Quote from: Rodal on 05/23/2015 09:37 pmQuote from: CW on 05/23/2015 09:03 pm...The equation should be Reaction + Thrust = 0 , under the premise that anyone even gives a darn about conventions anymore.I'm using D'Alembert's principle looking at Shawyer's diagram. His force convention does not follow any of the books I have in Mechanics (the fact that he has these two equal an opposite forces which should result in a body in equilibrium, hence having no acceleration).Let's say that we instead interpret Shawyer as you suggest. Then work out the bullet/gun split: one comes up with the accelerations having different signs which I agree is a more conventional view. If one consistently follows the same convention all the way through, for the bullet and the gun to both have real positive masses, then one ends up with the same result I have above that the mass of the Big End is the negative of the mass of the Small End and that the Total Mass of the EM Drive must be zero, according to Shawyer.I fear that the available documents from Mr. Shawyer are unusable for any reasonable discussion. Judging by the available reports of a number of groups telling that something seems to or is going on, I feel that Mr. Shawyer might have found something by sheer coincidence. It reminds me of the logical implication that tells us that starting from a wrong premise, any conclusion is possible - even the right one.
Quote from: CW on 05/23/2015 09:03 pm...The equation should be Reaction + Thrust = 0 , under the premise that anyone even gives a darn about conventions anymore.I'm using D'Alembert's principle looking at Shawyer's diagram. His force convention does not follow any of the books I have in Mechanics (the fact that he has these two equal an opposite forces which should result in a body in equilibrium, hence having no acceleration).Let's say that we instead interpret Shawyer as you suggest. Then work out the bullet/gun split: one comes up with the accelerations having different signs which I agree is a more conventional view. If one consistently follows the same convention all the way through, for the bullet and the gun to both have real positive masses, then one ends up with the same result I have above that the mass of the Big End is the negative of the mass of the Small End and that the Total Mass of the EM Drive must be zero, according to Shawyer.
...The equation should be Reaction + Thrust = 0 , under the premise that anyone even gives a darn about conventions anymore.
It seems like we are trying to apply Newtonian physics to something that may be a quantum level effect.But the thought had occurred to me that when Iulian got less thrust in the down direction, something else might come into play besides hot air. If the force is between his D.U.T and the floor, the difference might be caused by the inverse square law. By eyeball, the big end to floor distance increased by about 125-150%
Impedance?locus of 50+j0
But the thought had occurred to me that when Iulian got less thrust in the down direction, something else might come into play besides hot air.
What the EmDrive thruster does is to produce a force, which we call the thrust, in one direction. This is a force that you can measure. If you put your hand against the end plate that's producing the thrust you'll feel it pushing against you. And, as with all machines that follow Newton's principles, it will therefore accelerate in the opposite direction. So this is not a reactionless thruster, because those things just don't exist outside of science fiction, but it is a propellantless thruster.
Interesting. Maybe he should have a chat with Professor Woodward
This is a supposed quote from Roger Shawyer that I have copy/pasted.QuoteWhat the EmDrive thruster does is to produce a force, which we call the thrust, in one direction. This is a force that you can measure. If you put your hand against the end plate that's producing the thrust you'll feel it pushing against you. And, as with all machines that follow Newton's principles, it will therefore accelerate in the opposite direction. So this is not a reactionless thruster, because those things just don't exist outside of science fiction, but it is a propellantless thruster.It's a little confusing, but i hope i am not stating the obvious in saying that the interesting part is where he says that if you put your hand against the end plate that's producing the thrust you'll feel it pushing against you. That is, your hand is pushed away from the plate. In which case its behaving similar to a common or garden rocket. i.e. mass is thrown out the back end of the frustrum, which you feel bouncing off your hand. Momentum is conserved, and the frustrum goes in the opposite direction.As LasJayhawk suggested, it might be an idea if Iulian varied the distance between the thrust plate and the floor if more tests are done.
Totally off topic, but on a lighter note, if they can get the beams to turn ~90 degrees, does that mean that your laser rifle can shoot around corners?
(I step out of the playroom for one minute and ....)....At least on a cylindrical (ie symetrical) cavity, dropping the wall still integrates to a zero Poynting vector over the far-field sphere. No interesting ramifications yet.
This is a supposed quote from Roger Shawyer that I have copy/pasted.QuoteWhat the EmDrive thruster does is to produce a force, which we call the thrust, in one direction. This is a force that you can measure. If you put your hand against the end plate that's producing the thrust you'll feel it pushing against you. And, as with all machines that follow Newton's principles, it will therefore accelerate in the opposite direction. So this is not a reactionless thruster, because those things just don't exist outside of science fiction, but it is a propellantless thruster.It's a little confusing, ...
OK, they get some very interesting, non-intuitive solutions to Maxwell's equations: nondiffracting spatially accelerating solutions, with a Poynting vector that displays a turn of more than 90 degrees. Let's say that this would involve an acceleration of the EM Drive when the Poynting vector >0But then when the Poynting vector < 0 shouldn't the acceleration be in the opposite direction?And aren't we back to the same situation we are with standing waves in a cavity? : even with standing waves we have a non-zero Poynting vector, the problem is that it keeps switching direction back and forth at a frequency twice as high as the electromagnetic field frequency.It seems to me like we need a nonlinearity (at least 2nd order) in order for the Poynting vector average to be different from zero.Or we need a thermodynamic loss that will produce an energy flux preferentially in one direction
...Dropping the wall (ie letting it propagate) turns out to be the same as replacing w/ a dielectric resonator. That, of course, radiates. It "looks-like" the frustum resonator might radiate in a preferred and mode-dependent direction w/ or w/o the Poynting vector integrating to zero over the complete sphere (after all, antennas do it)...
I'm pretty sure he is not saying that he has actually felt the thrust with his hand from any of his exiting drives, but instead that were you to build a model with higher thrust you could feel it