  # Capiert

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486

## Everything posted by Capiert

1. Well done! But if the proportionality to distance is a (correct) mass_squared; instead of (only mass) NOT squared, then you have a definite difference that can NOT be pushed_away (=denied!) (& hid under the carpet). The numbers (resulting) for mass_squared versus only mass (NOT squared). Btw. That ("squared" versus NOT_squared (mass)) is math; NOT physics. Math_Physics at the most. Quite right. But with what (=which?) method you choose, to describe, determines your outccome. Choose momentum & then you are probably (quite) right. Choose Energy & then it can go wrong when mass changes in the collision. Choose mass*Energy & then you might get it right (again), when dealing with changing mass. The choice is (really) yours (NOT mine). (I'( woul)d call that bias.) ..What? (Please explain yourself. "NOT" what?) Quite right. Mass_squared does NOT exist in either of those (mom or Energy) equations, (& because they are inferior, e.g. limited). Mass_squared is a new concept from Ewert (1996) & it does seem to work right=correctly, e.g. better, such as in problems where the mass changes. & the cool (=neat) thing about it (=mass_squared) is m*E, mass*Energy (m*E) is VERY similar to (only) Energy so you hardly (=barely) need to change a thing. It'( i)s very easy. Simply multiply mass by Energy & you might (=should) get the correct answers. That'( i)s over_simplified of course. What sort of problem am I making up? I see definite problems with Energy that can NOT be over_come (otherwise). Dark_Energy is a real eye_sore. But the so_called energy loss in an inelastic (=NON_elastic) collision is most prevalent (=obvious). & I am NOT making that 1 up. YOU are! The Energy NEVER left the system. & I can tell you where it is. I (wrongly) said that you "lost" -1 [J]; but that is wrong only because you guys condition me into your same bad habits. (& I fell for it, as naive as I am.) You haveN'T lost a thing! Take a look at (conservation of momentum, COM) mom1+mom2=mom3. That'( i)s a perfectly balanced equation. Left_side (is mom1+mom2) is "before" the collision & right_side (is mom3) is "after" the collision. Square both sides & it is still balanced (=perfect)! (mom1+mom2)2=mom32. But that produces mom12+mom22+2*mom1*mom2=mom32. That +2*mom1*mom2 belongs to "before" the collision (=left_side of the equation); NOT "after" the collision (right_side of the equation). Again: That +2*mom1*mom2 belongs to the "before"_ side of the collision equation; NOT the "after"_side of the collision equation. Putting the 2*mom1*mom2 on the "after" side of the collision equation means it "needs" to be NEGATIVE. I.e. -2*mom1*mom2. NON_elastic collisions might seem to have lost energy on the (heavier) result; but they have NOT lost a thing (in that sense); you need only observe the math to correctly determine what has happened. & (thus) you physicists (have) preach(ed) an untrue gospel (=NONSENSE!) Sorry! But more SORRY for your (misled) followers.
2. (amount (of mass)) But it (that mass) is distributed differently (into the (mass) terms) so that it is only (1 mass, now e.g.) together. & that ((different) togetherness) (now) has a different affect on the math answer. are now 1 total mass m3=m1+m2 which is now larger than either 1 of its initial components (e.g. either: m1; or m2) & If mass is squared, then its parts produce different (math, number, value) results compared to its (=the total mass) sum squared. m32=(m1+m2)2 m32=m12+m22+2*m1*m2 <---But look at that! m32#m12+m22 Mass squared does NOT simply add only the squared terms; because the (NON_squared term) +2*m1*m2 is missing! I hope you get my drift. (What I am "trying" to tell you.) Such naive math (just) can NOT work, (sometimes).
4. That'( i)s how you (would) do it. E.g. Just so you do NOT have to bother. But why should I (try to) believe you. You have given me NO proof, with your inability NOT to bother (attitude). Are all scientists so lazy (like Minkowski hinted about Physicists)? (Surely NOT!) Why should (some of the kinetic) energy leave that system? (& I DON'T accept warm, acoustic, excuses either.) (I'd like to see (simple) tangible measurements.) Oh! Abracadabra (then). (It's a mystery!) That sounds like a boring disinterest in science e.g. trying to know. A half hearted attempt to throw a few things together. You either: know; or (else) you DON'T, & you obviously DON'T, because you give me useless excuses. Sorry, other people can try to be more thorough. You DON'T even give the effort. If you (were to) say: those answers can NOT be found; then there must be a reason why. (Oh we are too feeble, (at) attempting, (to) zero_speed, =zero results. It's more difficult that c.) Disinterest is NO excuse. You also avoid commenting (up)on the (=my) initial_kinetic_energy KEi (perhaps because you habitually evade it by subtracting it away). My syntax includes KEi. (What is your syntax, if mine is NOT an extended (syntax)?) All 3 (named KEs) KEf=KEi+KEd are "kinetic_energies" ((meaning) NOT your "the"(what? _unknown), NOT mentioned f form) syntax). You can clearly see that (they are kinetic_energies) in my syntax "KE" with a subscript. There is NO difference: meaning a KE is a KE, whatever its subscript is. The KEd can (equally) accelerate a(ny) mass from zero (speed) to a (new) final_speed vf, which would finally have its own KEf(new)=KEd equal to that kinetic_energy_difference KEd. I DON'T see why you try to sell a KEd distinction (away) from any other KEsubscripted just because you do NOT know what (else) KEd is (or could be). E.g. Even though you only want KEf to be "the" (only) kind of KE (possible). It is absurd to say: KEd is NOT a kinetic_energy simply because it is NOT "the" final_KE KEf=m*(vf2-0)/2 which uses the mass m multiplied by half the final_speed squared vf2 but "subtracted by zero(_squared)"! The initial_kinetic_energy KEi=m*(vi2-0)/2 is also a kinetic_energy (just like KEf is) because its half the initial_speed squared vi2 but is also "subtracted by zero(_squared)"! The universal KE_difference formula KEd=m*(vf2-vi2)/2 is the most universal KE "definition"! There you can (=may) use any reference(_frame) speed vref=vi (below c, that) you want to be your reference speed (e.g. at rest, when identical to the initial_speed vi). If I have 7 oranges (analogy KEf) & subtract 4 (oranges, analogy KEi), then I would expect out, 3 (oranges, analogy KEd) like any reasonable thinking person. NOT grapefruits or "lemons"! (or other hogwash). That's only common sense, which seems to be missing here (in (what some people call) science). You DON'T (even) have a clue where the energy has gone, you DON'T know what it is (e.g. called, other than "difference") & yet want to be called scientists. (& you want me to believe you?) Modern physics is like modern art, anything goes, even junk. All that matters is who (e.g. what ego) has the say. I suspect you mean, the initial_speed terms (e.g. KEi) are subtracted anyway; (&) so why bother. ? Taken from a different perspective, of: if the 2 masses are on Earth & the Earth is rotating let'( u)s say vi=~1 [km/s] eastwards ((just) to keep things simple, at where they are on the Earth's surface); then they are still moving ~1 [km/s] although they appear to you as at rest. (& that KEd did NOT leave the system.) What seems (as) "at rest" is an optical delusion (of) for both: observer; & an object, having the same (=identical) speed. (E.g. Even though they are separated by a distance d.) In reality (e.g. the universe), (we know) everything is moving. Meaning NOTHING is (really) static (=at rest, with zero speed). (Everything has a speed difference wrt some other (moving) object (reference, frame).) Your "choice" of reference(_frame) (e.g. of same speed as the observed object) (help) determines whether you want to be: deceived ((in)to think(ing): ) an object is at rest (when it has the same speed as the reference_frame); or NOT! **(Sorry! (Yes) "I") Modified. (Or do you mean your quote is dishonest? Which I (rather) doubt, in preference for the former.) How else should I add (extra) comments of mine into your text? I bracketed it, to distinguish it from your original text. Dishonesty was NOT intended; only clearness of the discussion (was intended), (before getting lost again in(to) confusion between syntaxes). Should I use different brackets?
5. (Our) Confusion arises from different syntax: mine; versus yours. I agree. & they do speak English. Yes, I think we can assume so. (& It's nice to hear it is fine.) You still have NOT commented on KEi. (final_) KEf What kind of Energy? It's NOT chemical, thermal, NOR potential etc. So what KIND of energy is that? & also, only a particle could have had that energy when it was by=with the particle. If a particle had lost that KEd then that particle has decelerated & is now moving slower, than it was (before the loss). But it was lost to another particle (forced to accelerate, E.g. Newton's 2nd Law). KEd can only be had by a particle (or object) e.g. mass. The KEd formula can discuss (either): the same mass m; or (else) it can discuss 2 different masses. But (the) K(inetic )E(nergy) d(ifference) can NOT be transferred without a mass (as mediator, or transporter). So what you said, e.g. that KEd does NOT belong to a particle, (simply) does NOT (seem to) make sense (to me). KEd can only be transferred by (a) mass, (instead). But: The bigger, more important, question (to ask) is: to which mass does KEd belong to; & when? (before versus after, a(n elastic) collision). E.g. Which mass lost the -KEd? & which mass gained the +KEd? NO mass? Then NO energy! You can NOT describe (e.g. formulate) energy without mass. (& charge is intimately related to the inverse "mass", as in charge to mass ratio.) motion_ final_ I think we can fictively equate KEd to various other masses & speeds just to get equivalents (or equivalence). (But that'( i)s modelling, e.g. NOT physical.) I find your next example (description) (non_elastic collision) rather interesting. Let (their masses, be) m1=1 [kg], & m2=1 [kg]. Let their initial_speeds, be vi1=1 [m/s], & =j0*1 [m/s], j0=1 (0°) vi2=-1 [m/s]=j2*1 [m/s], j2=-1 (180°) with initial_KE's KEi1=m1*(vi12-02)/2=1 [kg]*((1 [m/s])2-0)/2=0.5 [kg*m2/s2]=j0*0.5 [J] (0°) KEi2=m1*(vi22-02)/2=1 [kg]*((-1 [m/s])2-0)/2=1 [kg]*((j2*1 [m/s])2-0)/2=j4*0.5 [kg*m2/s2]=j4*0.5 [J] (360°) Their final_speeds, are vf1= 0 [m/s], & vf2= 0 [m/s] with final_KE's KEf1=m1*(vf12-02)/2=1 [kg]*((0 [m/s])2-0)/2=0.5 [kg*m2/s2]=j0*0 [J] (0°) KEf2=m1*(vf12-02)/2=1 [kg]*((0 [m/s])2-0)/2=0.5 [kg*m2/s2]=j0*0 [J] (0°) with KE_differences KEd1=m1*(vf12-vi12)/2=1 [kg]*((0 [m/s])2-(1 [m/s])2)/2=-0.5 [kg*m2/s2]=j2*0.5 [J] (180°) KEd2=m1*(vf22-vi22)/2=1 [kg]*((0 [m/s])2-(-1 [m/s])2)/2=1 [kg]*(-(j2*1 [m/s])2)/2=-j4*0.5 [kg*m2/s2]=j6*0.5 [J] (540°). Both (Energy) losses (together) are -1 [J], each -0.5 [J]. KEd1=KEf1-KEi1=-0.5 [J] KEd2=KEf2-KEi2=(-)3*0.5 [J] KEd3=KEd1+KEd2=-0.5 [J]+(-)3*0.5 [J], ~-1 [J]. How then can you claim that their (kinetic) Energy was NOT destroyed (=annihilated)? Are you claiming out of (good) "belief"? (E.g. Scientific religion. E.g. Hoping the theory will justify every "odd" detail in the end?) A reasonable person could NOT claim that, (when) seeing the facts. (Seeing is believing.) (You must (=most probably, surely) have good faith, (brother)!) kind (=type) of (moving (=motion) energy) Although we can loose energy (by using the minus sign symbol "-"), ((but) into NOTHING!). Because the math (e.g. (final_speed) "squaring") does NOT allow negative( value)s. I have (seen the inelastic example above), & it looks convincing (& puzzling, at the same time). It does NOT fit, like indefinite integrals (do). I DON'T want to argue with you there (not a value associated with any one particle, or even anything having motion.) But I suspect a closer examination (in detail) will allow associating to particles, & of motion. 3 indefinite integrals begin to stand (=argue) against (a) calculus working regardlessly. Serious mathematicians know calculus does NOT "always" work right. they have been successful at isolating those few quirks; but they CAN'T quite put their thumb on why to eliminate them completely. They simply set up warnings, DON'T do. With those Jack in the boxes, I'm curious what else (for problems) could be found awaiting us for: in the future. Btw. What is half of infinity? Calculus works with infinitesimal limits, hoping to make the problem so small so it disappears. The problem is the proportioning should stay the same, instead of approximating that it is gone. Yes, with your help we have identified a problem that could be cleared. Thank you. That'( i)s the way we learned it in school. & looking at it (now) it seems like a (more/most) general description: dealing on a macroscopic level/perspective; rather than a microscopic infinitesimal view. I'm sorry, but I have to disagree with you there. That is how you interpreted my text's syntax. You assumed I was talking about KEf; but I was always talking about KEd, instead. I DIDN'T use the d subscript as is typical for many variables in Physics. E.g. (duration) time t=tf-ti Eg. distance d=df-di E.g. speed v=vf-vi. When you measure a distance you rarely say it is a distance_"difference" of e.g. 100 [km] from 1 big city to another. You just say the distance is ..; & forget about the fact that it is a difference, too=also. Why the (complexity &) inconsistency in your syntax? Answer: Because normal people DON'T talk that way. You guys have to turn everything into a unique exception rather than talk like normal human beings. I mean "sometimes" there are advantages to the (few extra) details; but NOT always. The equation is KEd (NOT your (agreed upon) KEf); & as KEd=m*(vf2-vi2)/2 that equation is NOT wrong but instead absolutely correct. ∆KE=KEd It should have been obvious (excluding typo errors), that KEf#m*(vf2-vi2)/2 could NOT possibly be (correct) as KEf because it was missing (the term) -m*(vi2)/2; instead of being exclusively KEf~m*(vf2)/2 using vf2; without -m*(vi2)/2 losses. That is, unless you have found some(thing else) new, breath_taking news that we should (all) be cautious (about), for. Btw. I must comment that I was rather enthusiast at Bufofrog's original comment, (but unfortunately wrongly) assuming he might have found an interesting (new) problem that we could work on, e.g. he was thinking outside of the box. But instead, it only turned out to be (boring) syntax misunderstandings. But at least we got that cleared too. Thanks gang (=team)!
6. I suspect the confusion arises (between us) from the difference between my syntax versus your syntax. I have extended the syntax description details beyond yours. Your "the" (what?) KE, is my "final"_KE KEf. My KEd is the (KE_)difference of 2 ((perhaps) different) Kinetic Energies. In this case (final_KE minus initial_KE) KEf-KEi. (We can thank Swansont for helping us point out that 1.) I (would) find it (quite) peculiar if you would deny that the difference between 2 (different) KE's is NOT a KE (itself), because of: (1.) conservation of Energy COE: i.e. energy can NEITHER be created; NOR destroyed, & so the parts must add (up, together); & (2.) per definition_name kinetic_energy is (=means) the energy of "motion". There speed &/or acceleration apply. Delta_KE=KEd must be a KE (even if it is NOT your "the" KE, whatever "the" is suppose to be. Disclaimer: I am only trying to be thorough, with the(=my, algebra_)syntax.) If a mass object maintains a constant_speed (for that brief_time that "you" measure (it); instead of (versus, compared to) following its complete history as to how it (ever) acquired its (constant_)speed in the 1st place=originally), then, its initial_KE KEi (which is NOT ZERO, (&) which you seem to want to ignore (maybe) just to makes things difficult(?), but which I have already taken into account (of)) will be equal to its final_KE KEf & (but, thus) its KE_difference KEd=KEf-KEi will be (exactly) zero. (The analogy is taken directly from my speeds(_syntax) for the speed_difference vd=vf-vi. That is NOTHING new! (Except for (now) adding d subscripts (for "difference"), where they were NOT before.) Swansont should know that for linear acceleration the average_speed, is va=vi+vd/2 & his final_(instantaineous)_speed (equivalent), is vf=va+vd/2. E.g. vf=vi+vd I have NOT yet recognized that that algebra will NOT do. Meaning it (=the algebra) is still valid (as far as I can see). Really? Yes! Jerks are also accelerations, (similar to collision accelerations). Whether they (accelerations) are linear(?), I doubt it(!). Unfortunately there I suspect you assume too much for what is actually happening instead of measuring. Am I right? And yet you use average speed, which makes assumptions as well. Quite right. (NO bout a_doubt_it.) No, really, that's not how it's pronounced. They are pronounced "Kinetic Energy" and "Potential Energy" and they are sometimes equal to each other. "keep" & "peek" are (only) how I have nicknamed the formulas (acronymns if you will) as an easy reminder. NOT intended to offend anyone('s physics, indeed). Nothing theoretical about it. It's the value of the speed at a particular time. How then do you "measure" it (=instananeous_speed) exactly, especially for collisions? Relying on math (e.g. calculus) sounds pretty theoretical to me. But it's how the example worked that you gave. When you solve a problem, you use the appropriate physics for the problem, so if the speed is constant, you can use an equation for constant speed. And there are lots of problems where speed is constant, or that's a reasonable approximation of a situation. Please recognize (=acknowledge) my initial_KE Ki syntax into the problem. That'( i)s what it is there for, i.e. (it was designed) for such problems with (initial) constant_speed. If you can show that math is inconsistent that's a purely mathematical issue and has nothing to do with physics. If physics is using such math, then DON'T expect it (=physics) to hold itself together. I would (rather) say, then it (=that purely mathematical issue) should NOT have to do with physics (but unfortunately (does &) is messing it (=physics) up). Universities live & breathe calculus. (They love it!) I choke on it. In the example given they were the same, because speed was constant. Changing the parameters of someone else's example and then complaining about a problem that arises is not an argument made in good faith. Sorry (I was so slow), I must 1st grasp the problem before I can understand (what is happening) in order to tackle it (later). Stating what (examples) I know helps get me nearer e.g. Later then excluded. Stated otherwise: I had (wrongly) assumed you could follow me or my syntax, thus I could NOT understand where a problem was, if any. They can be, but the point was that if they aren't your equation quite obviously fails. It's wrong. But you're ignoring that. Nobody else is fooled by the distraction. Again, (when using my KEi, which I had thought you were already aware of) my 02:40 comment described what else my equation could also do. If you see that (extra) as ignorance, well then I'm sorry (for the misunderstanding). (But) I still do NOT see that my equation has failed. Simply take the (=my) KEi because it equals the (=my) KEf (which has (=is) your KE). The KEd has (=is) NOTHING! That has obviously succeeded, don't you think? To summarize the syntaxes: Mine: Yours: KEd delta_KE KEf "the" KE KEi ? I hope I have answered that sufficiently to your satisfaction(s).
7. Constant speed is NOT how the universe is working. & I am looking for a more general, (but) simpler description. Constant speed is the exception. (E.g. "Special" Relativity.) Interaction is the (higher, more dominating) rule. (Actually more basic=fundamental.) & there is NOTHING which says acceleration must always be linear. I get the idea "average" speed is an eye_sore for you because you prefer your instantaneous speed, instead. But that (instantaneous speed) is theoretical, & I suspect sometimes misleading. (I DON'T want to be lulled (or else tempted) into a false security. A(n unexpected surprise) jack_in_the_box (e.g. delayed) time bomb is NOT my idea of fun (for accuracy in extrapolation).) Calculus works "perfectly" (=flawlessly, NO errors) for "linear" acceleration because it is based on an (exact) "average"; (e.g. 2 triangles making a rectangle using a diagonal (line)) but I have my doubts for NON_linear (line) examples & irregular curves. E.g. Sigma_summing & (versus) the integrals (integrating, integration) should give the same (=identical) answers; but DON'T always. I would prefer to develop something more simple(?) & experimental, (e.g. averages), based on algebra, instead. But I also notice(d) that algebra has its limits (& errors) too. (So I can NOT trust algebra completely, either.) But I have NOT had enough chance to pinpoint exactly what those flaws=weaknesses are. (It happens so rarely, like a flipped coin standing on its edge.) I suspect it(s (math) flaws, or errors) has to do with (math) syntax compromises when dealing with negative exponents for totally=completely (symmetrically) reversible (e.g. recoverable?) math. E.g. A "negative" exponent should mean "rooting" as the reverse (operation) of exponentiation; (instead of a divided exponentiation). (Not to mention the meaning of the exponent zero, then.) & then there is the theme of sequence (history). (Math is a language & its sentences are equations (the phrases formulas).) But that'( i)s (all) too speculative right now. Too early to say (anything useful). -- You are content with that you have learned some (of th(os)e complexity's) math rules, & use them. I am NOT. I am more interested in finding the reasons why some (math) things (fail) do NOT work sometimes (as we (would) wish); & (then) try to fix=repair them. Experimenting with equations of motion & (so_called) mass are a good place=way to start, at seeing & discovering how this universe works. Geometry as well. (E.g. The "basis". Real proof!) NOT for me. I CAN'T follow you there, because "getting" (e.g. interacting) from initial_speed to final_speed (when they (speeds) do NOT equal) is acceleration. Making them (2 speeds) equal is an exception, because (that is why) they have different names so that they do NOT have to be the same. The (2 speeds) "can" be quite different. I'm aimed (=intended) at collisions, or jerks! (E.g. Unifying) the changes. Maybe you can tell me what I "should" be saying (or have NOT said) in my syntax description for initial (versus final) speeds? Because what I say (just) does NOT seem to land to you (all) as intended. Energy is (all) about (linear) acceleration (i.e. gravity). That was the reason for the vis_viva (living force, ~2*KE) controversy; & why Gravesande "abandoned" Descartes & Newton's momentum. (& that is the reason why we live with that mess NOW. i.e. Dark Energy ERRORS.) The speed is NOT constant for acceleration(s), such as (for, in) collisions where m*E is transferred. KE=PE (pronounced "keep", versus PE=KE pronounce "peek") (e.g. Conservation of Energy COE) might work (well) for (accelerating) a single (moving) mass that stays the same (sized mass); but (it, KE=PE) does NOT (always) work well when that mass interacts (& is accelerated) by other various sized masses (&) of different speeds. For them (collisions, interactions), m*E works better, because it is more general, universal, uniting conservation of momentum COM, & Energy COE together into 1 formula. I guess I am an early bird. (DON'T forget the worm.) What then is KE? Forced distance? I guess you mean I can NOT rely on gravity's force as constant. So its a poor example. ? Because we are talking about a constant velocity. That would be the reference frame that measures the objects velocity at 10 m/s. That seems like an indirect answer by trying to be (more) general. E.g. The Earth, ground or surface that you are standing on. Isn't a "reference frame" rather fictive imaginary. I'm talking about a real (physical) world.
8. Undoubtably average_power. I was NOT aware that James Watt used instantaneous power. He used how fast a (180 lb (force) ~82 kg (mass)) "horse" walking (continuously) a circle, using yards (=3 feet=0.9144 m) & seconds (time), could wheel up water from a well against (the horse's) gravity's weight (806 N). ~737 Watt (rounded to 740 W for 75 kg). 746 W would have needed 75.59 kg (~167 lb) at 1m/s. I question that math was any great deed of calculus. Watt determined that a horse could turn a mill wheel 144 times in an hour (or 2.4 times a minute). The wheel was 12 feet (3.7 m) in radius; therefore, the horse travelled 2.4 × 2π × 12 feet in one minute. Watt judged that the horse could pull with a force of 180 pounds-force (800 N). So: Watt defined and calculated the horsepower as 32,572 ft⋅lbf/min, which was rounded to an even 33,000 ft⋅lbf/min. Watt determined that a pony could lift an average 220 lbf (0.98 kN) 100 ft (30 m) per minute over a four-hour working shift. Watt then judged a horse was 50% more powerful than a pony and thus arrived at the 33,000 ft⋅lbf/min figure. https://en.wikipedia.org/wiki/Horsepower Then the "speed" should be constant. But the real (more probing) question is from where or what "speed" (reference) are you using to measure the "speed" you are observing. (Obviously, your own speed is the reference, there.)
9. Continued, (from my previous) That speed_change is the acceleration e.g. to or from a collision, but once that collision (duration e.g. t<1 , stops=) finishes the(n that) acceleration stops; so that friction decelerates the ball (further) to a (slow, much longer duration, e.g. t>1) stop. But, an Earthquake (e.g. an abrupt change in the Earth_plates' (rotational) speed) could shake the (table &) balls (in)to motion. Done. How would you describe (your) KE? (..without problems). That was mine (=my description). Identical speeds does NOT mean zero KE. It only means zero KE between the 2 objects that have the same speed. Constant speed does NOT mean zero KE. In fact any speed (helps) indicate a KE. (E.g. If you (also) know the mass(es), too.) But any (observed) speed(_difference) v is a (speed_)difference v, i.e. that uses a reference (initial_)speed vi (as basis), to observe the (final_)speed vf=vi+v. (But) we do NOT see the (final_)speed vf because it has the (initial_)speed vi in it, which we exclude. E.g. The Earth's_rotation_speed vi~1000 [m/s] eastwards remains invisible (to us) because everything (static) around us is (also) moving at (approx.) that same speed. How can we solve your problem? P.S. When Earthquakes happen 1 of the problems is the ((tectonic) plates') rotation slows down & then continues to speed up; repeatedly. That kind of (kinetic) Energy can bring down skyscrapers (crashing), (it's so powerful!) (m*PE=m*KE) m*m*a*d=m*m*(vf2-vi2) only equates the accelerated distance with the speed change of a mass; NOTHING more. a*d=((vf2)-(vi2)).
10. Why is (your) initial_speed vi (=vref) NOT zero? You have said the 10 kg mass is "moving" at (vf=vi+v=0+v=) 10 m/s, so what is the reference that you used?
11. Yuk! What a messterpiece. I'm trying to simplify (to algebra) so (even) a 5 year old can understand (Trump mentality, noted (by reporters) for its advantages), but it explodes & gets scattered into so much complexity (some interesting though), (really avoiding the question, with (fake) substitutes) (unfortunately demonstrating how little was understood, in some cases). When the cat'( i)s away the mouse will play. To get back on track. On 6/3/2022 at 2:50 AM, joigus said: What's the speculation here? Capiert: The (basic) question was whether the (Jame's Watt's) Power_equation's (forced) speed is a speed_difference v=vf-vi (No?, but most commonly claimed?) or the average_speed va=d/t (Yes? New). The algebra seems to confirm (the later, & itsself), (Unfortunately opening the question why the calculus did NOT do the job (right, for that question), which I DON'T really want to discuss (calculus yet). Maybe for later.) (I guess) I was a little sloppy because gravity's weight force was intended only as an example of force; NOT all forces in general as Swansont put it. Here is an example of using a specific equation and trying to apply it in general. Power applies to more than the situation of an object falling under gravity, so one cannot make a general claim that force is weight, since there are other forces. If you use weight, then the equation will only apply to a falling object, and also only if you can assume that this is a constant force. Should (better be) Power P=F*d/t is the Force F=m*a, multiplied by the distance d per time t. The average_speed va=d/t is the distance d per time t E.g. Let the (linear) acceleration a=g be gravitational free_fall's acceleration, & the force F=Wt be the Weight Wt=m*g, & the distance d=h height (fallen). P=Wt*h/t. I did NOT say that (an object moving at a constant velocity has zero KE); you did. If initial & final speeds are the same then the KE (between them) is zero. E.g. For 2 objects=bodies. E.g. For 1 object=body the (same, single) mass has NOT be accelerated (e.g. faster) from its initial speed; or e.g. it has NOT been accelerated from its final speed. Take your pick. I (still) do NOT see your problem. E.g. Same speeds (for same mass(es)) means NO kinetic_energy(_difference) KE, but KE is a difference (KEd=KEf-KEi) of (the 2) KE's; although I think Swansont will want to correct me there. No it does NOT. The equation uses 2 (different) speeds even if 1 of them (speeds) is allowed to be zero. A difference of KE's means that you are subtracting 2 KE's. So your components are KE's. E.g. KEd=m*(vf2/2)-m*(vi2/2). KEf=m*(vf2/2). KEi=m*(vi2/2). vf=vi+v vi=vf-v v=vd=vf-vi. Please show me the error (if so). As far as I know that math (syntax) works (for me). Maybe you have a task, example where I can use it? My concept (syntax) of KE (=KEd=KEf-KEi) is (a (KE) difference, &) similar to yours, but (differs in that it) is (also) extrapolated thru to (both) the initial_KE KEi=KEf-KE & the final_KE KEf=KEi+KE. (But Swansont commented=identified that is delta_KE.) What I'm saying is, the initial_speed vi is [often] invisible (for same speed objects); but NOT zero! Even if objects=bodies seem static; they are (really) moving! Same speed objects can NOT accelerate each other (because they do NOT collide with each other). Again PS: Very interesting. It looks like you did NOT get it (=the (2) perspective(s)). How can anything moving, NOT have KE? If vf = vi (then that) means you can NOT see the(ir) motion. (..when compared). The 2 objects (seem to) stand still, (perhaps) with a constant (separation) distance. E.g. 2 billiard balls on a (billiard) table (although the world=Earth is turning). Our reference(_speed) vi could be the Earth's rotation speed e.g. ruffly ~1000 [m/s] eastwards. The (2) balls are NOT going to do anything (e.g. (they are NOT going to) collide) with each other because they stay still (wrt the billiard table, on Earth); but (both (balls)) are rotating with the Earth_speed vi~1000 [m/s] eastwards (i.e. wrt the Earth)! Their KE would NOT become obvious until they collide with something else moving at a completely different speed. E.g. a cue hitting a ball. ((Please) allow me to exaggerate.) E.g. If the cue travelled ~-1000 [m/s] e.g. westwards (to hit the ball) to compensate against the Earth's rotational speed (so that the cue would seem like zero speed wrt the Earth's center). KE, (e.g. the speed_(energy)_change) can only be [acquired=] "received from", or else "transferred to" another mass('s motion).
12. Yes. I agree that there is NO such thing as conservation of Energy (COE); although conservation of mass*Energy m*E might exist. COE is a FAKE, because it often FAILS, although NOBODY has the guts to kick it OUT; they all still (perhaps naively) defend it (as traditional BRAINWASHING, e.g. NOT to upset things=tradition). Conservation of ENERGY does NOT deserve to be mentioned except for its (crude) history e.g. development of Physics. (That is:) Wasted time, considering my (just=immediately) previous comment. But what do you mean by extend the time frame? Distance d (x,y,z), d=P2(x2,y2,z2)-P1(x1,y1,z1), is the difference between 2 position(s points): e.g. P2(x2,y2,z2) minus P1(x1,y1,z1) ; (where P1(x1,y1,z1) is the origin(al starting point) which determines the positive direction). e.g. x=x2-x1 y=y2-y1 z=z2-z1. Please explain. (e.g. "contrary" to "self-referential" frame). (What is that?) (E.g. A Point source?) (Is ego supreme?) Everything exists "in" the universe; NOT OUTSIDE of it (=the universe). Everything is "a" part of the universe (=connected); NOT "apart" !. (E.g. I am in the universe; NOT independent from the universe). (Please give me a clue (as) to help understand you better.) But if I understand you correctly then the "initial" kinetic_energy KEi might help fix=remedy things; for the total=final kinetic_energy KEf=KEi+KE. E.g. Kinetic_Energy KE=KEf-KEi is the difference between final (kinetic_energy KEf) & initial (kinetic_energy KEi) similar to speed(_differrence vd vd=v) v=vf-vi where the subscripts are: final f & initial i. ("you can tease out the difference between") I'( a)m sorry (but) you have lost me (there). I can NOT imagine any speed without direction. "I DON'T know where we'( a)re going Captain; but (at warp_speed) we are getting there in a he(ck) of a hurry!"-Scotty. Direction (e.g. angle) is also relative. E.g. to a line. E.g. 2 points (P1 & P2, each an x,y,z); but the 1st (point) must be established from the 2nd, to complete that relation (for positve, (versus negative) direction) i.e. relativity. I will assume (with (the word) map) you are implying x,y,z (e.g. wrt to some other reference (of similar structure, e.g. Ref, e.g Frame)). I (can still) consider a speed(_difference) v (=vd), to be (composed of) components vx, vy, & vz. (But) I do NOT use instantaneous_speed; I use average_speed va=d/t, instead. I(' am sorry, I) DON'T follow (you). My (speed_difference) v is a mixture of x,y,z speeds. Where is the problem? (I suspect you are adding unneeded complexity; where NONE is needed. Am I right or wrong? If wrong please explain.)
13. We are assuming they are moving on parallel lines, in the same direction. Can (=May) I pacify that argument(?) by saying: (I acknowledge) an "initial" Kinetic_Energy KEi=m*vdi*vai where the intial_speed vi=vi-0 is (simply) extrapolated from a predacesser speed_difference with its maximum_speed being (only) that excluded initial_speed vdi=vi-0 (whatever zero=0 speed should be e.g. relative to something else('s motion_speed)). vai=(0+vi)/2 (analogy (similar to)) va=(vf+vi)/2). e.g. minus zero, where zero is that (next, smaller) reference. That is all done so because KE is (already) relative to its (own) initial_speed vi. Thus the kinetic_energies can be added (sequentially).
14. Please explain that problem. Kinetic_Energy is (already) relative (with respect) to the initial_speed vi. (-c<)vi<c can be (almost) anything less than light's_speed (+/-)c. That means the initial_speed vi is excluded in that (amount of) "kinetic_energy('s)" change of speed. E.g. 2 masses moving at the same speed (wrt each other), have NO speed_difference v=vf-vi (wrt each other), thus a constant distance is maintained (=kept) between them. Only if a speed_difference v exists between them can they affect each other in e.g. a collision, to change the other's speed e.g. via Newton's 3rd law, (equal & opposite) reaction, Repulsion. E.g. Assuming an (EM_)Field wrt (a decreasing) distance, to ((elastically) repulsively) bounce. The catch there is electric_repulsion is inversely proportional to the radial_distance "squared"! That is NO longer a linear relation; but instead exponential (wrt distance)!
15. (James Watt's, mechanical) Power P=F*d/t, va=d/t is the Force (e.g. Weight Wt=m*g) F=m*g multiplied by (e.g. the height h=) distance d per time t. That works out to, Power P=F*va is the (e.g. weight Wt=m*g) Force (F=m*a) multiplied by the average_speed va=d/t=h/t; instead of F*v=(p^2)/(m*t) which is the momentum_squared (p^2)=mom^2=(m*v)^2 (for the mass m; multiplied by the speed(_difference) v=vf-vi (of final_speed vf minus the initial_speed vi)); divided by both: mass m & time t. (I.e. F*v is definitely NOT Power; although it might seem similar.) Force F=mom/t (=p/t) is the (change in) momentum mom=m*v per time t. Thank you for asking. (I thought NOBODY would dare, (for at least 3 days).) James Watt's (definition of) Power is (the rate of doing work), where he defined Work(_Energy) WE=Wt*d as lifting weight Wt (=m*g, Force F=m*a, let (linear_)acceleration a=g) to a specific height (h=d distance). (We call that (kind of work(_energy)) Potential_Energy PE=m*g*h.) That (work_(energy)) done (with)in a specific amount of time t is (his) Power P=WE/t. Converting work_energy WE=KE to (moving) kinetic_energy, we get KE=m*v*va, where with the speed(_difference) v=vf-vi & the average_speed va=d/t=h/t that gives KE=m*(vf-vi)*(vf+vi)/2, or KE=m*(vf2-vi2)/2. Disclaimer: (To me) it looks like somebody goofed on (producing, e.g. creating) James Watt's (formula) syntax which needs "average" ((for the) speed). I suspect most people missed that (detail). (Easily found with (simple) algebra.) I hope that answers your question. Btw. Even, the (Ewert's 1996, universal) conservation of mass*Energy m*E=mom*moma m*E=(m*v)*(m*va) m*E=m*m*v*va which correctly proportions the mass_squared m2 with (respect to) the (single, non_squared) height h in m*PE, (indirectly denying ( https://en.wikipedia.org/wiki/Julius_von_Mayer Julius Robert von Mayer (25 November 1814 – 20 March 1878) (remarkable) 1841 conservation of Energy, as a(n absurd) bunder=Fake, due to lack of mass); (but) produces similar (confirming) results m*PE=m*KE m*m*g*h=m*m*v*va. (Why we are (supposedly) allowed to cancel mass m & (then, attempt to) maintain the (mass's, (fallen)) height h is a riddle to me. (..because..) It'( i)s physically wrong. Meaning it (=cancelling mass) can fail (the proportioning relation to height)! But does NOT always. Of course (naturally) everybody has heard, of, conservation of mass. ?) Noether's Theory is NOT going to help you, if you CAN'T even get the basics right=correct. It's (=Noether's theorem: which uses distance (e.g. (Work_)Energy WE=F*d, e.g. PE=m*g*h); instead of (per) time (e.g. (average_)momentum moma=m*d/t=m*va) is) based (most probably) on NOTHING (but (maybe (crumbling, unreliable)) NONSENSE). ? But more interesting might be where that (Noether theorem) went wrong, (when the simplest of algebra can prove an error that, that theorem did NOT) e.g. why you guys & gals put all your eggs (e.g. hope(ful assumptions)) in that 1 basket. (Why do things simply; when you can do them (more) complicated? (..so NOBODY can see (thru) the ERRORS)? NOBODY is PERFECT. NOT even me. Why NOT strive for error reducing methods, instead?) E.g. (Correct is, that:) Fallen height distance is (=must be (made)) wrt mass_squared m2; NOT mass m(1) ONLY. https://en.wikipedia.org/wiki/Conservation_of_energy Main article: Noether's theorem Emmy Noether (1882-1935) was an influential mathematician known for her groundbreaking contributions to abstract algebra and theoretical physics. The conservation of energy is a common feature in many physical theories. From a mathematical point of view it is understood as a consequence of Noether's theorem, developed by Emmy Noether in 1915 and first published in 1918. The theorem states that every continuous symmetry of a physical theory has an associated conserved quantity; if the theory's symmetry is time invariance then the conserved quantity is called "energy". The energy conservation law is a consequence of the shift symmetry of time; energy conservation is implied by the empirical fact that the laws of physics do not change with time itself. Philosophically this can be stated as "nothing depends on time per se". In other words, if the physical system is invariant under the continuous symmetry of time translation then its energy (which is the canonical conjugate quantity to time) is conserved. Conversely, systems that are not invariant under shifts in time (e.g. systems with time-dependent potential energy) do not exhibit conservation of energy – unless we consider them to exchange energy with another, an external system so that the theory of the enlarged system becomes time-invariant again. Conservation of energy for finite systems is valid in physical theories such as special relativity and quantum theory (including QED) in the flat space-time. If we take again P=F*va, swap sides F*va=P, /F va=P/F, va=d/t P/F=d/t, *t*F both sides produce energy P*t=F*d, P*t=E & F=mom/t E=(mom/t)*d, rearrange E=mom*d/t, va=d/t E=mom*va, mom=m*v E=m*v*va, is the Kinetic Energy KE=m*v*va. We do NOT need Noether's theorem to show the "connection" between Force (F=P/va) & Power (P=F*va) or distance (d=t*P/F) versus time (t=d*F/P) as average_speed va=d/t (=P/F). https://en.wikipedia.org/wiki/Julius_von_Mayer It (German knighthood) for caloric, Mayer was the first person to state the law of the conservation of energy, one of the most fundamental tenets of modern day physics. The law of the conservation of energy states that the total mechanical energy of a system remains constant in any isolated system of objects that interact with each other only by way of forces that are conservative. Mayer's first attempt at stating the conservation of energy was a paper he sent to Johann Christian Poggendorff's Annalen der Physik, in which he postulated a conservation of force (Erhaltungssatz der Kraft). However, owing to Mayer's lack of advanced training in physics, it contained some fundamental mistakes and was not published. ..(Mayer) examined experimentally; for example, if kinetic energy transforms into heat energy, water should be warmed by vibration Since he (=Mayer) was not taken seriously at the time, his achievements were overlooked and credit was given to James Joule. Mayer almost committed suicide after he discovered this fact. He spent some time in mental institutions to recover from this and the loss of some of his children. Several of his papers were published due to the advanced nature of the physics and chemistry. He was awarded an honorary doctorate in 1859 by the philosophical faculty at the University of Tübingen. His overlooked work was revived in 1862 by fellow physicist John Tyndall in a lecture at the London Royal Institution. In July 1867 Mayer published "Die Mechanik der Wärme." This publication dealt with the mechanics of heat and its motion. On 5 November 1867 Mayer was awarded personal nobility by the Kingdom of Württemberg (von Mayer) which is the German equivalent of a British knighthood.
16. Isn't (height h=) distance d divided by time t an average_speed va=d/t?
17. But (if you want) you could create a(ny) new "thing" unit, (depending on how you wanted it to be(come)). I suppose if you divided your stew into 3 equal portions, then it would be 3*[portions] (of that stew). I.e. That'( i)s wrt (only) that stew itself (with only 3 ingredients & their own proportions, on the spot, so to speak) NOT anything else. E.g. NOT necessarily 3*[bowls or cups etc]. The new unit [(equally_divided)_portion] can be found by converting all 3 units to any same desired_unit & then dividing by 3. That would be your (new) "anything" (unit). E.g. Some other (irrational?) factored unit of your (common) desired_unit. It'( i)s only a conversion (method).
18. Pronounced: "six foot, 3 (inches)". (Also, please notice the singular (1st) unit (foot, NOT feet) although the number (6), is >1; until we (might) get stuck in details by stating the 2nd (number's) unit(s (as more than 1, e.g. 3)).) That looks like 2 answers, stuck (=connected) together. E.g. 6'+3". Where the (empty) space (between them (both)) represents a virtual "plus", + (symbol). But why is there NO (empty) space between the (number) 6 & (unit) '=[feet] or [foot] (symbol); & between the (number) 3 & (unit) "=[inch] (symbol). The SI convention seems to use the empty space between number & units (Notice the s (on units) for both: singular; or plural or more) as a virtual multiply=multiplication. E.g. 6 [m]+3 [m]=9 [m], represents 6*[m]+3*[m]=9*[m]. There is also a subtle difference between infinitive, e.g. stone (in a quarry, to build a castle) versus more_than_1 e.g. plural or more=many, e.g. (made of) stones. E.g. Should we say, the building is 5 [meter] high? (infinitive). E.g. NOT 5 [meters]. (Plural or more). We often say 20°C (twenty degrees Centigrade, please notice: NO (empty) space at the °). But 300 K (three_hundred Kelvin (infinitive); NOT Kelvins (many).) Btw When I look at this (= all that (grammar) needed for the math) I am considering artificial_intelligence (e.g. program(ming)) too. (=NOT two, nor plur(e)al (~for crying out load, at reality).) & (I am) considering what sort of math (algebra) would be needed to incorporate such mixed units (singular or more). Algebra is perfect equality (e.g. balance); but NOT so with grammar that is brought in to distinguish finer differences. Our brains recognize those discontinuities (=differences).
19. Did anyone suggest that they might be? Yes, John. Added (verb) & addendum (noun) are (probably) NOT exactly the same; but it (=the added_onto method) suggests (to me), (that its meaning is) going in that (similar) direction. E.g. A hang_on, (that can be) added on(to almost anything).
20. 3?! I've only got 2 [feet]! & no (way, back)[yard]. (in the back :-). (Again, & again. So I guess there is a (small?) language problem (obstacle, of incompatibility) with the grammar('s singular versus plural "s", etc); versus the math('s algebra); & I suppose, the units' acronyms(' short_forms) help us (out) there, (at least) a bit, by ignoring the plural(s). E.g. 5 * meters = 5 [m]. Or unit meter(s)=[m].
21. How How (then) did you think about it (=their connection, relation to each other) (if NOT (as) multiplication of number & unit)?
22. Are number(_value)s multiplied by the unit? (Surely NOT added.) 1 * meter=1 [m].
23. Please explain. vi included: is the general equation. But without vi: is a limited (specific example) formula, that can NOT work for all cases. What I mean is: (vi was missing from the "general" equation, (NOT (just a) formula.) How do I know when I have the general equation, (if or when I start with only (limited (specific example) formula) fragments)? I'm searching for the general formula. I (attempt(ed) to) maintain vi (initial_speed) to try to NOT get lost ((&) for other things (=projects, concepts, extrapolations)). (You (may) think) you do NOT need it (=vi), (?) fine(!); but I do. E.g. vi remains an invisible (hidden (excluded)) term for you(r) speed(_difference) v=vf-(vi). It's always there.
24. I'll do my best! Thanks. (I got a good chuckle (out of that).) Yes! That is my intention. You obviously have NOT read this thread's file. It's NOT complicated! It's easy. Please explain. vi included: is the general equation. But without vi: is a limited (specific example) formula, that can NOT work for all cases. Locked threads DON'T allow quoting. Maybe (other_options) sharing can help linking, a bit.?