Author Topic: Waverider suborbital transit system?  (Read 2055 times)

Offline AlanSE

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Waverider suborbital transit system?
« on: 02/21/2016 03:21 am »
Suborbital point to point travel as a competitor to jet aircraft travel is something that we still need to recognize as a possibility, particularly as reusable first-stage boosters coming into reality, and as low-acceleration lifting bodies like the Dream Chaser refuse to go fully extinct. I think it's an opportunity equal to orbiting space hotels, even if that means neither possibility is going to be a smash hit in the near term.

I'm feeling torn about the differential calculus of this matter. If L/D (lift over drag) is greater than 0, is our mathematical model of suborbital hops still correct? Could we be missing something important?

Some background on suborbital hops: If you are traveling a substantial distance via a suborbital flight, the optimal trajectory goes high at apogee. Higher than LEO. This is because in a short-distance approximation the optimal angle to launch a projectile is (of course) 45 degrees. As you lengthen that distance further, the optimal angle bends down, until you get to the exact other side of the planet, at which point the optimal angle is flat, nearly exactly a LEO, trajectory. At that point, there's only a small energetic difference between going into orbit and re-entering on the other side of Earth.

Exactly what L/D could be for a lifting-body is up for debate. We have to be careful to cite hypersonic (not subsonic) values for the purpose of this problem. That is, probably 1.0 for the Shuttle, maybe 2 for the Dream Chaser, I'm not exactly sure. Some people have made the case that waverider designs can have absurdly high values like 6. This is still much lower than typical subsonic aircraft, but hypersonic is very different physics.

The Notion of Orbital-y Flights

So I'm curious, if you can get L/D of 2, or 4, or something like that, what does the optimal trajectory look like for different point-to-point suborbital trips on Earth? Probably (?), those nearly-flat trips that go from London to Sydney will get substantial fuel savings from upward lift from grazing low orbital altitudes. This lowers the target coasting height, and sets the optimal speed to some amount less than orbital. Critically, the optimal trajectory would seem to demand some slow-burn thrust for all of the flight up until the coast-down for landing.

Now, if this is starting to sound kind of like hypersonic air-breathing flights, that's cool and all, but that's not what I'm talking about. I'm speaking totally in the domain of reaction engines. Among the optimal trajectories, there is surely a "gulf". On one side, there is the fastest air-breathing trajectory that makes sense. On the other side, we have this orbital-y thing defined by substantial contribution to the lift from v^2 / R_earth. That is to say, there is a trajectory that balances reaction engine thrust, with aerodynamic lift, with "false" lift due to the recession of the horizon.

There seems to be a paper on this subject:

Study of Waverider-based Point-to-Point Suborbital Rocketplane, JAXA, 2012

I find myself blocked by a paywall here, but I don't find the abstract promising regarding my core question anyway. If they had calculated any parameters of an optimal trajectory, I would think they would have put some basic parameters in the abstract. I'm sorry if I'm missing some very important prior literature which covers the subject. I'll link it up if I find any.

My Prediction

Short flights will still need to be in an arc pattern. There's just no way that the lift can hope to pay for the substantial acceleration that it would involve. Either lift would be almost 100% from aerodynamic forces, or the rockets would do all of the work entirely.

At some point, a cross-over will happy, and all trajectories will follow the orbital-y pattern. Without considering L/D, some of these would have otherwise had substantial departure from atmosphere, but with constant burn, they will be much more fuel efficient with some non-trivial mix of orbital acceleration versus aerodynamic lift.

The Calculus

The problem should actually be pretty simple, if you can just consider it in a differential sense for cruse. All trajectories will probably follow the same cruse pattern. You'll be flying totally horizontally, so the only challenge is to figure out the optimal speed, and by extension you will know the balance between acceleration and aerodynamic lift. Altitude also comes from that mix. Then the only question is - for what set of trajectories is this more optimal than the classic solution to suborbital hops? And by how much?

As an alternative hypothesis, maybe the best amount of drag is no drag. Maybe there's a cutoff, where you need a certain minimum L/D value to make it payoff to reduce cruising altitude. Maybe this value is impractically high? As a separate alternative hypothesis, maybe the ballistic trajectories will never be favorable past some L/D value? The answer could be several different things, and I don't see it anywhere else, so I'm posting here.

Offline Jim

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Re: Waverider suborbital transit system?
« Reply #1 on: 02/21/2016 10:44 am »
Suborbital point to point travel as a competitor to jet aircraft travel is something that we still need to recognize as a possibility, particularly as reusable first-stage boosters coming into reality, and as low-acceleration lifting bodies like the Dream Chaser refuse to go fully extinct.

Not really.  It may happen but it is not going to be a competitor.

Offline Vultur

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Re: Waverider suborbital transit system?
« Reply #2 on: 02/21/2016 06:15 pm »
I don't think it would ever compete with jets for "ordinary" travel, but there might be some specialized things the ability to get anywhere on the planet in an hour would be useful for.

One possibility is that such vehicles crossing the "edge of space" would be regulated under different rules than ordinary aircraft... this might be desirable for those who could afford it (likely very rich businesspeople whose time is very valuable)... skipping the time-consuming security measures for an ordinary flight.

Offline Alf Fass

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Re: Waverider suborbital transit system?
« Reply #3 on: 02/21/2016 06:56 pm »
Along side their APT Black Horse concept Zubrin and Burnside-Clapp had a concept they called the Black Colt/Black Yearling, which had a sub orbital point to point capability.

http://www.ai.mit.edu/projects/im/magnus/bh/analog.html (towards the bottom of the page)

It's getting a bit old now, but it does look at L/D and used an atmospheric skip-bounce capability to travel long range.
When my information changes, I alter my conclusions. What do you do, sir?
John Maynard Keynes

Offline Jim

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Re: Waverider suborbital transit system?
« Reply #4 on: 02/21/2016 06:57 pm »

One possibility is that such vehicles crossing the "edge of space" would be regulated under different rules than ordinary aircraft... this might be desirable for those who could afford it (likely very rich businesspeople whose time is very valuable)... skipping the time-consuming security measures for an ordinary flight.

The opposite is more likely.  Such flights would more scrutinized security wise since there is more a potential to bring bad items or people into different places.

Offline AlanSE

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Re: Waverider suborbital transit system?
« Reply #5 on: 02/21/2016 09:08 pm »
I think I got it! I won't post the math, I'll just post my answer.

 - (L/D) - lift to drag ratio
 - theta - the angle that measures distance in terms of the arc (in radians) of the triangle that goes between the center of the Earth, the departure point, and the destination point

The following equation is valid only if (L/D) is greater than 2 theta.



So you can now make some case-in-point examples. If you're traveling 90 degrees, then the minimum L/D you need to make atmosphere grazing make sense will be 3.14. If L/D is 4, then you can save 20% of your Delta V. Not too shabby. However, if you're journey is 180 degrees, then you'll need 6.28 L/D ratio, and this is really pushing the envelope.

My understanding is that the L/D ratio is insufficiently large, then any amount you lower altitude into the atmosphere will be a penalty.

This is also not (yet) accounting for the alternative of a truly ballistic trajectory. There is a separate equation for that, which depends on theta and nothing else. For any given angle, we will need to check that this number doesn't yield a lower Delta V, in which case, a waverider gives no particular benefit. In the case of 90 degrees, that's 40% of orbital velocity.

So yeah, there's a pretty narrow window for which this notion makes sense, although that mostly fits with my intuition. The trip would have to be very long, and the L/D ratio would have to be very high. But this is a very simplified mathematical view. In reality, there will be more factors at play.

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