In Atmosphere Ion Drive(Assess Viability For Spacecraft)

1. Jul 27, 2011

FireStorm000

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 thermo-sphere (space shuttle/ISS orbit)

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/mT. 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?

Last edited by a moderator: Sep 25, 2014
2. Jul 27, 2011

Mech_Engineer

3. Jul 27, 2011

Staff: Mentor

I added a direct link to the YouTube video in the OP. It's a MythBusters video "Lifter Anti-Gravity Myth has been BUSTED"

Last edited by a moderator: Sep 25, 2014
4. Jul 27, 2011

enigma

Staff Emeritus
No air, no thrust.

5. Jul 27, 2011

FireStorm000

This is indeed the ionocraft from myth-busters. 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(MeG/r)
E =mV2/2 + mg$\Delta$h
E/m = MeG/(2h+2re) + g$\Delta$h
h=100km
E/m = 6.67e-11*6e24/(2*100000+2*6378000) + 9.81 * 100000 = 3.09e7 + 981000 = 3.09e7J/kg
h = 1000km
E/m = 6.67e-11*6e24/(2*1000000+2*6378000) + 9.81 * 1000000 = 2.71e7 + 9.81e6 = 3.69e7J/kg
$\Delta$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. Jul 28, 2011

Staff: Mentor

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 http://en.wikipedia.org/wiki/Atmosphere_of_Earth#Thermosphere" suggest that the atmospheric pressure is 1-10million 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 http://en.wikipedia.org/wiki/Nuclear_thermal_rocket" [Broken].

EDIT: Regarding the problem embedding youtube videos I was having the same problem. I reported it and found out how to fix the problem https://www.physicsforums.com/showthread.php?p=3425170#

Last edited by a moderator: May 5, 2017
7. Jul 28, 2011

enigma

Staff Emeritus
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. Jul 31, 2011

FireStorm000

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($\rho$) 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)

FD=CD$\rho$AV2
dm/dt=$\rho$AV
(Already I'm not liking that mass flow rate only increases by V1, while drag increases with V2...)
$\Delta$KEIon = 1/2m(VExhaust2 - V2)
P = .5 dm/dt VEx2-.5 dm/dt V2
VEx = sqrt( 2P/ (dm/dt) + V2 )
FThrust=dm/dt VEx
FThrust=$\rho$A ( sqrt( 2PV/($\rho$A) + V4 ) - V2 )
FTotal=$\rho$A ( sqrt( 2PV/($\rho$A) + V4 ) - V2 - CDV2)

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 $\rho$:
Density of air $\rho$ = .000288kg/m3
Coefficient of Drag CD = .1
Area A = 100m2(2X space shuttle area)
Power P = 30,000,000W(1 nuclear fission reactor from Boomer sub)
V = Mach 6 = 2058m/s
FT = 13,797N - 12,197N = 1,600N
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Altitude = 86km (the maximum I can find density for)
Density of air $\rho$ = .000006kg/m3
Coefficient of Drag CD = .1
Area A = 100m2 (2X area of shuttle)
Power P = 30,000,000W
V = orbital velocity = 7800m/s
FT = 3662N - 3650N = 12N
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Altitude = 86km (the maximum I can find density for)
Density of air $\rho$ = .000006kg/m3
Coefficient of Drag CD = .1
Area A = 50m2 (area of shuttle)
Power P = 30,000,000W
V = orbital velocity = 7800m/s
FT = 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 CD 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. Jul 31, 2011

FireStorm000

You are correct; the equations in that post only show the amount of energy needed to achieve a given change in orbit. The ones in my post immediately above this one do take into account atmospheric density to give thrust.

This would definitely be something for very specialized roles where carrying extra fuel has it's own problems. I don't think it's useless though. There should be a region where it beats conventional ion drives.

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)

You're awfully quick to shoot down my ideas without providing any math or links to support your position. I think my post above shows clearly that this propulsion method could in fact counter drag with thrust to spare. If you do the math, you'll find that mass flow rates are actually higher than conventional ion drives:
$\rho$AV=(.000288kg/m3)(50m2)(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, http://http://www.digitaldutch.com/atmoscalc/" [Broken] is the density calculator I used.

Last edited by a moderator: May 5, 2017
10. Aug 1, 2011

enigma

Staff Emeritus
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, ultra-rarefied 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 1e-10, 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.

Last edited: Aug 1, 2011
11. Aug 1, 2011

Mech_Engineer

You've posted a simplistic equation for calculating flow rate through an orifice using density, area, and speed; but did you calculate the power, current, or voltage that would be needed to achieve that kind of thrust with a EHD lifter? All I've seen is a 30MW number, but nothing to back it up...

Last edited by a moderator: May 5, 2017
12. Aug 1, 2011

BobG

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. Aug 1, 2011

Mech_Engineer

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
The article also states:

So, ignoring the fact that the drift velocity of the ions must be much greater than the speed at which the vehicle is moving w.r.t. the air it is immersed in (could be a killer), if the force is 1 gram-force (9.81e-3 N) at sea level, and if we assume that the lifter's thrust is proportional to the gas density, that means that:

• At 100 km: 3.95e-6 N/W
• At 200 km: 1.52e-9 N/W
• At 300 km: 7.01e-11 N/W

So, for a 30MW reactor (assuming the efficiencies we've stated even hold, probably not...) that means you'll get:

• At 100 km: 118.4 N
• At 200 km: 0.046 N
• At 300 km: 0.0021 N

So, considering the weight of a 30MW reactor, this is on track to not working at all.

14. Aug 1, 2011

FireStorm000

It appears you are right that you couldn't achieve orbit with this design. Performance falls off too fast with increasing velocity. Still, I'm curious is there might be other areas where ion-craft could be useful.

The viability of nuclear fission in space is a whole new discussion all together. Suffice it to say I give it a better chance of working than you do.

I don't believe there's anything wrong with my equation for mass flow rate; I'm picturing a ion-craft where the high voltage is applied to the hull itself. That is, work is done on all particles in any vicinity of the hull. If you want current and voltage that's fairly simple application of electrostatics. For the applications we are discussing, air can be assumed to be 100% ionized, simplifying calculations greatly.
First, recall the change in kinetic energy of a particle is:
$\Delta$KEIon = 1/2m(VExhaust2 - V2)
and
P = .5 dm/dt VEx2-.5 dm/dt V2, where P is the power (I*E) supplied to the craft's engines
we find that:
VExaust = sqrt( 2P/dm/dt + V2 )
and thus,
$\Delta$V = sqrt( 2P/dm/dt + V2 ) - 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=$\Delta$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 $\epsilon$ to avoid ambiguity with velocity:
W = $\epsilon$ * qElementary
Setting work equal to delta KE; let n be a number representing the average ionization (avg # of missing electrons)
1/2m(VExhaust2 - V2) = $\epsilon$ * n * qElementary
And solving for $\epsilon$
1/2mparticleAvg(2P/dm/dt + V2 - V2) = $\epsilon$ * n * qElementary
1/2mparticleAvg * 2P/dm/dt = $\epsilon$ * n * qElementary
$\epsilon$ = .5 * (mparticleAvg / ( n * qElementary)) * 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 * qelementary / mparticleAvg * dm/dt
~~~~~~~~~~~~~~~~~~~
Checking our work, we should find that P=I$\epsilon$
P = (.5 * (mparticleAvg / ( n * qElementary)) * 2P/dm/dt)(n * qelementary / mparticleAvg * dm/dt)

and indeed we find that P=P, so our work checks
~~~~~~~~~~~~~~~~~~~
Now plugging in numbers; let's choose the 86km altitude again:
$\rho$=.000288kg/m3
V=2000m/s
mAvgParticle = the weighted average of the masses of nitrogen and oxygen molecules = .79 * 14 * 2 + .21 * 16 * 2 = 28.84amu = 4.789e-26kg
qel=1.6e-19C
n = 1.2 (I'm taking a shot in the dark here)
$\epsilon$ = (mparticleAvg / ( n * qElementary)) * P/($\rho$AV)
$\epsilon$ = .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 shock-wave 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.

A subsonic wing can achieve .01 CD, so .1 is in the range of sane numbers; not saying it's necessarily accurate for a satellite, but it's about what would be necessary for this to have any chance of working at all.

Also, thanks for the numbers; those look much more accurate.

I have my doubts as to the usefulness of that page to this discussion. As stated above, the only reason that line needs to be at 30kV is to ionize the air, not to accelerate it. Because charge on an ion is very large in comparison to mass, a relatively low potential can cause meaningful thrust. That thrust statistic from Wikipedia assumes that you have to invest the ionization energy into every ion you want to ionize, whereas at speed and altitude, solar radiation and supersonic shock have done the ionizing for you; efficiencies should be orders of magnitude higher at altitude. My equations are derived from assumptions relevant to this discussion, while that may not necessarily be true of the equations on the wiki page.

Also in that article:
So the lifter is at the lower end of efficiency for the ionization reasons I already mentioned.

I also have another idea for a potential use of this technology. I'll crunch the numbers and post again.