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 >
!!!! I reread this and now I find something else perplexing. How do the ions manage to make their way to the lower electrode if the force between the ions and the air is equal and opposite? Harry
