univeral theory Posted January 24, 2013 Author Posted January 24, 2013 Your equation fails dimensional analysis, it's safe to assume it's wrong and meaningless. If I were you, I would start By quoting areas where I want to challenge univaso theory based on dimensional analysis and provide feasible evidence to challenge that area. Other wise; “….what a man means by a term is to be found by observing what he does with it, not by what he says about it” Percy w. Bridgeman(one of the greatest celebrities of dimensional analysis).
Klaynos Posted January 24, 2013 Posted January 24, 2013 (s/1/+ .-(√s) )=1 Where S = E = energy. Mass distance distance / (time time)/1/sqrt (mass distance distance / (time time)) cannot equal 1, the dimensions do not cancel out. This is confounded by you having pulled it out of the air. The only thing that can be fairly analysed is false the rest is word salad, not science.
univeral theory Posted January 25, 2013 Author Posted January 25, 2013 Mass distance distance / (time time)/1/sqrt (mass distance distance / (time time)) cannot equal 1, the dimensions do not cancel out. This is confounded by you having pulled it out of the air. The only thing that can be fairly analysed is false the rest is word salad, not science. (s/1/+ .-(√s) )=1 this is the equation for the quantun steps of anergy conservation (Q2). infact iam just wondering where did you derive such an interpretation with reference to the meaning of my equation. please quote me interpreting this equation to mean what you are saying or elaborate from where did you derive such a meaning with reference to how i defined the equation. Just to make it clear, there's many many people who think they have some universal theory, but they all fail the test of scientific testing, even string theory, so it likely doesn't even matter if somehow anyone is right because it could never be tested and therefore never be used for anything. For a theory to qualify a TOT, it is an obvious scientific and philosophical justification that it must not only depend on laboratory experiments for its proof. This is because even the accuracy or inaccuracies of lab tests depend on how it is exploited. And because even the living experiences of our ordinary behaviors are part of TOT as it is also part of them, it is just a matter of relevant interpretation to confirm it. The specific need of lab is to test and confirm the predictions made by this theory from specific ends of existence that is out of reach by our ordinary daily experience but not to determine its usefulness. This is because even the being of every thing and its life style must be an already benefit of the claims of this theory. You just need to look all around you and make critical interpretation or the interpretation of all that around your environment. Wherever there has been classical work (W=Fd),univaso work has been there inform of (W=gv2). Wherever there has been gravitational constant, univaso gravitation has been found there. Wherever there has been the constant speed of light (c2),univaso theory has been there. Whenever there is Planck constancy (h), univaso theory is there. Whenever there has been albrecht giese mass ,univaso theory is there. Wherever there is space or time, univaso velocity is there. Whenever there is Andy pibernick’s Q2 univaso theory is there-to mention but a few. But if this is not enough to demonstrate laboratory testability of such a framework, then in scientific faith the twenty first century labs would claim no scientific civilization at all or else, they would come out publicly and flamoyantly to denounce the pedigree of the authors and agitators of the above refferenced work as iam also ready to denounce this theory if it fails any feasible test.
Klaynos Posted January 25, 2013 Posted January 25, 2013 You said s was energy, I converted to dimensions which is what you do in dimensional analysis, if the sides of the equation do not match it is wrong. Your equation is wrong.
univeral theory Posted January 27, 2013 Author Posted January 27, 2013 (edited) The the cutting edge is definitely testable which is why its the "cutting-edge", because it's the most recently confirmed knowledge. If we assume things are true just because they are hypothetically logical then we might as well assume the Earth as wrong because we can't see that it's found from standing on it. Besides, you are trying to define the universe, but in order to have proof of what your saying in regards to the universe only being manifested in whatever specific manner, you would have to be there at the beginning of the universe to confirm that, because then you would know that because there is nothing outside the universe (because it wasn't created yet) that it is the only possible explanation, which isn't going to happen unless you found time travel, which is also impossible because it violates conservation laws. When it comes to intellectual proceedings, the cutting edge begins with conceptual reduction to practice. 2. When it comes to confirmation, we don’t need to be there at every phenomenon to make a confirmation, we only need to check the feasibility study of it and then confirm its validity or relevance to the phenomena. 3.And when it comes to time travel, time is never a conservable content but just a measure of the responsiveness of a given conservable content (energy) and thus time is not an independent dimension (phenomenon) of existence. And as a dependent variable, it operates with in the limits of a variable it depends on and those limits are the limits of its travel. Beyond those limits there is its beginning and its end. All laws of conservation describe the responsiveness of energy due to its displacement thus conservation has a beginning and an end. Beyond the limits of conservation there is energy in its virgin sense. So beyond time there is energy. And if we stand on that energy, we can make a perfect feasibility study of our universe through its time travel. Edited January 27, 2013 by univeral theory
univeral theory Posted January 28, 2013 Author Posted January 28, 2013 Please now properly address the point made by Klaynos that at least one of your equations fails dimensional analysis, or admit that you have no idea what dimensional analysis is, or why it is important in this context. In complex phenomena modeling, we need the most economical and most efficient methods of expressing a given phenomena in relation to the mathematical simplicity of solving that phenomena. According to the principle of similitude, a given measurable phenomenon can be expressed equivalently through different units of measurable dimension and the results of all these dimensions reflect the same phenomena. All measurements can be represented by numerical digits. And the value of a given numerical digit depends on the unit measurement of a given dimension that it represents. Thus; the numerical value of 10 is greater than 1, but 1 can be equal to or greater than 10 depending on the unit measurement of a given dimension that it represents. For example; 1centilitre equals 10 milliliters and 1 deciliter is greater than 10 milliliters. Thus the central mechanism of dimensional analysis is the degree of result equivalence arrived at through measuring all the dimensions that reflect the same phenomena. According to Wikipedia, “in physics and all science, dimensional analysis is the practice of checking relations among physical quantities by identifying their dimensions. The dimension of any physical quantity is the combination of the basic physical dimensions that compose it”. According to Bridgman, “The principle use of dimensional analysis is to deduce from a study of the dimensions of the variables in any physical system certain limitations on the form of any possible relationship between those variables. The method is of great generality and mathematical simplicity”. This is all about converting one thing to another. Such that; Given the definition of a physical quantity, or an equation involving a physical quantity, you will be able to determine the dimensions and SI units of the quantity Given an equation, you will be able to determine if the equation is dimensionally correct or incorrect. My friend quoted our quantum frames of energy conservation equation - Q2 = (s/1/( + .-(√s) )=1, which according to him that it fails dimensional analysis (which according to me I think he just missed out some fundamental points of either dimensional analysis or the Q2 equation). Let me re-illustrate this with the univaso equation of work which is W=gv2 (where w=work, g = gravitation and v2= velocity square) that the Q2 equation does not fail dimensional analysis. W=gv2 implies that there are two basic dimensions of work which is gravitation and velocity square. And according to dimensional analysis’s principle of similitude, the mathematical measurement of gv2 must be equivalent to W. their equivalence can be checked by the Q2 equation, such that; (W/1/g . v^2 (√W) )=1 And, ∂( W/1/g . v^2 (√W) )=1+or-1 And even if it is analyzing the dimensions of a given physical quantity say 9km/h, both the basic dimensions of speed (time and distance) still match at both sides of their function. Remember; Time is the measure of energy responsiveness to the displacement range of its Q2 as a function of cycle’s responsiveness. Distance is the measure of energy responsiveness to the displacement range of its Q2 as a function of location responsiveness. Such that; 1). (displacement range/1/displacement cycles . diplacement location (√displacement range) )=1 Or ; 2). (9km/h/displacement cycles . diplacement location (√displacement range) )=9km/h. -2
univeral theory Posted January 29, 2013 Author Posted January 29, 2013 I strongly encourage you to investigate dimensional analysis. It is not complicated and can be easily used for showing that concepts are incorrect. Such as those presented here. “I found that I was fitted for nothing so well as the study of truth; as having a nimble mind and versatile enough to catch the resemblance of things(which is the chief point) and at the same time steady enough to fix and distinguish their subtle differences….” Francis bacon “Think things, not words” or else, raise a question where you feel you don't understand well.
SamBridge Posted January 30, 2013 Posted January 30, 2013 “I found that I was fitted for nothing so well as the study of truth; as having a nimble mind and versatile enough to catch the resemblance of things(which is the chief point) and at the same time steady enough to fix and distinguish their subtle differences….” Francis bacon “Think things, not words” or else, raise a question where you feel you don't understand well. If your equation cannot generate the same units on both sides of the equation, then it doesn't make sense. Think about it, does it really make sense if I say "kilograms = beauty"? That's what you're doing if you don't have dimension analysis, units are just as important in physics and without keeping track of them we wouldn't be able to have all the knowledge we have and if they do not generate the same unit on both sides of the equation then something's wrong. Either you aren't using the right things for the equation or something is fundamentally wrong with what your using. 1
Klaynos Posted January 30, 2013 Posted January 30, 2013 I found that I was fitted for nothing so well as the study of truth; as having a nimble mind and versatile enough to catch the resemblance of things(which is the chief point) and at the same time steady enough to fix and distinguish their subtle differences. Francis bacon Think things, not words or else, raise a question where you feel you don't understand well. I see you're not interested in doing science but telling stories. But for the record your new equation using work also fails dimensional analysis and is therefore wrong.
univeral theory Posted January 30, 2013 Author Posted January 30, 2013 If your equation cannot generate the same units on both sides of the equation, then it doesn't make sense. Think about it, does it really make sense if I say "kilograms = beauty"? That's what you're doing if you don't have dimension analysis, units are just as important in physics and without keeping track of them we wouldn't be able to have all the knowledge we have and if they do not generate the same unit on both sides of the equation then something's wrong. Either you aren't using the right things for the equation or something is fundamentally wrong with what your using. This is the last equation from post #81 9km/h/displacement cycles . displacement location (√displacement range) =9km/h. I think we are having 9km/h at both sides of the equation. And i think "kilograms ≠ beauty"but “kilograms= kilograms” It is derived from the Q2 equation located at #58 equation 3 which is s/(+ .- √s)=s
Klaynos Posted January 30, 2013 Posted January 30, 2013 This is the last equation from post #81 9km/h/displacement cycles . displacement location (√displacement range) =9km/h. I think we are having 9km/h at both sides of the equation. And i think "kilograms ≠ beauty"but kilograms= kilograms It is derived from the Q2 equation located at #58 equation 3 which is s/(+ .- √s)=s You've ignored all the other quantities in the equation.
imatfaal Posted January 30, 2013 Posted January 30, 2013 This is the last equation from post #81 9km/h/displacement cycles . displacement location (√displacement range) =9km/h. .../ snipped so you have [latex]\frac{\frac{distance}{time}}{\frac{1}{time}}.distance.distance^{1/2}[/latex]
univeral theory Posted January 31, 2013 Author Posted January 31, 2013 You've ignored all the other quantities in the equation. Dimensional analysis: There are two basic concerns of dimensional analysis in its activity area; 1. Conversion 2.Similitude Such that; a given phenomena can be converted into different dimensional framework and the phenomena remains the same. For example; 30minuts can be converted into ½ an hour, or 1800seconds and the phenomena described remains the same (30minuts). Now I appeal to any reader to first concentrate on example 1 below, if it is wrong please be free to challenge it and if it is right then proceed to another step of reading: Example 1: when we say that 9 = 6+3, with dimensional analysis we are correct. Reason being, that we have just converted 9 into another dimensional framework of 6+3 and the quantitative phenomena which we are describing (9) remains the same. Example 2: supposing another person advances our 6+3 into another dimensional framework and says that; 9/+ .-(√9 )=6+3, in dimensional analysis he would be right. Reason being, that he has just advanced the dimensional framesork of6+3 to the dimensional framework of 9/+ .-(√9 ) but the phenomena described would be the same with out even loosing any quantity phenomena from any side of the equation. Perhaps our task here is to solve the equation of 9/+ .-(√9 )= 9/-3 . 3 =6.3 or 6+3, thus the equation yields absolutely the same phenomena from both sides of the equation. Example 3: and supposing another person dimensional framework of the equation is 9/+ .-(√9 ) = 9, in dimensional analysis, it is a matter of checking whether he is right at both sides of the equation through conversion. And here we shall have 6+3=9 which is the same as saying that 9=9. I would prefer that we first check the above motioned scenario before proceeding to any other complicated phenomena with details of physical relevance.
Ophiolite Posted January 31, 2013 Posted January 31, 2013 I have proceeded as far as Example 1. It is incorrect. The equation you have given, 9 = 6 + 3, is dimensionless. It does not possess dimensions. Your numbers have no assigned units. This is what it means to say something is dimensionless. Dimensional analysis involves checking that the units on one side of the equation match those on the other side. You have already been told this. It is apparent that you do not understand and, further, that your 'hypothesis' is built upon non-existent foundations.
univeral theory Posted February 1, 2013 Author Posted February 1, 2013 I have proceeded as far as Example 1. It is incorrect. The equation you have given, 9 = 6 + 3, is dimensionless. It does not possess dimensions. Your numbers have no assigned units. This is what it means to say something is dimensionless. Dimensional analysis involves checking that the units on one side of the equation match those on the other side. You have already been told this. It is apparent that you do not understand and, further, that your 'hypothesis' is built upon non-existent foundations. dear ophiolite sorry for my disturbances,but iam just struggling to understand some fundamental application of dimensional analysis from your point of view. Does your comment try to mean that pure arithmetical or algebraic equationsthat are not assigned with any physical measurable units have no physicaldimension? And thus they are dimensionless un less attached to a given physicalquanty? and further they are not subject to dimensional analysis?
univeral theory Posted February 2, 2013 Author Posted February 2, 2013 That is my understanding. And as such, when we translate the equation like1+0=1 or 6+3= 9 into the context of physical phenomena, by substituting the arithmetical variables with physical variables and we say for example that; W=F. d, Where; W= work F = force d = displacement just forexample Does the equation become dimensional? And can the dimensions cancel out?
Klaynos Posted February 2, 2013 Posted February 2, 2013 And as such, when we translate the equation like1+0=1 or 6+3= 9 into the context of physical phenomena, by substituting the arithmetical variables with physical variables and we say for example that; W=F. d, Where; W= work F = force d = displacement just forexample Does the equation become dimensional? And can the dimensions cancel out? Let us consider f=ma If the mass is 3kg and the acceleration is 5m/s/s then to say f=3*5 is technically not correct, units are important! f=3kg * 5m/s/s - is correct.
univeral theory Posted February 6, 2013 Author Posted February 6, 2013 Are you familiar with dimensional analysis? Dear klaynos; I can now appreciate the concerns of dimensional analysis and particularly special thanks go to you klaynos. Your concerns about the conventionality of dimensional analysis have made me discover some mistakes that I made in the definition of the Q2equations (the quantum frames of conserving work equation), and at the same time realizing the relevance of univaso theory equations to dimensional analysis. Mistake and correction: From the equations of the quantum frames of conserving work Q2, that is; [latex] \frac{\frac{s}{1}}{+.-(\sqrt{s})} = 1[/latex] Or [latex] \frac{s}{+.-(\sqrt{s})} = 1[/latex] or any other Q2 equation, s is not energy but s = w = work 2- And M=EX2, where M is mass, E is energy and X2 is the conservation square. E is not energy in its absolute form, but E= h (plank constant). Thus M=hx2 or M=h/r .c -as we all know that theconstant speed of light c2 is the physical limit of energy conservation due to the visible spectrum, so it turns out that r = radius of the two opposite particle that constitute the same quantum unit (for example electron and positron or specifically matter and anti matter) and c is the speed of light. Univaso theory and dimensional analysis: Step 1, You remember very well that; we said that E= W.Q2,where E is energy, W is work and Q2 is the quantum frames of work conservation. And as W.Q2 = g.v2,this implies that E= g.v2; where g is gravitation and v2is the velocity square. Step 2, We all know that gravitation (g)= m.a where m is mass and a is acceleration. This implies that E =m*a* v2.and because a* v2 are physical derivatives of displacement due to material change, this implies that a* v2= displacement and thus E=M.d;where d is displacement. Step 3, We also know that Energy is measured in units of Joules. And that; 1 Joule = 1kilogram*meter2 / second2. Mass is measured in units of kilograms and displacement is measured in units of meter/second. Thus according to dimensional analysis, E=M.d is dimensionally true or sound or in particular; E= W.Q2 is dimensionally sound due to its equivalence to E=M.d when converted to units of E=Md.
Klaynos Posted February 6, 2013 Posted February 6, 2013 av^2 is not equal to the displacement. v^2 / 2a is equal to displacement in certain circumstances. I didn't look too closely at the rest of it but your initial equations are still not sound.
univeral theory Posted February 7, 2013 Author Posted February 7, 2013 av^2 is not equal to the displacement. v^2 / 2a is equal to displacement in certain circumstances. I didn't look too closely at the rest of it but your initial equations are still not sound. corrections E=MVf Where;E is energy, M is mass and Vf is velocity field Step 1, You remember very well that; we said that E= W.Q2, where E is energy, W is work and Q2 is the quantum frames of work conservation. And as W.Q2 = g.d, this implies that E= g.d; where g is gravitation and d is displacement. Step 2, We all know that gravitation (g)=m.a where m is mass and a is acceleration. This implies that E=m.a.d. and because a.d form the same group of velocity field(Maxwell field) in field theory, this implies that a.d= velocity field and thus E= MVf. Step 3, We also know that Energy is measured in units of Joules. And that; 1 Joule = 1kilogram*meter2 / second2. Mass is measured in units of kilograms and Vf is measured in tesla and particularly units of meter/second. Thus according to dimensional analysis, E= MVf is dimensionally true or sound or in particular; E= W.Q2 is dimensionally sound due to its equivalence to E=MVf when converted to units of E=MVf. Thanks.
Klaynos Posted February 7, 2013 Posted February 7, 2013 Energy = mass * velocity is still not dimensionally sound. I think this demonstrates quite well why derivation is required and not just pulling things out of the air.
swansont Posted February 7, 2013 Posted February 7, 2013 Since E and W are both in units of energy, Q^2 is a unitless quantity ad has units of v^2, not v There is no way to get from Tesla to meters/sec
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