Selenian Boondocks

“Program manager John Shannon said Tuesday it costs $200 million a month to keep the fleet flying.”

This is why President Bush and Sean O’Keefe knew that we would have to bring the shuttle program to an end in order to have any hope of going forward with NASA’s use of space. Michael Griffin knew it. President Obama and General Bolden know this too.

Source: Money key to more space shuttle flights

Continuing on this vein, an article today in The Space Review: “Costs of US piloted programs”

“each day spent onboard by an ISS crewmember costs about $7.5 million (compared to $5.5 million for Skylab.)”

From NASA administrator Charlie Bolden:

“I find great comfort in knowing that President Obama has seen fit to put his faith in us to develop a game-changing strategy in our four mission areas, and that he has given us a $6 billion plus up on our FY10 budget as a show of support and trust. I fully believe in the plan that this budget has allowed us to set out for NASA’s road ahead, and unlike many of our detractors, I do believe it will very likely allow us to reach exploration destinations sooner and more efficiently than we would have been able to while we were struggling to develop the Constellation Program.”

I completely agree with this statement.

I predict that regardless of the outcome of SpaceX’s inaugural Falcon 9 launch, nobody is going to change their opinion. If it’s successful, Ares-huggers will suddenly begin to understand the concept that a single successful flight doesn’t prove anything about a vehicle’s overall reliability (while most on the pro-commercial space guys will start sounding like NASA guys post Ares-IX).

If it fails, commercial space people will switch back to “it was only a test” mode while to Ares-huggers, it will prove, prove, prove that all commercial vehicles (including those with existing proven track records) are all death traps. After all, imagine the national security risk of flying our astronauts on private launch vehicles! I mean, if we’re going to turn LEO crew transportation over to the private sector, we might as well all start learning Chinese and reading the little Red Book, cause them Commies are going to come and sap and impurify our precious bodily fluids.

This is my last update to the payload fraction calculation, I promise.

When I was learning how to use mass-estimating relationships (MERs) at Georgia Tech, our focus was on reusable launch vehicles, and most of our MERs came from NASA Langley, where my professor had once worked. When it came to much of the reusability aspects of the spacecraft, the MER tended to depend on the entry or landing mass of the spacecraft rather than the gross mass or the propellant mass, as I have previously defined.

Systems like thermal protection, wings, aerosurfaces, and landing gear all tended to scale with entry or landing mass. So to capture these effects in order to help calculate performance of reusable vehicles, I introduce another non-dimensional term. I use the letter epsilon to describe the entry-mass-sensitive mass items divided by the entry mass. The difference between the entry mass and the dry mass depends mostly on whether or not the payload is intended to return with the spacecraft, so I include a jettison factor (f_jett) whose value is one if all the payload is jettisoned and zero if none of the payload is jettisoned before entry.

I employ the now-familiar substitution for propellant mass:

And simplify:

Which as you can see is the same as the previous expressions if epsilon is zero.

A simple way to understand this expression is to think about it in terms of one. One is the most payload fraction you could have–if all of your spacecraft is payload. But some fraction has to be propellant, according to the rocket equation. So you start out with your final mass fraction (FMF). From that point, which will always be less than one, you subtract lambda times the propellant mass fraction. Then you subtract your phi term, which depends mostly on your engines. Finally you subtract epsilon times your final mass fraction. If you have anything left over, you have a payload fraction. If you were going to jettison your payload before reentry, then the denominator gets a little smaller than one and your payload fraction improves a bit. But it can never improve a payload fraction that is less than zero.

If your payload fraction is less than zero, then you had better go change something to clean things up. You better use a better Isp to improve final mass fraction, or better tankage or propellants to improve lambda, or better engines to improve phi, or better TPS or wings or landing gear to improve epsilon. Because if the numerator of the payload fraction is less than zero, you’ve got no reason to build your rocket.

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…

Now that I’ve gotten the math and derivations out of the way, let’s us the payload fraction expressions in a real-world example.


Let’s say you work for the chief technologist of NASA, and he’s thinking about sending humans to Mars. He’s considering whether or not to invest in a seemingly-promising new technology: nuclear thermal propulsion. He’s intrigued by the higher levels of specific impulse that you can achieve with nuclear thermal propulsion, nearly twice that of chemical, but he knows that it will cost billions to develop and test. He wants to know if the technology improves the payload enough to make it worth developing, so he asks you to do a study. He says:

“Assume that you have a nuclear thermal rocket engine with an Isp of 850 seconds and a chemical engine with an Isp of 460 sec. You have a heavy-lift launch vehicle that will put 80 metric tonnes into LEO. How much more payload will the nuclear thermal rocket get over the chemical rocket?”

You being a diligent engineer point out that you need a bit more data to do the analysis, so your boss tells you that you can make some more assumptions.

The NTR has a vacuum thrust of 15000 lbf and a weight of 5000 lbm with an Isp of 850 sec. It uses hydrogen with a density of 71 kg/m3 at a mixture ratio of zero. The chemical engine is based on an RL10 burning LH2 and LOX at a mixture ratio of 5.5 with an Isp of 460 sec. The RL10 has a vacuum thrust of 22,000 lbf and weighs 370 lbm. The LOX has a density of 1142 kg/m3.

For both vehicles, he tells you that you can assume that the thrust structure weighs 0.3% of the total vacuum thrust, and that the LH2 tank has a factor of 10 kg/m3 and the LOX tank is 14 kg/m3. The ullage in both tanks is 3%. You can “rubberize” the engines so that any particular thrust you need to get the stage thrust-to-weight can be calculated. Otherwise you’ll get weird effects from integer numbers of engines.

Since the delta-V to do a trans-Mars injection (TMI) burn from LEO varies from opportunity to opportunity, he wants you to run a sweep of DVs from 3800 m/s to 4400 m/s, incremented by 200 m/s. He’s also unsure of the initial thrust-to-weight that the injection stage should have in LEO, so he tells you to run a sweep from 0.2 to 1.0, incremented by 0.2. With four values of delta-V, five values of thrust-to-weight, and two different engine technologies for each case, he figures that you’re going to be pretty busy for the next few weeks designing forty different trans-Mars injection stages.

Little does he know that you’re a Selenian Boondocks reader, and that you think this would be a good chance to use the payload fraction derivation to simplify your workload substantially. So you reluctantly agree to take on this “huge” analysis effort, and with your head down trudge out of his office.

Meanwhile, you get to your office and call your wife and tell her not to worry, your beach vacation is still a go, and that you’ll be able to finish the analysis he wants by the afternoon.

Within about ten minutes, you’ve finished your analysis and have the results in front of you:

You are somewhat surprised. Although at higher levels of delta-V the NTR stage has more payload than the chemical stage, it is not nearly the improvement over chemical that you would have first expected by looking at the much higher value of Isp. And the NTR only has more payload fraction than the chemical stage at low values of initial thrust-to-weight. You know that the initial T/W can’t be too low, or else the stage will incur large gravity loss penalties in the form of higher DV. Using two separate perigee burns to do TMI might reduce this somewhat, but that will subject the crew to two extra passes through the Van Allen radiation belts, as well as bring a “hot” nuclear thermal core back within close proximity of the Earth’s atmosphere for the next perigee burn, and that might cause somebody some heartache.

You note that at initial T/Ws of around 0.6 and greater, the chemical stage actually has BETTER payload fraction than the NTR stage. There are two reasons for this–one is hydrogen and the other is engine thrust-to-weight. The hydrogen propellant of the NTR has a really low density, so you get very large tanks and a significantly greater tank penalty than the chemical case. But even worse is the wretched thrust-to-weight ratio of the NTR engine (3.0) versus the chemical engine (~60). With such a low thrust-to-weight, getting the required initial stage thrust-to-weight is very penalizing. So if the initial T/W is not kept low, then the NTR stage can’t beat the chemical stage for payload performance, and there’s really no reason to spend billions to develop it.

But all of this you keep to yourself for just now. Your boss expects you to be cranking the numbers for a few weeks, so you keep the door shut and let him keep thinking that.

(in case you’re curious, here’s the spreadsheet–took me about ten minutes to write)

In the last post, I attempted to calculate a basic expression for the propellant-mass-sensitive term (lambda) and in this one I will attempt to do the same thing for the gross-mass-sensitive term (phi). In so doing, I will hopefully be able to show how a number of key factors in the rocket design affect the payload fraction. We begin by reiterating the definition of the gross-mass-sensitive term:

Then I make the assumption that this term consists only of two things—the engines and the thrust structure. This assumption will have to be modified for different designs. For instance, if you have a vertically-launched rocket that is taking off from landing legs, like a moon lander, then those landing legs are gross-mass-dependent. Or if you have a winged horizontally-launched rocket on Earth, then the wings and landing gear are gross-mass-dependent. So alter this assumption according to your needs.

Then I commit a sin against the SI system of units by switching from a mass ratio to a weight ratio. This is done to get everything in terms of forces rather than masses. You’ll see why I did this in just a second.

Now I make the assumption that all of the engines on the stage are the same kind of engine, and assume that I can multiply the number of engines (n) by the individual weight of the engine (W_engine). I also assume that the thrust structure weight is proportional to the total thrust that the thrust structure will ever feel, which is the total vacuum thrust. This proportionality factor between the weight of the thrust structure and the vacuum thrust I call f_TSW.

Now, in an effort to get the vacuum thrust in terms of the engine weight, I’m going to replace vacuum thrust with the total number of the engines, multiplied by the individual engine weight, multiplied by the vacuum thrust-to-weight ratio of the engine. That should give me a substitute value for the total vacuum thrust.

Now I can group some terms and simplify things a bit.

I need the gross weight of the vehicle. What should that be? Well, let’s assume that the vehicle has to have some initial overall thrust-to-weight ratio. If it’s sitting on the surface of the Earth and it’s meant to launch, then that value had better be greater than one! Actually, it better be a bit more than that or the vehicle’s not going to accelerate. So I’m going to assume that the overall vehicle thrust-to-weight ratio at liftoff is something that we’re going to specify in the design, and that by knowing that value and the initial thrust (not the vacuum thrust) of all the engines, we can calculate the gross weight of the vehicle.

Now something really important happens. The number of engines (n) is sitting in both the numerator and the denominator and cancels out of the expression altogether. I can’t tell you how happy I was to find this result when I first tried this derivation! I had had this hunch that phi would depend on the number of engines, but it turns out that it didn’t. It only depended on the vehicle’s initial thrust-to-weight ratio and the initial and vacuum thrust-to-weight ratios of the engine. And the thrust structure factor of course. But that cancellation means that phi becomes something that can be calculated for each engine type rather than for the number of engines.

Three values of thrust-to-weight and the thrust structure factor give you phi. Isn’t that amazing? If you’re doing a calculation for a rocket stage that operates entirely in space, then the initial thrust-to-weight value for the engine IS the same as the vacuum value, and you only need two thrust-to-weight values.

If you assumed that your thrust structure factor was negligible, which often isn’t such a bad assumption, then the expression would just be:

This actually wouldn’t be a bad place to stop, assuming that you knew the initial and vacuum values of engine thrust-to-weight. But if you didn’t know initial engine thrust-to-weight you can calculate it, or thrust-to-weight at the desired altitude (which would determine the back pressure) by using this variation of the expression. I find this version especially useful when I’m trying to do air-launch calculations, because the ambient pressure is neither vacuum nor sea-level.

When you use the expression this way, you can “hit” the engine for pressure losses in the atmosphere. If you know the vacuum thrust, and the exit area, and assuming you can calculate the ambient pressure by knowing what altitude you are at, you can figure out the initial thrust-to-weight ratio and use the expression effectively.

Now as I mentioned earlier, if you’re designing a vehicle that has other gross-mass-sensitive terms, like landing gear or wings or landing struts or whatever, don’t forget to tack them onto the end of this expression so that their effect makes its way back into the payload fraction calculation.

Like with lambda, here’s some examples to get you started:

The upper group of engines are assumed to start at sea level, and the lower group of engines are assumed to operate entirely in vacuum.

In my last two posts I’ve been talking about calculating payload fraction of a rocket using the mass ratio from the rocket equation and some vehicle parameters that have been sensitive to propellant mass and gross mass. To use these parameters successfully, it would be helpful to have some idea what they should be for different designs. For instance, we all know that hydrocarbon fuels are more dense than hydrogen fuel, so how would that affect that parameter? To try to answer these questions better, I’ve taken the previously-defined lambda term and created a derivation of what its value ought to be in different circumstances.

Let’s assume that the propellant-sensitive mass consists of only two things—the fuel tank and the oxidizer tank. In the expressions I abbreviate fuel tank as FT and oxidizer tank as OT.

When I was at Georgia Tech, my professor gave us some mass-estimating relationships (MERs) for propellant tanks, but they were all based on volume. For instance, they would tell you to estimate the mass of a hydrogen tank as some number of kilograms per cubic meter of tank volume. So to use these mass-estimating relationships, I rewrite the expression in terms of a factor (f) for the fuel and oxidizer tanks and their volumes.

The volume of the fuel and oxidizer tanks will simply be the mass of the fuel or oxidizer they contain divided by the density of the fuel or oxidizer. To account for the role of ullage (extra volume) in the tank, we throw an ullage factor in the denominator, to make the tank have a little more volume than it would otherwise have if it was 100% full of fuel or oxidizer.

With expressions for fuel and oxidizer volume computed, we substitute these expressions back into the overall expression for lambda. The ullage term is collected and moved to the denominator.

Now the expression has fuel mass, oxidizer mass, and overall propellant mass terms in it. But we know that these terms aren’t actually independent. The fuel mass plus the oxidizer mass equals the propellant mass. And the oxidizer mass divided by the fuel mass gives us the mixture ratio (MXR). I use MXR in the expression because I already used MR for the mass ratio, and I want to limit confusion as much as possible.

Thanks to mixture ratio and the propellant summation, we can calculate fuel and oxidizer mass entirely in terms of propellant mass and mixture ratio.

Substituting these definitions for fuel mass and oxidizer mass back into the lambda expression, we get something that looks complicated but is on its way to being simplified.

With propellant mass showing up in the numerator and in the denominator, it cancels out nicely. The mixture ratio terms moves to the denominator, and we get a nice compact expression.

There’s some really nice aspects to this simple expression for lambda. Propellant mass has been completely removed from the equation, which means you don’t need to know how big or small your rocket is to calculate lambda. You need to know the mixture ratio (MXR), which is determined by whatever rocket engine you choose. Choosing the rocket engine also chooses the fuel and oxidizer, which lets you plug in their densities. Assuming you have an idea what ullage would be and what the factors for the fuel tank (f_FT) and oxidizer tank (f_OT) would be, you can calculate lambda pretty quickly.

To help get things started, here’s a table of tank factors based on both real and conceptual tank designs, from Langley Research Center. Remember to make sure that you’re using the same units for tank factor and propellant density (kg/m3 or lbm/ft3).

And here’s some different calculations I did based on a couple of different engines (including a nuclear thermal engine, which has no oxidizer at all and a mixture ratio of zero) showing the effects of propellant selection and mixture ratio on lambda.

As I prepared for this post tonight, I realized that I wasn’t really modifying the rocket equation at all–I have been using the rocket equation and a summation of mass terms to find the payload fraction, which I consider an especially useful value to know.

Furthermore, if you read my previous post, you probably figured out pretty quickly that all of the dry mass in the rocket doesn’t correlate to the propellant mass. That’s a pretty good guess for very large rockets with large delta-V’s, but as you get smaller, the assumption really starts to fall apart.

One of the big masses in a rocket outside of the propellant tanks are the engines themselves, and they really don’t correlate to propellant mass at all. They correlate to the gross mass, because the engines are typically sized to give you some particular value of thrust-to-weight when you light them up. So I went ahead and extended the previous derivation of payload fraction to include this important feature.

So let’s define some terms. Lambda is the relationship between the vehicle dry mass that correlates to propellant, and phi is the relationship between vehicle dry mass that correlates to gross mass.

We plug these definitions into an equation for gross mass, and I simplified things by just going ahead and combining the propellant-sensitive mass into (1+lambda)*propellant_mass. Then I went ahead and replaced propellant mass with the expression from the rocket equation that related propellant mass to gross mass, and now I had the equation entirely in terms of payload mass and gross mass, which is exactly what I was after.

Then it was a matter of doing the algebra to simplify things down until I had an expression for payload fraction like I did before.

Take a look at the final equation. You’ll note a few things. The first one is that if phi is zero, then the expression is just the same as the one I derived before. The second one is how lambda and phi impact the payload fraction differently. You can see that the effect of lambda is reduced somewhat by the mass ratio having one subtracted from it, but that phi has no such reduction.

This is because in the case where you had a vanishing small impulse, MR would be very close to one. And one minus one would be zero and the impact of lambda would be almost eliminated. But because phi is multiplied by one, its effect is still present even if the impulse was very small. This is because any rocket whose engines were sized to deliver a particular thrust-to-weight at ignition would still pay a mass penalty for those engines, even if it just briefly turned them on and then turned them off.

In an upcoming post I will show how to calculate lambda and phi from other vehicle design parameters, like mixture ratio, tank mass per unit volume, propellant densities, thrust-to-weight ratios of engines and vehicles, and so forth.

When I was an undergrad, I spent two summers interning on the X-33 program at the Lockheed Martin Skunk Works in Palmdale, California. It was a fantastic experience and I got to meet with and work with some wonderful people on a very exciting program. Plus I got to live in the Mojave for two whole summers!

But the X-33 program, as we know, failed. Then I went to grad school at Georgia Tech and studied under Dr. John Olds in his Space Systems Design Lab. We worked on highly-reusable hypersonic airbreathing space transportation systems. Mostly we studied them for NASA headquarters. But I couldn’t help but wonder where had X-33 gone so wrong?

After graduating from Georgia Tech, I got a job at NASA Marshall Space Flight Center in March 2000, and I worked for Dr. George Schmidt in the Propulsion Research Center. It was another wonderful place to work, and I was surrounded by people who were studying antimatter, or plasma rockets, or nuclear reactors–all kinds of interesting subjects.

At that time there was a bit of a “fad” making the rounds at NASA HQ, and it went something like this: we needed to figure out how to do a roundtrip Mars mission in a year or less. Someone had decided that a year was all the human body could take, or the public would keep interest it, or something like that.

I protested, pointing out the astrodynamics of the situation would make such a mission nigh under (very nigh unto) impossible. But there was another “fad” making the rounds at NASA HQ that was lulling them into a false sense that it could be done.

They called it “abundant chemical”, and it was based on the notion that if you could wave your arms and imagine that all the chemical propellant you could ever want was somehow waiting for you in low Earth orbit, then you could just build a big honkin’ rocket and go to Mars and come back just as fast as you wanted to. No one was super clear about how all this propellant got to LEO, and the most popular approach seemed to involve huge, poorly-defined guns or sling-a-trons of some sort that would somehow make it happen, but my boss George had a different idea.

He tried to derive a simple variation of the rocket equation that would show the folks pushing this “abundant chemical” idea that it really didn’t matter if they could assume unlimited propellant, that if the delta-V of the mission was too high (and the one-year Mars mission certainly fell in that category) that all the propellant in the world couldn’t do it.

Now, for a quick review, the rocket equation is derived rather quickly by integrating the differential change in velocity (dv) on an object with mass (m) by the expulsion of a differential amount of mass (dm) at an exhaust velocity (ve).

The result is the amount of change in velocity (delta-V) that can be expected from the expulsion of some fraction of mass at the given exhaust velocity. The rocket equation can also be rewritten so as to tell you for a given delta-velocity and exhaust velocity, what the mass ratio of the rocket (MR) will be in that situation.

For instance, if you had a delta-V of 7300 m/s and an exhaust velocity of 4500 m/s, the mass ratio predicted from the rocket equation would be 5.06, in other words, your vehicle would be 5.06 times more massive at the beginning of the delta-V maneuver than at the end.

So let’s imagine a rather simple type of rocket. We’ll say it consists of only three things: propellant, structure, and payload. We’ll assume that the initial mass of the rocket is all three of these together, and that the final mass of the rocket is just the structure and the payload–that all of the propellant was used up in the maneuver. We’ll also assume that we know the delta-V and Isp of the rocket, so that we can calculate the mass ratio.

Now here’s the part where George did something that got my creative juices flowing, many years ago. He proposed that we imagine that the structure is some fraction of the mass of the propellant. He called this fraction “lambda”, which is a pretty common Greek letter people use when they’re talking about some structural fraction in the rocket equation. George did up a spreadsheet showing how for some practical value of lambda, you would be limited on how much delta-V you could deliver, even if you had lots of propellant.

I started playing around with the equations and was curious if a closed-form solution might be possible that would relate the payload mass (what we’re after) to the gross mass of the vehicle in the first place. With lambda defined, you can proceed to go and use it to replace the structural mass in this modified rocket equation.

With a little more algebraic mastication, you can simplify things down to just payload mass and propellant mass.

To go further, you need to be able to define propellant mass some other way, and then substitute that definition into this expression. Fortunately, the original rocket equation (assuming you know mass ratio) can be solved another way to give you propellant mass.

Then this definition can be substituted into the equation for propellant mass, giving you the expression in terms of only payload mass and gross mass, which is what I was after in the first place. Things simplify quite nicely.

And finally there it is. An expression for payload fraction of a rocket, defined as the payload mass divided by the gross mass, with expressions for lambda and mass ratio embedded into the equation.

This equation can be very useful, because if you look at the numerator, you can imagine that there is some value of lambda that makes it zero for any given value of mass ratio. That would be the structural lambda at which point there would be no mass for payload. Finding it is very easy by setting the numerator to zero and solving for lambda.

At last I was beginning to get insight into my original question, which is “why didn’t the X-33 work?” For a single-stage-to-orbit vehicle, burning LOX/LH2 propellant at about 450 sec Isp, you can calculate the mass ratio from the rocket equation. Then you can throw the mass ratio and lambda into the expression I derived to get an idea of what kind of payload fraction you could expect.

In the ten years or so since I first did this work, I’ve taken these derivations much further, and I plan to share with you ever-extended derivations of this sort in upcoming posts.