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In Atmosphere Ion Drive(Assess Viability For Spacecraft) 
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#1
Jul2711, 06:42 PM

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For starters, watch this video to see what I'm talking about. It has been done, so I know it can work, but I want to assess viability for maneuvering while in the thermosphere (space shuttle/ISS orbit)
You can skip to 2:20 http://www.youtube.com/watch?v=Jk2GGoMJ7NU So based on that video, it's obvious that is you create a strong enough electric field, you will generate a force on the air, so my question is what range of altitudes, power, and orbiter mass would make this viable for maneuvering? To avoid relating the discussion to specific spacecraft, we can discuss in terms of power density, where power density is a measure of the total generating capacity of the spacecraft divided by the total mass; W/kg; P/m_{T}. For a flying nuclear fission reactor, expect at most something on the order of 1200W/kg, 100W/kg on an actual fission powered spacecraft. For a solar powered craft near earth such as the ISS, expect about .07W/kg So how would I go about figuring this one out? 


#2
Jul2711, 07:04 PM

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I can't see the video here at work, but I hope you didn't link to the infamous "lifter in a vacuum" video...
http://www.physicsforums.com/showthread.php?t=205133 http://www.physicsforums.com/showthread.php?t=211080 


#3
Jul2711, 07:14 PM

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#4
Jul2711, 07:39 PM

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In Atmosphere Ion Drive(Assess Viability For Spacecraft)
No air, no thrust.



#5
Jul2711, 08:35 PM

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I'm not sure why the video isn't loading. (do the youtube embed tags work?)
This is indeed the ionocraft from mythbusters. To clarify, I am not talking about using this to launch a spacecraft, but rather for maneuvering once already in orbit, IE, from 100km up. Once there could it, for example, be used to transfer from LEO to a higher orbit? Further, I am not saying this works in a vacuum. I'm wanting to take advantage of the fact that earth orbit (hell, even space itself) isn't a true vacuum. There is measurable atmospheric density even where the ISS orbits. It isn't much, but it could be enough. I want help to crunch the numbers to find out how well this could work, if at all. Some basic math: If you're already in orbit at 100km, then you have an orbital energy of: V = sqrt(M_{e}G/r) E =mV^{2}/2 + mg[itex]\Delta[/itex]h E/m = M_{e}G/(2h+2r_{e}) + g[itex]\Delta[/itex]h h=100km E/m = 6.67e11*6e24/(2*100000+2*6378000) + 9.81 * 100000 = 3.09e7 + 981000 = 3.09e7J/kg h = 1000km E/m = 6.67e11*6e24/(2*1000000+2*6378000) + 9.81 * 1000000 = 2.71e7 + 9.81e6 = 3.69e7J/kg [itex]\Delta[/itex]E = +6.03e6J/kg So, once you're in orbit at 100km, it only takes 600kJ/kg to change to a 1000km orbit, as compared to 3.09MJ/kg to get to the lower orbit. We could then make a very loose estimate that that change would take: E=P*t E/m=P/m*t t=(E/m)/(P/m*efficiency) So at the longer end of the time frame, a solar powered craft at .1% efficiency t~=3000yrs For a nuclear reactor at 5% efficiency: t~=1day So in my conclusion, this could be viable. Much more so with fission or fusion power. 


#6
Jul2811, 06:33 AM

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I'm not knowledgeable on the mathematics so forgive me if I'm wrong but your equations don't seem to take into account how thick the atmosphere will be in orbit (as far as I can make out you've just assumed an efficiency for converting energy to momentum). Sure you can pump that much energy in but how much of an effect is it going to have in the thermosphere? I don't have time to look for a good reference right now but according to wiki an average oxygen molecule can travel over a kilometre before hitting another gas molecule and these lecture notes suggest that the atmospheric pressure is 110million times less than that at sea level.
If you were allowed to place nuclear reactors in orbit then there would be better things you could do with it than try to power an ion lifter. You could use a simple nuclear thermal rocket or even use it to power an ion thruster or VASIMR. I'm sceptical of nuclear fusion in space, we've been experimenting for decades on Earth and haven't been able to build useful reactors that doesn't cost billions. If you could leap the technological hurdles to put a fussion reactor in space you could either use it the same way as I've just suggested for a fission reactor or build a fusion rocket. EDIT: Regarding the problem embedding youtube videos I was having the same problem. I reported it and found out how to fix the problem http://physicsforums.com/showthread.php?p=3425170# 


#7
Jul2811, 09:31 AM

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You aren't going to be able to pull enough thrust from the extremely rarefied atmosphere to overcome drag, J2, and solar flux effects.
Ion drives carry their own fuel, have an extremely heavy power system, and are still only able to pull an ounce of thrust with mass flow rates on the order of mg/sec. You're proposing doing the same thing with mass flow rates of pico to femtograms per second, if that. It's not going to work. 


#8
Jul3111, 10:15 PM

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I do accept that this method of thrust would either not work at all, or be limited to a number of very specialized roles where carrying additional fuel simply isn't an option.
Regardless, I really do want to either prove or disprove this mathematically. I need a value of air's mass density([itex]\rho[/itex]) from about 100km up, but I can't seem to find it. With that, the rest of the math falls into place: (assume all air is ionized at this speed and altitude; choose the craft as your reference frame) F_{D}=C_{D}[itex]\rho[/itex]AV^{2} dm/dt=[itex]\rho[/itex]AV (Already I'm not liking that mass flow rate only increases by V^{1}, while drag increases with V^{2}...) [itex]\Delta[/itex]KE_{Ion} = 1/2m(V_{Exhaust}^{2}  V^{2}) P = .5 dm/dt V_{Ex}^{2}.5 dm/dt V^{2} V_{Ex} = sqrt( 2P/ (dm/dt) + V^{2} ) F_{Thrust}=dm/dt V_{Ex} F_{Thrust}=[itex]\rho[/itex]A ( sqrt( 2PV/([itex]\rho[/itex]A) + V^{4} )  V^{2} ) F_{Total}=[itex]\rho[/itex]A ( sqrt( 2PV/([itex]\rho[/itex]A) + V^{4} )  V^{2}  C_{D}V^{2}) There, we now have an expression describing total force in the direction of motion. Choosing some reasonable numbers, lets see what it comes out to: Altitude = 60km gives [itex]\rho[/itex]: Density of air [itex]\rho[/itex] = .000288kg/m^{3} Coefficient of Drag C_{D} = .1 Area A = 100m^{2}(2X space shuttle area) Power P = 30,000,000W(1 nuclear fission reactor from Boomer sub) V = Mach 6 = 2058m/s F_{T} = 13,797N  12,197N = 1,600N ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Altitude = 86km (the maximum I can find density for) Density of air [itex]\rho[/itex] = .000006kg/m^{3} Coefficient of Drag C_{D} = .1 Area A = 100m^{2} (2X area of shuttle) Power P = 30,000,000W V = orbital velocity = 7800m/s F_{T} = 3662N  3650N = 12N ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Altitude = 86km (the maximum I can find density for) Density of air [itex]\rho[/itex] = .000006kg/m^{3} Coefficient of Drag C_{D} = .1 Area A = 50m^{2} (area of shuttle) Power P = 30,000,000W V = orbital velocity = 7800m/s F_{T} = 3509  1825 = 1684N Conclusions: Thousands or even tens of thousands of newtons of thrust is certainly not insignificant. Lower in the atmosphere, and at lower velocities, this can provide very significant thrust, without having to carry any fuel at all. Lowering the coefficient of drag, and decreasing the cross sectional area of the craft would be the two major design points for any craft employing this propulsion method. Halving the area halves the drag with negligible effect on thrust. Halving the C_{D} also halves drag. Basically, you should end up with a big, long, flying, pointy, submarine looking thing in space. Questions? Comments? Did I miss anything major? 


#9
Jul3111, 10:34 PM

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Also, thanks for the help on the video. If it's possible, I'd like th fix my first post (the edit button seems to have gone away) [itex]\rho[/itex]AV=(.000288kg/m^{3})(50m^{2})(2000m/s) = 28.8kg/s That's orders of magnitude more than the picograms/s you assumed. So please give my ideas a chance. also, this is the density calculator I used. 


#10
Aug111, 10:10 AM

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Launching a nuclear engine is not a feasible solution. We can't even get people on board with building ground based ones. Even if you could get past the politics, submarine engines weigh over 100 tons just for the housing. That's not including the 80 tons of coolant or 800 tons of ancillary equipment. For reference, the empty shuttle weighs 80 tons give or take. An A380 weighs 650. So, assume you get something immense enough to lift all that off the ground. I'll estimate it weighs 10 to 20,000 tons. Nuclear power generates a slew of heat. For a 30 MW reactor, ill estimate 70 MW worth. In a sub, you dump that into the surrounding ocean. You're going to be swimming through superheated, ultrararefied plasma. The only way to get rid of that heat will be football field sized radiators.
Now, the density of air up around 100km is around 1e10, 100,000 times less than your calculation. You're back down to milligrams per second, but to get there you need a 50 m^2 effective area. I highly doubt that the thrust from the little ion engine mythbusters scales anywhere close to linearly. Even if it did, you're getting ounces of thrust to push a craft that weighs thousands of tons. All this while the thing is still suborbital. If you don't circularize the orbit in one rotation, you're going to slam back down due to gravity losses. Really, and I mean really really. It's not going to work. 


#11
Aug111, 11:07 AM

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#12
Aug111, 01:37 PM

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Drag coefficients for typical satellites range from 2 to 4, but that's because there's virtually no effort to make them aerodynamic. I think it's reasonable to say you could decrease the drag coefficient if that was an issue, but 0.1 seems a little optimistic.
Find a version of Wert's Space Mission Analysis and Design (SMAD) if you want good tables for atmospheric densities. Good is relative because atmospheric density varies and is unpredictable. But at least SMAD's charts are reasonable up to 1500 km, at which point the atmospheric density is so low it's not worth considering for spacecraft design. For their charts, atmospheric density is about 4.61 x 10^7 kg/m^3 during solar min and 5.1 x 10^7 kg/m^3 during solar max at 100km 1.78 x 10^10 kg/m^3 during solar min and 3.52 x 10^10 kg/m^3 during solar max at 200km 8.19 x 10^ 12 kg/m^3 during solar min and 3.96 x 10^11 kg/m^3 during solar max at 300km In other words, the density is decreasing rapidly. 


#13
Aug111, 02:51 PM

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It is possible to calculate the thrust put out by an EHD lifter, but some of it is a little "fuzzy" for less than 1 atm air. According to Wikipedia:
http://en.wikipedia.org/wiki/Ionocraft
So, for a 30MW reactor (assuming the efficiencies we've stated even hold, probably not...) that means you'll get:
So, considering the weight of a 30MW reactor, this is on track to not working at all. 


#14
Aug111, 06:29 PM

P: 169

First, recall the change in kinetic energy of a particle is: [itex]\Delta[/itex]KEIon = 1/2m(V_{Exhaust}^{2}  V^{2}) and P = .5 dm/dt V_{Ex}^{2}.5 dm/dt V^{2}, where P is the power (I*E) supplied to the craft's engines we find that: V_{Exaust} = sqrt( 2P/dm/dt + V^{2} ) and thus, [itex]\Delta[/itex]V = sqrt( 2P/dm/dt + V^{2} )  V Now recalling the definition of electric field, and definition of voltage, and definition of work: E = F / q;F = E * q E =  dV/dr W=[itex]\Delta[/itex]KE=integral(F,dr,0,r) Integrating force over distance on a charged particle gives work done: W = 1 * E * q * r W = V * q Changing the variable V(voltage) to [itex]\epsilon[/itex] to avoid ambiguity with velocity: W = [itex]\epsilon[/itex] * q_{Elementary} Setting work equal to delta KE; let n be a number representing the average ionization (avg # of missing electrons) 1/2m(V_{Exhaust}^{2}  V^{2}) = [itex]\epsilon[/itex] * n * q_{Elementary} And solving for [itex]\epsilon[/itex] 1/2m_{particleAvg}(2P/dm/dt + V^{2}  V^{2}) = [itex]\epsilon[/itex] * n * q_{Elementary} 1/2m_{particleAvg} * 2P/dm/dt = [itex]\epsilon[/itex] * n * q_{Elementary} [itex]\epsilon[/itex] = .5 * (m_{particleAvg} / ( n * q_{Elementary})) * 2P/dm/dt ~~~~~~~~~~~~~~~~~~ To find current, realize that current is "charge in motion". 1 amp is a coulomb of charge passing a given point in 1 second. In other words, we can speak of current as a charge flow rate. dq/dt = k dm/dt; where k is the ratio charge per unit mass dq/dt = I = n * q_{elementary} / m_{particleAvg} * dm/dt ~~~~~~~~~~~~~~~~~~~ Checking our work, we should find that P=I[itex]\epsilon[/itex] P = (.5 * (m_{particleAvg} / ( n * q_{Elementary})) * 2P/dm/dt)(n * q_{elementary} / m_{particleAvg} * dm/dt) and indeed we find that P=P, so our work checks ~~~~~~~~~~~~~~~~~~~ Now plugging in numbers; let's choose the 86km altitude again: [itex]\rho[/itex]=.000288kg/m^{3} V=2000m/s m_{AvgParticle} = the weighted average of the masses of nitrogen and oxygen molecules = .79 * 14 * 2 + .21 * 16 * 2 = 28.84amu = 4.789e26kg q_{el}=1.6e19C n = 1.2 (I'm taking a shot in the dark here) [itex]\epsilon[/itex] = (m_{particleAvg} / ( n * q_{Elementary})) * P/([itex]\rho[/itex]AV) [itex]\epsilon[/itex] = .2598V I = 1.15e8A As crazy as that sounds, I think it's right. The shear amount of charged particles flowing ac cross the spacecraft at those velocities means currents are astronomical. At 2000m/s, a 30MW reactor only allows you to apply a potential of 1/4V. That's why this supplies so little thrust at high velocities; there just isn't enough power to sustain a strong enough electric field. The advantage of operating at these higher velocities and altitudes however is that energy need not be spent ionizing the air. The supersonic shockwave and solar radiation has already done that. I think ionizing the air is where 99+% of the energy is spent in a normal ion lifter. Also, thanks for the numbers; those look much more accurate. Also in that article: I also have another idea for a potential use of this technology. I'll crunch the numbers and post again. 


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