A while back I wrote my article "Can Giant Airships Slowly Accelerate to Orbit Over Several Days" . It lead to a lot of discussion and as usual whenever you discuss John Powell's ideas, there were physicists and aeroengineers saying it is impossible. So anyway as part of this I had an especially interesting online discussion over many days with James Fincannon on the Space Show blog and much of this article comes out of that discussion. Unlike many of those who get involved in these discussions on the topic, he actually did many detailed calculations and shared them on the blog. We had a lively debate there, and in the process I came up with some interesting new points about JP's ideas which I want to share here, along with James' lift calculation. This is a post for those of you who like figures and calculations :).

Incidentally JP himself liked my previous article, writing in his blog:

"There’s a article about JPA at Science 2.0. It’s a good article. A few minor errors (Dark Sky Station will be at 140,000 feet not 180,000 feet and we never charge for PongSats). However overall it really captures what we’re up to."
(I have now corrected those errors)

This shows the idea, one of his airships in the vacuum of space.

As Jeff Foust, editor and publisher of The Space Review wrote back in 2004, when JP Aerospace were still working as part of the USAF,

"It would be easy to dismiss ATO [Airships to Orbit] as simply an interesting idea—or worse—but for the fact that some elements of the system are already being built. Under an Air Force contract JP Aerospace has been developing the NSMV [Near Space Maneuver Vehicle, now called Ascender], developing several subscale vehicles as well as a full-sized one, over 50 meters long. The full-scale NSMV is currently in a hangar in Texas, with flight tests slated to begin later this month. The Air Force is not necessarily interested in the ATO  concept—they see the NSMV as a potential high-altitude reconnaissance platform—but their support has paved the way for continued development of the system."

So to start with, there's one basic point we need to be clear about. His airships don't literally "float into space" all the way as the title of his book, "Floating to Space", might suggest. They do float up to a very high altitude of 200,000 feet. But not all the way.

From their highest neutral buoyancy altitude, according to his ideas, they slowly accelerate forward like a plane, and in the process rise a bit higher, and at this stage the aerodynamic lift starts to counteract some of the weight of the airship, but most of it is still supported by buoyancy. At this point the airship is operating like the Airlander 10, a hybrid airship / lifting body which is actually slightly heavier than air but is able to fly with a combination of lift and buoyancy. Though the orbital airships do this at a far higher altitude of course, starting from 200,000 feet instead of ground level.

When they reach the speed of sound, they are still flying at a level where they have some assistance from buoyancy. By the time the airships reach Mach 5, however, he says that they are way above the buoyancy level. So

  • To start with it just uses buoyancy
  • As it gets faster it uses lift combined with buoyancy
  • Finally it just uses lift like an airplane

The buoyancy is very useful. It saves lots of fuel in the early stages and he couldn't get to those high altitudes and speeds without going through these earlier stages first. But once he is going at hypersonic speeds, it's no longer worth trying to use buoyancy to keep the airship flying. If he flew at an attitude low enough to keep it in the air just through buoyancy or to get significant assistance from it, the amount of drag would be so huge as to make that impossible.

At least that's how I understand his plans. Also the airship has to get to Mach 23 before it reaches orbital velocity so for nearly all the flight to orbit, from before Mach 5 through to Mach 23 at more or less constant acceleration, it would use lift just like a conventional airplane - though with a radically different shape.

When you think about it like this you may then think that the design surely can't work as it doesn't look aerodynamic. How can something as bulky looking as this produce significant lift, enough to lift its own weight?

But aerodynamics can be a bit deceptive at times. So anyway that's what he claims it can do, and that's what we will look at. 

I found a number of rather simple mathematical calculations that I think help with clarity of thinking in this topic.

There is almost nothing published on this in the literature. There is JP's book and his blog posts, but they don't go into a lot of engineering detail. Nobody else seems to be working on detailed calculations for kilometer scale orbital airships traveling at hypersonic speeds in the upper atmosphere where the pressure is only a tiny fraction of a millibar. At least I couldn't find anything in a literature search. So anyway here are a few ideas I'd like to share. 

MAIN POINTS

  • Buoyancy is not enough to carry the payload to 200,000 feet unless the airships are over three times the dimensions JP gives. 

There is no way to get his figures to work with a 6000 foot airship as he says it can carry 20 US tons to orbit as payload, and he also says it is neutrally buoyant at 200,000 feet. James Fincannon's buoyancy calculation show that JP's orbital airship has enough buoyancy lift for 2.6 tons at 200,000 feet (with density 0.000254864 kg / cubic meter and pressure 0.1776 millibars according to the 1976 standard atmosphere calculator). That's for everything including fuel, skin, payload, life support, airbeams, etc.

It has to be much larger than 6,000 feet to carry 20 tons (a bit over 18 metric tons) to orbit.

So, he is saying it can carry to orbit more then seven times the amount it could lift to 200,000 feet through buoyancy for the entire airship including its skin, fuel, life support etc. There is no way to tweak these figures to make it work no matter how hard you try. So far nobody has said they have seen any flaws in this argument.

  •  Feet as meters 

Unless we are missing something, if the airships are to be neutrally buoyant, it seems that wherever he has "feet" in his orbital airship diagrams, then if we read "meters", the numbers are about right for the buoyancy. That's my interpretation of the situation. It seems a natural thing to happen - he is using the same diagram now that he shared in his book and he may simply have never redone the calculations since the USAF days. James thinks that it is up to JP to correct it if he made a mistake and doesn't want to speculate about what happened.

Anyway whatever the reason, the figures just don't work for feet. But they work perfectly for meters. It would be the end of this article right now if I read those figures as feet. So, I'm just going to read feet as meters in all his diagrams from now on - for the orbital airships and for the transatmospheric airship.

That makes his transatmospheric airship 2.1 km and the orbital airship 6 km if it is going to be able to carry 20 tons to orbit. It would still be the same mass, less than 100 tons including skin, payload, life support, engine, lifting cells, etc (it can't be any heavier than that), but much larger.

  • Or target lower buoyancy level of 113,000 feet

The other way forward is to make his airships neutrally buoyant lower in the atmosphere. The figures work for neutral buoyancy at around 113,000 feet with an atmospheric pressure of around 6 millibars and density around 0.009 kilograms per cubic meter. In that case, I wonder if one could just scale everything down. If he could build his dark sky station at 60,000 feet or a little higher with the air pressure above the Armstrong limit, then humans could survive there with just an oxygen mask without a full body pressure suit which could simplify construction and safety.

Anyway that would be a major change in his plans.

Keeping the same height for the dark sky station and changing feet to meters is the minimum change we can do to continue the discussion. So I'll take that as the starting point for the rest of this article.

  • If it is buoyant at 200,000 feet, the skin must be micron thick, and you can't use the JAXA balloon thickness of 3.4 or 2.8 microns

The lower buoyancy figures in this calculation make the skin more of an issue because it makes the skin ten times larger in surface area for the same amount of buoyancy. To be buoyant at 200,000 feet,, the skin  has to be very thin to compensate. I work out that it needs to be about one micron in thickneess, or a third of the thickness of the thinnest balloon fabric used by JAXA for the transatmospheric airship at 1.2 km in length, which only needs to reach Mach 5 so has a smaller fuel fraction.. It could perhaps work with a somewhat thicker skin for the orbital airship at 6 km in length, but a one micron thickness skin makes it all much easier for that size also.

If you read feet as meters, his 1,200 foot transatmospheric airship would have 111 kg of lift and with a one micron skin would use 43 kg for skin + lifting cells assuming lifting cells have same mass as the skin. It would be more comfortable with a 2 micron skin, but that 111 kg doesn't give you much to work with for everything else. The 6,000 foot one would need around a 0.55 micron thickness skin, and with that thickness, it could carry 460 kg to orbit assuming 2000 ISP motor and thrust to drag ratio at least 2.

This is not an impossible target for the future, as the JAXA skin is way stronger than is needed with a breaking strain of 400 kg per square centimeter. The main question would be whether a thinner film can be made with sufficiently uniform thickness. This is especially so for the transatmospheric airship at 1.2 km - if you used the JAXA skin then it would be heavier than air at 200,000 feet, if you had a similar surface area of skin for the lifting cells as the outer skin.

  •  The time to orbit depends on the lift to drag ratio, based on some very simple assumptions. 

The formula is 7,844 * L / 9.80655 secs assuming a lift to drag ratio of L and at most 50% of the thrust used to offset drag and a target orbital velocity of 7,844 meters per second.

More generally if the thrust to drag ratio is R then the acceleration is (R-1) * 9.80655 / L and the time to orbit is 7,844 * L / (9.80655 * (R-1) ) secs.

  •  Also, from the time to orbit of three days, you can work backwards to find the lift to drag ratio assuming that no more than 50% of the thrust is used to offset drag. 

The formula is just found by solving for L in the previous calculation.

So L = t * 9.80655 / 7,844 where t is the time to get to orbit in seconds.

More generally,  L = t * 9.80655 * (R-1) / 7,844 where R is the thrust to drag ratio.

From that calculation, JP seems to be targeting a lift to drag ratio of over 300 putting in three days in seconds into the right hand side of that formula. So is that possible? If the lift to drag ratio is as good as the best gliders, 70 then it should get to orbit in 15 hours or so, assuming the engine uses no more than half its thrust to overcome drag. It would cirumnavigate the Earth about 5.5 times in the process.

  • A small overpressure from a shockwave can support the airship's weight 

Also. I work out that if the orbital airship as a hypersonic waverider has an overpressure of just 0.0366 millibars beneath it, that's enough extra lift just from the pressure difference to offset the entire weight of the airship. I don't know what that means for the lift to drag ratio but it is at least a rather different regime from normal flight.

  •  A 2000 ISP engine needs only a bit over 50% of the mass for fuel, no matter how long or short the time to orbit, assuming at most 50% of the thrust is used to overcome drag.

The other main point is that I work out that if he has a 2000 ISP engine with 50% of it used to offset drag, then he should use a bit over 55% of the mass for fuel, more like a commercial jet, or the lunar module taking off from the Moon than a normal launch from Earth. This is a straightforward application of the rocket equation.

If you see any errors in these calculations however minor, be sure to say! So now, down to the details:

LIFT TO DRAG RATIO CALCULATION

This is a simple argument I found recently which rather simplifies thought about orbital airships. This calculation is for the final stage, when it is flying as an airplane, which takes up most of the time to get to orbit. At this point, it’s the lift to drag ratio that matters most. 

Though JP Aerospace don't seem to have published estimates of the lift to drag ratio of their airships to orbit, I found a way of estimating the lift to drag ratio from the time taken to orbit. The calculation is rather simple and gives an easy "in" to the whole question of whether you can accelerate to orbit slowly and you can apply it to any spacecraft that accelerates to orbit.

It doesn't matter what you are using for propulsion for your spacecraft or airship - or what configuration it is, rocket, scramjet, skylon, airship, none of that matters. The only assumption in this calculation is that it's accelerating to orbit using lift all the way until it reaches orbital velocity.

In detail, our assumptions are that

  • It uses lift like a conventional airplane rather than using thrust to offset its weight (like a rocket) or buoyancy.
  • It uses a fixed percentage, we'll assume 50%, of its thrust to offset the drag.
  • It has a fixed lift to drag ratio throughout the flight (in practice of course this varies so then we can think of the figure used as its average lift to drag ratio)

EXAMPLE - HOW LONG WOULD A SPACECRAFT TAKE TO REACH ORBITAL VELOCITY, IF IT HAS A LIFT TO DRAG RATIO OF 4 AND FLIES AERODYNAMICALLY ALL THE WAY

So let's start with an example to show how it works, one that doesn't have the controversy of an orbital airship, Skylon.  Suppose, for instance, it has

  • A fixed lift to drag ratio of 4
  • It uses half of its thrust to offset the drag

Skylon when traveling at Mach 3.508 can achieve a lift to drag ratio of 3.583, so not quite as good as this, (see page 13 of this article), while at Mach 2.702 it can achieve a ratio of 4.495, rather better than 4. So this depends on how fast it is going. They don't give a figure there for Mach 5. So this is a simplification but it gives an idea of how the idea works

Techy note: gs of thrust

I'm going to talk about "gs of thrust" here, treating g as a unit of gravitational force instead of acceleration, where Earth gravity is 1 g of force.

This is for ease of exposition. I know that normally you think of gs as a measure of acceleration, but it's convenient here to use it as a measure of thrust and simplifies the exposition so much, and I can't think of a better way to do this, and it is sometimes done this way.

So, by a 1 g thrust, I mean, a thrust that equals the force of gravity on the spaceship. In free space one g of thrust would accelerate your spacecraft at 9.80655 m/sec². It's also enough thrust to hover against gravity at standard gravity.

Of course the heavier your rocket, the more thrust you need, to achieve a force equivalent to the effect of gravity. Also as the spacecraft gets lighter it will be able to throttle down its engines to continue at the same level of thrust in gs. So when I talk here about the airship traveling to orbit at a constant x gs of thrust, then - that means it throttles down like that as it uses up its fuel.

So, with the lift to drag ratio of 4, that means that the amount of drag is the same as a quarter of its weight (by definition of lift to drag ratio), so it has to use thrust of a quarter of gravitational force, let's call this "a quarter of a g", just to offset the drag.

Then, since we also assumed that it uses half the thrust to offset the drag, it accelerates the spaceship by the same amount, a quarter of a g. 

That's enough information to calculate the time to get to orbit as we have the acceleration, and the required delta v to reach orbital velocity.

As for the final velocity, it depends on the altitude of the target orbit when his airship first reaches orbital velocity. His transatmospheric airship reaches Mach 5 at 350,000 feet or about 100 km. So it will be at that level or higher. That makes the target orbital velocity of 7.844 km / sec. It's not much different at 200 km. (7.844 versus 7.843, see online orbit calculator). So let's use 7.844 km / sec as our target.

Then using v = at, we can deduce the time it will take to reach orbital velocity as v/a = 7,844 / (9.80655/4) = 3199 seconds or about 53 minutes.

So, Skylon, it seems, would take a little under an hour to reach orbital velocity if it were to use 50% of its thrust to offset drag and use lift all the way. It would need half a g of thrust, half of that used to offset drag. This is a simplification in the case of Skylon as it actually only flies like a plane for the first part of its flight, and then it accelerates like a rocket using thrust to offset its weight. But it gives the basic idea.

By a similar calculation, if Skylon used only 25% of its thrust to offset drag, still assuming a lift to drag ratio of 4, then it would need to use a full 1 g of thrust and accelerate at 3/4 g, and would get to orbit in around 18 minutes.

So this depends a lot on what percentage of its thrust is used to offset drag. The more thrust the better once it has enough to overcome drag, so whether it can do this will depend on its thrust to weight ratio.

GENERAL FORMULA FOR TIME TAKEN TO REACH ORBIT, GIVEN ARBITRARY LIFT TO DRAG RATIO

That's our general formula too. We'll assume 50% thrust to offset drag in these calculations with the airship accelerating constantly. Then if the lift to drag ratio is L, this makes the acceleration 9.80655 / L.

So then the time to get to orbit will be

7,844 / (9.80655 / L)

= 7,844 * L / 9.80655 secs.

More generally if the thrust to drag ratio is R then the acceleration is (R-1) * 9.80655 / L and the time to orbit is 7,844 * L / (9.80655 * (R-1) ) secs.

I'll use links to google calculator for the calculations so you can check them easily.

So now, suppose that our orbital airship can somehow manage a lift to drag ratio of 70 like the best glider, the ETA glider, and again uses 50% of its thrust to offset drag. Then it would need 1/35 g of thrust, and it would get to orbit in 7,844 *70 / 9.80655= 55991 seconds, or about 15 hours 33 minutes. (You can use this online tool to help convert seconds to hours and minutes quickly).

At an acceleration of say 1/70 g, then the distance traveled is s = 1/2 a t^2 = 0.5*(9.80655/70) * 55991^2 meters or 219,596 kilometers. The circumference of the Earth is around 40,075 km, so during the flight it would circle Earth about 5.5 times. Reality check: at its target height of 200 km then the orbital period is 1 hour 28 minutes 30 secs so that figure is reasonable.


ETA glider photographed by Britta Pulz. This is a glider with a remarkable lift to drag ratio of 70 : 1. 

If a spacecraft could achieve a lift to drag ratio as high as this, with hypersonic flight, and could keep up a continuous thrust of 1/35 g, it could reach orbit in 15 hours 33 minutes using half of its thrust to offset drag. 

WORKING OUT THE LIFT TO DRAG RATIO FROM THE TIME TAKEN TO REACH ORBIT

So, then you can also work backwards too, and from the time it takes to get to orbit, work out its (average) lift to drag ratio.

Let's try that for Skylon first.

It doesn't fly all the way to orbit, but according to its user manual (page 5), it takes 694 seconds (about 11.5 minutes) to reach Mach 5.14, or 1.763 km / sec. So now, assuming it uses half its thrust to overcome drag, then you have

1,763* L/ 9.80655 =694

So L = 694*9.80655/1763 = 3.86

So that would suggest it averages a lift to drag ratio of 3.86. Of course it probably uses less than 50% of its thrust to overcome drag. So the lift to drag ratio would be more than that, so the number looks reasonable.

So now let's try it with the orbital airship. John Powell estimates that it would take three days for his airship to get to orbit. So, if he uses at most half of the rocket power to counteract drag, then by a similar calculation, we have

7,844 * L/ 9.80655 = 3*24*60*60

So L = 3*24*60*60*9.80655 / 7,844 = 324.

So to achieve such a slow acceleration to orbit, his airships need a rather fabulous lift to drag ratio of 324. Or more than that if it uses less than 50% of its thrust to overcome drag. Not saying that's impossible but that seems to be what he would need to aim for.

Of course, it could be that he has a lower lift to drag ratio than that, and is using a lot more than 50% of the thrust just to offset drag. But if you find yourself doing that, would it not make more sense to use a more powerful engine, even if it meant increasing the amount of mass? Double the number of engines?

I find that a lift to drag ratio as high as 324, more than four times higher than the ETA glider, a bit hard to believe. However his airship is flying as a hypersonic waverider. Also, the density of the airship is so low, it's probably not that far off the density of the shockwave itself, and the airship, he says, is designed to ride on top of the shockwave. So, perhaps it might be a bit hard to apply the hard got knowledge of conventional hypersonic flight and even subsonic flight to this regime.

DISCUSSION

He says on page 111 about lift to drag ratios:

"Sailplanes have a ratio of over 60 to 1. The Space Shuttle has a L/D ratio of just over one to one. THe L/D ratio for hypersonic aircraft is thought to be around five. The orbital airship will need a L/D ratio much higher than that. It is important to note that the L/D barrier was calculated for conical, or cone shaped vehicles. We simply don't have the data to correctly model other shapes of hypersonic vehicles. Early tests suggest that lift to drag ratios much higher than those needed for ATO [Airship To Orbit] are possible. "

He doesn't give any figures though.When he says the early tests suggest L/D ratios much higher than those needed for ATO are possible - what L/D ratio does he think ATO needs, and what's his reasoning for them? Does he mean over 300:1 or does he think ATO is possible for a L/D ratio much smaller than that, and if so, how and why? That would be my main question for him here.

If these calculations are right, then - if he can only achieve an average lift to drag ratio of, say, 10, that would take 8056 seconds to get to orbital velocity or about two and a quarter hours, assuming no more than half the thrust is used to offset drag. If he achieves a lift to drag ratio of 40, then he could get to orbit in 10 hours. The thrust to weight ratio of the rocket in that case would only need to be 1 : 20. With a lift to drag ratio of 70, it would be 15 hours 40 minutes and a thrust to weight ratio of 1 : 35.

So, there seem two main possibilities here.

  • His airship can achieve far higher lift to drag ratios even than our best gliders and the thrust to weight ratio can be less than 1 : 150
  • Or, his airships can get to orbit in a matter of hours rather than days, with a rather higher thrust to weight ratio, 1 : 35 for L/D of 70 down to 1:5 for L/D of 10.

What you think about all this depends a lot on what lift to drag ratio you think he can achieve.

If he can only achieve a lift to drag ratio of 2, for instance, then he would need one g of thrust, if half of it is used to overcome drag. Enough thrust to support its own weight against gravity and it's not far off becoming a rocket.

This was the main point on which our views differed in my debate with James Fincannon. He calculated a lift to drag ratio of around 2. It was based on a simplified model, not using details of the shape, the upturned nose, flying as a wave rider, or the drag reduction technology installed in its nose. But he found that discouraging enough that he thought lift just wouldn't work as a way to get to orbit, and felt it would have to stay at its neutral buoyancy altitude instead. That of course leads to vast amounts of drag, and it's hard to do anything about that.

Without results from wind tunnel tests, there isn't much you can do at least by way of proof, beyond expressing differences in views there. But I was much more optimistic that it might be possible to achieve higher lift to drag ratios than him. I explain some of the reasons here later on.

Anyway, in the next section, I'm assuming a lift to drag ratio that's reasonable, significantly greater than 2.

AIRSHIP OPTIMAL ALTITUDE

In this section I hope to show that if he can achieve an airship with a lift to drag ratio significantly greater than 2, then the airship won't face excessive drag but instead will simply rise to whatever level in the atmosphere reduces the drag enough to give it the necessary lift, at its rated lift to drag ratio for that velocity.

To see how it works, suppose for instance he only achieves an average of a 4 to 1 lift to drag ratio like Skylon during the hypersonic flight phase.

It will help make the calculations simpler to use round figures here. Also I'm more used to working in metric units so will use meters, newtons, metric tons (1000 kg) etc instead of feet, pound force, and US tons.

So, let's suppose he has an airship with its weight measured in force units as 1,000,000 newtons (a little over 100 metric tons in mass). Then it would fly at whatever altitude gives a drag of 250,000 newtons. The amount of drag of course goes down as you ascend to higher altitudes where the air is less dense.

If it flies just a smidgen lower than that, at an altitude with a drag of 250,250 newtons, say, at its current speed, and it is using enough thrust to overcome the drag and fly forwards, then by a lift to drag ratio of 4 to 1, it has a lift of  1,000,100 newtons. Of that, only 1,000,000 newtons is needed to support the weight of the airship, so it would have an excess upwards force of 100 newtons. This would accelerate it upwards at a ten thousandth of a g. So it would gently rise up until the drag returns to that level of 250,000 newtons.

So, as he flies, as the airship accelerates to higher and higher speeds, although the drag would increase at whatever altitude it is flying at, the lift will also increase above what is needed to support its weight, so the airship will automatically rise in response until the lift to drag ratio equation balances again. For this to happen the pilot has to keep it angled at whatever angle optimizes the lift to drag ratio.

So for instance, during Skylon's stage in its flight when it is using lift to stay aloft, if its lift to drag ratio is 4 at a particular stage of its flight, then we can deduce from this that its drag will be a quarter of its weight as measured in force units. We don't need to know anything about the altitude it flies at, to say this. Of course in practice the lift to drag ratio is going to vary at least a bit depending on altitude, and there will be some lag too as the drag increases before it rises to the necessary altitude, so this is a first approximation. But if the ratio is reasonably constant then it will give a good idea of how much drag to expect.

Similarly if he can achieve a 10 to 1 lift to drag ratio, it will automatically rise until the drag is reduced to 100,000 newtons. If he can get to a 100 to 1 lift to drag ratio, the equilibrium altitude achieves a drag of only 10,000 newtons.

If you think about it this way, you don't need to take account of the air density or work out what altitude it is flying at. All you need is the lift to drag ratio.

This is why I was able to work out the average lift to drag ratio from the time taken to get to orbit, and vice versa, without knowing anything about the particulars of the airship, or even the altitude it flies at.

Though he doesn't say much about altitude, his idea for his transatmospheric airship is that it would achieve Mach 5 at 350,000 feet. I think it's reasonable to assume his orbital airship would achieve hypersonic speeds at a similar height and orbital velocity at a higher altitude than that.

The atmosphere thins out very quickly. At 200,000 feet it's 0.177603 millibars. At 280,000 feet it's 0.0034 millibars. The standard 1976 atmosphere model doesn't go up to 350,000 feet. However, from the diagram here, it looks like you get down to around 0.0001 millibars. So the atmospheric pressure is 1,776 times less at 350,000 feet than it is at 200,000 feet. Since the drag is normally proportional to the density of the air then that means that when the airship is traveling at its minimum altitude for hypersonic flight, it experiences less than a thousandth of the drag at 200,000 feet.

Note, though hypersonic flight by convention starts at around Mach 5, it doesn't necessarily follow that the airships experience the other effects of hypersonic flight such as high heat loads, or that you have molecular dissociation or ionization of the air. That's because they would be flying in a hard vacuum with hardly any air.

BUOYANCY CALCULATION

James Fincannon worked out the buoyancy for JP's 6,000 foot airship, which he says in his book can take a payload of 20 US tons to orbit, or 18.14 metric tons, from the published dimensions and from JP's statement that it would be neutrally buoyant at 200,000 feet. It turns out that it can't lift anything like that much of a payload, if the calculations are correct.

We can double check that, by using a simple over estimate. Let's estimate its volume as a tube of diameter 300 feet and length 12,000 feet. Let's treat the hydrogen as massless.

Then the volume is 12,000 * PI *150^2 cubic feet which works out at less than 850 million cubic feet or just over 24 million cubic metres. This will of course be a huge over estimate but that's the idea for this calculation. We also know the density of air at 200,000 feet, at 0.000254864 kg / cubic meter, for the  standard 1976 atmosphere model (atmospheric pressure 0.1776 millibars), so plugging that in, that means it can't possibly lift more than 24 million * 0.000254864 / 1000 = 6.11 tons through buoyancy, which has to include its own weight. So we already see that there is no way it can lift 18.14 tons to orbit.

James works out the volume more exactly, using the figures from JP's diagram, and he also of course doesn't treat the hydrogen as massless, so he subtracts the mass of the hydrogen when working out the lift. He gets a volume of 0.95* 380,250,000 ft^3 (= 361,237,500) or 10,229,106 m^3 and a lift of 2.6 tons in part 5 of his calculations.

If it is neutrally buoyant at 140000 feet, the height of the dark sky station, that makes things a little better, 0.00261346 kg / cubic meter. That makes the buoyancy lift 26.73 tons. But that leaves only 8.59 tons for the airship, skin, life support, structure etc.

To make it work comfortably he needs around 35 times the density, or 0.0089 . He would have to make it neutrally buoyant at 113,100 feet with an atmospheric pressure of around 6 millibars.

So how can this have happened? There is some mistake somewhere. Of course it could be in our calculations but they are so simple, that it's hard to see how that's possible.

Well, you do get the right answer if you read JP's feet as meters in his orbital airship diagrams. If you take our calculations and multiply James' figure by the number of cubic feet in a cubic meter, or 35.31467, that would change his 2.6 tons into 91.8 tons which then becomes much more plausible as a way of lifting 20 tons into orbit, taking account of the mass of the skin, life support, lifting gas, solar cell deposited layer etc.

It is something that could happen. If the Mars Climate Orbiter can crash on Mars because of a mix up of imperial and metric units, then it's no shame for the same to happen here. Maybe similarly it was some kind of a communications glitch?

Whatever the reason, it seems to me that whenever JP has feet in his dimensions of the orbital airship, we should read that as meters. If we do that then the amount of lift due to buoyancy is enough. That would increase the size of his already large 6,000 foot orbital airship to six kilometers in length.

It's the same for his transatmospheric airship. He says here

""The TransAtmospheric Ascender will be our first airship to reach space. The vehicle is 2100 feet long and will have a crew of three. It will have a peak altitude of 350,000 feet and a maximum velocity of Mach 5. Technically it is a hypersonic waverider.""

If those really are feet, then you have to take James' figure for the 6,000 foot long airship of 2.6 tons and scale it down by (2100/6000)^3 to get a mass of 111 kilograms as the most he could carry if it is buoyant at 200,000 feet. That wouldn't be enough lift to carry the mass of two average adults, never mind the rest of it.

But if the value is in meters, you get a more realistic 91.8*(2100/6000)^3 = 3.94 tons of lift for the ship, fuel and payload. It's volume then, scaling down James Fincannon's figure, is 361,237,500 * (1200/6000)^3 = 2.89 million cubic meters

You may throw your hands up in dismay at this point. But remember, it's very light for its size, at only 100 tons. After all weather balloons can be pretty huge and weigh little.

I don't think we should dismiss the idea just because the orbital airships would be huge

ACCELERATING TO ORBIT - FUEL EFFICIENCY AND SPECIFIC IMPULSE

This section is by way of background. A rocket accelerates to orbit in minutes The airships would take days according to JP's plans. However in both cases whether they can get to orbit depends on how much thrust the rockets produce, and how fuel efficient they are. 

The fuel efficiency is measured using its "specific impulse". This is often abbreviated as Isp or just written as ISP. This says how much mass of fuel is needed for a given change of momentum. It's defined as the ratio of the change in momentum to the mass of the fuel used to achieve that change.

If it's a conventional rocket and you measure the fuel used in units of mass such as the kilogram, and it's flying in a vacuum, then the ISP is just the exhaust velocity. If you fly the same rocket in an atmosphere, the ISP will be the same, but the actual exhaust velocity may be different. In that case it's the "effective exhaust velocity" which is defined as the ratio of the thrust to the mass flow rate.

However you may be puzzled at first to find out that the ISP is most often given in seconds. You don't often think of seconds as a way of measuring velocity! Well the way they do that is to measure its weight in newtons or pound force, instead of its mass in pounds or kilograms.When you divide a momentum by a force in newtons, you get a number measured in seconds.

Anyway the main thing you need to know is that you get from the specific impulse in seconds to the exhaust velocity in meters per second by multiplying by the standard gravity of 9.80655 - if you use meters as your distance measure. That's because one kilogram exerts a force (weight) of 9.80655 newtons in standard gravity.

Techy aside on standard gravity: Earth's gravity varies slightly depending where you are, reduced in equatorial regions because of the Earth's spin, counteracting it, also less as you get higher and varying depending on whether or not you are above a gravity anomaly, so they use a standardized gravity for calculations like this.

The Saturn V had a specific impulse of 260 seconds. So multiplying by 9.80655, this means the exhaust velocity was about 2.57 km / sec.

If you could somehow arrange for the fuel to be expelled at a velocity of ten times the velocity for the Saturn V, then it would be as easy to get to orbit from Earth as it was for the Apollo astronauts to take off from the Moon. You would only need around half of the mass of your spaceship to be fuel. As an example, if we assume a rocket with ISP 2000, starting mass 2600 kg, then the final mass on reach orbit is 1609 kg for ideal rocket equation. with less than half its mass used to get it into orbit. You can try out other ideas with this online rocket equation calculation

There is no way you can carry all the fuel for a Saturn V on an airship. So you need a higher ISP, at least if you need to have the fuel on board, and use the exhaust of a rocket style engine for thrust. And this actually is JP's plan, as we'll see. He says he has an engine in development that can accelerate his airships, much more slowly than the Saturn V, but with around ten times the specific impulse, so that it uses far less fuel.

Now ion thrusters can achieve much higher exhaust velocities. But they also need huge amounts of power. ESA is exploring a new type of ion thruster that in theory can achieve an ISP of 19,300 seconds - so an effective exhaust velocity of 190 km / sec, using xenon atoms. That's for "a single 20 cm diameter 4-grid ion thruster". See this paper. Also this page about their design. However it would need 250 kW power for 2.5 N of thrust. which is only enough thrust for 255 grams (2.5/9.80655) to hover against gravity.

You can get much more thrust and also high specific impulse using either anti matter or nuclear fission or fusion. The Orion pulse rocket design from the late 1950s which worked by exploding nuclear bombs one after another to get into orbit had a theoretical specific impulse of 10,000 to a million seconds, and would have been powerful enough to carry its own weight into orbit. Details here.

Another way is to use beamed power from the ground to a rectenna. This was the subject of Kevin Parkin's thesis in 2006. He calculated that you can achieve an an ISP of 700–900 seconds and a thrust to weight ratio of 50–150 by beaming microwave power to your spacecraft, which gives plenty of margin to get a payload into orbit. (You need a thrust to weight ratio of 1 to hover against gravity). See also Dmitriy Tseliakhovich in his thesis in 2010 and see also Jordin Kare's talks to the Space Show about power beaming.

JP Aerospace therefore have two main ways to overcome this problem of getting into orbit, assuming they don't use nuclear power or anti matter, and assuming we don't find some radically new form of propulsion like the EM drive.

  • They can beam power or collect vast amounts of solar power and somehow use ion thrusters powered by all that power.
  • Or, they can build a rocket engine that somehow has a much higher ISP than a conventional rocket engine.

The second one is what they are going for. They do have ideas to use solar power, but this is just to assist the dirty ion thruster.

It would need to have a higher thrust to weight ratio than a conventional ion thruster but it doesn't have to hover as that's done through buoyancy and then later on by lift. It needs a better thrust to weight ratio than an ion thruster, but it doesn't have to have a thrust to weight ratio greater than 1. So that's a rather unusual specification. Such an engine would be useless for a conventional rocket because it couldn't take off from Earth. Meanwhile, for an ion thruster the main requirement is to be as fuel efficient as possible, and so to have as high a specific impulse as possible, so it wouldn't be optimal for an ion thruster either. But it's what they need for this application.

He claims to have achieved such a rocket, with an ISP of 2000. It's called the symphony engine and it works by mixing the ions of an ion thruster with unionized gases.

He saves a lot of power by using the combustion of the fuel itself to ionize it. Whenever you light a fire, the flames consist of plasma and so they have ions within them which could be accelerated using an ion thruster. So that's his idea, and by accelerating those ions he also accelerates the burnt fuel that they are mixed in with as well.

So, I covered this in detail in the last article. One way or another he needs a high ISP motor to have any chance at all, at least, if the airship carries all its own fuel to get to orbit. You may have different ideas about whether he can do that.

However, supposing he can achieve a 2000 ISP motor, which he says he is well on the way to achieving with his engine tests, then would that be enough to get his airship into orbit? Well that leads us to another fun and neat calculation we can do.

WET TO DRY MASS CALCULATION

So anyway it is easy to work out the wet to dry mass ratio for a 2000 ISP motor that's using half of the thrust to overcome drag. We need to use the rocket equation. But it's simple to adjust it to allow for the need to offset drag. If you use half the thrust to offset drat, it has the same effect as doubling the delta v with an assumption of no drag. http://www.quantumg.net/rocketeq.html

So I just input double the delta v to orbit ( which we are taking as 7,844 m/s), or 15,688 m/s into the ideal rocket equation, assuming an isp of 2000, mass of, say 100 tons. It turns out that the dry mass you get to orbit is 44.9 tons. So you need 55.1 tons of fuel. Here is an ideal rocket equation calculator to check the calculations.

Techy aside. Note:, they have a default of 9000 meters per second for Earth to LEO in that calculator. That's due to atmospheric and gravity drag. The "Gravity drag" is the extra acceleration needed for a rocket to rise vertically out of the atmosphere when the optimal trajectory would be to accelerate horizontally. For instance Robert Braeunig did a Saturn V launch simulation in 2010, revised 2013 which he used to work out the gravity and atmospheric drag for the Saturn V. He worked out that it needed 9.194 km / sec delta v to get to orbit, of which 1.743 km / sec is lost to gravity drag and only 48 m/s to atmospheric drag (it also gained 390 m/s from the Earth's rotation). That makes a total of 1.791 m/sec drag out of 9.194 km /s or 19.48% lost, mainly due to gravity drag. See also stack exchange discussions here and here.

The atmospheric drag is low for a rocket as the atmosphere doesn't provide that much of a barrier. It's equivalent to thrusting through 10 meters thickness of water vertically and the rocket is designed to be long and thin to minimize drag, unlike a plane, because there is no need for lift. The reason atmospheric drag is the main issue for the orbital airship is because it is using drag to offset lift, so can't be streamlined like a rocket. It is also traveling horizontally or nearly so, and the distance traveled through the atmosphere is far greater, but this is less of an effect. For instance, the density of the air is 1.225 kg / m3 and there are 10,300 kg per square meter. So that means the atmosphere is equivalent in mass to an atmosphere of constant pressure of 1 bar of height 8.4 km. We calculated that with a lift to drag ratio of 70, the distance traveled through the atmosphere is 219,596 kilometers. So , supposing they achieve that much of a ratio and it is able to fly at 350,000 feet at a pressure of 0.0001 millibars then 219,596 km corresponds to the equivalent of flying through an equivalent of 2.2% of the mass of Earth's atmosphere. So, at least going by JP's figures for its operating altitude, then it actually travels through far less atmosphere than a rocket in total. If it was streamlined like a rocket that would have only a tiny impact on the delta v. But it isn't and what's more it has to fly at an angle sufficient to increase drag to the point where the lift to drag ratio gives it enough lift to support its weight.

From this, it would seem that the fuel is not going to be a problem if he can achieve a 2000 ISP engine and he can achieve the other things necessary to get to orbit.

This depends on the details of how his engine works. If it needs massive amounts of power to achieve that 2000 ISP thrust then that would change things. Even then, he does have a lot of solar power available. He would just need to cover a fraction of the square kilometers of his airship to achieve . If you have, say, 100 watts per square meter (probably would be more than that), that's a hundred megawatts per square kilometer. That depends on low cost robust high quality thin film solar cells however, and you have to transport the power from the cells to the engine, and so on.

Anyway I discussed this in the last article. .

STRENGTH AND STABILITY OF THE AIRSHIP

First, often physicists in the online conversations assume that they are launched from ground level. No, they are not. That would be impossible for these lightweight kilometer scale airships. The plan is to build the "Dark Sky Station" at 140,000 feet and to construct them there (so named because it is high enough in the atmosphere to have dark skies).

It's true that their smaller airships are launched from ground level but these are only ever used to shuttle up to the dark sky station, The orbital airship would fly a up from the dark sky station, and so can be made of lightweight materials and doesn't have to withstand gusts of wind etc. It would be neutrally buoyant up to 200,000 feet.

The outer envelope of the airship would be made of a very thin film. The JAXA balloon used a polyethylene film only 3.4 microns thick which they developed in 1998. At a density for high density polyethylene of 0.975 grams /cm3, that suggests a density of around than 0.00033 grams per square centimeter, or about 3;3 tons per square kilometer of surface area. It's amazingly strong though, with a breaking strength of 400 kilograms per square centimeter, so might be overkill though for his airships. Also it's from 1998 so they may have better materials now.

So the outer skin doesn't need to be that heavy. Let's try an over estimate first. I'm not sure of its total surface area. But if we just use the area of a cylinder - estimate it as a cylinder 12,000 meters long and 300 meters in diameter, then this is surely more than is needed. That gives a total surface area of PI*12,000*300 / 1,000,000 square kilometers or about 11 square kilometers. On the other hand as an under estimate, use a flat two sided sheet of area 12,000*300- 300*300 (for the overlap between the two rings approximating as a square), which gives it an area of double that, or 7 square kilometers (more exactly, 7.02 square kilometers). So it's somewhere between 7 and 11 square kilometers.

At a density of 3.3 tons per square kilometer, the skin's total mass would be somewhere between 23 and 36 tons. That's quite a lot out of just under 100 tons of total mass, especially as it is not taking account of the lifting cells, likely to at least double those figures and remember that with a 2000 ISP engine, more than half the mass will be fuel. Since then they have reduced the thickness to 2.8 microns for their successful test in 2013.

How thin can it get? If he can use a skin less than a third of that thickness of 3.4 microns, a little over a one micron thick, that would be better, giving it a mass of between 7 and 11 tons for the skin. Inside he'd have the lifting cells too, and those would easily double the mass or more. With his airship at just under 100 metric tons total, with 55.1 tons of fuel as we calculated it (assuming 50% thrust to offset drag and a 2000 ISP engine), and 18.15 metric tons of payload, then if the skin plus airbags were 20 tons, say, it would leave only 2.45 tons for everything else.

If we could get down to half micron thickness skin, even better, then we gain 11 tons for the skin, and the numbers change to 13.45 tons for everything else which is rather better. Remember the JAXA skin is very strong, with a breaking strain of 400 kilograms per square centimeter. The fabric of the orbital airship has equal pressure inside and out, and the lifting cells inside it would have an internal pressure of only 2 millibars as we'll see.

Those are very much back of the envelope calculations, but it shows the idea. The skin is likely to take up a lot of the mass of the airship even at the six kilometer scale. A thin skin, less than a micron in thickness, seems to be essential for getting his idea to work.

If we try it for his transatmospheric airship, then the area scales down as the square, and the volume as the cube. So the fraction of the buoyancy needed for the skin will get larger. Now the surface area is between 0.86 and 1.35 square kilometers. The total buoyancy is 3.94 tons (which we calculated above). If the skin has a mass of three tons per square kilometer, as for the JAXA balloon, and we have a similar mass for the lifting cells inside, then this now is impossible, as the total of skin plus fuel already goes beyond its lifting capacity.

If we can have a skin only one micron thick, and if the lifting cells also mass one metric ton, then around half of the buoyancy would be needed to lift the mass of the skin and cells, and we would be left with around two tons for the fuel plus crew, life support and everything else. For a transatmospheric airship then Mach 5 is 1.715 km / sec. So assuming 50% to offset drag (of course he could improve on that) it would require only 16% as the fuel fraction, using the rocket equation with ISP 2000 and delta v 3430 meters per second, so that makes it around 640 kg of fuel. That might well be feasible.

Aside: What about his diagrams dimensioned as feet? Well for a 6,000 foot airship, that's 1.8288 km, so to have a reasonably comfortable allowance for the skin to get to orbit, then we can just scale down all the figures for the 6,000 meter airship (for the same percentage of lift as skin, the skin volume has to scale with the airship volume so the skin scales linearly along with the other dimensions). So then, it would need a skin thickness of about 0.55 microns, for a reasonable payload of about 20% of the 2.6 tons or 460 kg to orbit ,which might perhaps be useful for a robotic early test vehicle.

For his 1,200 foot airship, or around 366 meters, then just to be buoyant at 200,000 feet the skin has to be very thin. Buoancy lift 111 kg. Just scaling down the dimensions and using our lower bound its wings would be 18.3 meters wide. So that makes the area at least 2*(2*366*18.3- 18.3*18.3), or at least 26,000 square meters. Double that again to take account of lifting cells and it's over 50,000 square meters. A one micron thickness skin would be 0.0001 grams per square cm or around one gram per square meter. So that would make the mass of the skin 50 kg. With a fuel fraction of 16%, it still leaves some extra capacity. The total would be 43.24 kg for everything else including the engine. You could double that if you have a skin of thickness perhaps a fifth of a micron with total mass including lifting cells of 10 kg saving 50 kg on the skn. It needs a very low mass engine and a very low mass everything else.

Even at his Dark Sky station at 140,000 feet that works out as density 0.00261346 kg / meter cubed so a buoyancy lift of 26.66 metric tons which is going to be far too little to carry 18.14 metric tons to orbit. It leaves 8.52 metric tons for everything else. The skin alone plus lifting cell skins would take up about four or five tons of that if we assume 3.4 microns thick skin. If we assume 2.8 microns thick skin. .

So, unless I've made a mistake in these calculations, his transatmospheric airship would seem to depend on ultra thin materials, thinner than for the JAXA balloons at the stage they have reached so far in their development. If the skin can be as thin as one micron or thinner then it may be doable depending on the mass needed for everything else.

JP Aerospace are a company with a long view. So maybe the strong thin films they need will be available when they are ready to develop the transatmospheric airship.

Anyway, more about his design. He has hydrogen in the lifting cells and nitrogen in the gaps between the cells and the skin. His airship is also made up of airbeams for structural strength.

He talks about his hydrogen cells unrolling from a roller as they expand, and has tested that already in his balloon flights. By the time it reaches its maximum neutral buoyancy altitude, then the hydrogen balloons would be fully extended.

Above that, If I understand right, then after that, he is just relying on the skins of the internal spherical balloons to hold them in against a near vacuum once in space, as for the echo balloons, Their internal pressure would be only 2 millibars if there is equal pressure inside and out at 200,000 feet so it's not such a huge challenge to hold that in against a vacuum. The region between them and the outer skin is filled with nitrogen gas which he then vents as it ascends.

He providing significant buoyancy at that point, indeed he could just bleed out the hydrogen altogether and use a pure nitrogen balloon once he is going so fast that only lift is needed to stay aloft. It would depend how much of the mass of the airship is hydrogen at that point. The hydrogen would help if he needed to abort though, he'd have enough hydrogen to stay buoyant as it slowed down, so I think for safety reasons also he wouldn't want to bleed out the hydrogen - unless he had lots of it to spare. But if he has it in compressed state, it's no great benefit to bleed it out anyway, he needed to have the same mass in compressed form to make it safe to bleed it out from the balloons.

He needs equalized pressure inside and out for the outer skin V shape, but the lifting cells inside could be spherical.

He has many design features to make his airship stable. The cells prevent the lifting gas from bunching up in the front, with 20 lift cells per arm (page 110).

He has a kevlar deck along the length of the wing at the top, and a keel along the bottom made of carbon composite truss type structures and inflated air beams. That's from page 114 of his book. He then has inflated ribs that run along the inside of the leading and trailing edges of the wings to maintain its airfoil shape. He also has a carbon fibre strip running from nose to tail all along the V shape leading edge of the outer V, and an aluminium cap to stiffen the nose which also acts as a mounting for the electrical drag reduction systems.

It is possible to maneuver it - it doesn't have to be agile and is controlled by gas shifting and vectored thrust. For docking with the dark sky station then it would use the smaller first stage airship as a tugboat. All this is on pages 114-5 of his book.

WHAT ABOUT TARGETTING A LOWER ALTITUDE TO INCREASE BUOYANCY LIFT?

The other way forward is to make his airships neutrally buoyant lower in the atmosphere. If we make them buoyant at 140,000 feet instead, then this increases the buoyancy lift of his airship to 26.66 metric tons and with a skin of 0.652 square kilometers. which at 2.73 tons per square kilometer (if we assume the thinner 2.8 microns skin) would be 1.78 tons for the skin, and assume the same amount for the cells, likely to be an under estimate, and taking away 18.14 metric tons for the payload leaves 23.1 tons for everything else. But assuming 55.1% of the buoyancy lift as fuel we need to find another 14.69 tons so that leaves 8.41 tons for everything else including air beams, drag reduction technology, engine and life support. It seems a big ask.

To do it comfortably he needs 35 times the buoyancy which give the total buoyancy lift 100 tons. So then he needs air density of around 35*0.000254864 = 0.0089 kilograms per cubic mete. The figures work for neutral buoyancy at around 113,000 feet with an atmospheric pressure of around 6 millibars and density around 0.009 kilograms per cubic meterr (standard atmosphere calculator). In that case, I wonder if one could just scale everything down. If he could build his dark sky station at 60,000 feet or a little higher with the air pressure above the Armstrong limit, then humans could survive there with just an oxygen mask without a full body pressure suit which could simplify construction and safety.

However I'm not going to take that any further here as it is a radical change of his plan.

CAN HE ACHIEVE REASONABLE LIFT TO DRAG RATIOS AT ALL, EVEN 4 TO 1?

At first it would seem almost impossible. His airship looks so unaerodynamic compared to the planes and gliders we are used to, more like a balloon or an airship than a plane. If we read feet as meters, its 138 meters high! High as a skyscraper. How can that have a reasonable lift to drag ratio?

Well, first it is flying at a very low atmospheric pressure. At 0.0001 millibars at his target height of 350,000 feet for Mach 5, then a height of 138 meters then scaling the reference area by 0.0001/2 then it's the same drag as for a scaled down airship of height of 138*sqrt(0.0001/2) = 0.976 meters flying with an atmospheric pressure of 2 millibars.

(That's by the drag equation D = Cd * A * .5 * ρ * V^2 - where ρ there is the density. So the drag depends linearly on the area and also on the density. Reduce the density and increase the area by the same factor and the drag remains the same.)

So the drag is similar to that of an airship with a nose of height just short of a meter, flying at 2 millibars, and the whole airship reduced in dimensions in proportion of course. The drag is similar to that on a v shaped airship that is 42.5 meters long with wings 2.1 meters wide and one meter high with atmospheric pressure of 2 millibars.

Or scaling to full atmospheric pressure, then the ratio is sqrt(0.0001/1000). So the drag on the entire airship is the same as you'd experience on a small model 1.89 meters long, with wings 9.5 centimeters wide and 4.4 centimeters high. And that small amount of drag is then spread out over an airship that is 6 km long and 138 meters high with wings 300 meters wide.

It no longer seems so exceessive does it?

Also, most of us probably don't think of airships as capable of using lift, but aerodynamics can be surprising sometimes. Hybrid airships do use lift.

For example, the Airlander 10 is a lifting body, and look how tall and fat it is for its length? 

Airlander 10 photographed by Airwolfhound - photos from its test flight on 11th May 2017 here.

Compared to it, then JP's orbital airship is positively slim! In its proportions at least, of course the whole thing is scaled up to a larger size in all directions.

The Airlander 10 is heavier than air, though only marginally, but it does depend on lift to stay aloft, and that still lets them calculate a lift to drag ratio. This paper calculates the lift to drag ratio as 3.8:1 when flying at an angle of twelve degrees, see figure 12.

A lot of that apparently comes from its rather small fins.

It's a completely different situation of course, slow flight at atmospheric pressure compared to hypersonic flight in a vacuum, but there aren't any examples of hypersonic airships to use. We tend to think in terms of airplanes rather than airships as they are the only things we have that go hypersonic at present.

So, this is just to show that you can have airships which are tall, yet have rather surprisingly useful lift to drag ratios. You might not necessarily expect the Airlander to have useful lift from our experiences of normal planes.

His airship would go much faster than this, but also in air that is far less dense than for a normal airplane. Small details in the design of the orbital airship, even tiny fins, could make a big difference to the amount of drag, and the lift to drag ratio.

He doesn't have a rigid skin or a stressed skin. You might think again, from analogy with ordinary aircraft, that he has to have a rigid skin to go at hypersonic speeds, or at least a stressed skin. But there's a precedent here,

You might think again, from analogy with ordinary aircraft, that he has to have a stressed skin to go at hypersonic speeds. But instead he has an unstressed skin - and it's not even pressurized from within like a Zeppelin but has equal pressure inside and out.

The Skylon however uses a very similar approach - on a smaller scale of course. It’s girder-like with a thin glass ceramic outermost shell, which is just a heat resistant covering and doesn’t take any stress at all. This is similar to the design of a zeppelin or a small plane. Unstressed skin over rigid girders.

Structure of the Skylon - internal truss framework made from carbon fibre reinforced plastic composite held together with Kevlar ties. It has aluminium propellant tanks suspended inside it. Covering that, it has a thin outer aeroshell of a high temperature silicon carbide fibre reinforced glass ceramic material. For details see page 2 of this report

This ceramic outer skin is black, which is why Skylon is shown that colour in most of the artist renderings.

JP's airship has a similar design, except the girders are airbeams to reduce the weight. Although it goes at hypersonic speeds, it's in a hard vacuum.

It does this with an airship which a huge amount of drag at lower altitudes. That drag is actually an asset if he can achieve a reasonable lift to drag ratio, as it would lift his airship up to altitudes where the forces acting on it are much lower. Since drag depends linearly on the density of the air, then anything that can fly so high in the atmosphere has to have huge amounts of drag at lower altitudes. For instance. at any given speed, it has over a thousand times as much drag at 200,000 feet compared to its drag at 350,000 feet. It's because it has so much drag at low altitude, combined with an aerodynamic design which is intended to have a large lift to drag ratio, that it isn't limited to the maximum altitude of an airplane, its Kármán line is higher

Looking at a worst case scenario for the drag, suppose it has only a 4 to 1 lift to drag ratio for a 1,000,000 newton airship, so that it has 250,000 newtons of force acting against it by way of drag. But that's spread over a vast surface area.

For a "back of the envelope" calculation, let's call the surface area over 270,000 square meters (reading feet as meters again) - there I just multiplied the 6000 meters by 2 and by the width of 300 meters, and subtracted 300 * 300 for the overlap between the two wings, as an estimate for the area for only one side. It under estimates because the cross section is oval, not flat,

That would make the force on the skin due to the drag at hypersonic flight around 0.926 newtons per square meter. It's the same effect as resting around 926 grams on a square meter of the material. Remember the JAXA skin can withstand a breaking strain of 400 kilograms per square centimeter. It would seem to be way overkill for this application, given equal pressure inside and out. There would be more force due to lift, and that would be of course the full 1,000,000 newtons, which then works out at 3.7 newtons per square meter or equivalent to resting 370 grams on a square meter of material.

Of course it wouldn't be spread out uniformly and there would be areas with much more force acting on them. However we do have plenty of leeway, including those airbeams spanning the ship, and parts of the airship that have to withstand stronger forces like the nose and leading edges, he says would be reinforced with carbon fibre.

HYPERSONIC WAVERIDER - CAN HE INCREASE THE HYPERSONIC LIFT TO DRAG RATIO ABOVE THE VALUE OF 4 FOR SKYLON

His airship is optimized not just to reduce drag, like a normal airship, but to maximize aerodynamic lift. It's actually optimized for maximum lift during hypersonic flight, he tells us, as a hypersonic wave rider (page 109 of his book).

"When the orbital airship reaches Mach three, it enters the low end of the hypersonic phase of the flight. This is where the vehicle has been tuned to fly. It's where it really shines. The airship is now a hypersonic waverider. A hypersonic waverider is a vehicle that takes advantage of the hypersonic wave. When traveling at hypersonic speeds, large shockwaves form around the airship. The shape of the wing is designed to ride on top of the shock wave. The orbital airship is the surfer of the upper atmosphere"

So what is a "hypersonic waverider". I was not familiar with the concept until reading it in his book and must confess to know knowing how it works in detail. There is a lot of publiched literature on it but most is very technical and assumes you know what a waverider is. But it's a well established idea. Here is one description:

"Simply put, a waverider is any vehicle that uses its own shock wave to improve its overall performance. The concept is probably most familiar from the world of water sports-motorboats ride the bow wave they create to reduce friction drag and surfers similarly ride the crests of waves. An aircraft traveling at Mach 1 or higher also produces a wave, a shock wave of air. If the aircraft is tailored correctly, it can be designed to ride this wave to produce greater lift, less drag, greater range, and overall improved performance. The reason waveriders are only practical at higher Mach numbers is due to the fact that the shock wave must remain close to the surface, a quality of hypersonic flow. A shock creates greater pressures in its wake, and the shock lying close the lower surface of the aircraft results in a large pressure force increasing the lift on the vehicle. This idea, which will be explored further in later sections, is known as compression lift and is the primary benefit of the waverider concept."

Here is the wikipedia page about it with lots of links. It depends on "compression lift" which is described here


Described in wikipedia as:

" At the time, Nonweiler was forced to use a greatly simplified 2D model of airflow around the aircraft, which he realized would not be accurate due to spanwise flow across the wing. However, he also noticed that the spanwise flow would be stopped by the shockwave being generated by the aircraft, and that if the wing was positioned to deliberately approach the shock, the spanwise flow would be trapped under wing, increasing pressure, and thus increasing lift."

It's described in this paper as:

"It is generally taken to refer to a wing which generates a shockwave attached to the leading edges and generating lift in a high-pressure area below the vehicle."

Which basically explains the idea as I understand it. When a plane is traveling slowly at supersonic speeds then the shock wave doesn't come near the body. But as it gets faster then the shock wave is pushed back against the body. The idea of a wave rider is to trap that shock wave underneath its body so creating an excess of pressure under the plane which causes lift.

It doesn't generate lift from nothing. It's a more efficient configuration which takes the high pressure air which is caused anyway by flying at hypersonic speeds and traps it so it gets tucked up beneath the plane and this then means that the high pressure area - which normally is a long distance above / below the plane, is actually adjacent to it, which then generates lift from the pressure difference between the air below the plane and the air above it. For this to work you need zero to minimal flow from below to above the aircraft.

That is not how lift is normally generated of course. This relative pressure dfference between the shock wave trapped below the plane and the thinner air above it acts in addition to the normal lift due to dynamic pressure from the velocity of the air hitting the plane.

This provides extra lift at any altitude so increases the lift to drag ratio, and also reduces drag, as the shock wave normally slows it down - letting it fly at a higher altitude. By increasing the lift to drag ratio, it reduces the drag. It can only do that at hypersonic speeds, fast enough so that the air that hits the skin flows next to the skin. John Powell says of his airship that it can only do that above Mach 3.

So there, remember that his airship requires only 3.7 newtons per square meter to offset its weight (taking its weight in force units as 1,000,000 newtons), and is very low density, with the lifting bags inside pressurized to 2 millibars. Since the pressure from the shockwave would be acting upwards only, it doesn't need to be much to offset the weight of the airship. It doesn't have to be as much as the neutral buoyancy pressure of 2 millibars at 200,000 foot that its internal lifting cells will have to hold in if they keep the same amount of gas within as it rises to orbit.

If the over pressure from the shock wave was dense enough, it could offset all of its weight as lift. So let's work that out. Atmospheric pressure is 10.3 metric tons per square meter, or around 101,000 newtons per square meter. So his 3.7 newtons corresponds to an overpressure of 0.0366 millibars. Just a tiny overpressure in the shockwave would be enough to offset the entire weight of his airship.

Note. This is not how lift normally works. But I think if I understand it right, then it may be part of how the extra lift of a hypersonic waverider works.

Skylon uses a similar approach to achieve higher L/D ratios. See page 16 of this paper about Skylon:

"Figure 19 presents the pressure field around Skylon as viewed from the top, at different Mach numbers. As the Mach number and spaceplane altitude are increased and the dynamic pressure is decreased, relative pressure near the aft end of the Skylon fuselage increases, which in turn favorably affects the airframe force coefficients by both increasing CL and reducing CD."

There CL is the lift and CD is the drag. So the increased relative pressure beneath the aft end of the fuselage reduces drag and increases lift.

Techy note, it's not the dynamic pressure due to the speed of the air hitting the airship, but the relative pressure, due to difference in density of the air below and above the airship. It's an effect that increases as the speed increases, as they say: " As the Mach number and spaceplane altitude are increased and the dynamic pressure is decreased, relative pressure near the aft end of the Skylon fuselage increases," I think this is only a significant effect for hypersonic flight.

It's the same basic idea as the waverider, that an overpressure beneath the plane causes extra lift and reduces drag, so increasing the L/D ratio. However if you read it in contect, in this case the overpressure comes from the nacelle plumes not from the shock wave.

However there is a proposal called "Vulcan" for a waverider version of Skylon. It would be able to achieve a L/D ratio of over 7 from Mach 3 all the way through to Mach 8.5 which is where the table ends. See Table 3 of this document: It also gives cites to earlier work. This also confirms that Skylon is not a waverider - but it shows how much of a difference the waverider approach makes, at least doubling the L/D ratio for hypersonic speeds.

It's by Kay Runne, Aerospace engineer with several aerospace companies and with DLR who has now retired and is writing about waveriders in his retirement. His home page, which links to the document is here, It's not the same design, just using engines based on the Sabre design with some modifications to let it operate at higher Mach numbers.

COULD THE ORBITAL AIRSHIP ACHIEVE BETTER THAN A DOUBLING OF THE L/D RATIO AS A WAVERIDER?

So what about the orbital airship? Could it achieve an increase in L/D ratio of more than this typical doubling using the waverider effect?

It's not so much the volume density of the airship that I think might make a difference to the waverider effect and perhaps make it even more significant than it is for normal hypersonic flight - as the very low areal density over its lower surface that would make a big difference to the waverider effect. That's just an intuition - that if the overpressure is able to offset much of the weight of the airship just through relative pressure surely that would have a significant effect on the L/D ratio. However, I don't know how to estimate L/D ratios based on waverider overpressures or how to estimate the overpressure in the first place.

If anyone reading this is an expert in hypersonic waveriders, please comment and correct if I've got anything wrong here! And if you have any idea about some way to take this further, do say!

The conditions here don't seem particularly extreme. If you look at it another way, then it's like firing a gas at a pressure of 0.0366 millibars at Mach 5 over a thin film in a vacuum pressurized to below 0.0001 millibars with the skin able to withstand a breaking strain of 500 kilograms per square centimeter if we assume the JAXA skin though it would probably be thinner than that.

Now, I don't know how that translates into a lift to drag ratio. But it does suggest I think that we are in a regime here that is very different from the normal situation of an airplane or even a normal airship. Conventional ways of thinking could easily lead us astray here.

DRAG REDUCTION

The nose of the airship is titled slightly up which he says "modifies the hypersonic wave and allows most of the vee wing to fly at a flatter angle, reducing drag".

Then, there's an aluminium cover to the nose, on which he'd mount the drag reduction machinery. He would have plenty of power from solar power, at least for as long as it is in sunlight.

Examples of it here: See Figure 6 Electrohydrodynamic drag control.

And here: Plasma Injection for Hypersonic Blunt-Body Drag Reduction

The drag reduction seems to work by moving the position of the shock wave away from contact with the "nose" of the airplane. I can imagine that might well make quite a difference to such a blunt nosed vehicle as an airship traveling at hypersonic speeds.

So, he plans to use this technology to elevate the shock wave away from the nose, then it hugs the airship below the wings, and it is designed so that the shock wave goes below rather than above the airship as it flies, which adds to lift and reduces drag. At least that's his idea, and he describes it clearly in his book.

WIND TUNNEL

It's very hard to test hypersonic models in a wind tunnel because the speeds are so high. The USAF operates a facility Hypersonic wind tunnel 9 which can simulate hypersonic flight up to Mach 14, and it can simulate flight altitudes of up to 173,000 feet and flight times of up to 15 seconds.

Most research is done with tiny models in shock tubes or shock tunnels or expansion tunnels. JP faces even more of a challenge if he wants to test the actual regime his airships would fly in, because he would have to simulate hypersonic speeds at 350,000 feet and higher.

He does have a tiny high speed "shock tube" type wind tunnel, mentioned here on the SpaceShow, capable of tests up to Mach 3.8 so far, was designed for Mach 4 but can't reach it yet.

I had one idea here, just sharing it in case it's of interest. His 2,000 ISP motor which he has in development would have an exhaust velocity of 19.6 km / sec. That's more than twice orbital velocity. Since he needs to simulate hypersonic speeds, but in vacuum conditions of only 0.0001 millibars, simulating 350,000 feet - could he do that simply by firing his rocket motor through a tube in a very hard vacuum chamber?

It's technically possible. Supposing he could have access to the Space Power Facility, a big ask of course, then it can get down to a little over 0.000001 millibars, so two orders of magnitudes harder than he needs. It's 100 feet in diameter and 122 feet high. There would be room there for rather large models, though of course, not anything approaching the size of his orbital airship.

Visitors inside NASA's Space Power Facility. This can achieve vacuums of a millionth of a torr (a bit over a millionth of a millibar). Could a large hard vacuum chamber be used for simulating hypersonic flight in near vacuum conditions by setting up a tunnel and just firing his 2000 ISP "dirty ion thruster" at very low thrust, if that's possible, to simulate the relative velocity of the air for his high velocity travel through a vacuum? Just my own idea here :).

CENTRIPETAL FORCE

By the time he reaches 6 km / sec he gets more than half the weight of his airship counterbalanced by centripetal force, so could move higher in the atmosphere with the same amount of aerodynamic lift. And by the time he reaches orbital velocity - then he may be spiraling out quite high in the atmosphere, but only for the last few hours towards the end of the journey I'd imagine because it is going up as the square of the velocity, over more or less fixed radius, a little over the radius of the Earth for the centripetal force calculation.

MAINTENANCE AFTER BREACHES IN THE HULL

Holes in the outer skin won't matter much as the pressure is equal inside and out, so as long it is non rip, maybe it just ends up having holes in it.

With the lifting cells inside, then it would matter, even though they are only at 2 millibars pressure. But you wouldn't need duplicates of all of them. How about a few backups stored in one place, with some method to move them automatically to the place where they are needed and to replace the leaking balloon in situ, and return it? It could even if ingenious perhaps compress the leaking balloon in such a way that the hydrogen is transferred to its replacement?

I did have one idea about maintenance of the outermost shell without having to deflate the airship completely, his Dark Sky station shaped like a starfish - there'd be room for a V shaped orbital airship under a pair of its arms - so then they might have access to it from above, from air beams running along the arms and then ropes down from them. It seems a bit precarious, but if you used telerobotics instead of humans to inspect and repair it might work.

TESTING HIS WAVERIDER CONCEPT IN SPACE

I had one idea about this,though it may seem obvious. The waverider concept is a bit like the idea of a "boost glide" which the Apollo command module did during re-entry, "half bouncing" of the Earth's atmosphere to slow down its speed of re-entry. A spacecraft can do a complete skip, bouncing back into space and then do multiple skips before re-entry. But it can also enter just on the edge of a skip trajectory and turn that into a long glide through the upper atmosphere.

Anyway - that lead to the idea - his waverider would work just as well during re-entry - as a way to slow down the re-entry and reduce temperature loads and stay higher longer. So - why not test the re-entry first? That would be a far simpler construction as he wouldn't need to have engines, after all the command module didn't have engines.

He could start with small models - but still far larger than you can test in even the best facilities on Earth. He would need to pay for a flight to orbit, but maybe he could start with a small model small enough to package into a cube sat? And inflate after it reaches orbit, to only 2 millibars similar to his orbital airship?

Then he could test it and find what its lift to drag ratio is in practice. Of course, this is expensive. But especially with costs going down, it may be within reach. Similar to the Planetary Society project to send a solar sail satellite into space.

AIR BREATHING

James also suggested the idea of an air breathing engine which we discussed. His idea is something to do with using the air for propulsion for electrical propulsion. But due to a misunderstanding I thought he meant an air breathing engine like Sabre using the oxygen for combustion.

Anyway it was a productive misundestanding and perhaps it might be interesting to look at that idea. I'd say that this makes no sense for an engine with ISP 2000 because you have a decent fuel fraction anyway - and I don't see how all the extra infrastructure to make it air breathing, a major technological challenge at 0.0001 millibars, is worth doing.

However, it might make a lot of sense if the ISP is only, say, 222 (before you take account of the air breathing benefits). Especially if you can use hydrogen, requiring a ninth of the fuel mass (instead of hydrogen + oxygen molecular mass 2 + 16 you have just the hydrogen, molecular mass 2).

That means that you need a ninth of the mass flow, to achieve the same change in momentum of the spacecraft, so multiplying the ISP by 9.

Hydrogen of course has its problems, that it has to be kept liquid, or else, that you have to have strong balloons to hold that much hydrogen.

However, he has a lot of space inside his airship. Suppose 50% of the mass is hydrogen, targeting an ISP of 2000 as before, with a volume of 2.89 million cubic meters for his 1.2 km transatmospheric airship, lifting just under 4 tons to 200,000 feet, Then, suppose 2 tons of that mass is hydrogen fuel, in a space of 2.89 million cubic meters. Or about a ton to every 1.445 million cubic meters. So a density of one kilogram to every 1,445 cubic meters. Or a density of 0.00069 kg / cubic meter.

At atmospheric pressure hydrogen has a density of 0.0899 kg / cubic meter.

So this would give it a pressure of 1000*0.00069/0.0899 or 7.7 millibars. That doesn't seem hard to contain even in a vacuum or outside pressure of 0.0001 millibars (his probable Mach 5 altitude), as his lifting cells have to hold in 2 millibars anyway (neutral buoyancy pressure at 200,000 feet) - unless he pumps the hydrogen out of them for storage.

So he could just keep the hydrogen fuel as over pressure in the cells. The volume would be a bit less than the amount I used there, as he has spherical cells inside a non spherical V shaped airship, but still, we have plenty of margin here. .

So, it's not the hydrogen fuel storage that would be the issue. Rather the air breathing. I think I'll just have to be "hand waving" there.

If he uses air breathing, he can achieve his 2000 ISP easily. Skylon has a specific impulse of 40,000 to 90,000 newton seconds per kilogram in air breathing mode (That's from their manual table 2 page 6) because of the way it can use oxygen from the atmosphere. So, dividing by 9.80655 to convert to seconds - well not really much point in working it out to four figures, it's over 4,070 seconds ISP.

There, they are probably using the turbojet effect as their Sabre engine has a spinning turbine, which is how turbo jets achieve such high ISPs of around 3,000, using the engine to accelerate inert air backwards.

So, air breathing could easily achieve ISP 2,000 or more. If he could achieve an ISP of 4,000 then that reduces the mass fraction to orbit even more. It's then got a fuel fraction of only 33%, by this online rocket equation calculation again, inputting 15688 for the delta v (assuming 50% of thrust used to offset drag).

The Sabre engine has way too much thrust of course. It's able to accelerate 325 metric tons of spacecraft + payload with a lift to drag ratio of around 4. That's more than three times the mass of the JP orbital airship. But if you scale it down it would have the right level of thrust. Suppose that you target a lift to drag ratio of 70. Then the Sabre would need its thrust reduced by 3.25 * 35 or a little over 100 assuming 50% thrust to drag ratio. As a ballpark figure. I don't know what the thrust depends on. If it is the intake area then that's a 1/10 scale model. If it is the volume, a 1/5 scale model. Either way, a scaled down Skylon would do the trick. That is - if it can be scaled down (probably can) but more importantly - if it can achieve air breathing at 0.0001 millibars.

However, I doubt that you could just as it were "drag and drop" a Sabre engine into the orbital airship. How could he funnel the air at such low pressures sufficiently to feed something like Sabre? I am not sure it is worth going into details here - to do that I'd need to be an aerospace engineer, it's not my discipline.

I think it might well be quite a challenge for an aerospace engineer too, to design an air breathing engine to work at an atmospheric pressure of 0.0001 millibars at Mach 5 and presumably decreasing pressure at higher speeds - the 0.0001 millibars is for his operating altitude of 350,000 feet which he gives for the transatmospheric airship when it reaches Mach 5.

But I think it's worth a mention in case it suggests something to someone.

ECONOMICS

This is not my strong point but just to say something about it.

I think it's hard to know if it would go the way of the airplane (huge financial success), or the airship (has some commercial value today) or the hot air balloon (of entertainment value but not of much else by way of commercial value). It might also depend on technology we don't have yet, to make it profitable, as with the early planes.

At the moment, they are like the Wright brothers before their first flight,. If they get as far as their transatmospheric airship, people will really start to pay attention. That would be a bit like the Wright brother's first hop - where they show they can actually get to hypersonic speeds. I don't know if that is possible.

The airship able to ascend to 140,000 feet from ground level seems almost within reach and they are getting to higher altitudes with each test. If I understand right, it seems to be a scaling issue that if they can build it large enough then they could reach that height. At that point, it will be like a new version of the airship they built for the USAF over a decade ago in 2004. Beyond that, nobody else is attempting anything like it, so it will be interesting to see how far they get.

The challenges seem formidable. But then they also seemed so with early flight and many other inventions, and we regularly get inventors achieving things that seem impossible on the face of it. For instance Skylon's SABRE rocket engines have now been demonstrated as able to cool the incoming air from 1000 °C to -150 °C in a hundredth of a second, and do so what's more without any problems of condensation and freezing chunks of ice.

I would hardly credit it, that it was possible, if it weren't already demonstrated and verified by independent observers - and the result leading to millions of pounds of funding from ESA and then from the UK government. The technology is secret, though we know it involved 3D printing to make components that can't be made in any other way.

So though it seems impossible, I don't know if it is seemingly impossible in the sense of the SABRE engine or really impossible. There is nothing to go on in the peer reviewed science journals - as they haven't published details like that, which again is a case similar to Skylon. There is no paper you can read that explains in detail how the SABRE rocket engine works.

So is the JP case like that, that it's possible but details not published for commercial reasons like Skylon, and also perhaps not fully worked out, or is it really impossible?

OTHER IDEAS

First the reason for my interest in this project isn't really that I think JP Aerospace are necessarily going to be the way that we succeed in reducing costs to orbit eventually. It's a long term project, and we will just see how far it goes and where it goes. They say they haven't come across any show stoppers yet. Even just getting these Ascenders to higher and higher altitudes is interesting, and when they start doing hypersonic glider tests and eventually the transatmospheric airship, they will be going into very new territory.

It's also intellectually interesting, and it's asking questions about a regime of flight that we simply have never explored in practice at all. So it will be interesting to see how practice lines up with theory and what new inventions are made while exploring this regime.

I think we should pursue all the possible avenues and after all JP is a private company, so they are free to follow whatever line they want to and are not depending on tax payer's money so they don't have to justify what they are doing to the rest of us. So, I'm cheering them on, to see how far they get and what they achieve. If I was in the US and within reach to do so, I can imagine wanting to volunteer to help them. It's such a fascinating project.just for what they have achieved already and should achieve in the near future.

Also, what they produce could be part of the mix at some point.

Meanwhile, Skylon I think can lead to low cost transport to orbit over a shorter timeframe, as soon as the mid 2020s.

There's also the idea of the orbital tether - which can be built with present day materials (unlike the space elevator) and also can get you into space much more quickly than a space elevator, and would pick up a Mach 12 or faster space planet and boost it direct to orbit in one go.

Skylon with its lower costs to orbit could help us build an orbital space tether by bringing its materials to orbit at low cost. That's the Hypersonic Airplane Space Tether Orbital Launch (HASTOL) System. It assumes a payload of 14 metric tons delivered to a speed of about Mach 12.
http://www.tethers.com/papers/HASTOLAIAAPaper.pdf

It's quite massive but not extraordinarily so, from 30 to 100 times the payload mass, so, up to 1,400 metric tons therefore, for their assumed payload of 14 tons, and made of conventional materials ,not needing new materials., they say there that

"The total mass of the space tether plus the Tether Central Station typically will be 30-200 times the payloads being handled. Most of that mass ratio requirement is driven by the fact that the tether system must mass considerably more than the payload it is handling, so that, upon pickup of the payload by the tether, the payload will not pull the space tether system down into the atmosphere. Thus, the advent in the future of better tether materials with higher strength at higher temperatures will not be used to lower the tether system mass significantly, but instead will be used to increase the tether safety margins, lifetime, and system performance, by allowing payload pickup at lower altitudes and lower speeds, thus decreasing the performance requirements on the hypersonic airplane portion of the system."

Skylon transitions to rocket propulsion at Mach 5.14, but it would need less rocket fuel to achieve Mach 12 from Mach 5 than to achieve Mach 22, all the way to orbital velocity, itself. Maybe it might even make sense for it or a similar space plane to continue air breathing through to Mach 12 and then get picked up by the orbital tether?

CONCLUSIONS FOR THE ORBITAL AIRSHIPS

Well, if those figures are right, the main conclusions are

  • His airships need to be rather larger than his figures suggest. It seems that we need to read the dimensions on his figures as meters rather than feet - if we do that, the numbers are about right. If this is correct, his orbital airship would be 6 km long and the transatmospheric airship 2.1 km long
     
  • He needs to have a very high lift to drag ratio of over 300 to 1 if he takes 3 days to get to orbit.
     
  • It's hard to say whether that's possible because the regime of an airship at Mach 5 in a vacuum at 0.0001 millibars flying as a hypersonic waverider on top of a shock wave in the vacuum is so unfamiliar and untested. Especially so, given that an over pressure of only 0.03 millibars would be enough to offset the weight of the airship.
     
  • If he can't achieve a lift to drag ratio anything like 300:1, if we suppose he achieves less than 70:1, then there seems no advantage in taking days to go to orbit rather than hours.
     
  • If he can achieve a 2000 ISP engine then (assuming 50% of the thrust used to offset drag), the fuel fraction would be only 55.1%, not unlike that for a commercial jet and it seems feasible to get to orbit. It's less than that if he uses less than 50% of thrust to offset drag.

    This all depends on whether he can indeed achieve 2000 ISP using only chemical combustion, and harvesting ions and accelerating them, and on how much power this needs. If he needs more power, he has hundreds of megawatts of power available from the surface area of the airship but this needs low cost robust thin film solar cells ideally only hundreds of nanometers thick deposited on the skin.

For more background, see my "Can Giant Airships Slowly Accelerate to Orbit Over Several Days"

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PLEASE CORRECT IF YOU SEE ANY MISTAKES IN ANY OF THIS

I hope you enjoyed these calculations. I know from experience how easy it is to make a mistake in something like this. If you see anything wrong with any of this, however small, or however major, please say.

And - what do you think about it? Do you have your own ideas and other calculations? Any other neat way at looking at the problems?

I welcome any comments, suggestions etc. Thanks!