There is no "pull" of gravity, only the PUSH of flowing ether!
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From: NoEinstein on 23 May 2010 17:08 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. But the DYNAMIC ones only have a FORCE equal to the LESSER resistance of the two. 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. 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 -
From: NoEinstein on 24 May 2010 10:35 On May 14, 10:33 am, PD <thedraperfam...(a)gmail.com> wrote: > Folks: For over two weeks, now, Google hasn't been reporting recent updates on this post. Therefore, I am re posting the original essay, but with the number "2" at the end. I hope that "thinkers" among you will leave comments. And I especially hope that Google will get their computers working right. PD, the Parasite Dunce, doesn't understand mechanics. He "dabbled" in H. E. Particle physics, then, taught High School science. No such teacher is the final word on anything. And those decades-old texts aren't authoritative, either. My ASSIGNMENT for PD: (1.) Define MOMENTUM, both in words and by formula. Explain what each variable is, and explain the UNITS of the results after using the equation. (2.) Give a clear and concise definition of the Law of the Conservation of ENERGY. (3.) Concisely define Kinetic Energy, and state the UNITS. (4.) If work can be done without there being both a force AND a matching resistance, explain WHY. (5.) The KE equation in most texts is KE = 1/2mv^2. Since the distance of fall of dropped objects increases parabolically with respect to time, please explain how your... 'WORK is proportional to distance' notion isn't a 100% exaggeration of the KE. > > 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: John Park on 24 May 2010 11:16 NoEinstein (noeinstein(a)bellsouth.net) writes: > On May 14, 10:33=A0am, PD <thedraperfam...(a)gmail.com> wrote: >> > Folks: For over two weeks, now, Google hasn't been reporting recent > updates on this post. Therefore, I am re posting the original essay, > but with the number "2" at the end. I hope that "thinkers" among you > will leave comments. And I especially hope that Google will get their > computers working right. > Go away. --John Park
From: spudnik on 24 May 2010 14:39 dood, you are *so* nonsequiter; what is the language in which you are most literate?... well, if the answer is English, we can grok that! thusNso: that is a cool graph. what is the quantity of CO2, exuded by volcanos, typically?... does anyone have a quantification of the amount of CFCs, they emit? the main question is about the "observed" temperature, due mostly to interpretations "in the anthropocene" (formerly known as, the Holocene interglacial -- you was then !-) > The following graph plots both and you can see which is larger. http://solar-center.stanford.edu/sun-on-earth/Climate_Change_Attribut... http://solar-center.stanford.edu/sun-on-earth/Climate_Change_Attribut... --Pi, the surfer's canonical value -- except no other! http://wlym.com
From: PD on 25 May 2010 18:46
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 - > > |