RE: starship-design: Hull Materials

[wharton@physics.ucla.edu (Ken Wharton)]

First of all, here's a web site with a lot of x-ray data.  
Transmissions, flourescence yields, etc.

http://www.chem.pwf.cam.ac.uk/~slms100/bkxrays.htm



>It should be possible to at least deflect charged particles with some 
sort
>of magnetic field, but a lot of these particles are going to either be
>neutral or simply to massive to deflect in time so the hull has to be
>capable of absorbing quite a bit of impact for long durations. This 
means
>high melting points, high thermal transmissivity or a least high
>emissivity, extremely hard and yet ductile enough to remain "tough for
>several years of use.

There actually won't be much neutral matter, I don't think; interstellar 
space is a pretty good plasma.  But true, there might be some 
micrometeorites with only a few electrons missing and a huge mass/charge 
ratio.  Those will be a problem, and they'll make lots of radiation and 
heat when they hit.

>We almost need something as tough as the inside of a nuclear reactor 
for
>the hull! Another problem that Timothy started to get into is secondary
>radiation. We have to screen against alpha and beta particle radiation
>caused by gamma ray collision with our own shield. Probably need to
>consider a fuel tank forward design to keep as much of the mass in 
front of
>the crew as possible.

This is very important, and I think the way to minimize it is to have 
the highest-Z materials on the outside, getting progressively smaller Z 
as you go in.  More in a sec.

>Which brings us to deceleration, NOW the shields need to be at the BACK 
of
>the ship, not the front...

This is especially problematic when you consider that you need holes in 
the back for the engines to spew out propellant.  How could we 
conceviably have shielding there??

>> My theory and tables mainly talk about mu/rho [gm/cm^2] (mass 
attenuation
>> coefficient) when considering X-ray shielding. For low X-ray energies 
the
>> photo electric absorption by the K,L,M electronshells seems to be the
>> dominating factor. It looks like we once again need tables to know 
what's
>> best.
>>
>>
>> Timothy
>
>I don't have anything like the tables you are describing...

g/cm^2 is areal density; the number of atoms an xray will encounter is 
proportional to the density (g/cm^3) times the thickness of the material 
(cm).  Multiply these and you get g/cm^2; that is the important paramter 
here.  So no matter what you make the shielding out of, to have the same 
stopping power it will weigh the same no matter what material you use.  
Tungsten is very dense, so the shielding wouldn't have to be as thick, 
but it weighs just as much as a thicker deuterium shield with the same 
g/cm^2.

Knowing the surface area of the shield, we can therefore calculate the 
total mass of the shield needed to stop a given energy xray.

But once you "stop" an xray, the energy doesn't just disappear.  It 
changes form.  We have to stop all the secondary radiation as well...

Okay - so how do the xrays dissapate their energy?  (I'm writing this 
into my thesis right now, so I've been learning about it lately)  Photo-
electric absorption is indeed dominant at low energies.  K-shell 
ionization is the one to worry about; L and M energies are going to be 
comparitively low.  Even the electron that is kicked out by photo 
ionization will have a small stopping distance compared to the K-alpha 
xrays.  Alpha particles will be even less of a concern.  The energies of 
the secondary k-alpha xrays are determined by the Z of the atom; from 
50eV in Li to 100KeV in Uranium.  That's why you want to have the 
higher-Z materials on the outside.  If you dump lots of energy into k-
alphas in an outer Tungsten layer, for example, you can attentuate the 
secondary 60KeV xrays in further lower-Z layers without worrying about 
creating more high energy xrays.  If the last part of the shield is 
Deuterium ice, the worst secondary radiation will be the 500eV k-alpha 
line from oxygen, which isn't horrible.

Bottom line, you want to turn the high energy xrays into as many low-
energy xrays as possible.

Another problem is that ionization cross-sections start dropping once 
the xray gets much more energetic than the k-alpha energy.  You start 
dumping energy into compton scattering, in which the atom only partially 
absorbs the x-ray, as well as electron-postiron pair-production over 
1.5MeV.  The strong 500KeV gamma line that might result from positron 
annihilation could be a problem, but I suppose that's still better than 
the orginal MeV-pluse xrays.

At very high x-ray energies (GeV?) you might have to worry about nuclear 
distentigration, fission and all that.  Unfortunately, you want that to 
happen in low-Z elements to limit the released energy.  Below Nickel, I 
think, fission products have more energy than their parent.  So if we 
start doing Tungsten fission on the outer hull, that will merely add to 
the total amount of energy that we need to shield.  So maybe we want a 
low-Z outer layer, then the high-Z layer, and then progressively smaller 
Z's toward the inside.  Unless we think there won't be many GeV xrays.


>I think the alloy of Tungsten was with chromium and iron. There is some
>promising new work in "intermetallics" which might yield even better 
long
>term performance. So far however, most of the intermetallic research 
has
>been with Aluminum for turbine blades with a sustained operating
>temperature around 300 C for only a few thousand hours of operational 
life.
>We need on the order of 500 - 900 C for tens of thousands of hours for 
hull
>materials and 2,000+ C for drives.

Yes; material will determine peak temperature, so that's important too.  
Most of the impact energy might go into temperature rather than x-rays.  
The outer material will obviously be the most important, so the most 
thought should go into that.

And then there's always cost...

As for doppler shift, I don't think our final design is going to go fast 
enough to matter.  But if we are going that fast, I recommend picking a 
destination that doesn't involve travelling in the galactic plane.  
That's where most of the interstellar x-rays come from, so if we were 
travelling at right angles to the plane of the Milky Way, there wouldn't 
be much of an upshift.

Ken