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You're so invested in this hobby that you can't admit you're wrong. Okay, if it's so easy to do TODAY, then why doesn't GWB run down to radio shack and buy all the lasers he needs, throw 'em into orbit on a Delta rocket and scrap the missile to missile interceptor he's building now? Common sense would tell you that the atmosphere plays a HUGE role in dissipating laser energy. It's not hard to see or understand.

You're so invested in this hobby that you can't admit you're wrong.

Its not a hobby.

Okay, if it's so easy to do TODAY, then why doesn't GWB run down to radio shack and buy all the lasers he needs, throw 'em into orbit on a Delta rocket and scrap the missile to missile interceptor he's building now? Common sense would tell you that the atmosphere plays a HUGE role in dissipating laser energy. It's not hard to see or understand.

Because an antimissle system needs to delivery energy to the target, not just high peak intensity as is the case for a blinding weapon.

This probly wont help but it looks complicated.

Knowing the size of the focused spot is helpful in calculating energy density at the work surface.

For a fundamental mode (TEM00) beam:

S = ( 4l / p ) × ( F / D)

where:

In performing a laser weld, optics to focus the laser beam to the desired size are necessary.

S = Focused Spot Diameter

l = Laser Wavelength

F = Focal Length of Objective Lens

D = Diameter of Laser Beam

For a multimode beam:

S = F × f

where:

F = Focal Length of Objective Lens

M = Laser Beam Divergence

If one assumes the part to be welded as a semi-infinite solid, with a constant incident heat flux, then the temperature distribution as a function of depth into the material is given by:6

T(x,t) = (2E/K) × [(kt/p)½× exp(-x2/4kt) - (x/2)erfc(x/2(kt)½)]

where:

T(x,t)=Temperature at a distance x below the work surface, at a time t after start of constant heat input

E = constant heat flux input

K = thermal conductivity

k = thermal diffusivity

x = depth below surface

t = time after start of heat flux input

erfc = complimentary error function

and at the surface (x=0), the temperature rise will be:

T(x,t)x=0 = (2E/K) × (kt/p)½

....sorry......