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From: PD on 1 May 2010 11:04 On Apr 30, 10:13 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On Apr 30, 3:40 pm, PD <thedraperfam...(a)gmail.com> wrote: > > Dear PD, the Parasite Dunce: I, sir, am King of the Hill in science. > If you would like for the readers to see some "textbook definition" > which you claim is more valid than my F. & W. Standard College > Dictionary, then copy and paste your definition for the world to see. > *** Put up or shut up, PD! *** You've done nothing to even hint that > you have objectivity in scienceonly empty bluster. NoEinstein Good grief. OK, I'll come part way. You do some work too. Go to the library and ask for Giancoli, Physics, any edition more recent than than the 4th. See sections 2-2 and 2-3. In my copy, that's pages 21-23. There, I have made the search bonehead simple for you. All you have to do is 1) Vacate your chair 2) Take your butt to the library 3) Open the book to the pages I mentioned 4) Read > > > > > On Apr 30, 2:31 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On Apr 30, 10:39 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > Dear PD, the Parasite Duncewho's WAY over-his-head arguing SCIENCE > > > with me!: My dictionary defines "displacement" relative to science, > > > thusly: " 'Astron' An apparent change of position. ... (4.) The > > > relation between the position of an object at any time and its > > > original position." Note: Displacement ISN'T defined as the change in > > > location of an object since the beginning of that second! > > > Displacement simply means how far an object has traveled since the > > > start of the experiment, or the DISTANCE traveled. And that distance > > > isn't limited only to expressing an object's velocity. NoEinstein > > > I'm sorry, John, but you're going to have to look at a physics book, > > not a dictionary of common usage, for the definition of a term as used > > in physics. For example, a field in physics does not typically contain > > corn or wildflowers. > > > In physics, the velocity over an interval is del(x)/del(t), where > > del(x) is the displacement over the interval del(t). So if del(t) is > > the preceding second, then del(x) is the displacement during that > > preceding second. You will find this very clear definition boxed and > > highlighted in just about any high school physics book. > > > Since you claim to have gotten a degree in architecture, you would > > have had to pass an exam at one point demonstrating your understanding > > of that definition. You either have forgotten it in the intervening 92 > > years, or you never learned it. > > > In either case, since you don't even understand the definition of > > velocity and displacement as used in introductory physics, you'll > > understand why I'd be very nervous about any structures that you have > > designed, and I'm glad your city in South Carolina declined your > > advice on the design of its public structures. > > > > > > > > > On Apr 30, 5:38 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > On Apr 28, 10:50 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On Apr 28, 9:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > On Apr 27, 10:16 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > On Apr 26, 9:42 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > > > On Apr 26, 1:30 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > > Dear PD, the Parasite Dunce: Your denseness makes you incapable of > > > > > > > > > understanding that NO force is required to cause an object to COAST. > > > > > > > > > I do understand that. I've told you that. > > > > > > > > Dear PD: Fine; remember that you just said there is no force to cause > > > > > > > COASTING! > > > > > > > Of course, John. That is Newton's First Law, first laid out by > > > > > > Galileo. Perhaps you've heard of them. > > > > > > > > > > The velocity of a falling object at the end of second one COASTS all > > > > > > > > > the way through till the object strikes the ground. The velocity at > > > > > > > > > the end of second two does the same thing, and etc. The INCREASE in > > > > > > > > > velocity per second is uniform (as you agreed two replies back). It > > > > > > > > > is the COASTING carry-over velocity which causes the shape of the free- > > > > > > > > > drop curve (distance vs. time) to be a parabola. > > > > > > > > > Nevertheless, the work is the increase in energy each second. > > > > > > > > The work is the product of the force times the displacement.. > > > > > > > > "Work" is the product of the force x distance moved ONLY for objects > > > > > > > that aren't: in free fall; in outer space; or which don't have a > > > > > > > resisting force matching the applied force. > > > > > > > That's just plain wrong, John. There is no such restriction. Work is > > > > > > defined as the product of force and displacement under all conditions. > > > > > > Dear Dunce: Of course you prefer defending things that are wrong or > > > > > incomplete. Try telling a construction foreman that you should be > > > > > paid for "pushing" a wheelbarrow that has no weight and is > > > > > frictionless. No resistance = zero work done! That's kind of like a > > > > > boxer aiming for the jaw, but only hitting air. There was no transfer > > > > > of KE, or work, there. In the case of your jaw, I wish work was > > > > > transferred. Ha, ha, HA! > > > > > Oh, dear. And all the textbooks are so clear on the distinction > > > > between physical, mechanical work and what a common laborer feels in > > > > his arm. Once again, John, you muddle things up by confusing loose, > > > > everyday language with the precise definitions that are given in the > > > > textbooks you supposedly studied in order to get your degree. > > > > > > > In fact, if you will look in a freshman physics book, you will see > > > > > > this definition of work applied to objects in free-fall. > > > > > > Coriolis's wrong definition of kinetic energy: KE = 1/2mv^2 doesn't > > > > > involve a force nor a distance, PD. > > > > > All this time I've been humoring > > > > > you when you say that work is involved in computing KE. > > > > > Oh dear. Look up the work-energy theorem, John. Good heavens, you are > > > > especially clownish this morning. > > > > > > Only the > > > > > VELOCITY determines the KE for a unit mass. My correct kinetic energy > > > > > equation is: KE = a/g (m) + v / 32.174 (m). Please note that neither > > > > > FORCE nor "displacement" are part of either equation. You love the > > > > > "work" equation, because you've never done any. You are a just couch > > > > > potato and a Parasite Dunce. > > > > > > > It does no good to just make up stuff, John. It's much better to check > > > > > > up on the meanings of words before wildly guessing or just making > > > > > > things up. > > > > > > > > The force on dropped > > > > > > > objects is just their static weight acting continuously until the > > > > > > > object hits the ground. The distance traveled times the force in a > > > > > > > single second is a COMPLETE work computation for that one second. > > > > > > > Of course. That is the definition of work. Note that you agree that > > > > > > the distance traveled times the force is indeed the work done on a > > > > > > dropped object. Just a second ago, you said this expression doesn't > > > > > > apply to dropped objects, and now you say it does. Confused, John? > > > > > > > > In > > > > > > > second two, that same work would get done, again; > > > > > > > On the contrary, in second two, the distance traveled is three times > > > > > > as far, and so the work is not the same at all. > > > > > > > Note that the work is defined as the force ACTUALLY applied times the > > > > > > distance ACTUALLY moved, and there is no subtraction of a component > > > > > > from the previous interval. You'd like it to be different, but it just > > > > > > isn't so. > > > > > > > > and in all the > > > > > > > remaining seconds. The thing you don't understand is: The velocity at > > > > > > > the end of second one will COAST all the way to the ground if the > > > > > > > force of gravity is "cut off" at that point in time. So, in say a > > > > > > > four second fall, the KE will be 2W, or two weight units. At the end > > > > > > > of seconds two, three, and four, there will still only be 2 weight > > > > > > > units of force, KE or "work" done! But there will have been 16 times > > > > > > > as much DISTANCE traveled. Since 15/16ths of the distance fallen has > > > > > > > zero pounds of associated force, the latter distance is from COASTING, > > > > > > > and thus has NO associated work! > > > > > > > > All well-written equations must explain what the variables are, and > > > > > > > state the "limiting conditions" under which such equation is valid. > > > > > > > Yes, indeed, John, which is why you should go back to picking up > > > > > > materials where those variables are very clearly defined, rather than > > > > > > you and I bickering about them here. I'm telling you the definitions > > > > > > of those variables, and they aren't what you think they are. To settle > > > > > > the dispute, you need only check a third and independent reference. > > > > > > > > Your... "work" equation does neither. The next time you accelerate > > > > > > > your car, ease off the gas and let the car increase its... > > > > > > > "displacement", as you say. > > > > > > > John, if I ease off the gas and go at constant velocity, the > > > > > > displacement is not increasing. It stays the same in successive > > > > > > seconds. > > > > > > Wow Folks: PD just said that an automobile that is COASTING covers no > > > > > distance! Can you readers not see that PD is a loony bird? NE > > > > > Uh, no, John, that is not what I said. Good heavens, is your reading > > > > comprehension that bad? > > > > > > That's what constant velocity MEANS, John, because velocity> is displacement per unit time. It's only when something is > > > > > > *accelerating* (such as in free fall), that the displacement changes > > > > > > from second to subsequent second. I don't know why this is so hard for > > > > > > you. Are you slow? > > > > > > The Work equation is: W = fd. That d is the total distance at the > > > > > point at which the work is to be computed, NOT the unit rate of doing > > > > > the work. > > > > > The distance d is the distance the object has covered during the > > > > interval during which you are computing the work. Note that there is > > > > no stipulation that the d should be the value with the coasting > > > > component from the > > ... > > read more »
From: NoEinstein on 1 May 2010 21:25 On May 1, 11:00 am, PD <thedraperfam...(a)gmail.com> wrote: > Dear PD, the Parasite Dunce: You just said that "physics isn't determined by logic". Of course, you would think that! That's because you don't know HOW to reason! Einstein got physicists believing that ILLOGIC is where the most... I.Q. is. Since you understood nothing taught to you in physics (the right stuff nor the WRONG), you figured your strength was to fight anything and everything that wasnt COOKBOOKED from some out-of-date, McGraw-Hill, Jewish publication. Tell me, PD, WHO on this EARTH is a qualification to confirm YOUR ideas about science? Anyone who understands math, and knows what the Law of the Conservation of Energy requires, will immediately confirm that Coriolis and Einstein had no earthly idea that KE and 'E' must not be exponential equations, but LINEAR equations (or additive). Since you don't think COASTING increases an object's distance of travel, it is YOU, not me, needing others to confirm your stupidity! Ha, ha, HA! NoEinstein > > On Apr 30, 10:05 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On Apr 30, 3:34 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > Dear PD, the Parasite Dunce: "We" (you and I) aren't having a > > discussion about science. You simply take the anti-thesis of any > > science truth, knowing that there are some naive readers who won't > > know the difference. It may sound 'high-and-mighty' for you to keep > > referring to... the experimental evidence, and the 'textbook' > > definitions, but you NEVER paraphrase a possible counterargument. You > > only claim that there is 'something', somewhere that disagrees with > > me. And you expect me to go look that up. > > Yes, indeed, because physics is not something that is settled by > puffed-up posturing and debate. > It is not something that is determined by force of logic. > You may be confusing physics with philosophy. > > Ultimately, the truth in physics is determined by careful and > independently confirmed experimental measurement. > That body of experimental evidence is documented and available to you. > It is referred to in textbooks, and references to it have been made > here to you. > > So yes, you are expected to look it up. > > ANYBODY doing physics is expected to look it up. > > > > > Folks, PD is the deep thinker (sic) who said that atomic decay is a > > "chemical reaction". And just today, he said that a car which is > > COASTING isn't increasing its "displacement". He has just proposed > > that... "displacement" is only apt to calculating, or measuring, an > > object's unit velocity. And since the unit velocity of the car > > doesn't change, he claims that coasting isn't increasing the distance > > of travel of the car. Can't most of you see how little PD cares about > > truth and logic? Does he think everyone but him is a fool? > > > *** Tell us this, PD: How many science experiments, of any kind, have > > YOU designed, built, and successfully tested? > > Are you sure you want to ask this question? My professional history is > as an experimental physicist, and my record is public. > Please don't puff yourself up as a songwriter when talking to a > professional musician. > It's not smart to put on airs as an expert on law when talking to a > judge. > > > I've made two most > > definitive tests which support the LOGIC that Coriolis's KE equation > > is not only WRONG, its so obviously in violation of the Law of the > > Conservation of Energy, that no experiments are needed, at all, to > > disprove: KE = 1/2mv^2; nor to similarly disprove E = mc^2 / beta. > > For you, a proof is only valid if it involves experiments which you > > have never cited, nor paraphrased, and definitions that you claim are > > in textbooks, but which you never quote. > > Two comments: > 1. Your experimental results will be worth something when confirmed by > an independent investigator. That is how it is done in science. Until > then, you are a self-feeding loop. > 2. Yes, I expect you to look up textbooks, as they are easy to find > even in your local library. I'm assuming that you are not under house > arrest, you aren't bedridden, that you have bus fare to get you > downtown, and that you are capable of reading when you get there. I'm > also assuming that you are not so pathologically lazy that you refuse > to budge your butt from your chair. > > > > > I recently told you that I had suspected that the readers agreed with > > my correctness our yours by two to one. But in light of your recent > > statements of utter stupidity, that number is probably closer to ten > > to one! > > This is just like you, to suspect something is true without a single > shred of tangible evidence. It's your style. > > > > > *** No scientist on Earth has more credibility than yours > > truly. *** If any think that they do, I would love for them to go > > head-to-head with me, so that I can kick their asses into solar > > orbit. Like those purported scientists, you, PD, dont have a leg, > > nor a stump to stand on. NoEinstein > > > > On Apr 30, 2:18 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On Apr 30, 10:29 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > Dear PD: Some readers, who don't know either of us from Adam, may > > > > think that your sidestepping of science is credible. An attack on... > > > > the messenger (me) is a quick put-down that you had to have learned > > > > (tongue-in-cheekha!) very early won't work on me. If the regular > > > > readers of my posts and replies got to vote, they'd probably say that > > > > I'm beating you in the "one-up-manship" by a two to one margin. But > > > > you're still around because you won't stay on any discussion long > > > > enough to get the life squished out of your... 'science'. I enjoy > > > > knowing that you haven't won; can't; and won't win, PD. That > > > > qualifies you as a looser; doesn't it? NoEinstein > > > > I'm fascinated by this idea you have of winning or losing. > > > > We're having a discussion about physics. I'm explaining to you what we > > > know matches experiment, and what the definitions of the words are > > > that are used in physics, what the equations mean, and how that is > > > exemplified in measurements, and the fact that none of what we're > > > talking about is beyond 7th grade science level. > > > > You on the other hand seem to be more worried about winning some kind > > > of battle or contest, and to you winning means: > > > - that you talk longer than anyone else, ensuring that you always have > > > the last word > > > - that no one can *force* you to believe what 7th graders have no > > > difficulty understanding > > > - that no one can *force* to you stop talking > > > - that you stick by your guns, no matter what, regardless of how > > > stupid it starts to sound even to you > > > - disparaging your respondents by calling them negativists and other > > > assorted names > > > - that you have offered a retort to every single response to your > > > posts. > > > > By that metric, someone who firmly believes that 17+4=32, and who > > > insists on this long after the last person has walked away, and who > > > insists that 2nd grade math teachers are obviously wrong, and who is > > > proud that no one has been able to get him to stop saying 17+4=32, and > > > who calls people who believe otherwise to be ninnies and brainwashed > > > -- well, by golly, in your eyes that person has won something. > > > > Of course, 17+4 is not 32, but the IMPORTANT thing, you see, is > > > winning, not being right. Isn't that so? > > > > As for attacking you, you'll pardon me if I'd decline to hire you to > > > be the architect for a doghouse. I'm sure you understand my reasons > > > why.- Hide quoted text - > > - Show quoted text -
From: NoEinstein on 1 May 2010 21:33 On May 1, 11:04 am, PD <thedraperfam...(a)gmail.com> wrote: > Nice "try" PD: Like I've told you a hundred times, PARAPHRASE, or copy, what you want me to read. You, an imbecile, don't qualify to tell me (who's off the top of the I. Q. chart) what I should do. You can only dream that I would care to follow your instructions, in any regard. NoEinstein > > On Apr 30, 10:13 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On Apr 30, 3:40 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > Dear PD, the Parasite Dunce: I, sir, am King of the Hill in science. > > If you would like for the readers to see some "textbook definition" > > which you claim is more valid than my F. & W. Standard College > > Dictionary, then copy and paste your definition for the world to see. > > *** Put up or shut up, PD! *** You've done nothing to even hint that > > you have objectivity in scienceonly empty bluster. NoEinstein > > Good grief. OK, I'll come part way. You do some work too. > Go to the library and ask for Giancoli, Physics, any edition more > recent than than the 4th. > See sections 2-2 and 2-3. In my copy, that's pages 21-23. > There, I have made the search bonehead simple for you. All you have to > do is > 1) Vacate your chair > 2) Take your butt to the library > 3) Open the book to the pages I mentioned > 4) Read > > > > > > > > On Apr 30, 2:31 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On Apr 30, 10:39 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > Dear PD, the Parasite Duncewho's WAY over-his-head arguing SCIENCE > > > > with me!: My dictionary defines "displacement" relative to science, > > > > thusly: " 'Astron' An apparent change of position. ... (4.) The > > > > relation between the position of an object at any time and its > > > > original position." Note: Displacement ISN'T defined as the change in > > > > location of an object since the beginning of that second! > > > > Displacement simply means how far an object has traveled since the > > > > start of the experiment, or the DISTANCE traveled. And that distance > > > > isn't limited only to expressing an object's velocity. NoEinstein > > > > I'm sorry, John, but you're going to have to look at a physics book, > > > not a dictionary of common usage, for the definition of a term as used > > > in physics. For example, a field in physics does not typically contain > > > corn or wildflowers. > > > > In physics, the velocity over an interval is del(x)/del(t), where > > > del(x) is the displacement over the interval del(t). So if del(t) is > > > the preceding second, then del(x) is the displacement during that > > > preceding second. You will find this very clear definition boxed and > > > highlighted in just about any high school physics book. > > > > Since you claim to have gotten a degree in architecture, you would > > > have had to pass an exam at one point demonstrating your understanding > > > of that definition. You either have forgotten it in the intervening 92 > > > years, or you never learned it. > > > > In either case, since you don't even understand the definition of > > > velocity and displacement as used in introductory physics, you'll > > > understand why I'd be very nervous about any structures that you have > > > designed, and I'm glad your city in South Carolina declined your > > > advice on the design of its public structures. > > > > > > > > > > > On Apr 30, 5:38 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > On Apr 28, 10:50 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > On Apr 28, 9:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > > On Apr 27, 10:16 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > > On Apr 26, 9:42 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > > > > On Apr 26, 1:30 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > > > Dear PD, the Parasite Dunce: Your denseness makes you incapable of > > > > > > > > > > understanding that NO force is required to cause an object to COAST. > > > > > > > > > > I do understand that. I've told you that. > > > > > > > > > Dear PD: Fine; remember that you just said there is no force to cause > > > > > > > > COASTING! > > > > > > > > Of course, John. That is Newton's First Law, first laid out by > > > > > > > Galileo. Perhaps you've heard of them. > > > > > > > > > > > The velocity of a falling object at the end of second one COASTS all > > > > > > > > > > the way through till the object strikes the ground. The velocity at > > > > > > > > > > the end of second two does the same thing, and etc. The INCREASE in > > > > > > > > > > velocity per second is uniform (as you agreed two replies back). It > > > > > > > > > > is the COASTING carry-over velocity which causes the shape of the free- > > > > > > > > > > drop curve (distance vs. time) to be a parabola. > > > > > > > > > > Nevertheless, the work is the increase in energy each second. > > > > > > > > > The work is the product of the force times the displacement. > > > > > > > > > "Work" is the product of the force x distance moved ONLY for objects > > > > > > > > that aren't: in free fall; in outer space; or which don't have a > > > > > > > > resisting force matching the applied force. > > > > > > > > That's just plain wrong, John. There is no such restriction. Work is > > > > > > > defined as the product of force and displacement under all conditions. > > > > > > > Dear Dunce: Of course you prefer defending things that are wrong or > > > > > > incomplete. Try telling a construction foreman that you should be > > > > > > paid for "pushing" a wheelbarrow that has no weight and is > > > > > > frictionless. No resistance = zero work done! That's kind of like a > > > > > > boxer aiming for the jaw, but only hitting air. There was no transfer > > > > > > of KE, or work, there. In the case of your jaw, I wish work was > > > > > > transferred. Ha, ha, HA! > > > > > > Oh, dear. And all the textbooks are so clear on the distinction > > > > > between physical, mechanical work and what a common laborer feels in > > > > > his arm. Once again, John, you muddle things up by confusing loose, > > > > > everyday language with the precise definitions that are given in the > > > > > textbooks you supposedly studied in order to get your degree. > > > > > > > > In fact, if you will look in a freshman physics book, you will see > > > > > > > this definition of work applied to objects in free-fall. > > > > > > > Coriolis's wrong definition of kinetic energy: KE = 1/2mv^2 doesn't > > > > > > involve a force nor a distance, PD. > > > > > > All this time I've been humoring > > > > > > you when you say that work is involved in computing KE. > > > > > > Oh dear. Look up the work-energy theorem, John. Good heavens, you are > > > > > especially clownish this morning. > > > > > > > Only the > > > > > > VELOCITY determines the KE for a unit mass. My correct kinetic energy > > > > > > equation is: KE = a/g (m) + v / 32.174 (m). Please note that neither > > > > > > FORCE nor "displacement" are part of either equation. You love the > > > > > > "work" equation, because you've never done any. You are a just couch > > > > > > potato and a Parasite Dunce. > > > > > > > > It does no good to just make up stuff, John. It's much better to check > > > > > > > up on the meanings of words before wildly guessing or just making > > > > > > > things up. > > > > > > > > > The force on dropped > > > > > > > > objects is just their static weight acting continuously until the > > > > > > > > object hits the ground. The distance traveled times the force in a > > > > > > > > single second is a COMPLETE work computation for that one second. > > > > > > > > Of course. That is the definition of work. Note that you agree that > > > > > > > the distance traveled times the force is indeed the work done on a > > > > > > > dropped object. Just a second ago, you said this expression doesn't > > > > > > > apply to dropped objects, and now you say it does. Confused, John? > > > > > > > > > In > > > > > > > > second two, that same work would get done, again; > > > > > > > > On the contrary, in second two, the distance traveled is three times > > > > > > > as far, and so the work is not the same at all. > > > > > > > > Note that the work is defined as the force ACTUALLY applied times the > > > > > > > distance ACTUALLY moved, and there is no subtraction of a component > > > > > > > from the previous interval. You'd like it to be different, but it just > > > > > > > isn't so. > > > > > > > > > and in all the > > > > > > > > remaining seconds. The thing you don't understand is: The velocity at > > > > > > > > the end of second one will COAST all the way to the ground if the > > > > > > > > force of gravity is "cut off" at that point in time. So, in say a > > > > > > > > four second fall, the KE will be 2W, or two weight units. At the end > > > > > > > > of seconds two, three, and four, there will still only be 2 weight > > > > > > > > units of force, KE or "work" done! But there will have been 16 times > > > > > > > > as much DISTANCE traveled. Since 15/16ths of the distance fallen has > > > > > > > > zero pounds of associated force, the latter distance is from COASTING, > > > > > > > > and thus has NO associated work! > > > > > > > > > All well-written equations must explain what the variables are, and > > > > > > > > state the "limiting conditions" under which such equation is valid. > > > > > > > > Yes, indeed, John, which is why you should go back to picking up > > > > > > > materials where those variables are very clearly defined, rather than > > > > > > > you and I bickering about them here. I'm telling you the definitions > > > > > > > of those variables, and they aren't what you think they are. To settle > > > > > > > the dispute, you need only check a third and independent reference. > > > > > > > > > Your... "work" equation does neither. The next time you accelerate > > > > > > > > your car, ease off the gas and let the car increase its... > > > > > > > > "displacement", as you say. > > > > > > > > John, if I ease off the gas and go at constant velocity, the > > > > > > > displacement is not increasing. It stays the same in successive > > > > > > > seconds. > > > > > > > Wow Folks: PD just said that an automobile that is COASTING covers no > > > > > > distance! Can you readers not see that PD is a loony bird? NE > > > > > > Uh, no, John, that is not what I said. Good heavens, is your reading > > > > > comprehension that bad? > > ... > > read more »- Hide quoted text - > > - Show quoted text -
From: spudnik on 1 May 2010 21:41 so, you applied Coriolis' Force to Quantum Gravity, and **** happened? > read more » thus: with only the "trivial" solutions on the curves o'Fermatttt, it sounds like a "necessary but insufficient" proof; PdF certainly could have done it. > I have been interested in the odd and even aspect of FLT , and when Cn = 1. May I have your reference? DRMARJOHN thus: so, your coinage of pi(a,b) is the same as pi(b) - pi(a); now, can you say thr proof as a wordprolemmum? --Light: A History! http://wlym.takeTHEgoogolOUT.com
From: NoEinstein on 1 May 2010 22:01
On Apr 27, 10:16 am, PD <thedraperfam...(a)gmail.com> wrote: > Dear PD, the Parasite Dunce: Several times before you have referenced Newton's ERRANT F = ma. Most equations that contain a "mass" can be changed to be a UNIT mass of one pound (or whatever). The "textbook" definition of MOMENTUM is F = mv. The latter mass can also be changed to be a unit mass of one pound (or whatever). SO... Since both equations are forces, set the right half of the two equations to be EQUAL, or: ma = mv. Since the masses are both one pound unit masses, then, the resulting equation says: ACCELERATION = VELOCITY! Even an imbecile like you, PD, should realize that velocity, (or say) feet/ sec, isn't the same as feet/second EACH second! Ironically, I was studying for college physics when I realized the conflict between those two equations. That same week, I concluded that the entire chapter on mechanics was screwed up. Newton' "Law", in words, says: For every uniform force, there is one and only one associated acceleration. The correct equation for that should have been F = a, provided, of course, that the relationships between those two variables are stipulated, or are included in a less generalized equation. The equation for MOMENTUM, F = mv, is correct! For objects in free fall, or objects that are accelerating, the correct kinetic energy formula is my own: KE = a/g (m) + v / 32.174 (m). The latter replaces both KE = 1/2mv^2 and E = mc^2 / beta. What contributions have YOU made to science, PD? Ha. ha, HA! NoEinstein > > On Apr 26, 9:42 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On Apr 26, 1:30 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > Dear PD, the Parasite Dunce: Your denseness makes you incapable of > > understanding that NO force is required to cause an object to COAST. > > I do understand that. I've told you that. > > > The velocity of a falling object at the end of second one COASTS all > > the way through till the object strikes the ground. The velocity at > > the end of second two does the same thing, and etc. The INCREASE in > > velocity per second is uniform (as you agreed two replies back). It > > is the COASTING carry-over velocity which causes the shape of the free- > > drop curve (distance vs. time) to be a parabola. > > Nevertheless, the work is the increase in energy each second. > The work is the product of the force times the displacement. The force > remains constant throughout the drop. The displacement in the second > second is three times that what it is in the first second. Therefore > the work increases each second, which means that the *increment* of > energy in each second is not uniform but steadily increasing. You'll > note that even if you remove the coasting component, this persists. I > don't know why this is hard for you. > > > > > In spite of what you suppose some G. D. formula says, there can be NO > > work performed unless there is a resisting force! > > That is incorrect, John, and I don't know where you ever got that > impression. > Newton's second law tells you this. F=ma. You perhaps have seen it > before. > If there is an impressed force on an object, and an equal and opposite > resisting force, then there is no net force on the object. This means > the F in F=ma is zero. Then the acceleration a must be zero. This is > Newton's 2nd law, to remind you. > > This is clearly not the case with a falling object, where the > acceleration is not zero, and so there is a net force. This net force > does work. > > > And since you > > suppose that the work done is increasing semi-parabolically (as would > > match KE = 1/2mv^2), then, the resisting FORCE must be increasing semi- > > parabolically, too. However, the only force countering the force of > > gravity is the INERTIA of the object dropped, and that never changes! > > The CORRECT formula for the kinetic energy of dropped objects is: KE = > > a/g (m) + v / 32.174 (m). And that formula increases LINEARLY, not > > parabolically. Both Coriolis and Einstein were wrong to think that a > > linear input of energy (velocity) will produce an exponential increase > > in KE. Doing so violated the Law of the Conservation of Energy. > > > So, the readers will know: PD, the Parasite Dunce has never made a > > +new post in the three plus years that I have been visiting > > sci.physics. > > That's a lie, John. You're just incapable of using usenet properly to > find them. Your incapacity is not my problem, and it doesn't give you > license to lie from your ignorance. > > > > > I copy some of my expertly explained posts, below. > > NoEinstein P. S.: In particular, see the two posts with the > > *** ... ***. > > > Where Angels Fear to Fallhttp://groups.google.com/group/sci.physics/browse_frm/thread/8152ef3e... > > Last Nails in Einstein's Coffinhttp://groups.google.com/group/sci.physics.relativity/browse_frm/thre... > > *** Pop Quiz for Science Buffs! ***http://groups.google.com/group/sci.physics/browse_frm/thread/43f6f316... > > An Einstein Disproof for Dummieshttp://groups.google.com/group/sci.physics/browse_thread/thread/f7a63... > > Another look at Einsteinhttp://groups.google.com/group/sci.physics/browse_frm/thread/41670721... > > Three Problems for Math and Sciencehttp://groups.google.com/group/sci.physics/browse_thread/thread/bb07f... > > Matter from Thin Airhttp://groups.google.com/group/sci.physics/browse_thread/thread/ee4fe... > > Curing Einsteins Diseasehttp://groups.google.com/group/sci.physics/browse_thread/thread/4ff9e... > > Replicating NoEinsteins Invalidation of M-M (at sci.math)http://groups.google.com/group/sci.math/browse_thread/thread/d9f98526... > > Cleaning Away Einsteins Mishmashhttp://groups.google.com/group/sci.physics/browse_thread/thread/5d847... > > *** Dropping Einstein Like a Stone ***http://groups.google.com/group/sci.physics/browse_thread/thread/989e1... > > Plotting the Curves of Coriolis, Einstein, and NoEinstein (is > > Copyrighted.)http://groups.google.com/group/sci.physics/browse_thread/thread/713f8... > > Are Jews Destroying Objectivity in Science?http://groups.google.com/group/sci.physics/browse_thread/thread/d4cbe... > > The Gravity of Masses Doesnt Bend Light.http://groups.google.com/group/sci.physics/browse_thread/thread/efb99... > > KE = 1/2mv^2 is disproved in new falling object impact test.http://groups.google.com/group/sci.physics/browse_thread/thread/51a85... > > Light rays dont travel on ballistic curves.http://groups.google.com/group/sci.physics/browse_thread/thread/c3d7a... > > A BLACK HOLE MYTH GETS BUSTED:http://groups.google.com/group/sci.physics/browse_thread/thread/a1702... > > SR Ignored the Significance of the = Signhttp://groups.google.com/group/sci.physics/browse_thread/thread/56247... > > Eleaticus confirms that SR has been destroyed!http://groups.google.com/group/sci.math/browse_thread/thread/c3cdedf3... > > NoEinstein Finds Yet Another Reason Why SR Bites-the-Dust!http://groups..google.com/group/sci.physics/browse_thread/thread/a3a12... > > NoEinstein Gives the History & Rationale for Disproving Einsteinhttp://groups.google.com/group/sci.physics/browse_thread/thread/81046... > > There is no "pull" of gravity, only the PUSH of flowing ether!http://groups.google.com/group/sci.physics/browse_thread/thread/a8c26... > > > > On Apr 26, 11:05 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On Apr 24, 2:29 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > Dear BLOCKHEAD PD, the Parasite Dunce: You have only a one neuron > > > > brain (encased in concrete). So, to you 'experimental evidence' > > > > matters even when the conclusions of such are WRONG. > > > > I'm sorry, John, but if you think that experimental results make no > > > sense because it conflicts with your common sense, then your common > > > sense is what's wrong. > > > > > IF as you say > > > > (sic) the KE of falling objects accrues non-linearly (sic), then the > > > > INPUT energyfrom the force of gravitymust be non linear, too > > > > (sic). > > > > The input of energy comes from work. There is more to work than just > > > the force. Recall the work is the *product* of force and displacement.. > > > So it is entirely possible for the force to be linear and the work to > > > be nonlinear, or the force to be constant and the work to be non- > > > constant. This is really not complicated, John, and 7th graders have > > > no difficulty with it, so I don't know why you have such a problem > > > with it. > > > > > NOTE: You must agree to that statement if you accept that the > > > > Law of the Conservation of Energy is correct. Agreed? Then, tell me, > > > > PD, what about the UNIFORM force of gravity is non linear? You've > > > > already agreed that the VELOCITY of falling objects is increasing > > > > uniformly in simple accelerations. Newton's Laws of Motion state that > > > > a uniform force will cause one and only one associated acceleration.. > > > > If the acceleration is... 'g', then the uniform FORCE causing the > > > > acceleration is the unchanging static WEIGHT of the falling object. > > > > Yes, indeed. But the work is not the force alone. The work is the > > > *product* of the force times the displacement. > > > In the first second, a gravitational force of 2 lbs will cause a rock > > > to cover 16 ft, if it starts from rest. In the next second, the same > > > gravitational force of 2 lbs on the same rock will cause the rock to > > > cover an additional 48 ft. > > > So you see, the work done on the rock, which is the amount of energy > > > that gravity supplies to the rock, is three times higher in the second > > > interval compared to the first interval, even though the force stays a > > > constant 2 lbs. > > > > > You typically escape from the above statements of truth by digging > > > > into your dusty textbooks. > > > > Nope. Real experiments, done in freshman labs. > > > > > You then SHOEHORN the errant mechanical > > > > definition of 'work' into the dropped object results. The latter are > > > > errant simply because the equation doesn't clarify that the 'distance > > > > of travel' is indicative of... 'work done' ONLY if there is a FORCE > > > > being applied against a RESISTANCE which is equal and opposite. > > > > That simply isn't correct, John. If there were a resistance force that > > > were equal and opposite, then the net force on the object would be > > > zero. Newton advised us in the late 1600's that the net force is the > > > product of mass and acceleration (F=ma, surely you've heard of it), so > > > that if the net force is zero, then the acceleration is zero. So a > > > dropped rock that is accelerating cannot possibly have zero net force > > > on it. In fact, NOTHING that is accelerating can have a net force of > > > zero acting on it. > > > > > That > > > > means that when the force increases, the RESISTANCE increases by the > > > > same amount. > > > > > For dropped objects, the only force causing the one rate of > > > > acceleration is the object's static weight. And the only RESISTANCE > > > > is the object's INERTICthat exactly matches the static weight, > > > > applied FORCE. Since the distance of fall with respect to time isn't > > > > LINEAR, but parabolic, then, the ENTIRE non linear component of the > > > > distance of fall has to be due to COASTINGbecause there isnt an > > > > associated increase in either the applied force, OR the resistance. > > > > > Folks, PD majored in high energy particle physics. I majored in > > > > architecture and STRUCTURAL ENGINEERING. Of those majors, which one > > > > would likely be the most proficient in understanding... the > > > > applications of FORCES? > > > > Well, John, since the introductory courses that- Hide quoted text - > > - Show quoted text -... > > read more » |