Moon Landing Hoax - The Space Suit

Debunk it? It's just Flat Earth woo-woo, as is everything else on that ridiculous Flat Earth website that you linked to. If you can't see that for yourself then that is up to you, matey. Are you really that gullible?

It's not my job to educate you, that was the job of whatever passes for an education system on whatever planet you might live on.

Thanks for the gratuitous insults, by the way, very mature.

a reply to: AthlonSavage

Spacesuits cool themselves by means of ice sublimator where water is trickled onto a series of plates

In the vacuum of space the water evaporates forming an ice layer on the plates - water circulating through the
tubes of the cooling garment is passed through the heat exchanger in the sublimator giving up its heat before being recirculated

Here is a detailed explanation of how the life support unit works

www.hq.nasa.gov...

a reply to: oldcarpy

Flat earthers/moon hoaxers are special breed of stupid

Willing to swallow the ridiculous pseudo science nonsense and parrot it back without verifying the information is right or not

Most reveal a total ignorance of the facts and basic science

a reply to: firerescue

Too true. Generally dishonest, too.

Nasa is trying to say the heated water from their cooling was subliminated via ice on coils and extinguished from their packs in transition from water to vapour gas.

A vaccum is a perfect insulator for heat. A vaccum is surroudning suits.
Imagine now the analogy of a themo flask which holds hot coffee at 200 F which is 93.333 C.

Just imagine that heat for a moment. We have all experience summer days in high forties 120F. Now double that in your imagination to consider what cooling system surroudning the body could protect a person from that.

Imagine the inside of the thermal flask coming up with a technology to expell the heat from the flask to keep it maintained at 75F where the flask always wants to return quickly to 200F.

The space suit pack is designed with technology of having water and iced coils to subilimate the heated water.
That's not going to work. The whole system which is inside the suit and pack is going to get very hot quickly in the vaccum of space and in no time be just like the flask holding t he hot coffee.



Electronic and electrical components can be designed to run that hot but not people.
The moon missions are hoaxes in terms of being blended parts of fiction and fact. The people walking on the moon part if fiction.

a reply to: AthlonSavage

The Earth doesn't over heat despite being surrounded by space. Check out what heat actually is.

a reply to: AthlonSavage

You seem to have a fundamental misunderstanding of reality. 200F is not all that hot, I can literally go to wal-mart right now and for about $10 buy a set of oven mitts that would have no trouble insulating me from something that's 200F yet you don't think a multi-million dollar space suit would be up to the task? Do you have any evidence for anything you say or are you just blindly repeating what you believe on faith alone?

Rough calcs I made below but appear accurate enough to indicate the suit will need to be able extinguish 4200 Joules of energy per second surrounded by the insulative effect of vacuum. That's 4.2Kw compare that with 100watt house light globe. Go place a light globe in a small metal box and see how hot it gets inside.




The solar constant (GSC) is a flux density measuring mean solar electromagnetic radiation (solar irradiance) per unit area. ... It is measured by satellite as being 1.361 kilowatts per square meter (kW/m²) at solar minimum and
Average body surface area for adult men: 1.9 m2.

sun covers half body so use 1m2.
1.361 kilowatts = 1361watts per m2
thats 1361 x3600 = 4899600 joules per hour.
The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram °C
Human body is 80Kg = 80000.
Its body temp interior 37 deg C
therefore 80000 x 4.186 x 37 = 12390560 Joules
12390560 + 4899600 Joules =
Per hour 17290160 Joules of heat energy needs to be managed.
divide by 3600 seconds per hour 4802 watts (Joules/sec)
Presuming human body mainly water
The specific heat capacity of water is 4,200 Joules per kilogram per degree
If the space around suit is perfect insulator will need to rid averge at least 4200 Joules of enery per second to maintin a regular body temp.

That's 4.2Kw compare that with 100watt house light globe. Go place a light globe in a small metal box and see how quickly hot it gets inside.

a reply to: AthlonSavage

You don't understand how the PLSS works. It works through evaporation and sublimation to remove heat. Both of which are phase changes. The heat capacity of water is irrelevant.

Some of the water released by the PLSS instantly vaporizes. Depending on its temperature it takes between 597 calories (at 0ºC) and 540 calories (at 100º) per gram to vaporize water. At the "minimum" setting on the PLSS the water would be at about 80ºF so it would consume about 560 calories per gram. Now, these figures are for an atmospheric pressure of 1 bar. At lower pressures more heat is consumed, the lowest reference I could find is for .02 bar and is 588 calories (at 100º).

The water that did not vaporize became frozen on the surface of the sublimator. The heat absorbed by this process of freezing and subsequent sublimation is greater. It requires the 80 calories to required to freeze it, plus the 560 calories required to evaporate it, for a total of 640 calories per gram of water.

www.engineeringtoolbox.com...

imagine a reverse analogy its winter and you have your coat on and sitting out side in zero degree. There is no blowing air its relatively calm but you still feel very cold even though you have two jumpers on beneath the coat. You run an extension chord out of a remote power point to runs a 1kw heater to keep you warm.You decide its not warming you sufficiently and change it to an outdoor camping stove has delivering 4 k W of power. Your happy now to rest the night next to that. Thats outputing 4000 watts a sec.

This gives you a ball park idea of how much energy you need to sustain yourself happily when deviating significantly from a normal temperatures ranges 25 to 30 deg celcius where your happy in T shirt and shorts.

Its a reverse analogy because think 100 degree celcius instead of 0 degrees celcius. The first scenario requires source of energy to prevent overheating whereas as the second is requiring input source of energy to prevent becoming too cold.

The amount of energy being consumed in both analogies works out to about 4kw per sec.
The cals do not need to be accurate just in the ball park range to show what is feasible and what is not.

originally posted by: AthlonSavage

sun covers half body so use 1m2.
1.361 kilowatts = 1361watts per m2
thats 1361 x3600 = 4899600 joules per hour.
The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram °C
Human body is 80Kg = 80000.
Its body temp interior 37 deg C
therefore 80000 x 4.186 x 37 = 12390560 Joules
12390560 + 4899600 Joules =
Per hour 17290160 Joules of heat energy needs to be managed.
divide by 3600 seconds per hour 4802 watts (Joules/sec)
Presuming human body mainly water
The specific heat capacity of water is 4,200 Joules per kilogram per degree
If the space around suit is perfect insulator will need to rid averge at least 4200 Joules of enery per second to maintin a regular body temp.

you put in all that effort, but why is the suit white? and why are your calcs assuming a black body object?

a reply to: AthlonSavage

The cals do not need to be accurate just in the ball park range to show what is feasible and what is not.

The calcs do need to calculate the correct factors however.

The only energy required to prevent overheating is the energy used to circulate a small amount of water. The physics of vaporization take care of the rest. The only energy required for heating is the energy provided by the astronaut's metabolism.

The PLSS cooling system is (and was) a brilliantly simple and elegant system.
www.hq.nasa.gov...

originally posted by: choos

originally posted by: AthlonSavage

sun covers half body so use 1m2.
1.361 kilowatts = 1361watts per m2
thats 1361 x3600 = 4899600 joules per hour.
The specific heat of water is 1 calorie/gram °C = 4.186 joule/gram °C
Human body is 80Kg = 80000.
Its body temp interior 37 deg C
therefore 80000 x 4.186 x 37 = 12390560 Joules
12390560 + 4899600 Joules =
Per hour 17290160 Joules of heat energy needs to be managed.
divide by 3600 seconds per hour 4802 watts (Joules/sec)
Presuming human body mainly water
The specific heat capacity of water is 4,200 Joules per kilogram per degree
If the space around suit is perfect insulator will need to rid averge at least 4200 Joules of enery per second to maintin a regular body temp.

you put in all that effort, but why is the suit white? and why are your calcs assuming a black body object?

He also doesn't understand that heat can be removed through radiation. Any place without direct sunlight will radiate heat infrared radiation will always be emitting from half your suit even in sunlight. Somehow he believes there is no way for heat to be removed. If that were true we could stop planes from showing up in infrared be handy for stealth fighters.

Just so you know the suit is designed to reduce absorption of IR radiation while at the same time boost its radiation.

a reply to: dragonridr

I guess he never sat in front of a camp fire.

Hot in front, cold in back.

Energy = mass x change in temperature x specific heat of water.
The specific heat of water is about 4184 joules per 1 degree temperature rise per kg.
So energy required to boil at 100 Deg C is 50 kg x 65 degrees C * 4184 (joules /kg x deg C)
= 13598000 joules.

To find an approximation of your machine's wattage, fill a microwave-safe liquid measuring cup with 1 cup cold water. Microwave on High and keep an eye on it, noting how long it takes for the water to come to a boil…
2 minutes (120 seconds): 1,000 watts
1 cup = 1/4 litre (0.25 kg)
120 seconds of x1000 watts = 120000 joules
13598000/120000 joules = 113.31
113.31 lots of 2 minutes
226 minutes to boil a man

13598000/ (226 minutes x 60)
So removing 1002 joules per second of heat over that time. or 1000 watts etc.

Quite incredible however is the longest apollo lunar excursions went for over 7 hours and suits have 279Watt-hour battery (i.e 279 x 60 *60) = 1004500 joules capacity. Seems quite underated for running the suits cooling system.

This is a more relaxed calculation as it caps temp rise to 100 deg C because. In fact sunlight hits the moon's surface, the temperature can reach 260 degrees Fahrenheit (127 degrees Celsius).

The only thing I can think of is mabey its not as hot on the moon as they tell us.

a reply to: AthlonSavage

again you are assuming that white suit is a black bodied object... a perfect black bodied object no less.

a reply to: AthlonSavage

Energy = mass x change in temperature x specific heat of water.

You are not considering the latent heat of vaporization.

So energy required to boil at 100 Deg C is 50 kg x 65 degrees C * 4184 (joules /kg x deg C) = 13598000 joules.

Starting from what temperature? But isn't the boiling temperature of water dependent upon pressure?

Quite incredible however is the longest apollo lunar excursions went for over 7 hours and suits have 279Watt-hour battery (i.e 279 x 60 *60) = 1004500 joules capacity. Seems quite underated for running the suits cooling system.

The only part of cooling that involved battery power was a small pump, used to circulate the water.

In fact sunlight hits the moon's surface, the temperature can reach 260 degrees Fahrenheit (127 degrees Celsius).

How long does it take to do so? Did the suits have the same albedo as the Moon's surface? Do you think the suits were not insulated?

a reply to: toocoolnc

I agree, but if that's the case then we should be able to go as close to the sun as that thing did a few years ago that was filmed "refueling" from the sun, that right? ... at least unitl we entered into some air like 'mater,'