Moonwards

kim holder

17:43

so, i was thinking... maybe a reasonable estimate of the radiation dose within a dugout-style habitat, with the radiation blind beam design, could be to model it as two solid blocks of x dimensions either side of gap y, within a cuboid area of dimensions z.

or something similar to that... the repeating gaps have to be considered in the calculation, but the above i can imagine creating a spreadsheet for. Doing that with those gaps i can't...

in fact maybe it could be set up in Geogebra, though it would be a bit awkward, as the results in the xy plane would need to be multiplied by the factor for the xz plane, i guess...

and so a matrix for each would need to be set up

Hohmannfan

You want to model radiation from random directions in the beam design?

kim holder

yes. the first estimate i did was rough, and pretty labor intensive because i just plain drew it and measured the lines by hand. :P

and now i realize the radiation model i was basing it on probably assumed the dose i used over a half-sphere, not a full sphere like i used.

Hohmannfan

Do you have the dimensions for the beams and gaps? If so, I can do a calculation.

kim holder

well, doubling that previous dose puts it up to 0.07 Sv/yr, roughly, and i'd like to get it down to where it was before, at 0.35

but in a larger sense, i'd like to express the concept in a more convincing way, because i continue to feel it works.

if you set up the calculation, can it be adapted to output a figure if the beams, space, materials are changed?

Hohmannfan

That is my intention, to do it in a way that can handle a change in dimensions.

kim holder

17:58

i provisionally sketched in beams that are 6m tall, instead of 4, and closed the gaps to 75 cm, instead of 1 m, between beams 50 cm wide.

@Hohmannfan that would be awesome :)

Hohmannfan

What is your intended distance to the ceiling? That matters too.

kim holder

yeah, it is actually pretty high even in the first dugout, higher than shown, but it has multiple floors...

(higher than shown in that quick sketch up there)

it is 15 m from the floor to where the beams start. and i'm ignoring their arch shape. and let's see...

Hohmannfan

OK, I can send something over in a few days.

kim holder

30 x 50 m

do you think doing it that way is a valid approach?

Hohmannfan

30 x 50 x 15 m is the cuboid dugout volume?

kim holder

18:06

yes. it has slanted walls, but that probably isn't worth considering

now that i am putting in placement of crew quarters, there are ramifications i'm thinking about a lot more

Hohmannfan

Just a thought: Corners just have 1/4 of a hemisphere.

kim holder

?

Hohmannfan

3/4 of the hemisphere is blocked by solid regolith.

kim holder

for the quarters? they are in the hall, and i've added more protection now that i've reflected on how much of the outside they can see

and i think i'll go even further and recess them into the wall a couple of meters, though i don't want to recess them completely because i think they would get claustrophobic

the first hab has changed a fair bit, i've just gotten caught up in fixing stuff and haven't posted it

i can't put quarters for 40 people all in corners, but they are doing pretty well now, i think

the dining area is in a corner...

kim holder

18:33

still looks pretty ugly, but those cubicles are a decent size to bunk in, especially when a lot of facilities are shared by everyone

i added that blocky thing outside the entrance on the crater slope side, the slanted part will reflect in the sky, including the Earth. between that and the low wall, most of the radiation is blocked now, though there remains a thin strip that is a direct view to the outside.

the dugout side is a lot more awkward because of the backscattered radiation off the floor

Hohmannfan

If you stand in the middle of the 30x50x15 box, at the floor, the visible part of the sky is only 41.5% of the hemisphere. That is before you add the beams.

kim holder

that's a good start :)

Hohmannfan

I had to look up some sphere trig first, but now I am ready to start. Geogebra is kind of messy in 3D though:

kim holder

should i send you the doc with the figures i used for radiation through different thicknesses of material?

oh, it can handle 3d? i didn't know that, that really helps...

Hohmannfan

kim holder

18:47

gee... so straightforward...

Hohmannfan

I would need those radiation figures if you need a more complex model than block-not-block.

I think I am going to opt for JS for this problem.

kim holder

yeah, it makes a big difference. it is the thing that makes it so awkward. thicknesses less than a pretty large amount are worse than standing outside

that's why i'm not sure that the model without gaps works, unless you can include a factor for them

i'm so close to this, i didn't even think to mention that :P

Hohmannfan

What is the worseness as a function of thickness here?

kim holder

Hohmannfan

If I have understood this secondary radiation right, it first rises before it shrinks.

kim holder

18:56

yes, it rises a lot. very nasty.

Space Environments and Planetary Civil Engineering

►

Hohmannfan

What is this unit for thickness? g/cm^2

kim holder

that video talks about it starting at 39: 50

@Hohmannfan yes

Hohmannfan

Ah, is it density*thickness?

kim holder

they are using bulk density of 1.3 g/cm2 for loose regolith piled on top of a hab

Hohmannfan

And the beam density is?

kim holder

18:59

yes. i postulated the high iron, high titanium regolith melted into basalt would have a density of 3.1 g/cm2

maybe that should be dropped to 3

sorry, 3 g/cm3

and 1.3 g/cm3

for the loose regolith used in the model

Hohmannfan

The secondary radiation thing makes it more complex than I thought. Definitely JS stuff.

kim holder

yep. and we are also not considering backscatter, which i think is legitimate because the model this is all based on was for the open surface, and would have included that

in the model i used, i took angles with less than 129 cm of stuff between a person and the outside as the pink in the notes i have shown up there, because it was just way to much to hand measure in the crude way i had

red was less than 161 cm, yellow was 226, green 323

Hohmannfan

As calculating the protection properties of every part of the hemisphere is too complicated to do analytically, I am going to create a simulation, ray tracing random rays.

kim holder

ooooo

Hohmannfan

It is not possible (I think) to do it in another way without significant effort.

kim holder

19:08

tell me about it : /

there might be a way to do it in blender, if you know how to write the python

i can't quite figure out how to ask about that

i mean, i have dim memories of a js course that didn't even get this far, then i got into other things and never really used it

articles.adsabs.harvard.edu/cgi-bin/…

^ main reference doc

@Hohmannfan you see how i did that? i hooked you with a simple version of the problem, and then i added in all sorts of complications once you'd expressed interest. :D

unwittingly, of course :]

Hohmannfan

19:25

Case 1: ray hits nothing.
Case 2: ray hits top of beam, going all the way to the bottom.
Case 3: ray hits top of beam, escaping through the side of it, continuing to 1 or 5
Case 4: ray hits side of beam, going all the way to the bottom.
Case 5: ray hits side of beam, escaping through the side of it, continuing to 1 or 5

kim holder

yeah, that was what made me want to skip the gaps and fudge them in later in some reasonable way...

Hohmannfan

This is, despite what I thought, doable.

[Hoh goes coding].

kim holder

well, i hope it is interesting and enlightening :)

me, i go back to fixing ignorant mistakes in my early models now : /

Hohmannfan

20:14

Project started as "dugout.js" in the js folder at GitHub.

kim holder

cool. :) i looked at it for a bit. i get some of it, though i'm very rusty.

um, i have a slight problem which hopefully doesn't affect this calculation...

the angle of the sun to the colony is 5.5 degrees

with arches so high, if i leave them vertical almost no sun will arrive directly. i could reflect it in...

Hohmannfan

This only sends a bunch of random rays, and outputs how much beam material each of them passes through. It is the first step.

kim holder

i take it is the hard step

Hohmannfan

yeah.

I simplified the roof to be flat.

kim holder

but of course...

Hohmannfan

20:27

I have actually not tried to run any part of the code yet. Time for testing!

kim holder

so, i think, just to complete that thought, both sides of the arches have to be mirrored

because as soon as i contemplate tilting them slightly to align with the sun, that just sounds like a bad idea

that would only work well for part of the day anyhow

5.5 degrees is more than i thought

Hohmannfan

I know some of the radiation is going to be reflected too, but that is impossible to calculate accurately.

kim holder

yes, apparently setting supercomputers to crunch it still hasn't yielded results they are very confident in

but i do think reflection is part of the figure in the paper i cited

Hohmannfan

OK, now I have to run the script in a browser. The problem with that is that it is likely to crash. Expect me to be gone for a few minutes.

kim holder

hehehe, it's all part of the fun :)

Hohmannfan

20:39

Another simplification: The ray hits randomly on the beams, not at a fixed quantified range. That means the simulation is not valid for a single point. It is more like an average for a small region.

Not very important though.

kim holder

probably better really

i'm not a point

Hohmannfan

Yeah, Humans are not points, and they move around.

That is kind of interesting actually: The base is going to have "hot" and "cold" spots, like a microwave. Be careful to not place your bed in a "hot" spot of the room.

kim holder

yes, i've been realizing that more and more

it might even be a good idea to paint areas accordingly

the beds will be in good shape though

Hohmannfan

The pattern repeats roughly every 1.25 m, but apart from that, the properties are hard to predict.

That "painting the floor according to the amount of radiation" idea is pretty cool.

kim holder

the dose is lower under beams?

Hohmannfan

20:49

Close to them, yes, but at a distance, who knows?

kim holder

yep. it's pretty complicated.

Hohmannfan

Proposed modelling of protection: radiation increases linearly to 9x intensity at 150 g/cm^2, and decreases linearly to 0 at 600 g/cm^2

kim holder

mmm. you will have to go higher than that - the 600 g/cm^2 figure in that presentation slide was showing the point at which radiation levels fall to what they are outside on the surface

oh no - actually that is about 500 g/cm^2

but the graph above that is not plotted, and must tail off

annoyingly, they specify what the minimum acceptable shielding thickness is, but not what level of radiation that represents

no, actually, that doesn't help either...

how did i deal with this last time?

Hohmannfan

First result: (the code is fresh, so do not thrust the numbers yet) only 2.2% of all the random rays passing through the beams get less than 600 g/cm^2

The overall radiation is about 16% of what you get compared to no beams.

kim holder

oh - pretty good then. so, you take the fall off with thickness as linear, right?

yes - you just said :]

but i think that should be pushed up, 600 isn't considered safe

i found the other ref i used

gah, but it doesn't give actual radiation figures either. why on earth don't they do that?

so i guess the thing would be to extrapolate that graph by extending a line that where it would be 1000 on the x axis is essentially 0, in a straight line between there and the last plotted point at 500.

Hohmannfan

21:08

OK.

That 2.2% number is bullshit. Forget it.

kim holder

so it decreases to 3/5 of its last value at the 700 g/cm^2 mark, which is what they say is the minimum acceptable long term shielding which would be 0.012 Sv/yr by their scale (which is off)

hahahah

Hohmannfan

Is this model OK?

kim holder

i was just asking myself that. because one would assume most of the curve is close after point C, which would be very helpful

Hohmannfan

This model gets 19% compared to no beams.

kim holder

21:24

when they say 0.012 Sv/yr, they mean to say 0.12, i think, because at 0 shielding the level they show is 0.015, which isn't right. It should be 0.15. That is the level on the lunar surface (maybe, according to one guess)

Hohmannfan

Are the other numbers in that graph off too? That is really important.

kim holder

0.05 Sv/yr is the level acceptable to the board governing radiation worker exposure. If we guess that is conservative, which it probably is, and double it, the 700 g/cm^2 figure is still a hair too high

@Hohmannfan i assumed they were all off by the same amount, and slid the decimal over one

Hohmannfan

Ah, that is good.

kim holder

because it is a very weird mistake, and rather unsettling in a reference document, but i can come up with no other explanation

sci.esa.int/Conferences/ILC2005/Manuscripts/…

^ that is the other ref doc used in the video, and it says on page 7 near the bottom maybe 0.25 Sv/yr on the surface. But there is debate about that, 0.15 is still a lot higher than earlier estimates

Hohmannfan

If I have not committed any horrible mistake, the conclusion of the simulation is that your beam design removes about 80% of the radiation.

kim holder

21:30

so 0.03 Sv/yr, say

that's pretty good

Hohmannfan

@kimholder No, that is what the beams block. Remember that the sides of the cuboid takes a lot too.

More like 0.012Sv/yr

kim holder

oh yeah - well that's peachy :)

Hohmannfan

That is for the location in the middle of the building, at the bottom of the hole.

Your mileage may vary.

kim holder

right. close to the beams, in the gap, is it much higher?

Hohmannfan

In between the beams?

kim holder

21:35

yes

Hohmannfan

...2 hours later

kim holder

:)

well, it is quite reassuring to have this analysis. i would love to have a version to play with

is it done enough to run then?

Hohmannfan

I can add tweaking instructions to the script.

Pretty much done, yes.

kim holder

i'll download it and try to make it work. it's good for me, don't feel obliged to walk me through it in detail.

Hohmannfan

OK, the latest version is up now.

kim holder

21:41

ok, thanks :)

Hohmannfan

Interesting: If the beam height is 3.4m instead of 6, there is no point is shielding at all.

kim holder

huh. yes, that is interesting

Hohmannfan

And increasing it to 7m makes the radiation drop to 8%. Less light though.

kim holder

yes, i'm wondering about the light

at the height they are, none of it will be direct

Hohmannfan

What is the latitude again?

kim holder

21:44

which might be good, actually

4.5 south, and the moon is 1.5 degrees to the ecliptic.

so 6 degrees to 3 degrees i take it

i thought about angling the sides of the beams in just a bit, thinking maybe it would help them function as radiators during the night

Hohmannfan

Hmm. I wonder if the 5.145° inclination to the ecliptic matters.

No, it does not.

kim holder

thermal control i haven't even gotten into yet, but i hope to use the mass of the big boulders of Gibraltar to keep it in a reasonable range even if there were no other controls

so a reduced gap is probably a good thing. If say 85% of what falls between the beams enters, and it is 3/5 of the amount on the surface because of the beams, that is still about 650 W/m2 during the day

Hohmannfan

But a reduced gap also mean a much higher loss even at small changes in the solar angle.

3 degrees is 40% loss (without mirrors). even at a gap of 0.75m at a beam height of 6m.

kim holder

hm

i was fudging in the mirrors when i put 85% enters

Hohmannfan

Mirrors is a good idea. Then a variation > 10 degrees does not matter.

kim holder

21:58

still... since there seems to be space to lower their height safely, maybe it would be a good idea

though i imagine they couldn't be shortened more than 50 cm or so before the rise in radiation is problematic

in fact i'm wondering if it would be a good idea to use reflectors near dawn and dusk

blender is going to hate the number of mirrors i'm using :)

in fact i'm pretty sure my machine isn't capable of rendering that, even if i knew how to do it properly. i'll have to figure out a way to represent that

and when i say 'mirrors', of course i mean aluminum foil

Hohmannfan

And when you say "aluminum foil" you mean the most awesome material in the world.

kim holder

it should be on the lunar flag :)

oh... a lunar flag...

Hohmannfan

The Moon needs a flag. And it should definitely include aluminium foil.

kim holder

alright, i should do some of the boring but really much more pressing work of fixing all the blender models...

at least it should look quite cool when i'm done

and now i will confidently put in arches with the dimensions described

Hohmannfan

Is the mirror rendering issue a RAM problem?

kim holder

22:10

yeah, on the gpu

especially since they are each reflecting directly into another mirror, doing that pesky infinity thing

Hohmannfan

Uh? So it is not RAM, but a processor problem.

kim holder

well, the graphics card has its own memory, and as i understand it blender really uses that

which is what i really meant

but considering this is a setup with, let's see, 82 mirrors facing each other, it would likely be a gpu problem too

Hohmannfan

Rendering-over-night: 20gigs of swap, 12 hours of gpu processor time.

"Oh wait, I wanted that point of view rendered instead!"

kim holder

22:30

view from near floor level at the outlet of the hall. fans became noisier than i've almost ever heard them. took only a minute though...

so, that's what the beams would really be like

in proportions, that is

Hohmannfan

It does look good.

But it is 13000000 kg

kim holder

there is an awful lot of material in those beams, but i don't think setting up enough stations to produce them would be hard

Hohmannfan

And you have to dig a 22500 m³ hole too.

kim holder

meh, the initial set of blocks can be placed and tensioned so the hab can be filled with a partial atmosphere, and then the rest can be set on top.

oh, that doesn't even count the hall :)

the dugout will be easier. i figure trench with the lenses, melting a furrow that you make deeper with successive passes

Hohmannfan

You can start small, and add more beams and dig a larger hole later.

kim holder

22:39

then bore with the fiberoptic solar concetrator, lay microcharges of explosives, and break up the regolith for removal

Hohmannfan

I have worked on removing 6 m³ of rock before. This is almost 4000 times more.

kim holder

its doable to go big, and it is critical to success. it is expensive but it is worth it to guarantee success.

it is mostly powder there. and you couldn't melt it with the sun. and you aren't a machine.

Hohmannfan

2/3 :P

kim holder

and i assume you didn't loosen it with explosives

Hohmannfan

No, solid granite.

kim holder

22:44

well, solid granite is a bit different, i think

@Hohmannfan :)

Hohmannfan

I am still not very sure about the consistence of regolith.

kim holder

it packs very hard. in many ways this is an advantage.

packed as tight as it will go, it is still 50% voids

if you melt it, the melt will sink into it. but once you have cleared away the loose stuff, what is left is a stable face unlikely to collapse

there will be a lot of rocks mixed in. there will be challenges. personally, i'm reassured by the reliable presence of 80 to 100 kW of heat focused to a tight spot available all the time for 2 week stretches.

Hohmannfan

Orbital mirror + the lunar landscape = the world's largest canvas.

kim holder

heehee.. that's sort of scary

Hohmannfan

If you have a 100 kW beam constantly pointed at the hole for 10 years, you have still only supplied 1/4 of a kW/h per kg.

kim holder

22:55

most of it isn't melted though. that is to create the trench around the perimeter so the charges laid in a grid in the middle can be restricted to that area and not damage the shape of the walls

and why on earth would you only have one?

you are pointing a lens. there should be at least 20 or 30

Hohmannfan

100 kW is a 73 m² lens. 20 to 30 of them?

kim holder

sure. that's 9 x 9 m, wafer thin each one

Fresnel lenses

Hohmannfan

This hall is perhaps too big.

kim holder

it has to be this big. if it seems too expensive, the position would be that we are presenting something farther in the future, but it has to be big. The main point here is to present what is possible if we aren't hemmed in. What the prospects are that make it worth it.

if you can't show that, there is no compelling reason to go. If you can, the resources will appear in due course

and the calculations are complicated. some fraction of 1% of the regolith is melted in this excavation scheme, i don't know how much.

and there is also the question of how much you could get an excavator to do through brute force, powered by the nuclear reactors beaming power to it as microwaves

an excavator with a rock ripper on earth can advance mighty quickly

and there are those huge boulders there so it can get purchase instead of tipping over

getting the very beginning of the colony well explained is the hardest challenge. it is probably better to do the first hab instead and then work backwards to scale the equipment, power, and supplies needed to achieve that using the best option

as near as i can tell, this is the best option for something of this kind, and something of this kind is the only thing that would convince people to move there and live there

kim holder

23:59

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