Payload Fraction Example Proof

Continuing with our story from last time…

The next day, your boss pokes his head in your office and asks:

“How’s those forty trans-Mars injection stages going?”

He notices that you’re checking out scuba-dive sites in the Caribbean for your upcoming vacation with your feet up on the desk, and comes into the room with the blood rising to your face. In your defense, you blurt out that you’ve already done the analysis!

He, somewhat increduously, demands to see the results, so you show him the spreadsheet. He’s less than impressed.

“I thought you were going to design each stage! I need pictures and layouts of these things, with lists of mass and volumes and so forth…you’re showing me a little number. Furthermore, I’ve got the Ohio and Nevada congressional delegations breathing down my neck to send billions of dollars to Stan Borowski to develop an NTR that he promises will get us to Mars faster. You’ve got to show me more detail for this. And how do you even know that your equations are correct?”

Now on the defensive, you offer to try to quickly verify two of the comparison cases for your boss. He looks through your results and decides to pick a comparison point in the middle of the trade space: 4000 m/s delta-V and 0.5 initial thrust-to-weight. Your spreadsheet quickly predicts that the NTR stage will have a payload fraction of 0.3828 and that the chemical stage will have a payload fraction of 0.3955, with a ratio of the two of 1.033, but your boss wants to see proof that your equations are correct.

So how do you go about turning these expressions into masses and volume and graphics?




This time, it took considerably longer than ten minutes, but you showed that your equations match up with reality (at least to the resolution of your initial assumptions). You can see, physically, that the NTR stage is much larger than the chemical stage, even though both carry roughly the same payload. The NTR stage needs almost 6 engines to meet the T/W requirements, while the chemical stage only needs 4. The real difference between the two stages, from a mass perspective, is in the engine weight. You can see that the total engine weight on the chem stage is 1480 lb, while on the NTR stage it is 29,333 lb. This is a staggering difference, due almost entirely to the wretched thrust-to-weight ratio of the NTR engines. This could also lead to another problem. There will be every incentive to try to remove weight from the NTR engines, and with six engines in close proximity, it is almost certain that there will be a lot of neutronic leakage from one engine to another. This means that the engines won’t be able to be controlled individually, but will have to be controlled as a group. It may not be possible to shut one of them down in flight. It also means that they might have to be tested as a group which will drive costs up like crazy.

From the propellant side of the house, the NTR LH2 tank is much larger than the chem tanks, and the mass devoted to tankage on the chem side is 1554 kg, whereas the NTR tank is 4426 kg. Those big NTR LH2 tanks might cause you to hit the volume constraint on your launch vehicle before the mass constraint is reached.

Your boss is happy, for now. He’s got the numbers to show that for 4000 m/s and 0.5 T/W, the NTR stage only has a few more percent payload, and that’s not worth paying the billions to develop it.

But unfortunately for you, now he knows you can get this kind of preliminary analysis work done much faster, so he’s loaded you down with all kinds of new analyses. And that beach vacation is looking further and further away…

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MS, nuclear engineering, University of Tennessee, 2014, Flibe Energy, president, 2011-present, Teledyne Brown Engineering, chief nuclear technologist, 2010-2011, NASA Marshall Space Flight Center, aerospace engineer, 2000-2010, MS, aerospace engineering, Georgia Tech, 1999

About Kirk Sorensen

MS, nuclear engineering, University of Tennessee, 2014, Flibe Energy, president, 2011-present, Teledyne Brown Engineering, chief nuclear technologist, 2010-2011, NASA Marshall Space Flight Center, aerospace engineer, 2000-2010, MS, aerospace engineering, Georgia Tech, 1999
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62 Responses to Payload Fraction Example Proof

  1. Robert Clark says:

    Kirk, I accidentally copied the above comment to this thread. You can delete it here. I already posted it to the intended “Using Payload Fraction Expressions in an Example” thread.

    I’m also interested in what you would get if you chose dense chemical propellants such as kerosene/LOX rather than LH2/LOX. What’s important is that you can get about 3 times the propellant for the same weight of the tanks. This is key because the tankage weight makes up the greatest component of rocket vehicle dry weight, at least for chemical propulsion. This propellant-carried improvement greatly swamps the lowered Isp numbers. So you have a vehicle of the same dimensions at only a small increase in dry mass due to the small increase in engine mass because of the greatly improved engine T/W ratio, but at *markedly* increased payload capacity.

    Bob Clark

  2. Kirk, hmm.. I’m wondering if you drew those graphics by hand or if you have a tool that takes numbers and spits out graphics.. in which case, how about making a web version 🙂

  3. I did it by hand in Powerpoint. It took longer than anything else about the post.

  4. john hare says:

    One of the points I’m taking from these posts is that you can follow the methods to check your own ideas instead of trying to get others to do it for you. Being the innovative person you are, you could set up the spreadsheets as Kirk has laid out to look at dozens of concepts your after you get it loaded.

  5. Yes, use the equations to do your own calculations. That’s why I spent so much time showing you how I got them.

  6. Kirk, you’ve got me thinking about a game where you design space vehicles with absurd accuracy to do a mission to Mars. You can choose to start building a vehicle with existing technology, even building something you cant afford with the expectation that you can wrangle more funds from Congress, but every election cycle there’s a greater than 80% chance that your program will get canceled. Or you can invest in technology but you’ve gotta demonstrate the technologies or all you get is paper studies on unrealistic junk. If you get stuck you can ask for advice from a variety of futurists and pundits, who recommend everything from developing absurdly optimistic plasma propulsion, to returning to the Moon for some reason, to just ignoring the radiation environment of deep space.

  7. When I was a kid, I used to play this game called “Earth Orbit Stations” on my Apple IIc (with 128K of RAM!) and it was so much fun for a future space junkie like me. You built space stations, selling products to earth and gently ratcheting up the price you charged to make enough money, to do the research, to develop the technology, to accomplish the mission that you had picked out at the beginning. A number of years ago I contacted the author of the game and asked him if he might be interested in revising it, but when I mentioned I had no funding the email conversation stopped.

  8. Josh Cryer says:

    I’m confused by “total engine weight.” In the graphic you have lbf, in the post you say lb, are you calculating some sort of phantom weight of the engine given the thrust-to-weight ratio?

  9. Terraformer (a.k.a Tobias Holbrook) says:

    What happens when you use a denser propellent on your NTR, such as Ammonia?

  10. Robert Clark says:

    Kirk, you might want to try the Orbiter Space Flight Simulator.
    I haven’t tried it yet but according to this discussion forum on it you can create space missions following space rocket equations:


    Bob Clark

  11. Josh, total engine weight is what the engines would weigh in a 1-g gravity field. What they weigh on the surface of the Earth.

    Terraformer, use the equations and find out for yourself.

  12. Robert Clark says:

    Fair enough, John.

    The calculation is quite easy to do. Anyone one can do it once you have the specifications for a LH2-fueled vehicle. You can then calculate what would be the payload you could carry if you switched to kerosene-fueled engines.
    Two *very* key points need to be kept in mind, though: first, as Kirk’s professor pointed out, propellant tank mass scales by volume, *NOT* by the mass of the material contained. This means that the same size and *same mass* tanks can hold about 3 times as much kero/LOX as LH2/LOX. This is extremely important because the propellant tanks make up the single biggest component of the dry weight of a rocket, typically 30% to 40%, even more than that of the engines.
    Secondly, dense propellant engines such as kerosene engines typically have thrust/weight ratios twice as good as hydrogen ones. This is key because switching to kerosene means your fuel load and therefore gross mass will be greater. But because of the kerosene engines better T/W ratio, the increase in engine weight will be relatively small.
    Many people get the second of these points. It’s the reason why first stages generally use kerosene or other dense propellant for example. However, the first one most people are not as familiar with. But it’s the more important of the two because the increase in propellant being carried *far* exceeds the increase needed to overcome the lowered Isp of the dense propellants.
    As I said this is an easy calculation to do. But many people simply won’t do it. They have been so conditioned to think that Isp is the most important thing that the assumption is hydrogen must be used for an SSTO. It probably doesn’t help matters the fact that the gross mass becomes about 3 times as great with the dense propellants. Gross mass has been frequently used as the measure of the cost of a launch vehicle. But this is actually a poor measure to use. The reason is propellant cost is a trivial component of the launch cost to orbit. More important is the dry mass and complexity of the launch vehicle for the payload that can be orbited. Then what’s important is switching to a dense propellant allows *multiple* times greater payload at the same sized and similarly massed vehicle.

    To show how simple this type of calculation is I’ll show it can be done just from the information Kirk provided in the table, you don’t need the full spreadsheet.
    Use the info for kerosene/LOX density and the various specifications for kero/LOX engines available on this page:


    On that Astronautix page are listed some kero/LOX engines. I’ll choose the NK-43 for its high Isp and good thrust/weight ratio:


    There are some other engines you might want to try it on that have higher Isp’s though their T/W ratio is not as good.
    Now for the NK-43 the oxidizer to fuel ratio is 2.80. In Kirk’s table the total volume of the propellant tanks is 141 m^3. From the densities of kerosene and LOX on the Astronautix page, you can calculate this volume can hold a total of 144,940 kg of kerosene/LOX, with 38,140 kg in kerosene and 106,800 kg in LOX. Note that the mass of the tanks stays the same even though this is nearly 3 times the mass of propellant because propellant tank mass scales by volume, NOT by the mass of the propellant contained.
    The initial specifications were for system thrust/weight of .5. The NK-43 has a vacuum thrust of 394,539 lbf. so this will quite likely suffice, but let’s calculate the gross weight, keeping the same payload. The fuel load is 97,911 kg higher, and the engine weight is 723 kg higher. The thrust structure weight scales by engine thrust so it is now at 538 kg, 418 kg higher. So the gross weight is increased to 179,052 kg, 393,914 lbs. So the system T/W is even slightly above 1.
    To calculate the delta-V we need also the dry weight. From the gross weight and propellant and payload weight in Kirk’s table we can calculate the original dry weight as 2,347 kg. Then the new vehicle dry weight coming from the increased engine weight and thrust structure weight is now 3,488 kg.
    Let’s calculate the delta-V now. Keeping the same payload mass means the mass at the end of the burn will be 3,488 kg + 30,624 kg = 34,112 kg. Then the delta-V is : 346*9.8*ln(179,052/34,112) = 5,622 m/s, well above the required 4,000 m/s. This means we can increase the payload. I’ll take it as 60,624 kg, nearly doubling it. Then the delta-V becomes 4,007.8 m/s.
    So the switch to dense propellants allowed us to nearly double our payload.

    Bob Clark

  13. Bob, don’t co-op the TMI stage example into an SSTO example. Use the original expressions and create an SSTO example. It’s not that hard.

  14. Jonathan Goff Jonathan Goff says:

    We’re still talking past each other.

    First, I know that tank mass scales mostly with volume. I’ve said this three or four times now. I know that for the same mass of propellants, the LOX/LH2 tanks will be around 2-3x as big, or for the same volume of tanks, LOX/Kero will hold 2-3x the propellant load. It should be pretty clear to you by this point that while I accept this point, I don’t think it makes your overall argument.

    Second, a minor quibble. If you look at the tankage factors Kirk provided, LH2 tankage actually weighs a bit less *per unit volume* than LOX tankage–about 71%. It has a lot lower density, and thus takes up more volume, but historically LH2 tanks are lighter per unit volume than LOX. So, even though the bulk density for LOX/LH2 is about 3x as bad as LOX/Kero, but the overall lambda you get is only 2.4x, *based on existing data from flight tanks*.

    Third, I agree that cost scales a lot more with dry mass than GLOW for ground-launched stages, and thus while LOX/LH2 definitely has a performance advantage, it’s price advantage might be a lot less if a LOX/Kero vehicle can compete. For the question of an SSTO, if tank and engine mass was all that mattered, LOX/LH2 would still win on pounds of payload per pound of GLOW, but would lose slightly on pounds of payload per pound of dry mass…but that point is academic if with LOX/Kero once you factor in reusability features it wipes out the payload entire.

    Fourth, for an in-space stage, the argument you seem to be making once again is based on tank size. Yes, if you made an orbital stage that used LOX/Kero that was the same *volume* as a LOX/LH2 stage, it could carry twice the payload. It also requires 3x the propellant mass. For a terrestrial stage you might have a point. On the ground, propellant is actually cheap, and its the dry mass that matters, so yes for a ground launched stage, having a smaller GLOW per unit payload isn’t anywhere near as important. But for a space stage, until the cost in $/lb to LEO comes down substantially, IMLEO matters a lot. Take the Centaur stage. It costs roughly $30M for the hardware according to my numbers. But in order to launch a full propellant load for a Centaur, you’re talking about ~$225M at current launch prices. Getting double the payload for 3x the IMLEO is not a good deal for in-space stages. Even if the cost per lb were to drop precipitously, reusing the in-space stage hardware keeps you in a situation where the cost of the propellant in LEO still dominates the transportation cost for rocket-based vehicles.


  15. Robert Clark says:

    Kirk, my preliminary remarks were speaking in general including the SSTO.
    However, the actual calculation was for your Mars injection vehicle with required 4,000 m/s delta-V and a .5 initial thrust/weight ratio. I wanted to show you can get a greater payload by using dense propellants.
    To create an SSTO I would want to use mass estimating relations that include more details such as for example the rocket’s fairings that would be needed when flying through the atmosphere, and also the avionics and wiring weights, which are not trivial for a large rocket.
    Therefore for that purpose I would prefer adapting that online lecture by Dr. Akin of the University of Maryland:

    Mass Estimating Relations.
    ENAE 483/788D – Principles of Space Systems Design.

    Here he describes a SSTO, though expendable, hydrogen-fueled rocket. Then with the included required system weights beyond the engines and propellant tanks, you can see you can get *4 to 5 times* greater payload by switching to dense propellants by using the fact you can hold 3 times the propellant with the tank weight staying the same and get twice as good T/W ratio with only a small increase in engine weight.

    Bob Clark

  16. Bob, I’m going to post a newer derivation of the payload fraction expression that accounts for dry-mass-sensitive terms soon. When I ran an SSME-based LH2/LOX SSTO against a NK33-based RP1/LOX SSTO and the LH2 vehicle had three times better payload fraction.

    SSTO sucks in any case.

  17. Bob,
    You keep missing my point that no, you can’t get 3x as much LOX/RP1 in the same weight of tanks. You only get 2.4x. And taking the absolute highest T/W ratio LOX/RP-1 engine (most LOX RP-1 engines are a lot closer in T/W ratio to LOX/LH2 engines than the NK-33/43) and comparing it with the typical T/W ratio for a LOX/LH2 engine seems a little disingenuous. Also, did you actually run the numbers using those iterative factors that Akin came up with in those papers, or are you still just handwaiving?


  18. Robert Clark says:

    Kirk, I would like to see that derivation you ran that gave a higher payload in the LH2 case compared to the kerosene case. Did you include the fact that you can get approx. 3 times the fuel load in the same size, and same mass, tanks with kerosene? Did you include the fact the kerosene engines have approx. twice the T/W of the hydrogen ones?
    I was having an email discussion with a rocket engineer who had done trade studies on SSTO vehicles. He said his trade studies showed LH2 offered better payload to orbit than kerosene. But it turned out his studies did not include either of those two factors! When he included them, he got the same result I did, that the kerosene case offered better than *5 times greater* payload to orbit than the hydrogen case.
    In that expendable hydrogen-fueled SSTO vehicle of Dr. Akin in his lecture, the payload is 5,000 kg, and the vehicle variously weighs 10,000 kg – 14,000 kg in dry weight. This is common in orbital vehicles where the payload might be 1/2 to 1/3 the total dry mass of the vehicle.
    But switching to kerosene the vehicle dry mass is only slightly increased, by greater engine weight, but the payload can increase *4 to 5 times*. In this case the payload can greatly exceed the vehicle dry weight! No other launch vehicle has ever had this quality of the payload exceeding the total vehicle dry weight. This is an important parameter because the dry weight is a key indicator of the cost of the launch system. And being only a single stage using simple kerosene rather than hydrogen makes it low cost on the complexity scale as well.
    This is no mere theoretical exercise. In a following post I’ll show that the Falcon 1 first stage mass ratio is so high that switching to a more high performance (Russian) engine than the Merlin turns the Falcon 1 first stage into an expendable SSTO vehicle.

    Bob Clark

  19. You’ll see the results soon. I account for engine thrust-to-weight and increased propellant density. The LH2 vehicle is much lighter in gross mass about the same in dry mass.

  20. Robert Clark says:


    That article by Dr. John C. Whitehead of Lawrence Livermore National Lab is a real eye opener:

    Single Stage To Orbit Mass Budgets Derived From Propellant Density and Specific Impulse.
    John C. Whitehead
    32nd AIAA/ASME/SAE/ASEE Joint Propulsion Conference.
    Lake Buena Vista, FL July 1-3, 1996

    He discusses the historical tank weight trends for hydrogen, kerosene, and LOX. By the historical data the propellant mass to tank mass ratio for kero/LOX is about 100 to 1. For LH2/LOX it’s about 35 to 1. So you can hold about 2.86 times as much kero/LOX as LH2/LOx by this data. You would still get significantly better payload in the Mars injection vehicle than by using hydrogen.
    The online lecture I mentioned by Dr. Dave Akin of the University of Maryland also does some historical tank sizing trends:

    Mass Estimating Relations.
    ENAE 483/788D – Principles of Space Systems Design.

    On page 7 he gives the propellant mass to tank mass ratios:

    MLH2 Tank = 0.128*MLH2
    MLOX Tank = 0.0107*MLOX
    MRP1 Tank = 0.0148*MRP1

    by his data. So using the masses of kerosene and LOX I found that you could hold in the same volume tanks as Kirk’s, the total mass of the kero/LOX tanks would be by Dr. Akin’s trend values: 0.0107*106,800kg + 0.0148*38,140kg = 1,707 kg. So the same size kero/LOX tanks would weigh about 153 kg more than the Kirk’s LH2/LOX tanks. So just subtract 153 kg from the payload you got by switching to kero/LOX. You still get about twice as much payload with the dense propellant.

    The Russians are very adept at getting good performance from kero/LOX engines. If you’re going to use a engine to get to Mars you ought to use the best available. The 460 sec Isp Kirk assumed for his hydrogen engine is no slouch either for hydrogen engines.

    I did the calculation for the SSTO using the data from Dr. Akin’s online lecture to get 4 to 5 times greater payload with kero/LOX than the hydrogen used in Akin’s lecture. It’s quite easy to do using the numbers he provides. Give it a try.

    Bob Clark

  21. Martijn Meijering says:

    Note that ratios of payload fractions are not necessarily a good predictor of cost. With expendable spacecraft it’s construction costs that dominate. With reusable in-space vehicles ratios of *propellant fractions* might be a better indicator.

  22. Martijn Meijering says:

    Or perhaps IMLEO per *ticket*.

  23. Tom Mattison says:


    Trans-Mars-injection doesn’t have signficant gravity losses in the traditional launch-from-ground sense, because since you are in orbit already you can apply all the thrust perpedicular to gravity. So there is little reason to use a high thrust to weight ratio. At 1 m/s^2 acceleration (about 0.1 gee), it takes 4000 seconds or 66.66 minutes or 2/3 of the LEO period to get your 4000 m/s of delta-v. You could go a few times lower in acceleration before you really are in the ion-engine regime of spiraling out. So the right optimization (for either chemical or nuclear) is much lower thrust than the ground launch case. Why pay any penalty for engine mass that you don’t need to avoid gravity losses?

    Chemical engines have such terrific thrust to weight ratios that you don’t notice the penalty for being “over-engined.” But nuclear-thermal has awful thrust to weight, so you really want to go low-thrust.

    Another optimization for TMI or other space-to-space scenarios but not launch from ground is staging via multiple drop tanks. Use a single engine, but multiple small fuel tanks, and drop them as you empty them. You want the nuke at the end of a long stick anyway (SS Discovery from 2001) to minimize the shielding mass, and you put the drop tanks along the stick, with springs to push them aside when empty.

    The drop tanks could just be bubbles designed for 0.1 gee acceleration, that you fill in orbit from a stronger tank that can survive launch from the ground. And anyone with any sense will have filled his bubble drop tanks from the orbital propellent depot anyway!

  24. Pete says:

    “you can’t get 3x as much LOX/RP1 in the same weight of tanks. You only get 2.4x. And taking the absolute highest T/W ratio LOX/RP-1 engine (most LOX RP-1 engines are a lot closer in T/W ratio to LOX/LH2 engines than the NK-33/43) and comparing it with the typical T/W ratio for a LOX/LH2 engine seems a little disingenuous.”

    LH2 is typically used on upper stages where mass matters far more and the development and operating costs are typically much higher. I am suspicious that this might be biasing LH2 T/W and LH2 tank weight numbers significantly.

    For example, if the same effort that was placed into developing the SSME had been placed into developing an equivalent kerosene rocket engine, what would its T/W and ISP be? I would ask a similar question with regard to LH2 tanks verse kerosene tanks.

  25. Trans-Mars-injection doesn’t have signficant gravity losses in the traditional launch-from-ground sense, because since you are in orbit already you can apply all the thrust perpedicular to gravity.

    No, that’s not the case at all. The thrust is applied at perigee, tangential to the gravity vector.

  26. Tom Mattison says:

    Hi Kirk

    I agree that perigee is the right place, but it’s smarter to thrust along the existing velocity (always tangent to the orbit, perpendicular to gravity at perigee, and nearly perpendicular to gravity for a nearly circular orbit). If the delta_v is parallel to v_initial, the sum is v_initial + delta_v, with final v^2 = v_initial^2 + delta_v^2 + 2*v_initial*delta_v. If the delta_v is perpendicular to v_initial, the final v^2 = v_initial^2 + delta_v^2. The final kinetic energy is 0.5 * m * v^2, and that’s what you want to maximize. I am neglecting the change in gravitational potential energy during the boost, but that’s not a huge effect except when the acceleration is super-low, like an ion drive.

    So I still claim that gravity losses are nearly a non-issue for boosting from LEO to GEO or the Moon or Mars or any other planet, unless you are down to a few percent of a gee.

  27. Robert Clark says:

    Kirk, thanks for emailing me that spreadsheet showing the results of a hydrogen, kerosene, and NTR comparison for a SSTO. I assume you won’t mind me discussing the results here.
    First, for the hydrogen-fueled SSTO you got a 3% payload to gross mass ratio. This is comparable to what you can get with multi-stage rockets, so isn’t bad at all.
    You get a lower percentage of the gross mass for the kerosene case because of the higher fuel load with the dense propellant, but the ratio of the payload mass to the vehicle dry mass is about the same, about 1 to 3. This I argue is a more important parameter because propellant costs are a trivial component to the costs of orbit. The kerosene-fueled case would also be cheaper since kerosene systems are cheaper than hydrogen ones.
    However, I’ll argue in a following post you can even do better with the kerosene case than your numbers imply.

    Bob Clark

  28. Bob Steinke says:

    As other people have pointed out, an NTR stage would optimize payload fraction at a lower thrust-to-weight than a chemical stage, which requires a higher delta-v due to the longer burn. So instead of comparing NTR to chemical at the same thrust to weight and delta-v, the best comparison would be an optimized NTR stage to an optimized chemical stage where the two have different thrust to weight and delta-v. This would require more complex trajectory analysis, which is probably what your boss was expecting to take several weeks.

  29. Bob, I don’t think that’s a fair comparison at all. For less T/W the chemical stage will improve as well. You can’t drive the T/W too low or the gravity losses mount dramatically. Earlier comments about the T/W being able to go really low are simply in error.

  30. Tom Mattison says:

    I googled “ion engine gravity loss” and the first thing that came up was

    Rocket and Spacecraft Propulsion: Principles, Practice, and New Developments, by Martin Turner.

    This confirms my assertion original assertion:

    “Chemical rockets used in orbit transfers can be very accurately assumed to have no gravity loss, because the burn is so short and is at right angles to the gravitational field.”

    But it goes on to quantify my caveats about spiraling out (with a graph no less on page 212). For accelerations 10 gee, there is negligible gravity loss (much less than the 10% you would get for a vertical launch from the ground). For 1 gee (where there would be 100% loss for vertical launch”) it looks like about 7% delta-v penalty for orbital transfer. For 0.2 gee, the lowest case considered in the blog post (and which wouldn’t get you off the ground at all) it’s a 20% delta-v penalty for orbital transfer. That’s a bit worse than my intuition, but it’s not a total show stopper. At 0.1 gee it’s a 30% penalty (factor of 1.3), at 0.01 gee, it’s a factor of 1.8, and it asymptotes at a factor of 2.3 for very low acceleration.

    But I think it’s still true that if 1 NTR engine gave you 0.2 gee for a factor of 1.2 penalty in delta-v, you would not choose to go to 5 or more engines to get up to 1 gee to get the penalty down to 1.07. But I also agree that going down to a peewee NTR engine for 0.01 gee isn’t going to be sensible either.

    I also agree with what is implicit in all of this, which is nuclear-thermal’s enticing exhaust velocity or Isp comes along with an awful engine mass penalty that makes it hard to come out ahead. And since they are really hard to test on the ground with current (sane) radiation dispersal standards, I don’t see NTR winning in the forseeable future.

  31. Robert Clark says:

    Kirk, for the calculation of the payload of a kerosene-fueled SSTO derived from the Excel sheet you sent, I again refer to the paper by Dr. John C. Whitehead of Lawrence Livermore National Labs:

    Single Stage To Orbit Mass Budgets Derived From Propellant Density and Specific Impulse.
    John C. Whitehead
    32nd AIAA/ASME/SAE/ASEE Joint Propulsion Conference.
    Lake Buena Vista, FL July 1-3, 1996

    This is a *great* paper. It’s full of important and useful information on every page. For instance I was initially puzzled in your spreadsheet why the propellant densities calculated from your given propellant mass and tank volume values were consistently low. I then realized you were assuming there was 3% less mass in your tanks than expected because of ullage. I hope this section from the spreadsheet formats correctly:

    engine SSME NK-33 RD-180 NTR

    desired payload kg 10000 10000 10000 10000
    gross mass kg 314015 619435 1027163 270072
    prop mass kg 274038 579474 960898 174806
    fuel mass kg 39148 160965 258306 174806
    oxidizer mass kg 234890 418509 702592 0
    fuel tank volume m3 568.4 204.9 328.8 2538.2
    ox tank volume m3 212.0 377.8 634.3
    fuel tank mass kg 5684 2868 4603 25382
    oxidizer tank mass kg 2757 4911 8245

    Ullage is the space that has to be left in propellant tanks to allow for evaporation. However, the Whitehead paper shows that for actual rockets instead of the lost ullage mass being a few parts per hundred it’s only a few parts per *thousand*. He discusses this on page 5, which you can see here:

    So I won’t include this small amount in the calculation.
    However, the most important reason why you got a low value for the payload that could be carried in the kerosene case is because of the low value of the propellant to tank mass ratio you used for the *kerosene* case, while you did use about the right value for the hydrogen case.
    For the hydrogen case, you can calculate from your spreadsheet it’s: 274,038kg/(5,684kg + 2,757kg) = 32.5 to 1. Divide this by .97 to put back the over amount you subtracted for ullage to get a ratio of 33.5 to 1. This is pretty close to the value of 36 to 1 that Whitehead gives on p.4 of his paper. See page 4 of the paper here:

    However, for the kerosene fueled NK-33 engine, which I’ll focus on, you can calculate the propellant mass to tank mass ratio as: 579,474kg/(2,868kg + 4,911kg) = 74.5 to 1, which by correcting the ullage number gives 76.8 to 1. This is well off the value Whitehead gives of 103 to 1.
    Such values for these ratios are based on historical data. But you want to use one of the better values, relying on more recent rockets, because for a SSTO you expect the technology will at least be as good as for some of these more recent rockets. The value of over 100 that Whitehead gives for example is based on the fact that with the modern alloys such as the aluminum-lithium tanks on the shuttle ET, the ratio for LOX alone is the rather high 118 to 1, due to aluminum-lithium alloys high strength at lightweight. The kerosene alone ratio will be similarly improved over that of the ratios for the older rockets. So 103 to 1 is more realistic.
    Now, in your analysis you were trying to see is what propellant sizes and dry mass sizes you would need for a *fixed payload capacity* for the different cases. But that’s not what I want to do. What I want to show is that for a rocket of similar physical dimensions as the hydrogen case and only relatively small increase in dry mass, the payload will be *markedly* higher in the kerosene case. So I’m not going to assume the payload has to be the same 10,000 kg.
    So first let’s calculate how much kero/LOX you can hold in the same volume tanks you had in the hydrogen case of 780.4 m^3. Using the mixture ratio for the NK-33, and the densities of kerosene and LOX you gave, you can calculate you can hold 800,120 kg of kero/LOX propellant in the same size tanks. Note that is far above the propellant amount you gave for the NK-33, where you made the tanks be smaller, probably to force the only 10,000 kg payload.
    The tank mass for this heavier amount of propellant will be 7,768 kg, using the 103 to 1 propellant mass to tank mass ratio. Note this is actually slightly less than the tank mass you gave using the much smaller amount of propellant. This is because of the much better propellant to tank mass ratio that Whitehead gives.
    Now let’s calculate the engine mass. The engine mass will have to be increased to account for the increased propellant load. The weight of propellant in pounds is 1,760,264 lbs. To be able lift this as well as just the original entry mass, dry mass plus payload, of 39,961 kg, 87,914 lbs., at a 1.25 initial thrust/weight ratio you’ll need 7 of the NK-33’s.
    What would actually be required is somewhat less than this number but I’ll take it as seven because the engine weight and thrust structure weight is increased and I expect to also be able to increase the payload. The extra engine weight brings the dry weight to 32,610 kg. The seven NK-33’s produce 2,380,000 lbs, thrust which at a thrust structure factor of .003, requires a thrust structure weight of 3,245 kg. This brings the vehicle dry weight to 33,259 kg.
    Now let’s calculate the delta-V we can get using a payload of 10,000 kg and the NK-33 vacuum Isp of 331 s:
    331*9.8ln(1 + 800,120/43,259) = 9,635 m/s.
    Now as you indicate in your spreadsheet the delta-V required for orbit using a dense propellant such as kerosene is less than that required for hydrogen. This is due to the fact the dense propellant vehicle can get to altitude more quickly. For the kerosene case it’s 8,900 m/s compared to 9,200 m/s for the hydrogen. So we see our kerosene case now has an overabundance of delta-V. Then we’ll increase our payload to bring the entry mass to 55,000 kg. The delta-V is then:
    331*9.8ln(1 + 800120/55000) = 8,901 m/s

    Then our payload has increased to 21,741 kg, more than twice the value for the hydrogen case at a relatively small increase in dry weight.

    Bob Clark

  32. I didn’t indicate less DV to orbit for the kerosene case for any other reason than to show you that it still lost out to the hydrogen case.

    This discussion of SSTO is getting increasingly tiresome I must admit. I think SSTO is a huge waste of time.

    Also, you’ll forgive me if I don’t assume that being from the Lawrence Livermore Lab means that you’re an expert in rocket design.

    Bob, why are you so determined to make the kerosene case better than the hydrogen case? I sent you the spreadsheet so you could play with it and see that it’s not so.

  33. Robert Clark says:

    OK. For anyone interested I’ll be posting the calculation for how the Falcon 1 first stage can become an expendable SSTO on sci.astro.

    Bob Clark

  34. Brad says:

    #30 Tom
    That is very interesting information.

    When it comes to the issue of orbital transfer and burn duration, I thought it would be interesting to look at some real world examples to see how they compare. The Atlas V Centaur upper stage, in the single engined version, has a burn time exceeding 15 minutes. The chemical Mars stage exemplar has a burn time of only 9 minutes. That suggests the chemical Mars stage is over engined.

    Of course by over engining the chemical stage, and then having the NTR stage match the same T/W as the chemical stage, the NTR stage pays an unnecessary penalty which masks NTR’s true payload advantage over chemical.

    Comment #28 is right on the money
    Bob Steinke, “…an NTR stage would optimize payload fraction at a lower thrust-to-weight than a chemical stage, which requires a higher delta-v due to the longer burn. So instead of comparing NTR to chemical at the same thrust to weight and delta-v, the best comparison would be an optimized NTR stage to an optimized chemical stage where the two have different thrust to weight and delta-v.”

    Cutting 8,800 kg of NTR engines from the Mars NTR stage exemplar leaving two engines remaining, would increase the ‘burn’ time to 30 minutes and shift most of the 8,800 kg directly to payload. I note that during the age of nuclear engine development burn times closer to 1 hour were assumed and NTR engines were developed to meet that standard.

  35. I can see that NTR lovers will keep making sorry excuses for that engine until I run an integrated injection trajectory and show them how much it really does suck.

  36. Robert Clark says:

    Kirk, that paper of Whitehead has importance that goes beyond just SSTO vehicles. For instance, in your spreadsheet your numbers indicate the propellant mass to tank mass ratio for LOX alone is 85 to 1. However, Whitehead shows with modern alloys this is actually 118 to 1. So this has importance for your chemical propulsion Mars vehicle as well even using hydrogen as the fuel.
    But why should you want to limit yourself there? You did these extensive calculations taking into account several different variables, why not take into account different propellant combinations as well *using the most up to date data*?
    The fact is kerosene is not even the best dense fuel to use for a SSTO. Several other dense fuel combinations are explored in this article:

    Alternate Propellants for SSTO Launchers
    Dr. Bruce Dunn
    Adapted from a Presentation at:
    Space Access 96
    Phoenix Arizona
    April 25 – 27, 1996

    Several dense fuels listed give better payload to orbit than kerosene. And it is notable all the ones in the list including kerosene give better payload than hydrogen.
    But since some of them give *markedly* better payload to orbit than hydrogen with a SSTO, it is quite likely they will also give better payload with your single stage Mars vehicle as well.

    Bob Clark

  37. Bob, those tankage numbers in the spreadsheet are meant to be indicative, not fixed values. I went to all the time and effort to do the derivations and show the equations so that you can plug in whatever numbers you desire, so long as you can justify them.

    Also Bob, you’ve posted that link about a dozen times. I think we’re getting it now. You can stop posting it. Whoever’s going to read it is going to read it.

  38. Kirk,
    Re: NTR. Some of the “DUMBO” concepts and their more recent derivatives looked like they would’ve been capable of much higher power densities (and thus T/W ratios) than the more traditional NERVA-type engines. Still fairly theoretical, so numbers should be taken with a grain of salt, but they were looking at T/W ratios competitive with LOX/LH2 engines (in the 50-70 range).

    Running your spreadsheet though, it still doesn’t look like a game-changer (about a 30-40% better payload fraction across the board), until you start going to higher delta V’s. Like if you also had it do the Mars capture burn or something instead of doing aerocapture.

    I’m not 100% with you yet that NTRs make absolutely no sense, but they do definitely seem to be a hard sell compared to other technologies.


  39. People need to show specifics of these supposed high-thrust-to-weight NTR concepts. I may think Borowski and his crew are wasting their time, but at least they show their work. I have a hard time believing that an NTR can ever have a good thrust-to-weight ratio, because heating of the working fluid is inherently a surface-area phenomenon involving heat transfer rather than a volumetric deposition of heat into a combusting fluid mixture. Then you get into the issues of shielding mass and it keeps getting worse and worse. Shielding alone would probably condemn an NTR to a low thrust-to-weight ratio.

  40. john hare says:

    It might be possible to get decent thrust to weight with decent Isp. The tether rocket and solar thermal concepts I posted might eventually do better than projected NTRs. . I wonder how many concepts by real researchers are floating around that might get serious improvements.

  41. John, please suggest how that might be done. Greater surface area or less shielding? What’s going to give?

  42. Martijn Meijering says:

    Ha, you finally said the magic word: tethers. In the short run LOX/LH2 will remain king for anything that departs directly from LEO and in the long run tethers are very promising, augmented by LOX/LH2 for crew.

  43. Bob Smith says:

    Tom Mattison, that is indeed a great find. Am I correct to interpret those numbers as spiraling out from LEO? If so, it would be informative to contrast with numbers that start from GTO or an earth-escape trajectory or similar. Starting from higher earth orbits the gravity penalty is much lower. It may well make sense to use high thrust/weight (i.e. chemical) to get to a GTO-like or earth escape orbit and only then start spiraling from there to Mars using low thrust/weight thermal or electric propulsion. Indeed, planetary probes like the Dawn mission work like this today.

  44. Tom Mattison says:

    Hi all

    Let’s 1. Give Kirk some applause for educating us all about the value of parametric design and 2. Giving us all an opportunity to do some learning of our own, then 3. Put this thread to rest. (Not that we shouldn’t stop thinking and learning, just maybe do it on a different thread).

    Bob Smith 43, yeah, I thought that was a really neat graph too. I intend to dig into the references to see what it means, but I haven’t done that yet. It’s probably (rocket-eqn delta-v for the mass ratio you really need) over (rocket-eqn delta-v for the mass ratio with no gravity loss) for LEO to Earth-escape or to Earth-escape+Hohmann-to-someplace. At least that’s what I would publish.

    My guess is that even the chemical TMI example would optimize to low thrust from a cost and reliability viewpoint, even if the great T/W of chemical engines means that you could slightly increase the payload for fixed mass in LEO by adding engines to reduce the finite gravity loss. Pretty clearly for the nuclear case the engine mass penalty pushes you to low thrust until the spiral-out gravity losses are painful, perhaps at somewhat lower T/W than in Kirk’s table. And since nuclear engines are vaporware anyway, it would take a huge potential performance advantage to make it worth the development effort, and that just doesn’t seem to be there.

    Maybe if you needed to get a robotic payload to Pluto pronto, so you wanted a high delta-v, and didn’t care about radiation, nuclear would win. But it’s not going to get any of _us_ to Mars!

  45. Bob Smith says:

    I don’t understand why we keep seeing these analyses starting from LEO. Nobody going to high earth orbit or deep space actually does that, because there’s a big delta-v penalty to circularize and then de-circularize your orbit. If you’re going to GEO, you stage at GTO, if you’re going to deep space, you never go into a circular orbit first. And as we see here, if we are using a low thrust/weight orbital transfer stage we also incur a big gravity loss penalty by first going to LEO and starting it from there.

  46. You most certainly start from LEO, or some LEO-esque orbit. That’s where your launch vehicle puts you before the trans-Mars injection burn. Every communications satellite has a brief LEO orbit before it does an injection burn to GTO. It might be only 15 minutes long, but it’s a LEO.

  47. Bob Smith says:

    Comsats launch straight to GTO. It is wrong to say they go via LEO because they don’t circularize there. Analyses of missions to higher earth orbits or deep space that have the mission start from LEO have large errors for this reason. With low thrust/weight ratio the error is even larger due to the added gravity losses of spiraling out from LEO which are far lower if we stage to the low thrust/weight stage in a higher earth orbit or at earth escape.

  48. No, comsats do not launch straight to GTO. If they did, then their lines-of-apsides would not be equatorial and they would not be able to execute a combined plane-change/circularization burn when they reached apogee to insert into GEO. They launch to a temporary LEO, coast to the equator (descending node), and then conduct the injection to GTO. This is how they manage to have an equatorial line-of-apsides.

  49. Bob Smith says:

    Most comsats launch from the equator and that maneuver is trivial. Even if we are doing something very suboptimal like launching to GEO from 28 degrees that maneuver is energetically not LEO because it is not circularized. So the analyses that start from LEO are quite wrong.

    And of course this doesn’t address the large error in a gravity loss estimate that assumes starting the orbital transfer stage from LEO instead of where interplanetary spacecraft actually start low thrust/weight ratio stages from, higher earth orbit or earth escape trajectories.

  50. No they don’t Bob. You seem awfully confident in a bunch of stuff that just ain’t so. Most comsats launch out of French Guiana or Baikonur Cosmodrome, and from Baikonur the plane change maneuver is quite a bit deal. You keep getting facts wrong, and show your work if you’re going to assert that I’ve made a big error. I show my work.

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