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Posted

Your limitation is going to be supplying energy to the laser. If you want to melt through steel, which is going to take about .5 J/gK, and that will require getting it to about 1300-1500 ºC, so that's maybe 700J/g and you want to melt through tens of grams, so maybe 20kJ of absorbed energy is required. If the reflectivity is 50% and laser efficiency is 50%, you need to supply a minimum of 80kJ of energy. You could do that with a 1 kg battery. The discharge rate so you can melt the steel quickly will be the limitation, since I've ignored any heat losses from the target, and that rapidly becomes a bad assumption.

  • 2 weeks later...
Posted (edited)

so what would be the specs of such a power source?

ie. amps/watts etc.?

and what would you say the weight/volume of the most compact/light of a power sourse such as this one useing todays technology in a releastic situiation?

ie.acounting for power loss and cooling

Edited by cipher510

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