If you can levitate yourself to orbit, you shouldn't have to deal with re-entry speeds any higher than current normal speeds. At least not inherently due to ME propulsion.

Yeah, but none of that's due to using Mach Effect propulsion, which is IIRC the premise Sith was arguing from/asking about. It's kind of vaguely phrased so I could be wrong.

Dr. Fierro says: F=dP/dt is wrong, but that F= Mo*dV/dt is correct. This seemed an extraordinary statement to me because they ought to be identical statements, so I was looking for an explanation.I was thinking if you start from Mo*V (the instantaneous definition of momentum) rather than than P (momentum), and then you differentiate WRT time to get F, you have started with the assumption that Mo does not change. Of course that eliminates any change in Mo. You've pulled it out of the integral! It seemed circular to me.So I was hoping Dr. Fierro would explain why it is wrong to start from P rather than Mo*V, but perhaps I am not asking clearly because he appears to actually be starting, prima facie, from the assumption that V is the proper place to start rather than P.

Quote from: cuddihy on 12/06/2010 04:37 pmDr. Fierro says: F=dP/dt is wrong, but that F= Mo*dV/dt is correct. This seemed an extraordinary statement to me because they ought to be identical statements, so I was looking for an explanation.I was thinking if you start from Mo*V (the instantaneous definition of momentum) rather than than P (momentum), and then you differentiate WRT time to get F, you have started with the assumption that Mo does not change. Of course that eliminates any change in Mo. You've pulled it out of the integral! It seemed circular to me.So I was hoping Dr. Fierro would explain why it is wrong to start from P rather than Mo*V, but perhaps I am not asking clearly because he appears to actually be starting, prima facie, from the assumption that V is the proper place to start rather than P. cuddihy:By definition P=Mo*V. The central issue is about the four-force definition: dP/dt or Mo*dV/dt? The equality only holds when you assume that Mo does not change, in which case we are dealing with a CLOSED physical system. If we allow Mo to change, the physical system becomes an OPEN one and the four-force will depend on the definition you use.The problem with the first definition (based upon dP/dt) is that, as a mathematical fact, in the Newtonian limit (c --> infinity) it predicts for zero external force that the particle's acceleration is frame dependent when Mo is allowed to change, so that the Galilean invariance is lost.

Thanks, Paul and to Dr Woodward as well. Lots to ponder...

Quote from: cuddihy on 12/10/2010 10:15 amThanks, Paul and to Dr Woodward as well. Lots to ponder...Cuddihy:While you are pondering here is a couple of links to Wolfgang Rindler's book on Special Relativity, first edition that was first published in 1985. His four-force definition is in Chapter-5.http://catatankuliah.com/2010/08/14/wolfgang-rindler-introduction-to-special-relativity/ Rindler's improved 2nd edition, published in 1991, can be found at Amazon.com but it will set you back ~$55.00. http://www.amazon.com/Introduction-Special-Relativity-Wolfgang-Rindler/dp/0198539525/ref=sr_1_1?s=books&ie=UTF8&qid=1291998555&sr=1-1Woodward tells me that the 2nd edition has a much clearer discussion of the four-force than the first edition.

Guys. I have not been following this thread too closely so I'm not up to speed with much but a few questions.1. Would/could ME lead to wheel less vehicles, as in the movie I ROBOT ?2. Would/could it be classed simply as Anti Gravity ?3. As my main interest is Mars, would ME be a solution to landing large/heavy objects ?Cheers for the season to ALL.Mick.

Quote from: MickQ on 12/22/2010 03:02 amGuys. I have not been following this thread too closely so I'm not up to speed with much but a few questions.1. Would/could ME lead to wheel less vehicles, as in the movie I ROBOT ?2. Would/could it be classed simply as Anti Gravity ?3. As my main interest is Mars, would ME be a solution to landing large/heavy objects ?Cheers for the season to ALL.Mick.2. no. The Woodward effect does not significantly cancel or even shield gravity in an observable manner, only inertia.

Quote from: cuddihy on 12/22/2010 04:05 amQuote from: MickQ on 12/22/2010 03:02 amGuys. I have not been following this thread too closely so I'm not up to speed with much but a few questions.1. Would/could ME lead to wheel less vehicles, as in the movie I ROBOT ?2. Would/could it be classed simply as Anti Gravity ?3. As my main interest is Mars, would ME be a solution to landing large/heavy objects ?Cheers for the season to ALL.Mick.2. no. The Woodward effect does not significantly cancel or even shield gravity in an observable manner, only inertia.hmmm... your answer is correct, but its does gives the impression you are implying it can shield inertia on the whole vehicle... to be more clear to MickQ, the Woodward Effect only messes with inertia on the particles that it uses to propel the craft.

I assume that how much force it can generate is still open to debate, and even them are trying to discover precisely how much force they can get out of the device/effect.I think it all depends of how much mass differential they can create between the "pushed" and the "pulled" particles... and how fast they can "push" them.

So the better question to ask is, in terms of these physics, what is the art of the possible given existing materials?