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Homing In On Laser Weapons
Bloodmoon1 writes "I just came across this article at GlobalSecurity.org that gives a very good summary of the current status of solid-state lasers as weapons. It gives you a good idea of where the JSF Laser system is at and just how much time, effort, and money has went into this project. Also has some basic, but very sufficent, explanations of some of the science behind the technology."
The answer to your question is called AN/TPQ-36 and AN/TPQ-37 "Firefinder" target acquisition radars. We've had them for 20 years - the -36 is designed to track mortar shells and the -37 other types of artillery (though IIRC, the -37 has all the functionality of the -36).
They are very effective. They calculate the location of the firing tubes, and that information is passed to artillery units tasked to provide counterbattery fire (usually MLRS rocket artillery). This all happens very quickly - 30 seconds to a few minutes' time.
I know this because Tyler knows this.
I would like to know how such a weapon will acquire/track/target an incoming projectile. (That was not sarcasm; I really would like to know.) Mortar rounds generally travel in a high parabolic path - think of the St. Louis arch. Larger artillery shells - such as those fired from a battleship - follow a flatter trajectory. The targeting system would have to acquire a small incoming object, predict the path it will follow, and fire within a few seconds. That looks like a daunting task.
It's a solved problem. The Sea Wolf point defence system can shoot down 4.5-inch shells as well as supersonic missiles. Sea Wolf was first deployed in combat in 1982. Of course, you are likely to run out of missiles before they run out of cannon ammo, but maybe you can buy enough time to hit them with an Exocet.
Warships are expensive, so a lot of money has been spent on ways to protect them!
In principle, yes. In practice, no. If you were to put a very high quality coating of silver (for visible wavelength lasers) or gold (for IR lasers) on your missile, in principle you could reflect 95 to 98% of incident light. Special optical coatings can result in >99% reflectance, but only over narrow wavelength ranges.
In other words, if the enemy knows the wavelength at which your laser operates, he can reduce the effectiveness of your laser weapons. For ground based installations, this still isn't a big problem--you just need a laser that's an order of magnitude more powerful, and you can cook even the reflective coatings on the other guy's missiles. I've done research work involving lasers in both physical chemistry and medicine, and I've seen a number of purportedly highly-reflective optical elements get toasted by a powerful enough beam. Also, high-quality optical coatings usually aren't meant to handle the stresses (physical and thermal) experienced by your typical missile (ballistic or tactical).
~Idarubicin
Okay, I'll bite. Where does E=mc^2 come into this? I've worked with lasers for a number of years, and I have yet to see any of my lasing medium converted directly to energy. Lasers operate by kicking atoms into an excited state (usually an excited electronic state) and then emitting light when excited atoms relax back to ground state.
For the record, small lasers don't require "gigawattage" to operate. I have a laser pointer that runs on one AA battery--I'll be giving a talk using it in a couple of hours. A laser designed for a weapons application would be larger. Still, I could assemble a carbon dioxide laser that could start fires from several hundred feet away and still be light enough to carry--and operate for a while on a moderately hefty battery back.
Granted, I couldn't destroy missiles with it, but the article discussses lasers that are mounted on aircraft or vehicles, or are part of fixed installations. You don't need a large power supply for even an extremely powerful laser if it only fires the very short pulses (microseconds or nanoseconds) that would be most useful for military purposes.
~Idarubicin
"A kilowatt is 3,600,000 joules, 10 kilowatts in respect is 36,000,000."
What are you talking about? A kilowatt is a measure of power, and a joule is a measure of energy. The two are not directly comparable without a time factor thrown in. Do you mean a kilowatt hour is 3,600,000 joules?
By your calculation, lightning is 280-2,800 kilowatts (0.3-2.8 megawatts). As we all know, lighting is more in the range of 1.21 gigawatts (humor intended, but general priniciple remains the same). It's not like lightning strikes last for an hour.
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"To put this in prespective, the adverage person uses 64,800,000 joules a month, or 18 kilowatts... So for every time they fire this baby, they are blowing 50-100 bucks....
They essentially are what cause the blackouts in California."
What the fuck are you talking about? This causes the blackouts in California, not some sergeant flipping the switch on $100 of electricity.
Kevin Fox
if you hit a mirror with a powerful enough beam of laser light, the small fraction of light that's absorbed (no such thing as an ideal reflector) will rapidly ablate the mirror coat, and then you're screwed. we have problems with this with our pulsed IR laser at work - we need solid polished aluminum mirrors with heatsinks on the back, ad that's for a 500mJ, 500ns pulse; they don't last that long, either. a 100KW IR laser will vaporize pretty much anything that's not *perfectly* reflective, i.e. anything we can build with current technology.
Facts do not cease to exist because they are ignored. - Aldous Huxley