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From: NoEinstein on 12 May 2010 23:46 On May 10, 10:57 am, PD <thedraperfam...(a)gmail.com> wrote: > Folks: In arguments over FORCES, the engineers and architects, who deal in real world problems, trump the HEP Physics majors who only know about the small and the insignificant. If a 250 pound linebacker hits you with a stated velocity, you will experience the action of a weight and velocity proportional FORCE. And all forces are in POUNDS, only! NE > > On May 9, 7:36 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On May 7, 12:47 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > PD, the Parasite Dunce, is hamstrung to the science formulas in > > textbooks. He, and perhaps some of you, cant fathom that simple > > logic can invalidate many of those. > > Ah, I see. Keep in mind that you've already AGREED with the facts as I > stated them below, and all I did was do use simple arithmetic to find > the ratios. Are you now recanting the facts? If so, which ones below > do you take issue with? Is it the values of the displacement? Is it > the values of the velocity? Is it the constancy of the force? Or do > you find that there is an arithmetic error somewhere? Or are you > claiming that if the arithmetic result is in conflict with your common > sense, then arithmetic must also be fundamentally in error? > > > The most common error in many > > equation types, is to allow the units of proportionality factors to > > be included in the units of the results. > > > The interesting TV show, MythBusters, enjoys crashing things or > > hitting things. I was amazed to hear those guys state that the > > calculated forces of impact are such-and-such foot-pounds. > > This is why TV shows are not to be taken as good references, John. > > > Foot- > > pounds? The latter is, actually the term for MOMENT, or the tendency > > of a force to cause a rotation about some fulcrum or point of pivot. > > The laughable units for MOMENTUM thats shown in many textbooks is: > > pound-feet/sec (sic). > > Which textbooks show those units of momentum? > > > That might be logical, because momentum is = > > mass times velocity, or mv. However, the v, in this case, is part > > of a proportionality fraction that becomes unit-less. To explain: > > > If a mass, like say, a 250 pound linebacker, has a velocity in some > > direction, and you are standing in his path, you will be hit by a > > force. Since that force is dependent on how heavy the linebacker > > is, and how fast he is moving just before hitting you, then the > > correct way to write the equation for momentum is F (or force) = mv. > > No, sir. You've just made that up. It does not appear in any > textbooks. > > > In many texts, the letter p is substituted for force or reactions. > > No sir. On this you are simply mistaken. The letter p is used to > denote MOMENTUM, not force, in physics. I really don't care if you > found somewhere in an Steel Handbook where a force is labeled with a > p. In the formula p = mv, none of those variables denotes force. None. > > > In > > fact, the letter p is used in all of the beam analysis equations in > > the AISC Steel Handbook that must show a point load or force. When I > > told PD that the equation for momentum is F = mv, stated in POUNDS, he > > accused me of lying. > > Yes, indeed, you are lying. > > > Because pwhich means FORCE, > > It does not. You are lying. > > > fis different > > alphabetically PD supposes that momentum must be a different > > animal than force. But, as usual, he is wrong! > > > Newtons second law states: A continuous, uniform forcewhen applied > > to a frictionless and unrestrained bodywill accelerate the body in > > the direction of the force, and in proportion to the force. The > > equation for Newtons law thats usually shown in texts is: F = ma. > > The momentum formula, thats in the above paragraph, is a close cousin > > to Newtons when it comes to measuring the force of impact, because > > the aspect of acceleration which quantifies the expected force to be > > delivered is the instantaneous VELOCITY right before the object (like > > the 250 pound linebacker) impacts. > > Sorry, but you've just made that up, and it is wrong. Newton was > right. You are wrong. > > > > > A 250 pound linebacker traveling 8 feet/second (1/4th of the g > > velocity increase) will deliver a force (f or p) of 62.5 pounds. > > No, John, that is just plain wrong. You don't have the foggiest idea > how to use Newton's equations. > > Here's a way to tell, John. According to you, if a baseball pitcher > throws a baseball at a catcher at the same speed, then the force > delivered to the catcher will be the same, whether the catcher "gives" > with his hand or not. Any 8-year-old boy will tell you this is wrong. > > You've used Newton's 2nd law wrong all these years, and you still > passed exams enough to be certified as an architect? I'm horrified, > John. All your customers should be horrified as well. > > > > > That > > would be exactly how hard a 250 pound linebacker would hit if he was > > accelerating 32 ft./sec. for ¼ second. A proviso is that he not > > continue to accelerate once the other player is impacted. > > > Since kinetic energy is the force-delivery potential of falling > > objects, as well as for objects traveling at any set velocity, Ive > > determined that KE and momentum are interchangeable terms, with THIS > > important exception: Objects that are RESTRAINED, but being acted on > > by a potentially propulsive force, will have the latter propulsive > > force ADDED to the force of impact. My mathematically and > > experimentally verified formula for kinetic energy is: KE = a/g (m) + > > v / 32.174 (m). The a/g will be 1 for objects being acted on by > > Earth gravity. So, the KE of objects being restrained prior to > > release is already one weight unit, even before any downward motion > > happens! The v / 32.174 (m) is the same unit-less PROPORTIONALITY > > factor that is in the momentum equation. Heretofore, the masses were > > required to be converted to SLUGS (32 pounds) in order to find the > > force. My velocity-variant fraction is more intuitive and doesnt > > require an explanation of usage below the equation. It is the > > omission of the conditions of usage, or applicability, that cause the > > proliferation of errant equations. The way I was able to master > > equations was to express what those say in clear English. When > > different users of equations have different ideas what the variables > > and the constants mean, there can be big trouble. > > > In most cases, my New Science will make the equations simpler and more > > intuitive. Does anyone, other than PD, fault that? NoEinstein > > > > On May 6, 9:23 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On May 5, 12:36 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > PD: The L. C. catalogue card number is: 5241857. (look on page 19). > > > > Here's the response to my query at the Library of Congress: > > > The LCCN you entered [ 5241857 ] was not found in the Library of > > > Congress Online Catalog. > > > Are you lying, John? > > > What's the ISBN? > > > > > Also, my The Wiley Engineer's Desk Reference, by Stanford I. Heisler, > > > > on page 94, says momentum = mv. > > > > That is different than F=mv. Momentum is not force. > > > > Moreover, this is not a good definition of momentum, though it is a > > > useful approximation for engineers, not suitable for physics. > > > > > A scripted style of the "m" is used > > > > to differentiate from "mass". That book errs by saying that the > > > > "units" is: (mass)-feet/secondwhich is bullshit! > > > > And yet you would have me trust this Wiley Engineer's Desk Reference, > > > when you don't believe it yourself. When are you going to support any > > > of your assertions, John, other than blustering about what comes out > > > of your own head? > > > > > Momentum is > > > > measured in pounds! It is velocity proportional, and that is a > > > > simple, unit-less FRACTION NE > > > > > > On May 5, 2:56 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > On May 4, 2:53 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > PD loves to extrapolate things into unworkability, so he can claim > > > > > > everything was invalid. MOMENTUM is: F = mv, expressed in pounds. > > > > > > He'll find that same equation (but not the correct units, pounds) in > > > > > > most textbooks. NE > > > > > > No, I won't, John. That equation F=mv is not listed in most > > > > > textbooks. > > > > > When you can clearly identify which title you think DOES have that > > > > > listed, then I can look for myself. > > > > > As it is, since you obviously have problems reading an understanding a > > > > > single sentence from beginning to end, I have my doubts. > > > > > > > > On May 4, 1:07 pm, af...(a)FreeNet.Carleton.CA (John Park) wrote: > > > > > > > > > PD (thedraperfam...(a)gmail.com) writes: > > > > > > > > > On May 3, 10:07=A0pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > > >> Dear PD: =A0A thin "College Outline Series" book (that fits into the > > > > > > > > >> bookcase behind my computer chair) entitled "Physics", by Clarence E > > > > > > > > >> Bennett, states on page 19: "G. =A0Momentum and Impulse. =A0(1.) =A0Momen= > > > > > > > > > tum > > > > > > > > >> is defined as the product of the mass times velocity (mv)..." =A0The > > > > > > > > >> letter F is used for momentum, because the equation defines forces. =A0= > > > > > > > > > =97 > > > > > > > > >> NoEinstein =97 > > > > > > > > > > Oh, good grief. John, what is the ISBN on this book? I'd like to > > > > > > > > > secure it to look at it. > > > > > > > > > From what it is you just told me is in it, if I can verify that you > > > > > > > > > can indeed read it correctly, it is a horrible, horrible booklet and > > > > > > > > > should be burned as worthless. > > > > > > > > > To quote the Spartans on a quite different occasion: If. > > > > > > > > > I can't help noticing that the actual quoted passage is reasonable and > > > > > > > > the inference about forces is purely in NE's words. > > > > > > > > Exactly. > > > > > > > > For what it's worth, momentum's *definition* is not mv, either. > > > > > > > Electromagnetic fields have momentum, but this expression certainly > > > > > > > does not work for them. The formula works for a certain class of > > > > > > > matter-based objects traveling at low speed, and that's it. > > > > > > > > PD- Hide quoted text - > > > > > > > > - Show quoted text -- Hide quoted- Hide quoted text - > > - Show quoted text -... > > read more »
From: NoEinstein on 12 May 2010 23:49 On May 11, 9:14 am, PD <thedraperfam...(a)gmail.com> wrote: > PD, the Dunce, never, ever discusses SCIENCE! NE > > On May 11, 7:30 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On May 5, 11:43 am, PD <thedraperfam...(a)gmail.com> wrote: > > > PD: Energy IN must = energy OUT. Since KE = 1/2mv^2 can't meet that > > requirement, then it is 100% in violation of the Law of the > > Conservation of Energy; and no 'consensus' of physicists (ha!) who say > > otherwise, can change that fact! NE > > But it does meet that requirement. I showed you exactly how, just the > other day. > It seems you are slipping, NoEinstein, no longer able to remember what > was said the day before. > So each day is brand new to you. You could hide your own Christmas > presents. > It's a shame you've slipped into senility, but it does give me pause > on how much effort to expend on a serious reply to you. > > > > > > > > On May 4, 6:39 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On May 3, 11:51 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > Dear PD, the Parasite Dunce: IF, as you've just said, everyone knows > > > > that the KE equation (KE = 1/2mv^2) is inconsistent with the Law of > > > > the Conservation of energy, then you've just agreed that the former is > > > > WRONG! > > > > But I didn't say that, John. I said that the KE equation above is > > > completely CONSISTENT with the Law of Conservation of Energy. > > > > I think I've isolated the source of your great difficulties, John. You > > > cannot comprehend the meaning of a single sentence that you read. Did > > > you understand THAT? > > > > > The physicists whom YOU know may not be concerned, but the > > > > Laws of Nature are very, very mad, indeed! NoEinstein > > > > > > On May 1, 8:25 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > 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! > > > > > > Well, it's because physics is a science, which means that it invokes > > > > > the scientific method, and it determines truth by experimental test, > > > > > not by logic. That is taught to 4th graders. Were you absent that day, > > > > > or did you determine in the 4th grade that your science teachers > > > > > didn't know what they were talking about and you realized then that > > > > > all of scientific truths could be determined by logic? > > > > > > > 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). > > > > > > I'm sorry, John, but just about everyone except for you knows that the > > > > > Law of Conservation of Energy is completely consistent with the > > > > > expressions for kinetic energy and total energy. It seems to be only > > > > > you with the problem. Shouldn't that be a flag to you? > > > > > > If everyone in the world points to the same animal and calls it a > > > > > zebra, and you call it a penguin, does that make you a world-class > > > > > genius or a world-class fool? > > > > > > > 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- Hide quoted text - > > - Show quoted text -... > > read more »
From: NoEinstein on 12 May 2010 23:51 On May 11, 9:15 am, PD <thedraperfam...(a)gmail.com> wrote: > PD: ISBN numbers must be bought. Since Barnes and Nobel sold the books in their own stores, they needed nothing "international". NE > > On May 11, 7:36 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On May 7, 12:47 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > Dear PD, the Dunce: You take any TRUTH; generalize it to absurdity; > > then claim that the truth is wrong. Actually, the only thing wrong is > > your (sidestepping) generalizations into absurdity! NoEinstein > > This from the man who can't find the ISBN number of a book, and can't > accurately copy down a Library of Congress catalog number. > > > > > > > > On May 6, 9:23 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On May 5, 12:36 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > PD: The L. C. catalogue card number is: 5241857. (look on page 19). > > > > Here's the response to my query at the Library of Congress: > > > The LCCN you entered [ 5241857 ] was not found in the Library of > > > Congress Online Catalog. > > > Are you lying, John? > > > What's the ISBN? > > > > > Also, my The Wiley Engineer's Desk Reference, by Stanford I. Heisler, > > > > on page 94, says momentum = mv. > > > > That is different than F=mv. Momentum is not force. > > > > Moreover, this is not a good definition of momentum, though it is a > > > useful approximation for engineers, not suitable for physics. > > > > > A scripted style of the "m" is used > > > > to differentiate from "mass". That book errs by saying that the > > > > "units" is: (mass)-feet/secondwhich is bullshit! > > > > And yet you would have me trust this Wiley Engineer's Desk Reference, > > > when you don't believe it yourself. When are you going to support any > > > of your assertions, John, other than blustering about what comes out > > > of your own head? > > > > > Momentum is > > > > measured in pounds! It is velocity proportional, and that is a > > > > simple, unit-less FRACTION NE > > > > > > On May 5, 2:56 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > On May 4, 2:53 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > PD loves to extrapolate things into unworkability, so he can claim > > > > > > everything was invalid. MOMENTUM is: F = mv, expressed in pounds. > > > > > > He'll find that same equation (but not the correct units, pounds) in > > > > > > most textbooks. NE > > > > > > No, I won't, John. That equation F=mv is not listed in most > > > > > textbooks. > > > > > When you can clearly identify which title you think DOES have that > > > > > listed, then I can look for myself. > > > > > As it is, since you obviously have problems reading an understanding a > > > > > single sentence from beginning to end, I have my doubts. > > > > > > > > On May 4, 1:07 pm, af...(a)FreeNet.Carleton.CA (John Park) wrote: > > > > > > > > > PD (thedraperfam...(a)gmail.com) writes: > > > > > > > > > On May 3, 10:07=A0pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > > >> Dear PD: =A0A thin "College Outline Series" book (that fits into the > > > > > > > > >> bookcase behind my computer chair) entitled "Physics", by Clarence E > > > > > > > > >> Bennett, states on page 19: "G. =A0Momentum and Impulse. =A0(1.) =A0Momen= > > > > > > > > > tum > > > > > > > > >> is defined as the product of the mass times velocity (mv)..." =A0The > > > > > > > > >> letter F is used for momentum, because the equation defines forces. =A0= > > > > > > > > > =97 > > > > > > > > >> NoEinstein =97 > > > > > > > > > > Oh, good grief. John, what is the ISBN on this book? I'd like to > > > > > > > > > secure it to look at it. > > > > > > > > > From what it is you just told me is in it, if I can verify that you > > > > > > > > > can indeed read it correctly, it is a horrible, horrible booklet and > > > > > > > > > should be burned as worthless. > > > > > > > > > To quote the Spartans on a quite different occasion: If. > > > > > > > > > I can't help noticing that the actual quoted passage is reasonable and > > > > > > > > the inference about forces is purely in NE's words. > > > > > > > > Exactly. > > > > > > > > For what it's worth, momentum's *definition* is not mv, either. > > > > > > > Electromagnetic fields have momentum, but this expression certainly > > > > > > > does not work for them. The formula works for a certain class of > > > > > > > matter-based objects traveling at low speed, and that's it. > > > > > > > > PD- 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: Timo Nieminen on 13 May 2010 02:29 On Wed, 12 May 2010, NoEinstein wrote: > On May 7, 5:29 pm, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > On May 8, 5:57 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On May 7, 2:21 am, Timo Nieminen <t...(a)physics.uq.edu.au> wrote: > > > > > Dear Timo: On the one hand you compliment me; on the other you chide > > > me for not having all of the numbers at my fingertips. > > > > No, I tell you that all of these numbers are available on www, so you > > don't even need to go and look in a book. When you sit at your > > keyboard, the relevant numbers _are_ at your fingertips. > > > > It's an obvious test. Since you claim your theory explains reality, I > > expected that you would have compared the two - your theory and > > reality - and checked if they agree. My mistake - you don't seem to > > have done this. > > > > You give a clear and easily checked statement: > > > > > Since gravity is > > > directly proportional to photon emission (not gravitons, which dont > > > exist), then it is the luminosity and the temperature of the light > > > that determine the gravity of stars. > > > > But before it's worth checking this, you should clarify: > > > > (1) By "directly proportional", you mean: gravity = (constant) times > > (photon emission)? That's is, linear proportionality. Or do you mean > > something else? > > > No. "Directly proportional" means: Double the luminosity, and you > double the gravity. Or double the surface area, and you double the > gravity. This is clear. Linear proportionality it is, then. > Your word (constant), is actually a variable fraction. But > I suppose that for a given luminosity and star surface area, the > gravity would be a single (constant) value. But this isn't clear. If (constant) is actually a variable, then do you have linear proportionality? (As a technical note, "luminosity" doesn't mean "surface brightness", but "total brightness", i.e., double the surface area while keeping everything else the same, and you double the luminosity. Anyway, it's time to check against reality. Take a well-known close binary system like Sirius. From their mutual orbit, the "known" masses are 2.02M for Sirius A and 0.98M for B (M = solar mass). So, gravitationally, they differ by a factor of 2. How do the luminosities compare? For A, we have 25L (L = solar luminosity), and for B, 0.026L. Factor of 1,000 difference in luminosity, but only a factor of 2 difference in "mass" as measured from the orbit. This isn't at all close to your prediction. Since you're highly intelligent, and a careful and logical thinker, it isn't likely that you made any error in proceeding from your theory to your prediction. Therefore, it's likely that your theory is wrong. > > (2) What do you mean by "photon emission"? What you mean by "photon" > > might not be what conventional science means by "photon". How is > > "photon emission" related to radiated power? Since the bolometric > > luminosity is the total radiated power, is there any _further_ > > dependence on temperature beyond its effect on the bolometric > > luminosity. (If talking about visual luminosity, then, yes, the > > bolometric luminosity depends on the visual luminosity and the > > temperature.) > > > The wavelength of the light (color) determines how many photons are > being emitted in a given, say, second. Gravity, actually, depends on > how efficiently the trains of photons 'pump' ether back into space. I > can't say, with certainty, that a wavelength of light that's half as > long will be exactly twice as efficient moving ether, out. > Experiments will have to confirm the efficiency for various > wavelengths. So, to test you claims of stellar gravity versus luminosity, one should compare stars of the same temperature. Sirius A and B aren't identical, but close. Sirius A is A1V, Sirius B is A2. Close enough, considering the x1,000 ratio of luminosities. > > > At room temperatures gravity is mass proportional, and matches > > > Newtons law. There has to be an object-size threshold that DENIES > > > mass in favor of surface area and temperature. > > > > This is new. You didn't say anything about this that I saw before. > > > Thanks! You've just admitted that you are a regular reader of my > science posts. And you are observant enough to realize that > 'reasoning' is taking place even as I write a reply. You, better than > most, should understand why I don't need to go running to books, by > others, to find answersI give things my own best shot, first. > > > > > I suspect that a > > > heated Cavendish ball will have gravity somewhere between the room > > > temperature, and the white hot. > > > > "Suspect" isn't good enough for the experiment to be worthwhile. It's > > directly connected to the following point which you didn't address in > > your reply. Do note that this is absolutely essential for the > > experiment to be worthwhile (as you will no doubt already know, since > > this is a simple matter of logic and analytical ability). > > > Timo: The Cavendish experiment is conducted in a room with a high > ceiling. There is air around the balls to influence the twisting > speed. And the radiation of the heated ball(s) would be reflected > back by the colder walls. In outer space, there would not be air to > both drag the balls and to intercept the radiant energy. I can > virtually guarantee that the Cavendish isn't a perfect analogy to the > gravity of stars that are very, very hot. > > > > To repeat the question: if a Cavendish experiment _doesn't_ detect a > > greater gravitational force, what does that mean for your theory? > > > Send me photographs of the experiment, and etc. It could be that the > heated balls might change the torsion characteristics of the wire > being more-so toward the end of the experiment, after the wire has had > time to get hot. The answer to your question requires that the > experiment be valid, for hot balls. M-M got nil results. > Understanding the reason for that failure took over a century until I > came along. I'd want to do a... post mortem on everything. The experiment can be designed to try to achieve a desired accuracy. If you could be bothered saying how accurate the experiment needs to be, then perhaps something could be done. Since you keep refusing to say (and surely you must know how large the effect should be, or at least capable of quickly deducing, from your theory, how large the effect should be), there isn't any point in trying the experiment. If you have two identical balls, and heat one to be double the absolute temperature of the other (i.e., to about 330C; yes, easy to heat it more, but let us keep this simple), the hot ball will radiate 16 times as much power, with peak emission at double the frequency (i.e., half the wavelength). How much stronger will the gravitational force be? If you can be bothered applying your mighty intellect, and actually provide a quantitative answer, perhaps an experimental test might be worthwhile. Without such an answer, useless. > > > > > Consider this: If you can heat one ball white hot, > > > > > and you DO detect a greater gravity, youve confirmed my theory. > > > > > > It would _support_ your theory, not confirm it in any absolute sense. > > > > If one tries this and _doesn't_ detect a greater gravitational force, > > > > would that mean your theory is wrong and it's time to forget it and move > > > > on?- Hide quoted text > > > Probably not. The LOGIC of gravity being flowing ether answers too > many of the century's old questions about the Universe. NoEinstein Well, if your theory is perfectly OK with a null result, why would a positive result support your theory? Surely your statement above means that the experiment is completely irrelevant to your theory. Or, since the luminosity/gravity predictions of your theory appear to fall at the first astronomical hurdle, perhaps it's more that your theory is irrelevant to reality.
From: PD on 13 May 2010 11:01
On May 12, 9:01 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 5, 12:23 pm, PD <thedraperfam...(a)gmail.com> wrote: > > PD: What, exactly, is Newton's Second Law useful for? In essence it > says that the uniform, continuous force that's necessary to cause a > given acceleration must be the same fraction (up or down) as the > target acceleration relates to 32.174 feet per second, each second. > Or, in order to produce, say, a 64.348 feet/sec., each second > acceleration, there must be two pounds of force acting upon each one > pound of the object's mass. Or... a doubling of the continuous force > will cause a 32.174 feet per second, each second increase in the > acceleration, IF the initial acceleration was 'g'. Newton's second > law wasn't very useful, was it. Oh, good heavens. No, it does not say that, even in essence. And if you have no idea what Newton's 2nd law is good for, then good heavens, you learned nothing in school. Nothing. > > The force and the acceleration must be in this proportion: F = v / > 32.174 (m). Oh, good grief. No. Geez, you are denser than a stack of National Geographic magazines. > I say 'v', rather than 'A', because the compared > fractions have the same unites in both the top and bottom of the > fraction. So, the units cancel out. That leaves just the numeric > portion of the fraction. All accelerations use the "per second" > velocity as the datum point. So, in essence, you are comparing two > velocities at "one second". That's why it says "v/32.174" instead of > "A/32.174". > > V/32.174 (m) = the MOMENTUM of the object! No again. This has no bearing whatsoever on the concept of momentum used in physics. You simply manufacture things in your head and apply words to them that are already reserved for something else. > That is the latter half of > my correct kinetic energy equation which replaces Coriolis's KE = 1/2 > mc^2; and Einstein's E = mc^2 / beta, or : *** KE = a/g (m) + v / > 32.174 (m). Note: Having a unit mass in any equation doesn't require > that there be a "work" calculation for moving the mass. The only > thing needed is to know that the VELOCITY, and the force will be in > the proportion as given by my equation. NoEinstein > > > > > On May 5, 2:30 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On May 4, 11:33 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > PD: And the point of your 'addition' extrapolation is? Your science > > > notions are shallow enough without implying that I have disavowed > > > common math. If Einstein had known how to do simple mathnowhere in > > > evidence in his (mindless) equation physicsperhaps the dark ages of > > > Einstein wouldn't have lasted so long. NoEinstein > > > You made a general statement that if something is generally accepted, > > then that is a sign that it is nearly certainly WRONG. > > > Now you don't seem so sure. > > > You don't want to disavow common math, but you are certainly willing > > to disavow common, grade school mechanics like Newton's 2nd law. And I > > want to point out again that this has nothing to do with the "dark > > ages of Einstein", since Newton's 2nd law has been around for 323 > > years! You've decided that all of physics since Galileo and Newton are > > the dark ages! Einstein has nothing to do with your complaint. > > > PD > > |