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From: NoEinstein on 6 May 2010 22:09 On May 5, 12:34 pm, PD <thedraperfam...(a)gmail.com> wrote: > No imbecile, not even you, PD, instructs me to do anything! NE > > On May 5, 2:48 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On May 4, 11:44 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > On May 3, 10:07 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On May 3, 12:02 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > On May 1, 9:01 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > On Apr 27, 10:16 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > Dear PD, the Parasite Dunce: Several times before you have referenced > > > > > > Newton's ERRANT F = ma. > > > > > > Ah, excellent, just so it's clear. You're problem then isn't with > > > > > Einstein and the physics of the 20th century. It's with all of physics > > > > > since the 1600's. Basically, it's just ALL plain wrong, everything > > > > > that is taught to schoolchildren from the 3rd grade on. And you, in > > > > > your infinite genius, have discovered this by the power of reason.. > > > > > > > Most equations that contain a "mass" can be > > > > > > changed to be a UNIT mass of one pound (or whatever). The "textbook" > > > > > > definition of MOMENTUM is F = mv. > > > > > > I'm sorry, but that equation appears in no textbook anywhere. > > > > > If you disagree, cite the textbook and the page number. > > > > > > > The latter mass can also be changed > > > > > > to be a unit mass of one pound (or whatever). SO... Since both > > > > > > equations are forces, > > > > > > First of all, you just said it was an equation for momentum (though > > > > > you got it wrong), not a force. > > > > > > Good heavens, John, you've gotten confused two equations for two > > > > > different quantities, you can't even get one written down right and > > > > > you call the other one wrong. > > > > > > You're a mental case, John. > > > > > > > set the right half of the two equations to be > > > > > > EQUAL, or: ma = mv. Since the masses are both one pound unit masses, > > > > > > then, the resulting equation says: ACCELERATION = VELOCITY! Even an > > > > > > imbecile like you, PD, should realize that velocity, (or say) feet/ > > > > > > sec, isn't the same as feet/second EACH second! > > > > > > > Ironically, I was studying for college physics when I realized the > > > > > > conflict between those two equations. That same week, I concluded > > > > > > that the entire chapter on mechanics was screwed up. Newton' "Law", > > > > > > in words, says: For every uniform force, there is one and only one > > > > > > associated acceleration. The correct equation for that should have > > > > > > been F = a, provided, of course, that the relationships between those > > > > > > two variables are stipulated, or are included in a less generalized > > > > > > equation. > > > > > > > The equation for MOMENTUM, F = mv, is correct! For objects in free > > > > > > fall, or objects that are accelerating, the correct kinetic energy > > > > > > formula is my own: KE = a/g (m) + v / 32.174 (m). The latter replaces > > > > > > both KE = 1/2mv^2 and E = mc^2 / beta. What contributions have > > > > > > YOU made to science, PD? Ha. ha, HA! NoEinstein - Hide quoted text - > > > > > > - Show quoted text - > > > > > Dear PD: A thin "College Outline Series" book (that fits into the > > > > bookcase behind my computer chair) entitled "Physics", by Clarence E > > > > Bennett, states on page 19: "G. Momentum and Impulse. (1.) Momentum > > > > is defined as the product of the mass times velocity (mv)..." The > > > > letter F is used for momentum, because the equation defines forces. > > > > NoEinstein > > > > Oh, good grief. John, what is the ISBN on this book? I'd like to > > > secure it to look at it. > > > From what it is you just told me is in it, if I can verify that you > > > can indeed read it correctly, it is a horrible, horrible booklet and > > > should be burned as worthless. > > > > If this is what you learned physics from in your architectural > > > studies, then I have absolutely no doubt that you and your firm are on > > > thin legal ground. > > > > PD- Hide quoted text - > > > > - Show quoted text - > > > The AUTHOR and the Title are all you need. NE > > Then let me make sure we're talking about the same title, because > Clarence E. Bennett has written the following: > Physics Problems and How to Solve Them (1958, 1959, 1960, 1968, 1972, > 1985) > College Physics (College Outline Series) (1962, 1972) > Physics Without Mathematics (College Outline Series) (1949, 1953, > 1960, 1970) > New Outline of First Year College Physics (1944, 1946, 1948) > An Outline of First Year College Physics (College Outline Series) > (1937, 1943) > Physics (1952, 1954) > First Year College Physics (1954) > Descriptive Physics (1945) > > As you can see, it's important that I know more about the particular > title you own. The ISBN is either in the frontmatter or is printed on > the back of the paperback. It's a 10-digit number right next to the > letters I-S-B-N. Can you do that, John? > > PD- Hide quoted text - > > - Show quoted text -
From: BURT on 6 May 2010 22:16 On May 6, 6:34 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 5, 5:22 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > Dear Timo: Since physicists have been looking for the "missing mass" > in the Universe for decadesand not found itits occurred to me that > it was the estimates of the masses of the stars and galaxies that was > wrong. The observed red/blue shifts imply the rotational speed of the > arms of galaxies. Since physicists can calculate the centrifugal > force that must be countered by the "central" gravity source, over- > estimating the mass of stars would exaggerate the centripetal force > needed to keep the stars from flying away. I also realized that it > isn't 'just' the central gravity holding the galaxies together, it > also includes the EFFECTIVE central gravity of all of the stars, > combined. Think of that as being similar to having two equal size and > mass binary stars rotating about their common centerhalfway between > the two stars. Though there is no "mass" at the center, the two stars > orbit as though there is a mass there. > > Since gravity is "distance proportional" (actually inverse > proportional) stars that aren't on a 'diameter' line, can still help > to keep the whole thing from flying outward. What then is flying outward? The whole of a gravity must be an influence. > That would be like > having lots of people hold hands to form a circle. If there is a > 'flying out' force, the tension (gravity) in their arms will keep the > circle together. NOTE: I strongly suspect that 'physicists', who > aren't structural engineers (like was my training), neglected to > consider the CIRCULAR routes of gravity, which could be 50 plus > percent of what is holding the galaxies together! > > Timo, a good way to 'estimate' the gravity sensitivity needed, is to > search for the accepted missing mass in the Universe (99%?); divide > that in half (due to the circular paths of gravity), yields 44.5% that > is unaccounted for. Since both the mass and the gravity force are > about equal in the fly out predicted by Newtons errant equation, > there would only need to be a 22.25% under-estimate of the gravity of > the stars, and a corresponding 22.25% over-estimate in the mass of the > stars. > > The gravity of a star is proportional to the surface area (not the > mass) and the surface temperature. It isnt proportional to the > internal temperatures, at all. Not counting solar flare temperatures, > determine the surface temperatures of different size and color stars. > Of course, that will be a plasma which you certainly cant do a > Cavendish on. The change in gravity that you seek is probably > linearly proportional to temperature. Assume the Earth to be zero > temperature. Find what percentage of the stars surface temperature > that you can achieve without melting the balls. My guess is you can > get about 10% of the typical surface temperature. 10% of 22.25% means > that you are hoping to detect a 2.225% increase in the gravity of > the balls. If you only heat the larger ball, increase its > contributing gravity by 2.225%, and leave the other ball(s) as they > were. > > The sensitivity of a well-designed Cavendish can probably verify the > gravity within .5% So, if you can run the experiment at all, the > results sought should be within the sensitivity! > > *** However, this just occurred to me: The gravity of every possible > star attractionnot just the circular and the cross-diameterwill be > helping to hold the galaxies together! As proved by the Andromeda > Galaxy (that has a zone without stars next to the center), Black Holes > have zero gravity. So, the multi-paths of gravity, taken together, > must be capable of holding, say, the Milky Way together without > needing a super-massive black hole (sic) at all! > > My theory, counter to Newtons Law of Universal (sic) Gravity, > states that the gravity of a star is directly proportional to the > temperature-determined, photon emissions over the entire surface area > of the star (without needing to consider the mass). But the > centrifugal force of stars orbiting the galaxy is directly > proportional to the MASSES of the stars. Timo, because of what Ive > just reasoned your Cavendish may not be sensitive enough. Until > someone does an every star gravity weave calculation for, say, the > Milky Way, I dont know if there is a 22.25% underestimate of star > gravity, or a 5%. Consider this: If you can heat one ball white hot, > and you DO detect a greater gravity, youve confirmed my theory. I > suppose that a temperature-corrected Law of Universal Gravitation > could take decades to validate. But that isnt all bad! Simply by > understanding that temperature affects gravity, can begin being used > to design, say, gravity drive spacecrafts. And Ive got THOSE on my > to-do list! NoEinstein > > > > > > > On May 5, 6:27 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On May 4, 7:59 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > Dear Timo: Obviously, you want an 'out'. You were so insistent that > > > the numbers be right, and the math be done before performing > > > experiments, that I assumed you were wanting ME to do all of those > > > things. > > > There is very little point in doing the experiment without knowing how > > big the effect is supposed to be. If I know in advance that our > > Cavendish apparatus isn't sensitive enough, there isn't any point in > > trying the measurement using it. But this is simply a very elementary > > idea in experimental physics, so you will already be thoroughly > > familiar with it. > > > Which is why your continued reluctance to actually say how large the > > expected effect should be is truly baffling. Why, even a typical > > physicist - an imbecile by your exalted standards - would be able to > > do the quantitative prediction without too much trouble, perhaps with > > a few hours of work. A genius of your level should have no trouble at > > all, and certainly shouldn't take more than a few hours. Why would it > > even take that long? Likely some tens of minutes at the most would > > suffice. > > > But once again, you refused to answer the question. Obviously, the > > hotter the balls, the larger the effect, but _how much_ larger? Don't > > you know? Can't you be bothered doing a calculation which should take > > well under an hour for somebody of your claimed ability when it would > > lend tremendous and convincing support to your theory? > > > > But if you are wishing to do the experiment, yourself, why > > > didn't you say just that? > > > I did say just that. But it seems Mr Genius can't understand plain > > English, or simply can't be bothered reading, because he's too busy > > writing irrelevant essays to distract attention from having had an > > error of billions of dollars per year pointed out. > > > But because you showed so abundantly that you're a rude and abusive > > arsehole, it clearly isn't worthwhile donating one's time, effort, and > > equipment to you. > > > Besides, you can't be bothered providing information - trivial for one > > of your intellect to provide - that's necessary for experiment to be > > useful, so again, it isn't worthwhile doing it. > > > Getting concrete experimental support for your theory would be vastly > > more convincing than endlessly repeating waffle, but you seem to > > prefer the waffle. Enjoy!- Hide quoted text - > > - Show quoted text -
From: NoEinstein on 6 May 2010 22:23 On May 5, 12:36 pm, PD <thedraperfam...(a)gmail.com> wrote: > PD: The L. C. catalogue card number is: 5241857. (look on page 19). Also, my The Wiley Engineer's Desk Reference, by Stanford I. Heisler, on page 94, says momentum = mv. A scripted style of the "m" is used to differentiate from "mass". That book errs by saying that the "units" is: (mass)-feet/secondwhich is bullshit! Momentum is measured in pounds! It is velocity proportional, and that is a simple, unit-less FRACTION NE > > On May 5, 2:56 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On May 4, 2:53 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > PD loves to extrapolate things into unworkability, so he can claim > > everything was invalid. MOMENTUM is: F = mv, expressed in pounds. > > He'll find that same equation (but not the correct units, pounds) in > > most textbooks. NE > > No, I won't, John. That equation F=mv is not listed in most > textbooks. > When you can clearly identify which title you think DOES have that > listed, then I can look for myself. > As it is, since you obviously have problems reading an understanding a > single sentence from beginning to end, I have my doubts. > > > > > > > > On May 4, 1:07 pm, af...(a)FreeNet.Carleton.CA (John Park) wrote: > > > > > PD (thedraperfam...(a)gmail.com) writes: > > > > > On May 3, 10:07=A0pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > >> Dear PD: =A0A thin "College Outline Series" book (that fits into the > > > > >> bookcase behind my computer chair) entitled "Physics", by Clarence E > > > > >> Bennett, states on page 19: "G. =A0Momentum and Impulse. =A0(1.) =A0Momen= > > > > > tum > > > > >> is defined as the product of the mass times velocity (mv)..." =A0The > > > > >> letter F is used for momentum, because the equation defines forces. =A0= > > > > > =97 > > > > >> NoEinstein =97 > > > > > > Oh, good grief. John, what is the ISBN on this book? I'd like to > > > > > secure it to look at it. > > > > > From what it is you just told me is in it, if I can verify that you > > > > > can indeed read it correctly, it is a horrible, horrible booklet and > > > > > should be burned as worthless. > > > > > To quote the Spartans on a quite different occasion: If. > > > > > I can't help noticing that the actual quoted passage is reasonable and > > > > the inference about forces is purely in NE's words. > > > > Exactly. > > > > For what it's worth, momentum's *definition* is not mv, either. > > > Electromagnetic fields have momentum, but this expression certainly > > > does not work for them. The formula works for a certain class of > > > matter-based objects traveling at low speed, and that's it. > > > > PD- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: Timo Nieminen on 7 May 2010 02:21 On Thu, 6 May 2010, NoEinstein wrote: > My theory, counter to Newtons Law of Universal (sic) Gravity, > states that the gravity of a star is directly proportional to the > temperature-determined, photon emissions over the entire surface area > of the star (without needing to consider the mass). Measurements of the "mass" of stars in binary systems are really measurements of the gravitational force of stars in binary systems. If you're right, a plot bolometric luminosity versus measured "mass" of stars in binary systems should give a straight line (within experimental error). Since you're obviously smart enough to have realised this long ago, and are also obviously smart enough to have checked this yourself, what was the result? The "mass", as measured from binary orbits, is available for many stars (including nearby ones such as Alpha Centari A and B, Sirius A and B), and the relevant information is readily available online, so I suppose I could check this myself if you don't care enough to provide the result (or didn't care enough to bother checking something so trivial). If it isn't a directly proportional linear relationship, what would that mean? > Timo, because of what Ive > just reasoned your Cavendish may not be sensitive enough. Until > someone does an every star gravity weave calculation for, say, the > Milky Way, I dont know if there is a 22.25% underestimate of star > gravity, or a 5%. So, you don't know? Why not apply your mighty intellect and provide the answer? > Consider this: If you can heat one ball white hot, > and you DO detect a greater gravity, youve confirmed my theory. It would _support_ your theory, not confirm it in any absolute sense. If one tries this and _doesn't_ detect a greater gravitational force, would that mean your theory is wrong and it's time to forget it and move on?
From: PD on 7 May 2010 09:08
On May 6, 8:42 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 5, 10:24 am, PD <thedraperfam...(a)gmail.com> wrote: > > Dear PD, the Parasite Dunce: If you have "other" supporting evidence > for Lorentz (ha!), paraphrase it! You are all bluster and no > substance! NoEinstein John, as I said, there are SCORES of independent experiments that have all provided experimental evidence. You might as well be asking for a paraphrased summary of the support for Newton's laws of motion. If you want to understand the depth of the experimental support, then you're going to have to immerse yourself in the OVERWHELMINGLY MASSIVE documentation of that support. That's the only way to truly convince yourself. > > > > > On May 4, 7:22 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On May 4, 11:38 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > Whatever the truth is, PD contorts it. "Rubber Rulers" has no > > > supporting experiment! Lorentz, the imbecile, used RR to 'explain' > > > the nil results of M-M. Then, supposed scientists say that M-M > > > SUPPORTS Lorentz! Where are the brains, and WHERE is the scientific > > > method! NE > > > Of course there are supporting experiments, John. You seem to be under > > the impression that the MMX was the only experiment ever done to test > > relativity and that the whole of relativity rests on this one > > experiment, so that if you somehow fault the MMX, then all of > > relativity falls. > > > Nothing could be further from the truth, John. Relativity has been > > tested in scores of experiments, all independent of each other. > > > > > On May 3, 9:43 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > On May 3, 11:51 am, PD <thedraperfam...(a)gmail.com> wrote:> On May 1, 8:25 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > On May 1, 11:00 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > Dear PD, the Parasite Dunce: You just said that "physics isn't > > > > > > > determined by logic". Of course, you would think that! That's > > > > > > > because you don't know HOW to reason! > > > > > > > Well, it's because physics is a science, which means that it invokes > > > > > > the scientific method, and it determines truth by experimental test, > > > > > > not by logic. > > > > > > Dear PD: WHERE was the "scientific method" when Lorentz proposed his > > > > > ANTI-ENGINEERING, "rubber ruler" explanation for the nil results of M- > > > > > M? > > > > > Lorentz's proposal was subject to experimental test, NoEinstein. > > > > That's how science works. > > > > And what on earth makes you think that this stuff is "anti- > > > > engineering"? > > > > Perhaps you don't know that engineers make use of relativity in their > > > > designs whenever it is needed? If it's anti-engineering, why are > > > > engineers happy to use it as needed? > > > > > > And where was the scientific method when both Coriolis and > > > > > Einstein wrote energy equations that were exponential, and thus in > > > > > violation of the Law of the Conservation of Energy? > > > > > Those energy equations have also been thoroughly tested in experiment, > > > > John, exactly as I was stating. You on the other hand are trying to > > > > rule them out with your bandy-legged logic, rather than considering > > > > independently verified experimental tests. > > > > > > When the truth be > > > > > known, PD, is this low I. Q. flunky who compensates by constantly > > > > > faulting his superiors. He has never stated a single contribution > > > > > that he has made to science. For one who devotes so much time to... > > > > > 'science' shouldn't PD have... "something" to show for it? > > > > > NoEinstein > > > > > What do you think I should have to show for it, John? > > > > > PD- Hide quoted text - > > > - Show quoted text - > > |