There is no "pull" of gravity, only the PUSH of flowing ether!
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From: BURT on 15 May 2010 02:05 On May 14, 8:45 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > On May 15, 11:14 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On May 14, 7:46 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > For example, double the absolute temperature of the balls; easy to do, > > > just heat to about 330C. 16 times the radiated power, at double the > > > peak frequency, as compared with a room temperature ball. How much > > > larger should the gravitational force be? > > > Send me a drawing of the relative size, weight and spacing of the > > balls. I can approximate the gravity, especially, if I know how close > > the big ball is to the closest one on the beam. There will be a more > > pronounced gravity effect if the balls are fairly close togetherlike > > six inches. > > No, I'm not asking what should the result be in a particular Cavendish > apparatus. I'm asking about the gravitational force between two balls, > and how large your predicted temperature dependence is supposed to be. > > You say: > > > Once you get that, THEN you can > > start doing the higher math to figure the likely increase in gravity. > > but this is what needs to come _first_. It is very useful to have some > idea of how large some effect you're planning to measure is. If > somebody asks you to help measure the length of something > (accurately), you want to know whether to bring along a long tape > measure, a metre ruler, vernier calipers, a micrometer, or something > else. > > How large is your predicted effect? Given this, one can see if > existing apparatus is sufficient, or if not, what apparatus is needed. > > > Air turbulence will be the limiting factor for how > > close. > > It is possible to do experiments in vacuum. Is it necessary to do the > experiment in vacuum? Thus the question, how large is the effect > supposed to be? > > > Also, tell me the metal type you > > are using, and how hot you can safely heat the entire ball. Is 330C > > the highest you can go? > > Of course 330C isn't the hottest possible. 330C is double room > temperature, convenient for giving a quantitative prediction (16 times > radiated power, at 2 times the peak frequency). > > If the experiment should yield a large enough effect at 330C (or even > lower), then it might well be better to do it at a lower temperature. > Less convection, etc. How large should the effect be at 330C? This is > very useful information in order to determine the best temperature to > try the experiment at. > > Why should it take "higher math" to give a quantitative result? What > higher math? Of course, since you yourself have spoken many times of > your superior math skills, it shouldn't be a significant obstacle to > you even if "higher math" is required. Even with only a fraction of > your ability, it should be possible to do this easily enough, surely > in less than an hour. So why not just say how large the effect is > supposed to be? How does the aether push speed up in increasing strength of gravity? Mitch Raemsch
From: John Murphy on 15 May 2010 03:20 On 15 May, 07:05, BURT <macromi...(a)yahoo.com> wrote: > On May 14, 8:45 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > On May 15, 11:14 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On May 14, 7:46 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > For example, double the absolute temperature of the balls; easy to do, > > > > just heat to about 330C. 16 times the radiated power, at double the > > > > peak frequency, as compared with a room temperature ball. How much > > > > larger should the gravitational force be? > > > > Send me a drawing of the relative size, weight and spacing of the > > > balls. I can approximate the gravity, especially, if I know how close > > > the big ball is to the closest one on the beam. There will be a more > > > pronounced gravity effect if the balls are fairly close togetherlike > > > six inches. > > > No, I'm not asking what should the result be in a particular Cavendish > > apparatus. I'm asking about the gravitational force between two balls, > > and how large your predicted temperature dependence is supposed to be. > > > You say: > > > > Once you get that, THEN you can > > > start doing the higher math to figure the likely increase in gravity. > > > but this is what needs to come _first_. It is very useful to have some > > idea of how large some effect you're planning to measure is. If > > somebody asks you to help measure the length of something > > (accurately), you want to know whether to bring along a long tape > > measure, a metre ruler, vernier calipers, a micrometer, or something > > else. > > > How large is your predicted effect? Given this, one can see if > > existing apparatus is sufficient, or if not, what apparatus is needed. > > > > Air turbulence will be the limiting factor for how > > > close. > > > It is possible to do experiments in vacuum. Is it necessary to do the > > experiment in vacuum? Thus the question, how large is the effect > > supposed to be? > > > > Also, tell me the metal type you > > > are using, and how hot you can safely heat the entire ball. Is 330C > > > the highest you can go? > > > Of course 330C isn't the hottest possible. 330C is double room > > temperature, convenient for giving a quantitative prediction (16 times > > radiated power, at 2 times the peak frequency). > > > If the experiment should yield a large enough effect at 330C (or even > > lower), then it might well be better to do it at a lower temperature. > > Less convection, etc. How large should the effect be at 330C? This is > > very useful information in order to determine the best temperature to > > try the experiment at. > > > Why should it take "higher math" to give a quantitative result? What > > higher math? Of course, since you yourself have spoken many times of > > your superior math skills, it shouldn't be a significant obstacle to > > you even if "higher math" is required. Even with only a fraction of > > your ability, it should be possible to do this easily enough, surely > > in less than an hour. So why not just say how large the effect is > > supposed to be? > > How does the aether push speed up in increasing strength of gravity? > > Mitch Raemsch The aether is a medium that will accommodate any postualte that suits it! -- Harbinger.
From: BURT on 15 May 2010 04:15 On May 15, 12:20 am, John Murphy <london.accommodation.homest...(a)googlemail.com> wrote: > On 15 May, 07:05, BURT <macromi...(a)yahoo.com> wrote: > > > > > > > On May 14, 8:45 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > On May 15, 11:14 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On May 14, 7:46 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > > For example, double the absolute temperature of the balls; easy to do, > > > > > just heat to about 330C. 16 times the radiated power, at double the > > > > > peak frequency, as compared with a room temperature ball. How much > > > > > larger should the gravitational force be? > > > > > Send me a drawing of the relative size, weight and spacing of the > > > > balls. I can approximate the gravity, especially, if I know how close > > > > the big ball is to the closest one on the beam. There will be a more > > > > pronounced gravity effect if the balls are fairly close togetherlike > > > > six inches. > > > > No, I'm not asking what should the result be in a particular Cavendish > > > apparatus. I'm asking about the gravitational force between two balls, > > > and how large your predicted temperature dependence is supposed to be.. > > > > You say: > > > > > Once you get that, THEN you can > > > > start doing the higher math to figure the likely increase in gravity. > > > > but this is what needs to come _first_. It is very useful to have some > > > idea of how large some effect you're planning to measure is. If > > > somebody asks you to help measure the length of something > > > (accurately), you want to know whether to bring along a long tape > > > measure, a metre ruler, vernier calipers, a micrometer, or something > > > else. > > > > How large is your predicted effect? Given this, one can see if > > > existing apparatus is sufficient, or if not, what apparatus is needed.. > > > > > Air turbulence will be the limiting factor for how > > > > close. > > > > It is possible to do experiments in vacuum. Is it necessary to do the > > > experiment in vacuum? Thus the question, how large is the effect > > > supposed to be? > > > > > Also, tell me the metal type you > > > > are using, and how hot you can safely heat the entire ball. Is 330C > > > > the highest you can go? > > > > Of course 330C isn't the hottest possible. 330C is double room > > > temperature, convenient for giving a quantitative prediction (16 times > > > radiated power, at 2 times the peak frequency). > > > > If the experiment should yield a large enough effect at 330C (or even > > > lower), then it might well be better to do it at a lower temperature. > > > Less convection, etc. How large should the effect be at 330C? This is > > > very useful information in order to determine the best temperature to > > > try the experiment at. > > > > Why should it take "higher math" to give a quantitative result? What > > > higher math? Of course, since you yourself have spoken many times of > > > your superior math skills, it shouldn't be a significant obstacle to > > > you even if "higher math" is required. Even with only a fraction of > > > your ability, it should be possible to do this easily enough, surely > > > in less than an hour. So why not just say how large the effect is > > > supposed to be? > > > How does the aether push speed up in increasing strength of gravity? > > > Mitch Raemsch > > The aether is a medium that will accommodate any postualte that suits > it! > > -- > Harbinger.- Hide quoted text - > > - Show quoted text - In that case it is blank slate like string dimensions are. Mitch Raemsch
From: tadchem on 15 May 2010 18:10 On May 7, 3:57 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 7, 2:21 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > Dear Timo: On the one hand you compliment me; on the other you chide > me for not having all of the numbers at my fingertips. Im pleased > that you know so much about astronomy. That was a major hobby of mine > through high school. I saw things like stars that werent round, > and I never realized that most were galaxies. I would lie in bed at > night wondering what the mechanism of gravity is. I heard that the > many colors of light related to the temperature of the source. It > was by reasoning, alone, that Ive concluded that all photons are > identical regardless of the temperature. The only temperature variant > in light is the spacing of the photons. Photons carry some ether away > like hobos on a train, which eventually jump off. Since gravity is > directly proportional to photon emission (not gravitons, which dont > exist), then it is the luminosity and the temperature of the light > that determine the gravity of stars. > > At room temperatures gravity is mass proportional, and matches > Newtons law. There has to be an object-size threshold that DENIES > mass in favor of surface area and temperature. I suspect that a > heated Cavendish ball will have gravity somewhere between the room > temperature, and the white hot. > > I had concerns with my theory that very cold planets would have > limited infrared emissions. Then, I realized that it is the SUN that > provides the photons to the colder planets, that keeps the gravity > forces going. Block off that solar energy from a planet, and it will > go flying out on its tangent. Ive realized that the estimates of the > masses of the planets are probably wrong. Kepler had the Laws of > Planetary Motion exactly right, if one substitutes apparent mass, > for the object mass. > > Timo, you probably know that my mind isnt a compendium with every bit > of data youre seeking. Until about grade 5, I had an ear > photographic memory. Later, I hated courses requiring memorization > (history), favoring, instead, courses that require analytical ability, > like math and science. I would be happy to assist you in quantifying > gravity. But most of my realizations about the Universe have come > from accepting the data of others, and making my theories not be in > conflict. Now, since Ive opened up this gravity-can-of-worms to > include temperature, there isnt any raw data to readily aid me in > finding the answers, no matter how smart I am. > > I hope you can select some little corner of that problem and attack > it, objectively. Keep me apprised of your progress. Its possible I > might have a spark of inspiration that could help you. *** Because of > you, I realized that every star in a galaxy gets tugs from every other > star. Successfully calculating the STRUCTURE of that unifying force > will require a super-computer. I suspect that those resultstogether > with my temperature determinant gravitywill show that ZERO mass > (gravity) is needed at the galactic centers. And since the Big Bang > never happened, there is no dark matter, whatsoever, needed to hold > the Universe together. > > Im busy trying to save the USA via my New Constitution. The way most > people think about science, having that man who disproved > Einstein write a constitution may not give me very many Brownie > Points. But I invite any of you who are interested to go to Political > Forum and read: Start the Revolution! Government is out-of-touch > with the People! Talk-it-up! NoEinstein (AKA John A. > Armistead) > > > > > > > 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?- Hide quoted text - > > - Show quoted text - So how does a siphon work? Tom Davidson Richmond, VA
From: NoEinstein on 15 May 2010 18:39
On May 14, 10:33 am, PD <thedraperfam...(a)gmail.com> wrote: > Dear PD: Nowhere in any high school physics text does it say that work can be done when pushing against ZERO resistance. Don't you recall agreeing that a hockey puck sliding many feet across slick ice isn't increasing in KE simply because a "displacement" has occurred? You, sir, are fighting for your crumby "science" life. I'm still King of the Hill, and you are a lying, sidestepping fool. Your legacy is that you are one of the most pathetic people on planet Earth. NoEinstein > > On May 14, 3:13 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On May 7, 6:07 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > PD hasn't quoted any authoritative source showing that WORK is in any > > way involved in calculating KE. > > Oh, yes, I have, John. You don't seem to remember anything that was > told to you the day before. > Do you like easy to read pages? Here's one for students at West > Virginia University:http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Work/WorkEngergyTheor... > "The energy associated with the work done by the net force does not > disappear after the net force is removed (or becomes zero), it is > transformed into the Kinetic Energy of the body. We call this the Work- > Energy Theorem." > > > And he hasn't quoted any > > authoritative source saying that "work" can be done simply by > > COASTING, against no resistance! > > The definition of work is in high school books, John. > > > And he certainly can't explain how > > 'gravity' could possibly 'know' the velocities of every falling object > > (like hail from varying heights) and add the exact semi-parabolic KE > > increase to each. > > Doesn't have to, John. The force is not solely responsible for the > increase in energy. The work is. The work is the product of both the > force and the displacement. That's how the work increases in each > second. It's simple, John. Seventh graders can understand it. I don't > know why you're so much slower than the average 7th grader. > > > > > In short, PD is a total, sidestepping FRAUD! And > > 95% of the readers know that he's a fraud! NoEinstein > > > > On May 7, 3:16 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On May 7, 9:12 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > PD, you are a LIAR! Never ONCE have you explained why KE = 1/2mv^2 > > > > isn't in violation of the Law of the Conservation of Energy. Until > > > > you do (and you CAN'T) everyone will know that you are just an air- > > > > head FRAUD! NoEinstein > > > > Oh, but I have. If you really need to have it explained again, I ask > > > you this time to print it out. > > > > The law of conservation of energy says that any change in the energy > > > of a system must be due solely to the work done on the system. > > > > The work is the force acting on the object times the displacement of > > > the object. So any change in energy of the object must be due solely > > > to this work. > > > > In the case of a falling body released from rest, we'll look at the > > > increase in the kinetic energy, which must be due to the work done by > > > the only force acting on the body -- gravity. If the increase of > > > kinetic energy the body has at any time is accounted for by the work > > > that was done on the body during that time, then we know that the law > > > of conservation of energy has been respected. > > > > In the first second, the body will fall 16 ft. In the next second, it > > > will fall an additional 48 feet. In the third second, it will fall an > > > additional 80 feet. During these first three seconds, the force has > > > remained constant, so that it is the same in the first second, the > > > second second, the third second. The speed increases linearly, so that > > > it is falling at 32 ft/s after the first second, 64 ft/s after the > > > second second, and 96 ft/s after the third second. > > > > Now, let's take a look at the work. The work done since the drop, > > > after the first second, is the force of gravity times the > > > displacement. This is mass x g x (16 ft). So this is how much kinetic > > > energy the object has after one second. Now, in the second second, > > > we'll add more work, in the amount mass x g x (48 ft), since that's > > > the displacement for the next second. This increases the kinetic > > > energy of the body, so that it now has kinetic energy mass x g x (16 > > > ft + 48 ft) = mass x g x (64 ft), and that number is four times bigger > > > than it was after the first second. Now, in the third second, we'll > > > add more work, in the amount mass x g x (80 ft), since that's the > > > displacement for the next sentence. Since energy is conserved, this > > > added energy must add to the kinetic energy of the body, so that it > > > now has kinetic energy mass x g x (64 ft + 80 ft) = mass x g x (144 > > > ft), and that number is nine times bigger than it was after the first > > > second. > > > > Now, it should be plain that the kinetic energy is conserved, since > > > the only thing that has been contributing to it is the work done in > > > subsequent seconds. We lost nothing, and we added only that which > > > gravity added. The energy is conserved. > > > > It should also be apparent that the kinetic energy is increasing in > > > the ratios 1:4:9. > > > Meanwhile, the velocities are increasing linearly, in the ratios > > > 1:2:3. > > > > Now, any fourth grader can see that we've completely conserved energy, > > > losing track of nothing, and yet the kinetic energy is increasing as > > > the square of the velocity. 1:4:9 are the squares of 1:2:3. > > > > There is no violation of conservation of kinetic energy, and yet KE is > > > proportional to v^2. > > > > Now, don't you feel silly that a 4th grader can understand all of > > > this, but you've never understood it? > > > > > > On May 6, 8:54 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > On May 5, 11:43 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > OH? Then please explain, PD, how a UNIFORM force inputthe static > > > > > > weight of the falling objectcan cause a semi-parabolic increase in > > > > > > the KE. Haven't you heard?: Energy IN must = energy OUT! > > > > > > NoEinstein > > > > > > I have explained this to you dozens of times. I gather that you do not > > > > > remember any of those posts, and you do not know how to use your > > > > > newsreader or Google to go back and find any of those dozens of times > > > > > when it has been explained to you. > > > > > > I surmise that you are slipping into dementia, where each day begins > > > > > anew, with any lessons learned the previous day forgotten. > > > > > > I don't think it's a good use of my time to explain the same thing to > > > > > you each day, only to have you retire at night and forget it by > > > > > morning, do you? > > > > > > PD- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > - Show quoted text - |