Reusable Rockets to Orbit

GW Johnson · 2011-12-06 09:12:38

Joshnh4h asked about his rocket trajectory. I haven't digested all of this. Old guy, slow. Also about limited trajectory distance over which ramjet adds energy: true enough for vertical launch, that's what worries me about the attractiveness of it. For HTO staged aircraft, the range spent in the air is quite long: that's why it looks so good.

Hop said something about a turbojet/ramjet aircraft. That looks pretty good, and could very well be integrated into the same engine. It would be one step past the J-58's that powered the old SR-71. Those had a 25% of airflow bypass after stage 3 compression, straight to the afterburner duct, meaning 75% of the air still went through the turbine core. Limited to about M3.8 max because of blading temperature problems at both ends of the spool.

On the other hand, if you rigged a 100% bypass capability that you could modulate 0-100, from the inlet straight to the afterburner duct, you could shut down the core turbine (to protect it, and just run the afterburner as a ramjet. Should be capable to M5 or 6.

Depending upon the arrangement, a ramjet pod or nacelle need not be complex or expensive, and can be tough as an old boot. With some sort of recovery built onto it, items like that should be more recoverable than the rocket stages we have so much trouble reusing. I'm thinking a stowed swing wing down the side, skids and a nose wheel, and R/C controlled glideback to a runway.

A ramjet airplane for HTO would be even easier. No integrated engines, just parallel-burn with separate rockets and ramjet. It's just stick-and-rudder flying. The ramjet is your fuselage. Your fuels and propellants go in the wing, the rockets go in the wing strakes, and the pilot goes in the inlet centerbody spike. The payload goes on the back.

GW

JoshNH4H · 2011-12-06 10:34:49

Hope-

You admit this mistake and then ignore the correct numbers.
Again, the difference between 8.1 and 7.9 km/s is 1.6 MJ/kg.
This is NOT the same as the potential energy difference between earth's surface and a 300 km altitude, 2.81 MJ/kg.

As I suppose you missed in my last post, I mentioned that the difference was presumably due to the difference in orbital speed between a 0 km orbit and a 300 km orbit. I still do not see why this number has any bearing on a rocket launching from Earth's surface to LEO; could you please explain your reasoning to me? That is probably the best way to go about this. Chances are we're not going to get anywhere at all unless we start listening to what each other says, given that we both obviously have reasons to think that we are correct.

Wrt the thrust-to-weight of a rocket, when did I ever suggest that gravity drag was insignificant? For my model of Delta-V to orbit, both could be about a a km/s or two and the final number would still end up being what it is. The relation between the two is of course not quite as linear or direct as you portray it, but among chemical fuels, H2/LOX does have a significantly lower T/W than other fuel combinations. Even H2/LOX isn't too bad, though, given for example the Space Shuttle Main Engine, with a T/W in the 70s.

GW- Given the increase in total Delta-V to orbit caused by a vertical firing followed by horizontal firing trajectory, it's possible that it will be in the atmosphere for some period of time. The trajectory would probably look something like this, which is the same image that I posted before. Ideally they wouldn't spend any more time in the atmosphere than was necessary, but there are potentially significant gains from a ramjet if you can make it just as reusable as the rest of the craft (Or more, that would be totally fine too ) and (This seems to be the bigger issue to me) be reintegrated with the rest of the rocket at launch in a cheap and safe manner, reliably enough that we could reuse it a hundred or a thousand times.

Terraformer · 2011-12-06 10:53:45

Hmmm... so a ramjet-turbojet integrated engine is feasible? I presume it could be used to fly the carrier plane back after it's launched it's payload? Of course, if you can build it into an SSTO spaceplane...

It certainly begs the question of why Reaction Engines are working on the SABRE engine - surely the mass benefit of cutting out a few tonnes of engine mass don't warrant the extra complexity?

I imagine such an Spaceplane would fly up on jet power to about 20km, before shutting down the jet and lighting the rockets. Total required delta-V could come to say 9km/s, with the first 2km/s being jet power. Perhaps a mass ratio of 10 could be achieved. If the craft is 20 tonnes and delivers a 10 tonne payload, a payload fraction of around 3 percent could be achieved, though you're talking about a GLOW of 300 tonnes... assuming a runway could be found for it, and the average cost of the fuel comes to 250 dollars a tonne, we're talking about 7.5 dollars per kilogram in orbit in fuel costs. If the price is double this, it's still ridiculously cheap.

Hop · 2011-12-06 12:49:29

JoshNH4H wrote:

As I suppose you missed in my last post, I mentioned that the difference was presumably due to the difference in orbital speed between a 0 km orbit and a 300 km orbit.

Which is different from what you wrote earlier:

JoshNH4H wrote:

To go from the surface of the Earth to 300 km requires an input of 2.94 MJ/kg. The force for this energy ha to be applied straight up. Therefore, as per implicit convention, we use the equation Esp=.5*v^2 where Esp is in J/kg to calculate a velocity of 2,425 m/s (It will actually be slightly less than this because I did not account for the difference in gravity between . If you do the energy calculations, the difference in energy between 8.1 km/s and 7.1 km/s is very similar to that number

Which is flat out wrong. First some minor nitpicks: the potential energy is 2.81 MJ/kg, not 2.94 MJ/kg. And the 7.1 km/s is a typo, it should have been 7.9 km/s.

More importantly, the energy difference between 7.9 km/s and 8.1 km/s is nowhere near 2.81 MJ/kg.

A good response would have been for you to admit you're wrong.

That you then say something else and pretend you never wrote the above is very disappointing.

Terraformer · 2011-12-06 13:01:55

I wrote:

though you're talking about a GLOW of 300 tonnes... assuming a runway could be found for it,

Well, according to wikipedia...

The 747's maximum takeoff weight ranges from 735,000 pounds (333,400 kg) for the −100 to 970,000 lb (439,985 kg) for the −8.

So no problems with the runways then, assuming we're allowed to use them. Apparantly the A-380 has a maximum GTOW of over 650 tonnes... I think such a craft should be able to find somewhere to takeoff and land. We might have to use the additional GLOW to make it so sturdy that it just needs refueling and checking out before it can fly again - 50 tonnes of structure rather than 20, with a 10 tonne payload - but that will still bring costs down to ridiculously low levels.

JoshNH4H · 2011-12-06 14:21:40

Hop- I didn't need to go out and say that I was wrong about that one thing because it was obvious given that I did not contest your number. It's a lot more productive to think about what is right than who is right. I proceeded to offer an explanation as to why the difference in specific kinetic energy of delta V was different from the change in specific potential energy, and why that further supports the point that I am actually arguing, which is that my method is a reasonable way to look at the components of the delta V to orbit and, frankly, yours is neither reasonable nor right.

JoshNH4H · 2011-12-06 14:27:57

Terraformer- According to Zubrin in Entering Space, airlines generally look at costs around three times the cost of fuel. I think we might be looking more at about 5 times, at a guess, depending how many times you can reuse it, what the unit cost is, and how complex launch operations are.

On the other hand, rockets use a lot of fuel. Actually, HDA is quite expensive as far a rocket fuels go. Perhaps I should reconsider.

Oh also, where did you get a delta V of 9km/s from? I've been using 9.4, and flying through the atmosphere with a ramjet is just goimg to make it higher.

Last edited by JoshNH4H (2011-12-06 14:38:25)

Terraformer · 2011-12-06 15:21:35

Josh - actually I just pulled that figure from nowhere. But you're going 2km/s @20km, so you have to add maybe 200km altitude and 6km/s velocity, so the required delta-V shouldn't be as high. Remember, you're lighting the rockets off at near vacuum conditions - most of the (pretty much negligible) air drag is gone by then. The more inefficient parts of the launch will be born by the ramjet, which is the most efficient engine, so...

I'm assuming, of course, that the rockets are designed properly for reuse. That means scrapping this notion of stripping mass down to the utter minimum.

JoshNH4H · 2011-12-06 15:34:40

Terraformer-

I agree that the benefits of a ramjet are certainly worth considering, and that they would result in some instances in a simplification of operations. But that's best quantified in terms of a lower Mass Ratio, as opposed to an artificially low Delta-V.

Terraformer · 2011-12-06 16:12:23

Well, 9km/s isn't really that artificially low - you can get benefits like that from switching to denser fuels already. Bear in mind that you don't need to oppose gravity using thrust for the first 2km/s, so gravity drag is going to be slightly less anyway. But even if you use 9.3km/s, you still get such mass ratio's (actually, lower when I did the calculation - to add 2km/s using a system with an effective exhaust velocity of 9km/s only requires a mass ratio of 1.25).

So, say we're moving at 2km/s, 20km above ground. The vertical delta-V we need to reach 300km is 2.4km/s. The horizontal delta-V we need is 6km/s. If we neglect gravity drag, we only need to add 6.5km/s acting diagonally to provide this. The Saturn V had gravity loss of 1.5km/s, but I don't know how much our spaceplane will have. If it's 1km/s, then we need the rocket engines to provide us with 7.5km/s delta-V. If they are Methane-LOX rockets with an Isp of 380 (if you are going to quote Zubrin, then to Zubrins numbers I shall go) we need a mass ratio for this stage of the ascent of 7.5. 1.25*7.5 gives a total mass ratio of 9.4, so the extra payload can be used to strengthen the rocket even more. I don't want to be shelling out lots of cash on repairing it each launch.

Perhaps, GW, a small craft with say 500kg payload could be built and tested to demonstrate the technology, massing no more than a small jet?

Hop · 2011-12-06 21:09:27

JoshNH4H wrote:

Wrt the thrust-to-weight of a rocket, when did I ever suggest that gravity drag was insignificant?

Your propellant table looks at ISP and density. No mention of thrust to weight. If you regard gravity drag as significant, your tables should also look at T/W.

JoshNH4H wrote:

For my model of Delta-V to orbit, both could be about a a km/s or two and the final number would still end up being what it is.

So 2.4 km/s for potential energy and 2 km/s for gravity loss? Sounds like you're still trying to get a vertical velocity vector with magnitude around 5 km/s.

JoshNH4H wrote:

The relation between the two is of course not quite as linear or direct as you portray it,

As I portray it? Recall that the first stage going straight up and coming back down to the same launch pad is your image.

You imagined this first stage making a 5.25 km/s vertical velocity vector which is added to a 7.9 km/s horizontal velocity vector to get 9.4 km/s hypotenuse for your total delta V budget.

I stuck with your image since it was familiar to you. I noted a vertical ascent incurs gravity loss.

A 9.8 m/s^2 vertical acceleration is killed by gravity. It gives you 0 km/s. But this acceleration still consumes propellant, so you have to include the vertical a * t lost to gravity in your delta V budget.

Your mistake is assuming the lost vertical a * t gives a vertical velocity vector that you can add to the horizontal velocity vector. Doesn't work when velocity vector has magnitude zero.

Gravity loss can be arbitrarily large. You can make it as big as you want by increasing payload mass thus reducing thrust to weight ratio.

If you do a slow vertical ascent over 612 seconds, gravity loss is 6 km/s. Does that mean you can turn sideways, do an 8 km/s horizontal burn, and claim the 10 km/s hypotenuse as your delta V budget? No. Your total delta V budget would be 14 km/s.

Actual ascent trajectories typically start vertical and turn horizontal over time. If your acceleration vector is alpha degrees from horizontal, cos(alpha)|a| is the magnitude of the horizontal component and sin(alpha)|a| is the magnitude of the vertical component. The magnitude of the net vertical component would be sin(alpha)|a| - g + (w^2 * r), where w is angular velocity in radians and r is distance from earth's center. g is GM/r^2, of course.

In this case, the vertical acceleration component lost to gravity also doesn't add to vertical velocity. Even with this more accurate and complicated flight path, your model is wrong.

Last edited by Hop (2011-12-06 21:13:09)

JoshNH4H · 2011-12-07 20:18:30

First, a quick reply to Terraformer on the topic of Methane Vex, then I'll address Hop's points Hop- You may not have time to read this whole post. Whether you do or not is obviously up to you. But please, if you only read one thing, skip down to the next line of bold text and read the stuff that follows.

I use Zubrin as a source when there are no others. Specifically, when factual and experimental information is not available, one of his conjectures is at least a reasonably informed (if occasionally optimistic) assessment of the situation when it comes to rocket technology, which is his original career field. For the Isp of a methlox rocket, I took a look at the list of engines fueled by it on Encyclopedia Astronautica. 375 s vacuum seems reasonable. 345 s trajectory averaged is actually starting to seem a bit pessimistic; 355 s might be more reasonable.

Hop-

I'm really glad you made that reply. It's a good starting point for actually getting somewhere.

First: Isp and Density are much easier to quantify than the thrust-to-weight ratio, especially given that some of the propellants I'm looking at here have never been fired in a real rocket engine. Further, T/W is much more dependent on engine design than it is on the actual fuel being used. The theoretical limit on engine T/W is dependent more on the strength of your materials than it is on anything else, but it is my understanding that that is not what limits it in most rocket engines. Given that even among fuels that have been fired in engines, they have been developed to differing levels in different places, I could not realistically just look at existing engines and approximate a T/W from there. How would you have me come up with a measure of T/W based on the fuel that was reliable enough to be used? I can't think of one. Further, while T/W of the engines is important, it is not in my opinion as important as Mass Ratio or Density, in terms of selecting a fuel for the rocket.

My comment about "it is not so linear as you portray it" was with regards to the density of fuel vs. T/W of the engine; you implied (or at least, it seemed to me that you were implying) that there is a relatively direct relation between the density of the fuel and the T/W of the rocket burning that fuel. I do not believe that this is true. At the most, fuel density is one factor among very many which affect the T/W of a rocket and a rocket engine.

Now, I can see why you would be so vehemently opposed to my method of calculating the ⌂V to orbit if I were saying what you seem to think that I am, based on your last post. However, the argument and method described there is not what I have been describing for the past page or so of posts. Perhaps it would be helpful for you if I were to give you an idea of my thought process.

Since I first learned what ⌂V was, I pretty much implicitly assumed that the ⌂V required to get to orbit was composed of several separate components, each of which could just be added together to get the total ⌂V to orbit. In other words, given that:

Da=Air Drag
Dg=Gravity Drag
Vu=Velocity needed to get to orbit altitude
Vo=Velocity of Orbit

I would have told you that:

⌂V1=Da+Dg+Vu+Vo

And I believe that this is the perspective taken by a significant part of the space community. More recently (in fact, soon before I wrote the beginning post in this thread), having learned some physics and some math (I was plugging the rocket equation into a calculator for about a year before I actually learned what a natural log was other than being a function of some kind), I took another look at it. Knowing that force, and therefore impulse and velocity, are vectors, I assumed that they could be added together as with other vectors. In this case,

Vv=Da+Dg+Vu
Vh=Vo

and

⌂V2=sqrt(Vh^2+Vv^2)

But you pointed out in this post, quite correctly, that this was predicated on the assumption that the rocket always had its force directed along the direction of the hypotenuse of the triangle which you drew. In my very next post, at the very beginning, I agreed that I was wrong. If you would like, I can provide links and quotes.

[Most important part of this post, e.g., Hop, if you're losing patience with me you can skip everything up to here and I won't mind that much]

Because my original method of calculating ⌂V was wrong, I had to figure out another one that worked, and actually accounted for the way in which real rockets actually launch. I found this graph and posted it in this post, about a week ago. Looking at that, I came to the same conclusion that you did in your last post, which is quite a reasonable one. As you stated it:

Hop wrote:

Actual ascent trajectories typically start vertical and turn horizontal over time.

This means that you can neither add the Vertical and Horizontal ⌂Vs using linear addition (which would involve always firing your engines either straight up or precisely horizontal) nor can you use vector addition, because that would involve a constant angle, which is not correct. So, I came up with the theory, suported by logic and evidence, that:

The real delta V of an actual rocket launching into Low Earth Orbit will be somewhere in between ⌂V=⌂Vv+⌂Vh and ⌂V=sqrt(⌂Vv^2+⌂Vh^2), which are the results of linear addition and vector addition, respectively.

However, the launch trajectory of a rocket is quite complicated and can't necessarily be modeled with any function. Even if there were a function that modeled it to an acceptable degree of accuracy, I do not have sufficient data to derive it. Therefore, we instead have to find something where we can say "Well, this is somewhere in the middle, and it's going to be a reasonably close approximation even if it won't be exact."

My response to this was to do the geometric average of ⌂V1 (the ⌂V resultant from doing Vh+Vv) and ⌂V2 (The ⌂V resultant from sqrt(Vv^2+Vh^2)). In combination with the expressions for ⌂V1 and ⌂V2 above, this would give an equation for the ⌂V to orbit of:

⌂V=sqrt(⌂V1*⌂V2)

Given that we're dealing with more complex curves that are far from linear, I think that the geometric mean is better than the arithmetic mean (e.g., ⌂V=(⌂V1+⌂V2)/2) for approximating the ⌂V to orbit.

---End the really important part of the post--- (I still have interesting stuff to say, though, I promise!)Good theory. How about I test it out? Say

Vo=7800 m/s
Vu=sqrt(2*2.81e6)=2,370 m/s
Dg=~2/3*300 s*9.8 m/s^2=1,960 m/s*
Da=150 m/s

*Gravity drag depends to a significant degree on trajectory, as you mentioned. I assumed 5 minutes (e.g. 300 s) to get into orbit, and a weighting factor of 2/3. This weighting factor accounts for the fact that the effective acceleration due to gravity goes down as the horizontal velocity goes up. This factor would be exactly 1/2 if it followed a straight line trajectory with constant acceleration to orbit. This is not the case, so the weighing factor will be more than that.

Vv=Vu+Dg+Da=4,480 m/s
Vh=Vo=7,800 m/s

⌂V1=12,280 m/s
⌂V2=8,995 m/s

⌂V=10,510 m/s

Hmm, looks like my model needs to be reviewed. I suppose it shouldn't be a surprise that a rocket would go for a path similar to the one assumed for ⌂V2, given that that path has a lower ⌂V.

GW Johnson · 2011-12-07 23:38:29

I once built several 2-D trajectory codes out of the concept below, when I was young. One of them was for vertical ballistic launch, the rest were quite different problems. It matched real ballistic (nonlifting) flight performance very well. I might even still have it, but none of these modern PC's and Windows versions now supports the advanced BASIC language that I wrote it in. It was meant for a DOS machine, not Windows. The last Windows variant that still could execute it in its DOS emulator was Windows 98. (I still have a working antique like that. Every time I power it up, I wonder if it will die this time.)

Here's the concept:

Along the flight path forward acts thrust, which you can set as a user-defined set of inputs. Also, drag in the reverse direction, which varies with speed squared and density, and as an empirical coefficient as a function of Mach (different for each stage). The flight path makes an angle to the horizontal, but starts out nearly vertical, usually only about half a degree off vertical. Weight acts vertically downward, but that is related to thrust by the massflow rate, and by the stage configuration. As you climb, gravity really isn't constant, either. (What I wrote was flat-Earth 2-D, but the same stuff can be programmed into spherical-Earth coordinates.)

You can use a simple forward-stepping estimate, if you choose the time step small enough. Long ago we had to be more sophisticated (such as Runge-Kutta), but with the advent of 486 PC's that was simply no longer necessary. Just calculate thrust, drag, and weight at the current time, and ignore their change over the short interval. Keep it really short, though, like under 0.1 second.

When you talk about "gravity loss", you are really talking about the effects of that non-constant vertical weight vector, integrated over time.

GW

GW Johnson · 2011-12-07 23:41:36

For crude estimates, you can factor up the orbital velocity by about 10-15%, as the effective delta-vee required to overcome gravity and drag with a clean, fast-ascent vertical launch vehicle. Good enough for 2, 3, even 4-stage vehicles to zeroth order. To do any better than that requires real trajectory code calculations.

GW

JoshNH4H · 2011-12-08 14:56:13

GW- that sounds a lot more comprehensive. If I have some time, maybe I'll give a go at writing it in MATLAB code (read: the only programming language I know that could conceivably actually do this). That would presumably give pretty good results, though I'm not quite sure I know enough about the physical effects involved to approximate the actual delta-V to orbit of a rocket using this method (Specifically drag). For example, how does the coefficient of drag change with velocity?

If I succeed in writing the code, I would imagine that pretty much any desktop with XP or later would be able to handle it. MATLAB code can be run either in Matlab, which can be bought for about $90, or in Freemat, which is legit, free, and just as good (I use it in my engineering classes, and I've never had a problem).

Rune · 2011-12-08 15:12:57

Nice catfight... ^^ It still amazes me how polite you can be at times, Josh. Must drive trolls crazy, you read like a patient monk. Now if I can add my two cents to increase the general confusion:

I think what is really driving Hop nuts if that your approach, however good or bad your results are, is conceptually flawed, because it doesn't take into account time. I mean, he insists and insists about gravity loss, and he's right about that, it IS a function of time. It could be minimal if your rocket shoots up at 20G's, or it could multiply delta-V several times if you choose to go up slowly and majestically, like the alien ships in sci-fi shows.

Thus, total delta-v to orbit, by definition, is at least in part a function of flight time, and therefore T/W. Figure out your flight time, and you have your gravity loss, no need to estimate (only since mass is a variable, so is gravity loss/s, and therefore getting the gravity loss involves integrating a differential equation, no two ways about that). If your model doesn't take that into account, as Hop says, try again. Or fix the T/W of your model so it only works for a given acceleration profile and stick with a given gravity loss. I'd suggest the Saturn V's, since it's rather pessimistic (T/W 1.18 at liftoff, barely enough to crawl away from the tower), and widely available.

Oh, and T/W of an engine, for the same engine, IS proportional to the fuel's density (a few engines can run several fuels, russian experiments with methane on kerolox engines jump to mind). Or to be more precise, with it's molecular weight, and directly proportional at that. The rest (which you can approximate to a fixed coefficient and not go too wrong about it) is the engine cycle that you use, and you weight efficiency (a function of the chosen material, and those don't really change that much)... with clever designing assumed, of course. That's why H2/LOX has such low T/W no matter what engine cycle you use.

On the actual topic of the thread, reusable rockets to orbit... I think we should open up a new one to discuss Musk's idea for turning F9 into F9"R"... I suppose you have seen the video and heard him talk about it? It looks... well, it made me grin like a stupid. And the words "like god and Robert Heinlein intended them to" jumped to my mind. Also, why has no one actually tried to do something like that before?

Rune. Now will you both get mad at me for obviously misinterpreting your argument? Most probably.

Hop · 2011-12-09 16:37:53

GW Johnson wrote:

When you talk about "gravity loss", you are really talking about the effects of that non-constant vertical weight vector, integrated over time.
GW

This my model:

All these things are constantly changing

a the gross acceleration vector
alpha, the angle from horizontal
ah, the horizontal component of a
av, the vertical component of a
gravity changes with altitude
and so called centrifugal force grows with the (horizontal velocity)^2/r

Integrating all that is beyond my abilities. But I did break it into chunks and do a Riemann sum. Saturn V launched from a 100 km altitude.

For most of it I kept the acceleration vector vertical enough to meet or exceed the net downward pull. Over time it gets more horizontal.

Rune wrote:

...he insists and insists about gravity loss, and he's right about that, it IS a function of time. It could be minimal if your rocket shoots up at 20G's, or it could multiply delta-V several times if you choose to go up slowly and majestically, like the alien ships in sci-fi shows.

You're describing well what I have in mind, perhaps more clearly than my attempts.

Rune wrote:

Thus, total delta-v to orbit, by definition, is at least in part a function of flight time, and therefore T/W. Figure out your flight time, and you have your gravity loss, no need to estimate (only since mass is a variable, so is gravity loss/s, and therefore getting the gravity loss involves integrating a differential equation, no two ways about that).

Wish I could solve it as an integral. But I've been modeling by brute force numeric method (see spreadsheet above).

Rune wrote:

Oh, and T/W of an engine, for the same engine, IS proportional to the fuel's density (a few engines can run several fuels, russian experiments with methane on kerolox engines jump to mind). Or to be more precise, with it's molecular weight, and directly proportional at that. The rest (which you can approximate to a fixed coefficient and not go too wrong about it) is the engine cycle that you use, and you weight efficiency (a function of the chosen material, and those don't really change that much)... with clever designing assumed, of course. That's why H2/LOX has such low T/W no matter what engine cycle you use.

I'm not sure of these. I can see how molecular weight is related to density of a gas, thrust, and ISP. But I believe most of the propellant densitiess Josh lists are for liquids. Just comparing the density of water and kerosene suggests to me this doesn't work for liquids.

Rune wrote:

On the actual topic of the thread, reusable rockets to orbit... I think we should open up a new one to discuss Musk's idea for turning F9 into F9"R"... I suppose you have seen the video and heard him talk about it? It looks... well, it made me grin like a stupid. And the words "like god and Robert Heinlein intended them to" jumped to my mind. Also, why has no one actually tried to do something like that before?

Musk's Grasshopper videos are very exciting. For the first stage, I am wondering if it could land on an island east of the launch site. That way the first stage could pick up some horizontal velocity and still be recovered relatively unscathed.

I am skeptical, though. The video seems to indicate some of the re-entry velocity is shed by aerobraking and some by using propellant as reaction mass. Using propellant to slow down adds to the delta V budget and makes an already difficult mass fraction even more difficult. At this point I'm giving Musk less than even odds of pulling this off.

JoshNH4H wrote:

Vo=7800 m/s
Vu=sqrt(2*2.81e6)=2,370 m/s
Dg=~2/3*300 s*9.8 m/s^2=1,960 m/s*
Da=150 m/s

For Vu you're using difference in potential energy between 0 and 300 km altitude. If you are going 300 km straight up (as in your original scenario), this would be accurate.

But an orbit's energy is the sum of both kinetic and potential energy. And a more practical trajectory would pick up kinetic energy along the way.

Let's say our rocket achieves a circular orbit at 100 km altitude. It doesn't want to stay there as the tenuous atmosphere at that altitude will degrade its orbit over time. How much will it take to reach a more enduring 300 km altitude circular orbit? The potential energy difference between these two orbits is 1.84 MJ/kg. But the difference between their total energies (that is potential and kinetic energy) is exactly half that -- .92 MJ/kg. Rather than achieving an additional 200 km altitude with a vertical burn, it's better to do a horizontal perigee burn followed by another horizontal apogee burn to circularize.

In reality a rocket might achieve a fraction of orbital velocity at 100 km. But even in this case, it still has kinetic energy.

Your method of assessing energy to reach an altitude relies solely on potential energy and ignores kinetic energy.

Last edited by Hop (2011-12-09 16:41:01)

Rune · 2011-12-10 10:16:30

Hop wrote:

I'm not sure of these. I can see how molecular weight is related to density of a gas, thrust, and ISP. But I believe most of the propellant densitiess Josh lists are for liquids. Just comparing the density of water and kerosene suggests to me this doesn't work for liquids.

But Hop, as soon as those liquids hit the combustion chamber, they stop being liquids. Actually, the molecular weight you have to look at is the one of the combustion products. Water vapor+CO2 weighs much more than water vapor alone. You have to average them stoichiometrically (does that word exist in english?), of course, but I'm sure you'll get the idea. What is kerosene's chemical formula, by the way? CxHy, I know, but what is x and y? Lazy to look it up... ^^'

Hop wrote:

Wish I could solve it as an integral. But I've been modeling by brute force numeric method (see spreadsheet above).

Well, that's how computers integrate... adding triangles, it's also called. Still, looking through that spreadsheet would feel more like homework than I am ready to tackle on a saturday afternoon... I'll just assume you got it right.

However, if you care to get a differential equation out of it as a function of time, I might try and solve it to get you an integral to work with, if I can do it. The subjects I have passed imply I should be able to... and it would give Josh a perfectly valid response to his dilemma, both conceptually and practically. Plug in values for your rocket, solve for x and all of that.

Hop wrote:

I am skeptical, though. The video seems to indicate some of the re-entry velocity is shed by aerobraking and some by using propellant as reaction mass. Using propellant to slow down adds to the delta V budget and makes an already difficult mass fraction even more difficult. At this point I'm giving Musk less than even odds of pulling this off.

Well, you have to keep in mind as soon as it stages, the weight of the thing drops like hell. Given the weight fraction of the first stage (I believe it is the best in the business, actually), just a little fuel left over would give it quite a decent mass ratio as a single stage rocket. Let's say I'm cautiously optimistic about it. Landing it is just a matter of choosing you launch site if the horizontal movement is a problem, which you would have to run a lot of numbers to state confidently.

I have more doubts about what would happen to the second stage's mass ratio if it is built to go through full reentry a lot of times with minimal refurbishment, and I thing GW would agree with me here. Also, the nozzle trick and the VERY deep throttling implied seems tricky to pull off without a big weight/complexity penalty... I'd just ditch the nozzle and install additional dracos to land it. But who knows? He has a lot of smart people looking into it, and they seem confident it would work. Just reusing the first stage in that way would be a huge thing.

Rune. This post would be more a lot more helpful if I got into it the effort you guys put into yours...

Hop · 2011-12-10 13:37:23

Rune wrote:

But Hop, as soon as those liquids hit the combustion chamber, they stop being liquids. Actually, the molecular weight you have to look at is the one of the combustion products.

Yes, the combustion product is a hot gas. H20 when you burn hydrogen and oxygen. A mixture of H20 and CO2 when you burn hydrocarbons and oxygen.

With longer hydrocarbon chains you have more CO2 and less H20. Thus better thrust but lower ISP.

What that has to do with the density of liquid propellants, you still haven't shown.

Rune · 2011-12-10 14:27:20

Ok, you got me there, I should have said you have to look at the density of the exhaust products to figure out relative T/W between different engine fuel choices. Well, actually at their stoichiometrically averaged molecular weight, not the density, that's also incorrect. Whatever else I said, disregard as wrong.

Rune. But wait, haven't I already said that? Like in the post you actually quote.

Hop · 2011-12-10 16:46:43

Rune wrote:

Rune. But wait, haven't I already said that? Like in the post you actually quote.

Which quote? There's this:

Rune wrote:

Oh, and T/W of an engine, for the same engine, IS proportional to the fuel's density

And then there's this:

Rune wrote:

Actually, the molecular weight you have to look at is the one of the combustion products.

I've agreed with the latter.

Rune · 2011-12-10 23:52:11

So you like to rub people's faces in their mistakes. I had forgotten about that. The full quote of my first, already acknowledged to be wrong twice, post goes:

Rune wrote:

Oh, and T/W of an engine, for the same engine, IS proportional to the fuel's density (a few engines can run several fuels, russian experiments with methane on kerolox engines jump to mind). Or to be more precise, with it's molecular weight, and directly proportional at that.

My mistake was saying fuel instead of propellant. Yes, that's the only actual mistake, if I get as serious as you. But I also acknowledged it is far more misleading than that. You happy now?

Rune. It's a shame you are actually right on this one, 'cause now I want to argue the opposing point.

Last edited by Rune (2011-12-10 23:53:24)

louis · 2011-12-11 06:39:22

Hop,

I have a question for you:

How much more fuel and propellant would be required for a direct shot, in percentage terms, compared with a "sling shot approach" (is it at "opposition"?).

I'm interested in that question because it seems to me to have great importance for the growth of the colony. Restricting the colony to a once every two years' connection, seems v. inflexible.

I think once the colony is established, it can devote a lot of effort to creating rocket fuel/propellant and will probably be able to build primitive rockets to get the fuel to Mars orbit. From there it can be used for the journey to Earth and possibly fuel can be ferried to Earth Orbit.

It is important whether the additional mass required for a direct shot is 50% or 500% or 5000%

JoshNH4H · 2011-12-11 11:28:52

Wow, lots of posts to respond to. First, a quick question to Hop that would clear things up for me a lot:

Rune says that Gravity Drag is a function of the time it takes to get to orbit. You'll see no argument on this one from me. You say that you have to consider orbital velocity when calculating gravity drag. Is your argument that because the velocity required to be in orbit is the strongest determining factor, you have to take into account the velocity required to get into orbit when calculating gravity drag?

If that's your argument, it could have been much more succinctly stated. Also, it kind of misses the point of what I was doing, which was to come up with a rough, quick, and dirty way of understanding how the different components of delta-V combine to get that needed to get into orbit. I'm talking quick, dirty, and 5-10% error, a tool to understand rather than a scientific analysis of the factors involved. A calculator for the delta-V required to get to orbit sounds extremely useful, though. I don't have time to write a program for it at the moment, given finals, but I have a six week winter break with only a few important things to do, so I'll do it then (okay, well one really important thing to do and then several less important ones).

If your argument is that you have to somehow add orbital velocity to gravity drag, that is absolutely nonsensical.

Rune-

With regard to the proportionality between molecular mass of the products and Isp, do you think you could provide some kind of source (even if it's a textbook or something that we can't access) for that? It' not that I don't believe you or even that it doesn't make sense, but it's the kind of thing where a source would be useful. For the record, the equations would seem to imply that T/W would actually be proportional to the square root of the molecular mass, because Isp is proportional to 1/sqrt(M), and the mass going through the chamber at any time is directly proportional to M, unless there's some factor that I'm missing.

In general, I do find Musk's proposals to be quite exciting. Actually, I didn't really even think about the long-term potential of reusable vehicles to orbit until SpaceX's announcement that they intended to take them seriously.

By the way, Rune, the chemical formula for a saturated hydrocarbon is, in general, CnH(2n+2). For kerosene, n is generally between 6 and 16, but it's a mixture so you can't get an exact molar mass. You can approximate to a value that is very much "close enough" with an empirical formula of CH2.

Also, Rune, I find that when talking with someone who treats constructive scientific discussion as a contest of intellect, the best way to win isn't necessarily to be right, but to be levelheaded and reasonable. That makes the other person look silly regardless of who is technically the most right. Plus, flames rarely get anyone anywhere.

Louis- I don't quite understand the relevance of that question to this thread. We're talking about reusable rockets from Earth's surface to LEO, not LEO to Mars surface. Further, that question would be extremely easy for you to answer yourself by simply Plugging numbers into this calculator. It's easy. It uses almost no bandwith and will not crash your computer. You do not need to do any math at all. It's as simple as putting in "4250" in the dv box, "450" in the isp box, 1 in the M1 box, and clicking "recalc" next to the M0 box. You should not need anyone to do this for you. The value that appears in the box is the ratio of fully fueled mass to empty mass of your rocket to go from LEO to Mars.

You literally do not need to have any skills to use this calculator. You are an intelligent person, I guarantee you can do it and then you would not need us to do the calculations for you.

Rune · 2011-12-11 17:20:31

With regard to the proportionality between molecular mass of the products and Isp, do you think you could provide some kind of source (even if it's a textbook or something that we can't access) for that? It' not that I don't believe you or even that it doesn't make sense, but it's the kind of thing where a source would be useful. For the record, the equations would seem to imply that T/W would actually be proportional to the square root of the molecular mass, because Isp is proportional to 1/sqrt(M), and the mass going through the chamber at any time is directly proportional to M, unless there's some factor that I'm missing.

Well, I'm afraid I can't help you, since I got it out of thin air and a bit of thinking. My reasoning went something like this: you always have the same propellant flow (by volume) through the same engine and nozzle, so mass flow in the exhaust is directly proportional to molecular mass. Therefore, T/W is directly proportional if the speed of the exhaust remains constant. For chemical rockets, that's not really very far off reality, but you could always adjust for the difference in exhaust speed... isp1/isp2 should be the correct factor to adjust with. But of course no source is going to use such a crude method... I guess I could cite anything about conservation of momentum? It's also where you derive the rocket equation from... XD

How much more fuel and propellant would be required for a direct shot, in percentage terms, compared with a "sling shot approach" (is it at "opposition"?).

"Brute force" trajectories would take about as much delta-v as is the difference between the orbital speeds of mars and earth, so about 29.8km/s (for earth) - 24km/s (for mars) = 5.8km/s, without taking into account entering and leaving either planet's orbit (whichever orbit you choose to park in, the delta-v requirements vary).
As to how much it takes to actually go into orbit and leave and such, I will refer you to this handy picture, with the most common delta-v's in cislunar space and mars. I got all the numbers from wikipedia, I'm that lazy.
For comparison, the "standard" minimum energy Hohmann is about 1.5km/s, again without the departure and capture taken into account. You should have enough data with that to plug in the dV in the calculator, and get the mass ratio of pretty much any trajectory between the earth, moon and mars. Just add the dV of all the steps your imaginary ship takes, and then plug the number in the calculator like jumpboy says.

Rune. Informed common sense almost never fails. Emphasis on "almost"... big one.

Last edited by Rune (2011-12-11 17:21:32)

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