Capiert Posted July 14, 2018 Posted July 14, 2018 (edited) The electromagnetic force of a D'Arsenval (spring_loaded) meter Fiv=P^0.5 is proportional to the rooted electrical_power Piv=I*V (small postscripts used) for current I voltage V, resistance R. Isn't that in conflict to James Watt's definition of mechanical_power (caution: small postscript m used) Pm=F*va (not to be confused, with the m) for (Newton's) force F=m*a mass m acceleration a, average_speed (velocity) va=(vi+vf)/2 initial_speed (velocity) vi final_speed (velocity) vf. (I mean:) Both forces Fiv=F produce mechanical displacement that can be equated. --- (Sorry for the sloppy syntax, but yours confuses me too.) Re_done that would be: The electromagnetic force of a D'Arsenval (spring_loaded) meter F_I*V=sqrt(P_I*V) is proportional to the rooted electrical_power P_I*V=I*V (small postscripts used) for current I voltage V, resistance R. Isn't that in conflict to James Watt's definition of mechanical_power (caution: small postscript m used) P_m=F*v_a (not to be confused) for (Newton's) force F=m*a mass m acceleration a, average_velocity v_a=(v_i+v_f)/2 initial_velocity v_i final_velocity v_f. (I mean:) Both forces F_I*V=F produce mechanical displacement that can be equated. Edited July 14, 2018 by Capiert
John Cuthber Posted July 14, 2018 Posted July 14, 2018 Once the needle has swung into place, all the power delivered to the meter is dissipated as heat so, if you only look at mechanical energy, it doesn't seem to be conserved.
Capiert Posted July 14, 2018 Author Posted July 14, 2018 (edited) 22 minutes ago, John Cuthber said: Once the needle has swung into place, all the power delivered to the meter is dissipated as heat so, if you only look at mechanical energy, it doesn't seem to be conserved. Isn't it (=the meter) dissipating heat whether it (=the needle) swings or not, while electricity flows? I.e. It (=the needle's deflection) "costs" (continuous) power (=E/t, energy E, per time t), not just a once_only energy to keep that needle in place (beyond zero). Isn't there something wrong with James Watt's (mechanical power) definition? E.g. It's not universal. It does NOT include the static (equilibrium) case. Edited July 14, 2018 by Capiert
John Cuthber Posted July 14, 2018 Posted July 14, 2018 A shelf does not need power to hold up a book. In principle, it's possible to make a meter entirely from superconductors so the needle would stay in place without the supply of any power.
swansont Posted July 14, 2018 Posted July 14, 2018 5 hours ago, Capiert said: The electromagnetic force of a D'Arsenval (spring_loaded) meter Fiv=P^0.5 is proportional to the rooted electrical_power Piv=I*V You have written these as an equality, but say it's a proportionality. (The units are wrong for it being an equality) Quote Isn't that in conflict to James Watt's definition of mechanical_power (caution: small postscript m used) Pm=F*va Why should mechanical power and electromagnetic power have the same equation? (plus, one is a proportionality) 5 hours ago, Capiert said: Isn't it (=the meter) dissipating heat whether it (=the needle) swings or not, while electricity flows? I.e. It (=the needle's deflection) "costs" (continuous) power (=E/t, energy E, per time t), not just a once_only energy to keep that needle in place (beyond zero). Isn't there something wrong with James Watt's (mechanical power) definition? E.g. It's not universal. It does NOT include the static (equilibrium) case. If the needle isn't moving, no mechanical work is being done.
Capiert Posted August 15, 2018 Author Posted August 15, 2018 (edited) On 2018 07 14 at 5:46 PM, John Cuthber said: A shelf does not need power to hold up a book. Atomic repulsion (power?) is what keeps the atoms away from each other. But you'( a)re right, the bookshelf doesn't have an electrical bill (after it was made; & transported into place. Those costs (seem to) fall away). Quote In principle, =(in) theory (=where'( i)s the practice?) Quote it's possible to make a meter Are you sure you are talking about a D'Arsenval meter here? I was not aware he had superconductors in his time. Quote entirely from superconductors so the needle would stay in place without the supply of any power. Please show me 1, (that) you mean. But I'd hate to pay for that power bill (for making those superconductors (& that they) work =stay cold & magnetized). Edited August 15, 2018 by Capiert -1
Strange Posted August 15, 2018 Posted August 15, 2018 (edited) 7 minutes ago, Capiert said: =(in) theory (=where'( i)s the practice?) I *ca&n't! U n d e ... rst (an)yth(i)n[g] u w-r-i-t-e Edited August 15, 2018 by Strange
Capiert Posted August 15, 2018 Author Posted August 15, 2018 On 2018 07 14 at 8:06 PM, swansont said: You have written these as an equality, but say it's a proportionality. An (exact) equality (=equation) can state a proportionality, but a ruff idea proportionality cannot always state an equality. On 2018 07 14 at 8:06 PM, swansont said: (The units are wrong for it being an equality). That'( i)s my (1, major) complaint. The units totally conflict, although the concepts (excluding the units) are proportional. Something is very wrong (with the definitions)! On 2018 07 14 at 8:06 PM, swansont said: Why should mechanical power and electromagnetic power have the same equation? Simply because we are dealing with displacive force. That is the link, the basis, for equating. That can be translated into pressure (multiplied by area). Let me ask it so: wouldn't it be better (=an advantage, as less bother, simplification) if electromechanical power had the same (=only 1) equation? On 2018 07 14 at 8:06 PM, swansont said: (plus, one is a proportionality) Stating 1 as a proportionality was the only way I could state the problem (without getting into too much difficulty, & conflicts). On 2018 07 14 at 8:06 PM, swansont said: If the needle isn't moving, no mechanical work is being done. That's Watt's definition. I challenge it, because it (still) costs. 12 minutes ago, Strange said: I *ca&n't! U n d e ... rst (an)yth(i)n[g] u w-r-i-t-e Thanks for your emogies.
Strange Posted August 15, 2018 Posted August 15, 2018 4 minutes ago, Capiert said: That'( i)s my (1, major) complaint. Is there something wrong with the keyboard on your computer? 4 minutes ago, Capiert said: Thanks for your emogies. I don't know what "emogies" are, but I don't think there were any. Just a pisstake of your idiotic writing style.
Capiert Posted August 15, 2018 Author Posted August 15, 2018 1 minute ago, Strange said: Is there something wrong with the keyboard on your computer? Yes.
Strange Posted August 15, 2018 Posted August 15, 2018 Just now, Capiert said: Yes. Well, I'm relieved it isn't something more serious. Why don't you replace it.
Capiert Posted August 15, 2018 Author Posted August 15, 2018 (edited) 1 hour ago, Strange said: Well, I'm relieved it isn't something more serious. Why don't you replace it. I'm not rich like you. It's in a laptop that overheats while in internet (& hangs in slow motion); & I don't know how to get the hackers out, either. I'm attached to it. 1 hour ago, Capiert said: Quote If the needle isn't moving, no mechanical work is being done. That's Watt's definition. I challenge it, because it (still) costs. Watt defined with the speed horses could pull weights. But if the weights were too heavy, they wouldn't budge (=move, a bit); & the horses would get exhausted, wasting their (bio_chemical) energy, on getting "NO WORK" done. Facit: Watt's definition does NOT give us the whole picture. & your cosmologists are still trying to make sense of energy, because they report their stupidity not knowing what "dark energy" is (as much as 95%?, in the universe). (A few years back it was only ~75%?) I'm trying to tell you the culprit is energy's definition. ENERGY DOES NOT FIT with momentum. You can NOT serve both conservation "laws?" at the same time. Somewhere along the line you are going to get hash=chaos=trash=garbage non_sense (if you do)! Edited August 15, 2018 by Capiert
Strange Posted August 15, 2018 Posted August 15, 2018 (edited) 55 minutes ago, Capiert said: I'm trying to tell you the culprit is energy's definition. As always, the problem is with your lack of knowledge of physics. 55 minutes ago, Capiert said: Watt defined with the speed horses could pull weights. Nonsense. Quote Watt (unit of power). — The watt is the power which in one second gives rise to energy of 1 joule. https://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf#page=52 55 minutes ago, Capiert said: Facit: Watt's definition does NOT give us the whole picture. This is not a "facit". 55 minutes ago, Capiert said: yourcosmologistsarestilltryingto make sense of energy, because theyreport their stupidity not knowingwhat "dark energy" is Ignorant nonsense. 55 minutes ago, Capiert said: (as much as 95%?, in the universe). (A few years back it was only ~75%?) Nonsense. " It turns out that roughly 68% of the universe is dark energy." (https://science.nasa.gov/astrophysics/focus-areas/what-is-dark-energy) Quote ENERGY DOES NOT FIT with momentum. Nonsense. Quote You can NOT serve both conservation "laws?" at the same time. Nonsense. You are saying that physical systems will behave differently with changes in time and space (Noether's Theorem). This is not what we observe. We can also measure conservation of energy and momentum directly. So all the evidence shows that you are wrong. 55 minutes ago, Capiert said: Somewhere along the line you are going to get hash=chaos=trash=garbage non_sense You are the only one getting that. Edited August 15, 2018 by Strange
Capiert Posted August 16, 2018 Author Posted August 16, 2018 2 hours ago, Strange said: As always, the problem is with your lack of knowledge of physics. (I feel) Resignation ((but) accepting your criticism (as your good intent)). 2 hours ago, Strange said: Nonsense. Let's try again: (James) Watt defined ((horse_)power): P=F*va with the (average) speed va, horses could pull weights Wt=m*g=F. (I'm sorry you missed your history lesson (if you did?); or was I too brief (& that's why you did not understand (me))?). I'll assume the later. I suspect a specific quotient (instead) might have fit Newtonian (physics) structure better (tendencial). 2 hours ago, Strange said: https://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf#page=52 Yes, mechanical versus electrical definitions clash. It would be easier (avoiding conflicts) to maintain the electrical units Ampere*Voltage as maybe A*V; instead of (re)naming them Watts W, (unlike) as the committee did. The unit Watt has unfortunately been double defined (e.g. electrically & mechanically), & that is more than double trouble! Theoretically, the 1st definition receives priority (in honour=respect) to be maintained, while the 2nd (named usage) must (=should) be stopped. 2 hours ago, Strange said: This is not a "facit". Watt's definition does NOT give us the whole picture, & that is 1 aspect (point of view) indicating its weakness (in this argument). How would you describe that point? 2 hours ago, Strange said: Ignorant nonsense. Your cosmologists are still trying to make sense of energy, because they report their stupidity not knowing what "dark energy" is. (The exception can define some rules.) Your Nasa article below confirms they don't know "what" they are talking about (if you'll excuse the pun). 2 hours ago, Strange said: Nonsense. " It turns out that roughly 68% of the universe is dark energy." (https://science.nasa.gov/astrophysics/focus-areas/what-is-dark-energy) Add the +27% dark matter (related to energy by E=m*(c^2)) & that adds to 95% dark (universe). 2 hours ago, Strange said: Nonsense. So the expansion of the universe has not been slowing due to gravity, as (scientists wished) everyone thought, it has been accelerating. No one expected this (although I did, (but) why don’t they mention me?), no one knew how to explain it (but I did (with gravitational push*). Is that quote suppose to be scientific propaganda=brainwashing?). But something was causing it. *I predicted Hubble’s constant starts small, increases, & (then) tapers off as we progress outwards from the (universe’s) center. Scientists originally did NOT know if it was a straight line, curved: up; or down (=taper off). Energy is Leibnitz’s (math) physics (NOT Newtonian). Momentum is Newtonian physics. Maybe you should (also) deal with the real McCoy? Theorists still don't know what the correct explanation is.. Their solution is called “dark energy”. More is unknown than is known. We know how much dark energy there is because we know how it affects the universe's expansion. Other than that, it is a complete mystery. So they are pleading their (own) stupidity, & attempting to put everyone in the same boat. roughly 68% of the universe is dark energy. Dark matter makes up about 27%. Together that (& mass as energy) constitutes 95% total (of the universe). 2 affects, both dark=unknown why or what. Other than that, the article (is friendly &) explains well. Thanks. Another explanation for how space acquires energy comes from the quantum theory of matter. In this theory, "empty space" is actually full of temporary ("virtual") particles that continually form and then disappear. But when physicists tried to calculate how much energy this would give empty space, the answer came out wrong - wrong by a lot. The number came out 10120 times too big. That's a 1 with 120 zeros after it. It's hard to get an answer that bad. So the mystery continues. At least they are honest, (thus) making themselves credible. Now if Swansont says (or continues to say) Physics has NO problems; & that only I have the problems, then I guess he was not (really) aware of that 1. The article confirms you guys need help, because you’ve got big problems you have NOT solved. If I believed you 10 years ago, it’s all wrong today. How do I know what you now declare, is not wrong, & will not be declared outdated, again? Who had expected a 180° change, in views? (Some) TV documentaries are (now) suggesting physics may need to be newly redefined. (& just guess who has been working on that?) A last possibility is that Einstein's theory of gravity is not correct. But if it does turn out that a new theory of gravity is needed, what kind of theory would it be? I don’t want to get (too) “pushy” (t)here. Btw the book on the (book)shelf (mentioned earlier) is being pushed up (& maintained) at that (bookshelf) height, by the bookshelf (itself). The maximum pressure is on the lowest side (of the book). 2 hours ago, Strange said: Nonsense. You are saying that physical systems will behave differently with changes in time and space (Noether's Theorem). I'm sorry I am not familiar with him. Or is that an acronym for no_ether? 2 hours ago, Strange said: This is not what we observe. We can also measure conservation of energy and momentum directly. Would you mind showing me how. I can't make it always work mathematically. How can you? 2 hours ago, Strange said: So all the evidence shows that you are wrong. ? 2 hours ago, Strange said: You are the only one getting that. Your Nasa paper (above), also indicates the non_sense. I can't take the "credit" for being the only 1 that gets non_sense from the physics theory.
Strange Posted August 16, 2018 Posted August 16, 2018 5 hours ago, Capiert said: (James) Watt defined ((horse_)power): (D_i)d) (hE)? As you claim this, my starting assumption has to be that it is not true. But, even if he did, so what? That is not the modern definition. 5 hours ago, Capiert said: Yes, mechanical versus electrical definitions clash. No they don't. 5 hours ago, Capiert said: The unit Watt has unfortunately been double defined No it hasn't. But even if it had been, it wouldn't matter because the two definitions are equivalent. And note: "watt" not "Watt". (Details mater in science, which is why you are doomed to failure.) And you do realise watts are a unit of power, not energy? There is no law of conservation of power. 5 hours ago, Capiert said: How would you describe that point? Ignorant nd incoherent gibberish. 5 hours ago, Capiert said: Your cosmologists are still trying to make sense of energy, because they report their stupidity not knowing what "dark energy" is. Nonsense. Not knowing what dark energy is does not mean we don't know anything about energy. That is like saying that because we don't know what dark matter is we don't know anything about chemistry or physics. 5 hours ago, Capiert said: Add the +27% dark matter (related to energy by E=m*(c^2)) & that adds to 95% dark (universe). That isn't what you said. You can't pretend to be correct by lying. But if you are so proud of you ignorance of cosmology, I suggest you start a thread to show it off, rather than polluting this one with even more irrelevant and ignorant ramblings. I really don't know why you post on a science site: you are almost totally ignorant, you are unwilling to learn (because you believe your ignorant guesses trump years of theoretical and experimental science) and you have a very poor opinion of scientists. Perhaps you should find a more appropriate forum. Perhaps one about sewage engineering. 5 hours ago, Capiert said: I'm sorry I am not familiar with him. Emmy Noether was a woman. And of course you don't know about her, I wouldn't expect you to. You would need to study science, which you are unwilling to do. You prefer to just make up random nonsense. 1
swansont Posted August 16, 2018 Posted August 16, 2018 11 hours ago, Capiert said: An (exact) equality (=equation) can state a proportionality, Yes, it can, but the issue here is that when you actually use an equals sign, it means you are saying they are equal. It does not mean proportionality. 11 hours ago, Capiert said: That'( i)s my (1, major) complaint. The units totally conflict, Not when you do things correctly. 11 hours ago, Capiert said: although the concepts (excluding the units) are proportional. Something is very wrong (with the definitions)! No, what is wrong is the way you are understanding this. 11 hours ago, Capiert said: Let me ask it so: wouldn't it be better (=an advantage, as less bother, simplification) if electromechanical power had the same (=only 1) equation? We are describing how nature behaves. We don't get to tell nature how to behave. So if it takes more than one equation to do the former, then that's how it is. Nature is a tad complicated.
Capiert Posted August 17, 2018 Author Posted August 17, 2018 (edited) 19 hours ago, swansont said: Yes, it can, but the issue here is that when you actually use an equals sign, it means you are saying they are equal. It does not mean proportionality. What happens when I declare they are equal (as well, just to present my dilemma)? I mean I'm assuming you can understand my problem. (I "can" equate both (as mechanical) forces.) How should I proceede (acceptably)? I have applied rooted_power to a D'Arsenval meter; & I have weighted the needle with weights (without electrical power) & found (calculating the torques' force) that they (both: electrical (rooted_power) force & weight force) are "directly" proportional. I claim no great precision in my own measurements, you are the experts there. But the effect is obvious (to me, experimentally; & mathematically). I recommend you to (also) investigate (to convince) yourselves (also), if possible. Quote Not when you do things correctly. No, what is wrong is the way you are understanding this. I do not see the error you imply. (Enlighten me.) (My problem is more, getting "you" to understand my perspective & observations. It's often easier to explain to a simple person on the street, than it is to scientists.) I hope the above can explain my perspective. Quote We are describing how nature behaves. Me too. Quote We don't get to tell nature how to behave. I have tried to tell "you" what I observed. Quote So if it takes more than one equation to do the former, then that's how it is. Nature is a tad complicated. (Prof) Maxwell started with 20 equations. The telegraph operator Heaviside compacted them (20) into only 4. Amazing things are possible. Especially simplification. (Nature often uses similar blueprints. There is often a similar general scheme (to be found (in nature)).) Edited August 17, 2018 by Capiert
swansont Posted August 17, 2018 Posted August 17, 2018 8 hours ago, Capiert said: What happens when I declare they are equal (as well, just to present my dilemma)? If they are not actually equal, then the equation will be wrong, and any conclusions you draw will be invalid. 8 hours ago, Capiert said: I mean I'm assuming you can understand my problem. (I "can" equate both (as mechanical) forces.) How should I proceede (acceptably)? I have applied rooted_power to a D'Arsenval meter; & I have weighted the needle with weights (without electrical power) & found (calculating the torques' force) that they (both: electrical (rooted_power) force & weight force) are "directly" proportional. But are they equal? IOW, is it possible to do both mechanical work and electrical work? 8 hours ago, Capiert said: I have tried to tell "you" what I observed. You tried to tell us that one equation should work for all situations. How is that an observation? 8 hours ago, Capiert said: (Prof) Maxwell started with 20 equations. The telegraph operator Heaviside compacted them (20) into only 4. 4 equations is not 1 equation, and none of the four deals with mechanical systems. So there are even more than that.
Capiert Posted August 17, 2018 Author Posted August 17, 2018 (edited) 4 hours ago, swansont said: If they are not actually equal, then the equation will be wrong, and any conclusions you draw will be invalid. But are they equal? Yes. They are equal. (Is that too difficult to believe?) Quote IOW, is it possible to do both mechanical work and electrical work? (I suspect the answer (to that question) is yes.) What do you mean (exactly) by electrical work? (E.g What formula are you referring to?). I know I can achieve the same amount of spring displacement (angle distance), with either: weights; or electricity (as rooted power). Their affect is identical=the_same (to the meter's spring, Hooke's law F=-k*x). Similar forces, produce similar displacements, thus I achieve identical results with 2 different methods. Quote You tried to tell us that one equation should work for all situations. Yes, I believe that is true, I have been convinced it is possible. Quote How is that an observation? In that it contradicts your statement that 2 equations must (always) be used (separately), instead. Quote 4 equations is not 1 equation, True. Quote and none of the four deals with mechanical systems. If you insist waves' motion is not mechanical.?) Quote So there are even more than that. (Most likely, at lot for me to still do (if possible, at all?).) I stated Maxwell versus Heaviside as a tendancy (analogy) (towards simpification, from a non_PhD (Heaviside) to their wonder (e.g. at least Maxwell)); but Oliver's uncle was Kirchhoff whose law pertains to the meter's functioning, as well. (If that's the connection (relavance) you are searching for, from my statement, as to why I ever mentioned something like that, at all.?) Naturally my (own, (formula) compression) ratio is not 20:4, but instead only 2:1. Edited August 17, 2018 by Capiert
swansont Posted August 17, 2018 Posted August 17, 2018 16 minutes ago, Capiert said: Yes. They are equal. (Is that too difficult to believe?) When the units don't match, yes, it is difficult to believe. 16 minutes ago, Capiert said: (I suspect the answer (to that question) is yes.) You only suspect? If I send current through a resistor, there is electric power dissipated, but no mechanical work. If I push a cart, it's mechanical and no electrical. So, how can there be one equation? 16 minutes ago, Capiert said: What do you mean (exactly) by electrical work? (E.g What formula are you referring to?). Curious you should ask, since you introduced the topic. Electrical power, P=IV, mechanical P=Fv. Multiply by time to get the energy. 16 minutes ago, Capiert said: I know I can achieve the same amount of spring displacement (angle distance), with either: weights; or electricity (as rooted power). Their affect is identical=the_same. Similar forces, produce similar displacements, As I think we discussed, there is no work being done while there is no motion. You have too many examples to know which one you are referring to, and you tend to leave out sufficient detail to have a useful discussion 16 minutes ago, Capiert said: thus I achieve identical results with 2 different methods. Yes, I believe that is true, I have been convinced it is possible. In that it contradicts your statement that 2 equation must (always) be used (separately), instead. I don't trust that you have done a sufficiently careful experiment, since you don't seem to understand that there is only mechanical work done while the needle is moving. And since there is no motion once the needle stops, and thus no work, how can you possibly think that the mechanical work(or power) equation can be applied? 16 minutes ago, Capiert said: True. If you insist waves motion is not mechanica.(?) EM wave motion is not mechanical
Capiert Posted August 18, 2018 Author Posted August 18, 2018 (edited) 7 hours ago, swansont said: When the units don't match, yes, it is difficult to believe. I believe you! The units are crazy! Quote You only suspect? Yes. The answer depends on how things are defined (versus how they are derived=self_defined, instead); & which basis of reference(d definitions should be used for trouble_shooting). I question both of James Watts definitions: mechanical_power P_W=F*va; & work_energy WE=F*d (as possible culprits of the problems). Thus, I should isolate both & NOT_use (=avoid) them as my (starting) basis. Instead I should choose another definition (that is more reliable) on which to rely (on) just (in case) to be careful & see what gets derived (out, from that (other or new) usage). In my case I have used the 2 (simple) quantities: (Newton's) force; & displacement (angle) distance. E.g. If a units problem occurs, then there is always the question: where (=at which end, or perspective) should we tackle the problem (1st)? What could be corrupt? & what could be used, instead that is (more) reliable? Quote If I send current through a resistor, there is electric power heat (=random mechanical energy, in 6 directions: +/-x,y,z) Quote but no mechanical work moved mainly in only 1 direction (.e.g. angle). Quote . If I push a cart, it's mechanical and no electrical. But if the cart is braked, or too heavy when you push, then it will (also) NOT move (thus no mechanical work, even if you are using a bull_dozer e.g. to push a(n extremely large) tree, or mountain of granite that does not budge (a bit), although your gasoline gets wasted trying all day till the gas tank is empty (although paid, for that chemical energy). James Watt's (2) definitions (P & WE) do NOT account for what does NOT get moved! How can you rely on that deficit when dealing with (a declaration of) conservation of energy? It's ridiculous! Quote So, how can there be one equation? Via a (self_defining) derivation. Quote Curious you should ask, since you introduced the topic. Electrical power, P=IV, mechanical P=Fv. Yes. Those told me how you are looking at things, so I can orientate better as to finding the problem (in the communication gap). You do NOT see any problem with energy so you trust it completely, & automatically use it (instinctively); while I do the opposite avoiding energy as much as possible with the suspicion that it is corrupt & unreliable (in some cases). Quote Multiply by time to get the energy. If I multiplied force by time, instead I would get momentum, which I would prefer (& trust more). I'd trust Newton's force & momentum; over Leibnitz's energy, & James Watt's power. Quote As I think we discussed, there is no work being done while there is no motion. Yes, that exclusion (of motion) in (James) Watt's definition of work (energy) & power is the (deficit) problem. That deficit won't (always) allow a total account for registering energy conservation. Intuitively, we know we can loose energy without being able to account for it (=the lost energy) with Watt's method. E.g. bull dozer against a mountain all day getting nothing done except getting warm=hot. Here I (hope I) have shown why. Quote You have too many examples to know which one you are referring to, and you tend to leave out sufficient detail to have a useful discussion. I hope that (above) helps better. Quote I don't trust that you have done a sufficiently careful experiment, I do not brag about any of my experimental results (precision), they are only done to orientate me, for what tendancy is to be expected (based on my algebra). Quote since you don't seem to understand that there is only mechanical work done while the needle is moving. As I have said that's the weakness in (James) Watt's definition. His definition is unacceptable (for me). It does NOT fit in well enough with Newton's math (for me). Quote And since there is no motion once the needle stops, and thus no work, how can you possibly think that the mechanical work(or power) equation can be applied? I suspect you are right there. I must drop the usage of mechanical work (energy) & attempt to exclusively use (Newtonian) momentum (& force); instead of (work) energy (& James Watt's mechanical power). Otherwise I'm only asking for future problems with such unreliable definition as those from James Watt. Quote EM wave motion is not mechanical. Please explain a bit. Isn't mechanical, a motional (=moving) mass study (analysis, e.g. measurements). Don't waves have momentum mom=m*v? Edited August 18, 2018 by Capiert
swansont Posted August 18, 2018 Posted August 18, 2018 11 hours ago, Capiert said: I believe you! The units are crazy! The units are fine. You appear to simply not understand what's going on. 11 hours ago, Capiert said: Yes. The answer depends on how things are defined (versus how they are derived=self_defined, instead); & which basis of reference(d definitions should be used for trouble_shooting). I question both of James Watts definitions: mechanical_power P_W=F*va; & work_energy WE=F*d (as possible culprits of the problems). And the answer to your question is that no, the equations are fine, when used within the constraints of how they may be applied. Which can possibly be understood by studying some physics. 11 hours ago, Capiert said: Thus, I should isolate both & NOT_use (=avoid) them as my (starting) basis. You should use them when the conditions allow their use. 11 hours ago, Capiert said: Instead I should choose another definition (that is more reliable) on which to rely (on) The physics definition is reliable, if you understand the physics. If not, all bets are off and you will likely end up with nonsense. 11 hours ago, Capiert said: just (in case) to be careful & see what gets derived (out, from that (other or new) usage). In my case I have used the 2 (simple) quantities: (Newton's) force; & displacement (angle) distance. Displacement angle is not the same as displacement distance. 11 hours ago, Capiert said: E.g. If a units problem occurs, then there is always the question: where (=at which end, or perspective) should we tackle the problem (1st)? What could be corrupt? & what could be used, instead that is (more) reliable? The equations are very reliable. The problem lies with the person using them, not understanding how to use them. Then it's like a neophyte using a power tool. Unfortunate things can happen.. 11 hours ago, Capiert said: heat (=random mechanical energy, in 6 directions: +/-x,y,z) Energy does not have a direction. It can, however, have a gradient. Heat is energy transfer owing to a temperature gradient. 11 hours ago, Capiert said: moved mainly in only 1 direction (.e.g. angle). But if the cart is braked, or too heavy when you push, then it will (also) NOT move (thus no mechanical work, even if you are using a bull_dozer e.g. to push a(n extremely large) tree, or mountain of granite that does not budge (a bit), although your gasoline gets wasted trying all day till the gas tank is empty (although paid, for that chemical energy). James Watt's (2) definitions (P & WE) do NOT account for what does NOT get moved! Yes.You have to use the equations that apply to the problem, rather than using ones which don't. If you need to account for thermodynamic events, you must apply thermodynamics. If you don't then simple mechanics will work for such a problem, e.g. you could simply be accounting for work done by a frictional force, rather than having to worry about heat and temperature. It depends on the problem. And, of course, if you have current and resistance, etc., you will need to use electrical equations. But if these effects are absent, you would not. It depends on the parameters of the problem. 11 hours ago, Capiert said: How can you rely on that deficit when dealing with (a declaration of) conservation of energy? It's ridiculous! That energy is conserved can be derived. It is one of Noether's theorems. 11 hours ago, Capiert said: Via a (self_defining) derivation. Then derive it. 11 hours ago, Capiert said: Yes. Those told me how you are looking at things, so I can orientate better as to finding the problem (in the communication gap). You do NOT see any problem with energy so you trust it completely, & automatically use it (instinctively); while I do the opposite avoiding energy as much as possible with the suspicion that it is corrupt & unreliable (in some cases). It is true that one cannot always experimentally measure all of the energy in a system, so conservation of energy is not always useful to apply, even if it is conserved. In an inelastic collision, for example. One cannot easily account for the kinetic energy that is converted into other forms. (i.e. collisions don't tend to conveniently take place inside of calorimeters) 11 hours ago, Capiert said: If I multiplied force by time, instead I would get momentum, which I would prefer (& trust more). OK. Solve some problems involving thermodynamics with conservation of momentum. Go ahead. Try it. Or even a block sliding down an inclined plane with friction. Or an elastic collision, without involving energy. That, too, has limits on practicality. 11 hours ago, Capiert said: As I have said that's the weakness in (James) Watt's definition. His definition is unacceptable (for me). Fortunately you are not the arbiter of acceptance in physics. 11 hours ago, Capiert said: Isn't mechanical, a motional (=moving) mass study (analysis, e.g. measurements). Don't waves have momentum mom=m*v? EM waves have no mass.
Capiert Posted August 19, 2018 Author Posted August 19, 2018 (edited) On 2018 08 18 at 3:07 PM, swansont said: The units are fine. You appear to simply not understand what's going on. Unfortunately (I suspect) you might have missed my point? The electrical "watt" (definition), conflicts (severely) with the mechanical value in (=by) an exponential amount. I.e. According to (my) experiment(s), using Hooke's spring law, equating forces, for the same (=identical) angular displacement. I can NOT verify the committee's (electrical) definition (because it is way off!). I do not get the correct (electrical) values (when compared to mechanical values) using their assumption (=definition). I do get correct (electrical & mechanical, corelation) values using my formula Fiv=(I*V)^0.5, (=F=ma=Wt=m*g) instead. That is why the(ir) (electrical) units are crazy! (According to my formula derivations (the (electrical) units for), current multiplied by voltage should be (mechanical) watt_squared, instead of (only) watt units.) & I don't know how to explain that (discrepancy, problem) to you otherwise. Maybe you can help? Quote And the answer to your question is that no, the equations are fine, when used within the constraints of how they may be applied. Which can possibly be understood by studying some physics. I think you have made your point, that (our earthly) work energy WE=F*d (concept) is a very limited (=restricted) basis for studying the total cosmic system (accurately) (because it lacks the static case, & can only document moving things). It's useful for studying some (e.g. moving) things, but not everything (e.g. static wrt to our own (earth's) speed), thus useless in those cases). Quote You should use them when the conditions allow their use. I however, am looking for a more universal formula instead of those (limited) restrictions. Quote The physics definition is reliable, if you understand the physics restrictions (=limits) Quote . (That sounds (a bit) like using a crutch, as a sports car, for the (human) race.) Quote If not, all bets are off and you will likely end up with nonsense. (I think) I've made some sense out of the chaos. Quote Displacement angle is not the same as displacement distance. Quite right. But do you have a problem with (using) a (D'Arsenval_meter's) spiral_spring (versus linear_spring) & seeing the (for me, obvious) correlation, & (my self_allowed) substitution? Either: weights or electrical rooted_power produce identical deflection angles (confirming Hooke's law F=-x*k, angularly F=-theta*k). Quote The equations are very reliable. As you can (or will) see, there are discrepancies between the electrical versus mechanical watt values. So at least 1 of them is doubtful. Quote The problem lies with the person using them, not understanding how to use them. Quite true & I propose the committee's personal concerning the electrical concerns might need to brush up a bit on the facts. To my person, I am always learning. Quote Then it's like a neophyte using a power tool. Unfortunate things can happen.. I agree, as we have seen. e.g. my errata thread. Quote Energy does not have a direction. I would have said, (heat) energy('s randomness) could have a net direction of all directions, thus an exclusive single direction is NOT discernible. Quote It can, however, have a gradient. Heat is energy transfer owing to a temperature gradient. Yes. Quote Yes. You have to use the equations that apply to the problem, rather than using ones which don't. If you need to account for thermodynamic events, you must apply thermodynamics. If you don't then simple mechanics will work for such a problem, e.g. you could simply be accounting for work done by a frictional force, rather than having to worry about heat and temperature. It depends on the problem. And, of course, if you have current and resistance, etc., you will need to use electrical equations. But if these effects are absent, you would not. It depends on the parameters of the problem. I tend to agree with you (improvising where possible). Quote That energy is conserved can be derived. It is one of Noether's theorems. Could you (please) summarize that for me that I could comprehend it better (that it would stay better in my memory)? Quote Then derive it. OK. It's a repeat (from above). Newton's force Fn=m*a Wt=m*g Hooke's (spring) force Fs=-x*k then angularly Fa=-theta*k2 electrical (rooted power's) force Fiv=(I*V)^0.5 Equate Newton's force (gravitationally as weight) with the electrical force because the angular displacements are identical gives Wt=Fiv m*g=(I*V)^0.5 Should I go further? Quote It is true that one cannot always experimentally measure all of the energy in a system, so conservation of energy is not always useful to apply, even if it is conserved. In an inelastic collision, for example. If I may say(?), during an elastic collision there is a brief period (=duration of time) when the collision is non_elastic. ? Quote One cannot easily account for the kinetic energy that is converted into other forms. (i.e. collisions don't tend to conveniently take place inside of calorimeters). What do you mean (there) by conveniently? Are you saying, many collisions are very complicated? I see that KE fails to account for everything, & that some of the other (so_called converted energy) forms might be (unjustifiable) guesses=assumptions. Quote OK. Solve some problems involving thermodynamics with conservation of momentum. I don't see the problem with putting things on a momentum basis, but I don't know the thermodynamics you are looking for. (Could you please tell me the formulas you want & mean?) Momentum is mom=m*v mom=m*((vi^2+2*h*g)^0.5)-vi) with speed_difference v=((vi^2+2*h*g)^0.5)-vi) initial speed (velocity) vi, & mass m. The only problem there is sequentially dealing with negatives under the root sign, correctly, when squaring both sides. Please give me a simple thermodynamic problem (to get started), that needs the momentum conversion. If you give me enough hints maybe I can convert (for you)? I do not know if I can, because I don't know enough of your thermodynamics (syntax & concepts you exclusively use) to evaluate if I can. Quote Go ahead. Try it. Or even a block sliding down an inclined plane with friction. I know (next to) nothing about them (inclined planes & friction). If you help get me started (with the energy description setup), maybe I could (also) convert that to momentum for you? Quote Or an elastic collision, without involving energy. For (starting with) a non_elastic collision The total momentum of the system is (also) mom3^2=mom1+mom2, ^2=square both sides mom3^2=(mom1+mom2)^2, gives mom3^2=mom1^2+2*mom1*mom2+mom2^2, rearranged that'( i)s mom3^2=(mom1^2)+(mom2^2)+(2*mom1*mom2). So you (can) see there is some continuity. Assume mom4^2=-2*mom1*mom2. If mom1 & mom2 are the initial momentums, then mom3 & mom4 are the final momentums; where initial mass1 m1=m3 mass3 final, & initial mass2 m2=m4 mass4 final. I.e. Odd postscript numbers are for the same mass. & even postscript numbers are for the same mass. Please choose some values, & we can (try to) extract some answers. Before=After collision (mom1^2)+(mom2^2)=(mom3^2)+(mom^4^2). Quote Fortunately you are not the arbiter of acceptance in physics. EM waves have no mass. If a wave has momentum (e.g. impulse) then I would say it also has a very small mass m (=mom/c) value based on the momentum formula (& wave_speed c=v), only as a number. I mean I can equate. Edited August 19, 2018 by Capiert
John Cuthber Posted August 19, 2018 Posted August 19, 2018 34 minutes ago, Capiert said: The electrical "watt" (definition), conflicts (severely) with the mechanical value in (=by) an exponential amount. I.e. If that was true then the people who make power stations would have noticed. If your writing style was more readable I might bother to look for other problems in your post.
Capiert Posted August 19, 2018 Author Posted August 19, 2018 (edited) 1 hour ago, John Cuthber said: If that was true then the people who make power stations would have noticed. I suspect they didn't because they (attempt to) use constant voltage(s) V. Piv^0.5=((V^2)/R)^0.5=V/(R^0.5) =I*(R^0.5). Metering (for billing) is indicating current flowing. Quote If your writing style was more readable I might bother to look for other problems in your post. Edited August 19, 2018 by Capiert
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