Low-cost-reusable vehicle design-FICTION *2* - last topic got borked

JimM · 2004-06-08 14:03:47

Hey Jim, I know i'm guilty of getting in a huff with you about spaceplanes, but "comic or pathetic" and "grasping at straws" is extremely... poor form.

Well yes, I have been getting a little T'd off. The guy just does not know when he's beat... not by me, mark you, but by the physics, which I did not invent and he does not understand.

Sometime I do not suffer fools as gladly as you're supposed to in polite company. So, sorry.

(BTW, I'll conceed "comic or pathetic" might have been a bit strong in the circumstances. I'm not so sure about "grasping at straws".)

ANTIcarrot. · 2004-06-08 17:55:12

When concord flies at mach 2 it grows in length. This is due to the airframe being heated. This heating is a side effect of air resistance, and air resistance consists mostly of moving air sideways (and downwards for lift) so the vehicle can move forwards. The heat is produced because the inelastic collision between concord and the air in front of it.

The atmosphere heats up because the kinetic energy of the projectile turns into heat, and as it does some of that heat radiates or is conducted to the atmosphere.

What's more, this expansion (and therefore heating) remains constant even when concord is flying at a fixed speed at fixed altitude. Since it's flying at a constant speed, it's not loosing kinetic energy, and so the loss of kinetic energy cannot be the source of aerodynamic heating, which is what a literal interpretation of what you said, JimM.

See. I can play the "you're using the wrong words" game too.

As to knowing about things, I studies aerospace at university. Hence I sometimes use terms which might not seem clear to you, since you did not. Interactions for example is another word for drag, used because there are lots of forms of drag, and referring to drag as a sole source of drag can get dragging; or tiresome. And confusing for that matter. I also sometimes use simplifications and approximations, in the same way everyone else here uses Newtonian equations, even though we all know Einstein's equations are more accurate.

the LARGER shield (the second) will decellerate much more rapidly and therefore will momentarily be much hotter than the smaller one. (This is the exact opposite of what you would have us believe.)

I'm afraid I must disagree with you in part. Firstly re-entry doesn't quite work that way, secondly you are confusing heating energy with rise in temperature, and thirdly you seem to be failing to take the larger area into account. In no particular order:

Yes, a larger heat shield for any given mass will slow the spaceship down. This is what I have said all along. Yes, as a result of being slowed down, the vehicle will experience a larger heat transfer than a smaller craft. However that heat energy will be spread over a far wider area. In fact, because heat-transfer is proportional to the velocity of the vehicle, and not any measure of size, for any given area (ignoring the hideously complicated issues of airflow over the heat-shield) the heat transfer rate should be equal in both designs.

But even though the kJ/s/m^2 rate remains constant the larger vehicle is still shedding speed faster due to it's larger size. So as we've both previously agreed it will slow down faster, and therefore be exposed to the heating effects of re-entry for a shorter period of time. Because the heat-shield itself has been exposed to roughly equal re-entry temperatures, but for a shorter period of time, its temperature will have increased to a lesser degree than a vehicle with a lower heat-shield area : mass ratio. Similarly the large spacecraft can use materials with a lower melting point than the smaller one.

Think of it as moving your finger though a candle flame. Do it quickly and even though you've being exposed to the same temperature and environment, you don't absorb any significant amount of heat and so it doesn't hurt. Do it slowly and you do absorb a significant amount of heat and it does hurt.

In summary, heating is proportional to aerodynamic air resistance which is proportional to the square of the velocity and area; and not deceleration. And by using large shield/mass ratios we can reduce heat shield temperature rise. Not I expect you to believe me. But perhaps you'd be inclined to believe a Mr H. Julian Allen of the Ames Aeronautical Laboratory; cera 1950.

http://www.1903to2003.gov/essay/Evoluti … ...h19.htm

ANTIcarrot.

JimM · 2004-06-09 01:47:31

What's more, this expansion (and therefore heating) remains constant even when concord is flying at a fixed speed at fixed altitude. Since it's flying at a constant speed, it's not loosing kinetic energy, and so the loss of kinetic energy cannot be the source of aerodynamic heating, which is what a literal interpretation of what you said, JimM.

No. You still havn't got it.

The aircraft is loosing kinetic energy, and that is the source of the heating experienced by it. The surrounding atmosphere is heated by the aircraft, not the other way round. However the aircraft's kinetic energy is continually replenished by the new kinetic energy supplied by the thrust of its engines, which is why it retains constant velocity.

ME: ...the LARGER shield (the second) will decellerate much more rapidly and therefore will momentarily be much hotter than the smaller one. (This is the exact opposite of what you would have us believe.)
YOU: I'm afraid I must disagree with you in part. Firstly re-entry doesn't quite work that way, secondly you are confusing heating energy with rise in temperature, and thirdly you seem to be failing to take the larger area into account.

I have not confused heat energy with temperature. However you have, consistantly.

I have not failed to take the larger area into account. You have not either, but you have somehow managed to invert the actual consequences, consistantly.

I'm fed up with this nonsense. You say you studied aerodynamics at university. How did the university do that without teaching any physics?

(Advice reminder: stop digging.)

ANTIcarrot. · 2004-06-09 08:59:55

I'm not going to write a ten thousand word iron cast essay which you couldn't nitpick, because you'd no doubt follow form and not read it JimM, so I'll simplify my arguement:

NASA AGREES WITH ME.

And in this matter of basic physics, I agree with them, every other space agency, and the rest of the aerospace industry. In fact, much to my surprise (and quite possibly his own) GCNR agrees with me! As did Euler. The faster you slow down during reentry the less your heat shield heats up.

http://www.1903to2003.gov/essay/Evoluti … ...h19.htm

At the end of the day you can make as many digging comments as you like because I can prove the aerospace industry is on my side of this debate, and every derogatory insult you make proves you can't find a single person or page on the entire internet that agrees with you. Must be lonely in your camp.

ANTIcarrot.

JimM · 2004-06-09 15:32:24

ANTIcarrot, in fact none of them agree with you. You are reading what you want to believe into what they actually say.

Just ask yourself two questions:

(1) If the heat comes from the atmosphere and not the vehicle, where does the atmosphere get its heat from?

(2) If heat is transferred from the atmosphere to the vehicle, why does the atmosphere heat up instead of cool down?

It is clearly no longer possible to have a serious discussion with you. I know I said to GCNRevenger that I would try to be more polite to you in future, but I'm sorry, your invincible ignorance would try the patience of a saint.

Go and have your foolish wrongheaded arguments with someone else in future.

Goodbye.

JimM · 2004-06-09 19:17:08

The amount of peak heating at any point on the shield does not drop proportionatly with surface area of the heat shield because the larger shield causes more rapid deceleration and faster generation of heat. "Its more complicated then that" would suffice.

Yes GCNRevenger, it is more complex than that, but I have been thinking about how to express this more clearly (that is, ‘clearly’ for normal people) and so going right back to the nature of kinetic energy I come up with the following…

E(k) = 0.5*M*(V^2)

Where…
E(k) is kinetic energy
M is the mass
V is the velocity
… of the projectile

This equation can be transposed to read…

V = (E(k)/(0.5*M))^0.5

… so, considering part only of the reentry path, and assuming atmospheric pressure is uniform during that part of the path, then the drag on a projectile will be a direct function of its cross-sectional area (X).

Hence, if X = 1…

deltaV1 = (1/0.5)^0.5 = 1.4142

However, if X = 2…

deltaV2 = (2/0.5)^0.5 = 2.0000

… and so on.

Thus deltaV is proportional to X^0.5.
Or if you prefer: deltaE(k) is proportional to X^2.

(where ‘delta’ is ‘incremental’, which can of course be a negative 'incremental')

In other words, the larger the heat shield the hotter it will get (for a time) everything else being equal.

QUED

Euler · 2004-06-10 01:06:47

Thus deltaV is proportional to X^0.5.
Or if you prefer: deltaE(k) is proportional to X^2.

Exactly. That is why vehicles designed to enter a planet's atmosphere are intentionally not very aerodynamic-- so that they can lose more of their velocity before they hit the dense parts of the atmosphere.

Think of it this way. The balloon and the bullet both have to lose the same amount of kinetic energy. The balloon spends more time in the atmosphere than the bullet. That means that the heating of the balloon happens more gradually. The balloon also radiates heat much better than the bullet, and the heat is distributed over more area. This means that the balloon will not get as hot as the bullet during reentry.

JimM · 2004-06-10 05:15:28

Thus deltaV is proportional to X^0.5.
Or if you prefer: deltaE(k) is proportional to X^2.
Exactly. That is why vehicles designed to enter a planet's atmosphere are intentionally not very aerodynamic-- so that they can lose more of their velocity before they hit the dense parts of the atmosphere.
Think of it this way. The balloon and the bullet both have to lose the same amount of kinetic energy. The balloon spends more time in the atmosphere than the bullet. That means that the heating of the balloon happens more gradually. The balloon also radiates heat much better than the bullet, and the heat is distributed over more area. This means that the balloon will not get as hot as the bullet during reentry.

Euler, please look at what I said again.

deltaE(k) ~ X^2

(I'm using '~' to represent 'is proportional to')

This means the balloon will decelerate more rapidly than the bullet and thus spend less time decelerating, turn kinetic energy to heat more rapidly, and so, for a period, will be hotter per square inch than the bullet in the following ratio ...

deltaE(k) ~ (X(balloon)/X(bullet))^2

... and NOT cooler as you suggest. In fact it will be a very great deal hotter than the bullet. (Remember, all this is predicted on the assumption that the balloon will not burst, melt, evaporate or otherwise escape its fate.)

To take an extreme example, suppose X for the bullet is 1 and for the balloon is 1000. Then since deltaE(k) ~ (X(balloon)/X(bullet))^2, it follows that deltaE(k) ~ (1000/1)^2 = 1,000,000. But the balloon's X is only 1,000 times greater than the bullet's, therefore the balloon will be 1,000 times hotter per square inch than the bullet!

Of course at the end of the day, both bullet and balloon will convert exactly the same quantity of kinetic energy to exact the same quanity of heat. But the balloon, not the bullet, will have done it quicker.

If you care to think about it, that is the principle everyone who ever used or might have cause to use a parachute relies upon. The quantity of kinetic energy involved is not sufficient to make heat a problem for parachutists, but the parachute enables you to slow down a lot faster than falling without one, so that you can reach the ground at a manageable velocity instead of conducting all your deceleration at the instant you collide with Mother Earth, which I believe is not very healthy for you.

ANTIcarrot. · 2004-06-10 05:23:17

DeltaV is a measure of force over time resulting in a change of speed. In this case that force is aerodynamic drag, which is proportional to cross sectional area, not the square root of cross sectional area.

http://www.cmflowmeters.com/Rotameter/d … force.html
And again:
http://www.deos.tudelft.nl/ers/operorbs … node5.html

Hence deltaV itself (which is proportional to drag and time) is proportional to the cross sectional area, not the square root of cross sectional area. You are correct in saying that delta(KE) is (if all things remain equal) proportional to CSA squared. But you still seem to be forgetting that not all of the kenetic energy goes into heating up the vehicle.

Orbital speed ~7000mps

Shuttle mass ~ 100,000kg

So shuttle KE ~ 2.45*10^12

Specific heat capacity for steel = 450 J/kg°C

So if the shuttle's kinetic energy was converted to thermal energy, and for arguement's sake the shuttle was made completely out of steel, it's tempreature after reentry would be:

T = 54,444 degrees celcius

ANTIcarrot.

JimM · 2004-06-10 08:36:48

DeltaV is a measure of force over time resulting in a change of speed.

Not in this universe.

I suspect that what you were fumbling towards is dv/dt which is rate of change of velocity, ie. acceleration, as expressed in differential calculus. I had avoided using calculus because I didn't think you'd understand it. I see I was right.

C = 54,444*C

Apart from…

(1) …the surprising absurdity that C equals 54,444 times itself.
(2) …the sudden and unexplained arrival of the velocity of light into these deliberations…

I can’t resist pointing out that…

(3) … you’re out by a factor of about 1,000.

Starting from the same point as you, with orbital velocity of 7000 m/sec and Shuttle mass of 100,000 kgs, and applying the good old formula…

E(k) = (0.5*M)*(V^2),

We get…

= (0.5*100000)*(7000^2)

… which turns out to be …

= 2.45*10^11 joules

From the mechanical equivalent of heat this becomes …

= 5.854*10^10 calories

(The specific heat of the Shuttle is irrelevant here, so long as it’s uniform. So for simplicity, let’s imagine it’s all made of water, specific heat 1.)
On that basis it turns out the Shuttle’s temperature will be raised by 585.4 degc (degrees centigrade), if spread uniformly, not the 54,444 degc you came up with.

Of course, this is to ignore the loss of heat by radiation, conduction and convention, principally to the atmosphere. It also ignores the fact that some parts of the vehicle, for instance the leading edges of the wings, will experience much higher temperature increases than others.

I really must stop coming back to upset you. But you are such a tempting target. Sorry, but you really are.

I’ll try to be good and not come back at you again, if you promise not to put up such obvious howlers that cry out to be shot down.

JimM · 2004-06-10 08:48:36

(3) … you’re out by a factor of about 1,000.

My typo. That should read, "you're out by a factor of about 100."

Sorry.

Euler · 2004-06-10 11:17:31

Of course at the end of the day, both bullet and balloon will convert exactly the same quantity of kinetic energy to exact the same quanity of heat. But the balloon, not the bullet, will have done it quicker.

Yes, the balloon initially decellerates more and recieves greater heating. That is a good thing, because it means that the balloon has a lower velocity when it enters the thick parts of the atmosphere. Both vehicles must lose all of their kinetic energy before they hit the ground. The bullet arrives at the ground first, meaning that it had less reentry time to lose that kinetic energy.

E(k) = (0.5*M)*(V^2),
We get…
= (0.5*100000)*(7000^2)
… which turns out to be …
= 2.45*10^11 joules

My calculator thinks that it comes out to 2.45*10^12 J.

ANTIcarrot. · 2004-06-10 16:01:21

My calculator thinks that it comes out to 2.45*10^12 J.

As did mine. Three times. This figure if converted to heat would produce the following tempreature rises if the shuttle was composed entirely of:

Steel: ~54,000C
Aluminimum: ~27,000C
Water: ~600C

ANTIcarrot.

JimM · 2004-06-10 16:41:33

My calculator thinks that it comes out to 2.45*10^12 J.

Woops! Sorry, you're right.

As a result, the vehicle would have acquired a theoretical temperature 5854 degc if none of it was dissipated by radiation, etc.

This figure if converted to heat would produce the following tempreature rises if the shuttle was composed entirely of:
Steel: ~54,000C
Aluminimum: ~27,000C
Water: ~600C
JimM is apparently not familiar with the idea of different materials having different specific heat capacities, and that water is well known for having one of the highest.

What's that got to do with the price of fish?

I brought specific heat into this discussion long before you did. The only way it matters is that it should be identical in and uniform through all examples. Specific heat is otherwise a total red herring here. The amount of heat generated by loss of kinetic energy is identical, whatever the specific heat of the object loosing the kinetic energy. I'd have thought that should be self-evident to everyone, even you. So your temperatures quoted above are 100% twaddle.

The reason for using H2O is that it's Sp. Ht. is 1, which simplifies out an irrelevance. And even then you're out by a factor of 10, BTW. It's 5,854 degc. (taking into account my earlier error--see above-- which you forgot to correct for)

RobS · 2004-06-10 17:13:20

I had read somewhere--perhaps in an Arthur Clarke book--that the kinetic energy of slowing something from orbital velocity to zero is sufficient to melt metals. This will be a serious problem with any heat shield unless it can be designed to throw most of that heat into the atmosphere. That's what the silicon shields do, as I understand it, because they don't conduct heat much; they can get incredibly hot, but they don't heat the underlying vehicle much, so most of the heat eventually goes into the air. A metal heat shield will conduct a lot of heat into the vehicle, which is a potential problem.

-- RobS

ANTIcarrot. · 2004-06-11 04:23:19

By way of explanation at the strart of this little diversion I wrote:

But you still seem to be forgetting that not all of the kenetic energy goes into heating up the vehicle.

Still not very good at reading other people's posts are you JimM?

The point is to prove that the vast majority of the kinetic energy shed during reentry does not go into heating up the vehicle. Hence there are other mecanisms at work. So unless you can proove that it's a simple linear relationship between KE lost and heat absorbed by the vehicle, you cannot say that higher rate of KE loss will automatically result in increased heating in either the heatshield or the vehicle, which seems to have been your key objection to high G reentry.

ANTIcarrot.

JimM · 2004-06-11 09:15:00

I had read somewhere--perhaps in an Arthur Clarke book--that the kinetic energy of slowing something from orbital velocity to zero is sufficient to melt metals.

I'm sure Clarke has written on this, but I could not off-hand say where.

OTOH, Robert Heinlein's "The Moon is a Harsh Mistress" describes how the rebelling lunar colonists make war on earth by showering down inert rocks launched from the moon by way of a mass driver. Each rock, entering the earth's atmosphere near vertically and thus having little opportunity to be braked before hitting the earth's surface, was like a nuclear explosion as its near-hyperbolic-velocity kinetic energy was instantly turned into heat.

JimM · 2004-06-11 09:43:35

The point is to prove that the vast majority of the kinetic energy shed during reentry does not go into heating up the vehicle.

One: Of course it does in the first instance, but then it is radiated off, etc., mostly into the atmosphere.
Two: But that was not what this was about. Your point, repeated over and over, was that the heat came from the atmosphere, mine was that on the contrary it came from the vehicle. It's gratifying to see you concede this to me at last.

…you cannot say that higher rate of KE loss will automatically result in increased heating in either the heatshield or the vehicle, which seems to have been your key objection to high G reentry.

However I see the missionary work with you is not yet finished.

It's not just my 'objection' to high-gee re-entry, but that of everybody else involved in the business too.

If it's high-gee stopping you want, why not do what Heinlein suggested (see my previous post) and keep on coming in at full tilt 'till you hit terra firma? That'll give you high-gee with a vengence.

ANTIcarrot. · 2004-06-11 15:02:19

Two: But that was not what this was about. Your point, repeated over and over, was that the heat came from the atmosphere, mine was that on the contrary it came from the vehicle. It's gratifying to see you concede this to me at last.

Want to bet? Okay. Let's take a quick trip down memory lane to a post you made on June 7th at ~10AM.

Look, you started this by saying--
Which will be subjected to the greater heating? The bullet. It's high density and efficient aerodynamics means it will slow down slowly, will be passing through high energy plasma for longer, and will absorb more heat.
But neither bullet or balloon 'absorb' heat from their surroundings.

Interesting how my phrase 'absorb more heat' became twisted in your mind into 'absorb more heat from their surroundings'. Or maybe not so strange as this isn't the first time you've done that.

I'm not going to debate this anymore because:
1) It's an engineering fact. Whether you believe in it or not is your problem.
2) You ignore every source I offer.
3) You misread my posts more often then you read them; often ignoring intended content to nitpick typos and play silly word games.
4) You're a rude and condescending git.

The last is of course subjective, the rest is not.

If this topic dies here, then I leave happy I have disproved every claim you've made about the DH-1; either though logic or by stumping you though the simple act of asking for proof.

ANTIcarrot.

JimM · 2004-06-11 18:19:53

Interesting how my phrase 'absorb more heat' became twisted in your mind into 'absorb more heat from their surroundings'. Or maybe not so strange as this isn't the first time you've done that.

I have not twisted your phrase. Why should I want to or have to do that? You have said it there once more, loud and clear, plain and simple, no twisting involved. If an object has ‘absorbed more heat’, where can it have absorbed it from? Itself? But for an object to absorb something from itself is an oxymoron, something that is self-contradictory or absurd, a linguistic and scientific nonsense.

Therefore, assuming you are not talking nonsense, it must follow that the expression ‘absorb more heat’ means quite specifically ‘absorb more heat from its surroundings’. There is no other rational or logical meaning or interpretation. Therefore, assuming you mean what you say, I have proceeded on that basis.

However unfortunately what you claim is ‘an engineering fact’ is not one; it is twaddle. The heat arises from the energy already contained within the object being transformed from kinetic to heat energy, which involves no energy whatsoever being ‘absorbed’ by the object.

I do not ignore your sources, but somehow you manage to misinterpret their meanings to suit your cause on a regular basis.

I do not misread your posts. The problem is that you do not seem to understand that carelessly written or badly worded text will not be automatically excused and corrected for by the reader, but rather be taken to be what you mean, even if it is not. By writing like this you (or anyone else doing so) risk a regular supply of serious misunderstandings and what I can only call own goals. I am not asking for perfect English prose, but if what you say can be misinterpreted or not understand because you have not expressed yourself clearly enough, you have only yourself to blame.

I have not once ‘nitpicked typos and played silly word games’. I've got better things to do.

If I’m ‘rude and condescending’, then I’m sorry but your invincible wrongheadedness has driven my patience to and beyond its breaking point.

If this topic dies here, then I leave happy I have disproved every claim you've made about the DH-1; either though logic or by stumping you though the simple act of asking for proof.

I for one hope it dies here; I’ve already indicated that I’m fed up. As for DH-1, I had rather assumed you’d gone quiet about that because it was so blindingly obvious even to you that you had comprehensively lost the argument for the DH-1 on every front.

Soyuz · 2004-06-22 08:15:32

On the X-prize forum John Carmack(http://www.armadilloaerospace.com]Armadillo aerospace) wrote:

orbital:
My pet plan for orbital has been "boosted SSTO" for quite some time. This is basically the plan described in "The Rocket Company" serial on Hobby Space -- a straight up / straight down booster, and a very high performance upper stage. I think there are strong engineering and operational reasons for this, even though it looks like an inefficient staging point at first glance.

Rxke · 2004-06-22 09:17:18

Hi, Soyuz,

something's wrong with the link, http://www.xprize.org/messageboard/view … rt=60]here it is...

Interesting post from Carmack all around, looks like he'll quit the X-Prize competition, and do something outside that framework...

Looks like he's fed up with the 'legalese' stuff, hindering his progress...

GCNRevenger · 2004-06-22 10:13:41

Well, then this clearly illustrates that Carmack should stick with making video games and not rockets... the up/down trajectory is patently useless for anything except tourists who want to go up a nose just above the atmosphere for a few minutes and down again.

To put "inefficent" into a clearer light, he would have to build a rocket of even larger scale than the flying skyscrapers of the Delta rocket models to deliver a useful payload, and thats burning LOX/LH. If he sticks with whimpy storables like catalyzed-peroxide then it gets truely monsterous from a thrust/weight perspective.

The DC-X rocket's ultimate goal, the DC-I, was 40m tall and 16m wide, and while this is slightly overkill for three passengers + pilot, it is the bare minimum for useful payload masses... and Carmack proposes a vehicle with around half or a third the efficency?

Soyuz · 2004-06-22 12:25:18

I agree that the cost estimates in rocket company may be a little to optimistic. But I think there are some advantages to the concept. At the moment a true RLV seems not possible. I think this may be a good step between current rockets and an RLV. I think you should look at it as X-33 on top of a DC-X (or DC-Y). The advantages are that you can drop the expensive Linear aerospikes of the X-33 and optimise the engines for vacuum. And you're past the atmosphere when you launch the "X-33". So you don't have drag. You can build a very simple first stage and you only have to worry about the second stage being very effective. There's the argument that you can do the same whit a balloon or a big plane. But i think the problems with designing such a balloon or airplane are gigantic. You have to have a huge balloon or airplane and they can't go as high as the first stage. You also need to have enormous amounts of helium. And i think the plane and balloon will be as expensive to maintain and build as a rocket as first stage.
If you put a X-33 (Or Venturi Star) on top of a DC-Y you probably will have a rather big vehicle. But i think Carmack will be happy if he can get three people in to orbit. So you don't have to make them enormous.
If he then can keep improving his second stage it may be possible to drop the first stage. And he will have a true RLV.

GCNRevenger · 2004-06-22 13:03:28

The problem there is, a reasonable cost estimate is so bad considering the very large development bill, that -no- tiny RLV like the TRC or what the AltSpace folks can make is profitable. The least ambitious one attempted, Kistler Aerospace's KH-1 is less then half finished and their company is bankrupt with heavy debt.

An airplane or balloon to assist launch would be very big, but so would a first stage that operates like the one in TRC or like Carmack wants to build, and thus would be very expensive no matter how simple it could be built. The upper stage would also have to be quite heavy, and it is questionable economics to go through all the trouble to try and recover it.

A two-stage RLV aproach is not in itself a horrible idea, but doing it in any way like TRC or how Carmack wants it is a horrible idea. Too much of the payload fraction, the mass of stuff you can deliver to orbit per-pound of rocket, simply takes too big of a hit when you do this up & down nonsense, its bad enough as it is if you just throw the first stage into the sea.

Oh, and on a side note, the X-33's aerospike engines are beautiful because they operate just as well at high altitudes as low altitudes, plus have engine-out capacity built into each module. They really are great, albeit expensive, if you don't mind a little extra complexity.

Right now I would put no stock in any AltSpace system that couldn't lift a minimum of 10MT into LEO, raise that to ~15MT for a robust manned vehicle, and right now none of them seem even close. There is just no use for a vehicle that gets half-way to orbit, and so nobody will pay for it to be built. It either lifts many tons to LEO, or it doesn't.

If it takes a three seater to orbit to make John Carmack happy, considering that his rocket needs to be tens of times larger, I can gurantee that he will be a sad man to the end of his days.

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#216 2004-06-11 04:23:19

Re: Low-cost-reusable vehicle design-FICTION *2* - last topic got borked

#217 2004-06-11 09:15:00

Re: Low-cost-reusable vehicle design-FICTION *2* - last topic got borked

#218 2004-06-11 09:43:35

Re: Low-cost-reusable vehicle design-FICTION *2* - last topic got borked

#219 2004-06-11 15:02:19

Re: Low-cost-reusable vehicle design-FICTION *2* - last topic got borked

#220 2004-06-11 18:19:53

Re: Low-cost-reusable vehicle design-FICTION *2* - last topic got borked

#221 2004-06-22 08:15:32

Re: Low-cost-reusable vehicle design-FICTION *2* - last topic got borked

#222 2004-06-22 09:17:18

Re: Low-cost-reusable vehicle design-FICTION *2* - last topic got borked

#223 2004-06-22 10:13:41

Re: Low-cost-reusable vehicle design-FICTION *2* - last topic got borked

#224 2004-06-22 12:25:18

Re: Low-cost-reusable vehicle design-FICTION *2* - last topic got borked

#225 2004-06-22 13:03:28

Re: Low-cost-reusable vehicle design-FICTION *2* - last topic got borked

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Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked

Re: Low-cost-reusable vehicle design-FICTION 2 - last topic got borked