LIFT TO DRAG RATIO CALCULATION

This is a simple argument I found recently which rather simplifies thought about orbital airships. It’s based on this idea of how the airship accelerates to orbit

• 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 well worth it because it saves so much fuel in the early stages when it is moving through the thicker lower atmosphere. But this lasts for only a small fraction of the total flight duration. The question of whether it can get to orbit depends on the final stage when it's flying as an airplane. 

When flying as an airplane then it’s the lift to drag ratio that matters most. So, 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.

It doesn't matter what you are using for propulsion for your spacecraft or airship - or what configuration it is, rocket, scramjet, skylon, airship, it doesn't matter what. The only assumption in this calculation is that it's accelerating to orbit using lift all the way until it reaches orbital velocity.

Our assumptions are that

Then 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 a g just to offset the drag.

Then, since 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.

Let's take its final orbital speed as 7900 meters per second (the details would depend on its altitude when it reaches orbital velocity).

Then using v = at, we can deduce the time it will take to reach orbital velocity as v/a = 7,900 / (9.807/4) = 3222 seconds or about 53 minutes.

So, Skylon, it seems, would take about 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 simplication 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.

If Skylon used only 25% of its thrust to offset drag, 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.

That's our general formula too. We'll assume 50% thrust to offset drag in these calculations and target orbital velocity of 7.9 km / sec. Then if the lift to drag ratio is L, then the time to get to orbit will be

7,900 * L / 9.807 secs.

I'll use links to google calculator for the calclations 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 2/70 g of thrust, and it would get to orbit in 7,900 *70 / 9.807 = 56388 seconds, or about 15 hours 40 minutes. (You can use this online tool to help convert seconds to hours and minutes quickly).

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.

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, we have

7,900 * L/ 9.807 = 3*24*60*60

So L = 3*24*60*60*9.807 / 7,900 = 322.

So to achieve such a slow acceleration to orbit, his airships need a rather fabulous lift to drag ratio of 322. 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 and is using more thrust to offset drag. But if you find yourself using more than 50% of your thrust to offset drag, 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 that, more than four times higher than our best gliders, 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.

LIFT CALCULATION

James Fincannon worked out the lift for his orbital airship based on its dimensions, and it turned out to be far too small for an airship of the dimensions that JP described for the masses he gives.

The calculation was rather simple. For the 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, then James worked out the lift from the published dimensions and from JP's statement that it would be neutrally buoyant at 200,000 feet, and his results just don't fit with this. He gets a mass of 2.6 metric tons as the maximum lift.

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 850,000,000 cubic feet or just over 24,000,000 cubic metres. This will of course be a huge over estimate but that's the idea for in this calculation. We know the density of air at 200,000 feet, at 0.000254864 kg / cubic metre, so that means it can't possibly have more thant 6.11 tons of lift, so no way it can lift 18.14 tons to orbit.

James works out the volume more exactly,using the figures from his 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, and he gets a mass of 2.6 tons in part 5 of his calculations.

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 metres in his orbital airship diagrams. If you take our calculations and multiply James' figure by the number of cubic feet in a cubic metre, 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.

So, well 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 lift is enough. That would increase the size of his already large 6,000 foot orbital airship to six kilometers in length.

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.