NASA Eyes Crew Deep Sleep Option For Mars Mission

astroengine writes: A NASA-backed study explores an innovative way to dramatically cut the cost of a human expedition to Mars — put the crew in stasis. The deep sleep, called torpor, would reduce astronauts' metabolic functions with existing medical procedures. Torpor also can occur naturally in cases of hypothermia. "Therapeutic torpor has been around in theory since the 1980s and really since 2003 has been a staple for critical care trauma patients in hospitals," aerospace engineer Mark Schaffer, with SpaceWorks Enterprises in Atlanta, said at the International Astronomical Congress in Toronto this week. "Protocols exist in most major medical centers for inducing therapeutic hypothermia on patients to essentially keep them alive until they can get the kind of treatment that they need." Coupled with intravenous feeding, a crew could be put in hibernation for the transit time to Mars, which under the best-case scenario would take 180 days one-way.

Comments

What will happen to their physical condition

[Meshach]

If they are just sleeping (or in whatever state they are in) will not their muscles deteriorate? After having no nourishment for several weeks most people will waste away to nothing.

Re:What will happen to their physical condition

[Anonymous Coward]

"Coupled with intravenous feeding, a crew could be put in hibernation for the transit time to Mars"

Re:What will happen to their physical condition

[Anonymous Coward]

Well, the article has the following text pretty much at the top:

"During interplanetary transit, the crew would receive low-level electrical impulses to key muscle groups to prevent muscular atrophy."

Re:What will happen to their physical condition

[Anonymous Coward]

I doubt they would waste away as their metabolic rate would be turned way down, but I imagine they would go through a significant period of physical difficulty when they wake up.

Re:What will happen to their physical condition

[wile_e_wonka]

Thanks for RTFA for me. Now I won't bother.

Re:What will happen to their physical condition

[sycodon]

My teenager sleeps all day but still can walk and talk when she gets up.

Re:What will happen to their physical condition

[Anonymous Coward]

That might take care of the muscle issue, but what about the problem with bones becoming brittle in low gravity? Are they planning to generate artificial gravity too?

Re:What will happen to their physical condition

[DittoBox]

This won't help with bone density loss, lowered heart strength, or a number of other issues.

Re:What will happen to their physical condition

[Holi]

your teenager is not dealing with the effects of microgravity.

Re:What will happen to their physical condition

[gcnaddict]

You'll lose most of that on Mars anyway. Reduced gravity :)

Re:What will happen to their physical condition

[Anonymous Coward]

IV feeding doesn't mean that your muscles are actually going to be being built. Unless you're using them, your body doesn't try to repair and build them up.

Re:What will happen to their physical condition

[TWX]

Maybe it's time to actually design a ship with a centrifuge in it, so that a lot of the effects of microgravity are mitigated...

Re:What will happen to their physical condition

[ShanghaiBill]

IV feeding doesn't mean that your muscles are actually going to be being built. Unless you're using them, your body doesn't try to repair and build them up.

When your body temperature is lowered, and your metabolism is reduced, you also reduce the physiological processes that cause muscle deterioration. Also you can "exercise" in your sleep by using mild electric pulses to contract your muscles.

Re:What will happen to their physical condition

[Anonymous Coward]

The oppressive force of adults isn't enough?

Re:What will happen to their physical condition

[LifesABeach]

Sounds workable. But what of lack of gravity? How does the sleeping one deal with bone loss?

Re:What will happen to their physical condition

[LifesABeach]

Maybe we should try for the Moon? It's a lot closer, and it would give us time to work out these types of issues?

Re:What will happen to their physical condition

[LifesABeach]

Your teenager is resting is not 9 months long; even though it feels like it.

Re:What will happen to their physical condition

[Kartu]

Well, yeah, but it's still a valid concern.
Weightlessness is a major problem, muscle atrophy, skeleton deterioration, both at rapid rate.

Re:What will happen to their physical condition

[Anonymous Coward]

Sounds workable. But what of lack of gravity? How does the sleeping one deal with bone loss?

That's why you send convicts first.

Re:What will happen to their physical condition

[BaronAaron]

This is also in TFA:

One design includes a spinning habitat to provide a low-gravity environment to help offset bone and muscle loss.

Re:What will happen to their physical condition

[ShanghaiBill]

But what of lack of gravity? How does the sleeping one deal with bone loss?

Lack of gravity causes your heart and bones to weaken. The heart problem can be ameliorated by putting your legs in a partial vacuum suction, pulling blood away from your body core. This simulates the pooling of blood in your legs while standing on Earth. Then you heart has to work to pump it back up. For the bones, I dunno, but likely the lower body temperature and reduced metabolism would reduce bone deterioration as well. So you would have done loss during deep sleep, but likely less than you would if you were awake.

A better long term solution is to genetically modify humans to make them better adapted to life in space.

Re:What will happen to their physical condition

[ShanghaiBill]

That's why you send convicts first.

Sending convicts has a secondary benefit of opening up more interesting plot lines for a movie version of the events. Image what Ron Howard could have done with "Apollo 13" if Jack Swigert (Kevin Bacon) had been a convicted murderer.

Re:What will happen to their physical condition

[krept]

Armageddon?

Re:What will happen to their physical condition

[Jane Q. Public]

Maybe we should try for the Moon? It's a lot closer, and it would give us time to work out these types of issues?

I'm with you on that.

Seems to me, the "cold sleep" option mainly solves the problems of crew space, resources, and radiation. Those are not small things.

A long-term space expedition must have room to move and exercise. That's a lot of size and mass. Then it needs food to promote exercise and waking function, and waste disposal to match. And THEN all that has to be wrapped in effective radiation shielding, which adds a lot more mass.

Eliminate the exercise, confine the crew to a small space, feed intravenously, and shield only that small part is FAR more efficient.

On the other hand, as many have pointed out, it comes with some serious cost as well.

I think landing on Mars would be a great accomplishment, but establishing a permanent moon base would be a vastly greater accomplishment.

Re:What will happen to their physical condition

[wasteoid]

Plus by the time the ship arrives at its destination, the good astronauts will be separated from the bad ones.

Re:What will happen to their physical condition

[Anonymous Coward]

That's why you send convicts first

Why, so they can break the law of gravity too?

Re:What will happen to their physical condition

[WillAffleckUW]

A lot of bone density loss is due to biological processes. Slow those down, slow down the bone density loss.

You can also wake them up and spin them in shift changeovers. Trust your friendly HAL 9000 to manage that.

Re:What will happen to their physical condition

[angel'o'sphere]

Why should any 'sleep option' solve any radiation issue?
WTF you always proclaim you had a clue about physics, another post of yours where it is clear: you have not!
Ah, you try to talk about shielding, face palm ... the volume you shield is irrelevant, the main hazard is the sun, which is 'behind' you. Actually, reliable shielding is impossible anyway. We are not talking about a nuclear reactor where one yard of lead or ten yards of water are a nice shielding.
Radiation in this case are atomic particles at relativistic speeds. Perhaps they get stopped by a 'shield' to fry you with 'Bremsstrahlung' instead.
WTF, many problems are 'inherent' problems, you can not get 'around' of them.
Perhaps you can find a compromise, sure. But I for my part rather stay awake and die consciousness in case of a solar storm than do the greatest endeavor of mankind in my sleep!

Re:What will happen to their physical condition

[WillAffleckUW]

More like cats which spend most of their day sleeping.

Re:What will happen to their physical condition

[BarbaraHudson]

"During interplanetary transit, the crew would receive low-level electrical impulses to key muscle groups to prevent muscular atrophy."

What about "that" muscle? Or is it going to be an all-women crew?

Re:What will happen to their physical condition

[khayman80]

Reliable shielding isn't impossible. Shielding of 4.41 tons/m^2 is sufficient. Putting the crew in hibernation does reduce shielding because otherwise the entire back side of the spacecraft (at least) has to be covered with 4.41 tons/m^2 of shielding. In hibernation, the crew could be closely packed and aligned with their feet towards the sun, reducing the required shielding area and mass.

Re:What will happen to their physical condition

[Jarik C-Bol]

Turns out, that is probably a LOT harder than we'd imagine. You know the 'stationary' bike they use on the ISS to keep in shape? turns out, its attached to the station in all sorts of weird special ways to keep you from shaking the station to pieces/rotating the station due to the forces the spinning wheel/pedaling action causes. If a exercise bike in space is that bloody hard, imagine what a ship with a multi-person hamster wheel will be like to engineer.

Re:What will happen to their physical condition

[Jane Q. Public]

Why should any 'sleep option' solve any radiation issue?

For reasons I explained but which you did not bother to read and understand.

I think it's hilarious that you blame me for your own failure to read.

Re:What will happen to their physical condition

[angel'o'sphere]

You fail to read as well,
like me you only knee jerk react
to the first three rows of a post.

Otherwise you had realized I addressed your "failed to read" argument a few lines further down :D

They where wrong nevertheless ... good luck.

Re:What will happen to their physical condition

[rtb61]

The flip side of that is toughening up the ship to provide protection between faults, emergencies, impacts and crew wake up time. How long it takes to crew to go from extended sleep to active functioning, in the movies, they always fast forward through this, likely reality is days, during which they will have to be exercising a lot to rebuild muscles.

What efficiency accept reality a place size limits on access to the space program, no taller than say 1.6m and that reduction really does make a saving in life support systems and overall size of systems.

The real constraint is how long, once you make it a really long slow trip, then the ship becomes big enough for a managed aquaponic system to provide sustenance and oxygen. Go for a long slow trip with a very large ship and conduct experiments on the way out. Likely reality is, when are going to have to based permanently on the moon before when can tackle a manned trip to mars.

Re:What will happen to their physical condition

[angel'o'sphere]

Hypothetically ...
In real live that is irrelevant. Regardless if your 4.41 ton/m^2 is right (sounds a retarded measurement, tons of what? Lead? Water?) The number you quote does not show up in the link :D

I never said shielding is impossible, but the question if one is hibernated for 9month versus awake for 6month versus in danger of "radiation" for either 6 or 9 or 12 months ... has not much to do with shielding.

As I said before: I had no problem being awake on such a journey, there are plenty of books to read while traveling.

I never would do that hibernated. Sounds like the "death by injection" penalty ... except you "believe" you wake up later.

Re:What will happen to their physical condition

[TWX]

I'm thinking something like a contrarotating set of cylinders, inner and outer, with the inner being the habitat. On top of that, there would be a series of weights located on the inside cylinder that could be automatically shifted as the weight in the habitat moves around, to keep it in balance.

I've built engines, and while they are statically balanced, and do not change balance once set up, they can be either internally balanced, where the weight is added or removed from the crank as needed, or externally, where the vibration dampener and torque converter or flywheel have weight added to make them balance. I don't see why one couldn't do that dynamically if the system were properly set up.

Re:What will happen to their physical condition

[khayman80]

NASA found that 441 grams/cm^2 of silicon dioxide (Moon dust) would be sufficient shielding, which equals 4.41 tons/m^2. Hibernation dangers and personal preference regarding books may vary, of course.

Re:What will happen to their physical condition

[angel'o'sphere]

Erm, your numbers still make no sense, as the real question is only the thickness.
In other words: the bigger the ship in diameter the more shielding you obviously need, but the thickness over that area would be rhe same.
So, how thick should such a ahield be? 2m? 5m?

Re:What will happen to their physical condition

[Anonymous Coward]

Screw that, send the middle managers and politicians! Given that most convictions are non-violent drug offenses, I'd rather have them than scum that spend their days dreaming up ways to violate the human rights of others.

Re:What will happen to their physical condition

[khayman80]

A hibernating crew could be closely packed and aligned with their feet towards the sun, reducing the required shielding area and mass at constant thickness. That's because only the hibernation chamber would need to be shielded, not the entire ship.

Re:What will happen to their physical condition

[Anonymous Coward]

A better long term solution is to genetically modify humans to make them better adapted to life in space.

Also known as 'evolution'. Let's start with the politicians.

Re:What will happen to their physical condition

[angel'o'sphere]

Yes, but we still don't know how big the shielding would be :)
Hence we can not judge if it makes any sense (shielding wise, and based on shielding, fuel wise)
And actually, you very likely wont align them with the feet to the sun. That makes no sense. If one gets hit by a particle into the foot, it will likely go straight through the whole body to the brain. It is much better to put the people perpendicular to the sun. If one gets hit somewhere the particle just goes out of the other side with much less damage. Unless you only have a single person in the craft the area with shielding would be just the same.

Re:What will happen to their physical condition

[khayman80]

If the main hazard is the sun, that requires thicker shielding on the sunward side. Minimum shielding mass would then be obtained by putting 4.41 tons/m^2 on the sunward side, which given moon dust density equals a ~2.4 meter thick shield on the sunward side. If the people are perpendicular to the sun, that shield is heavier. The people are awake and moving around, that shield is much heavier.

Re:What will happen to their physical condition

[Jane Q. Public]

You fail to read as well,

Unlike you, I didn't "fail" to read. The direction most of the dangerous radiation comes from is not irrelevant, BUT you seem to think that JUST because most comes from the sun, that's the only significant shielding needed.

Bullshit.

We already know better from experience. Why didn't YOU know that?

Re:What will happen to their physical condition

[Jane Q. Public]

As I implied elsewhere, when you minimize solar radiation you are eliminating most of the energetic radiation/particles, but by no means all. We already know this from Spacelab and ISS experience. If you ignore extrasolar energetic particles you're just being stupid. Unless you plan a 1-way trip. Which has been suggested.

Certainly most of the shielding should be between the sun and the crew. But it's not all the shielding necessary. And though the "other" shielding need not be as heavy, its area is much larger so it still contributes a lot to the overall mass of the vehicle.

Based on other arguments with khayman80, to be honest I would not trust him to build a bridge over a creek, much less a spaceship. That's just the truth.

Re:What will happen to their physical condition

[khayman80]

I never said we should ignore extrasolar particles. I was just showing that even using angel'o'sphere's assumption that the sun is the main hazard, the shielding mass decreases for a hibernating crew. In other words, I was defending you, Jane. Even though I can't be trusted to build a bridge over a creek.

But since you brought up those other arguments...

There is no reason to "guess" at my reasoning. I spelled it out quite clearly when we had our "argument" (which you lost). You do realize this is all going to be published, right? I warned you not just once or twice, but many times now. Every time you pull this kind of BS will be just another instance of widespread public knowledge of your dishonesty. [Jane Q. Public, 2014-10-03]

I have to guess at your reasoning because what you've said doesn't make any sense.

If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out".

Just no. If radiation goes in and comes right back out, we do not need to account for it, because then the NET amount of that particular radiation crossing your boundary is ZERO. A = A. You do know how to add and subtract, right? You know what a zero is, right? [Jane Q. Public, 2014-09-24]

I have to guess at what Jane meant by this, because it's not in equation form. In physics, statements in equation form are easier to analyze.

Draw a boundary around the (gray or black body) heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out

At steady state, Jane's power in = Jane's power out:

electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)

Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?

I am not going to get drawn into an argument that you have already lost. I repeat that the equation you show is for HEAT TRANSFER, not "radiative power out". You are just plain wrong about that and any heat transfer textbook will you so. ... [Jane Q. Public, 2014-10-03]

Once again, to calculate "electrical heating power" you need to use a heat transfer equation which accounts for power in and power out. That's because power in = power out through any boundary where nothing inside is changing. Once again, the equation Jane's using is only valid for "radiative power out" which is completely different than "electrical heating power". That's why I'm starting with the principle of "conservation of energy" and trying to understand what Jane's saying, in equation form.

Jane, if you don't agree with the "power in" and "power out" that I've tried to glean from your rants, just fill in the following blanks like I did. It'll be much faster than accusing me of dishonesty, fraud, and libel.

Jane's power in = ?
Jane's power out = ?

Re:What will happen to their physical condition

[Jane Q. Public]

We've been over this before, and you already know the answers I've given you. Stop being a grandstanding asshole. I don't have to keep repeating my answers every time you demand them. That's called ASSHOLE behavior, asshole.

You have already seen my calculations and my answers to all these questions. By bringing them up and demanding them AGAIN in a different forum, you are advertising your own dishonesty.

It didn't work. Don't worry, as I promised this will all be published when I find the time.

Re: What will happen to their physical condition

[Anonymous Coward]

If they're accelerating towards Mars they won't be weightless.

Re:What will happen to their physical condition

[excelsior_gr]

I suppose NASA will fatten the astronauts up and make them nice and chubby before sending them on a mission.

Re:What will happen to their physical condition

[Electricity Likes Me]

Or you know, Earth.

If we were going to practically do this, we'd be doing it here, in a hospital first. We'd have to take a bunch of people, and have them asleep for 180 days under the same conditions as the trick, and see what the effects - physical and psychological, actually were.

Re: What will happen to their physical condition

[Anonymous Coward]

That would likely negate the weight savings from putting the crew to sleep by ramping up the amount of fuel you need to carry for constant acceleration.

Re:What will happen to their physical condition

[Anonymous Coward]

If they are just sleeping (or in whatever state they are in) will not their muscles deteriorate? After having no nourishment for several weeks most people will waste away to nothing.

Depends, if all of your brain shuts down in cryosleep, me for instance, all "back of the bus and sh*t" I would be in good shape upon waking. (with Christina Cox on my..)

Animal side: at attention!

Re:What will happen to their physical condition

[Half-pint HAL]

Sounds workable. But what of lack of gravity? How does the sleeping one deal with bone loss?

First and foremost: slowed metabolism. Bone loss is caused when the bone metabolises bone material. A slowed metabolism metabolises more slowly.

Secondly, from TFA: "One design includes a spinning habitat to provide a low-gravity environment to help offset bone and muscle loss."

Re:What will happen to their physical condition

[Half-pint HAL]

the main hazard is the sun, which is 'behind' you.

That depends on your trajectory. The planets aren't in one straight line, remember.

Re:What will happen to their physical condition

[Half-pint HAL]

A hibernating crew could be closely packed and aligned with their feet towards the sun,

If you do that, you preclude the use of rotation as a simulation of gravity to deal with bone deterioration.

Re:What will happen to their physical condition

[angel'o'sphere]

Because you and others are talking about shielding from the sun, rofl.
You can not shield a space craft against super high energetic 'radiation' anyway. (Or do you want to try to clad the whole thing in a 5m, 6m, 10m ... Xm thick lead armor?)
So bringing this point up now is arguing for arguings sake only ... pointless.

Re:What will happen to their physical condition

[angel'o'sphere]

Yeah, but as soon as you stop burning the engine, you can turn the ship to point the shielding to the sun.
Perhaps it would make more sense to have a 'room' or the shielding itself, that can be rotated.
But I guess, as "exciting" such hibernating sounds, we are better of with a sustainable vasimir engine that allows to fly the whole way under acceleration. AFAIR it should be possible to accelerate with roughly 0.25g and reach Mars in less than 3 months.

Re:What will happen to their physical condition

[khayman80]

Centrifuges need to rotate no faster than 1 rpm to avoid inducing motion sickness. That's for long-term colonies, so maybe 2 or 3 rpm would be acceptable for astronauts selected for their resistance to motion sickness. Maybe even faster if they're hibernating the whole way. But regardless, the centrifuge would still have to be quite large.

If the centrifuge is inside the shielding, that makes the shield enormously bigger and heavier. Alternatively, only the hibernation/living chamber at the end of the centrifuge could be shielded. But that requires that the shielding mass be attached to the centrifuge, which vastly increases its required tensile strength. That's why the NASA study placed the colony's centrifuge inside a separate shield: if the shield rotates with the centrifuge then the centrifuge would have to be built out of carbon nanotubes. If the shield is separate then the centrifuge can be built out of aluminum.

Re:What will happen to their physical condition

[Anonymous Coward]

what part of "coupled with IV feeding" did you not grasp?

part of the long term issues would be muscle condition even ligaments and tendon going 180 days without use will be a problem but until they go to the planet surface they could be in null or micro gravity conditions which if done too long is an issue. They also have to shield them from solar f\radiation the whole time and Mars with almost no atmosheres and I have no clue if it has a magnetosphere -suspect its minimal if at all-is also crew issues they have to deal with.

Re:What will happen to their physical condition

[RockDoctor]

A better long term solution is to genetically modify humans to make them better adapted to life in space.

and which genes do you target for that, and what will the side effects be?

Rotational pseudo-gravity is long-term the only technique we've got. Unless you know of something better.

Re:What will happen to their physical condition

[Immerman]

Sonic stimulation perhaps? As I recall there've been some studies showing that it's not lack of gravity directly that causes bone loss, but lack of the shock waves traveling through your skeletal structure every time you take a step. And follow up studies showing that sonic stimulation can achieve much the same result.

Re:What will happen to their physical condition

[RockDoctor]

Shielding of 4.41 tons/m^2 [nasa.gov] is sufficient.

Hmmmm, My fist estimate on this sort of this is to look at the Earth's atmosphere. Living at the bottom of the atmosphere is good enough for a lifetime's shielding from interplanetary radiation. One atmosphere is equivalent to 10m of water depth (consult your diving manuals). That's 10 tonnes per square metre of protected area. Half of that amount - sounds reasonably credible. What material to use? Well, air is evidently sufficient (see "lifetime" above). Water is convenient - you'll need a considerable quantity any way. Some metals for taking out the slowed down and secondary radiations. But you'll need be needing metals anyway. I'd probably let the outer parts of it freeze to ice, for a degree of micrometeorite protection, and for the same reason you'll want several layers of it.

Re:What will happen to their physical condition

[khayman80]

Yep. That's the same sanity check used by that NASA study:

"Passive shielding is known to work. The Earth's atmosphere supplies about 10 t/m^2 of mass shielding and is very effective. Only half this much is needed to bring the dosage level of cosmic rays down to 0.5 rem/yr. In fact when calculations are made in the context of particular geometries, it is found that because many of the incident particles pass through walls at slanting angles a thickness of shield of 4.5 t/m^2 is sufficient."

Water could be an effective shield, and would be especially easy to apply and repair. Just melt it and let it freeze in place. That's how most of the lighthuggers in Revelation Space were shielded, as well as the starship in Songs of Distant Earth.

The only downsides I can think of would be the low tensile strength, so a water shield couldn't spin with a rotating ship, and the fact that if the ship overheats then its radiation shield sublimates away...

Re:What will happen to their physical condition

[megahurts.gr]

There are far easier ways to do it. For example, the entire ship can be a wheel, instead of having to be a ship with a wheel on it. And it does not even have to be shaped like a wheel. Tether the living compartment to a spent rocket stage, and let the two rotate around their common weight center which will be at some point along the tether line.

Re:What will happen to their physical condition

[megahurts.gr]

But that's all irrelevant to the article. Proper gravity for comfortable space travel is not the issue here; the issue is minimizing the cost of the travel. And sleeping astronauts travel a lot cheaper than woken ones. Plus, they don't need gravity. Unless someone is under the impression that proper gravity will help with the muscles of sleeping astronauts, which I highly doubt.

Re: What will happen to their physical condition

[spkay31]

Mars, like Australia but bigger and further away from civilization ðY

Re:What will happen to their physical condition

[khayman80]

All those lead thickness options are too thick by at least a factor of 5. Even if that NASA study is "retarded", RockDoctor just mentioned that Earth's atmosphere protects us with only ~10 tons/m^2. Since lead's density is 11 tons/m^3, a lead shield wouldn't have to be thicker than ~0.9 meter.

Re:What will happen to their physical condition

[khayman80]

If a VASIMR drive could sustain 0.25g acceleration, its fuel tanks would be enormous. It would also use a lot of power, requiring either a nuclear reactor or huge solar panels capable of supporting themselves at 0.25g.

But if it could be done, continuously accelerating at 0.25g to the midpoint then decelerating at 0.25g would result in an Earth-Mars travel time much shorter than 3 months. When Mars is closest to Earth, the travel time would only be 3.5 days. Even when Mars is on the other side of the Sun, the travel time would only be 9.4 days.

Re:What will happen to their physical condition

[RockDoctor]

The only downsides I can think of would be the low tensile strength, so a water shield couldn't spin with a rotating ship, and the fact that if the ship overheats then its radiation shield sublimates away...

I don't really see that as being an issue of any import. The marine industries have lots of experience of moving fluids around. We (I work at sea) uniformly stow and move fluids in multiple tanks of relatively small cross-sectional area. It reduces (as you seem to be worried about) "slosh" effects as the vessel accelerates, turns, decellerates etc. So in the spaceship context, you'd have multiple shells of (relatively) small tanks with plumbing between them and pump manifolds so that you can choose which tanks to fill at which time. You'd also probably need a set of plumbing for pumping steam around too (it's an effective way of moving heat).

Overheat such a radiation shield and at worst some of the liquid melts. Oh, you'd need to incorporate some expansion tanks for managing the volume changes on freezing / melting. Having spent enough nights of my life trying to clear sample lines plugged with ice, I've been looking at how to get around this for ages, and have a little dream of sample lines with strips of bubble-wrap type material along their interior to avoid splitting the damned things. The same sort of concept could accommodate the volume changes in your "ice shell shielding".

Rocket science, it ain't.

Re:What will happen to their physical condition

[khayman80]

That makes sense. I was thinking in terms of the lighthuggers in Revelation Space which are literally "glazed" in water ice because that's cheaper than setting up a system of tanks and pumps. That shield would be very easy to repair even after a collision with a "large" micrometeorite because there would be no infrastructure. Just melt more water and apply.

The system you describe would also be useful as an alternate (albeit temporary) way to dump heat in case the external radiators were damaged. I've been thinking about a similar setup, but using loop heat pipes instead of steam pumps because heat pipes don't have moving parts.

Re:What will happen to their physical condition

[Jarik C-Bol]

It won't help the muscles, but it will help the bones apparently. Bone becomes brittle and light in null G over longish periods apparently, so the gravity spin is to make it so your still strong muscles (from the electroshock therapy while in stasis) don't snap your bones like twigs when you try to walk when you get to mars. In theory. Its still all just numbers on a spreadsheet at this point.

Re: What will happen to their physical condition

[Anonymous Coward]

Not sure we want to encourage natural selection processes in our political class this way. Weightlessness, lack of gravity, used to spinning around more or less aimlessly. Wait ... Never mind. We aren't talking about bringing them home again, are we?

Re:What will happen to their physical condition

[angel'o'sphere]

You are mixing some stuff up :) actually lots of it.
Radiation absorption is not measured in tons per m^3.
It is measured by the likelihood that an incoming particle hits enough atoms/molecules of the absorbing material to be harmless, usually caught in that material.

Similar to half time of decay we use a 'half value layer' ... but likely you are right and there are not multiple meters needed.

Bad article: http://en.wikipedia.org/wiki/H...
A better one: https://www.nde-ed.org/Educati...

You basically have to weight how often a craft will be hit by really high energy particles.

Re:What will happen to their physical condition

[angel'o'sphere]

Erm, are you sure you don't mix up the travel time. I guess I mixed up the possible acceleration, but I'm pretty sure the vasimir inventors talked about a travel time of 3 months.

Anyway, the time in seconds would be calculated via: s = 1/2 a t^2. As we fly half of the way accelerating the other half decelerating, we use (2 * s) for the total. distance and solve the equation for t = sqrt [(2 * s) / (1/2 a)].

Now insert the approximated distance as s in _meters_ and the acceleration a in meter/sec^2 and you get the result as travel time in _seconds_.

Re:What will happen to their physical condition

[angel'o'sphere]

The german wiki article is talking about a travel time to mars of 39 days, btw. But is unspecific for what kind of craft (size, weight etc.)

Re:What will happen to their physical condition

[Jane Q. Public]

You basically have to weight how often a craft will be hit by really high energy particles.

But this is the whole point.

"Really high energy particles" do not come exclusively from the sun. While we can agree that MOST of them do.

So -- again I think we agree -- shielding in non-sun-facing parts does not have to be anywhere near as heavy. HOWEVER... regardless of whether you are referring to sun shielding or shielding from the rest of space, reducing the volume (and cross-sectional area) of the crew area that needs to be shielded can make a big difference in the mass. And, as you probably know, reduce the volume and mass of the crew area and the entire ship can be a lot smaller from start to finish.

That is all I was saying.

Re:What will happen to their physical condition

[khayman80]

If you think I mixed up the travel time, try calculating the travel time at 0.25g from Earth to Mars when it's closest to Earth at 55 million kilometers. Again, that only takes 3.5 days.

Re:What will happen to their physical condition

[khayman80]

You are mixing some stuff up :) actually lots of it.
Radiation absorption is not measured in tons per m^3.

If I'm mixing lots of stuff up, just explain how this NASA study was wrong to conclude that 4.41 tons/m^2 would be sufficient shielding.

Re:What will happen to their physical condition

[angel'o'sphere]

Because no one uses tons per m^2 to describe radiation absobtion.
A measure like that would imply the material used is irrelevant, which it is not. The correct material is the prime shielding factor.

Re:What will happen to their physical condition

[khayman80]

no one uses tons per m^2 to describe radiation absobtion.

Except NASA: "Passive shielding is known to work. The Earth's atmosphere supplies about 10 t/m^2 of mass shielding and is very effective. Only half this much is needed to bring the dosage level of cosmic rays down to 0.5 rem/yr. In fact when calculations are made in the context of particular geometries, it is found that because many of the incident particles pass through walls at slanting angles a thickness of shield of 4.5 t/m^2 is sufficient."

Re:What will happen to their physical condition

[angel'o'sphere]

That is a laymen explanation for people like you.

I linked you the 'half value layer' articles ... metric tons per square meter are irrelevant.

Relevant is how dense the material is and what its actual properties are to 'break' or capture cosmic rays.

A ton of water simply does not equal a ton of lead, even if you believe so after you got missleaded by that NASA article :)

Re:What will happen to their physical condition

[khayman80]

That is a laymen explanation for people like you.

What do you mean by that? What are "people like me"? "Laymen"?

Re: What will happen to their physical condition

[RockDoctor]

Your water chamber "floors" "ceilings" and "walls", plus several levels of interlinkage can be made to uniform design. It's a trick called "mass production". Given that, manufacture comes by robotic production, followed by assembly with humanoids to gasket between units and apply sealant.

Having most of your shielding as solid most of the time eases the sealing problems considerably. (Spend a year maintaining gas test equipment. You'll grow to love solids and the way they don't go through holes.)

These are issues of industrial design. Look at the way that shipbuilding costs and times have fallen as one - off craft design has been replaced by prefabrication (e.g. riveting plates together as opposed to building slices of a ship in a yard and welding them together). The same methodology changes are needed in space. If that implies a step change in launch costs and QC. .. well, it does.

Re:What will happen to their physical condition

[beastofburdon]

There is a major problem with that. What if they encounter intelligent life while they are out? The chances are very small that this will happen, but any chance is too high. If the first contact a highly advanced civilization had with us was with a politician we would be immediately exterminated to prevent our spread throughout the universe.

Re:What will happen to their physical condition

[angel'o'sphere]

Obviously, otherwise you would not insist that weight per square meter is a useful measurement for radiation shielding.
Yes, you linked an article ... however I linked you the math behind that problem :)

Re:What will happen to their physical condition

[khayman80]

What do you mean by that? What are "people like me"? "Laymen"?

Obviously...

Why are you making assumptions about who I am? If you clicked on my homepage, it would only take a few seconds to realize that you're wrong. But more importantly, it's not necessary or productive to accuse someone of being a layman. There's no reason to be nasty. Just discuss the science, and leave your assumptions about who the other person is out of it.

I linked you the 'half value layer' articles ... metric tons per square meter are irrelevant.

No. Metric tons per square meter = thickness * density, so if density is relevant then metric tons per square meter is also relevant.

Relevant is how dense the material is and what its actual properties are to 'break' or capture cosmic rays. A ton of water simply does not equal a ton of lead, even if you believe so after you got missleaded by that NASA article :)

The NASA article I showed you explicitly calculated the required shielding using silicon dioxide (Moon dust) as I've failed to explain. They're not saying a ton of water exactly equals a ton of lead, and neither am I.

no one uses tons per m^2 to describe radiation absobtion. A measure like that would imply the material used is irrelevant, which it is not. The correct material is the prime shielding factor.

No, density is the prime shielding factor. That means metric tons per square meter is a good first order approximation.

That is a laymen explanation for people like you. I linked you the 'half value layer' articles ... metric tons per square meter are irrelevant. Relevant is how dense the material is and what its actual properties are to 'break' or capture cosmic rays. A ton of water simply does not equal a ton of lead, even if you believe so after you got missleaded by that NASA article :)

Again, metric tons per square meter = thickness * density. That means the half-value layer should be inversely proportional to the shield's density. So if metric tons per square meter are relevant to the half-value layer, the half-value layer should be inversely proportional to the shield's density.

Did you try plotting those half-value layers against the inverse densities for concrete, steel, lead, tungsten and uranium? If you did, you'd notice that they're all close to a straight line. So metric tons of shielding per square meter is a good first order approximation.

Also, you claimed I mixed up the travel time, but you still haven't shown that my 3.5 day travel time to Mars at 0.25g is somehow wrong. What travel time did you get?

Re:What will happen to their physical condition

[angel'o'sphere]

Again, metric tons per square meter = thickness * density. That means the half-value layer [nde-ed.org] should be inversely proportional to the shield's density. So if metric tons per square meter are relevant to the half-value layer, the half-value layer should be inversely proportional to the shield's density.

Yes, on a first glance it should.
But in fact it does not. It highly depends on the material you use.

And to revert your argument: exactly all the stuff you correctly pointed out is why physicist use thickness and not weight. My point about layman, badly expressed, sorry, was: they use moon material as reference. And likely where more concerned about the total mass that involves, instead of proper arguing how to make a "perfect shield".

My point is: if you shield a reactor, mass is irrelevant, you take the cheapest per "needed" mass/volume material that fulfills the task.

In a spacecraft you likely want lowest mass ... and when we look at the moon as source, other issues like mining and launching ...

Nevertheless using the finally figured: "oh, with glass from the moon we can get it done with those masses" is incorrect for nearly every other material you can use.
E.g. glass with a low dosage of lead, weights perhaps 1% more but shield 40% better.

Also, you claimed I mixed up the travel time, but you still haven't shown that my 3.5 day travel time to Mars at 0.25g is somehow wrong. What travel time did you get?
I did not claim that. Why are people so obsessed with mixing up a simple statement, which was worded as an assumption, with a claim?
I recalculated it and you where right.
So my conclusion is that I remembered the "possible" or likely acceleration wrong. As I pointed out in another post: wikipedia (german, but left out all the calculations) speaks of 39 days, I remembered 30 days. However I was to lazy to recalculate what that implies for acceleration :D

Re:What will happen to their physical condition

[khayman80]

Again, metric tons per square meter = thickness * density. That means the half-value layer should be inversely proportional to the shield's density. So if metric tons per square meter are relevant to the half-value layer, the half-value layer should be inversely proportional to the shield's density.

Yes, on a first glance it should. But in fact it does not. It highly depends on the material you use.

Again, to a good approximation, those half-value layers are inversely proportional to the shield's density.

I plotted those half-value layers against the inverse densities of concrete, steel, lead, tungsten and uranium. The blue squares are for the iridium source, and the red circles are for the cobalt source. Since those points lie close to a straight line, radiation absorption is determined primarily by density. So metric tons per square meter is a good first order approximation, at least for those materials.

Re: What will happen to their physical condition

[khayman80]

I suspect that glazing a ship in ice would always be cheaper than building a network of chambers to act as a radiation shield. That's because it seems like improvements in robotic construction (prefabrication, etc.) could also be applied to the glazing process.

It seems even more likely that the repair costs of a "glazed" shield would be lower than a shield made out of water chambers. If a "large" micrometeorite blasts a chunk of a glazed shield away, you just send a robot with a water tank out to the hole and let it spray more water on the shield. If that micrometeorite hits a shield made out of water chambers, you have to repair or replace whatever chambers and pumps were damaged in the explosion.

Re:What will happen to their physical condition

[khayman80]

Oops, I actually plotted inverse density versus half-value layer thicknesses. This doesn't affect the conclusion, but here's a plot with a corrected file name.

Re:What will happen to their physical condition

[khayman80]

Oops, I actually plotted inverse density versus half-value layer thicknesses. This doesn't affect the conclusion, but here's a plot with a corrected file name.

Re: What will happen to their physical condition

[RockDoctor]

I suspect that glazing a ship in ice would always be cheaper than building a network of chambers to act as a radiation shield.

Vapour pressure is not your friend. At temperatures where water is liquid, it has a high enough vapour pressure (6.1173 millibars) that it will evaporate pretty rapidly. You'd need to cover your water glazing with something - probably something thicker than cling film. Sure, for patching, your repair robot can carry patches (I was applying a patch to my bike tube this afternoon - the parallel amuses me) to glue, loosely, over the hole before starting to spray the water. Maybe the water has ... glass fibre or wood pulp in it, to add mechanical strength. But that vapour pressure is going to be a problem.

Maybe you'd surround your outermost structure with bladders like this for the outermost protection ... engineering details. Velcro and/ or webbing straps to get them to stick together, but be repairable/ replaceable. A couple of sizes to fit all options ; it doesn't have to be a tight fit, just solid enough to do the job.

Re: What will happen to their physical condition

[khayman80]

Yes, liquid water's vapor pressure is a problem in vacuum. Your solution seems like a good idea: loosely glue a patch over the hole before spraying. After the water freezes, its vapor pressure also decreases. Then the patch could be removed and reused if the cost of dealing with the remaining sublimation loss is less than the cost of replacing bladders hit by micrometeorites.

The remaining sublimation loss could be minimized by keeping the glazed shield as cold as possible. At first it seemed like this would be easier in the outer solar system, but then I realized that the side of the ship facing the sun would probably be covered with solar panels anyway. Moving farther from the sun either requires larger solar panels, or a large cheap mirror to collect more sunlight.

So the most important variable is how much power the ship needs. I'm working on a simple design which has enough garden space to feed 4 people, and I calculated the power needed to light the garden as it would be lit on Earth. Assuming blackbody radiators at 0C (which puts a lower bound on the attainable interior temperature), dissipating that power requires that ~30% of the ship's surface not covered with solar panels would need to be covered with radiators.

And that's just the power needed to light the garden. So you're probably right: it would be better to cover the rest of the surface with bladders to reduce sublimation loss. Those bladders could be covered with radiators and individually connected or disconnected to the ship's interior via insulated loop heat pipes.

Also, I liked your pykrete link. Wood pulp is likely to be expensive in space, but glass fiber made from moon dust could probably be cheap.

Re: What will happen to their physical condition

[RockDoctor]

the cost of replacing bladders hit by micrometeorites.

Think back to the bike tube I was repairing. Why empty, move and replace a bladder for a relatively small hole when you can apply a patch in-situ? (I first met those bladders as non-potable water storage on a desert island off the coast of Tanzania. They're laid out on roughly cleared ground, and when you pump ten tonnes of water into one, it's not uncommon for a stone to rip the bottom. Since you're pumping water from the shuttle tanker, you need to fix the leak quickly, so the mud man used a pole and clip arrangement to pinch the leak closed. Since low pressures are involved, it's sufficient (of course, that bladder gets used first, so a glued patch can be applied at leisure).

The remaining sublimation loss could be minimized by keeping the glazed shield as cold as possible. At first it seemed like this would be easier in the outer solar system, but then I realized that the side of the ship facing the sun would probably be covered with solar panels anyway.

I think we got into this discussion talking about rotating ships, to provide midi-gravity. We know that microgravity requires a lot of effort to counteract, so ... you're going to need some major engineering reasons to not go down the spin-for-pseudo-gravity route. And on your general voyage (no, you don't design a vessel for only one voyage - craft design versus industrial production?) you are going to have a component of travel which is not radial to the Sun. Therefore, essentially all parts of the ship's surface are going to have alternating exposure to light and dark. So now, your mass production design moves to coating the whole of the ship with cheap-as-you-can-get solar cells. Which in the context of the design we're iterating would mean the bladders have solar cells on one side ("this side out"), and part of the hook-up includes plugging the solar cells into the vessel's power bus. Actually, revise that - the bladders aren't exactly lightweight, so including some power conditioning and a battery would provide you with options for powering condition-monitoring, condition reporting by wireless, maybe even corner-stretching propulsion to assist emplacement.

corner-stretching propulsion

Doh ! The terrestrial bladders I'm familiar with are rectangular, but lenticular shape may well be more appropriate. referring back to the small range of standard sizes - maybe also some range in shapes, but you do not want the ships stores to be carrying 35 different stock lines, half of which won't be used. And of course, all of the different models use the same fittings, electronics, fixings, etc (I was about to make a point about the different fire-hose fittings on last-months vessel compared to next week's vessel. But while looking for illustrations, I came across this page, which makes the point by reductio ad absurdam .)

or a large cheap mirror to collect more sunlight.

Now that's a point. A good one. Yeah, hanging a solar sail off an axial protrusion would boost your power production (90% exposure time instead of 50%) nicely, and help with the radial component of your velocity management too. At destination, hang a science package off the solar sail then cast it adrift - probably easier than attempting to recover.

Those bladders could be covered with radiators and individually connected or disconnected to the ship's interior via insulated loop heat pipes.

Hmmm, I'd keep the components as simple as possible. Take a close look at the design of the ISS (because I've seen those designs online ; other spacecraft will have the same issues) : the radiators protrude in one direction radial

Re: What will happen to their physical condition

[khayman80]

I think we got into this discussion talking about rotating ships, to provide midi-gravity. We know that microgravity requires a lot of effort to counteract, so ... you're going to need some major engineering reasons to not go down the spin-for-pseudo-gravity route.

Yes, centrifugal gravity seems like the only way to stay healthy in space. I pointed out that long-term colonies shouldn't rotate faster than 1 rpm in order to avoid inducing motion sickness. That imposes such serious tensile strength requirements that it seems like the shield can't spin with the ship unless the ship is made of carbon nanotubes.

Take a close look at the design of the ISS (because I've seen those designs online ; other spacecraft will have the same issues) : the radiators protrude in one direction radial to the Sun, but the solar panels are perpendicular to the Sun. If you rotate the system by 90 degrees, then the solar panels are useless and the radiators become heat absorbers. That's probably a large part of the reason for not rotating the ISS, but ... comments above about the effort needed to avoid the health problems of microgravity.

Yes, the ISS is a useful example. I'm proposing a modular design, where a sphere with interior radius of 10.7 meters has enough living and garden space to support 4 people. One sphere alone couldn't provide centrifugal gravity, but in that configuration the solar panels would be unfolded perpendicular to the Sun, and the radiators would be unfolded behind the sphere, radially away from the Sun.

But two spheres could dock and attach tethers at the top of each sphere. Then if they separate to a distance of 1800 meters, they could rotate at 1 rpm around their shared center of mass to produce 1g of centrifugal gravity.

If they're not going anywhere, their plane of rotation should probably be the ecliptic plane. Otherwise the Sun's orientation would change as they orbit the Sun. Each sphere's radiators could be attached to the tethers, parallel to the ecliptic plane so they never face the Sun.

During the docking procedure, each sphere's solar panel would be detached and remain at the midpoint between the spheres. They'd have to be able to move along the tether in case one of the spheres becomes heavier and moves the center of mass. The solar panels would be kept perpendicular to the Sun as the spheres rotate, so they'd have to be kept in place magnetically and transfer power to the spheres using induction or microwaves.

I still don't like relying on rotary joints, particularly coaxial ones. I'd use them where unavoidable, but I'd avoid them where possible. And in life-support, they'd scare me.

Yeah, me too. That's why I spent more time than I'd care to admit trying to think of a way to arrange the solar panels that doesn't require a special magnetic rotary joint. At first I thought the sphere's plane of rotation should have a surface normal that points directly at the Sun. That way the solar panels could be attached directly to the tethers on the side that always faces the Sun, and the radiators could also be attached directly to the tethers, but at 90 degrees so they never face the Sun. They could also be attached to the side of the sphere which never faces the Sun.

That might be an emergency configuration if the magnetic rotary joint fails, but the sphere's plane of rotation stays fixed as they orbit the Sun. That means that in 4 months the configuration will have shifted by 90 degrees, making the solar panels useless.

It would be too expensive to continually use fuel to keep the sphere's plane of rotation in place relative to the Sun. Maybe an electrodynamic tether could work, but I haven't looked at that possibility in detail.

And on your general vo

Re: What will happen to their physical condition

[khayman80]

That means that in 4 months the configuration will have shifted by 90 degrees, making the solar panels useless.

Oops. In 3 months, or one quarter of an Earth year, the configuration will have shifted by 90 degrees.

Re: What will happen to their physical condition

[khayman80]

I've been considering Hohmann transfer orbits because they only require thrust that's completely tangential to the Sun. In that case, the spheres' plane of rotation would be perpendicular to the ecliptic plane.

Sorry, this is ambiguous. Here's a better explanation. I've been considering Hohmann transfer orbits because they only require thrust that's completely tangential to the Sun. In that case, the spheres' plane of rotation would be perpendicular to the ecliptic plane, with its surface normal pointing along the (circular) orbital velocity vector.

Re: What will happen to their physical condition

[khayman80]

I spent more time than I'd care to admit trying to think of a way to arrange the solar panels that doesn't require a special magnetic rotary joint.

Maybe the solar panels could be physically attached to the midpoint, and arranged in a circle with the same surface normal as the plane of rotation.

Ordinarily this would result in no solar power, regardless of whether the spheres' plane of rotation has the same surface normal as the ecliptic plane (the "parked" configuration) or if its surface normal points along the orbital velocity vector (the "Hohmann transfer" configuration).

But a large cheap mirror could reflect sunlight onto the circular solar panel, eliminating the need for a special magnetic rotary joint, and the inefficiency of microwave or inductive power transfer.

The mirror could be held in place against solar pressure using VASIMR drives. When the spheres are under thrust, the total fuel needed to move the mirror should be negligible compared to the fuel needed to move the heavily shielded spheres. Moving the mirror independently would allow the spheres' plane of rotation to change without reconfiguring its solar panels each time.

Re: What will happen to their physical condition

[RockDoctor]

Woo, more in there than I'm going to try to deal with on a phone's screen - board. L8R

well who's

[Anonymous Coward]

going to watch the kettle? so to speak.

I imagine they would have to have one hell of an upgrade in remote control or assisted
intelligence to handle any emergencies.

~G

Re:well who's

[scotts13]

going to watch the kettle? so to speak.

I imagine they would have to have one hell of an upgrade in remote control or assisted
intelligence to handle any emergencies.

~G

One just has to be careful of the acronym used for the computers name, and assiduously avoid omnipresent red-glowing video eyes. Then you'll be fine.

Re:well who's

[BarbaraHudson]

going to watch the kettle? so to speak.

I imagine they would have to have one hell of an upgrade in remote control or assisted intelligence to handle any emergencies.

Why not have them awake and doing all sorts of science outside the immediate earth-moon environment? Kill 2 birds with one stone.

"But it will take more resources."

Better than Hal9000 turning them into corpsicles.

Sounds a bit risky

[jandrese]

The problem with this idea is that if anything goes wrong there's no hospital you can rush the people to, and there is always a risk of something going wrong when you start messing with biological systems like this. I suppose we are getting more data about the process regularly from hospitals, but NASA is going to want to do a lot of their own experiments first. I guess since we are nowhere near getting ready to launch the Mars mission it isn't too bad. They still have time.

Re:Sounds a bit risky

[mythosaz]

More or less risky than putting a team of men and/or women in a tin can and blasting them toward Mars?

No matter what, they're going to end up at least 6,778km from the nearest hospital. :)

Re:Sounds a bit risky

[Spy Handler]

Launch the mission in autumn. Send bears. Natural organic hibernation to Mars!

Re:Sounds a bit risky

[NotDrWho]

If anything goes wrong, they'll just wake up in a distant future where everyone is really stupid, or they're a delivery boy, or the Earth is ruled by damned dirty apes. Either way, hilarious hijinks and adventures will follow. Problem solved!

Re:Sounds a bit risky

[Anonymous Coward]

Or we could give them virtual memories of being sent to Mars.

Re:Sounds a bit risky

[TWX]

Yeah, one has to balance the effects of 6+ months one-way in a small metal box while awake and possibly having conflicts with fellow astronauts and cabin fever that one can't even go out on-deck to mitigate versus people that don't wake up again.

I expect that years of studies, including Earth-orbiting studies will be conducted before we ever send people to Mars this way.

Re:Sounds a bit risky

[Triklyn]

:) but in that case at least they're not also holding them at death's door.

there's a reason anesthesiologists make the big bucks. :)

death is no shy wallflower, you ask her to dance, you better be ready to get danced.

Re:Sounds a bit risky

[Triklyn]

if i were in charge i would give you all the monies to make this a reality... space bears :)

Re:Sounds a bit risky

[MozeeToby]

You're giving them too much credit. Yes, these ideas have been used to great effect in emergency rooms around the world: chilling someone for a few hours, even days in extreme cases can do wonders depending on the situation. Chilling someone for a few months? 18 months? I think I'll pass on that one, at the very least I'll wait a good long time while a few 10s of thousands of others try it first.

Re:Sounds a bit risky

[Anonymous Coward]

The obvious answer is put an android on the ship, whose primary function during transit is to watch out for the health of the crew and the ship. In case of serious issues the android wakes the chief medical officer and/or captain.

I nominate Brent Spiner. He hasn't been doing much, anyhow. A few bit parts on various television shows. I'm sure his agent would jump at the opportunity.

Plan B: We can turn it into a reality TV show, which could help fund parts of the mission. Think Big Brother, but in space. You keep two or three people awake for a few days at a time to let them scheme, overlapping their schedules. Every month, after you've cycled through everybody, they choose somebody to throw out the airlock. But of course, _everybody_ goes back into stasis first, because that's the humane thing to do. Everybody goes to sleep, and one person never wakes up.

Re:Sounds a bit risky

[Anonymous Coward]

If it lengthens life expectancy, count me in. It sounds like an awesome way to time travel to the far future. Keep me awake for a few years at a time, interspersed with 5 or 10 years of stasis.

Obviously I'd only do this if I were rich. I don't think I'd like to wake only to have to wash dishes for the next two years. Wash, rinse, repeat, etc.

Re:Sounds a bit risky

[dpilot]

It's going to be interesting for the test subjects. You don't really think that they're first going to use this on the way to Mars, do you? I would suspect that the first many-month tests will be right here on Earth, with continuous monitoring, and they'll probably build time up from the current week until they reach the target.

Then at some point they'll ship the "hibernaculum" up to the ISS for the next layers of testing. They'll probably again ramp the time up, looking for zero-G degradations. By the time anyone ships for Mars this way, it'll be well tested.

But here's the question - I get the impression that you get a better pay scale for being on-orbit. What will be the pay scale for sleeping for 9 months solid? After all it IS hazardous duty. And when someone goes up to the ISS to sleep for long periods will they get on-orbit pay?

Re:Sounds a bit risky

[mythosaz]

Putting people into a medically induced coma (or some sort of other suspension) isn't the trick. It's waking them up.

Re:Sounds a bit risky

[Khashishi]

An optimist, I see. I'm inclined to believe that the future is going to suck (overpopulation, resource conflicts, pollution, fuel shortage, wealth inequity), and I don't particularly care to travel there.

Re:Sounds a bit risky

[BarbaraHudson]

Even a month of total inactivity seriously impacts your health. You need to learn how to sit up again, how to walk, etc.

You might gain a few calendar years, but lose in actual minutes of life.

Re:Sounds a bit risky

[Half-pint HAL]

If anything goes wrong, they'll just wake up in a distant future where everyone is really stupid, or they're a delivery boy, or the Earth is ruled by damned dirty apes. Either way, hilarious hijinks and adventures will follow. Problem solved!

You forgot the scenario where Earth is under attack and no-one knows how to fly fighters any more.

Re:Sounds a bit risky

[Half-pint HAL]

Yes, but "inactivity" is different from "slowed metabolic rate".

Re:Sounds a bit risky

[Anonymous Coward]

Having been in a coma, its not so much the being under that scares me, its the nightmares of my ex wife calling to me from the bridge of the Event Horizon that creeps me out. (and barfing up all that water upon waking.)

Re:Sounds a bit risky

[Anonymous Coward]

Ha, you think this is the real me ... :) Well it is!

Alive but unable to perform work?

[trout007]

Sounds like most government employees.

I'm sorry Dave,

[Anonymous Coward]

I can't do that.

Say what?

[Anonymous Coward]

How is this innovative?
It's been a staple of science fiction for a very long time and as TFS states, we're already using it in hospitals. The only thing new here is that they're considering it as a viable option.

Re:Say what?

[Anonymous Coward]

Because it's in space, therefore we invented it to go to space. That's a central belief in geek space worship.

Re:Say what?

[Half-pint HAL]

How is this innovative? It's been a staple of science fiction for a very long time and as TFS states, we're already using it in hospitals. The only thing new here is that they're considering it as a viable option.

It's called "pushing the boundaries".

Roh-oh

[ambisinistral]

I beg you --- don't do it! As we know from SciFi movies, only bad things happen when astronauts wake up.

OK.

[Anonymous Coward]

A catheter and poop-sack for everyone.. And 6 months, that's one big sack!

Re:OK.

[LduN]

OR I just figured out a clever idea, that might piss of any aliens the encounter. Feed the astronauts beans, attach hoses to their butts nad use "natural gas" for thrust. In a perfect vacuum I'm sure farts do give some accelertion... think of it like an organic Ion engine

Re:OK.

[Anonymous Coward]

attach hosts to their butts nad

Wow. Those are some freaky aliens.

more details on the technology used.

[nimbius]

Nasa is using a combination appropach to this statis project. Whereas before drugs and temperature controlled environments had to be used, the far more economical approach of C-SPAN recordings of US Senator Robert Byrd are used to maintain a comatose like state. This is induced with a combination of John Kerry lectures and once astronauts must be awakened, the system automatically switches to arguments against climate change as presented by the congressional science committee.

storyline

[broadriver]

That has been a staple if science fiction, especially when the storyline is more about the destination then the trip.

Re:storyline

[TWX]

Oooh. Maybe they'll invent the Transporter next!

Even better idea...

[itzly]

Just leave the people at home, and send a robot to do the work.

Re:Even better idea...

[Anonymous Coward]

Just leave the people at home, and send a robot to do the work.

Those million robots are going to look pretty silly boinking each other trying to make baby humans.

http://science.slashdot.org/story/14/10/01/029232/elon-musk-we-must-put-a-million-people-on-mars-to-safeguard-humanity

Re:Even better idea...

[captaindomon]

That won't help us when the Earth stops supporting life, or a mega disease wipes out everyone on earth. We really need to start planning to migrate away from a single source of failure for our species. Human exploration of Mars is the first logical step.

Re:Even better idea...

[Anonymous Coward]

That won't help us when the Earth stops supporting life, or a mega disease wipes out everyone on earth. We really need to start planning to migrate away from a single source of failure for our species. Human exploration of Mars is the first logical step.

I often wonder why this is a reason for human exploration? So as an organism, I supposed invest all of this effort (taxes, smart people, materials) and it's unlikely that my genes will be the ones saved by such an endevour. While I won't discount the value of human cultures and histories, does it matter if we cease to exist? If an alien race finds our history isn't that just as good? Isn't it enough to leave something behind to say, "we were here?"

Think about your lives. I work daily as a drone for a company so that I have shelter and can eat. The most exciting thing in most people's lives is Sunday football. In the USA, people won't even vote in elections to guarantee a future, why bother with this?

Exploration is important. There is knowledge to be gained that can make life more enjoyable here on Earth, but to "save mankind?" I think not.

Re:Even better idea...

[itzly]

Even if Mars would support life, you could realistically only move a tiny portion of the human population over there, so your mega disease would still kill nearly everybody.

Re:Even better idea...

[WrongMonkey]

There are no conceivable circumstances where Earth would be less suitable for life than Mars. Even during the worst extinction level events, Earth was a paradise compared to Mars.

Re:Even better idea...

[Toshito]

So we just don't have to take a trip to anywhere now?

And it's not an argument against sending robot, it makes a lot of sense from a science point of view and I'm all for it.

But it's boring as hell. Last January I went to Paris for the first time, and let me tell you that it's way different than looking at beautiful pictures and exploring via Google Street View.

I would love to be able to go to the Moon or Mars someday... not that I think it will come to be in my lifetime sadly.

Re:Even better idea...

[Beck_Neard]

A great movie quote: "It's circular. You exist to continue your existence. What's the point?"

Re:Even better idea...

[Khashishi]

It shouldn't be _too_ hard to develop an artificial womb. Robots raising babies might be tricky, but everything is going to be hard when it comes to Mars.

Re:Even better idea...

[confused one]

You're just lacking in imagination is all. Of course, Earth's big advantage is having this stuff called "air", which contains this stuff called "oxygen". Otherwise, (besides Mars being smaller and colder and having no magnetic field) they're pretty similar.

Re:Even better idea...

[Khashishi]

I can conceive a few circumstances. Sun, red giant phase, will engulf the Earth. Life will probably be impossible on Mars, too, by that time, but if I were forced to choose a place to live between them, I'd choose Mars.

Re:Even better idea...

[WrongMonkey]

When the sun goes red giant, Mars is likely to be engulfed, too. Feel free to try again.

Re:Even better idea...

[BarbaraHudson]

Titan.

Re:Even better idea...

[Anonymous Coward]

That won't help us when the Earth stops supporting life, or a mega disease wipes out everyone on earth. We really need to start planning to migrate away from a single source of failure for our species.

No other planet supports human life anyway so what difference does it make if you stay on earth or go somewhere else?

As for "mega diseases" wear your spacesuite to wallmart and you'll be just fine.

Human exploration of Mars is the first logical step.

You can fart around on mars, play geologist and "search for life" or you can spend your time working R&D to build out necessary automation and infrastructure to facilitate fabrication of superconducting rings around the planet - establish a magnetic field - restore usable atmosphere.

Fart around with "humans" just because it sounds cool or do something that is actually cool...

Re:Even better idea...

[spiritplumber]

But they'll become a weird mix of Libertarians and Communists! http://en.wikipedia.org/wiki/V...

Re:Even better idea...

[turbidostato]

"That won't help us when the Earth stops supporting life"

A colony on Mars won't help _us_ when Earth stops supporting life, either. It might help those living in Mars, though.

"We really need to start planning to migrate away from a single source of failure for our species."

Yes, I also feel the dramatic feeling of "our species". But think a bit deeper about it. What's the hell with "our" species? What do _you_ eventually earn from Home sapiens still being over there in a thousand years or not?

Re:Even better idea...

[turbidostato]

"Otherwise, (besides Mars being smaller and colder and having no magnetic field) they're pretty similar."

So you make his point: worse case scenario, Earth won't be any worse than Mars, so it seems wiser to...
a) hope for the best: maybe Earth's worst case scenario doesn't happen
b) Only once worse case scenario you go afte the "terraforming" endevour, only here, in the Earth, instead of going to Mars to do the same in a worse planet: being shorter you will always have a harder day to sustain an atmosphere there than in the Earth.

Re:Even better idea...

[Khashishi]

There will be an intermediate stage. It might not last long on a cosmological scale, but still maybe millions of years.

sigh grow their own food

[Anonymous Coward]

Everyone is still under the delusion that all of the food will have to be carried from the start. How much food has the current space produced and people consumed?

Solve that problem first before even thinking of Mars extravagance.

will they wake up as blind astronauts?

[turkeydance]

the link to one posted here in 2011: http://news.discovery.com/spac...

Ooops oh my!

[buggsdummy]

What if they never wake up?

Re:Ooops oh my!

[roc97007]

What if they never wake up?

They'll be dead.

Re:Ooops oh my!

[stoploss]

What if they never wake up?

They'll be dead.

Unless they've gone Unix-style, because then they would be zombies.

I hate that the Linux kernel has no way to kill a process that refuses to accept the signal to die because it is in permanent iowait while simultaneously holding resource locks. Yes, I'm looking at you, Samba Team chumps who coded cifsclient. You have taken down too many systems and forced too many hard resets.

Necessity is the mother of invention

[Overzeetop]

Send 'em there first and they'll have a huge incentive to figure out the food conundrum.

A drastic solution for a minor problem

[Anubis IV]

So, we've saved 180 days worth of food and consumables for each passenger, but have done so at great risk to them. Okay, sure. Now, if we can just keep them in that state, we may not need the substantially greater amount of supplies that are necessary to sustain life once they actually arrive at their destination.

Re:A drastic solution for a minor problem

[Anonymous Coward]

Technically you don't save 180 days of food either, they still need to be fed intravenously. But you might be able to make the food like the goo in the Matrix.

Re:A drastic solution for a minor problem

[Anonymous Coward]

How do you know it's a minor problem? Most likely the scientists behind this aren't retarded and invest their time in at least plausible prospects that may give advantage.

The rocket equation says that to carry stuff to space you not only carry the stuff, you also carry the fuel to carry it, and the fuel to carry the fuel, once you add it up it may be a substantial mass that you need to propell depending on how many crewmembers you send. If there's potential to shave off part of the cost here, and there and in a few other places to reduce the cost there may be a chance to actually send people to Mars as right now nobody of importance cares about space faring missions as there is no cold war to force them to.

PS.
It's also certainly not 180 days worth of food, unless you intend to send a starvation/suicide mission.

Re:A drastic solution for a minor problem

[Anubis IV]

It's also certainly not 180 days worth of food, unless you intend to send a starvation/suicide mission.

That was actually exactly what I was getting at, and why I referred to it as a "minor problem". The major problem is how you send enough to keep the people alive once they get there. The minor problem is how you deal with the 180 days leading up to that. Yes, every little bit helps, but it seems odd to me that we'd suggest such a risky procedure for such a small savings when we have another problem that is so much bigger still.

Re:A drastic solution for a minor problem

[Anonymous Coward]

But you might be able to make the food like the goo in the Matrix.

I for one am down with liquified baby goo. Much less noise than from actual babies!

Cheapest way to Mars

[HughJazz]

This might be useful for transporting animal life but not humans. A astronaut coming out of a long term induced coma would have severe muscle atrophy. There is also extreme risk that an astronaut would stay in coma (with no medical facilities to deal with any complications) Makes for good science fiction but highly impractical any time soon.

A better approach to long space trips... one way travel. It greatly lightens the mass of the spacecraft. In addition, break up the mission. First send automated ships with supplies, habitat and tools. Then send colonists not tourists. If Mars colonies get big enough, they will eventually have the capacity to be build launch facilities of their own. These could be used both for a return to earth and missions deeper into the solar system.

The first few colonists to Mars will be responsible for setting up self-sustainable facilities that can be expanded using local resources. After that's done the chief Mars colonist requirement will be to be youth and fertility. The cheapest way, by far, to get humans on Mars: make them there.

HAL 9000: "I'm sorry, I'm afraid I can't do that"

[Anonymous Coward]

Captain: "Please re-animate the mars crew!"
HAL 9000: "Windows 420 refuses to boot in secure mode.",
                                    "Would you like to play a game of solitaire on Windows XP instead?"

Longterm use - tried out on humans ?

[burni2]

All information points to Torpor as a short term treatment option - indeed there are animals but those are adapted to that condition, humans are not.

The first set of problems that comes to my mind are kidney stones -> Solution catheter/bladder flushning -> next problem infections in the urinary tract due to catheters. Due to the urinary tract not being "flushed" regularly keeping the germs in the lower urinary system. This problem is also much more challenging for women.

Also the subjection of different germ kinds to the lower temperature needs to be taken into account.

Different germ populations have different temperature ranges were they show different reproduction rates. If the cold condition does not favour the reproduction rate that the lactic acid producing germs over the germs from No.2
this can lead to -> Vaginal flora will be less acidic = starting point for "unwanted/dangerous" germs from No.2

Don't think that when your body is in this "pseudo stasis"
germs are too, they aren't.

Re:Longterm use - tried out on humans ?

[radtea]

TFA says it has been used up to seven days in humans, so it's only a factor of ten or so to get a significant chuck of Mars transport out of the way.

In general, chemical reactions slow down with temperature, and while typical therapeutic hypothermia involves fairly high temperatures (~33 C) there may be room to reduce this considerably. Humans will never hibernate without a whole lot of physiological intervention, but it is far too early to say whether or not metabolic activity--including that of our commensal bacteria--can be reduced sufficiently to sustain a mission to Mars.

Re:Longterm use - tried out on humans ?

[wasteoid]

Instead of "think of the children" perhaps the new bleeding heart slogan should be "think of the vaginas!"

Re:Longterm use - tried out on humans ?

[Anonymous Coward]

Instead of "think of the children" perhaps the new bleeding heart slogan should be "think of the vaginas!"

Must I?

Most of them stink like they've never seen a bar of soap. Fantastic for the pee-pee, less so for the tongue (and nose).

Planet of rthe Apes

[Anonymous Coward]

I call Prior Art..

(the Original Movie, with Charleton Heston)

Trials Already Underway - World of Warcraft

[Anonymous Coward]

Sitting in a Low Metaboilic State, conserving energy.. on a couch.. in front of a Xbox or Dream Console.

Yep.. they got plenty of study data

They won't even have to worry about boredom or 'Space Psychosis' .. just a minor side effect of Homocidal tendencies during and after the mission.

Re:Trials Already Underway - World of Warcraft

[peragrin]

Xbox requires an always on connection. Hen they can't authenticate they can't play.

Yes I know Xbox one dropped that requirement. Sometimes at least. Many games still have it.

Re:Trials Already Underway - World of Warcraft

[WillAffleckUW]

If you get the PS4 version, you can use data gloves and run on a treadmill while you play.

oh, wait, that's Diablo. My bad.

Just Go Nuclear and Get There Quick

[thrich81]

At the risk of proposing simplistic answers to these technical questions (as per /. standard), I don't know why NASA isn't considering nuclear propulsion as their first choice for crewed missions to Mars. The nuclear thermal engines were investigated intensively and test articles tested and built in the 60's and were ostensibly cancelled only because there was no mission for them, not due to technical show-stoppers. Once you have a nuclear capability, trips around the Solar System become nearly routine. NASA should let Musk work on chemical rockets for his Mars trips and spend tax money on nuclear which the private guys can't do.

Re:Just Go Nuclear and Get There Quick

[Anonymous Coward]

the outer space treaty
http://en.wikipedia.org/wiki/Outer_Space_Treaty
prohibits the use of nuclear power in space.

Re:Just Go Nuclear and Get There Quick

[Anonymous Coward]

And pollute the vaccuum of space with all that radiation? Some of us have to breathe that stuff!

Re:Just Go Nuclear and Get There Quick

[Anonymous Coward]

NASA generally considers nuclear propulsion a politically and socially charged issue at this time and hasn't been willing to fund serious work on it in forever. There are plenty of NASA engineers who are totally up for the challenge but "we don't touch nuclear" is a high-level management decision.

Re:Just Go Nuclear and Get There Quick

[Garfong]

Which section says that? Searching the Outer Space Treaty for the word nuclear, I can only find prohibitions on nuclear weapons, not nuclear power. IIRC the USSR launched several small nuclear reactors into earth orbit.

Re:Just Go Nuclear and Get There Quick

[Beck_Neard]

Yes, some nuclear engines were tested and yes, none of them exactly blew up. But nuclear engines wouldn't make a Mars trip any less expensive or much shorter. It's estimated that a nuclear rocket would shorten the length of a Mars trip from 6 months to... 4 months. And this would come at huge increase in mission complexity and cost. Not worth it.

For exploring the outer solar system, though, nuclear rockets could have value.

Re:Just Go Nuclear and Get There Quick

[khallow]

No, it doesn't. Russia in particular has launched a number of nuclear reactors into space without violating the treaty.

Re:Just Go Nuclear and Get There Quick

[confused one]

Study I read said it could be done in 39 days using a 10MW VASIMR engine. Now, that was written by the group pushing the technology, and may have been a bit optimistic; but, we can do better than 4 months.

Re:Just Go Nuclear and Get There Quick

[angel'o'sphere]

I don't really get why american so often have the argument "it was written by a group of".
Is the USA really that retarded, I mean: do such groups really exist? How can they survive and be a problem? I don't get it!

Regarding a vasimir engine: the math is straight forward! WTF is the problem in realizing that this is a viable engine?

Re:Just Go Nuclear and Get There Quick

[Beck_Neard]

VASIMR isn't a nuclear rocket, and nuclear power in space currently falls far short of the required W/kg requirements.

Re:Just Go Nuclear and Get There Quick

[confused one]

I was politely trying to say that it was written by the company Ad Astra Rocket Company and they have an interest in promoting VASIMR. Yes the math is straight forward and I believe the technology works. However, I have seen (and worked for) companies that will write white papers that put everything in the most favorable light, even if the result is optimistic, and frankly not realistic. Overstating efficiencies, for example, by using the best case numbers observed in the R&D lab, and not the real-world numbers obtained from field trials. So, yes, such "groups" really exist.

Re:Just Go Nuclear and Get There Quick

[confused one]

I know VASIMR isn't a nuclear rocket (although at some field densities with the right fuels and energy input, the math shows there is the potential for limited amounts of fusion). VASIMR engines of that size will require a nuclear plant to power them. -- and I'm not talking about RTG's here. Frankly, this is territory we've not explored really, beyond a few early ground tests and small scale (10kW) fission reactors launched in the '60's and '70's (SNAP, RORSATs, Topaz) which mostly used thermoelectric conversion. That's not going to work for a 10MWe plant,

Re:Just Go Nuclear and Get There Quick

[Beck_Neard]

> (although at some field densities with the right fuels and energy input, the math shows there is the potential for limited amounts of fusion)

It's not hard to achieve fusion. What's hard is getting more energy than you put in. VASIMR can never produce a self-sustaining fusion reaction unless you're talking about gigawatt-scale power levels. And even then, there's no indication that it will actually work in practice (it most certainly won't, if decades of experience in plasma physics is anything to go by).

> Frankly, this is territory we've not explored really,

Except a whole lot of really smart engineers have explored fission reactors in space and could not come up with a reasonable plan for the type of power densities required. Look at, for instance, the JIMO project: http://en.wikipedia.org/wiki/J...

A Fission-powered VASIMR rocket to Mars that could get there in 2 weeks remains a fairytale. Again, though, VASIMR could be highly useful for other purposes, like increasing mission payload (instead of shortening mission length) or for missions to other solar system bodies.

Re:Just Go Nuclear and Get There Quick

[angel'o'sphere]

In Germany that would be considered "fraud" :D

Re:Just Go Nuclear and Get There Quick

[Anonymous Coward]

A constant one-g acceleration to the halfway point and then a one-g deceleration until you drop into orbit at Mars will take approximately 36 hours. Yepper, 36 hours. With such a propulsion unit you have true one-g gravity (meaning no micro-gravity deterioration); you're exposed to very little radiation; the mass of your life-support supplies (food, oxygen, water etc) is a small percentage of a 180 day voyage; the psychological trauma of being cooped up in a tin can are minimized and, should there be an emergency, a rescue mission is possible.

A side benefit of such a propulsion system is that unmanned orbital probes to the planets and their intriguing moons is possible in weeks rather than years. Cheaper missions, and more of them, to the outer planets will open up a golden age of planetary exploration.

Speed is the only solution to the long flight times. It won't be cheap or easy. It will probably take the better part of a decade to get to the first test flight. We need to get started.

Re:Just Go Nuclear and Get There Quick

[Chelloveck]

Public relations. You have to get the nuclear fuel to orbit somehow. What if the rocket blows up and scatters it everywhere? What if it makes orbit but is inoperable, and falls out of the sky? What if the Mars mission fails en route, and the ship comes back on return orbit and whacks into the Earth? What if terrorists take control via the Internet and use it as a weapon? It doesn't matter if the scientists and engineers say it can be done safely. Who trusts them, anyway? Think of the children!

Remember, large fractions of people still believe that cell phones cause cancer and vaccines cause autism.

Re:Just Go Nuclear and Get There Quick

[Immerman]

>What if the Mars mission fails en route, and the ship comes back on return orbit and whacks into the Earth?

Then you send a second mission to salvage it. You should have at least several decades before its orbit intersects Earth again.

As soon as you leave Earth's local space and start boosting yourself on a transfer orbit to Mars you increase the semi-major axis of your orbit around the sun, and with it your orbital period. On an efficient transfer orbit (the ~180 day trip) your new period will be partway between the original and destination inner and outer orbits, probably somewhere in the neighborhood of 500 days for Earth and Mars. So, 500 days later your ship will be very briefly in the same spot in Earth's orbit that it started from. Earth however will be ~150 days away. It will take many, many orbits before the two are again in the same place at the same time.

And if you're going for a faster transfer - well to do that you boost to a much more elongated orbit and then slow down again so you don't go sailing right past Mars out to Jupiter or Neptune's orbit (the faster you go, the further out the far end of your transfer orbit will be before you slow down again). And the larger the orbit the longer the period, you can easily be talking 5-10 years before your ship touches Earth's orbit again, and it will still take many, many passes before Earth happens to be reaching the exact same point in it's own orbit. Only now the passes are far less frequent, and we could be talking centuries before impact. Plus, since the orbit is extending far beyond Mar's there's a decent chance that the impact will be with Mars, Jupiter, something in the asteroid belt, etc. and hence much less of an issue for us.

Meanwhile its just orbital mechanics - we'll know within days of the failure exactly where the spacecraft will be the entire time, exactly when the eventual impact will come, and whether it will hit us or something else.

Six little words

[Anonymous Coward]

Open the Pod Bay Doors Hal...

Just send Submarine crews

[Anonymous Coward]

They are use to living in a tin can already. Yea sure they can't pop open a hatch now and then like in a submarine but they most likely already have the psychological mindset to tolerate it more than any other astronaut currently in service.

Re:Just send Submarine crews

[confused one]

A tour for a sub crew is typically 70-90 days.

What was it

[Falos]

FTA:

>An alternative to having the whole crew in stasis is to have one person awake for two to three days, then hibernate for 14 days. By staggering the shifts, no one person would be in stasis for more than 14 days at a time and one crewmember would be awake to monitor the ship, conduct science experiments and handle maintenance chores.

IIRC, one of the stories included in OSC's "Intergalactic Medicine Show" did something like this, but a crewmember wakes up to people missing or dead or something. Can't remember it well.

Re:What was it

[spitzak]

The awake guy is useful if HAL has to be disconnected.

Therapeutic torpor was first used in the 1960s.

[Anonymous Coward]

My dad worked on it at Ohio State U in the mid 1960s.

not a new idea

[Anonymous Coward]

The headline should've been: NASA Research Catches Up With the Movies.

Re:HAL 9000: "I'm sorry, I'm afraid I can't do tha

[Anonymous Coward]

The sixth member of the Discovery crew was not concerned about the problems of hibernation, for he was the latest result in machine intelligence: The H.-A.-L. 9000 computer, which can reproduce, though some experts still prefer to use the word mimic, most of the activities of the human brain, and with incalculably greater speed and reliability. We next spoke with the H.-A.-L. 9000 computer, whom we learned one addresses as "Hal."

HAL: This mission is too important for me to allow you to jeopardize it.
HAL: I know that you and Frank were planning to disconnect me. And I'm afraid that's something I cannot allow to happen.
HAL: Dave, this conversation can serve no purpose any more. Goodbye.

Use the more efficient ION-Thruster

[burni2]

http://en.wikipedia.org/wiki/I...
(successfully tested)

You also need to carry a big reactor = big mass (F=m*a) with you + propellant and thus combined with radiation protection problems for the crew and the inefficiency of the system if your mass gain(reactor+additionalshielding) outruns your win (2x specific impulse) over chemical rockets the system is out of question.

Like that "nuclear bomb drive". Sweet on the outside but bitter if you dig into the realisation problems.

Only 19 days

[JonahsDad]

If they use a Lyle Drive. http://en.wikipedia.org/wiki/S...

Re:Only 19 days

[angel'o'sphere]

You are aware that such a drive is SF?

Re:Only 19 days

[JonahsDad]

4th and 5th words after the title in the Wikipedia article.

Used to be you could reference Heinlein without worry that someone would misunderstand.

Obligatory quotes

[Bodhammer]

Drake: Hey, Hicks. Man, you look just like I feel.

Drake: They ain't paying us enough for this, man.
Dietrich: Not enough to have to wake up to your face, Drake.
Drake: What? Is that a joke?
Dietrich: Oh, I wish it were.

Apone: All right, sweethearts, what are you waiting for? Breakfast in bed? Another glorious day in the Corps! A day in the Marine Corps is like a day on the farm.
Every meal's a banquet! Every paycheck a fortune! Every formation a parade! I LOVE the Corps!

Hudson: Man, this floor is freezing!
Apone: What do you want me to do, fetch you some slippers?
Hudson: Ah gee, would you sir? I'd like that.
[Apone points at his face with his middle finger]
Apone: Look into my eye. Fall in, people!

Torpornauts!

[PapayaSF]

I hereby coin the word "torpornauts," which had zero Google hits when I checked.

Pros and Cons

[WillAffleckUW]

Pro: reduced cellular activity means less oxidation stress, less probable impacts from zero G (electro stimulation of muscles instead).

Con: Might not wake up. Then again, it's a one way trip, so not that much of a risk.

Con: Might provide delicious incubation location for hitchhiking space aliens looking to adapt cellular structures to take over Mars and Earth.

Pro: No annoying sales calls.

Con: Computer AI will decide you're a flight risk.

Re:Pros and Cons

[Immerman]

>Might not wake up. Then again, it's a one way trip, so not that much of a risk.

You seem to have a rather ridiculous idea of what "one way trip" means. It does NOT mean suicide mission, it just means you won't ever visit your hometown again. Virtually all the early American colonists came knowing it was a one-way trip - crossing the Atlantic was *expensive*, most people couldn't reasonably afford to do it more than once in their lifetime. Hell, many had to sell themselves into indentured servitude just to afford it, and nobody was buying servants with a ticket for the return trip.

Who will watch the watchmen?

[WillAffleckUW]

Depends. Let's say you have a landing crew of 20, and it's a one way mission. You can sleep 16 and keep 4 awake (good choice for sanity), with rotations every six months in staggered groups of two. Having only 2 awake usually leads to bad things, which is why a waking crew of 2 is part of every good SF space horror movie. If one goes crazy, everyone dies.

Re:Sounds a bit risky (bears in space)

[WillAffleckUW]

Might be better to send astronauts from northern and southern polar regions. More likely to be adapted to long periods of isolation by social customs, more likely to not mind it, better overall.

So, basically, Scots, Norse, Inikvuk (Yukon/NWT), Alaskans, Russians, Finnish, Laplanders, Peruvians, Chileans. Oh, wait, skip the Russians, they'll turn all the food into vodka.

Re:Sounds a bit risky (include androids)

[WillAffleckUW]

Plus, bonus, the android could have a sub mission to return alien life to Earth, using humans as hosts.

Obligatory references

[Anonymous Coward]

Doppelgänger/ Journey to the Far Side of the Sun Barry Gray composition Sleeping Astronauts Also used in the subsequent series UFO episode The Man Who Came Back.

intravenous feeding

[manu0601]

I wonder what happens to guts microbes with 180 days of intravenous feeding. If we fail to slow down their metabolism too, they will start to eat the host's bowels.

Jane/Lonny Eachus goes Sky Dragon Slayer

[khayman80]

We've been over this before, and you already know the answers I've given you. Stop being a grandstanding asshole. I don't have to keep repeating my answers every time you demand them. That's called ASSHOLE behavior, asshole. You have already seen my calculations and my answers to all these questions. By bringing them up and demanding them AGAIN in a different forum, you are advertising your own dishonesty. It didn't work. Don't worry, as I promised this will all be published when I find the time. [Jane Q. Public, 2014-10-03]

Jane, the answers you've given don't make any sense. That's why I'm asking you for a very simple equation describing the required electrical heating power. Again, filling in the following blanks would be be much faster than repeatedly calling me an asshole.

Jane's power in = ?
Jane's power out = ?

Re:Jane/Lonny Eachus goes Sky Dragon Slayer

[Jane Q. Public]

Jane, the answers you've given don't make any sense.

They don't make any sense to you. This much is obvious.

That's why I'm asking you for a very simple equation describing the required electrical heating power.

I repeat: I have already answered these questions several times. You have no legitimate purpose in asking them again somewhere else. And yes, repeating your questions here after they have already been answered is ill behavior on your part.

Re:Jane/Lonny Eachus goes Sky Dragon Slayer

[Jane Q. Public]

Draw a boundary around the (gray or black body) heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out

Just no. That is not even remotely what I meant, and I explained this to you clearly at least several times already. I have no reason to continue to re-explain it just because you keep asking.

Instead I'm going to repeat something else I have stated several times: pick up a textbook on heat transfer, and see what the accepted, textbook, "consensus" science says about it. Hint: they don't agree with you.

I don't appreciate this constant harassment over something that has been explained to you clearly many times over. If you truly still don't understand it, that is sad but it is also not my problem. A textbook might help.

Re:Jane/Lonny Eachus goes Sky Dragon Slayer

[Jane Q. Public]

Jane, mainstream physics is based on conservation of energy. That means power in = power out through any boundary where nothing inside is changing. If your textbook doesn't agree with that principle, it's either wrong or you're misinterpreting what it says. For instance:

I know how this works. Stop trying to be insulting. I'm not the one who got it wrong. YOUR answer (checked 3 different ways) violated conservation of energy. Not mine. Again: I checked both my work and yours. Your "answer" didn't even check out using your own heat transfer equations.

No Jane, you've misinterpreted your textbook. Energy is always conserved, so power in = power out through any boundary where nothing inside is changing. This isn't a "very rare exception". It's a fundamental law called "conservation of energy". Does Jane seriously think his textbook says that using a fundamental law like "conservation of energy" is "doomed to fail"?

I know energy is always conserved, you insufferable ass. I have already proved that my answer conserved energy and yours did not. Your constant blathering about it elsewhere (like here) does not change that.

Is that how Jane "derived" his incorrect equation that electrical heating power per square meter = (e*s)*T1^4?

Your insistence on "electrical heating power" is a red herring. Energy is energy. Your misguided attempt to include the power used to cool the chamber walls does not change that. Spencer stipulated "electrical power to the heat source". It is neither necessary nor called for to calculate the power used to cool the chamber walls in order to find the temperatures of the other bodies.

(e*s)*T1^4 is often called the "Stefan-Boltzmann relation", which is derived from the Stefan-Botlzmann radiation law, and which describes the relationship between thermodynamic temperature and radiative power output of a single gray body. I repeat: you can find this equation in heat transfer textbooks and I also showed you where it is in Wikipedia. Stop pretending ignorance about things I already explained to you clearly several times. I can only conclude that you're doing this in order to harass.

If so, that's kind of a boring mistake because "radiative power out" isn't just a fancy way of saying "electrical heating power". They're completely different. To find electrical heating power, Jane needs to use conservation of energy, where power in = power out. That results in a heat transfer equation, not just an equation for "radiative power out".

NO, it does NOT result in a heat transfer equation. There is no need to account for other, cooler bodies when calculating radiative power out. What, do you imagine that these cooler bodies are somehow "sucking" power away from the heat source? And that a warmer body (but still cooler than the source) "sucks" less power than colder ones do? That seems to be what you're saying here.

Just no. That's not the way it works, man. At steady-state, radiative power out can be calculated from temperature and emissivity alone. Other factors (such as heat transfer) are affected by nearby bodies, but radiative power out of a gray body at steady-state is related ONLY to temperature and emissivity. It's that simple, and claiming otherwise is just wrong.

I repeat: look it the hell up. I have not just one but 4 textbooks here, plus Wikipedia, plus the testimony of experts in the field of heat transfer. They ALL disagree with you. It's that simple.

Re:Jane/Lonny Eachus goes Sky Dragon Slayer

[Jane Q. Public]

For other readers (not for you, because despite your claims you've seen this already several times), from Wikipedia (edited here for clarity given Slashdot's character handling):

A body that does not absorb all incident radiation (sometimes known as a grey body) emits less total energy than a black body and is characterized by an emissivity, epsilon
j* = epsilon * sigma * T^4

In the above equation, using the "dot" notation which YOU pointed out to ME, j is energy and j* is power. This isn't just Wikipedia. It is very easy to find this relation in other sources as well:

Here is "A Textbook of Engineering Thermodynamics . The section on radiative power of a gray body:

Since all bodies are continuously receiving and radiating thermal energy, energy radiating from unit area (all this energy is absorbed by the black surroundings) = sigma * emissivity * T^4

The example goes on to express heat transfer between long co-axial cylinders using heat transfer equations similar to those we discussed before. But heat transfer is NOT the same as the radiative power of a SINGLE gray body at steady-state. Power out is a function of emissivity and temperature ONLY. Heat transfer from one surface to another requires 2 bodies, or 2 surfaces of the same body. But note that the equation for power out clearly implies it is independent of transfer to cooler bodies.

You can also find it here. In this case, note that it gives the equation for power output as distinct from radiation "loss" (heat transfer). BECAUSE THEY ARE TWO DIFFERENT THINGS. One is the power output of a SINGLE gray body at a given temperature. The other is radiative transfer to another body. One requires ONLY emissivity and temperature to calculate. The other involves 2 bodies.

Is this clear yet? Or are you going to continue to erroneously claim that radiative POWER output is dependent on the presence of cooler bodies? Do you really need more examples, or are you finally willing to admit you have been proved wrong? If you need more examples of this, you can find them with a quick search of the 'net. I just did, since I don't have a good way to link to my textbooks.

Re:Jane/Lonny Eachus goes Sky Dragon Slayer

[Jane Q. Public]

I've explained that net heat transfer = radiative power out - radiative power in, so of course they're not the same.

Your BS "explanations" are not informative to readers who actually want to be educated.

Once again, I'm just saying that "radiative power out" is different than "electrical heating power".

No, you aren't, because then your "explanation" re-introduces the dependency. Which is what I have been saying all along (and repeatedly): your methodology contradicts itself.

I'm not even going to bother answering the rest of your blather. Because your whole argument was PUT to rest weeks ago and your failure to understand that (or at least admit it) is rather like a zombie which hasn't quite realized it is dead yet.

I repeat: I have documented this all. I have the reputable and credible (and MAINSTREAM, "ACCEPTED") references which show you to be wrong.

For a while I thought explaining this in different ways would show you that you were wrong. But over time, I have come to accept that you simply won't admit it, no matter what. That's too bad, because I had really hoped you would listen to the actual accepted SCIENCE behind this, and further accept that it was right and you were wrong.

I no longer hold any such hope. I have myself come to accept that you are either a religious zealot, or a self-interested liar.

And I very seriously doubt that you were ever actually a physicist.

Re:Jane/Lonny Eachus goes Sky Dragon Slayer

[Jane Q. Public]

Seriously, "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.

You're just re-hashing old arguments that I've already shot down.

Why are you doing that, if your purpose is not dishonest?

Re:Jane/Lonny Eachus goes Sky Dragon Slayer

[Jane Q. Public]

Seriously, "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.

I didn't say they were the same. They don't need to be the same.

This doesn't cause "radiative power out" to depend on anything but its emissivity and temperature.

If you want to propose some relationship between "radiative power out" and "electrical heating power" then you need to use conservation of energy.

What "I propose" is the textbook answer to this question. It's not even "my" idea, as I clearly showed you just yesterday. YOU are the one going against "established" physics here. So I daresay it's up to you to prove your point, rather than arguing with me about it.

Which you will never do, because you're wrong. If you could actually show how the physics textbook idea of heat transfer was wrong, you would be world famous by now. Instead, you're arguing ineffectively with some person on Slashdot, about something every textbook on the subject, as well as other sources, say your are wrong about.

Instead of calling me a blathering religious zealot liar who wasn't ever actually a physicist, could you calmly explain why you disagree with this energy conservation equation?

I already did so, several times. What, do you honestly think that If I fail to refute this idea just one more time, it will somehow magically become correct?

The heat transfer scenario I presented, and my calculations of temperatures, were correct within a reasonable degree of precision. Yours, on the other hand, were not.

By what stretch of your imagination am I obligated to KEEP refuting your same, lame arguments? This is all old news now. You can read about it all again later, when I write this all up and publish it. In the meantime, if you want answers to these questions AGAIN, you can go back and read our prior discussion of the matter.

Re:Jane/Lonny Eachus goes Sky Dragon Slayer

[khayman80]

Typo: Jane, you're saying: electrical heating power per square meter = (e*s)*T1^4.

Re:Jane/Lonny Eachus goes Sky Dragon Slayer

[Jane Q. Public]

But the Stefan-Boltzmann law in your textbooks actually says:

radiative power out per square meter = (e*s)*T1^4.

Jane, don't you see how your equation for electrical heating power would only be true if "radiative power out = electrical heating power"?

This, from someone who keep saying "power in = power out"?

Of course I realize that. That is, TOTAL power out equals (power out per unit area) * area. When are you going to get it through your head that I'm not a moron?

The problem with your theory is that you have failed to show that electrical power in = anything BUT power out. It isn't heat transfer, as you have several times asserted. Heat transfer to a cooler body has NO relevance to the radiated power output of a warmer body at known temperature. And since it does not affect the power out, it does not affect the power in. QED.

You're trying to give me some crackpot story that the temperature of nearby bodies reduces the required INPUT power for the heat source. I understand what you're saying. I understood it from the beginning. You're just wrong, that's all. It would violate your own "power in = power out" rule. Which obviously you are not seeing, but which I saw right away.

Cooler bodies do NOT lend or transfer any net energy via radiation to warmer bodies. Period. Doing so would be a violation of the Second Law of Thermodynamics. Therefore, the only way a nearby cooler body could create some kind of condition of "less input power needed" for the warmer body, was via magic. You are proposing a magical idea, not physics. Because, again, this violates your "power in = power out" rule. If you draw your boundary around just the heat source itself, since there is NO NET RADIATIVE POWER COMING IN (which doesn't then just go right back OUT, yielding a net of 0), then the only way you can reduce your "electrical" power input is by violating the Second Law.

You're trying to play some kind of trick of adding the incoming radiation to the power output. But that's wrong. No NET incoming radiation is absorbed. Some may be absorbed, but it goes right back again, at equivalent radiant energy. But that power going back out again is not "added" to the object's radiant power, which is independent. Again, that would violate your "power in = power out" rule: you're counting it twice.

That is exactly WHY you can calculate power out of a gray body at steady-state with (e * s) * T^4. Because any incoming radiation is already accounted for. Which you aren't getting through your head, and so you're counting it twice.

And no, the cooler bodies don't "prevent" the hotter body from radiating exactly as much as it was radiating before. They don't "lend" their radiation to the hotter body.

ALL net energy flow in this system is from the center outward. There is no "backflow". It would violate the Second Law. And your "answer" for final temperature of the heat source did exactly that... you were "creating" something like 3kW (I forget the exact number now) from nothing.

And don't try to tell me you're calculating the TOTAL electrical power needed to both heat the source and cool the walls, because that would be a different experiment. Spencer stipulated "electrical power" to the heat source. He left power to the walls unstated, except to say that they are maintained at 0 degrees F. He did not say the power to the heat source AND to the walls was constant. He said the power to the heat source.

So if necessary, technically the power to the walls could vary, but not the power to the heat source. And if you're having a big issue with conservation of energy, that's probably where you're falling down.

You have kept trying to convince me that the cooler passive body somehow "holds the source power in" and thereby makes it hotter. But that's not the way it works. I repeat: EVERY textbook and online reference I've found -- and it's a significant list by now -- disagrees with you. Your own answer disagreed with you: it didn't balance the heat transfer equations, and power in <> power out.

Let' start with Kim Jong-un

[Anonymous Coward]

...Just to be proactive a bit...

Not needed for a Mars mission

[Anonymous Coward]

Torpor state might be needed or very helpful for a longer mission, but it isn't needed for a mission to Mars. The Mars Direct plan was developed about two decades ago and features sending a shipment of supplies ahead of time to Mars and manufacturing methane fuel on Mars for the return trip beforehand. One spaceship needs only carry the crew and supplies for the six month trip to Mars. The crew size could be three or two and they don't have to be huge people - how about two 100 pound women. The one way trip takes 180 days or so. A spaceship can carry sufficient supplies.

What we really need

[wringles]

is an Orion type spaceship. Not the lame Orion capsule that has been in design for the last years; what would really answer the problems of solar system exploration would be a fleet of honest to god atomic explosion powered spaceships. It would cut down travel time to a fraction of current estimates, cargo weight would become much less of a problem, hell you could even bring back those astronauts from Mars when the mission is done.

Jane/Lonny Eachus goes Sky Dragon Slayer

[khayman80]

I repeat: I have already answered these questions several times. You have no legitimate purpose in asking them again somewhere else. And yes, repeating your questions here after they have already been answered is ill behavior on your part. [Jane Q. Public, 2014-10-04]

Jane, you're still wrongly insisting that electrical heating power per square meter = (e*s)*T1^4. Once again, Jane's equation violates conservation of energy. That's why I'm trying to understand why you keep insisting it's correct. At first I thought you agreed that power in = power out, but that we only disagreed about which terms to include:

If radiation enters the boundary and goes right back out, we need to account for it entering and exiting. That's why there are separate terms for "power in" and "power out".

Just no. If radiation goes in and comes right back out, we do not need to account for it, because then the NET amount of that particular radiation crossing your boundary is ZERO. A = A. You do know how to add and subtract, right? You know what a zero is, right? [Jane Q. Public, 2014-09-24]

Jane's statement originally made me think that Jane is reasoning like this:

Draw a boundary around the (gray or black body) heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out

At steady state, Jane's power in = Jane's power out:

electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)

Jane, is that your equation for required electrical heating power? By "A = A", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?

But now it seems like our disagreement is even more fundamental:

I am not going to get drawn into an argument that you have already lost. I repeat that the equation you show is for HEAT TRANSFER, not "radiative power out". You are just plain wrong about that and any heat transfer textbook will you so. ... [Jane Q. Public, 2014-10-03]

This objection is completely different than Jane's "A = A" objection above, which at least seemed to acknowledge that we should start with the principle of conservation of energy, where power in = power out. But now Jane even seems to dispute that starting point.

I'm starting to suspect that Jane opened a textbook and found "radiative power out per square meter = (e*s)*T^4" and simply assumed that "radiative power out" is just a fancy way of saying "electrical heating power". Is that how Jane "derived" his incorrect equation that electrical heating power per square meter = (e*s)*T1^4?

If so, that's kind of a boring mistake because "radiative power out" isn't just a fancy way of saying "electrical heating power". They're completely different. To find electrical heating power, Jane needs to use conservation of energy, where power in = power out. That results in a heat transfer equation, not just an equation for "radiative power out".

Jane, if you don't agree with the "power in" and "power out" that I've tried to glean from your rants, just fill in the following blanks like I did. It'll be much faster than accusing me of ill behavior.

Jane's power in = ?
Jane's power out = ?

Or, explain why we shouldn't start with the principle of conservation of energy which results in a heat transfer equation. Or, (more

20 years too late, but still...

[Anonymous Coward]

...Kubrick got it right. Again.

Vazimir Engine?!?!?

[Anonymous Coward]

Slashdot has previously reported on former astronaut Chang Diaz' Vazimir engine getting our astronauts into Martian Orbit from Earth Orbit in 39 days! I don't understand the need to hibernate anyone unless you really really want to conserve those food/water resources. Since the "VAZ" screams across the heavens, it may take two weeks of aero taking to SLOW THE CRAFT DOWN. But still, that's only about 55 days there & 55 days back? Again, why hibernate?

Jane/Lonny Eachus goes Sky Dragon Slayer

[khayman80]

Just no. That is not even remotely what I meant, and I explained this to you clearly at least several times already. I have no reason to continue to re-explain it just because you keep asking. Instead I'm going to repeat something else I have stated several times: pick up a textbook on heat transfer, and see what the accepted, textbook, "consensus" science says about it. Hint: they don't agree with you. [Jane Q. Public, 2014-10-05]

Jane, mainstream physics is based on conservation of energy. That means power in = power out through any boundary where nothing inside is changing. If your textbook doesn't agree with that principle, it's either wrong or you're misinterpreting what it says. For instance:

I will do you a favor here, and say: don't bother to go calculating the energy, either. The problem is that an analysis of this kind, based on the assumption that power-in = power-out, is doomed to fail except in coincidental cases. Even conservation of energy can give very misleading results. The black body example I gave shows why your "energy conservation just inside the surface" won't work. Aside from just "view factor" and a few other things, a certain amount of the power in (often a very significant amount) just ends up going right back out, but you often don't see that in the formulas. Quote from one of my references, "Fundamentals of Heat and Mass Transfer", by Inropera, et al., 6th edition, 2006, p13. I have to type this in by hand from the book so any typographical errors are mine. Emphasized words have been capitalized.

Relationship to Thermodynamics

At this point it is appropriate to note the fundamental differences between heat transfer and thermodynamics. Although thermodynamics is concerned with the heat interaction and the vital role it plays in the first and second laws, it considers neither the mechanisms that profide for heat exchange nor the methods that exist for computing the RATE of heat exchange. Thermodynamics is concerned with EQUILIBRIUM states of matter, where an equilibrium state necessarily precludes the existence of a temperature gradient. Although thermodynamics may be used to determine the amount of energy required in the form of heat to pass from one equilibrium state to another, it does not acknowledge that HEAT TRANSFER IS INHERENTLY A NONEQUILIBRIUM PROCESS. For heat transfer to occur, there must be a temperature gradient and, hence, thermodynamic nonequilibrium. The discipline of heat transfer therefore seeks to do what thermodynamics is inherently unable to do, namely, to quantify the RATE at which heat transfer occurs in terms of the degree of thermal nonequilibrium. This is done via the rate equations for the three modes ...

Heat transfer requires a temperature gradient, and therefore thermodynamic non-equilibrium (as we established early on). I was hoping you would catch on that this also implies that power-in = power-out is not necessarily true, and in fact that is probably a very rare exception. Therefore, you aren't going to prove anything with this approach. I wanted to stop you before you wasted more of your time. [Jane Q. Public, 2014-09-07]

No Jane, you've misinterpreted your textbook. Energy is always conserved, so power in = power out through any boundary where nothing inside is changing. This isn't a "very rare exception". It's a fundamental law called "conservation of energy". Does Jane seriously think his textbook says that using a fundamental law like "conservation of energy" is "doomed to fail"?

Again, it really sounds like Jane opened a textbook and found "radiative power out per square meter = (e*s)*T^4" and simply assumed that "radiative power out" is just a f

Jane/Lonny Eachus goes Sky Dragon Slayer

[khayman80]

... Your misguided attempt to include the power used to cool the chamber walls does not change that. ... It is neither necessary nor called for to calculate the power used to cool the chamber walls in order to find the temperatures of the other bodies. ... [Jane Q. Public, 2014-10-05]

Good grief, Jane. Once again, I never attempted to include the power used to cool the chamber walls! In fact, I've repeatedly told you it's irrelevant. Once again, that's not what "power out" means. Months ago, after I asked Jane if he agreed that power in = power out, Jane misunderstood my question and responded:

... As long as the power used by the source and the power used by the cooler are constant as required, any relationship between them has no bearing on the experiment. [Jane Q. Public, 2014-08-02]

So I explained that "I've never even mentioned the power used by the cooler of the chamber walls... none of these equations has anything to do with the power used by the cooler. ... Jane's also wrong to claim that the power used by the cooler is required to be constant. ..."

I tried again a month later: "I've repeatedly failed to explain that the power consumed by the refrigerator on the outside is irrelevant. So obviously we'll have to agree to disagree about that."

I tried once again: "... Jane might think I meant power in = electrical heating power, and power out = cooling power of the chamber walls. If so, that's not what I meant, and I'm sorry for not being more clear. I take full responsibility. Just to be clear, power in = power flowing into the boundary in question, and power in = power flowing out of that boundary. ... any power used by the cooler is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant."

I tried yet again: "I've explained why the power used to set the chamber wall temperature is irrelevant. Any power used is simply being moved from some point outside the boundary to another point which is also outside the boundary. Because that power never crosses the boundary, it's irrelevant."

... Your misguided attempt to include the power used to cool the chamber walls does not change that. ... It is neither necessary nor called for to calculate the power used to cool the chamber walls in order to find the temperatures of the other bodies. ... [Jane Q. Public, 2014-10-05]

After I repeatedly explained that the power used to cool the chamber walls is irrelevant, it's bewildering that Jane accuses me of trying to include it.

... The problem is that an analysis of this kind, based on the assumption that power-in = power-out, is doomed to fail except in coincidental cases. Even conservation of energy can give very misleading results. ... power-in = power-out is not necessarily true, and i

Jane/Lonny Eachus goes Sky Dragon Slayer

[khayman80]

... heat transfer is NOT the same as the radiative power of a SINGLE gray body at steady-state. ... [Jane Q. Public, 2014-10-05]

I've explained that net heat transfer = radiative power out - radiative power in, so of course they're not the same.

... Power out is a function of emissivity and temperature ONLY. ... [Jane Q. Public, 2014-10-05]

I've repeatedly failed to communicate that I agree radiative power out is a function of emissivity and temperature only:

"Again, radiative power out is dependent only on emissivity and thermodynamic temperature. We don't disagree about that, despite your repetitive claims to the contrary."

Once again, I'm just saying that "radiative power out" is different than "electrical heating power".

... But note that the equation for power out clearly implies it is independent of transfer to cooler bodies. ... [Jane Q. Public, 2014-10-05]

I've repeatedly failed to communicate that I agree radiative power out is independent of transfer to cooler bodies:

"Once again, I agree that "power out" through a boundary drawn around the heat source is given by the Stefan-Boltzmann law."

Once again, I'm just saying that "radiative power out" is different than "electrical heating power".

... In this case, note that it gives the equation for power output as distinct from radiation "loss" (heat transfer). BECAUSE THEY ARE TWO DIFFERENT THINGS. One is the power output of a SINGLE gray body at a given temperature. The other is radiative transfer to another body. One requires ONLY emissivity and temperature to calculate. The other involves 2 bodies. ... [Jane Q. Public, 2014-10-05]

Exactly. Radiative power out is different than electrical heating power, because only electrical heating power goes to zero when the chamber walls are also at 150F. So electrical heating power involves 2 bodies, but radiative power out requires ONLY emissivity and temperature to calculate.

... are you going to continue to erroneously claim that radiative POWER output is dependent on the presence of cooler bodies? ... [Jane Q. Public, 2014-10-05]

I've never claimed that radiative power out is dependent on the presence of cooler bodies. Once again, I've repeatedly agreed that radiative power output doesn't depend on the presence of cooler bodies:

"I've been trying to tell Jane: we don't disagree about the equation for radiative power out."

Once again, I'm claiming that "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.

If you want to propose some relatio

Jane/Lonny Eachus goes Sky Dragon Slayer

[khayman80]

Once again, I'm just saying that "radiative power out" is different than "electrical heating power".

No, you aren't, because then your "explanation" re-introduces the dependency. [Jane Q. Public, 2014-10-05]

Seriously, "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.

This doesn't cause "radiative power out" to depend on anything but its emissivity and temperature.

If you want to propose some relationship between "radiative power out" and "electrical heating power" then you need to use conservation of energy.

Here's how to use the principle of conservation of energy. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out through any boundary where nothing inside is changing:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

Instead of calling me a blathering religious zealot liar who wasn't ever actually a physicist, could you calmly explain why you disagree with this energy conservation equation?

Jane/Lonny Eachus goes Sky Dragon Slayer

[khayman80]

You're just re-hashing old arguments that I've already shot down. Why are you doing that, if your purpose is not dishonest? [Jane Q. Public, 2014-10-06]

Dishonest? Shot down? Have you even considered the possibility that radiative power out might actually be different than electrical heating power?

For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same. Is saying that dishonest?

This doesn't cause "radiative power out" to depend on anything but its emissivity and temperature. Is saying that dishonest?

If you want to propose some relationship between "radiative power out" and "electrical heating power" then you need to use conservation of energy. Is saying that dishonest?

Here's how to use the principle of conservation of energy. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out through any boundary where nothing inside is changing:

electrical heating power + radiative power in from the chamber walls = radiative power out from the heat source

Instead of calling me dishonest, could you calmly explain why you disagree with this energy conservation equation?

Jane/Lonny Eachus goes Sky Dragon Slayer

[khayman80]

What "I propose" is the textbook answer to this question. It's not even "my" idea, as I clearly showed you just yesterday. YOU are the one going against "established" physics here. So I daresay it's up to you to prove your point, rather than arguing with me about it. Which you will never do, because you're wrong. If you could actually show how the physics textbook idea of heat transfer was wrong, you would be world famous by now. Instead, you're arguing ineffectively with some person on Slashdot, about something every textbook on the subject, as well as other sources, say your are wrong about. [Jane Q. Public, 2014-10-06]

Once again, your textbooks don't say I'm wrong. They just say that "radiative power out per square meter = (e*s)*T^4". Once again, I agree with that statement. So how am I going against "established" physics or arguing with "every textbook on the subject"?

Seriously, "radiative power out" is different than "electrical heating power". For instance, we agree that "radiative power out" stays constant even if the chamber walls are also at 150F, but "electrical heating power" goes to zero. So they can't be the same.

I didn't say they were the same. They don't need to be the same. [Jane Q. Public, 2014-10-06]

Jane, you're saying:

electrical heating power out per square meter = (e*s)*T1^4.

But the Stefan-Boltzmann law in your textbooks actually says:

radiative power out per square meter = (e*s)*T1^4.

Jane, don't you see how your equation for electrical heating power would only be true if "radiative power out = electrical heating power"? If you "didn't say they were the same" then why does your equation depend on them being the same?

Instead of calling me a blathering religious zealot liar who wasn't ever actually a physicist, could you calmly explain why you disagree with this energy conservation equation?

I already did so, several times. What, do you honestly think that If I fail to refute this idea just one more time, it will somehow magically become correct? The heat transfer scenario I presented, and my calculations of temperatures, were correct within a reasonable degree of precision. Yours, on the other hand, were not. By what stretch of your imagination am I obligated to KEEP refuting your same, lame arguments? This is all old news now. You can read about it all again later, when I write this all up and publish it. In the meantime, if you want answers to these questions AGAIN, you can go back and read our prior discussion of the matter. [Jane Q. Public, 2014-10-06]

Jane seemed to try to explain why he disagrees here by saying "A = A" and helpfully asking if I knew what a zero was. But as usual Jane refused to actually write down what Jane considered to be the "correct" energy conservation equation. When I asked what equation Jane meant, Jane said that wasn't it. So Jane's never written down an energy conservation equation around the heated source, which is the first step to calculating the required electrical heating power.

Here's how to use the principle of conservation of energy. Draw a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Since power in = power out throu

Jane/Lonny Eachus goes Sky Dragon Slayer

[khayman80]

... don't try to tell me you're calculating the TOTAL electrical power needed to both heat the source and cool the walls, because that would be a different experiment. Spencer stipulated "electrical power" to the heat source. He left power to the walls unstated, except to say that they are maintained at 0 degrees F. He did not say the power to the heat source AND to the walls was constant. ... [Jane Q. Public, 2014-10-07]

Again, I've repeatedly explained that the power needed to cool the walls is irrelevant, and that it isn't required to be constant.

The problem with your theory is that you have failed to show that electrical power in = anything BUT power out. It isn't heat transfer, as you have several times asserted. Heat transfer to a cooler body has NO relevance to the radiated power output of a warmer body at known temperature. And since it does not affect the power out, it does not affect the power in. QED. [Jane Q. Public, 2014-10-07]

Again, why does Jane think if something doesn't affect the power out, it can't affect the power in? For example, black body "power in" depends on the chamber walls even though "power out" through that boundary doesn't depend on the chamber walls.

Since we agree that "electrical heating power" goes to zero when the chamber walls are also at 150F, has Jane also noticed that "net heat transfer" also goes to zero when the chamber walls are also at 150F?

Isn't that a weird coincidence? So why does Jane keep using an equation that depends on "electrical heating power = radiative power out" without even writing down an energy conservation equation to try to justify that claim? Has Jane even considered the possibility that if he applied conservation of energy, he'd find that electrical heating power really is determined by net heat transfer, rather than "radiative power out" which stays constant even if the chamber walls are also at 150F?

If you draw your boundary around just the heat source itself, since there is NO NET RADIATIVE POWER COMING IN (which doesn't then just go right back OUT, yielding a net of 0)... [Jane Q. Public, 2014-10-07]

If there's no net radiative power coming in, that must mean all the "power in" from the chamber walls just goes back out. That would yield a net of zero. But as usual Jane didn't write down the power in = power out equation showing these terms before they supposedly cancel. Is this what you mean, Jane?

Draw a boundary around the heat source:
Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out

At steady state, Jane's power in = Jane's power out:

electrical heating power + radiative power in from chamber walls = radiative power out from source + radiative power from chamber walls, re-emitted back out (Jane's equation?)

Jane, is that your equation for required electrical heating power? By "NO NET RADIATIVE POWER COMING IN", are you saying "radiative power in from the chamber walls" = "radiative power from chamber walls, re-emitted back out"?

Jane/Lonny Eachus goes Sky Dragon Slayer

[khayman80]

Jane responds.