Michel Jullian wrote: > > ----- Original Message ----- > From: "Harry Veeder" <[EMAIL PROTECTED]> > To: <vortex-l@eskimo.com> > Sent: Thursday, February 22, 2007 8:04 AM > Subject: Re: [Vo]: Lifters > > > ... >>> ...Sigmond's derivation for the lifter thrust (or rather it's opposite >>> namely the force >>> exerted by the ions on the air > ... >> However, what about the force of reaction by the air on the ions? > > That's the thrust, and as I said, it's exactly the opposite vectorially to the > force exerted by the ions on the air calculated by Sigmond (they are equal in > magnitude: action=reaction). You see the recirculated charges are internal > parts of the lifter, just like the paddles are internal parts of the paddle > wheel boat, so any external force on them is a force on the lifter.
> To clear up a possible confusion, the forces we discussed wrt the tubular > lifter between the electrodes and the flying charges are all internal forces, > like one could discuss the internal actions between the paddles and the ship, > or the propeller and the helicopter. They are interesting as a way to > visualize what pushes the _electrodes_ up, but they cancel when you add them > all up (e.g. force of charges on cathode + force of cathode on charges = 0), > what really applies a net force to the lifter is the reaction of the medium. >> Unless this force exceeds the force exerted by the ions on the air >> the lifter will not rise. If it is less than this, the lifter is >> just an air pump. > > Not at all, they are equal in magnitude in all circumstances :) The lifter > will simply rise if the force exceeds its weight, in which case its > acceleration is (force - weight)/mass, as long as the aerodynamic drag remains > negligible as is the case in all practical lifters. > > Michel > Ok, now I understand the essentials of this explanation. However, concerning the tubular lifter, I would argue that the elevated tube when the lifter is _accelerating upwards is evidence that the _internal forces_ don't add up to zero. Harry
