You are not logged in.
No I meant with robots building robots: Robots building machines. For instance if you agree that robots now do almost all the car building. Now if you would put in a remote radio control and a computer and make it do stuff like the mars rovers are doing, then you have robots building robots.
Robots building machines, even other robots, are not the same thing as von Neuman machines.
The sort of robots that we have in car plants today are really pretty primitive. Indeed, they only just qualify as robots. They can’t ‘see’ or otherwise know anything about their environment. They can repeat the same sequence of actions over and over, but as soon as something unplanned happens they can either ignore the change and cause chaos, or switch themselves off. IOW, they are completely and utterly dumb.
The current crop of Martian Rovers are certainly a step forwards, but still they need constant help from us humans. They can’t decide for themselves where to dig a hole and then take or not a sample because they don’t know what’s interesting and what’s not, they can’t decide what route to take, they can’t go beyond the point we can see because they have no automatic way of adjusting their travel to take account of unexpected obstacles like rocks. And they sure as anything can’t replicate themselves, let alone turn themselves into machines that do other things. And of course, they can’t repair themselves. They’re good, don’t get me wrong. But they’re not that good.
Why is this impossible if its already happening? Only thing missing for the machines being build to be called robots is a telecontrol. You just need to feed in another design in to the manfacturing plant and there are your machines (robots).
I’m all in favor of optimism, but you can get too much of a good thing.
A machine that replicates itself is not so straightforward as you make it sound. For a start, it has to go prospecting and find the right suitable raw materials. Then it has to turn them into the right sort of refined materials needed to make every one of the components that goes into its make up. Then it has to manufacture all—every one—of these components. Then it has to assemble them all together into a twin of itself. Then it has to program its twin. And unless you want it to stop and wait for new instructions from earth (or even from somewhere else on Mars) every time it hits a snag in all these millions of millions of steps it has to be able to do all this on its own initiative or the whole business will take for ever. So forget remote radio control. And its computer will have to vastly more powerful than any made so far—and that’s just one of the things that will have to be replicated to make a true copy.
Of course, to carry out this replication process, it will be necessary—
(A) Either to have a machine that is, among other things, a minerals prospector and miner, universal materials refiner, universal component manufacturer, and so on,
(B) or a machine that can also manufacture other machines that are specialist prospectors, miners, refiners, component makes, etc., etc…. but of course it can do none of that until the first of all these are made so it can make them (it’s getting a little circular here, isn’t it?)
So, in the end—or rather, in the beginning, (B) does not work out; only (A) does, which means you need a sort of universal machine.
How far do you suppose we are away from having this sort of machine? 10 years? 25? 50? 100? 200? Never?
Because we sure don't have this technology now.
From what I gather the inside of the space shuttle is pretty cool. Else we would have fried astronauts
Uh, we have had fried asronauts (in Columbia).
And anyway it's a long, long way, temperature-wise, from room temperature (about 20 C, say) to -249.6 C, which is not all that far from absolute zero. The D2 would have to be some 270 C cooler than the inside of the Shuttle is when it behaves. That is no minor detail.
I don't understand why robots building machines is ST technology? Don't they build cars now?
Yes they do build cars and many other things, but that's not what I presumed you to mean when you said…
Just robots building other robots thats what needed.
Robots building other robots--in other words, von Neuman machines --is what you need to automatically build a factory (actually a process plant in the case of D2) on Mars. That is what is significantly beyond our abilities today.
In fact if we had von Neuman machines, all sorts of things would become possible like for example, asteroid mining to create virtually free SSPSs and the terraforming of Venus-- just to start with.
The thing is, we don't have them yet.
------
On top of that, if the boil-off rate is 1% a week the whole idea becomes completely impossable instead of just unfeasable.
Yes. I seems to me that's another reason why fusion reactors on earth would want to make their own D2. Apart from it being effectively free to them, it is clear that with H2/D2's boil-off rate, which applies on earth as well as heaven (sorry about that) it would be essential to create D2 for almost instant use; the stuff is effectively unstoreable over the medium term.
And the earth's oceans are awash with the stuff anyway; it's just rather more dilute than on Mars.
Anyway I'm curious to learn how a large low density cargo that must be kept below -249.6 C is going to be induced to reenter earth's atmosphere without exploding, imploding, blowing every pressure relief valve in sight, or otherwise failing to deliver itself intact to the ground?
Who says it's 5 times commoner on Mars than Earth? What is the basis of this belief?
Robert Zubrin, that's who. (Case for Mars p. 224) And, just about anybody else that knows anything about Mars
OK, thanks. I should have thought of checking 'Case for Mars'. (Although my google search-- look at http://www.agu.org/sci_soc/prrl/prrl001 … l0019.html --suggests that the quantity may actually be just 2 times commoner when you get to the Martian groundwater. And I see no information on this topic from the latest visitors to Mars this year yet.)
But that is not the main problem about this proposal, in my view.
It says that the boiling points of hydrogen and deuterium are slightly different, so just cool your electrolized hydrogen down to -249.6 C....
You make it sound like a dawdle. It is not. Engineering this sort of fractional distillation process plant to work is a major engineering effort on earth, never mind Mars.
We don't need electrolysis or liquid hydrogen much here on Earth, so extracting deuterium is very difficult for us.
What has our need for the stuff got to do with how difficult it is to extract?
When we have full-scale commercial fusion reactors, we will need deturium, and it will not be all that difficult or expensive when compared with making the stuff on Mars and shipping it back to earth. It will be produced though fractional distillation process plants that take a tiny fraction of the power from the reactor to deliver all the deuterium needed for it.
It would very silly to depend on deliveries from Mars, when you can make all you need on your doorstep. In fact, I suspect the reactor's waste heat would be more than enough to drive the distillation, meaning the real incremental cost/Kg would be zero.
So it would certainly be useful for Mars-based thermonuclear reactors, but that's about it. It's landed value on earth might well turn out to be almost zero.
I don't understand why you guys are talking about star trek technologies as it's not needed at this point.
....getting energy and resources that is the problem not ST nano tech.
Just robots building other robots thats what needed...
Unfortuinately, robots building other robots is at this time also ST technology, just to exactly the degree that nanotechnology is, because they are actually the same technology.
The feasability of self-replicating machines was first postulated and given a sound theoretical basis by John von Neuman (1903 - 1957), a Hungarian genius who, courtesy of Hitler, ended up at the Institute for Advanced Studies in Princeton alongside Einstein & co. This is why such devices are known as von Neuman machines.
Unfortunately for your proposal, and despite the huge advances in computing since the 1950's, real von Neuman machines stil remain some way away, somewhere over the current-day 'event horizon'.
Thus, the same is true of real nanotechnology because real nanotechnology depends on our being able to create tiny von Neuman machines. That would be the only practical way for it to come about, as you can see when you stop and think about the problem.
So if one idea is star trek technology, so is the other. On the other hand, neither is as futuristic as all that; just out of reach for the near-term future.
The same can be said for commercial fusion reactors.
I think we would be well advised, as Robert Zubrin has been, to base all our plans and proposals for traveing to, exploring, settling and exploiting Mars, on existing, established technology. It needlessly weakens our case by coming up with proposals that depends on some scientific or technical breakthrough that has not happened yet-- especially as this is really not necessary.
On Deuterium:
Who says it's 5 times commoner on Mars than Earth? What is the basis of this belief?
As I understand it, deuterium is used in heavy water to act as a moderator in a fission reactor. There are many suitable alternative moderators.
OTOH, deuterium with tritium is/will be used as a fuel in fusion rectors. The trouble is we don't have a viable fusion reactor technology developed yet, never mind a commercial (that is, profitable) one.
How would you transport deuterium back to earth?
-- As heavy hydrogen? If so, have you allowed for the cryogenic tankage requirements and the 1%/week boil-off rate? You better plan on getting it back here fast. No leisurely cruises back by solar sail or ion thruster, etc.
-- In the form of heavy water? Think of the weight penalty just getting the stuff off Mars for a start.
If you go for the D2O transport method, first you have to distill out the heavy water. That will require prodigious amounts of power and a huge distillation facility weighing thousands of tons at least-- and at least at first, that'll all have to be brought from earth. If OTOH you want to send D2, you'll need even vaster power and plant requirements on the Martian surface to refine it from the heavy water.
And anyway, are you certain Mars is going to give up its water so easily for your D2O plant, at the probable rate of thousands of tons a day, to make the whole thing worthwile in any case?
There must be an easier way to earn a crust or develop a planet...
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.
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.
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.
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: ~600CJimM 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)
(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.
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.
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.
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
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.
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.)
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".)
A ship during re-entry heats up because of interaction with the atmosphere, not because of any internal energy conversion., etc., etc., (more rubbish)
No. Stick to things you know about.
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. Your description of what you apparently imagine to be the physics of the Shuttle's reentry is so wrong it's either comic or pathetic.
Therefore in two theoretical reentry shields, one having 1m2 area and the other 100m2 area, the second will absorb 1% of the energy per square meter, and will not get as hot as the first. In fact, if they form part of equal weight space-craft, the second ship's heatshield will experience roughly 1% of the tempreature rise of the first.
No. If everything else is equal as we previously discussed (the mass and specific heat of the two are identical) then it should be obvious that --
-- 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.)
-- therefore, the larger the heat shield (everything else being equal) the higher the peak gee experienced during reentry, NOT the lower.
-- however, over the entire reentry process, both will emit IDENTICAL amounts of heat. They MUST do this, because ALL the heat comes from convertion of exactly the same quantity of kinetic energy.
-- the true function of a heat shield, in terms of the physics, it to act as a heat sink for the dissipating kinetic energy of the whole vehicle as it slows.
Newton is still right and you are digging yourself into a deeper and deeper hole. The first thing to do when you find yourself in a hole is-- stop digging.
-------------
PS: And do you seriously deny that traffic expands to meet supply when new roads are built??
That is not the normal purpose that a road is built for. I cannot say that political pressure never causes roads to be built for wrong reasons, or NOT built for wrong reasons. And I can't deny that somethimes there is SOME traffic growth due to the arrival of a new road. But that is 'spin-off' as it were. It would never be enough to justify the construction of the road on its own-- just as SOME people used Concorde to cross the Atlantic, but not enough to justify the project. (You're clutching at straws.)
Should we invest the few tens of billions needed to make a spaceplane for real for keeps, unlike the silly rocket/space/shuttle/thing, right now? No, but I think it is prudent to keep such a vehicle in the backs of Nasa's minds, - concept and componets not blueprints - working on the hard pieces, so it can be made ready with less effort down the road since we know for fact that regular rockets won't do and that a spaceplane will probably work.
Apart from the highly dubious proposition that the back of NASA's mind is a good place for anything that you want to work and be effective one day, I might just about go along with you here.
I admit it, I suspect the single most effective step forward for manned spaceflight would be to abolish NASA...
I agree that "build it and they will come" is a shakey business case, nor can SSPS compete with Earth-bound sources of energy... the only reason I am sticking up for a spaceplane is that someday I think there will be such demand (colonization, flights to and from) and that day will be sooner with the technology available for easy orbital access. Neither dinky rockets like DH-1 or now-and-then megaultrasuperboosters are suited... I used the Pan Am space example because that was one of the companies flying passengers into orbit on 2001 Space Odessy.
(I'll get back to SSPS and why you are wrong there later.)
Meanwhile your reason for sticking with the spaceplane now does not compute.
If we are agreed that (1) the demand does not exist yet, and (2) it will be cheaper and easier to do the job later as our technology in general gets better, and (3) it will only be possible to fund once the demand is perceptable even to boring people like bankers, etc., then trying to build a spaceplane now is basically daft, like trying to push things around a tabletop with a length of string-- or maybe trying to build a jumbo jet back in 1904.
------
For everything there is a season...
Another example of push string which we should all be familiar with is the way the number of cars on the road increases with the road space governments fill. Build a new road to speed up traffic? People buy more cars. This is of course driven by an interest in personal freedom (not waiting for public transport) personal choice (there isn't public transport) and keeping up with the Jones' Next Door. (They're destroying the enviroment 50% more than we are! We can't let them snub us like that!)
Nope. The philosophy behind building new roads is called "predict and supply". The engineers who decide where new roads go look at the existing and forecast traffic demand, and build roads to meet that demand. It is a crystal-clear example of demand-pull.
JimM, tempreature during re-entry is a function of speed and altitude. If all things are equal, and both the bullet and balloon begin reentry at the same speed and theoretical height, then how can one be subjected to higher tempreature plasma than the other?
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. They become heated by trading energy of velocity (kinetic energy) for heat energy. As the Second Law of Thermodynamics makes clear, all energy eventually deteriorate into its most chaotic form, which is to say heat. The shortest way to state that is, "entrophy maximises". But outside nuclear reactions which we are not discussing here (where still, mass-energy cannot be created or destroyed) energy can neither be created or destroyed. All it can do is change its form.
Thus I did NOT "point out the same amount of thermal energy gets transfered to both shapes." NO thermal energy gets transferred TO them. The thermal energy (more popularly called "heat") is "created" within them from kinetic energy as the object decellerates.
However english is (almost by design) an imprecise language, and that many meanings can be attached to one word. You took heat to mean energy apparently without considdering that it might have meant tempreature.
English can be as precise as you want it to be.The whole business of lawyers is built around that very fact.
And I did not for a moment confuse heat, energy and temperature. But you have and I fear still do.
Dismissing a concept based purely on wether you do or do not think the other person is using correct english (without reguard to the physics or maths) is a poor habit shared by politicians, preachers and GCNR.
I did not do this; I just took you at your word.
The trouble about using incorrect english is that it's a good way to lose an argument, perhaps especially in physics where the correct language to use is really mathematics, exactly because it is a highly precise subject. Serious discussions on physics topics all 'happen' in mathematics; all non-math discussions on physics are really just attemps to enlighten the non-mathematical.
I have started thinking about DH-1 and related alt-space efforts as being a "supply side" solution to humnaity getting out into space. Invent low cost Earth to LEO access (provide supply) and people will find things to do with that supply.
Wanna bet? (see below)
The other way round is to look at the "demand side" and think about whether we can locate a potential demand for space access that might be more price inelastic, or which can be funding by supplemental means besides taxpayer revenue.
Bingo!
Demand side is what I've been preaching for years.
That's why I believe SPSs are the way to create a huge demand-- indeed, a case for a huge manned presence in space that is logical and stands up to examination even by people like politicians, bankers and corporations that have no intrinsic interest in space per se.
That's also why I favor BDB as a way to lift huge payloads to orbit cheaply, but only when needed. It does not have to keep to a flight schedule or fly so many times a year to justify itself, like Shuttle was supposed to and spaceplanes, DH-1 and all the rest are supposed to too (but won't).
----------------
Demand-pull not supply-push, in other words.
To see the difference, take any small object, tie some string around it so about a foot or so is left free for you to hold, place it on a smooth surface and try to ...
(a) Push it away using only the string. (That's supply-push)
(b) Pull it towards you with the string. (That's demand-pull.)
Honestly now, which works better?
No, he was right. The ballon does not get as hot because the energy is spread out over more surface area. Actually, the deceleration phase does not happen that much faster for the ballon, it mainly just happens higher up in the atmosphere where the air is thinner.
No, you are both wrong.
The amount of heat to be absorbed is equal to the kinetic energy of the object, be it bullet or balloon. Kinetic energy is a function of velocity and mass. (Our old friend Sir Isaac Newton again.)
E(k) = 1/2M * V^2
where E(k) is kinetic energy
M is mass
V is velocity
We have already specified that the mass is identical in both balloon and bullet. As both are re-entering at identical velocities, it follows that...
E(k) Bullet = E(k) Balloon
End of argument.
In a dispute between instinct and physics, physics wins every time.
You are wrong.
----
The race is not always to the swift nor the battle to the strong, but that's the way to bet.
Imagine a bullet and a balloon reentering earth's atmosphere. We'll ignore heating effects for the moment and concetrate on heat itself. 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. The balloon has lousy aerodynamics, and low density and will slow down much faster, even if it's of equal weight to the bullet.
I think your physics is wrong.
If the bullet and balloon both have the same mass but the balloon has much the largest cross-sectional area, then the balloon will as you rightly say, slow down more quickly. But this does not mean it will absorb less heat. Bullet and balloon will both absorb the same. However, the balloon will experience the higher peak temperature during the decelleration phase because it will happen more rapidly than for the bullet.
Aerodynamic heating is a direct product of the resistance of the atmosphere to the object traveling though it. Technically, it is the transformation of kinetic energy K(e) to heat. It does not matter if this energy-change happens quickly or slowly, the final result will be the same.
(This in a idealised world where balloons do not burst, and bullets and balloons have the same specific heat, mass and shape-- say, a sphere.)