Project Roskosmos "тайга" Lunar Lander

N_Molson

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There are now tool compartements on the lander chassis. There could be a spare inside, not a big deal ;) Titanium tools would be perfect, both light and strong !

Edit : uh, of course, the chassis stays on the Moon... So I could clip an "emergency wrench" outside the cabin...

At worse, the LM could be operated with a broken door. The cosmonauts keep the suits on, and plug them directly on the ECLSS. Of course, manoeuvering the LM with a full Orlan suit on seems difficult, but it still should be possible to initiate an autopilot ascent sequence, and once in orbit, the SoyuzTMA can be the active vessel in the docking procedure (the BO would have to be depressurized then repressurized).

And we should list the contingency scenarios :

1) Once landed, the door is jammed, or the valve doesn't work. It could be opened from the inside with a mechanical emergency system. Or, more wisely, the LEVA would have to be scrubbed.

2) If we take this abort scenario caused by a jammed hatch, we can imagine that the docking mechanism is jammed too. Then the cosmonauts would have to manually depressurize/open the hatch, the one inside the TMA would depressurize the BO, then the 2 others open the BO EVA hatch from the outside.

3) Once landed on the Moon, the LM door is jammed while the 2 cosmonauts are on LEVA. They have to use tools to manually open the door from the outside.

Only scenario 3 requires an external manual opening. And if it's possible, I would let the door open during the EVAs. It would be interesting to know what was the procedure during the Apollo missions.

---------- Post added at 07:01 PM ---------- Previous post was at 05:16 PM ----------

Some more work on the hatch :

11_04_27_20-55-03_LM.jpg


The "grey disc" is going to be a mechanical manometer, with an animated needle giving the internal pressure.

I'd need to translate in russian :

"In case of emergency, loose the bolt to vent cabin atmosphere" :p

---------- Post added at 11:08 PM ---------- Previous post was at 07:01 PM ----------

Now features LiOH canisters replacement ; LiOH canisters become ineffective after 12 hours (their efficiency slowly decreases over the hours ; at some point the CO2 balance becomes positive). The replacement takes a few minutes, and you need to switch off the fan to do that. It consumes 1 canister from the inventory, and adds 5.5 kg of waste when the process is finished.

11_04_28_01-04-28_LM2.jpg
 
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mc_

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I'd need to translate in russian :

"In case of emergency, loose the bolt to vent cabin atmosphere" :p

При аварии открыть клапан, выпустить воздух из кабины.

Someone russian-speaking should check it, i'm not sure to completely understand your post - too much english text at once.

Emergency signs is not the thing where we would allow a typo or misunderstanding, right?
 

N_Molson

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Now I'm focusing on the crew management. Since I tried to make the cabin environnement as realistic as I could (I still have to code temperature), it would be interesting in terms of gameplay that those parameters affect the crew health.

So I'm going to use, for now, the good old "HitPoints" system. The idea : some actions will be available only if you have enough hitpoints. It could be something like :

100%-90% : Very good condition - All actions allowed

90%-80% : Good condition - All actions allowed, light penalty on (L)EVA duration

80%-70% : Fair condition - All actions allowed, moderate penalty on (L)EVA duration

70%-60% : Average condition : All actions allowed, heavy penalty on (L)EVA duration

60%-50% : Below average condition : (L)EVA is not allowed, light penalty on Water/Food/O2 consumption.

50%-40% : Bad condition - (L)EVA is not allowed, moderate penalty on Water/Food/O2 consumption.

40%-30% : Very bad condition - (L)EVA is not allowed, moderate penalty on Water/Food/O2 consumption.

30%-20% : Critical condition - Loss of the flight controls, unable to replace LiOH canisters, heavy penalty on Water/Food/O2 consumption.

20%-10% : Loss of consciousness - No actions available.

10%-0% : Dying - Urgent medical attention required (death unless a fast docking to the SoyuzTMA).

Some factors would remove hitpoints : too much CO2, not enough Water/Food available, cabin pressure/temperature, EVAs, excessive angular velocity...

The crew should be able to slowly resplenish hitpoints when the environnemental conditions are nominal, maybe at a faster rate when landed (lunar gravity helps to feel better). Age, Weight, Heartbeat uMMU parameters could be used to determine how well a given crewmember can endure harsh conditions.

That means a lot of coding, but I think it's possible to do, and that would add a nice challenge : take care of your crew, or face the consequences ! It would be shameful to land on the Moon with the crew too exhausted to perform any EVA :facepalm:
 

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I like that a lot. The stressors include hypoxia and hyperoxia, loss of sleep (how many hours the crew has had in between operating the craft's controls), being outside the thermal comfort zone, among others.
 

N_Molson

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how many hours the crew has had in between operating the craft's controls

Great ! That's the idea I was searching for to define rest ! :thumbup:
 

Urwumpe

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I would maybe have made fatigue as second dimension there, with low health status making you more easily susceptible to fatigue induced problems.

Like reacting slower or falling a sleep for a second.
 

N_Molson

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I would maybe have made fatigue as second dimension there, with low health status making you more easily susceptible to fatigue induced problems.

Like reacting slower or falling a sleep for a second.

Yeah, I considered this idea, but that would be for a v2.0 ; otherwise the add-on will not be released before next X-mas :lol:

My C++ skills are still basic and I'm currently fighting with "static_cast" that seems to act randomly...
 

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Yeah, I considered this idea, but that would be for a v2.0 ; otherwise the add-on will not be released before next X-mas :lol:

My C++ skills are still basic and I'm currently fighting with "static_cast" that seems to act randomly...

static_cast<> is usually as reliable as death and taxes. :lol:

The real magic happens when you have to select which cast is the fitting one.

reinterpret_cast<>? dynamic_cast<>? static_cast<>? const_cast<>?

I know the christmas problem too well (I am on the other side of it already, it isn't really hell, but heaven has less flames)
 
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N_Molson

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I've done some progress, but I lack of technical knowledge in a lot of areas :

- if there is no internal heating, what temperature would you get inside the module (provided there is a 300 mmHg atmosphere) ? It will be greater that 0 K because of the exposition to the Sun. And it seems that the Apollo 13 crew survived with all the heaters off.

- How much time (grossly) will it take for a 1 K temperature elevation inside the module ? Which heating power is required for a 11 cubic meters cabin ?

Idea for a next version : simulating the Moon shadow cone. Geometric calculation should allow that (Spacecraft distance to the Sun/Moon distance to the Sun).

Other idea : simulating a radio blackout cone.
 

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if there is no internal heating, what temperature would you get inside the module (provided there is a 300 mmHg atmosphere) ? It will be greater that 0 K because of the exposition to the Sun. And it seems that the Apollo 13 crew survived with all the heaters off.

no internal heating means also no crew and no electronics, heaters are for additional extra heat and often not needed.

Then the temperature should be the average between shadow side and sunlit side. The shadow side can be much higher than 0K, for example because of thermal radiation from the lunar surface.

- How much time (grossly) will it take for a 1 K temperature elevation inside the module ? Which heating power is required for a 11 cubic meters cabin ?

Depends on what you want to measure - air temperature? Then you need to calculate how much energy the air absorbs. Keyword is specific heat.
 

N_Molson

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Well, I found that the specific heat of oxygen at 300 mmHg is 0.6576 J/(g°C) ...

http://www.wolframalpha.com/input/?i=specific+heat+of+oxygen+at+300+mmHg

---------- Post added at 12:33 AM ---------- Previous post was at 12:20 AM ----------

And since I have 6003 grams of O2 in the cabin at 20°C :

6003 * 0.676 J/(g°C) = 3947.6 J/°C (joules per degree Celsius difference)

So I need 3947.6 Joules to raise the cabin temperature by 1°C ? Is that correct ?

And if I want to lower the temperature by 1°C, I need to remove 3947.6 Joules from the air ?

Since 1 Joule = 1 Watt/second, assuming a 100% efficient heating device, I need to power the device to 3947.6 Watts for 1 second, or 394.76 Watts for 10 seconds, or 39.476 Watts for 100 seconds ? (assuming no losses)
 
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Yes, that sounds correct. If you have 500W of energy flowing into the cabin air, you would need about 8 seconds to heat it by one degree(K or °C)
 

N_Molson

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OK, now according to the HIDH, a male human being, at rest, produces 500 BTU/h. (it's an average, of course)

...which gives us 527,528 J/h , or 146.5 Joules/s, or 146.5 Watts.

So, our two cosmonauts produce 293 Watts, and rise the cabin temperature by 1°C in 13.65 seconds ?

But there is something I'm obviously missing, because, since their body temperature is 37°C, they will never rise the cabin temperature over that ?

And also, the 1°C elevation should be faster from 0°C to 1°C than from 36°C to 37°C (which should take a near-infinite time ?)

I guess it is a matter of entropy, but I don't see how to calculate this... :hmm:
 

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So, our two cosmonauts produce 293 Watts, and rise the cabin temperature by 1°C in 13.65 seconds ?

Yes, that is correct. The same happens at the same speed if you put enough students into a class room.

But there is something I'm obviously missing, because, since their body temperature is 37°C, they will never rise the cabin temperature over that ?

There is no limit, except by heat transfer to outside. Your body could push the temperature way past 40°C, before you would die slowly by overheating.

Your body will regulate its own internal temperature to 37°C usually, but this system is not made for regulating the temperature of a spacecraft. If it gets too hot inside your body, your body will try to extract more heat to the environment actually, into the cabin air.
 

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Thanks, I begin to have a better understanding of basic thermodynamics ;)

So, assuming a perfect closed system where there are no losses and no exchanges with the outside, a heating power of a few Watts (a lightbulb) could raise the temperature of the atmosphere inside to hundreds of degrees ? Wow.

That being said, if I assign a "heating value" (losses caused by a limited efficiency in the conversion of electrical power) to each subsystem, I can get a "total internal heat output per second". That's not too hard.

To that "internal heat output", I should add an "external heat output", in fact the Sun warming the spacecraft hull, that radiates the heat to the internal atmosphere.

On the other side, I should define the amount of heat that the cooling system including radiators can dissipate in space each second.

And, finally, I should define how much energy is radiated to space each second, because of losses (the hull itself, at least the side in shadow, radiates heat into space).

To sum up (everything being expressed in J/s, or Watts) :

Total Heat Balance = (Internal subsystems + Crew + Heat from the Sun) - (Cooling Systems & Radiators + Losses through the Hull)

Now, the difficult part seems to define :

Heat from the Sun = 1315-1422 J/s * (1 - insulation efficiency)
Cooling Systems & Radiators = ? J/s
Losses through the Hull = ? J/s
 
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Urwumpe

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Remember that the heat of the sun should NOT reach the cabin at all actually. It does it slowly, because there is no perfect insulation possible, but actually cabin air and hull should be treated as separate volumes.
 

N_Molson

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... and it's why a material with low emissivity like polished aluminium (e = 0.04) is good.

MLI is even better but is too heavy to cover all the spacecraft.

Of course, the window can be a problem there, it could cause a severe greenhouse effect. So I guess it would be a good idea to provide an external shutter for it. And an internal panel too, to provide thermal insulation.

According to a NASA paper on the Orion spacecraft, the solar constant in the transit environnement is between 1315 and 1422 W/m².

If I take 1400 W/m² as a value, assuming the hull is made of polished aluminium (emissivity 0.04), I get a temperature of 886.4°K ! :blink:

If, to keep it simple, we assume that the temperature of the other side (in shadow) is about 10°K, we have an average of 438.2°K for the hull temperature, which seems acceptable.

Now I wonder how much of those 438.2°K, assuming a good insulation, could make their way to the cabin ? 1/10 ? 1/100 ? No idea :idk:

But I guess we should express it that way :

Energy from the Sun heating the cabin = 1315 to 1422 J/s (per exposed square meter) * (1 - insulation efficiency)

---------- Post added 05-01-11 at 12:50 PM ---------- Previous post was 04-30-11 at 07:54 PM ----------

Let's assume that the surface area of the spacecraft cabin is a 3*1.5 meters cylinder. (1.5 for radius)

I want to calculate the total amount of energy absorbed by the hull, when the spacecraft is pointed at 90° of pitch/yaw (reference : Sun).

The total surface area of the cylinder is 42.4 m². I remove the two "caps" on the top and the bottom ; and I get 28.3 m².

I assume that the hull is made of polished aluminium, which emissivity is 0.04 (almost a mirror), much like the Apollo spacecraft. Using the Stefan-Bolzmann equation, I get two results :

- The half of the hull exposed to the Sun has a temperature of 881.6 °K.
- The radiant flux of each exposed square meter of hull is 1370 W/m².

We will assume that the dark side of the hull is exposed to 2.7 K, the cosmic background radiation, and we'll neglect planets albedo for simplicity.

The average temperature of the cabin hull, assuming that we have an efficient heat transfer all over the hull (aluminium is highly conductive), is 439.45 K.

The part of the hull exposed to the Sun is 28.3/2 = 14.15 m².

Then the total radiant flux is 1370 W * 14.15 m² = 19385.5 W, or 19385.5 J/s.

In the case in which the spacecraft is facing the Sun (0° of pitch/yaw) we have : 1370 W * 7.05 m² = 9658.5 W or 9658.5 J/s.

These are the total energy inputs received from the Sun as heat.

Now I have to determine how much energy the "dark side" of the hull can radiate in vacuum.

... and this is where I get stucked :huh::hmm:

---------- Post added at 06:30 PM ---------- Previous post was at 12:50 PM ----------

Here's what I missed :

Net Radiation Loss Rate

If an hot object is radiating energy to its cooler surroundings the net radiation heat loss rate can be expressed as

q = ε σ (Th4 - Tc4) Ac

where

Th = hot body absolute temperature (K)

Tc = cold surroundings absolute temperature (K)

Ac = area of the object (m2)

σ = Stefan-Boltzmann constant

ε = emissivity


So :

q = 0.04 * 5.6703 * 10^-8 (W/m^2K^4) × (439.45 Kelvin - 2.37 Kelvin) × 42.4 m^2

q = 0.00004203 K^5 W

:sos:

EDIT : WARNING, MOST OF THE ABOVE CALCULATIONS ARE BASED ON FALSE DATA. HEAT AND TEMPERATURE ARE VERY DIFFERENT THINGS !
 
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N_Molson

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The temperature management finally is taking shape. I'm going to consider that each subsystem is 95% efficient, meaning it converts 5% of the electrical power it consumes in heat.

As pointed Urumpwe, unless you power everything down (during LEVA...), there is little chance you'll have to use the heater :)
 

Urwumpe

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The temperature management finally is taking shape. I'm going to consider that each subsystem is 95% efficient, meaning it converts 5% of the electrical power it consumes in heat.

Depending on the system - for electrical circuits like computers, it can easily be that 95% of the energy is converted into heat. Electro-motors are around 70% effective.
 
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