How much energy is necessary to power a truly destructive portable laser weapon?

Factual Questions
1

The “slice your car in two” from 10 meters away kind. How much power input do you need to do this with a laser capable of this feat?

2

Knowing absolutely nothing about physics, I’ll take a guess and say “Not Much”, since the problems seem to be more in our ability to focus the beam than our ability to pump raw power into it. It makes sense that a beam of light can theoretically be focused into an unimaginably small point, and it makes sense that the more it is focused and narrowed, the hotter it gets.

3

I wouldn’t use destructive and portable in the same sentence when talking about a laser. I once had the opportunity to play with a CO[sub]2[/sub] laser that could slowly cut through sheet metal. It used up most of a 19" equipment rack and needed a special dedicated AC circuit to supply its power.

When you read about lasers being used to shoot down missiles, you have to remember two things. The lasers are chemically fueled, using some very nasty and toxic chemicals. Missiles are relatively fragile. It doesn’t take much energy to disrupt the thin metal skin of a missile’s fuselage.

4

As you can see from the answers already, there is confusion over just what it means to be “truly destructive”? How long would you allow to cut the car in two from 10 meters? Like just waving a Star Wars light saber through it and having the chunks fall apart. Or would you allow more time? One second? Ten seconds? IANA physicist, but I would suggest that the power requirement for any one of those scenarios would be enormously different from the others.

5

Plus, why “portable”? The energy necessary to make a laser “truly destructive” by any standard is the same whether the laser can move or not. The question of portability relates to its convenience as a weapon or tool, not to the power it needs to accomplish a certain task.

6

I don’t have the numbers, but your power supply has to be able to provide enough power to produce a beam strong enough for your needs. Two major factors to take into account: inefficiencies and attenuation.

**Inefficiencies: **lasers are inefficient. Not all the power pumped into the lasing element comes out as part of the laser beam.

**Attenuation: **the beam will spread with distance, but not much. The air between the “muzzle” and the target, plus whatever crud is in the air, will attenuate it more.

Let’s say you need [pulling numbers out of ass]1 gigajoule [/pnooa] to cut a car in two. Your power supply has to provide 1 gigajoule plus enough extra power to compensate for the inefficiencies and attenuation.

You also have to provide cooling. A megajoule system gets hot enough to require a dedicated refrigerant system. So you also have to power that. And you have to be able to carry it: a pump, a heat-exchanger and a tank of coolant, such as a clear flourocarbon liquid. More power; more weight.

I’ve worked on a megajoule laser system that could, in theory, be made man-portable. I’ve been zapped by it. It felt like being lightly poked by hot needles. It was driven by a largish capacitor bank, which was in turn fed by a 10,000 V/1,000 A power line. It’ll take more than Duracells to power a useful manpack laser.

7

So… portable car cutters are not in the offing anytime soon?

8

I’m guessing that astro’s just about had it with people who park too close to his driveway.

9
10

For the sake of having a bit of fun with this, let’s see how much energy it would take to slice a car in two. Say we approximate a car as a cylinder of steel, 1 meter in diameter, 3 inches thick. How much energy would it take to sever all atomic bonds in a cross-section of that cylinder?

From there, we can see, laser inefficiencies and such aside, just how much energy we’d need to do it even if we had a perfect laser of some sort.

11

A discussion of laser cutting can be found here:

Note that the process involved here has the laser positioned very close to the material being cut, and also uses an oxygen jet to assist in burning away the material (same as in an oxy-acetylene lance).

You want to do it from some distance, and without the oxygen jet, so as a guess, I’d multiply everything by a factor of at least 10, and maybe 100. That means an electrical power input of between 240 kW and 4 MW. 240 kW is readily achievable by a car engine, and 4 MW is achievable (for short periods) by a drag car engine.

So it’s kinda doable in a car or truck mounted system.

OTOH, you could just use the oxy-acetylene setup that everyone else uses. Fits on a hand trolley.