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I have been reading up about the EM drive which supposedly allows for reactionless propulsion. Regardless of whether or not it is real, I was trying to do a back of the envelope calculation to see how fast it could go to gauge if it would even be viable as a propulsion system if it turned out to be real. I was expecting the number to be quite small and inconsequential compared to using the same energy to power a mass launcher or an even more efficient propulsion system. However, the number I came up with was mind boggling big. I double checked my work, ensured my units were being converted appropriately and I got the same answer. The only thing I can think of is that my assumptions are faulty; please review my work and help me figure out where I went wrong.

Starting conditions: The EM drive supposedly produces 0.72 newtons of thrust at 1000 watts of power as an absolutely best case number, several other numbers that are much smaller have been named, but I'm trying to give this the best chance. This works out to 1388.9 W/N. I found different sources quoting energy density for uranium ranging between 4.5E4 and 2.4E7 kWh. So I erred on the low end that 1 kg of uranium contains 4.5E7 Wh (not kWh) or is better expressed in terms of energy density as 4.5E7 Wh/kg (24 billion). Since the fuel is not used as reaction mass, the mass of the ship remains constant as the fuel is spent which makes the calculations a lot more straight forward. Since it remains constant I simplified it so that the mass of the fuel was vastly more than the weight of the engine, the reactor, and the hull of the vessel, thus I could ignore the effects the added mass would have.

The thrust I calculated to be (mass (kg) * energy density (Wh/kg)) / (thrust power (W/N) * hours (hours)) = (1kg * 4.5E7Wh/kg) / (1388.9W/N * 1hour) which comes out to 1.08E5 newtons of force. I then rearranged F=MA to get A=F/M which works out to 1.08E5N / 1kg = 3.24E4 m/S^2. I then wanted to calculate the final velocity, so I used Vf = Vo + A*T. Vo is my initial velocity which I set to be 0m/S, acceleration is previously calculated, and I converted my 1 hour into 3600 seconds so my final answer would have m/S as it's units. The final velocity Vf is equal to 1.1664E8 m/S, or 388.8 times the speed of light!

If I adjust my fuel weight and burn time, it always works out to being the exact same change in velocity, which makes sense. More fuel mass means I need more thrust, but I have more energy to dump into acceleration. Also, if I use more energy for less time it is functionally equivalent to using less energy over a longer period of time.

I played around with adding a fixed amount of weight to account for the reactor, engine, and ship, producing a horribly inefficient 1000:1 ship fuel ratio and still came up with a reasonably large velocity change of 1.165E5 m/S or about 39% the of the speed of light.

Please help me figure out if these numbers are legit, or if I made some glaring mistake.

Edit: For what it's worth, I believe that the likeliness of this being a real phenomenon is very unlikely. Some labs have found a small net force, others have found nothing. Of the labs that did reproduce it, the numbers varied widely. As was mentioned, real world effects are repeatable.

I was simply running some numbers to show that even if it did work it would be useless as a propulsion method due to its low thrust. Since I wrote this I have looked up the Ion propulsion drive on Wikipedia and found it had similar numbers (25-250 mN for 1-7kW of power).

CraigC
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    In the recent paper by White et.al. the quoted thrust is $1.2 \times 10^{-3},\mathrm{N/kW}$, which is 60 times lower than your figure (and is still almost 500 times the expected value for a ideal photon drive). If your figure represents a value from an earlier communication I'll go from resigned to there being nothing actually there when all is said and done to convinced that there is nothing actually there: you don't lose a factor of 60 if the thing you are chasing is real. – dmckee --- ex-moderator kitten Nov 30 '16 at 03:49
  • @dmckee I think that the bigger figures were from it running in air as the initial experiments were done in air -- so, obviously, some result of hot air rising around the thing as it dumps all that heat. –  Nov 30 '16 at 10:48
  • @tfb White et.al. (DOI: 10.2514/1.B36120) did ambient pressure tests as part of their systematic controls and write "The test article is tested in atmospheric conditions and under vacuum conditions. The impulsive thrust performance (in millinewtons per kilotwatt) is observed to be nearly the same." Which puts us in the strange position of having a careful and systematic investigation with an unexplained result, but one that is far less than earlier reports. – dmckee --- ex-moderator kitten Nov 30 '16 at 16:08

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Power equals force times velocity (P=F.v), or work is force times distance (W=F.s)

The trouble with fixed power input creating constant thrust regardless of velocity, is that if you moved fast enough (1389 m/s in this case), the power output would exceed the input, you would be creating energy out of thin air (not a big leap from thrust out of quantum fluctuations, I admit).

(the situation is different with normal rockets, they eject reaction mass. The "excess" energy at high speed comes from the energy they spent at lower speed to take that extra mass along)

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  • You realize that every rocket does this same thing. Work and kinetic energy are not the same in every frame of reference and that isn't a problem with the formalism. – dmckee --- ex-moderator kitten Dec 11 '16 at 22:10
  • @dmckee attach the 0.72N, 1kW thruster to a frictionless flywheel (with thrust in cross-radial direction). At t=0 and zero velocity, apply power. The thruster reaches v=2.778 km/s at t=T. If it delivers a constant thrust of 0.72 N per kW, then for t>T, the kinetic energy of the flywheel + fictional thruster will be larger than the total amount of energy put in (= t*1000 Joule). How is this a frame of reference issue? Or how would every rocket do that? – Previous Dec 29 '16 at 19:06
  • See http://physics.stackexchange.com/questions/230054/how-am-i-able-to-stand-up-and-walk-down-the-aisle-of-a-flying-passenger-jet and links therein. Work and kinetic energy are frame-dependent quantities. And you are comparing the work done in the (proposed) thruster's own frame with the change in KE measured in a different frame. That's an error even when applied to to non-exotic problems. As a non-exotic example, most solid rockets have approximately constant thrust for their whole burn so your (erroneous) math applies just as well to them. – dmckee --- ex-moderator kitten Dec 29 '16 at 19:35