There is no "pull" of gravity, only the PUSH of flowing ether!
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From: NoEinstein on 14 May 2010 20:15 On May 14, 10:33 am, PD <thedraperfam...(a)gmail.com> wrote: > Dear PD, the Parasite Dunce: Thanks for copying... "something". You pasted: "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." For starters: The most common KE is for dropped objects. There is no "work done" by the net force. There is only the uniform force acting to cause a linear increase in the KE from second ZERO. If the force (such as a rocket engine) is cut off, the KE at the last instant before the cut-off will continue as a constant KE. Your supposed Work- Energy Theorem is NOT the same as the Law of the Conservation of Energy. If you can, find any place that says that WORK can increase without there being a reaction LOAD. Zero LOAD = ZERO work done. Ha, ha, HA! 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 -
From: Timo Nieminen on 14 May 2010 20:45 On May 15, 9:42 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 14, 7:46 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > Dear Timo: Since my New Science has gravity being temperature and > surface area dependent, you can't "know" the correct masses of those > stars. The "new" mass of each must consider the color temperature and > the surface area. All you know for sure is: The color temperature; > the orbital period; and the eccentricity (wobble). You can't know the > diameter(s) of the stars, because such is obscured by their own > brightness. Michelson successfully used his interferometer to measure > very small angles, but the width of stars wasn't possible I dont > think. However, you should verify this. The diameters of both Sirius A and Sirius B have been measured. Sirius A by Hanbury Brown and Twiss, using the intensity interferometer. (Wikipedia has this being done at Jodrell Bank in 1959; I thought that this was done at Narrabri somewhat later - the Narrabri instrument is optical, Jodrell Bank was radio.) Sirius B has been imaged by Hubble. Both diameters are known. > Obviously, you are intrigued with trying to confirm my New Science > from astronomy rather than from a simple Earthly experiment(s). Be it > known: The Laws of Physics are the SAME all across the Universe. > Doing experiments, here, makes the most sense, I think. If your New Science is invalidated by already existing astronomical observations, then we can save a lot of time by not bothering to do experiments that aren't needed. Since checking your predictions against astronomical observations only takes a few minutes, why not do it? > You can measure what you "think" is the total brightness. But you are > actually only measuring the very tiny cone of light that has come all > the way from the star(s) to the photoelectric cell. Which is enough to tell you the total brightness, for an isotropically radiating star. If you think that there is some reason that the measurement might be inaccurate enough to save your theory, do exlain. Again, just go outside at night and see for yourself - Sirius A is very bright, and Sirius B is invisible to the naked eye. > The color of that > tiny beam will match the color temperature of the star(s), but won't > give you a true idea of the actual lumen of a star, nor its surface > area. There could, indeed, be valid indicators of the surface areas, > but I'm not yet privy to how those could be determined. The luminosity (i.e., total brightness) and the temperature are enough to give you the surface area, at least to reasonable accuracy, simply by approximating the star as a black body. How accurate is such a black body assumption? Quite enough so, which we can tell from looking at the spectrum. > I clearly implied that the total gravity of a star is the total > luminosity over the known surface area of the star. Yes, you did. Alas, 1000 times the luminosity doesn't appear to give 1000 times the gravity. Did you really not check this before? It's such a simple and obvious check, the data are readily available online, and the mathematics is trivial (not that you should need the mathematics to be so trivial, in light of your tremendous mathematical ability). I wouldn't claim to have a theory explaining everything without checking that it actually explains at least the easily checked things.
From: NoEinstein on 14 May 2010 21:12 On May 14, 10:33 am, PD <thedraperfam...(a)gmail.com> wrote: > PD, the STONEHEAD: WORK IS NEVER NONE WITHOUT THERE BEING A RESISTING FORCE. HAIL THAT IS COASTING (or has large components of coasting) DOESN"T HAVE A SEMI-PARABOLICALLY INCREASING RESISTANCE. So your theory is patently wrong!. Oh, you never have quoted the text, nor shown the formulas that you claim consider that COASTING objects are increasing in either KE or work potential. Ha, ha, HA! NE > > 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 -
From: NoEinstein on 14 May 2010 21:14 On May 14, 7:46 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > Dear Timo: This is in reply to the END of your previous comment: 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. Air turbulence will be the limiting factor for how close. Also, since I dont know the torsion characteristics of the wire, nor its length, I have little idea what the osculation frequency will be. Ideally, you should put exactly 90 degrees of initial twist in the beam, measured with highest accuracy. Its important that the wire stay perpendicular, and be twisted 90 degrees. I recommend that you install a surveyors plumb Bob in line with the axis of the wire so that you can be sure the beam wasnt pulled sideways when you rotated it. A longer wire will be more forgiving of the accuracy of the initial twist. TIME how long it takes for the twisted beam to come to rest. Keep as much air motion out of the room as possible. Do that with the unheated ball, and tell me the time required for the beam to come to rest, and the number of osculations. 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? That ball will loose heat to the air so quickly that the average ball temperature should be what is used. Do you have an infrared, photonic thermometer? If so, you need to record the ball temperature about once per minute during the likely fifteen minutes or so for the beam to come to rest. My guess is that the heated ball will come to rest in 80% to 90% of the time it took for the unheated ball. In a fifteen minute experiment, you should easily be able to record the quicker time. Once you get that, THEN you can start doing the higher math to figure the likely increase in gravity. If you can document everything very well; even repeating BOTH experiments, you should publish the results in a technical journal, so a dozen or more interested physicists can tackle the gravity implications. This is important basic science, that can have far ranging implications in understanding the Universe, and conquering the Universe! NoEinstein > > On May 14, 8:01 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On May 13, 2:29 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > What you arent considering (relative to your 1,000 fold gravity > > difference) is that those stars, with the masses that you state, > > arent rotating about the midpoint between their centers. > > The size of the orbit - which we can directly measure since this > binary system is very close to us - and the orbital period - which we > can also directly measure - together give us the combined mass of the > two stars. Where the centre of the orbital motion lies along a line > joining them tells us where the centre of mass is, and how that > combined mass is divided between them. The figures of m_A = 2M, m_B = > 1M come from this. So, you now know where the centre of motion is, > relative to the 2 stars. If, according to your theory, this > calculation should be done differently, do it. > > > And you > > arent including the surface area of each star in the equation. > > Brighter stars will have a larger surface area per unit mass. Also, > > though the TOTAL gravity of a star is equal to the product of the > > luminosity and the surface area, > > No, "luminosity" is the total emitted power. It already includes the > surface area. "Luminosity" = "surface brightness" times surface area. > As I said in the previous post. Sirius A radiates about 1000 times as > much power as Sirius B. This already includes the effect of surface > area. Yes, Sirius A has a much larger surface area - that's why it's > brighter. Go and look for yourself - Sirius A is the brightest star in > the night sky, and Sirius B isn't visible to the naked eye. It's a big > difference in luminosity. > > Look up the numbers for yourself, at your fingertips via www. > > > the fraction of the gravity thats > > holding two objects together is the illuminated area, or the > > percentage of the total stars light that actually hits the other > > star. It is the addition of photons to the facing sides of stars that > > allows the ether pressure on the opposing sides to hold the two stars > > together. Please re read my original post, There is no PULL of > > gravity; only the PUSH of flowing ether! > > You're the one who said that the gravity is proportional to the photon > emission. You didn't give any other usable quantitative model of the > strength of gravity. If the gravitational force should be found some > other way, perhaps you should have said so, and said how. All I did > was test the quantitative model that you gave me. If it's the wrong > model, why did you give it? If it's the wrong model, give the right > one. > > One star emits 1000 times as many photons as the other, yet only > appears to have twice the gravity. This is compatible with > conventional physics, including conventional theories of gravitation. > If it isn't compatible with your theory, then perhaps reality has cast > its vote, the only vote which counts in science. > > > The easiest way for you to confirm my theory would be to heat the > > larger ball in the Cavendish experiment as hot as possible. The > > torsion slowing should occur quicker with the hot ball than with the > > same ball cold. No other measurements are required. Do THAT > > experiment, and find that the heated ball has more gravity, and you > > can sit back and let the astronomers and scientists all over the world > > quantify the temperature-variant gravity! I, the generalist, provided > > the spark of inspiration. If others get to determine more of the > > specifics, they can share in the glory. NoEinstein > > As I keep asking, and as you keep refusing to say: How large is the > effect supposed to be? In other words, how sensitive does the > experiment need to be to detect it? There isn't much point in trying > it without knowing this. Since you claim that gravity is proportional > to "photon emission", for masses above some threshold mass (which you > haven't explained or given even an approximate value for yet), and > radiation by hot bodies is well known and understood, why can't you > say how large the effect should be? A simple calculation, surely, and > should be trivial for you. Why not just answer? > > 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?
From: Timo Nieminen on 14 May 2010 23:45
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? |