Capiert Posted May 9, 2016 Posted May 9, 2016 (edited) Work abstract Coe & com seem unreliable, (algebraically) they can NOT be confirmed; & a (new) cow seems to be the culprit (for explaining the milky way's problems). Mon 2016 05 08 06:51 PS Wi 12.1 C clouds & mist clearing Conservation of energy coe is often taken for granted but some math (algebra) derivations from momentum to kinetic energy (via binomial squaring) have not allowed confirmation. Results were such that simple (energy) addition was impossible sometimes, due to explicid complexer solutions, indicating the more basic momentum mechanics might be prefered as fundamental. To my amazement, momentum (conservation com) could "not always" be confirmed either, in preferance for simple energy addition. Facit: momentum failed to add (correctly) sometimes, while energy succeded; & visa versa for different (peculiar) circumstances. Unprepared, that neither (energy nor momentum) were always entirely reliable (100% of the time, for all examples), speeds were analyzed & corrected, but even that brought no guarantee. It occured to me then, that a severe (subtle) fundamental error existed (or must exist), & that made neither (E nor mom) acceptable for a 100% conservation law (title). Analyzing the physics framework further on such trival (math) peculiarities, led me to conclude that the basic definitions (stated) (e.g. for work, power & force..?) were not being followed (strictly) mathematically & could leed to such a disaster (collapse of physic's, momentum & particularly energy, concerning the (unknown=) dark energy (problem, dilemma) vs the 25% real ratio when compared to quadratic binomial results.) Convinced (to some extent), (& that my time was better used dealing with finding a solution directly (first), because the problem details could be found again later, it was) that the solution required a better basis for mechanics than the (unreliable) com & coe; (unstatisfied, as unsatisfactory) a compromise was made for the cow.-Muhh! Here is that (new) framework (derivation) solution, & perhaps (if allowed) with insight as to why the other 2 failed. Here are NOT examples, of the failures mentioned above. Are you interested? Dedication: May the goods be invoked for memory purposes only (games if you will), & (your) deficit of greek culture (hints of vocabulary), among others. Otherwise it doesn't look like you'll have a clue. -SheerLuck Homes. (Unlock my heart('s core). See my ideas, not my words. I won't always say it right. What's left is also an alternative. & visa versa.) P.S. Maybe we can write this as science fiction, a least you won't be disappointed, then. Edited May 9, 2016 by Capiert
swansont Posted May 9, 2016 Posted May 9, 2016 Work abstract Coe & com seem unreliable, (algebraically) they can NOT be confirmed; & a (new) cow seems to be the culprit (for explaining the milky way's problems). Mon 2016 05 08 06:51 PS Wi 12.1 C clouds & mist clearing Conservation of energy coe is often taken for granted but some math (algebra) derivations from momentum to kinetic energy (via binomial squaring) have not allowed confirmation. Results were such that simple (energy) addition was impossible sometimes, due to explicid complexer solutions, indicating the more basic momentum mechanics might be prefered as fundamental. To my amazement, momentum (conservation com) could "not always" be confirmed either, in preferance for simple energy addition. Facit: momentum failed to add (correctly) sometimes, while energy succeded; & visa versa for different (peculiar) circumstances. First of all, shorthand notations tend only to confuse the issue, especially when they haven't been defined. Second of all: examples? You are making bald assertions here.
John Cuthber Posted May 9, 2016 Posted May 9, 2016 "Coe & com seem unreliable," to whom? "(algebraically) they can NOT be confirmed" Actually, they can https://en.wikipedia.org/wiki/Noether%27s_theorem You can stop now Capiert, at least until you can show why she was wrong.
studiot Posted May 9, 2016 Posted May 9, 2016 Conservation laws and system definition can often be tricky. swansont That's the whole concept behind solving physics problems — you can define your system in such a way that a problem can be solved. What conservation means for momentum is not necessarily suitable for energy. Consider the following system. You have a totally empty box. (Posh term = control volume) and a source of moving particles of momentum M such that the transit time of one particle through the box is 2 seconds. If t is time then So at t = -1 the total mometum in the box is zero. If at t = 0 a particle enters the box the total momentum in the box at t =1 is M. At t = 2 the total momentum in the box is again zero. So to consider conservation you need to take time into account. Another situation, considering space, this time (pun intended) What is the momentum of a stationary object? So how does conservation apply at a stagnation point in a flowing fluid?
Capiert Posted May 11, 2016 Author Posted May 11, 2016 (edited) Conservation Of What?Nice of you to ask COW=Conservation of work Abstract summary Basic thesis: there is a (very stupid, primative rudimenty) math error in the (standard) work formula (till now) indicating work is not energy (Work#Energy) when compared to the stated definition of work (in words). It's so fundamental that it's significance might be missed. The formula does not agree with the worded definition. That has serious consequences for energy calibration in physics, producing very peculiar problems. It is definitely a math error. The error was discovered with algebra, & should be obvious (implied). A solution (compromise) is proposed. It's a peculiar kind of momentum using only the initial_momentum & half of the momentum_difference. (A few tricks & tips are given to help memorize the formulas.) The correct formula should look something like this. Work=m*h/t, m=mass, h=height, t=time (That is very fundamental! The most important thing. Most logical=obvious.) Did you get it? Where's the problem? The standard work formula does not look so simple. Energy does NOT look like that, at all! That's the problem! It needs the following to bend it right: Work=W/v, W=WE=Work's_Energy (please observe it's possesive 's , to distinguish it from work), v=speed_difference Work=moma=m*va, va=average speed Work=mom0+mom/2, mom0=initial_momentum, mom=momentum_difference=impulse. Work's_Energy (is not work!, it's energy instead! There is a difference!) W=WE=F*d Speed_difference v=v1-v0 Initial speed v0 Final speed v1 Please see the trashcan for more details. It's been put there because of ignorance. (Would somebody please help me recover it, & iron out the bugs? If any? I'm new. Nobody else seems to have noticed it before nor taken this issue seriously what it's all about. The error is so basic, that it's difficult to believe. I do not know how to correct it better than that.) I consider this theme very important, (high priority), & suspect it will help solve the dark energy issue. It is a very fundamental error affecting energy calibrations. Power as well. If you can grasp the above then you've got most of it. Signed GENERAL I. N. Formation. Cheers! Edited May 11, 2016 by Capiert
John Cuthber Posted May 11, 2016 Posted May 11, 2016 The biggest bug is the one I pointed out earlier. It is possible to prove mathematically that energy and momentum are conserved quantities You seem to have ignored this As I said earlier, please get back to us when you have shown the errors in this https://en.wikipedia.org/wiki/Noether%27s_theorem Unless you can do that there is no point in you posting anything. The second bug is this "Work=m*h/t, m=mass, h=height, t=time (That is very fundamental! The most important thing. Most logical=obvious.)" which is plain wrong.
Sensei Posted May 11, 2016 Posted May 11, 2016 (edited) Work=m*h/t, m=mass, h=height, t=time (That is very fundamental! The most important thing. Most logical=obvious.) kilograms * meters / second is unit of momentum.. Velocity is change of position of object, divided by time needed to do so: [math]v=\frac{x1-x0}{t1-t0}[/math] with units [math]\frac{m}{s}[/math] Momentum is velocity multiplied by mass of object. [math]p=m*v[/math] You can substitute and receive: [math]p=m*\frac{x1-x0}{t1-t0}[/math] [math]p=\frac{m*x1-m*x0}{t1-t0}[/math] with units [math]kg\frac{m}{s}[/math] Edited May 11, 2016 by Sensei
Capiert Posted May 11, 2016 Author Posted May 11, 2016 (edited) Thanks for the comments. Very appreciated I'm going to look into it. But at first glance, I see that you are using calculus, as your basis, & I am sorry to have to tell you it's corrupt. Please read Miles Mathis's analysis also for the Lagrangian. Btw The FEM finite element method & calculus both give different answers. Industry has dumped calculus in favor of the FEM, & for good reasons, I will too in favour of algebra. Industry & top consulting companies do NOT recommend calculus if you want stability without the fear of collapse. For building anything large & complex such as bridges or skyscrapers, large rockets, aircraft carriers, spacestations, microwave antenna designs, ..whatever. A few are ok, sometimes. But the risk is too large when life is at stake, or large production & financing. Calculus produces very different answers & has been proven to been unreliable because of that. I'm not interested in rolling dice with Noether's 1915 theorems, (so_called) proof methods, either. That's not an acceptable basis. It's unreliable. Sorry. If I get a chance to pin point her errors exactly, I will, but that's a needle in a haystack, & I can't promise anything. It seems futile for anything in that calculus direction. No reliability. Thus a waste of time. The cause for that, is calculus is only an approximation tool. If you need an exact, accurate answer, all the time, then calculus will NOT do! Newton knew it had errors, & did NOT want to release (=publish) it for over 20 years. It wasn't until Liebnitz came up with something similar & published, that the students of both began to question who found the method first supposing Newton had been robbed of the ideas by the other, & visa versa. That false presumption (from the (loyal) students (with anger & resentment) from both sides) finally led to the (bitter) conflict between both professors. On their own both professors had no need to argue. Science is an independent, reproduction (or replication) of similar methods. An interesting article exists in scientific american's biography of Newton. Miles Mathis('s articles) can point out for you, some of the weaker points of calculus analysis. Perhaps with some improvement proposals. Cheers! The biggest bug is the one I pointed out earlier. It is possible to prove mathematically that energy and momentum are conserved quantities (PS: ..if the method of proof is reliable, but since that is not the case we must reject that method.) You seem to have ignored this (PS: ..for good reasons.) As I said earlier, please get back to us when you have shown the errors in this https://en.wikipedia.org/wiki/Noether's_theorem (PS: ..that seems to be becoming embarassing.) Unless you can do that there is no point in you posting anything. (PS: But perhaps that is sufficient, for now?) The second bug is this "Work=m*h/t, m=mass, h=height, t=time (That is very fundamental! The most important thing. Most logical=obvious.)" which is plain wrong. Please explain. Work is truely not an energy, it is a type of momentum, as the definition states. Please take note carefully. work=m*h/t is the mass m moved to a distance d(=h height), within (=per) a (specific) time t [in seconds]. That is the definition of work (done). No other is allowed. (All others lie, deceiving their accomplishments). That is the calculation, Tesla used, for the Niagara falls, to get momentum out (as moving electrons, called current I), electricity; & the same (electrical) momentum (electricity) was used to pump (the same amount of) water (mass) back up the height, in (=per) time, to (test &) calculate the efficiency. I.e. How true(thful)=accurate the calculations were (to reality). I'll assume you did not know all that? Are there any other supposed bugs? Edited May 11, 2016 by Capiert -1
Sensei Posted May 11, 2016 Posted May 11, 2016 (edited) There is such experiment: scientists dropped one by one objects (or shoot and measured velocity of bullet with hi-speed camera) with mass m, to water below it, h distance, and measured increase of temperature of water (including normal decline of temperature). If you drop something from height h, potential energy is: E=m*g*h And water has specific heat ~4.1855 J/K*g (old definition of calorie). It means if you add 4.1855 Joule of energy, one gram of water will have temperature increased by 1 K (1 C). When there is no lost of energy by water: [math]dT=\frac{m_{object}*g*h}{4.1855*m_{water}}[/math] is increase of temperature of water. You cannot do this calculation with ordinary momentum of object p=mv, as target is not solid, but liquid, that gets splashed after hitting. It's "impossible" to read the all water drops momentum. Thermometer is appropriate tool for measuring change in substance. Similarly, definition of watt and energy has been done in electricity: P=I*U (power in Watts=J/s) but I*t=Q (charge in Coulombs) so P*t=I*t*U so E=Q*U Q is quantized every e=1.602176565*10^-19 C Q/e is quantity of electrons. If you put resistor with resistance R to water (which is typically called "heating element"), and there will be flowing through it current I, I=U/R, P=U^2/R=I^2*R, so there will be Q=I*1s charges per second. You will increase temperature of water, to some higher temperature. Pass current (read on ammeter), with voltage U (read on voltmeter), for time t (stopwatch), read temperature of water prior experiment and after experiment (thermometer). The higher voltage, the higher current, the more energy released by resistor, and the higher temperature increase. Above equations are just result of experiments. Experiments were earlier, equations further, to describe experiments. And being able to predict outcome the next time. If you know how much energy is flowing through heating element, you can predict time in which water will have right temperature, even without thermometer. The cause for that, is calculus is only an approximation tool. Calculus produces very different answers & has been proven to been unreliable because of that. Calculus is used all the time while calculating area and volume.. Edited May 11, 2016 by Sensei
John Cuthber Posted May 11, 2016 Posted May 11, 2016 (1) I see that you are using calculus, as your basis, & I am sorry to have to tell you it's corrupt. (2) Industry & top consulting companies do NOT recommend calculus (3) Calculus produces very different answers & has been proven to been unreliable because of that. (4) with Noether's 1915 theorems, (so_called) proof methods, either. That's not an acceptable basis. It's unreliable. Sorry. (5) If you need an exact, accurate answer, all the time, then calculus will NOT do! (6) Miles Mathis('s articles) can point out for you, (7) Work is truely not an energy, it is a type of momentum, as the definition states. (8) work=m*h/t is the mass m moved to a distance d(=h height), within (=per) a (specific) time t [in seconds]. (9) That is the definition of work (done). No other is allowed. (All others lie, deceiving their accomplishments). 1 Show some sort of evidence 2 Show me any statement by any competent engineer or scientist from any big company to back that up 3 No it doesn't. if you disagree- show us where you get two (or more answers from valid calculus. 4 It is, in fact, widely (in fact, practically universally) accepted. If you think it isn't reliable you need to show where it gets things wrong. 5 Yes it does- if you disagree show where it fails 6 He's a crank as shown here http://www.thunderbolts.info/forum/phpBB3/viewtopic.php?f=10&t=15094&start=225&sid=5229d20b2c0a74dc0ef38fa6a2cc7ade and on plenty of other sites. 7 It truly is a form of energy and not momentum- by definition 8 No it is not- apart from anything else, your idea has no meaning in zero gravity where you can't define "up" so you can't define height. 9 No. It's not the definition its some rubbish you made up because you don't really understand the underlying physics. If you claim the others "lie" then you need to show where they get it wrong. 1
swansont Posted May 11, 2016 Posted May 11, 2016 ! Moderator Note unless you start complying with the guidelines for speculations discussions, this is going to be closed. You need more rigor than claiming that accepted science is a lie. You need to back up your claims with evidence.
Capiert Posted May 28, 2016 Author Posted May 28, 2016 (edited) Please Wait! Under construction The simple (approximate) answer for the temperature change is in Energy: (yours) [math]dT=\frac{m_{object}*g*h}{4.1855*m_{water}}[/math] in momentum: (new) [math]dT=\frac{m_{object}\sqrt{2*h*g+v^2_0}}{m_{water}\sqrt{2*4.1855 J}}[/math] m1=drops mass "object" (m2=bath mass before, without drops) mt=m1+m2 total mass (of bath & drops) "water" Let g=Pi^2 [m/(s^2] v0=initial speed All values are positive for simplicity. Calibration: height h~420m is for 4.1855 [J/(g*C)] energy; & is also for 0.091 [N*s/(g*C)] momentum that's in Newton_seconds, per gram, per degree Celsius. I must admit that momentum looks small compared to the energy. The electrical momentum is a little tricky because of the (wrong) syntax (used everywhere). It's very confusing. A spring_loaded voltmeter, displays Hooke's law of "Force" (Fs=-x*ks, -x=negative displacement distance, ks=spring constant,) when it displays anything "linearly". (The spacing between the numbers is the same size or distance.) (Someone did not do their homework, & billions followed.) That means both voltage, & current scales, are displaying as "force" because they are "linear scales". (U=) V=Fv, I=Fi. (My use of V instead of U as symbol for voltage. U=V.) So their product (P=I*V=F^2) is really "force squared". Thus we must "root electrical power" P=I*V to (get some sort of mean or) "average force" alone, to use with "time t", to get momentum mom=F*t. Electrical power is really a power_squared but nobody knows that. It's correct symbol would be P(^2)=Pi*Pv=Fi*Fv. Nobody writes the squared symbol, because they don't know the problems. Current readings use the maximum force Fi=max, & voltage reading use the minimum force Fv=min. So somewhere in the middle of those 2 is what we use for the work_momentum force when we root standard electrical power P(^2)=I*V as (P(^2)=I*V)^0.5=Fiv. I hope you understand that. I'm not (really) allowed to let that info out. Too many people will object, because they do not know it (yet). It only confuses them. That was the answer to your problems. --- The following is extra notes, methods etc. Thank you for your beautiful: formula (wonderfully simple & transparent), presentation (thorough & brief, informative) & (friendly) appeal (for help,) (But "we" "can't"..). It's a very inspiring style for me. (Challenge, to master, because I notice a need to be fullfilled, that we do (=did) not have yet.) It's a good basis to derive everything (else) from. (I've (also) noticed: It looks like the (strange) old energy (4.1855 J/(C*g) calibration calculation) has ~-1% measurement difference (perhaps not error?), due to centrafugal effects, from the earth's spin (=rotation), decreasing gravity_acceleration g -~1% (=-0.96%). Such (added correction) values (are different) depend(ing) on latitude, from pole(s) to equator. Maybe the new standard, is (a comprimise, a ruff approximation, calculated simply) without earth's spin. That (new simple standard, unfortunately) then makes each city's lab location (e.g. lattitude, & height) look like they have different amounts (of extra measurement_)error seen as ~-1% less accuracy. Acceptable, not important. If the lab uses their own local g acceleration (calculation including, their centrafugal acceleration, too) then that error of no centrafugal acceleration will not happen. That (~1%) error can be eliminated.) PS: I've reworded your text (in my words, longer) so I can understand it better. Yours was fine (brief). If I have misunderstood anywhere, please tell me. There is an experiment: where scientists dropped objects one by one (e.g. they shot bullets (each with a mass m), into the water below a distance h and measured: their speed with a hi-speed camera; & the water's temperature changes (increase, decreased). PS: I assume, the bullet's energy is lost as friction (heat). The measured speed, was it the initial or final speed? When the bullet is shot, then the explosion (from the gun powder) also heats the bullet. The air also heats the bullet as friction, but also cools that bullet's heat like a fan. That's a very complicated experiment. How do you separate all those things? It's much easier, just to drop things, but if they fall thru air, & they are water drops, then we know, that when you blow air at anything wet (e.g. a wet person out from the swimming pool) that they cool. That experiment, is also almost impossible, to separate into the different effects. E.g. How big is the water drop that falls? How much water did it loose, as evaporation, into the air, while falling? How much did the water drop cool because of evaporation? It's also a room temperature & humidity saturation (effects) problem. Maybe we could measure the total mass later. ok=(I am now convinced the experiment is reasonable, after pondering through the (difficulties) impossible problems. It now seems possible for a solution. I doubted before & I recognize the start (PEKE) to (target's) goal (mom(entum)); & its method (mass). Tip: "Speed" represents the temperatures' energy &/or momentum.) If you drop something from height h, the potential energy is: PE=m*g*h (mom=m*v, impulse). Water has a specific heat of ~4.1855 J/K*g (old definition of calorie). It means if you add +4.1855 Joules (of energy) to 1 gram of water, then it's temperature will increase +1 C. When there is no loss of energy by the water: dT=T2-T1 dT=PE(object)/(4.1855J)*(m(water)). then the object's_energy vs the water's energy is (somehow) proportional to the temperature change. What I see (=understand, intuitively) depends on the mass. But it's an interesting problem. Very interesting. Really cool! (=neat!) [math]dT=\frac{m_{object}*g*h}{4.1855*m_{water}}[/math] [math]dT=\frac{m_{object}\sqrt{2*h*g+v^2_0}}{m_{water}\sqrt{2*4.1855 J}}[/math] is increase of temperature of water. You cannot do that calculation with ordinary momentum of the object p=mv, because the target is not solid. As a liquid, water splashes after being hit. It's "impossible" to read all the water drops' momentum. The thermometer is an appropriate tool for measuring changes in a substance. Similarly, the definition of watt and energy has been done in electricity: P=I*U (is power, in Watts=J/s) but I*t=Q (is charge in Coulombs) so P*t=I*t*U so Q*U=E. Q is quantized every e=1.602176565*10^-19 C. Q/e=N is a quantity (=number) of electrons. ? If you put a resistor with resistance R into the water (which is typically called "heating element"), and there will be flowing through it, then the current I=U/R, power P=U^2/R=I^2*R, so there will be Q=I*1s charge per second or Q/e=N (number of) "charges" per second. (I think you mean: Q "charge" (infinitive), vs Q/e (=number of) "charges" (plural), either per second.) You will increase the water's_temperature, to some higher temperature. Pass current (read on ammeter), with voltage U (read on voltmeter), for time t (stopwatch), read temperature of water prior to experiment and after experiment (thermometer). The higher the voltage is, the higher the current is, & the more energy is released by the resistor, & the higher the temperature increase is. The above equations are just results from experiments. Experiments were done first, & equations came later, to describe the experiments, & being able to predict the outcome for the next time. If you know how much energy is flowing through the heating element, you can predict the time in which the water will have the right temperature, even without a thermometer. Really? Cool! Edited May 29, 2016 by Capiert
Capiert Posted May 28, 2016 Author Posted May 28, 2016 Yes. Your lack of knowledge is incredible. Yes I agree. Is there anything wrong with the formulas, there?
Strange Posted May 28, 2016 Posted May 28, 2016 Is there anything wrong with the formulas, there? Almost everything you say is incorrect. The rest is incomprehensible. I can't be bothered to try and read through it all to see if, by pure chance, you have got one small detail right.
Capiert Posted May 29, 2016 Author Posted May 29, 2016 (edited) Almost everything you say is in ..Now "your" lack of knowledge, amases me! At least we have something in common. We're human. I don't think any of you have been capable of that calculation til now. Otherwise you would have done it. It was thought out to get the exact answer. The more exact answer is a little more complex but does not change anything significantly. Can you understand that, (at least). Sorry. Edited May 29, 2016 by Capiert
swansont Posted May 29, 2016 Posted May 29, 2016 What all does this have to do with conservation of momentum?
studiot Posted May 29, 2016 Posted May 29, 2016 Capiert I don't think any of you have been capable of that calculation til now. Otherwise you would have done it. It was thought out to get the exact answer. First you fail to reply to my polite post#3 Now you are insulting me by including me in some claimed universal failing, (which I'm sure others don't accept either).
Capiert Posted May 29, 2016 Author Posted May 29, 2016 (edited) Please Wait! Under construction. "Coe & com seem unreliable," to whom? Hi John. To me, because I've still got to find & dig up those examples where they (mom & Energy) swap unreliability. Hackers were on my old hard disk, later. Not accessible yet. Maybe I can remember, or else stubble on new 1's (ones) later. If I remember correctly, the problem(s) started with a few strange matrix calculations, using 2x2 mixed polarities. It required adding in a specific sequence to get the correct (desired) answer, using minus_squared (something Napier used, to maintain rotatation (angle) integrity for reverseability, my explaination. "-"=180 degrees, (-)^0.5=90 degrees, multiply polarity to add angles. E.g. Binomial squaring (3+4=5)^2 vs (4-3=1)^2 vs varying polaries, for quadratic (=4 terms) results. I think I remember which binder my notes are in. It has to do with (negative) polarity recognition & restoration: multiplying, rooting & restoration. Used for momentum collision calculations, e.g. 2 different gas moleclues, or things like 2 cars, etc, It uses an elastic non_elastic equivalence. A+B=C. It observes KE, momentum, & their conservation. Excel table. The numbers don't lie. Conservation works, sometimes. Rarely failing. What I tend to conclude is a quirk of polarity. (angle) Getting the polarity right, depended on majority values, for which of the 2 original numbers got the right polarity. The largest value dominated. Physicists generally right it off as statistics, but there seems to be a pattern to it. It's something like, have space_time as a pot of stew. Put your meat & vegetable in A & B. a vector & a scalar (up & face down). Mix it (e.g. multiply) & then try to get them out again right (per machine, Excel table, rooting) Rooting strips polarity, thus must manually reapply a synthetic (artificial) polarity. How can we apply the correct polarity from the exising (elastic) info? In other words what do we need to know, to correctly store polarity before (loosing) for the polarity so that when we do the multiply then root operation, we can find out which value should receive the negative polarity? The goal was a 100% reliable calculation. No guesswork. E.g. Murphy's law, (what could happen did happen) a fluke (that) failed (& mathematically broke the conservation law calculation). Why then, & not before. Say it differently. Is there a conservation of energy? Answer: Everybody say there is. Can we (model &) track it with excel? Excel isn't the only math program, but the task seems simple enough (to do) we're only looking at 2 masses (max 3..6, as before, after & composites=added a+b=c). Can we guarantee it. Answer: No. Does it depend on Excel? Answer: to see the problem, depends on excel's limitations (with polarity). Can we set up excel to take care of the problem correctly? Answer: we'd also need that syntax manually. If standard math isn't covering that polarity reversibility problem ("squaring rooting", "polarity ambiguity" "math hash" against conservation.), how can it be expected that a machine (software calculation) get it write? I think by now you might be getting my drift. I still need time to prepare some examples. Such basic errors have further reaching affects. They are not transparent. KE uses speed squared & so it is very difficult to track that energy transfered due to the ambiguity of squaring. Transfer implied conservation law. The table also shows energy anihilation. (Mathematically). Why? due to it's accuracy as far as that is concerned. The major arguement would be, energy is not a vector although speed & momentum are. Mass is not a vector. It's interesting however, that momentum & energy are 2 different ways for decribing speed of something (=mass). 2 different ways, for the same thing, speed. They can also tell us about the mass. However, both momentum & mass are completely different. They are only math methods. Tools. Our weakness in this problem is the math. (Perhaps syntax?) --- It's an overseeable problem, reduced to a minimum (no distractions). Allow me to reconstruct it. Mixed Polarity determinator: two numbers exist with absolute values1 & 4, one is negative. Which is negative? If you have a matrix (with two diagonal (product) pairs of positive (down to right) vs negative (down to left)) with 1 & 16 (raw data) -4 & -4, which (original) number (before multiplying) has the correct positive polarity (1 or 4)? (In other words, the other number is negative, which one?) Both give the value 9, whether +9, or --9 (Napier's syntax, minus squared). It's ambiguous (-1+4)^2=3^2=9 (1-4)^2=(-3)^2=--9 The standard practice, is to give up as impossible (to knack). Only absolute accuracry requires it. Thus they (everybody) say forget it. Programmed without, is unreliability. Originally discovered in energy vs momentum table calculations. Time & intent for an absolute are already invested. Intended as a universal tool for energy & momentum exporation (research). (Anything else is half_assed, concerning programing, life or death calculations. General usage.) Reliability is the highest order. Done right is finished. (We are Finish but this is not the end.-Happy Kerkeling, pun.) Everything else is a comprimise. To be(come) a needed reference, for (diagnotic) analysis (of energy). Considered possible to achieve, because so simple. Basic. Those assumptions are now doubtful. Due to the (mixed) polarity issue (theme). So without further delay. Recommended Operation (method) sequence*: ROMS. Using the squared binomial expansion FOIL=First (term) outside, inside, last (terms' pair), to build the (2X2) matrix, above (half) & below half. (x,y)=coordinates I've observed, the smaller purebred (either x,y matrix_coordinate) (1,1) or (4,4) (positions of the 4x4 matrix) is suppose to be cancelled first (as insignificant) by the negatives (x,y matrix_coordinates) either (1,2) &/or (2,1). *The larger purebed (x,y matrix_coordinate) (4,4) or (1,1) has priority, (& dominates), so the smallest (purebred) (1,1) or (4,4) should be operated on first (to be gone as trivial) by the negative (hybrids) (1,2 &/or 2,1). (Ref Mon 2006 03 13 12:19 PS) That's basically the jist of it. That's my problem. The dilema (=Die Laemer, German Lambs. Bah!) Please give me some time. "(algebraically) they can NOT be confirmed" Actually, they can https://en.wikipedia.org/wiki/Noether's_theorem You can stop now Capiert, at least until you can show why she was wrong. I don't want to be rude, but she is not my theme, nor method. That would be a sidetrack, off theme.If most of the witnesses say yes, I don't learn anything new. Only that strange & rare no, sets the gears working. Cheers.-SLH Edited May 29, 2016 by Capiert
John Cuthber Posted May 29, 2016 Posted May 29, 2016 I don't want to be rude, but she is not my theme, nor method. That would be a sidetrack, off theme. If most of the witnesses say yes, I don't learn anything new. Only that strange & rare no sets the gears working. Cheers.-SLH The whole basis of the thread- it's title- refers (unclearly) to the conservation laws. How can you say that the mathematical proof of those laws is not your theme? The point is that the conservation laws can be proven mathematically and that's the opposite of what you say. You are wrong. Explaining that is not a "sidetrack", it's a requirement of the forum rules.
Capiert Posted May 29, 2016 Author Posted May 29, 2016 Conservation laws and system definition can often be tricky. Sounds careful? Good! What conservation means for momentum is not necessarily suitable for energy.That gets me wondering, sceptical? maybe? Why? Consider the following system. You have a totally empty box. (Posh term = control volume) and a source of moving particles of momentum M such that the transit time of one particle through the box is 2 seconds. If t is time then So at t = -1 the total mometum in the box is zero. If at t = 0 a particle enters the box the total momentum in the box at t =1 is M. At t = 2 the total momentum in the box is again zero. So to consider conservation you need to take time into account. Good. Clear enough. Another situation, considering space, this time (pun intended)Good! What is the momentum of a stationary object?My instinct would say, temperature (ruffly, no calibration, nor etc). So how does conservation apply at a stagnation point in a flowing fluid?The stagnation point has lost a significant part of its momentum, to other parts of the fluid. ? Or it's lost that speed into rotation, vortexes, whirlpools. The question is naturally, to the slow down (loss). The whirls stop. Where did the speed go? Especially if it is not (gone) into heat (random) motion. I say, 2 colliding particles that stop (not bounce), becoming 1 total (mass, like bread doe) have cancelled their opposite speeds. Result, (& I know you're not going to like it) momentum cancelled (=destroyed); & if that can happen(?), then it can happen to another law (guess which one?) then energy can also cancel (e.g. like 2 waves, with same frequency, that meet out of phase (180 degrees, 1 inverted) from opposite directions). Thus energy is destroyable. & momentum is also destroyable. But since so much of both exists, then I must conclude they are both creatable, too. But how? As opposite pairs, simply by the reverse annalogy. The conservation law holds for the pairs (only) not the exclusive thing (e.g. wave). How is that done? Perhaps via 90 degree intervention. We need energy, to make energy (at 90 degrees). (?) Repulsion. A water wave can "push up", 90 degrees to its "horizontal motion", e.g. surfing. EMF electromagnetic force is at 90 degrees. But like charges repel too, as in radioactive decay. 180 degrees, into KE near light speed straight line motion beta particles, that bang around into collisions, ending random (zig zag) as then after, heat (motion=speed+acceleration, in varrying values).
studiot Posted May 29, 2016 Posted May 29, 2016 (edited) Congratulations, you seem to have applied you mind to the substance of what I was saying about conservation. Some points arising Quote Studiot What is the momentum of a stationary object? My instinct would say, temperature (ruffly, no calibration, nor etc). No, the momentum is zero by definition of the word stationary. As an aside, What is the temperature of a single isolated particle? Quote studiot So how does conservation apply at a stagnation point in a flowing fluid? The stagnation point has lost a significant part of its momentum,to other parts of the fluid. ?Or it's lost that speed into rotation, vortexes, whirlpools.The question is naturally, to the slow down (loss).The whirls stop. Where did the speed go?Especially if it is not (gone) into heat (random) motion.I say, 2 colliding particles that stop (not bounce),becoming 1 total (mass, like bread doe) have cancelled theiropposite speeds.Result, (& I know you're not going to like it)momentum cancelled (=destroyed); Why wouldn't I like it? That is exactly what happens. Play a horizontal fire hose jet onto a vertical wall. There is a stagnation point at impact where the fluid horizontal momentum is destroyed. However the energy is not destroyed since the fluid still possesses the potential energy due to pressure. This pressure is known as the stagnation pressure. You should note in this that energy and momentum are different types physical entities. Energy is a scalar and thus has no direction. Momentum is a vector and thus has direction. Whilst horizontal momentum is destroyed, vertical momentum is created. The condition for the destruction/ creation of momentum is the application of an external force. If you like it is diagnostic of such application. Edited May 29, 2016 by studiot
Capiert Posted May 29, 2016 Author Posted May 29, 2016 (edited) Please Wait! I always have problem programming quotes.I do not know why this sytem is not stable! Help!I've pressed multiquote & quote. "Coe & com seem unreliable," to whom? "(algebraically) they can NOT be confirmed"Actually, they canhttps://en.wikipedia.org/wiki/Noether%27s_theorem You can stop now Capiert, at least until you can show why she was wrong. Conservation laws and system definition can often be tricky. What conservation means for momentum is not necessarily suitable for energy. Consider the following system. You have a totally empty box. (Posh term = control volume)and a source of moving particles of momentum M such that the transit time of one particle through the box is2 seconds.If t is time thenSo at t = -1 the total mometum in the box is zero.If at t = 0 a particle enters the box the total momentum in the box at t =1 is M.At t = 2 the total momentum in the box is again zero. So to consider conservation you need to take time into account. Another situation, considering space, this time (pun intended)What is the momentum of a stationary object?So how does conservation apply at a stagnation point in a flowing fluid? The whole basis of the thread- it's title- refers (unclearly) to the conservation laws.How can you say that the mathematical proof of those laws is not your theme? The point is that the conservation laws can be proven mathematically and that's the opposite of what you say.You are wrong.Explaining that is not a "sidetrack", it's a requirement of the forum rules. John, that is your link, not mine. Please explain it (bit of math if needed), to me. I not responsible for you info. You have to explain it yourself, not me. I have not seen her proofs. (Perhaps would like to eventually.)Perhaps address your questions as y/n. Maybe I can answer they. I don't have to fall into every cobweb that falls my way. That wouldn't be productive.If I understand you correctly, you've prosed the question. Hey! Conservation laws have been proven mathematically with calculus (see link). Why aren't they ( those conservation laws working) always working? What's wrong? Where is the mistake. Obviously not in the the calculus.Here please see where you went wrong.According to us she did it right. Please can you help us see the problem better?If you're right what did she do wrong? Rigor weakness with constant C? Please tell us.If not I get that jist.If I were your foot doctor & said you don't look well today, you look a little dizzy & lobesided.Yes doctor fix me. Would you expect me responsible & trained enough to remove your (deep) brain cancer, for you? No you would do that yourself with enough clues. Your calculus is not (really) my job, sorry. It's not my area of competence, it's yours. Thus you are resposible for it (your area of competance) not me.The whole basis of the thread- it's title- refers (unclearly) to the conservation laws.How can you say that the mathematical proof of those laws is not your theme?Hi John. I mean it's not the weakness. My problem lies on a different basis, the reliability of the math rules, not their application.Your're dealing with using the tools. There appears to me to be some quirks with algebra (tool itself). ? </p></blockquote>The point is that the conservation laws can be proven mathematically and that's the opposite of what you say.<blockquote class='ipsBlockquote' >QuoteQuoteQuote<p>Fine. That's the level you're working at. You're not going to find an error at you're level.It does not address the problem. I hope you can see that.You are wrong.Explaining that is not a "sidetrack", it's a requirement of the forum rules.John, if you pull that 1 forward, we're not going to get anywhere.I think you are more interested in seeing what sort of calculations go screwy.Let's try to stay reasonable.I don't need a lot of calculus that tells me nothing right now.I'm looking for something that shows me differences, not just can show me differences.Any person can eat a meal, but which 1 (single 1) ate your dinner?We can deal with her proof with calculus, when we've "found" an error to work on, to backtrack & see why the proof goofed(?), as comparison.E.g. What does calculus address, what are it's limitations.At present, it's only a needle in the haystack, guessing at what we don't know.Please wait & be patient.If you need a formal replay that I have failed with your reply then hear it is (for now).I can't deal with it (yet). It doesn't help me solve the problem, including this 1. Yours.I've asked you for time, & I still have to prepare the FEM comparison for you too.I'm not trying to ignore you.But let's turn the table. Can you please explain to me what she did that I understand?Most of that link just goes thru my ears (right now).I can't even begin with it. Edited May 29, 2016 by Capiert
Strange Posted May 29, 2016 Posted May 29, 2016 My problem lies on a different basis, the reliability of the math rules, not their application. There is no problem there. You certainly haven't shown that there is a problem. So we can dismiss that part of your argument as being unsupported. Can you please explain to me what she did that I understand?Most of that link just goes thru my ears (right now).I can't even begin with it. If you don't understand it, how can you claim it is wrong? This is just another unsupported claim that we can ignore.
Recommended Posts