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Re: Close but no cigar? again

>I know. I promised no more remarks until I did my homework.
>The Stefan-Boltzmann law of radiation is P = s T^4, where s (the
>Stefan-Boltzmann constant, usually written as the greek lower-case
>letter sigma) does indeed have the value 5.67 E-8 W/(m^2 K^4).  A
>black-body surface absorbing a power of 7.5E5 W/m^2 (your number)
>will be heated to an ultimate equilibrium temperature Teq at which
>it will reradiate all the absorbed power.  Therefore,
>                   7.5E5 = 5.67E-8 Teq^4  .
>So Teq equals 1907 K, below the melting temperature of titanium
>(2073 K) and well below the melting temperature of tungsten.
>Unless I missed something, I believe you (I don't know who,
>because I haven't yet learned to read these quotes of quotes of
>quotes) equated power to energy in your calculation of no cigar.
>Rex Finke

I was just testing if the group was paying attention (as the teacher would say)

I presented this neat formula P=s T^4 but somehow I never used it. (?!?) I
guess I was amazed too much by the simplicity of the calculation, sorry for

It indeed seems to be even easier, namely just like Rex wrote above.

For those who are wondering what I calculated: the amount of energy needed
per square metre to melt the sail. This isn't necessary at all, so forget it.
Oh yeah, I told that the product of melting temperature and specific heat
was important, forget that too, only the melting temp. is important.

By the way, my latest remark about the increased surface area of a mesh
instead of a plate is still valid.

So finally we can conclude, bulls eye, two beer and some peanuts.