AE-8/AP-8 Questions

[Bob B.]

I've got a few more questions to which I welcome your reponse.

As I mentioned earlier, I'm trying to estimate the radiation dose received by Apollo astronauts during their transit through the Van Allen Belts.  I want to be reasonably accurate but I'm also willing to make some simplifying assumptions.  Here are some of the things I'm currently considering in my analysis:

► It looks like the only particles that will penetrate the spacecraft shielding are the highest energy protons.  Let's say I determine that only protons > 100 MeV have the necessary energy.  Of course by the time a 100 MeV proton has passed through the shielding it is no longer a 100 MeV particle; it has lost energy during its penetration.  In this case, is it a reasonable assumption to say it takes 100 MeV to penetrate the shielding, thus all particles that penetrate will have 100 MeV less energy than their original energy?

► I've found equations that give the penetration depth of high-energy protons in air and aluminum.  Of course the CM shielding consists of, in order of penetration, the epoxy resin heat shield, the stainless steel sheet/honeycomb, the fibrous insulation, and the aluminum sheet/honeycomb.  If I convert the penetration depth in aluminum to g/cm², is it reasonable to assume that an equal mass of some other material will have an equivalent stopping power.  It may not be exact but I think it will be close enough.  For instance, if I use the formulas to calculate the penetration of a 100 MeV proton and convert to g/cm², I get 8.8 in air and 9.8 in aluminum.  If air and aluminum are that close, then surely other materials such as epoxy resin and stainless steel will be similar.

► Although the numbers indicate that no electrons will penetrate the hull, I want the account for Bremsstrahlung.  I've found an equation that gives the fraction of the original electron energy that is converted to X-Rays.  Surely these X-rays scatter in all directions, so I think it is a valid assumption to say that 50% of the X-rays travel harmlessly back into space.  Do you concur?

► I haven't found much (or at least not much that I understand) about Bremsstrahlung for protons.  Most of the literature I've found seems to focus on electrons.  Is secondary radiation from protons a big issue that I need to account for?  If so, does anybody have any information or equations that will help me?

I'm continuing to work my way through this analysis, so I may answer some of my own questions or come up with new questions.  I appreciate any help you can give me.  I'll post a link to my article when it is finished so that I can get some critical feedback.

Thanks.

[Bob B.]

I just remembered another question I left off my list...

► I found a source that gave me the thickness of the two stainless steel facesheets sandwiching the SS honeycomb that makes up the structural base for the heat shield.  These facesheets are 1.9 mm (0.075") thick.  Despite much looking, I have not been able to find any source for the thickness of the two aluminum facesheets sandwiching AL honeycomb that makes up the inside pressure hull.  Does anybody know the thickness of the aluminum sheets?  If I can't find the answer, I'm going to have to assume something.  I could assume the same as the SS, but they could be thinner than that (I believe then LM hull was considerably thinner than 0.075").  Does anybody have any suggestions on what I should assume?

[Bob B.]

► It looks like the only particles that will penetrate the spacecraft shielding are the highest energy protons.  Let's say I determine that only protons > 100 MeV have the necessary energy.  Of course by the time a 100 MeV proton has passed through the shielding it is no longer a 100 MeV particle; it has lost energy during its penetration.  In this case, is it a reasonable assumption to say it takes 100 MeV to penetrate the shielding, thus all particles that penetrate will have 100 MeV less energy than their original energy?

I think I found the answer to my own question.  Roughly speaking, when the energy of a proton doubles, its penetration range goes up by a factor of about 3.4.  If the energy loss through the material were constant, then in the same distance that a 100 MeV proton loses 100% of its energy, a 200 MeV proton will lose only 1/3.4 of its energy, i.e. 200 x 1/3.4 = 59 MeV.  But it's even less than that because the rate of energy loss is not constant; it increases as the proton slows down.  I've found the rate of energy loss in graphical form, so I should be able to derive an equation.

[Trebor]

Hi Bob,
This might be of some use :
http://physics.nist.gov/PhysRefData/Star/Text/PSTAR.html

[Bob B.]

Hi Bob,
This might be of some use :
http://physics.nist.gov/PhysRefData/Star/Text/PSTAR.html

Thanks, that helped with my question #2 (below).  It looks like the range in g/cm² will be lower than aluminum in the heat shield and insulation, and greater than aluminum in the stainless steel.  When I average it out across the entire composite structure, it comes out pretty close to the aluminum value.  I think it is a fair simplification to calculate the aluminum value and assume that applies across the entire thickness.

► I've found equations that give the penetration depth of high-energy protons in air and aluminum.  Of course the CM shielding consists of, in order of penetration, the epoxy resin heat shield, the stainless steel sheet/honeycomb, the fibrous insulation, and the aluminum sheet/honeycomb.  If I convert the penetration depth in aluminum to g/cm², is it reasonable to assume that an equal mass of some other material will have an equivalent stopping power.  It may not be exact but I think it will be close enough.  For instance, if I use the formulas to calculate the penetration of a 100 MeV proton and convert to g/cm², I get 8.8 in air and 9.8 in aluminum.  If air and aluminum are that close, then surely other materials such as epoxy resin and stainless steel will be similar.

[JayUtah]

Despite much looking, I have not been able to find any source for the thickness of the two aluminum facesheets sandwiching AL honeycomb that makes up the inside pressure hull.

I found a drawing with a specification that suggests face thicknesses between 0.035 and 0.065 inch per face.  That seems too thin to me, so I'm hoping to double-check.  I know for control surfaces on high-end, high-performance airframes that's credible, but those are custom sandwiches.

[Bob B.]

I found a drawing with a specification that suggests face thicknesses between 0.035 and 0.065 inch per face.  That seems too thin to me, so I'm hoping to double-check.  I know for control surfaces on high-end, high-performance airframes that's credible, but those are custom sandwiches.

That's helpful.  It you find anything more, please let me know.

I'm well into my analysis and have a good portion of the web page finished.  For the part where I analyzed the area density of the hull, I used two methods.  I first added up the mass of all the layers using thicknesses and densities.  In the second method I took the mass of the heat shield (848 kg) and spacecraft structure (1567 kg) and divided by the surface area of the CM (36 m²).  Both methods yield very similar results - about 7 g/cm².

[JayUtah]

Both methods yield very similar results - about 7 g/cm².

...which is the published figure.

[VQ]

I found a drawing with a specification that suggests face thicknesses between 0.035 and 0.065 inch per face.  That seems too thin to me, so I'm hoping to double-check.  I know for control surfaces on high-end, high-performance airframes that's credible, but those are custom sandwiches.

Why does that seem too thin? In my industry, self-supporting ductwork is routinely fabricated out of metal in that thickness range (20-14 gauge) or thinner, even in aluminum.

[Bob B.]

I just finished writing my article...

Apollo 11 and the Van Allen Belts (an estimate of the radiation dose received)

I request proof readers.  I welcome proof reading of any kind - spelling, grammar, math, etc. - but what I really want is to have my logic scrutinized.  I want to make sure that I'm correctly understanding the problem; that I'm interpreting data, terminology and definitions correctly; that I'm making valid assumptions; that I'm using the right equations and solving them correctly; etc.  I understand that I may have oversimplified things in some cases, but that's OK.  This analysis is intended just to get into the ballpark, not to be balls-on accurate.  Criticisms are welcomed.

I'm going to be out of town and without Internet access for the next couple of days.  So if I don't immediately respond to your posts, please have patience.  I'll be back.

Thanks in advance for your help.

[Allan F]

You could mention en passant that the SM with it's 30+ tonnes shielded the CM from that direction very effectively, so that your calculations err on the high side, with relation to the dosis received.

[Bob B.]

You could mention en passant that the SM with it's 30+ tonnes shielded the CM from that direction very effectively, so that your calculations err on the high side, with relation to the dosis received.

I was going to say something about that but forgot.  I do mention the SM in the "Radiation Plan for Apollo" but I can probably work it into the part about "Spacecraft Shielding".

[JayUtah]

Why does that seem too thin? In my industry, self-supporting ductwork is routinely fabricated out of metal in that thickness range (20-14 gauge) or thinner, even in aluminum.

Because the structures fabricated from those sandwiches are meant to accept substantial additional loads.  There are some spacecraft chassis design methods that employ nothing but aluminum honeycomb as the primary structural elements.

[ka9q]

I just finished writing my article...

Outstanding work, Bob. I can't see any major problems at all, but if I go through it again I'll probably find some very minor nits.

Thanks so much for doing this.

[ka9q]

Here's another source of shielding that might be worth considering: the propellants (MMH and N2O4) for the CM reaction control system. Both contain fairly low-Z atoms that should be good at stopping charged particles. The tanks were arrayed around the outside of the pressure hull at the base of the CM cone (i.e., close to the RCS thrusters), forming a ring around the astronaut couches. These tanks were full until SM separation, i.e., until after both VAB transits. You'd have to do a geometrical analysis to see how much solid angle each one covers, but it clearly extends the shielding they got from the SM and its tanks behind them.

The SPS tanks were full on the way out and nearly depleted on the way back, but the tanks themselves (not to mention the SM structure, fuel cells, battery(ies) (A14 onward), H2 and O2 tanks, etc) undoubtedly provided a lot of shielding mass.

Where was the potable water tank? That could be another significant shield if it's close to the couches.