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Background: in my setting, there's a defensive directed-energy weapon installation that runs off of four gas turbine engines similar to the General Electric T64. Between the four of them, approximately 14 megawatts of power are produced. This power feeds a 14-megawatt directed-energy weapon.

14 megawatts is 14 megajoules per second.

According to the Atomic Rockets Boom Table, 14 megajoules is equivalent to about two thirds the detonation of an anti-tank mine, or about 3 and 1/3 kilogram of TNT. I imagine that this is pretty dangerous for an armored fighting vehicle to be hit with.

The problem is hitting it. Microwaves, for instance, don't really hit hard over long distances, and x-ray lasers tend to kill the operator as well as they kill the target.

What wavelength of electromagnetic radiation would be best suited for carrying 14 MW of power over ten kilometers in a manner that could cut a hole in a modern main battle tank?

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    $\begingroup$ In air? In a vacuum? (Those are very different situations.) (And you may want to be careful with cooling the weapon. Even if the conversion if 90% efficient, that still leaves you with 1.5 MW of waste heat which you need to get rid of. That is one Massive Radiator.) $\endgroup$
    – AlexP
    Oct 17, 2021 at 21:21
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    $\begingroup$ @AlexP In air. Radiator mass isn't a problem - it's a fixed installation built over a 10-meter cube of water which it uses as a heat sink. It'll take 4.186 TERAJOULES to raise the temperature of that mass by 1 degree, if my math is right. $\endgroup$
    – KEY_ABRADE
    Oct 17, 2021 at 21:26
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    $\begingroup$ Gigajoules not terajoules, I would think, but still plenty good enough if you can move the heat to the water. Remains the small problem of transmitting over ten megawatts of power in a narrow beam through air, which I have no idea how to solve. (Has to be a narrow beam, or else the tank will survive long enough to move out of the beam and possibly fire anti-battery rounds at you.) $\endgroup$
    – AlexP
    Oct 17, 2021 at 21:35
  • $\begingroup$ @AlexP You are correct. I messed up and replaced "a gram of water" with "a kilogram of water". Still, to raise the heat of the heat sink by ten degrees, the energy equivalent of the world's smallest nuke - the Special Atomic Demolition Munition, 10 tons TNT yield - is needed. $\endgroup$
    – KEY_ABRADE
    Oct 17, 2021 at 21:40
  • $\begingroup$ CO2 lasers which are in the infrared are routinely used in industry for cutting ferrous metals, but optimally, not sure - can't find the absorption spectra for steel, so not an answer. $\endgroup$ Oct 18, 2021 at 0:27

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Step one: absorption. Pick a wavelength that isn't strongly absorbed by the atmosphere. This roughly means any wavelength that some kind of animal can see, because those wavelengths are the one that propagate best through an atmosphere.

Here's a helpful-ish diagram from this paper.

atmospheric absorption of solar radiation from near-UV to far-IR

Most things that aren't visible are strongly absorbed by the atmosphere... generally oxygen on the left of the chart, and water vapor on the right

UV and longer wavelength IR (or far IR) are also somewhat undesirable because they readily cause atmospheric breakdown, generating bright streaks of plasma in the air which reveal your position and suck all the energy out of the beam.

Step two: scattering. Scattering is wavelength dependent. For Rayleigh scattering, the amount of scattered energy is proportional to the fourth power of the wavelength, which strongly recommends red and near-IR light for lasers used in an atmosphere.

Rayleigh scattering probabilities for the spectrum of visible light

There are various ways to ameliorate scattering with clever laser tricks, but as a starting point this tells you the key limitation.

Step three: range and emitter size. Your laser's focusing ability will be limited by many factors, but underpinning them all is diffraction. The diffraction limit can be modelled by $S \propto \frac{R\lambda}{D}$ where $S$ is the spot size at the target, $R$ is the range to the target and $D$ is the diameter of the final optical element (probably a mirror).

To kill your target, you want to maximise radiant flux at the target... the amount of power per unit area. By using a 500nm green laser instead of a 1000nm near-IR laser, you're reducing the spot size at the target by a factor of two. This decreases spot area by a factor of 4, which means that all else being equal, the green laser is four times more damaging than the infrared laser at a given range, or has the same damage capability at twice the range.

The final step, therefore, is to make all the necessary tradeoffs... increased range and radiant flux with shorter wavelengths vs increased energy losses due to scattering. This is strongly affected by the ranges you expect to fight at, and how big you're prepared to make the muzzle of your laser.

14 megawatts is 14 megajoules per second [snip] 14 megajoules is equivalent to about two thirds the detonation of an anti-tank mine, or about 3 and 1/3 kilogram of TNT.

The energy of a chemical explosive detonating is released in a very short period of time, meaning that the peak power level is very high. You haven't specified the pulse length of your laser... those 14MJ delivered over a whole second (peak power 14MW) will have significantly different effects to, say, 10 pulses of 1.4MJ each delivered over a millisecond (peak power 1.4GW).

Luke Campbell has written a laser damage calculator... whilst it looks particularly old-school and is slightly user unfriendly, it is quite powerful. I suggest you have a play with it to find out what sort of beam or pulse powers you need, how big a spot size is necessary, and then have a think about how big your laser has to be to generate that spot size at 10km.

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    $\begingroup$ How do I solve the equation to determine the beam diameter at the target? $\endgroup$
    – KEY_ABRADE
    Oct 18, 2021 at 11:16
  • $\begingroup$ "By using a 500nm green laser instead of a 1000nm near-IR laser, you're reducing the spot size at the target by a factor of two" That's the minimum spot size. I doubt that you'll ever be able to maintain it at this size over any distance if you can't control the propagation medium. Even when you can control the propagation medium, at that power density weird thing will start happening - at NIF, a 500TW picoseconds pulse heats the hohlraum so high it emits X-rays $\endgroup$ Oct 18, 2021 at 13:05
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Infrared.

laser https://news.usni.org/2020/05/22/video-uss-portland-fires-laser-weapon-downs-drone-in-first-at-sea-test

The issue for your endeavor is scatter. 10 km is a long way. Shorter wavelengths have more energy but they scatter more. Longer wavelengths are less energetic but less prone to scatter. Infrared uses long wavelengths that are absorbed by targets as heat, and that is what you want to melt a hole. Infrared is what the current military lasers use, although these are much less powerful than what you propose.

An interesting thing - certain frequencies in the infrared are less subject to scatter than others. I did not know that! Your laser should include all of these frequencies I think, on the theory of not putting all your eggs in one basket.

Navy Shipboard Lasers for Surface, Air, and Missile Defense: Background and Issues for Congress June 12, 2015

There are certain wavelengths of light (i.e., “sweet spots” in the electromagnetic spectrum) where atmospheric absorption by water vapor is markedly reduced.8 Lasers can be designed to emit light at or near those sweet spots, so as to maximize their potential effectiveness. Absorption generally grows with distance to target, making it in general less of a potential problem for short-range operations than for longer-range operations.

Lasers being developed for potential shipboard use produce light with wavelengths in the near-infrared portion of the spectrum. Sweet spots in this part of the spectrum include wavelengths of 0.87 microns, 1.045 microns, 1.24 microns, 1.62 microns, 2.13 microns, and 2.2 microns. (Other sources, such as the research paper cited in footnote 7, cite somewhat different figures for sweet spot wavelengths, depending in part on whether sweet spot is for water vapor alone, or for multiple sources of atmospheric absorption and scattering.)

But why near infrared? Why not far infrared? Is it reduced energy with the longer frequencies and no increased advantage as regards reduced scatter? I don't know the answer to that but would be interested if someone does.

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