There is no "pull" of gravity, only the PUSH of flowing ether!
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From: spudnik on 26 May 2010 15:55 there's only one possible cure for that kind of thing; sit back & read Shakespeare, aloud; switch-off with the KJV Holy Bible (translated by WS & co. .-)... unless, you hate Shakespeare more than the British Petroleum oligarchy. thusNso: that is not an "image of a photon," but an "artist's impression" of some heuristic statement by a theoretician (which is as much status as any of us can have, hereinat). the wave & particle are merely mathematical duals, not that the object in question is both at the same time, or neccesarily both at different times. > http://superstruny.aspweb.cz/images/fyzika/foton.gif thusNso: Death to the lightconeheads; long-live Minkowsksi! thusNso: "real-valued time" is why, we have quaternions; it's the "scalar" in Hamilton's lingo of vectors. now, you mentioned tensors, and that is apropos, because it is used for stress & strain, which are clearly irrereversible; perhaps, that is one of the first math-physics examples of it. thusNso: I haven't read _Disquisitiones_ in Latin, either, but there are good translations & it is highly recommended by the LaRouchies ... they should put it on their website, like they have *Les OEuvres du Fermatttt*, but you can look at some cool tutorials, in the meantime, at wlym.com. thusNso: I never read a word about Palin's hubbie's Seccesh "movement" in the Liberal Media (Owned by consWervatives) and that is sort-of the issue in AZ. I'm all for kids whose parents managed to sneak across the border & give birth, but I was taken aback by the "sense of entitlement" that the older kids have, about college (the DREAM Act; I stated to a group of them, that crossing the border is essentially a Mexican "rite of passage," and it is certainly not very dangerous as a proper hike, if you check the FAQs and maps & so forth from the Mexican goment (and those advocacy/ haven groups in the USA). well, it's either that or college *in* Mexico, or you'll probably be made to join a gang. La Raza d'Atzlan are openly racist, not just by their title; at least, that's the impression that I got, attending one of their meetings at UCLA, two or three years ago -- it's in their God-am constitution. of course, teh real problem is "free trade," and this is already here to roost; the little spill in the Gulf is being used by British Petroleum -- which is also the #1 driller in the Alaska North Slope, that Ted Palin works for -- to creata an "outsourcing" mandate to solve the problem, because we can't do it with our post-industrial cargo cult. well, screw it; read LaRouche, if you want to know the history with Lincoln and his "Spot Resolutions;" Cinco de Mayo should be a pan-american holiday! thusNso: Dear AG candidate Kelly; no change from Jerry Brown's '69 "platform," eh? it is intolerably stupid, insofar as we do need "fossilized fuels TM (sik)," to not get our share from our own "reserves." really, though, it is merely biomass, and the techniques have progressed since '69. Dubya's bro's ban offshore of Florida (and Louisiana) seemed like a tactical maneuver to support the oilcos' scarcity programme in our state. (why O why O why do folks believe, that the oilcos did not support the Kyoto Protoccol, which was just another cap'n'trade "free trade" nostrum, that Dubya'd have undoubtdely signed, if he had been told?) British Petroleum, the balls-out advocate of cap'n'trade, "Beyond Petroleum," is also the biggest company in the Alaska North Slope -- doesn't any body wonder, why no-one asked Palin about her BP-employed hubbie, and his Seccesionist ideals? one must take into consideration, with all of the hype about it, that oil comes out of the ground underwater in "seeps," under pressure. so, how much would come out, if BP et al ad vomitorium were not pumping like crazy? Waxman's current cap'n'trade bill just mandatorizes the huge, voluntary cap'n'trade since 2003 -- tens of billions in hedging per annum. what the Liberal Media (Ownwd by consWervative) don't talk about, is that he brought the first cap'n'trade bill in '91, under HW (who worked with Gore on the Kyoto cap'n'trade). what it amounts to, as Waxman basically admitted to, when he was at UCLA, is "let the arbitrageurs raise the price of energy, as much as they can in the 'free market' -- free beer, freedom!" a small, adjustable carbon tax would achieve the same ends -- as I even read "in passing" in a guest editorial in the WSUrinal, as well as from an "expert" in a UCLA seminar, but who said that it was (some how) "politically impossible" -- without being the Last Bailout of Wall Street (and the City of London). --mister Kelly, please, take me off of your list, Brian H. --Light: A History! http://wlym.com
From: NoEinstein on 29 May 2010 14:06 Dear Timo: Your bad attitude preceeds you. Like I just told PD, you don't qualify to have any sort of back-and-forth conversation with me. But if you have a valid science question, and if my answering it might be of interest to the readers, put your statement in the FIRST two sentences of your reply. Have you made that '+new post' yet? NE > > > > > > > A two year old kid pushing on a square marble block as tall > > as the kid, is applying a FORCE to the block, even though there is > > ZERO displacement. Objects which are being held in some high place > > prior to being dropped in an experiment, are actually already being > > acted upon by the steady force of gravity that's = the object's > > weight. The latter static weight is my contribution to understanding > > momentum-variant kinetic energy. My correct formula, which replaces > > KE = 1/2mv^2, and E = mc^2 / beta, is: KE = a/g (m) + v / 32.174 > > (m). An object falling for one second will have a KE (force in > > pounds) of two weight units; two seconds, three weight units; three > > seconds, four weight units; four seconds 5 weight units. Those values > > were confirmed in TWO KE experiments, including that $40.00 one which > > you could easily do yourself. But since your motive is to disagree, > > you'll never accept any science truth. NoEinstein > > > KE = 1/2mv^2 is disproved in new falling object impact test.http://groups.google.com/group/sci.physics/browse_thread/thread/51a85... > > Here's another simple derivation for you, so you can see how it is > inescapable. > There is a nice kinematic relationship that you will find in > engineering texts, as well as high school texts: (v_final)^2 - > (v_initial)^2 = 2*a*x, where x is the displacement, a is the > acceleration, and v_final and v_initial are final and initial > velocities. I can show you where this relationship comes from, if you > like. > > Now, we have the definition of work: > Work = Net force x displacement. > But we also know from Newton's 2nd law, which was written in the > 1600's, > Net force = mass x acceleration > So by arithmetic (please don't tell me this confuses you) > Work = mass x acceleration x displacement. > But we can recognize that (acceleration x displacement) from the > kinematic relation I mentioned above -- there it is on the right hand > side, next to the factor of 2. > So we also know acceleration x displacement = (1/2)(v_final)^2 - (1/2) > (v_initial)^2 > And so again by arithmetic > Work = (1/2)(mass)(v_final)^2 - (1/2)(mass)(v_initial)^2 > So you see? The right hand side is the change in the kinetic energy, > and here we have the law of conservation of energy: any change in the > kinetic energy is due to the work done. > > All we used in this derivation were three things, which you surely > must accept: > 1. The kinematic relationship mentioned above > 2. Newton's 2nd law: F=ma > 3. Work = force x displacement > and besides that, just the eensiest bit of arithmetic. > > Awfully hard to find fault with it, isn't it? > > > > > > > > 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- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -... > > read more »
From: NoEinstein on 1 Jun 2010 15:25 On May 25, 6:46 pm, PD <thedraperfam...(a)gmail.com> wrote: > Dear PD: Your comment, below, about equal and opposite thrust force and resistance force would be astute, if not for this fact: Objects at rest, waiting to be dropped, are already experiencing a downward force equal to their static weight. As soon as the retaining force is released, there is a dynamic imbalance that starts the object accelerating at 'g'. The downward force can never exceed the object's inertia. If 'a' is substituted for 'g', the acceleration could (perhaps) be greater, and the force would not be limited to the inertia. The latter is true ONLY for objects dropped near the Earth. My correct kinetic energy equation, KE = a/g (m) + v / 32.174 (m) works anywhere in the Universe, as long as the "Pound" of reference is that for 'g', or on Earth. NE > > On May 23, 4:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On May 20, 9:42 am, PD <thedraperfam...(a)gmail.com> wrote: > > > Dear PD: If you had ever taken a course in structural engineering, > > there are TWO distinct types of force interactions. The STATIC ones > > always have the opposing forces being equal. > > Yes, indeed, so the net force is zero, so the acceleration is zero. > Hence the word "static". > > > But the DYNAMIC ones > > only have a FORCE equal to the LESSER resistance of the two. > > What??? > > Let's take a simple example. There is a wooden block on a wooden > inclined plane, and the plane is tipped up until the block starts to > slide, accelerating. List the forces acting on the block, what their > origin is, what direction the force is pointing. > > > You've > > never realized that DYNAMICS limits the downward force on the falling > > object to be whatever the INERTIA of the object is. Since the inertia > > of a one pound mass will always be just one pound, there can never be > > an exponential KE increase, because the force and the resistance are > > equal. > > WHAT???? > If the force and resistance were equal, then there would be two equal > and opposite forces acting on the object, and you'd be back to a > static situation. Try again. > > See the block example. > > > > > Since your precious WORK would have to increase exponentially, > > that violates simple dynamics, because the distances traveled by > > falling objects differ so markedly from second to second. I hope that > > you are sitting down, PD, because the latter statement means that > > there can neither be an increasing Work nor and increasing KE, from > > just the accruing COASTING components, which you are so reluctant to > > acknowledge. The COASTING components of the distance of fall for all > > dropped objects KILLS your made-up science! NoEinstein > > > > On May 20, 7:40 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On May 18, 9:18 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > On May 17, 6:04 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > On May 17, 2:59 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > Dear PD, the Parasite Dunce: You have changed the subject AWAY from > > > > > > KE, > > > > > > No, I didn't. I responded directly to your post about ether flow on > > > > > muons. > > > > > > > because you can't find any place in any text that states: "Work is > > > > > > being done even if there is no resistance. (sic) The only requirement > > > > > > to have work is that there be a displacement. (sic) Thus, if a hockey > > > > > > puck slides twice as far across the ice, twice as much work was done, > > > > > > and there is twice as much KE in the puck, even if the ice is > > > > > > frictionless. (sic)." > > > > > > You are not paying attention. > > > > > Remember, PD: I am King of the Hill in science. It isn't "my job" to > > > > pay attention to you. > > > > :>) > > > Just make sure you tell yourself that each day in the mirror. If you > > > like, please add the line, "And I am the heir to the throne of the > > > kingdom, by birthright." You may also consider adding, if you are > > > feeling confident, "And I am irresistible to women." > > > > > > There is no work if there is no force present, even in the absence of resistance. > > > > > I correct you: 'There is no work if there is no force present', AND > > > > there is no corresponding resistance > > > > There is no work done if there is no net force, regardless whether > > > there is resistance or not. > > > > > > There is work if there is a force present, even in the absence of resistance. > > > > > Dear PD: Newton's Laws of Motion require: "For every action there > > > > must be an equal and opposite reaction." It is IMPOSSIBLE to apply a > > > > force... UNLESS there is a corresponding resisting force! > > > > Newton's 3rd law is not a statement about a resisting force. That is a > > > common mistake by high school students that is corrected in virtually > > > every textbook on the subject. For example, from the high school text > > > that I've been quoting from, page 133: > > > "One important thing to remember about action-reaction pairs is that > > > each force acts on a different object. Consider the task of driving a > > > nail into wood.... To accelerate the nail and drive it into the wood, > > > the hammer exerts a force on the nail. According to Newton's third > > > law, the nail exerts a force on the hammer that is equal to the > > > magnitude of the force that the hammer exerts on the nail. > > > "The concept of action-reaction pairs is a common source of confusion > > > because some people assume incorrectly that equal and opposite forces > > > balance one another and make any change in the state of motion > > > impossible. If the force that the nail exerts is equal to the force > > > the hammer exerts on the nail, why doesn't the nail remain at rest? > > > "The motion of the nail is affected only by the forces acting on the > > > nail. To determine whether the nail will accelerate, draw a free-body > > > diagram to isolate the forces acting on the nail.... The force of the > > > nail on the hammer is not included in the diagram because it does not > > > act on the nail. According to the diagram, the nail will be driven > > > into the wood because there is a net force acting on the nail. Thus, > > > action-reaction pairs do not imply that the net force on either object > > > is zero. The action-reaction forces are equal and opposite, but either > > > object may still have a net force acting on it." > > > > > That is > > > > like applying the "force" of a skyscraper in a marsh. The maximum > > > > static force that can ever be applied is determined by the supporting > > > > capacity of the marsheffectively ZERO. My thesis in Architecture > > > > was: "Float Foundations for Poor Soils". Essentially, I created > > > > structural boats under buildings to support those by the bouyancy of > > > > the marsh. For the record, I made an 'A' on that thesis. > > > > NoEinstein > > > > > > > The resistance on electrons imposed by the ether IS the force being > > > > > > measured in those early Lorentz experiments. > > > > > > Sorry, what "Lorentz experiments"? > > > > > Lorentz experimented, extensively, with trying to measure the velocity > > > > and the mass of electrons. Those were inside vacuum tubes, and were > > > > speeded up by electromagnets. But, strangely, the electrons > > > > encountered exponentially more resistance the closer the velocity came > > > > to 'c'. > > > > Reference, please. > > > > > The equation beta = 1 / [1 - v^2/c^2]^1/2 was written years > > > > before the M-M experiment. Lorentz shoehorned such to also explain > > > > (sic) the nil results of M-M. Lorentz was a mathematition who DABBLED > > > > in science. Expect bad results whenever that happens. > > > > > > > So, the very experiments > > > > > > you inquire about, only need the correct CAUSE, not a new set of > > > > > > experiments! > > > > > > A correct cause would be accompanied by calculations, indicating the > > > > > size of the effect expected due to this cause. Without those > > > > > calculations, you've got nothing. > > > > > PD, the drag on electrons due to the ether that is clumping in front > > > > is very close to Lorentz's Beta. > > > > Prove that. > > > > > The MATH is close to correct, but > > > > the cause is ether drag. > > > > Prove that. Show the derivation. Your bluff is called. > > > > > Note: Ether can drag electrons and massive > > > > objects, but it never drags photons! NoEinstein - Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: BURT on 1 Jun 2010 15:42 On Jun 1, 12:25 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 25, 6:46 pm, PD <thedraperfam...(a)gmail.com> wrote: > > Dear PD: Your comment, below, about equal and opposite thrust force > and resistance force would be astute, if not for this fact: Objects at > rest, waiting to be dropped, are already experiencing a downward force > equal to their static weight. As soon as the retaining force is > released, there is a dynamic imbalance that starts the object > accelerating at 'g'. The downward force can never exceed the object's > inertia. If 'a' is substituted for 'g', the acceleration could > (perhaps) be greater, and the force would not be limited to the > inertia. The latter is true ONLY for objects dropped near the Earth. > My correct kinetic energy equation, KE = a/g (m) + v / 32.174 (m) > works anywhere in the Universe, as long as the "Pound" of reference is > that for 'g', or on Earth. NE > > > > > > > On May 23, 4:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On May 20, 9:42 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > Dear PD: If you had ever taken a course in structural engineering, > > > there are TWO distinct types of force interactions. The STATIC ones > > > always have the opposing forces being equal. > > > Yes, indeed, so the net force is zero, so the acceleration is zero. > > Hence the word "static". > > > > But the DYNAMIC ones > > > only have a FORCE equal to the LESSER resistance of the two. > > > What??? > > > Let's take a simple example. There is a wooden block on a wooden > > inclined plane, and the plane is tipped up until the block starts to > > slide, accelerating. List the forces acting on the block, what their > > origin is, what direction the force is pointing. > > > > You've > > > never realized that DYNAMICS limits the downward force on the falling > > > object to be whatever the INERTIA of the object is. Since the inertia > > > of a one pound mass will always be just one pound, there can never be > > > an exponential KE increase, because the force and the resistance are > > > equal. > > > WHAT???? > > If the force and resistance were equal, then there would be two equal > > and opposite forces acting on the object, and you'd be back to a > > static situation. Try again. > > > See the block example. > > > > Since your precious WORK would have to increase exponentially, > > > that violates simple dynamics, because the distances traveled by > > > falling objects differ so markedly from second to second. I hope that > > > you are sitting down, PD, because the latter statement means that > > > there can neither be an increasing Work nor and increasing KE, from > > > just the accruing COASTING components, which you are so reluctant to > > > acknowledge. The COASTING components of the distance of fall for all > > > dropped objects KILLS your made-up science! NoEinstein > > > > > On May 20, 7:40 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > On May 18, 9:18 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On May 17, 6:04 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > On May 17, 2:59 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > Dear PD, the Parasite Dunce: You have changed the subject AWAY from > > > > > > > KE, > > > > > > > No, I didn't. I responded directly to your post about ether flow on > > > > > > muons. > > > > > > > > because you can't find any place in any text that states: "Work is > > > > > > > being done even if there is no resistance. (sic) The only requirement > > > > > > > to have work is that there be a displacement. (sic) Thus, if a hockey > > > > > > > puck slides twice as far across the ice, twice as much work was done, > > > > > > > and there is twice as much KE in the puck, even if the ice is > > > > > > > frictionless. (sic)." > > > > > > > You are not paying attention. > > > > > > Remember, PD: I am King of the Hill in science. It isn't "my job" to > > > > > pay attention to you. > > > > > :>) > > > > Just make sure you tell yourself that each day in the mirror. If you > > > > like, please add the line, "And I am the heir to the throne of the > > > > kingdom, by birthright." You may also consider adding, if you are > > > > feeling confident, "And I am irresistible to women." > > > > > > > There is no work if there is no force present, even in the absence of resistance. > > > > > > I correct you: 'There is no work if there is no force present', AND > > > > > there is no corresponding resistance > > > > > There is no work done if there is no net force, regardless whether > > > > there is resistance or not. > > > > > > > There is work if there is a force present, even in the absence of resistance. > > > > > > Dear PD: Newton's Laws of Motion require: "For every action there > > > > > must be an equal and opposite reaction." It is IMPOSSIBLE to apply a > > > > > force... UNLESS there is a corresponding resisting force! > > > > > Newton's 3rd law is not a statement about a resisting force. That is a > > > > common mistake by high school students that is corrected in virtually > > > > every textbook on the subject. For example, from the high school text > > > > that I've been quoting from, page 133: > > > > "One important thing to remember about action-reaction pairs is that > > > > each force acts on a different object. Consider the task of driving a > > > > nail into wood.... To accelerate the nail and drive it into the wood, > > > > the hammer exerts a force on the nail. According to Newton's third > > > > law, the nail exerts a force on the hammer that is equal to the > > > > magnitude of the force that the hammer exerts on the nail. > > > > "The concept of action-reaction pairs is a common source of confusion > > > > because some people assume incorrectly that equal and opposite forces > > > > balance one another and make any change in the state of motion > > > > impossible. If the force that the nail exerts is equal to the force > > > > the hammer exerts on the nail, why doesn't the nail remain at rest? > > > > "The motion of the nail is affected only by the forces acting on the > > > > nail. To determine whether the nail will accelerate, draw a free-body > > > > diagram to isolate the forces acting on the nail.... The force of the > > > > nail on the hammer is not included in the diagram because it does not > > > > act on the nail. According to the diagram, the nail will be driven > > > > into the wood because there is a net force acting on the nail. Thus, > > > > action-reaction pairs do not imply that the net force on either object > > > > is zero. The action-reaction forces are equal and opposite, but either > > > > object may still have a net force acting on it." > > > > > > That is > > > > > like applying the "force" of a skyscraper in a marsh. The maximum > > > > > static force that can ever be applied is determined by the supporting > > > > > capacity of the marsheffectively ZERO. My thesis in Architecture > > > > > was: "Float Foundations for Poor Soils". Essentially, I created > > > > > structural boats under buildings to support those by the bouyancy of > > > > > the marsh. For the record, I made an 'A' on that thesis. > > > > > NoEinstein > > > > > > > > The resistance on electrons imposed by the ether IS the force being > > > > > > > measured in those early Lorentz experiments. > > > > > > > Sorry, what "Lorentz experiments"? > > > > > > Lorentz experimented, extensively, with trying to measure the velocity > > > > > and the mass of electrons. Those were inside vacuum tubes, and were > > > > > speeded up by electromagnets. But, strangely, the electrons > > > > > encountered exponentially more resistance the closer the velocity came > > > > > to 'c'. > > > > > Reference, please. > > > > > > The equation beta = 1 / [1 - v^2/c^2]^1/2 was written years > > > > > before the M-M experiment. Lorentz shoehorned such to also explain > > > > > (sic) the nil results of M-M. Lorentz was a mathematition who DABBLED > > > > > in science. Expect bad results whenever that happens. > > > > > > > > So, the very experiments > > > > > > > you inquire about, only need the correct CAUSE, not a new set of > > > > > > > experiments! > > > > > > > A correct cause would be accompanied by calculations, indicating the > > > > > > size of the effect expected due to this cause. Without those > > > > > > calculations, you've got nothing. > > > > > > PD, the drag on electrons due to the ether that is clumping in front > > > > > is very close to Lorentz's Beta. > > > > > Prove that. > > > > > > The MATH is close to correct, but > > > > > the cause is ether drag. > > > > > Prove that. Show the derivation. Your bluff is called. > > > > > > Note: Ether can drag electrons and massive > > > > > objects, but it never drags photons! NoEinstein - Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > - Show quoted text - What is the strength of gravity?
From: NoEinstein on 2 Jun 2010 19:46
On May 25, 6:46 pm, PD <thedraperfam...(a)gmail.com> wrote: > Dear PD: Most structural engineering problems involve the use of statics, rather than dynamics, to solve the equations. In statics, a one pound force, matched head-on by a one pound resisting force, wont allow the point of reaction to go accelerating away. On more than one occasion, when Ive said that the force must be equal to the resistancefor falling objects, that involves dynamics, not statics. Ill explain the difference. Suppose that there is a one pound object that is the payload of a rocket with a one pound net thrust; i.e., the rocket thrust is one pound MORE than the weight of the rocket minus the payload. Under those conditions, Newtons Second Law says that the rocket will be accelerating at 32.174 feet per second, each secondbetter known as the acceleration of gravity, g. As is the case for one pound objects on the surface of the Earth, the uniform force of gravity will cause any such object to press against the ground with a force of one pound. The ground, in turn, must offer an upward resistance of one pound to order to maintain static equilibrium (zero motion). Einstein himself noted that a person in a windowless box thats accelerating at g cant tell whether the box is resting, stationary, on the Earth, or flying through space. The box will be experiencing the weight of the person, and the person will experience the support (force) offered by the box. As required by Newtons Third Law, the action (the persons weight) is exactly matched by the reaction the supporting force provided by the box. Those two are equal and opposite, exactly as in the solving of problems in statics. The design equations for the box would be identical to the ones used on Earth for static loading. The INERTIA of falling objects isnt a constant force thats equal to the static weight, but can have any value *from zero up to the static weight. If a one pound block is resting on frictionless ice, that block can be caused to slide at, say, one ounce of force. In order to experience the maximum inertia of one pound, it will require applying a force, via some point of reaction, that is capable of moving 32.174 feet per second in the first second. So, the inertial resistance of any mass is VELOCITY dependent, within the limits, *above. Consider that payload being accelerated by a one pound net thrust rocket. If the payload weight is one pound, the acceleration will be g. If the payload weight were somehow reduced to .5 pounds, while the net rocket thrust remains one pound, the acceleration would jump to 2g. If the payload weight, somehow, could be increased to two pounds, the acceleration would be cut to .5g. It is very well known that all compact, near-Earth falling objects are accelerating at g'. Since the uniform force of gravity, acting on the one pound object, always causes such to accelerate at g, then, the force must be matched by the inertia of the objectas required by Newtons Third Law of Motion. PD, you have repeatedly claimed that there can be a downward one-pound force of gravity, WITHOUT there being a resistance. But heres why that cant be: Gravity requires an object to act upon. An object that has no resistance, also has no inertia, and will, therefore, have zero mass. You, erroneously, claim that a force can remain at one pound without there being a resistance. Heres why that is never the case: A big black house fly is circling your den, and aggravating you. So, you roll a newspaper into a one pound bat and swing away at the fly. After several misses, you finally hit the pest. Your hand and the newspaper were traveling exactly 32.174 feet per second. You figure you just smashed the fly with a one pound force on its tiny body. But much to your surprise, the fly hits the floor, readjusts its wings, and flies away, apparently unharmed. The physics of the above is simple: The impact of the fly and the newspaper will never cause a force thats greater than the inertia of the FLYquite small indeed! Since the flys body is always functioning under its full weight, accelerating it to 'g isnt a killing force. The fly could be stunned, allowing you to smash it flat, but you can never do that with a bat of ANY weight that isnt traveling faster than 32.174 feet per second. If the fly had ZERO resistance or zero inertia, the newspaper wouldnt experience any force at allexactly as if you had missed hitting it at all. Learn this, PD: The available force of gravity on a one pound dropped object, is 100% expended in causing the linear increase in the kinetic energy, as is given by my equation, KE = a/g (m) + v / 32.174 (m). Each second of fall, there is a COASTING carryover velocity from the end of each of the previous seconds. Coasting is what causes the free- drop curve to be parabolic in shape, rather than linear. Since 100% of the force of gravity produces only a linear increase in KE, then, there is no net force available to act through the remaining distance of fall, to cause the work that you imagine is being done. You err, PD, by assuming that a one pound force is available for both the coasting portion of the curve, and the straight line portion. As for a hockey puck already sliding on frictionless ice, there is no force needed to keep the puck moving, or coasting. If you still suppose that there is then you are ignoring the Law of the Conservation of Energy. Ive taken this long amount of time to walk-you-through the physics, not because I expect you to ever get it, but because I want 97.5% of the readers to know, that I know, that I know! NoEinstein > > On May 23, 4:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On May 20, 9:42 am, PD <thedraperfam...(a)gmail.com> wrote: > > > Dear PD: If you had ever taken a course in structural engineering, > > there are TWO distinct types of force interactions. The STATIC ones > > always have the opposing forces being equal. > > Yes, indeed, so the net force is zero, so the acceleration is zero. > Hence the word "static". > > > But the DYNAMIC ones > > only have a FORCE equal to the LESSER resistance of the two. > > What??? > > Let's take a simple example. There is a wooden block on a wooden > inclined plane, and the plane is tipped up until the block starts to > slide, accelerating. List the forces acting on the block, what their > origin is, what direction the force is pointing. > > > You've > > never realized that DYNAMICS limits the downward force on the falling > > object to be whatever the INERTIA of the object is. Since the inertia > > of a one pound mass will always be just one pound, there can never be > > an exponential KE increase, because the force and the resistance are > > equal. > > WHAT???? > If the force and resistance were equal, then there would be two equal > and opposite forces acting on the object, and you'd be back to a > static situation. Try again. > > See the block example. > > > > > Since your precious WORK would have to increase exponentially, > > that violates simple dynamics, because the distances traveled by > > falling objects differ so markedly from second to second. I hope that > > you are sitting down, PD, because the latter statement means that > > there can neither be an increasing Work nor and increasing KE, from > > just the accruing COASTING components, which you are so reluctant to > > acknowledge. The COASTING components of the distance of fall for all > > dropped objects KILLS your made-up science! NoEinstein > > > > On May 20, 7:40 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On May 18, 9:18 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > On May 17, 6:04 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > On May 17, 2:59 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > Dear PD, the Parasite Dunce: You have changed the subject AWAY from > > > > > > KE, > > > > > > No, I didn't. I responded directly to your post about ether flow on > > > > > muons. > > > > > > > because you can't find any place in any text that states: "Work is > > > > > > being done even if there is no resistance. (sic) The only requirement > > > > > > to have work is that there be a displacement. (sic) Thus, if a hockey > > > > > > puck slides twice as far across the ice, twice as much work was done, > > > > > > and there is twice as much KE in the puck, even if the ice is > > > > > > frictionless. (sic)." > > > > > > You are not paying attention. > > > > > Remember, PD: I am King of the Hill in science. It isn't "my job" to > > > > pay attention to you. > > > > :>) > > > Just make sure you tell yourself that each day in the mirror. If you > > > like, please add the line, "And I am the heir to the throne of the > > > kingdom, by birthright." You may also consider adding, if you are > > > feeling confident, "And I am irresistible to women." > > > > > > There is no work if there is no force present, even in the absence of resistance. > > > > > I correct you: 'There is no work if there is no force present', AND > > > > there is no corresponding resistance > > > > There is no work done if there is no net force, regardless whether > > > there is resistance or not. > > > > > > There is work if there is a force present, even in the absence of resistance. > > > > > Dear PD: Newton's Laws of Motion require: "For every action there > > > > must be an equal and opposite reaction." It is IMPOSSIBLE to apply a > > > > force... UNLESS there is a corresponding resisting force! > > > > Newton's 3rd law is not a statement about a resisting force. That is a > > > common mistake by high school students that is corrected in virtually > > > every textbook on the subject. For example, from the high school text > > > that I've been quoting from, page 133: > > > "One important thing to remember about action-reaction pairs is that > > > each force acts on a different object. Consider the task of driving a > > > nail into wood.... To accelerate the nail and drive it into the wood, > > > the hammer exerts a force on the nail. According to Newton's third > > > law, the nail exerts a force on the hammer that is equal to the > > > magnitude of the force that the hammer exerts on the nail. > > > "The concept of action-reaction pairs is a common source of confusion > > > because some people assume incorrectly that equal and opposite forces > > > balance one another and make any change in the state of motion > > > impossible. If the force that the nail exerts is equal to the force > > > the hammer exerts on the nail, why doesn't the nail remain at rest? > > > "The motion of the nail is affected only by the forces acting on the > > > nail. To determine whether the nail will accelerate, draw a free-body > > > diagram to isolate the forces acting on the nail.... The force of the > > > nail on the hammer is not included in the diagram because it does not > > > act on the nail. According to the diagram, the nail will be driven > > > into the wood because there is a net force acting on the nail. Thus, > > > action-reaction pairs do not imply that the net force on either object > > > is zero. The action-reaction forces are equal and opposite, but either > > > object may still have a net force acting on it." > > > > > That is > > > > like applying the "force" of a skyscraper in a marsh. The maximum > > > > static force that can ever be applied is determined by the supporting > > > > capacity of the marsheffectively ZERO. My thesis in Architecture > > > > was: "Float Foundations for Poor Soils". Essentially, I created > > > > structural boats under buildings to support those by the bouyancy of > > > > the marsh. For the record, I made an 'A' on that thesis. > > > > NoEinstein > > > > > > > The resistance on electrons imposed by the ether IS the force being > > > > > > measured in those early Lorentz experiments. > > > > > > Sorry, what "Lorentz experiments"? > > > > > Lorentz experimented, extensively, with trying to measure the velocity > > > > and the mass of electrons. Those were inside vacuum tubes, and were > > > > speeded up by electromagnets. But, strangely, the electrons > > > > encountered exponentially more resistance the closer the velocity came > > > > to 'c'. > > > > Reference, please. > > > > > The equation beta = 1 / [1 - v^2/c^2]^1/2 was written years > > > > before the M-M experiment. Lorentz shoehorned such to also explain > > > > (sic) the nil results of M-M. Lorentz was a mathematition who DABBLED > > > > in science. Expect bad results whenever that happens. > > > > > > > So, the very experiments > > > > > > you inquire about, only need the correct CAUSE, not a new set of > > > > > > experiments! > > > > > > A correct cause would be accompanied by calculations, indicating the > > > > > size of the effect expected due to this cause. Without those > > > > > calculations, you've got nothing. > > > > > PD, the drag on electrons due to the ether that is clumping in front > > > > is very close to Lorentz's Beta. > > > > Prove that. > > > > > The MATH is close to correct, but > > > > the cause is ether drag. > > > > Prove that. Show the derivation. Your bluff is called. > > > > > Note: Ether can drag electrons and massive > > > > objects, but it never drags photons! NoEinstein - Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > - Show quoted text - |