scale the strength of the shield with the third power - but power generation capabilities also scale with the cube of size, so those two things are about a wash. But the fighter, being smaller, would have a more curved shield, so it would be stronger.

Moreover, with less space dedicated to other things (like hydroponics or living space) and more justification for having as much power as possible, a fighter would likely have more power per cubic metre, and thus an even stronger shield.

Why wouldn't this be the case? The justification doesn't need to be short and pithy; I need the explanation so I can feel good about the system more than I need to have players understand why it works the way it does.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – L.Dutch Jan 8 at 16:51
  • $\begingroup$ Can you fire your weapons while the shield is at full power? Could be that the shields have to be lower, if not off completely, in order to make full use of offensive capabilities. Think back to early aviation warfare, it could be that you have to fire "inbetween the propeller blades" ie the shield has to have some sort of oscilatting on/off feature which allows firing without damage to fighter but results in weakening of overall shield strength. $\endgroup$ – EveryBitHelps Jan 9 at 18:34
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    $\begingroup$ Why is curvature better? I would think that if a shield gets hit, it is hit. The only advantage of a smaller shield is a smaller chance of getting hit. But if it is hit, what is the advantage of greater curvature? $\endgroup$ – Tyler S. Loeper Jan 9 at 21:28
  • $\begingroup$ @TylerS.Loeper As mentioned by MongoTheGeek, it's not as great an effect as I had expected. But see that answer for a better explanation. $\endgroup$ – Spitemaster Jan 9 at 21:44
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    $\begingroup$ Suggest the distance from the emitter should scale effectiveness by inverse-square, not inverse-cube. $\endgroup$ – imallett Jan 9 at 21:59

33 Answers 33


Just make the dependence on curvature less sensitive than on the surface area of the shield and you should be fine.

Suppose the power of the shield emitter is proportional to the volume of the reactor powering it, and the strength of the shield inversely proportional to its surface area. Then if the radius of your ship is $R$, the strength of the shield is proportional to

$S\propto V/A \propto R^3/R^2 \propto R$

That is, without a curvature effect, scaling everything in your ship up will increase its shield strength. But, mathematically, curvature $C$ is proportional to $1/R$. So now if, say,

$S\propto \frac{V\sqrt{C}}{A}$, or maybe the logarithm of $C$, or something else slower than linear, you will still get an increase in shield strength with increasing ship size.

Another small trick you could exploit is to demand that there be a gap of a minimum size between shield and hull. For the sake of argument, let's say it's ten meters. Then a spherical fighter with hull radius 5m would need a shield radius 15m, but a 10m fighter would only need 20m of shield. Thus, doubling the size of the fighter increases the shield surface area by a factor of only about 1.8, instead of 4. Even a destroyer with radius 100m would get a benefit of 3.6 v 4 when you double everything except for that mandatory 10m gap. That would be another way to make big ships naturally have stronger shields, though the benefit diminishes for very large ships.


I would like to point out that distance from the generator doesn't have to go up if the ship is bigger. Having multiple shield projection points that make a local shield would be good enough to keep the shield close to its point of projection. So no inverse square rule law here.


It's simple math.

  1. Power. For simplicity's sake let's assume the bigger the engine, the more power you can generate (for example, a nuclear power plant can generate much more power than a nuclear submarine and stupendously more than a family car). So, engine size grow by volume, as you make the ship bigger both the volume of the ship and the volume of the engine increase by x^3 (cubed).

  2. Distance. Distance grows linearly by x.

  3. Surface area (what you imply by "curvature"). Surface area of a sphere grows by x^2 (squared).

It's easy to see that engine size trumps the other two factor. If it's not obvious it's easy to plug in some numbers to see it. For simplicity the following are just factors rather than any real units:

x           : 1, 2,  3,  4,   5,   6,   7,   8,   9
power       : 1, 8, 27, 64, 125, 216, 343, 512, 729
distance    : 1, 2,  3,  4,   5,   6,   7,   8,   9
shield area : 1, 4,  9, 16,  25,  36,  49,  64,  81

So a ship that is 9 times larger will need 81 times more energy to generate similar strength shield but can potentially install an engine that is 729 times more powerful.


protected by James Jan 9 at 21:59

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