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*To*: KellySt@aol.com, kgstar@most.magec.com, stevev@efn.org, jim@bogie2.bio.purdue.edu, zkulpa@zmit1.ippt.gov.pl, hous0042@maroon.tc.umn.edu, rddesign@wolfenet.com, David@InterWorld.com, lparker@destin.gulfnet.com, bmansur@oc.edu, DotarSojat@aol.com*Subject*: Re: Close but no cigar?*From*: T.L.G.vanderLinden@student.utwente.nl (Timothy van der Linden)*Date*: Thu, 14 Mar 1996 01:27:58 +0100

>> >> Power out: P=s T^4 >> > >> >Is this the proper blackbody radiation formula? Looks too simple to me. >> >> This is the Boltzmann equation, the Planck equation looks much more ugly and >> only tells the energy density per "delta wavelength". So one would need to >> integrate the latter (which doesn't seem easy). > >I think you will have to. Remember, the amount of IR will go up as the >temp goes up. so as the sail heats up, the amount of radiated energy will >increase also. or does your formula show that? I'm confused. The formula I used, shows that the amount of radiated energy per surface area (P) increases as the temperature (T) of the black-body goes up. It does not discriminate between the different wavelengths (which is why I used that formula in the first place). I don't see any reason why you want to know specifically how much IR is send out. But if you want to, you need to use Planck's formula. (Both Plank's and Bolzmann's equation have T to the power 4) >> Kevin suggested using Titanium because it is easy to mine on the moon. >> Since I haven't numbers about that I will use Wolfraam. > >if we can't make it work for tungsten (W) we can't make it work for >titainium (Ti) As I suggested, we could decrease the density of the radiation by a factor 10 (and increase the sail by the same factor) What I should note is that I used the surface of a flat plate. Since we use a mesh the effective area may be several times larger! Timothy

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