ApolloHoax.net
Apollo Discussions => The Reality of Apollo => Topic started by: Bob B. on September 04, 2014, 12:13:47 AM
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I've been looking at the AE-8/AP-8 Radiation Belt Models and I have some questions I'm hoping somebody can answer.
► My first question is rather straightforward but I just want to verify that I'm correctly understanding what I'm seeing. The fluxes appear to for the number of particles with an energy equal to or greater than a given value. For example, let's say the flux for protons => 10 MeV is 104 protons/cm2-s and the flux for protons => 50 MeV is 103 protons/cm2-s. Then am I correct in saying that the number of protons with energies between 10-50 MeV is equal to 104 - 103 = 9000 /cm2-s?
► The coordinates for particles within the belts are L and B/Bo. I understand the meaning of L and B/Bo but don't know to transform a known position in polar or rectangular coordinates to L and B/Bo coordinates.
L is pretty straightforward as it is simply radial distance from the center of Earth in Earth radii, which I assume is measured in the geomagnetic equatorial plane and not on a direct line to my known position. In other words, if I'm 3 radii out from the center of Earth along the equatorial plane and 1 radii above the equatorial plane, then my position is SQRT(32+12) = 3.16 radii from Earth's center. I assume my L coordinate in this case is 3 and not 3.16. Is this correct?
B/Bo is a much bigger problem for me. I understand that B/Bo is magnetic field strength normalized to the equatorial value, and I know how to calculate the value of Bo. What I haven't been able to figure out is how to find the value of B for a given physical location within the magnetic field. Say, for instance, we're at the previously given location of 3 radii along the equator and 1 radii above the equator. Surely there is a way to find the field strength, B, at that location. Is there a formula or some other means by which I can determine this? Any help provided is greatly appreciated.
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The following might be what I need regarding the value of B:
http://en.wikipedia.org/wiki/Dipole_model_of_the_Earth's_magnetic_field
I still welcome additional answers/comments.
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To answer your first question, yes.
[pause for laughter]
Which is to say, the model computes fluxes for the given energy and greater, so if you specify some low energy E1 and some high energy E2 in a single run, the flux values you get back for each energy is for that energy and greater. So yes, the flux for the energy range (E1,E2) is found by the subtraction you derived.
The B-L coordinates system makes grown men either cry, drink, or both. And for any practical purpose, no one uses the web interface or computes the McIlwain coordinates by hand. In fact, I'm pretty sure I would have to spend a morning with McIlwain's paper again before I could even remember how to do it. The practical way is to use additional computer programs from NSSDC at Goddard that automate converting from geodetic coordinates or orbital elements to B-L coordinates, and further automate stuffing long sequences of B-L coordinates into the AX8 models, which they embed.
I think Jim Vette wrote most of that code. NASA/Goddard repackages their software approximately every eight minutes, so I don't know off the top of my head what package it's in these days, or if you have to pay for it. Let me check.
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Thanks for your answers, Jay. If you can find out anything more about that NASA/Goddard software I would be very grateful.
I also happened across the following, which gives equations I might be able to use:
http://onlinelibrary.wiley.com/doi/10.1029/JZ069i023p05089/abstract
Although not defined in the abstract, the previously referenced Wikipedia article defines R as the radius and lambda as the magnetic latitude. One problem is that the image quality is so poor that I'm having trouble making out some of the text. I think I'm seeing:
(1) B = Bo (4 - 3 cos2(lambda))1/2 / cos?(lambda)
(2) R = L cos2(lambda)
(3) Bo = M/L3
The only number that I really can't make out at all is the exponent for cos(lambda) in the denominator of equation (1), which I show above as a question mark. Any idea what that is?
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Thanks for your answers, Jay. If you can find out anything more about that NASA/Goddard software I would be very grateful.
I also happened across the following, which gives equations I might be able to use:
http://onlinelibrary.wiley.com/doi/10.1029/JZ069i023p05089/abstract
Although not defined in the abstract, the previously referenced Wikipedia article defines R as the radius and lambda as the magnetic latitude. One problem is that the image quality is so poor that I'm having trouble making out some of the text. I think I'm seeing:
(1) B = Bo (4 - 3 cos2(lambda))1/2 / cos?(lambda)
(2) R = L cos2(lambda)
(3) Bo = M/L3
The only number that I really can't make out at all is the exponent for cos(lambda) in the denominator of equation (1), which I show above as a question mark. Any idea what that is?
I wonder whether this helps?
http://www.spacewx.com/Docs/AIAA-655-543.pdf
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Thanks for your answers, Jay. If you can find out anything more about that NASA/Goddard software I would be very grateful.
I also happened across the following, which gives equations I might be able to use:
http://onlinelibrary.wiley.com/doi/10.1029/JZ069i023p05089/abstract
Although not defined in the abstract, the previously referenced Wikipedia article defines R as the radius and lambda as the magnetic latitude. One problem is that the image quality is so poor that I'm having trouble making out some of the text. I think I'm seeing:
(1) B = Bo (4 - 3 cos2(lambda))1/2 / cos?(lambda)
(2) R = L cos2(lambda)
(3) Bo = M/L3
The only number that I really can't make out at all is the exponent for cos(lambda) in the denominator of equation (1), which I show above as a question mark. Any idea what that is?
It's almost certainly a 2. I know NSSDC has code to convert R-λ to B-L, but I can't remember for sure what the core equations are so I can't confirm that yours are they. However, the subexpression cos2 λ is reasonably ubiquitous as a normalization divisor.
This source https://www.spenvis.oma.be/help/background/magfield/rlambda.html gives Eqn (3) as
B = M R - 3 (1+3 sin2λ)1/2
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http://www.spacewx.com/Docs/AIAA-655-543.pdf
Eqn 8 seems to be stabbing at the uniform-dipole problem, so ... maybe?
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I wonder whether this helps?
http://www.spacewx.com/Docs/AIAA-655-543.pdf
It appears that equation 24 in that document is the same as the previously posted equation 1. Equation 24 shows the exponent that I couldn't read as 6. It's looks that might be the equation I need to do what I want to do, though I have some more reading to do before I'm convinced.
What I'm wanting to do is to calculate a better estimate of the radiation dose received along the trajectory of Apollo 11 for my web page,
http://www.braeunig.us/apollo/apollo11-TLI.htm
I already have all the R and lambda coordinates from my previous calculations, so all I need are equations to convert them to L and B.
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It appears that equation 24 in that document is the same as the previously posted equation 1. Equation 24 shows the exponent that I couldn't read as 6.
Equation 9 on Jay's linked page is your equation 1 (with a 6 exponent as above), isn't it?
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This source https://www.spenvis.oma.be/help/background/magfield/rlambda.html gives Eqn (3) as
B = M R-3 (1+3 sin2λ)1/2
(above equation edited by me, changing R - 3 to R-3)
That is almost the same equation given in the Wikipedia article, in which M is replaced by Bo
B = Bo R-3 (1+3 sin2λ)1/2
After doing a little rearranging, I currently have two equations,
B/Bo = R-3 (1+3 sin2λ)1/2
B/Bo = (4 - 3 cos2λ)1/2 / cos6λ
Of these I'm more inclined to believe that latter. The reason is because the online AE-8/AP-8 model (http://ccmc.gsfc.nasa.gov/modelweb/models/trap.php) allows a range of B/Bo values of 1 to 1000. The first equation doesn't yield values anywhere close to that. On the other hand, the second equation gives B/Bo = 1000 when λ = 69.3855o.
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It appears that equation 24 in that document is the same as the previously posted equation 1. Equation 24 shows the exponent that I couldn't read as 6.
Equation 9 on Jay's linked page is your equation 1 (with a 6 exponent as above), isn't it?
Yes it is. Rearranging equation (7) we have,
RL-1 = cos2λ
Substituting cos2λ for RL-1 in equation (9) gives,
(BBo-1)2 (cos2λ)6 - (4 - 3 cos2λ) = 0
Which when rearranged gives,
B = Bo (4 - 3 cos2λ)1/2 / cos6λ
I'm starting to feel pretty confident that this is the equation I want to use.
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I'm trying to reconcile Bob B's two equations:
B/Bo = R-3 (1+3 sin2λ)1/2
B/Bo = (4 - 3 cos2λ)1/2 / cos6λ
Using equations 6 and 8 from Jay's linked page, the first one seems to be missing an L3 factor. If you add it in and rearrange you get the second equation.
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I'm trying to reconcile Bob B's two equations:
B/Bo = R-3 (1+3 sin2λ)1/2
B/Bo = (4 - 3 cos2λ)1/2 / cos6λ
Using equations 6 and 8 from Jay's linked page, the first one seems to be missing an L3 factor. If you add it in and rearrange you get the second equation.
That first equation comes from this Wikipedia article (http://en.wikipedia.org/wiki/Dipole_model_of_the_Earth's_magnetic_field). Their definition of B0 is,
"First, define B0 as the mean value of the magnetic field at the magnetic equator on the Earth's surface. Typically B0 = 3.12x10-5 T."
which is not the same definition used in other sources. What Wikipedia is calling B0 is defined in other sources as M. These other sources define B0 as the value of B along the magnetic equator at a distance L (in radii) from the center of Earth, calculated by B0 = M/L3. At Earth's surface, B0 = M.
Therefore, the first equation is incorrect if we are to be consistent with our definitions. It should be rewritten as,
B = MR-3 (1+3 sin2λ)1/2
which is equation (1) in Jay's link. Substituting M = B0L3 we get,
B/B0 = L3 R-3 (1+3 sin2λ)1/2
(edited to add)
Eureka! That does it. Although it is not apparent at first observation, the following two equations are indeed equivalent and yield the same answers:
B/B0 = L3 R-3 (1+3 sin2λ)1/2, where R = L cos2λ
B/Bo = (4 - 3 cos2λ)1/2 / cos6λ
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Based on our conversations here, below is how I think it's done. Comments and criticisms are welcomed. Thanks to everybody who helped.
(http://www.braeunig.us/pics/MagField.gif)
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That seems graphically right, Bob.
As for the quantitative stuff, yeah I transcribed those equations completely wrong. I was late for Comic Con and didn't really have time to delve much further.
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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.
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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?
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► 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.
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Hi Bob,
This might be of some use :
http://physics.nist.gov/PhysRefData/Star/Text/PSTAR.html
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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.
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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.
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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.
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Both methods yield very similar results - about 7 g/cm2.
...which is the published figure.
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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.
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I just finished writing my article...
Apollo 11 and the Van Allen Belts (an estimate of the radiation dose received) (http://www.braeunig.us/apollo/VABraddose1.htm)
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.
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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.
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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".
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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.
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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.
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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.
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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.
Thanks and you're welcomed. I'm still finding my own nits and have already made a few minor edits.
Here's another source of shielding that might be worth considering: the propellants (MMH and N2O4) for the CM reaction control system.
I the propellant is included in what I refer to as "secondary shielding" since my calculation of secondary shielding is based on the total CM mass (which includes the propellant mass). I didn't get real specific about all that is included in secondary shielding because I'm largely ignoring it in my calculations. Since the hull by itself does such an effective job, I didn't find it necessary to write much about secondary shielding. However, I have changed the description from "equipment, instrument panels, etc." to "equipment, instrument panels, propellant tanks, etc."
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Here's an interesting thought experiment. Let's say you're a health physicist supporting an Apollo mission when it's unexpectedly hit by a severe CME. An abort is not viable. How could the astronauts best protect themselves until it's over?
We'll assume they're in the CSM.
The safest place would seem to be underneath the couches, as close to the protection of the SM as possible. If the radiation is directional, you'd reorient the SM. (The particles follow the sun's interplanetary magnetic field so you wouldn't necessarily put it directly between the astronauts and the sun.)
You'd also want to cover yourself with as much shielding as possible to protect against radiation coming through the sides of the CM, probably by strapping just about anything movable to the tops of the couches: food, trash, LiOH canisters, other supplies, etc. If the LM is attached, the PLSSes would probably make good shields, especially if their tanks are full. Water is especially good shielding, so loading it (or urine!) into bags seems like a good idea.
Anything else?
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This contingency was planned for. Anisotropic flux was indeed mitigated by orienting the CSM stack to place the bulk of the SM in the direction of highest flux. The crew carried a portable radiation detector/counter that they were supposed to use to find the most shielded place in the CM cabin.
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This contingency was planned for. Anisotropic flux was indeed mitigated by orienting the CSM stack to place the bulk of the SM in the direction of highest flux. The crew carried a portable radiation detector/counter that they were supposed to use to find the most shielded place in the CM cabin.
My article briefly mentions this in the section "Radiation Plan for Apollo."
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On the outgoing journey, the LM would shield quite well from the front, the SM from the back. That leaves the sides. So hide in the bottom compartment, stacking whatever loose supplies around the side?