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Doubts If ATO will be working
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Moderator
Joined: Thu Jun 03, 2004 11:23 am
Posts: 3745 Location: Hamburg, Germany 
Hello, Lourens,
you are assuming numbers where no numbers are given by JP Aerospace  so what about using a function instead of one or a few special numbers? This I preferred in the threads in the Financial Barriers section about the Accumulation of a financial base in detail as well as in the thread about the Costs of the CXV. Then you could apply a lower boundary of number(s) and an upper boundary and provide a table with ther results for the two boundaries and some values between them. This would  from my point of view  better fit into the situation that JP Aerospace didn't provide enough numbers. I am doubting if their ATO has a mass or weight of 100 tons and what the mass really will be. Additionally it may be that the wings of the ATO aren't filled with Helium or Hydrogen but contain nearly a vacuum. Dipl.Volkswirt (bdvb) Augustin (Political Economist) 
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Space Station Commander
Joined: Thu Oct 27, 2005 7:44 am
Posts: 707 Location: Haarlem, The Netherlands 
Hi Ekkehard,
My intention was to show that an airship could reach 60 km in principle. People were saying that that would be fundamentally impossible, because you would be "out of the atmosphere". If a 100tonne vehicle can do it, so can a heavier one, it just has to be bigger. Exactly how large the JPA one will be wasn't really my concern. Unfortunately, I made a mistake in my previous post: I conveniently rounded JP's 60 km to 40 miles. However, 40 miles is 64 km, and according to my slide rule, at 60 km, air density is 18 Pa, not 10 Pa. The exponential falloff in air pressure magnifies the roundoff error. That translates directly into a total weight of 180 tonnes rather than 100 for the same volume at 60 km instead of 64, or a 1.8 times smaller volume for 100 tonnes at 60 km. I redid the math, this time using 18 Pa at 60 km. I assume that the images in the ATO handout are more or less accurate, and that the craft is 2 km long, so we get two cylinders of about 2.1 km length and 130m diameter. That yields a volume of 5.6 * 10^7 m^3 (actually a bit less because of the overlap near the nose, and the arms are probably not cylindrical either, so consider this an orderofmagnitude estimate). At 18 Pa, the lift would be about 15 tonnes. I also never checked what the balloon itself would weigh. It appears that the material weather balloons are made of is in the range of 45 g/m^2. A 2.1 km by 0.13 km diameter cylinder would have a surface area of 8.7 * 10^5 m^2, and if made from that material, a weight of 39 tonnes. That's not good; the craft would weigh 78 tonnes, and we only have 15 tonnes of lift. Polyethylene balloons do about 6.6 g/m^2, for a weight of 11.5 tonnes. Then we still need lifting gas, engines, and payload. It might just work as a technology demonstrator, but if we want to get a significant payload into orbit it will have to be larger, or we will have to start accelerating at less than 60 km altitude. As for boundaries, I don't see why there would be any. You can get any weight up there as long as your balloon is big enough. Of course, there are practical limits to building such a huge craft, but I don't see any fundamental problems. I'm not sure making a vacuumfilled airship would help. You would save the weight of the lifting gas, but if you use hydrogen then it's only 5% of total weight, and that doesn't include solar heating. Additionally, the structure would have to be stronger to support the pressure difference, increasing its weight. Of course, if you want to take the whole thing to hypersonic velocities, you may have to make it a bit stronger than a zeropressure balloon anyway, and then it could be useful. I've also thought about accelerating to orbital speeds a bit more. Both lift and drag are proportional to air density and to the square of the velocity. Here is a little "Gedankenexperiment": For waveriders, it appears that lift/drag ratios are about 46 depending on design.Let's assume for a moment that ATO really is a waverider, weighs 15 tonnes, has a fixed L/D ratio of 5, and that at 37 miles up, it's weight is entirely supported by buoyancy. If you go up 3 miles, pressure drops to half of what it was, so you lose half of your remaining buoyancy, and what you've lost needs to be made up with lift. However, lift is halved with pressure (multiplied by 0.5), and so is drag. Lift needs to increase by 50% (multiplied by 1.5) to compensate for the lost buoyancy, so we need to go faster. In particular, v^2 needs to become 1.5/0.5 = 3 times as large, so our velocity has to be sqrt(3) = 1.7 times for every time we go up 3 miles. In the drag equation, v^2 becomes three times as large as well, but rho is halved, so we need 1.5 times as much thrust. That makes sense, because it keeps L/D constant. It seems we need a lot of thrust to go fast enough. There is a third factor beyond buoyancy and lift though. Orbital velocity at 100 km is a little less than 8 km/s. When you reach orbital velocity, you don't need either buoyancy or lift anymore, because you go fast enough that you "fall around the earth". You just need to overcome drag to keep that velocity. The centripetal force required to perform a circular motion is Fcent = m v^2 / r. Gravity delivers Fgrav = m g, so if g = v^2 / r then it balances out and you are in orbit. If you go slower than orbital velocity, you need some form of lift to stay at that altitude. Assuming lift is pointing up, and gravity and centripetal force are pointing down, we get Flift = Fgrav  Fcent = m * g  m * v^2 / r = m (g  v^2 / r) The radius of the orbit varies by 3% or so between 0 and 200 km altitude, so I'm going to cut a corner and consider it constant at 6500km (which would be a 122km orbit). g = 9.8 m/s^2, and m = 1.5 * 10^4 kg. That yields Flift = 1.5 * 10^5  2.3 * 10 ^3 * v^2 N = 1.5 * 10^4  2.3 * 10^4 * v^2 tonnes equivalent Plug in 8 km/s in the above and you get Flift = 0, while at 0 km/s Flift is 15 tonnes equivalent, that is, the entire weight of the vehicle. At 4 km/s Flift is down to 11 tonnes, and with an L/D of 5, we'd need 2.2 tonnes equivalent thrust. Note that that is still about an order of magnitude more than the giant ion drive array I mentioned in the previous post, but it is much much less than your average chemical rocket. So, as you go faster lift increases, drag increases and centrifugal force increases. As you go higher, lift decreases, drag decreases, and buoyancy decreases. So, can we adjust our speed and altitude in such a way that we fluently tranfer our weight from buoyancy to centrifugal force? And how much thrust would it take? What if we adjusted our altitude and velocity in such a way that the drag (and thus required thrust) was constant? Would that be possible without falling? And if so, can we do it at ion drive thrust levels? 
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Moon Mission Member
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361 Location: Austin, Texas 
Lourens wrote: according to my slide rule Lourens wrote: At 4 km/s Flift is down to 11 tonnes, and with an L/D of 5, we'd need 2.2 tonnes equivalent thrust. Note that that is still about an order of magnitude more than the giant ion drive array I mentioned in the previous post 
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Space Station Member
Joined: Tue Dec 07, 2004 6:50 am
Posts: 265 Location: UK 
Lourens, well done for actually doing some calculations on this...
proofbyauthority : "is the weakest form of proof, especially where that authority is not established " however I think you will find the L/D arguments are solid Partial inflation at 42km: Helium/hydrogen is kept in underinflated cells, the outer envelope can be inflated with air (at extremely low overpressure) to make it semirigid below 64km. Approximately one seventh of the mass of the vehicle is helium, it would be practically impossible to carry this into orbit at RT due to material stress and permeatation alone. Hydrogen and helium liquification during flight are also utterly impractical. If you calculate the maximum allowable wall thickness for the postulated vehicle it is shockingly low. Large structural elements in compression are not practical in such a low density vehicle, all rigidity has to come from members in tension derived from gas pressure. The forces on the vehicle do not become excessively high at great speed, they are essentially bounded to remain practically within than the true weight of the vehicle, at higher speed the centre of pressure shifts forwards but otherwise much remains unchanged. Hydrogen rather than helium: I'm pretty sure this has been discussed, lots of chemical rocket thrust does not fit well with what is known about ATO. Hydrogen was apparently mentioned with regard to use in a fuel cell or external burning by JP Aerospace, but it makes little sense. However we haven't even got into the whole flying at night thing yet where you need phenomenal amounts of backup power. Resistojets running off liquid hydrogen seem to be about the most appropriate technology for propelling an electrical ATO scheme, but they need more power than is realistically available and would need to work over a period of hours rather than days. Importantly the actual energy involved in hydrogen burning(and hence ISP) isn't significant compared to what you can do with solar cells over several days. Particle accelerator: You can think of an ion engine as a kind of neutral particle accelerator optimised for thrust. Ion engine ISP is hard to get *low* enough while maintaining efficiency, needing very little fuel is only an advantage if you are using an expensive one like xenon or are very mass constrained. If you are energy limited increasing ISP beyond that of more ordinary ion engines is just about the last thing you want to do. To achieve a given deltaV with the least amount of energy you want the exhaust speed to be similar to the deltaV needed. Things like hydrogen arcjets operate in a more appropriate ISP range for ATO than conventional ion engines. BTW in a frictionless model, the increase of altitude to 150km can add much less than 1350 m/s to the total deltaV needed. The more altitude is raised at higher speeds as oppose to most of the altitude being gained early on, the more efficient the process (effectively limiting gravity losses). The lower limit when the vehicle pops up to altitude only after burning all its fuel is only about 120m/s which can be figured out just from energy considerations. This turns out to not be an issue at all since drag is overwhelmingly more important. I'm skeptical about drag and lift during acceleration : As you have discovered, drag calculations optimistic or otherwise for a specific altitude and speed are not hard given just frontal area of the vehicle. On the drag side, in effect the vehicle cannot produce less drag than a optimised nosecone of equal frontal area. If you notice that ATO does not have sharp leading edges you can see it is a very high drag wing in the vast majority of the required working speed range. But drag calculations alone turn out to not tell you very much at all, they simply state that with a given thrust to maintain a given speed you must be above a particular altitude. It cannot on its own say if you actually can reach a given speed or altitude or sustain it. To do that you need some estimate of the maximum available L/D. With any estimate which is close to being reasonable it is *immediately* obvious nothing is going to reach orbit over several days. If you pair ATO with any ion engine you care to mention (or any orbit capable electrical engine) you can notice that the L/D needed is ordersofmagnitude away from the possible. Transferring from buoyancy to centrifugal force: Buoyancy and weight reduction due to centripetal force can have no overlap, there is big region inbetween where they have no real significance in calculations. Things are such that you cannot even go supersonic at an altitude where lift due to buoyancy force is bigger than lift that can be generated through drag (with any functioning wing moving in an gas atmosphere), so at Mach 3 where centripetal acceleration is negligible, the optimum (and necessary) altitude also involves negligible buoyancy lift. You need to gain a huge amount of speed before centripetal acceleration becomes important and there's no apparent way for that void to be crossed within what is known about ATO. BTW. your reasoning for need for increased thrust at higher altitude is not entirely rigorous because while the *proportion* of remaining buoyancy lost remains constant, the absolute amount of extra lift needed every 3 miles is actually decreasing, (not exponentially increasing, 50% more every 3 miles) If altitude is relative to 64km in meters and rho0 is the density at 64km, A the frontal area in m^2, Cd the drag coefficient...you get something approximately like this for maintaining constant speed and altitude. 0.5*Cd*A*rho0*0.5^(altitude/5000)*v^2 = drag = thrust L/D*thrust=weight*(1.00.5^(altitude/5000)) so L/D*thrust/weight=(1.00.5^(altitude/5000)) In practice the amount of thrust needed converges to a fixed upper limit... thrust=weight/(L/D)*(1.00.5^(altitude/5000)) While the velocity needed carries on increasing until the centripetal acceleration becomes large enough to be important. v=sqrt(thrust/(0.5*Cd*A*rho0*0.5^(altitude/5000))) Over small ranges the lift (and therefore thrust) needed can be thought of as increasing linearly with altitude. With thrust in the range of conventional ion engines ATO gets stuck below 1km of altitude increase and at a low speed. Using other engines to get going will beat that performance, but ion engines won't manage to maintain any increased speed or altitude until you are very, very close to orbital speed. There are so many areas where ATO without levitation falls short it is hard to focus on any one with any prolonged seriousness. ATO intuitively will not work under any reasonable assumptions of propulsion or aerodynamics. It appears to require new physics or magic to function. 
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Spaceflight Trainee
Joined: Mon Aug 29, 2005 6:37 pm
Posts: 23 Location: Lake Charles, LA 
Lourens wrote: ...I found the arguments unconvincing. Most specifically, nobody gave any numbers, it was all a lot of magic handwaving and proofbyauthority. I'm not a rocket scientist, but the question of whether an airship can get to 60 km on bouyancy should be solvable even by a lowly computer science student such as myself I figured. And I can speculate on the rest . Way to go, Lourens! I just love the sound of numbers crunching. You may wish to review the power calculations that gave your array size, though. They're not incorrect as far as they go, but, IMHO, you started with an unjustified assumption about the type of rocket engine used. The only thing we know so far is that JPA will not be using an ion engine. Ekkehard's suggestion about looking at where you need to be in a range sounds useful for addressing this issue. I think an ATO can get by with a substantially smaller array just by using an engine with a more favorable thrust:power ratio. Also, you'll want to look more closely into what happens to aerodynamic lift and drag in the supersonic and hypersonic regimes. Saddly, you won't get the luxury of assuming constant coefficients for lift and drag at those speeds. Hypersonic calculations are particularly hairy, because energy is lost to induced chemical reactions in the air on top of everything else that saps lift in the supersonic regime. If all that extra energy from more speed is going into ionizing the air stream and melting plastic, then it isn't going into more aerodynamic lift. If anything can make an ATO physically impossible, my guess is that hypersonic flight conditions will be it. Nihiladrem, I can't agree that the L/D arguments against the ATO are solid, because aerodynamic lift alone clearly never supports the vehicle’s entire weight. The only universal requirement for sustained flight is that the sum of vertical force components be greater than zero. Many of the assumptions necessary to get all the way from “Fy > 0” to “D/L > F/W” do not apply to an airship. An argument cannot be solid if its underlying assumptions are physically meaningless for the case in question. Lourens’ approach of only requiring Fy > 0 is much more promising. _________________ “The next generation of engineers, astronauts and scientist are not going to appear out of thin air. They need to be inspired and educated, and the best way to do that is to get them involved.”  John M. Powell 
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Moon Mission Member
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361 Location: Austin, Texas 
nihiladrem wrote: Ion engine ISP is hard to get *low* enough while maintaining efficiency nihiladrem wrote: To achieve a given deltaV with the least amount of energy you want the exhaust speed to be similar to the deltaV needed C M Edwards wrote: unjustified assumption about the type of rocket engine used. The only thing we know so far is that JPA will not be using an ion engine. Still, http://www.jpaerospace.com/atohandout.pdf does say Quote: The ion engine 120,000 foot flight test for the orbital airship will be flown in the next five months. Of course that statement is now over five months old. 
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Space Station Commander
Joined: Thu Oct 27, 2005 7:44 am
Posts: 707 Location: Haarlem, The Netherlands 
Wow, lots of reactions, and good ones too! First, I'm doing this in a lookupasyougo fashion. By the time I finish writing a post, I have 25 tabs open in Firefox, and 5 of them are pages about things I never knew existed before. I just looked up what a resistojet was. If you think I'm doing something wrong, I probably am. Please tell me, I love to learn.
If the ATO consists of two cylinders of 2.1 km long and 130m diameter, I'd say we have about 2.1 * 10^3 * 1.3 * 10^2 *2 = 5.5 * 10^5 m^2 available for solar cells on top. Mono or polycrystalline solar cells have a peak power output of about 120W/m^2 (that's peak, but we're above most of the atmosphere so we should be able to get close). That yields a total amount of power of 65 MW that we have available for propulsion. Ion drives will give us a thrust of 2.4 kN for that, or a measly 0.24 tonne equivalent. I think we can safely say that we're not going to be using ordinary ion drives. I'll have to look up other kinds of electrical drives. I like resistojets, I mean, going to the stars with a steam engine, how cool is that? . Anyway, maybe it would be better to figure out a mission profile first, and then see what kinds of thrust levels we need, what kind of propulsion system would be appropriate for that, and whether the abovementioned 65 MW is enough for it to work. Nihiladrem: you're right about the lost buoyancy vs. altitude increase. I'll have to rethink that and do an actual calculation. C. M. Edwards: I'll have a closer look at the graphs on the site I linked to earlier. 46 seemed like a safe lower limit for the L/D of a waverider, but at low speeds L/D seems to be higher, so maybe we can use that to our advantage. I will be spending a few hours in trains this weekend, so I'll have some time to try and figure out a speed/altitude profile, and how much thrust would be needed for that. And yes, I'll take my slide rule. Actually, I did most of the calculations for this using the GNOME calculator. I used the slide rule to visualise the air pressure gradient: I figured the left side was at 30 miles up, and the right side at 40 miles, and then you can easily see how it all works out. It helps intuition. Besides, no space project should have to do without a slide rule . 
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Moon Mission Member
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361 Location: Austin, Texas 
Lourens wrote: That yields a total amount of power of 65 MW that we have available for propulsion. 
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Space Station Member
Joined: Tue Dec 07, 2004 6:50 am
Posts: 265 Location: UK 
campbelp2002 wrote: I assume you mean electric power efficiency and not mass efficiency, because isp *is* efficiency in the normal rocket equation mass fraction sense. campbelp2002 wrote: It may be possible to find a break even point where the added mass of solar cells equals the saved reaction mass. I haven’t done that but I suspect it would be at an exhaust velocity well above the desired DeltaV. C M Edwards wrote: Many of the assumptions necessary to get all the way from “Fy > 0” to “D/L > F/W” do not apply to an airship. An argument cannot be solid if its underlying assumptions are physically meaningless for the case in question It is quite possible to handle L/D varying (principaly with velocity) but it would add nothing here but complication. Instead it is more productive to assume a very high L/D of 10(?) over the entire velocity range with something close to Lourens proposed vehicle, two 2.1km long tubes 130m in diameter. Frontal area somewhere around 0.38km2, Cd=0.55, mass 14000kg. Assume an air density at 60km of rho0=2.5e4 kg/m3 Lets see what velocity the vehicle has twenty km higher, air density and buoyancy is about 1/16 of its value at 60km. thrust=weight/(L/D)*(1.00.5^(altitude/5000)) thrust=14000*9.81/10*(1.00.5^(20000/5000))=12.9kN v=sqrt(thrust/(0.5*Cd*A*rho0*0.5^(altitude/5000))) v=sqrt(12.9e3/(0.5*0.55*380e3*2.5e4*1/16)) v=sqrt(7900) v=88m^s1 So there you have it, almost all the buoyancy gone at under 200mph, and needing over a tonne of thrust, try it yourself. If you wish you can correct L/D iteratively. If we had picked lower L/D the required thrust would be higher and also the velocity but to a lesser degree. The air density model is pretty crude here, density falling off somewhat too fast but even when you fix that you will find the general picture does not depend strongly on the details. Resistojet: Dammit I meant arcjet, though fairly similar in principle hydrogen resistojet runs a bit too cool for this purpose, you want to be aiming at an ISP of just >1000seconds under these conditions. Last edited by nihiladrem on Fri Nov 11, 2005 12:25 am, edited 2 times in total. 
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Moon Mission Member
Joined: Tue Feb 10, 2004 2:56 am
Posts: 1104 Location: Georgia Tech, Atlanta, GA 
Lourens wrote: It helps intuition. Besides, no space project should have to do without a slide rule . Absolutely. Which is why, as soon as I can afford the one I want down at the Engineer's Bookstore, I'm buying one. If for no other reason than to say I have actually used it. _________________ American Institute of Aeronautics and Astronautics Daniel Guggenheim School of Aerospace Engineering In Memoriam... Apollo I  Soyuz I  Soyuz XI  STS51L  STS107 
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Spaceflight Participant
Joined: Thu Mar 31, 2005 1:19 am
Posts: 67 
Hah! this topic is still alive! Hooray!
Ok, this probably doesn't make alot of difference, but in the dragless energytoorbit calculations, I think Lourens forgot to take into account that the mass*height energy is offset by the bouyancy of the ship. This would be difficult to do, though, b/c the bouyancy changes w/ altitude, as you know. I too am beginning to get worried about the feasibility of this beautiful idea. I think the main problem is how we are going to extract enough lift out of a limited amount of power to augment the bouyancy enough to get the ship to an altitude where the drag is low enough so that orbital velocity can be approached. The important thing to remember is that all we have to do is augment the bouyancy with lift, not displace it entirely. Wait! I got it! We have three kinds of lift that will come into play: aerodynamic, bouyancy, and centripital. The sum of these needs to be slightly greater than the wieght of the ship at all times, and we have to achieve that on a limited energy/propellant mass buget. It might be useful to make some starting assumptions about the amount of thrust we can produce. Given 30 MW, say, someone might find out for us how much thrust we can make. Then we could make a table of the amount aerodynamic lift we could get from that thrust, the amount of lift we could get from centripetal force by going to max velocity at that altitude, and also the bouyancy at those altitudes. I don't know how much difference it would make, but you might also want to make a column for the centripetal force at the speed for max aerodynamic lift. Anyway, then you can look at the sums of all three forms of lift at different altitudes, and play with the drag, L/D, thrust, power, and at what altitude it would begin to be efficient to switch from aerodynamic lift mode to 'max speed for max centripetal force' mode. In fact, I think that if you find such a point, the simulation proves that orbit can be achieved. I am thinking about different ways to produce aerodynamic lift besides flying that ungainly thing past the speed of sound. The basic concept of lift is to push as much air down as possible, and the lower the air velocity is, the better, b/c power=(mass flowrate)*(exit velocity)^2, while lift=(mass flowrate)*(exit velocity). The airplane method has the advantage that by flying thru lots of air, you get a very high mass flowrate so you can use low downward velocity of air to get good efficiency, but then you bring in too much drag. But it's not the only way. If you had 30 MW of power, figure out how much mass flowrate and velocity you would need to produce 50 tones of thrust. 
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Moon Mission Member
Joined: Tue Oct 05, 2004 5:38 pm
Posts: 1361 Location: Austin, Texas 
LukeSkywalker wrote: We have three kinds of lift that will come into play: aerodynamic, bouyancy, and centripital. F=mv^2/r where r is the radius of curvature of the circular path you are describing. http://phun.physics.virginia.edu/topics ... fugal.html 
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Spaceflight Participant
Joined: Thu Mar 31, 2005 1:19 am
Posts: 67 
Thanks, I realized that after I'd already posted and was out the door. What do you think of the table idea? Seems like if somebody has alot of time on their hands, they might be able to actually work that out in Excell or something. Shouldn't be too hard to go back and change the drag, L/D and thrust to find out what we'd really be up against.

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Space Station Member
Joined: Tue Dec 07, 2004 6:50 am
Posts: 265 Location: UK 
There's no real need for something approaching a table, just pick a few representative speeds. 5000m/s is a particularly hard one to cross as close to 60% of the gravitational weight has to be supported aerodynamically yet the speed is approaching that needed for orbit.

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Space Station Commander
Joined: Thu Oct 27, 2005 7:44 am
Posts: 707 Location: Haarlem, The Netherlands 
I figure it can be done analytically without too much trouble; it's just a bit of calculus. My main problem is calculating the absolute lift. L/D ratios are nice, but you can't add them to forces.
The general formula for lift is Fl = Cl * Al * rho * 1/2 * v^2 For a waverider, a value of 0.07 for Cl (see e.g. here, that's a 2.8 MB PDF) seems reasonable, but I'm not sure how Al is defined. Wikipedia has it as the "surface area of the lifting surface", but how do you determine that? Second, for drag the equation is similar, but there appears an Ad which is supposed to be the frontal area of the vehicle. And how do these things vary with the angle of attack? Or does convention dictate that they be kept constant, and that Cl and Cd are a function of angle of attack? Can anyone shed some more light on this? 
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