Posts Tagged ‘RLV’

I always had a different idea compared to the one Jon and Kirk posted, (Kirk Sorensen is now a contributor at Jon Goff’s place, I’m afraid having such top men in the same place might cause a awesomity criticality event). I assume this idea is probably found in some old NASA report from the sixties or seventies, like most things are.

Rendezvous is mostly a 4D problem: 3 space dimensions and time (some more if you take into account that proper attitude must be maintained as well, but that is assumed to be trivial). If you can take out two space dimensions, the problem should simplify greatly. This is possible with the following arrangement:

A boom at both the target and the vehicle, placed at right angles (and both at a right angle to the approaching vector). Basically, this should reduce positioning accuracy requirements hugely. The booms could be short barrels (even inflatable), or really long semi-rigid wires or composite girders or whatever. Depends on design aims.

When they contact, they will both slide until one boom is caught by a hook at the end of the other boom. Then one boom will slide through the hook until the hooks contact. From there on it’s a known geometry. You can reel in the boom, if it is flexible, or just slide it if it is rigid, and get both craft to a configuration you want, for either ordinary robot arm capture for berthing (as demonstrated by HTV, many station modules and Shuttle MPLM:s) or traditional docking (Soyuz/Progress/Shuttle/ATV).

This concept has some problems. For example whipping the target or the vehicle with an improper attitude / position boom. In the pictured boom configuration, approaches should have an offset always to one side. Alternatively one could have multiple booms. That way it wouldn’t matter on which side the rendezvous error would be. Also, the target could have a V shaped bow to avoid having the vehicle hitting dead center with a boom.

Another issue is if there is some kind of failure in the rendezvous, like too high velocity, the boom might rip off. That would result in a very dangerous object co-orbiting with the vehicles. This would be a very bad day for something like the ISS or a propellant depot.

One way to avoid this is to have the hooks have a mechanism to give way if the load gets too high. Another more outlandish is to have a weaker boom attachment in the vehicle. This would sever its boom and leave it hanging to the target in case of a problem.

This all was motivated to make unmanned rendezvous much easier to enable cheap propellant depot tankers. As all know, ATV and HTV are hugely expensive and high dry mass systems. Something like a Centaur or any basic “dumb” already existing restartable upper stage with just mostly a working attitude control system (including a star tracker) could be used instead, if some out of the box thinking is deployed. Most of the smarts should be in the target that is launched only once, but it can’t be the maneuvering party since it is very heavy. This system should get the best of both worlds.

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Armadillo flying to 600+ meters with a "mod". I say, it looks like the East German Sandman!

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Xombie NOW

Live stream just went up at http://qik.com/video/312581

they should be flying at 45 past whatever hour it is now in your time zone. Now on the pad loading propellants and helium.


And they did it! Congratulations! Also great accuracy.

The live cellphone video of the second flight was shot from quite close: http://qik.com/video/3126566

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Made huge strides in the last few days. A half-L1 done. They might be able to compete with Armadillo on L2, though I’m somewhat skeptical since they’re only going to assemble the new L2 vehicle soon.

Suddenly looks like there are two viable big VTVL sounding rocket companies!

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I’m quite that just right now. It will pass. Perhaps.

There’s been some discussion in various places about both NASA and potential future launch vehicles. Everything’s just so static in a large sense. Completely hopeless. I’ll throw in the towel for now.

Almost nobody has the required long attention span or patience to make any useful progress on the space front, and certainly not society itself.

The Players

USA is the only instance that is putting any significant money into doing anything new. And that’s wasted on the Ares rockets. ESA consists of a bunch of bickering countries, they’ve achieved some nice things but most of the people in the parttaking countries don’t even know they exist. No significant money spent on doing anything new, and what is done in Europe, is very often just me-too copying of American approaches. (Take Hermes as an example.) India is running with some crazy hypersonic stuff. China is doing intermittent Soyuz copy PR flights. Japan is doing something overcomplicated and abortive like they have always seemed to.

What are we left with? A bunch of US newspace companies with so little funding they won’t reach much in the next decade (Euro real newspace like SPL has zero funding at the moment). Scaled’s Spaceshiptwo is a dead end propulsion wise with the hybrids, and the air launching provides some scalability problems too. Maybe XCOR’s Lynx will fly some tourists to some altitude, and maybe there might be some X-racers. It won’t change stuff radically. The X-15 lessons were tossed to the trashbin too, to make way for the farces of NASP and X-33. Armadillo might fly something newish. So what? They don’t have enough money to even put turbopumps on the vehicle, resulting in ridiculous performance for orbital missions.

SpaceX? Forget it. It’s a rerun of Orbital Sciences Corporation, at best (and at the moment it looks much worse). No revolution, and evolution only very slightly.

COTS? Maybe something will actually fly, as it seems it has to try to pick up the mess that NASA put itself in with Ares and Orion. I’m not so well versed into the coming phases and how the politics will go. Both Lockmart and Boeing are in Ares/Orion so they don’t have such strong incentives to replace it with their own COTS solution flying on EELV on the short term. Depending how tightly they can keep their own ULA/EELV guys on a leash, and that has been shown to be ugly, people having gotten into trouble for what they have said on some web forums. NASA’s logical short term COTS alternative, a capsule on an EELV is thus self-censored.

But all this, even when happening in a good way, won’t change price to orbit significantly or enable real spacefaring.

What You’d Need

You’d need a refuel and go again reusable launch vehicle (RAGA RLV) that has turbopumps. No newspace company has money for that (and they are wisely using their little money on something else anyway). Besides, you’d in any case need multiple X-vehicles to develop the techniques like TPS or launch infrastructure and procedures to maturity so they could be operated with reasonable crew size and consistency. A launcher could be depended upon.

Human societies don’t seem to have capability to demand long term commitment to that technology development.

Environment Analogy

Same with the environment. If oil prices stay above 100 dollars, coal based petroleum will come soon and the synthesis already will produce massive amounts of CO2. New coal plants will be built too to produce cheap electricity to consumers who want it. Earth will change significantly with the resulting temperature rise.

No significant new energy producing or saving technology or international pacts will be seriously considered, never mind put into effect in the next ten years.

P.S. This post was written with the new Firefox 3. Hope it doesn’t muck up during publishing. Happy Midsummer. Looks to be rainy here.

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Jon Goff had an analysis (no equations or numbers though!) of an air-launched reusable SSTO idea, furthermore he promised to look at other reusable orbital launch vehicle ideas in the future. I thought I’d drop a few lines in regards to almost all architectures.

The summary is that big rockets have little drag losses going up but coming down they come too fast and need some tricks.

Drag in Ascent

The total negative impulse of drag during ascent is to a big order well approximated to be proportional to the frontal area of a rocket.

I_{drag} = \int C_d A \rho_{air}(t)v(t)^2dt \propto A

Drag loss as delta vee is a more ideal, but even more useful variable. It is simply the integral of drag force per mass.

\Delta v_{drag}  =\int \frac{C_d A \rho_{air}(t)v(t)^2}{m(t)} dt

Of course, mass at any point is directly proportional to takeoff mass.

m(t) \propto m_{glow}

And thus

\Delta v_{drag} \propto \frac{A}{m_{glow}} = \frac{A}{A \cdot l \cdot \rho_{rocket}}  = \frac{1}{l \cdot \rho_{rocket}}

So the profound result is that drag loss as delta vee is directly proportional to frontal area per liftoff mass, or in the case of a constant shape rocket, inversely proportional to rocket length l and rocket density rho.

It is NOT proportional to frontal area alone. This means that if you have a rocket that is, say, 50 m tall at takeoff, and you make an enlarged version and double it’s diameter while not changing the density, the drag loss delta vee stays the same. Drag loss grows in Joules or Newton seconds of course, but the rocket’s energy and total impulse grow as much at the same time, because the mass grows – both frontal area and thus mass are quadrupled.

What this of course means is that shorter rockets will always suffer from worse drag losses. Increasing density and decreasing the drag coefficient might help a little but probably not that much. Of course, in real life the acceleration profile and trajectory mean a lot too. Here is some data gleaned from this thread at nasaspaceflight.com and from “Space Propulsion Analysis and Design” by Ronald Humble.

  • Ariane A-44L: Gravity Loss: 1576 m/s Drag Loss: 135 m/s
  • Atlas I: Gravity Loss: 1395 m/s Drag Loss: 110 m/s
  • Delta 7925: Gravity Loss: 1150 m/s Drag Loss: 136 m/s
  • Shuttle: Gravity Loss: 1222 m/s Drag Loss: 107 m/s
  • Saturn V: Gravity Loss: 1534 m/s Drag Loss: 40 m/s (!!)
  • Titan IV/Centaur: Gravity Loss: 1442 m/s Drag Loss: 156 m/s

The Saturn V being by far the tallest has a much lower drag loss than the rest. Though it also had the slowest acceleration in the low atmosphere.

Drag in Descent

The very first post in the whole blog derived the analytical solution for delta vee in a vertical powered landing that takes into account the drag force. The total landing delta vee is:

\Delta v_{land} = v_{ter}(1+\frac{2g}{3a})

This means the system is sensitive to terminal velocity. And terminal velocity is determined by terminal mass, drag coefficient and frontal area:

m_{ter} g =  C_d A \rho_{air} v_{ter}^2


v_{ter} = \sqrt{(m_{ter} g) / (C_d A \rho_{air} ) }


v_{ter} = \sqrt{ (l \rho_{ter} g) / (C_d \rho_{air} ) }

Meaning that the terminal velocity is inversely proportional to the square root of the length of the rocket l and the density of the almost empty rocket, rho_ter. This means that if density is held constant (a reasonable first assumption), as well as shape (similar C_d), the terminal velocity is doubled when the rocket size (length) is quadrupled. And this means more delta vee needed for landing. It doesn’t matter if the rocket is lengthened and widened or only lengthened. If the constant density is held, the increased length (or depth) just means more mass per frontal area.

Frontal loading with various depths

So, rocket size presents a scalability problem for the powered landing. Bigger rockets have bigger terminal velocities and need more landing deceleration propellants.


What does this mean? Well, a rocket should be tall and dense for ascent to minimize drag losses. And it should be short and fluffy to minimize powered landing requirements. Slenderness or stubbiness is not necessary, but absolute tallness is what counts here, if we assume a constant cross section.

What can be done to help? Well, obviously, the rocket can have small frontal area in ascent but deploy a speed brake when descending to lower terminal velocity. From looking at TGV rockets’ materials, they already have thought of this. But it seems to me they are the first to actually do it. It’s weird that for example Armadillo or Masten hasn’t talked about speed brakes before. They could be very simple structures. The first idea that pops to mind self-evidently is a system that has brake plates on the surface of the rocket that are hinged at the top. During ascent the airflow pushes the brake to stay closed, but during tail-first descent the airflow raises the brake plate up until it hits a builtin stop at 90 degrees. Then it’s perpendicular to the airflow and exerts a maximum drag force. Quadrupling frontal area leads to halving of terminal velocity and thus halving of landing propellant needs.

Let us assume a 1000 kg empty vehicle and braking deceleration of 20 m/s^2 or 2 gees. Penalty is 0.3 and thus total delta vee 1.3 v_ter. Furthermore we can assume an ISP of 255 s or v_ex 2500 m/s.

Entering at 300 m/s (transonic at sea level), the needed braking propellant mass is about 1000 kg * 390 m/s / 2500 m/s = 156 kg. If we use a drag device to quadruple the area (double frontal diameter) the terminal velocity drops to half and propellant needed to 78 kg. Thus we can estimate that the drag device can be mass-useful if it weighs less than 78 kg. On one hand, it introduces additional failure modes, but on the other, it can be deployed pretty early and reduces timing and sensor criticality a lot by decreasing the terminal velocity.

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There was some excellent commentary in the last post, and that has partly been one inspiration in writing this post.

In the last part, the absolute minimum efforts by the launch vehicle making a vertical landing were determined. On top of these, there are lots of slightly more practical ones. This is still a one-dimensional vertical analysis.

Effects of a Time Delay

Let us see what a delay of t_delay before starting the main deceleration would do to a landing vehicle. The craft would travel t_{delay} v_{ter}=y_m extra distance before starting deceleration. That would mean that the actual reaching of zero speed would be y_m behind the original deceleration target. If we assume the original target was on ground level, then the new target would be underground. The difference between the targets is y_m=0.5 a t_m^2. (We used the trick of counting backwards in time from the underground target.) We can solve the imaginary underground time duration, t_m=sqrt(2 y_m/a). And the speed at ground level thus is v_m = t_m a = sqrt(2 y_m a) = sqrt (2 a v_{ter} t_{delay})

Example Vehicle

Our example vehicle, with v_ter=100 m/s and a=20 m/s^2 would stop in 5 seconds and 250 meters normally. If it traveled without any margins and would miss the engine ignition by one second (t_d=1 s) and 100 meters (y_m=100 m), it would impact the ground at over 60 m/s (130 mph, 220 km/h), or roughly two thirds of its original terminal velocity of 100 m/s. That would destroy it completely.

If we used a gentler and more lossy 10 m/s^2 deceleration (and started the engines higher accordingly, but still had the unfortunate one second delay), then the impact speed would be about 40 m/s. Still far too much.

Even a realistic 0.1 s of delay would cause a 20 to 14 m/s impact, depending on planned acceleration. Complete crash, remotely possibly survivable.

Musings about Results

The results might look unintuitive. That’s because in a constant deceleration, the most time is spent on the end, low speed distances, while most distance is covered with high speeds, early on. So if the deceleration is cut short even by a short percentage in distance, the time is cut short by a much bigger percentage, so the vehicle does not have enogugh time to brake much. Running out of a little distance you run out of a lot of time. This in total makes the vertical landing a very time sensitive problem!

It is also the reason why in the classical driving school example, even if you lock-brake and the car overshoots the pedestrian crossing by just a few meters, it was still going quite fast when you crossed it. The car’s speed doesn’t change much per meter when you start braking, but in the end half it changes a lot because it’s going slower and there are many more seconds per meter and thus the force has more time to effect. The drag is not dependent on velocity so in a sense it functions like a rocket engine. Accelerating with the car’s engine is the opposite as it generates constant power, not constant acceleration.

So, if we anyway design the vehicle to survive some vertical touchdown speed v_touch , then we can calculate the margin for the fast deceleration directly to pad in seconds. t_d = v_{touch} / sqrt (2 a v_{ter}). With our example vehicle it could be 4 m/s, equal to a drop from 0.8 meters height. Then the acceptable time delay would be 0.063 s or 63 milliseconds. That’s probably too small of a margin.

More Realistic Approaches

So, in reality, one can not have a direct high deceleration to pad. One could have multiple other approaches and ways to think about it. These are the most obvious to me:

  1. Put the the deceleration aim target (hover) some distance above the ground. Layne Cook told that was an approach the DC-X used.
  2. Stop the high deceleration at some determined velocity and switch to a low deceleration. For example at 10 m/s.
  3. Use some more elaborate forecast of the vehicle position and throttle accordingly to always have a plan to reach very slow speed at touchdown. I think Armadillo is using something like this approach.

These are handled below.


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There is a long standing dispute whether winged landing or vertical rocket powered landing is more efficient and overall better for reusable rockets on earth. 🙂
Powered landing can be looked at in a broad overview. The post documents what I doodled in my notebook some day in the summer.


Deceleration from terminal velocity to hover usually requires about 5-10% of total landed mass as propellants. The amount is directly proportional to the vehicle’s terminal velocity. Also, the higher acceleration, the better. The powered landing penalty fraction is N_{penalty}=\frac{2g}{3a}, so landing at 2 gee acceleration (3 gee felt) gives a penalty of 0.3. The required delta vee is \Delta v = v_{ter} (1+N_{penalty} ) where v_ter is terminal velocity. And the required impulse I=m \Delta v and propellant mass m_{prop}=I/v_{ex}.


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