3 ms·

> in the original paper, 500W is going somewhere, but there is no explanation of how, exactly, any current flows in the circuit I puzzled over that a bit too.

> in the original paper, 500W is going somewhere, but there is no explanation of how, exactly, any current flows in the circuit I puzzled over that a bit too. I think the answer is that as protons are shot out the negative side and electrons out the positive side, they create a field that counteracts the field between the grids. Maintaining the field strength thus requires continuously adding charge to the grids. In the hypothetical case of infinite grids (being charged by an infinite number of generators), the areal charge density on the grids would increase without bound, so that the charge density difference, and therefore the field, across the infinite plane halfway between the grids would remain constant. Exactly how this would translate to the finite case I'm less clear on, but I suppose the behavior would be similar, at least shortly after the drive was first turned on. Sooner or later, though, I think the voltage across the gap would increase to the point that any generator you could actually build wouldn't be able to push any more charge onto the grids.

amluto 7y ago
> Maintaining the field strength thus requires continuously adding charge to the grids. I'm not sure what you mean. The whole system is running at steady state. For simplicity, pretend that the spaceship is very massive or is otherwise accelerating at negligible rate so that the spaceship's frame is inertial. [0] In the spaceship's frame, there is some electron current density at all points in space, some proton current density at all points in space, and some distribution of charge carriers on the grids. If we further assume an infinite work function on the grids, then there is no field emission and no plasma charges are captured, so there is no transfer of charge between the plasma and the grids. But then there is no current on the grid, so the driver circuit outputs no power, and the engine can't possibly do any work. So either my analysis is missing something or the paper is wrong. Or both. [0] This could be achieved exactly by firing an equal-but-opposite-thrust conventional rocket in the opposite direction. It doesn't really matter for my argument anyway, I think. [1] Or the grid could be made of metal filaments wrapped in perfect insulators, but then there's an issue of charges building up on the outside of the insulators. In general, the grid will capture some charge from the plasma, and this will be a parasitic loss on the system. I would certainly believe that a pair of solid parallel finite metal plates flying at high speed through a plasma generates a force, but that's not what's being described in the paper.
- ScottBurson 7y ago
  Consider what happens when you've just turned the thing on and it has just barely expelled its first proton through the negative grid. As long as the proton is close to the grid, it may as well be on the grid: it cancels out the contribution of one electron to the field between the plates, which means the voltage across the gap is incrementally reduced. This reduction in voltage allows one electron to flow out of the generator into the grid. Of course, as the proton moves away, its contribution to the field between the plates diminishes at a rate depending on the ratio of its velocity to the radius of the grids. (This is why the infinite-grid case is easier to think about: the contribution doesn't diminish at all.) Eventually — and depending on the proton velocity and grid size, "eventually" might mean milliseconds — its contribution becomes negligible. At that point a steady state will have been reached, and (as you intuit) it won't be possible for the generator to push any more charge onto the grids without increasing its output voltage.

amluto 7y ago

> Maintaining the field strength thus requires continuously adding charge to the grids. I'm not sure what you mean. The whole system is running at steady state. For simplicity, pretend that the spaceship is very massive or is otherwise accelerating at negligible rate so that the spaceship's frame is inertial. [0] In the spaceship's frame, there is some electron current density at all points in space, some proton current density at all points in space, and some distribution of charge carriers on the grids. If we further assume an infinite work function on the grids, then there is no field emission and no plasma charges are captured, so there is no transfer of charge between the plasma and the grids. But then there is no current on the grid, so the driver circuit outputs no power, and the engine can't possibly do any work. So either my analysis is missing something or the paper is wrong. Or both. [0] This could be achieved exactly by firing an equal-but-opposite-thrust conventional rocket in the opposite direction. It doesn't really matter for my argument anyway, I think. [1] Or the grid could be made of metal filaments wrapped in perfect insulators, but then there's an issue of charges building up on the outside of the insulators. In general, the grid will capture some charge from the plasma, and this will be a parasitic loss on the system. I would certainly believe that a pair of solid parallel finite metal plates flying at high speed through a plasma generates a force, but that's not what's being described in the paper.

ScottBurson 7y ago
Consider what happens when you've just turned the thing on and it has just barely expelled its first proton through the negative grid. As long as the proton is close to the grid, it may as well be on the grid: it cancels out the contribution of one electron to the field between the plates, which means the voltage across the gap is incrementally reduced. This reduction in voltage allows one electron to flow out of the generator into the grid. Of course, as the proton moves away, its contribution to the field between the plates diminishes at a rate depending on the ratio of its velocity to the radius of the grids. (This is why the infinite-grid case is easier to think about: the contribution doesn't diminish at all.) Eventually — and depending on the proton velocity and grid size, "eventually" might mean milliseconds — its contribution becomes negligible. At that point a steady state will have been reached, and (as you intuit) it won't be possible for the generator to push any more charge onto the grids without increasing its output voltage.