[maks_holi]

What would you use this for? Frying Pakistan? Won't work, it's already a desert as it is.

[Yfclciak]

The problem would IMHO be finding concave mirrors. Do they even sell something like that?

[Yinekol]

You could use polished metal OTOH, but presumably it would reflect a narrower spectrum.

[HOTgirlsXXL]

How much metal would you want to melt? You can calculate the power needed from that, and from that the area you need.

[r5YOPDyk]

Well, if you have some figures we can do some calculations.

BTW. you asked about limiting factors. Heated metal would lose temperature through radiation, convection and conduction. They are proportional to the size of the thing you are heating and its temperature. I'm kind of curious if this can be done now too. Maybe tommorow I'll do a calculation and let you know what I got.

[ClaudeMarkus]

You're completely ignoring the heat loss from the metal.

[wrbwrbwrb]

You're also ignoring the fact that generally even "black" materials are not particularly good blackbodies. They generally have an emissivity of ~0.4 (this is spectrum dependent, however). I'd be willing to accept that in the visible range (i.e. the solar spectrum) your black piece of metal has an absorption coefficient of ~0.7. So any energy you hit it with, only 70% of it gets absorbed

Secondly, the heat loss goes as the stefan-boltzmann law. Your true blackbody emissivity is still probably ~0.4 at temperatures under 2000K (this is good, meaning that you're a more efficient heat absorber than you are heat emitter) and at temps above 500K radiative cooling is probably dominant (so we ignore convection, since most of the important cooling will take place in this range or higher anyway). So heat loss = 0.4*sigma*(T^4). sigma = 5.67*10^-8 W/(m^2*K^4). At 500K your heat loss is 1400 W/m^2. At 1000 K it's 23000 W/m^2. At 1500K it's 115000 W/m^2

[DoctorDulitlBest]

You're also ignoring the fact that you only heat one side of the ball. You should use as the are of the ball pi*r*r instead of 4*pi*r*r when you figure out your maximal effective mirror size. You still have to use 4*pi*r*r when you're talking radiative cooling, however.

Let a be the absorption coefficient
e be the emission coefficient
m be the mass of iron you wish to heat in kg
p be the mass density of iron in kg/m^3
s be the power per unit area of solar radiation at the earth's surface in W/m^2
c be the heat capacity of iron in J/(K*kg)
sigma be the stefan boltzmann constant
n be the number of maximally sized mirrors
and T0 be the ambient temperature (i.e. starting temp of metal) then a good approximation is:

dT/dt = [pi*(r^2)*n*s*a - 4*pi*r^2*sigma*(T^4)*e]/(m*c)

with r = (3m/4*pi)^(1/3)

unfortunately I have no idea how to integrate 1/(1-x^4) to get an analytic solution...

[TNgqZhLR]

I hate partial fractions.

[wrardymar]

There was a myth busters episode about this. They were trying to recreate Archimedes mythological death ray. No one could do it.

[MyOwnStyle]

Originally posted by Lul Thyme


yeah in the reals the 1/1+x^2 is some trig function or something.
It's probably in a basic table. I think it's an arctan or smething. Not sure.

[Goksiodiffeli]

I think it's an arctan or smething. Not sure. Yep.