Capiert

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?

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).

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.)

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).

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)!

(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.

Isn't (height h=) distance d divided by time t an average_speed va=d/t?

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).

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).

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).

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].

How How (then) did you think about it (=their connection, relation to each other) (if NOT (as) multiplication of number & unit)?

Are number(_value)s multiplied by the unit? (Surely NOT added.) 1 * meter=1 [m].

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.

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.?

Swansont: (Kinetic_Energy is already relative, 2020 11 23 16:26) What valid physics principle is this based on? Capiert: (Kinetic_Energy is already relative, 2020 11 23 16:28) Mechanics: Gravity's (free_fall, linear_acceleration a=g) fallen_height h=hf-hi=vi*t+g*t*t/2 for time t & initial_speed vi gives final_speed vf=((vi^2)+g*h*2)^0.5 Please explain: What value do you mean; & why? Maybe you missed something (important)? If you mean g=2*h/(t^2)-2*vi/t then please read (my files, in this thread) further. You do NOT sound completely connected. That is quite a natural reaction. But you have NOT made a suggestion how I can fix your sfn strikethrough software bug. No response to my question (request) as neither: yes; NOR NO. =NO! Makes it (=my request) sound (=seem) rhetoric. Yes, we (all) know that. So (I guess) I am (1 of) the 1st! to present the gravitational_acceleration (formula) g=2*h/(t^2)-2*vi/t as exactly so (=in its corrected form). P.S. (But) I can NOT believe I am the ONLY 1 who has ever done that, or tried. You (only) confirm that I can NOT find (exactly) this ("g=2*h/(t^2)-2*vi/t") in books. I have NOT seen that term (-2*vi/t, exactly in that position) in books. (This is a speculation thread thus I want to present (either) something new, or an improvement against a flaw or weakness.) My complaint is (from), Capiert: (Kinetic_Energy is already relative, 2020 11 23 16:28) How do we know that acceleration a=x/(t^2)? Better said: If I let (distance) x=h (height), instead; then: How do we know that acceleration a is (suppose to be) h/(t^2)? My answer: We do NOT know that (at all)! I do NOT get that formula, that you claim. (It is NOT possible (for me).!) Especially when the height h=d distance NEEDS a factor "2". g=2*h/(t^2)-2*vi/t". Anyone who uses your basis will get g#h/(t^2) h#g*(t^2). But a~2*x/(t^2) is the minimum requirement. E.g. x~a*(t^2)/2. & where is the vi*t term in that? NOWHERE! So (thus) it is NOT the (EXACT) general (equality) formula of "standard fare in any physics textbook.." NOR "an equivalent equation", thereof; BUT instead a distortable (mere) approximation. (I can NOT rely on it for (my) other derivations.) You may be happy with such carelessness; but I can NOT be. It (=That formula: whether e.g. a#x/(t^2), or e.g. x~a*(t^2)/2) is either an equation (=equality); or (else) it is NOT. & there it is definitely NOT (an equation). Swansont: (Kinetic_Energy is already relative, 2020 11 23 16:26) If I drop a mass from some height, how fast will it be moving when it hits the floor? My answer was, Capiert: (Kinetic_Energy is already relative, 2020 11 23 16:28) the final_speed is vf=((vi^2)+g*h*2)^0.5 . But you did NOT like that (answer), perhaps because its result is independent of mass m? (It (=That answer) does NOT NEED mass.) (But it is also new to me that algebra is NOT suppose to be math, if it is? You closed the thread. Perhaps you wanted some number( value)s to assist (confirming) your understanding of something you can easily do yourself; but doubted my ability. ? E.g. Mistakes happen. Typos.) I was only trying to answer your question, (preparing) while you closed. That (=your impatience) does NOT seem fair. (Especially when you said there is NO time_limit.) (Mordred also stated to (someone) NOT to answer if you do NOT know. I also wanted to check some details before I answered you. Thus my hesitation=delay. There are still (some) things I need to clear. However:) For that equation vf=((vi^2)+g*h*2)^0.5 we typically set the initial_speed vi=0 [m/s] if (=when) we only drop, g=9.8 [m/(s^2)], & e.g. let the height h=1 [m] vf=((0^2)+9.8 [m/(s^2)]*1 [m]*2)^0.5 vf=4.4 [m/s]. E.g. let the height h=2 [m] vf=((vi^2)+g*h*2)^0.5 vf=((0^2)+9.8 [m/(s^2)]*2 [m]*2)^0.5 vf=6.3 [m/s].

That surprises me because I can read the following (from above) The fallen height (taken from a graph), h=vi*t+g*(t^2)/2 is only 2 terms: the constant (initial_)speed term vi*t; & the accelerating term g*t*t/2. & also in my posted (above, 2nd of 2) .pdf's (above). https://www.scienceforums.net/applications/core/interface/file/attachment.php?id=23252 Surely you will be able to determine when that (equation) was posted. Good! (for that product). But now for the quotient -2*vi/t in g (alone). Someone to assist me in getting the WRONGLY strikedthrough text (above) to be NOT strikedthrough, anymore. Would you help? But to answer your question: Again, clearly -2*vi/t was NOT present in g (before my threads). I have NOT seen that term in books; but algebraically it is correct (as to how I gave it to you). Science should confirm itself, but it seems (to me) you (might) probably prefer NOT that the g equation be intact, when rearranged (e.g. if you leave out that term -2*vi/t)? (Is that possible?) I have only moved the acceleration g to the left side, & all other terms to the right side. That should give the correct g (equation). (Otherwise, the (general) equality is either destroyed, or distorted.) It'(=What I have done i)s NOT complicated.

From g. You obviously disagree, otherwise you would NOT have asked. Why NOT? It works. (As far as I know) I have shared the equation: the fallen height (taken from a graph), h=vi*t+g*(t^2)/2. Typical is NOT all cases, but instead most (e.g. a majority). For me it was an unknown that I wanted to find, e.g. experimentally. Can someone please help me with unstrikethrough my text (above)?

STOP! Under Construction Please wait! g=2*h/(t^2)-2*vi/t. It looks (to me) like that term -2*vi/t has been missing for a long time. Surely you ((should) already) know it; but do you use it? For (fallen) height h time(_difference) t initial_speed vi. The free_fall acceleration g~-(Pi^2) [m/(s^2)]+ac g=~-9.8 [m/(s^2)] is (instantaneous,) linear, & negative; & (please bear with me) (the bothersome details:) (=but it) (is) slightly) reduced by the (Earth's daily_rotational) centrifugal_acceleration ac=(vc^2)/r having the (squared) Earth's (surface) circumferential_speed vc=cir/T with (the Earth's) circumference cir=2*Pi*r using ((your) Earth's) per radius r (position) & (sidereal_)day (time_)period T~23 56 [min] 4 [sec] in seconds; (instead of 24 in seconds). --- Disclaimer: That's about all there is to it (=g, overview), when (also) observing your location & position. (Air friction would have to be considered; but I'm NOT going to bother for in(side) a vacuum.) E.g. Naturally (all) those (extra) details (for the centrifugal_acceleration ac, etc.) can be determined. However, I'( a)m concerned (here) with (mostly only) the initial_speed's vi term. (E.g. A term that would (also) occur (for you) in the ((total) work_)energy WE=F*d if you ((were to) also) included the (integration) limits from zero to the initial_speed; instead of exclude it (=initial_speed, as a different reference (frame, choice)). (Which can be determined, graphically with a plot: speed versus time. The (plot's) area is the distance.) It's easier (for me) to use positive instead of negative (plot) values. --- Freefall: If I let a pebble (stone or ball) fall (in a vacuum) (from rest, (with) initial_speed vi=0 [m/s], at time t0=0) --- CAUTION: From here on this website's software has WRONGLY Strikedthrough the rest of my text, automatically. Please see the files until someone helps me undo the Strikedthough. Please would someone help me fix this? Marking & reclicking Strikethrugh does NOT undo Strikethrough like it would for Bold, italics or underline. What is wrong with your Software? --- then in (time(_difference) t1=)1 second it will (have) fall(en) the height(_difference) h1=-4.9 [m] & have the final_speed vf1=-9.8 [m/s]. (I could also reverse that & say: If I let a stone (pebble) fall (from a height h1=)4.9 [m] (wrt the ground), -4.9 [m] (to the ground), then it will hit the ground in (time t1=)1 second & impact (ruffly) at a (final_) speed vf=-9.8 [m/s]. We know freefall is "linear" acceleration, so in time(_difference) t2=2 fallen_height h2=-9.8 [m] & final_speed vf2=-19.6 [m/s]. etc. E.g. Instead, we could have used the 1st final_speed (vf1) (from the 1st (time_(difference)) second (t1)) as the 2nd initial_speed vi2=-9.8 [m/s] (e.g. as if thrown) & allow the pebble to restart (accelerating) as if from zero in initial_speed vi1=0 [m/s], but for: (only) the 2nd (time_(difference) t(2-1)=1) second; its (=the pebble's) additional fallen_height h(2-1)=-4.9 [m]; & additional final_speed vf(2-1)=-9.8 [m/s]. So, the (total) fallen_height h2=h1+h(2-1)=-4.9 [m]+(-4.9 [m])=-9.8 [m]; & the (total) final_speed vf2=vf1+vf(2-1)=-9.8 [m/s]+(-9.8 [m/s])=-19.6 [m/s]. The fallen height (taken from a graph), h=vi*t+g*(t^2)/2 is only 2 terms: the constant (initial_)speed term vi*t; & the accelerating term g*t*t/2. The speed(_difference) v=g*t is the (freefall) acceleration g multiplied by time t. But the accelerating term g*t*t/2=(v/2)*t is "half" of that speed(_difference) v(=vf-vi) multiplied by time t h=vi*t+(v/2)*t h=(vi+v/2)*t, & va=(vi+v/2) is the average_speed h=va*t. The average_speed va=h/t is the fallen height h(=d distance) per time(_difference) t. (The final_speed vf=vi+v vf=((vi^2)+g*h*2)^0.5 ) The (linear) average_acceleration ga=h/(t^2). <----That'( i)s Important! So simple, NO other terms! ga=va/t ga= The freefall gravitational "average_(linear)_acceleration", (&) is ga=vi/t+g/2, swap sides vi/t+g/2=ga, -vi/t g/2=ga-vi/t, *2 then the (instantaneous) linear freefall gravitational acceleration, is g=2*(ga-vi/t). Note: Syntax (my) ga=g_a (yours). Those are the conversions to & from linear instantaneous versus average: speeds; & accelerations. Again: The average_speed va=h/t va=vi+(v/2). The (linear) average_acceleration ga=h/(t*t)=va/t ga=vi/t+g/2. The (average_)time t=h/va is the height h per average_speed va; or t=(h/ga)^0.5 the rooted: height h; per (linear) average_acceleration ga. It'( i)s that simple. (Algebra.) 2021_05_24_2011_ Linear_acceleration's_Average_speed__diagram__2021 05 24 2038 PS Wi.pdf 2021_05_24_1702_Gravity g's, missing term_2021 05 24 2130 PS Wi.pdf

I received the telegraph (message) instructions: "go east 3 strides, then go 4 strides north. Where are you (now)?" Answer: 5 strides east. Signed Virtual.

I sympathize with you. I've been wracking my brains on this stuff for years trying to make some sense of it. I guess you missed the point Sensei. My (perspective_reversal) calculations show that the Rest_mass is inherent, (meaning it'( i)s) (already) in the KE('s reversed perspective). I don't need to add it (rest_mass) randomly because somebody thought it was forgotten & needed. That's right, mass is conserved, conservation of mass com. The mass does NOT change (in my calculations), only the speeds (change). & I haven't done (=changed) anything; (except) only the perspective. & I get (Fitzgerald)_Lorentz_contraction similar results with (only) algebra. So I have to ask: Why do you think you (might) have to add rest_mass additionally (extra) ((randomly) out of the blue ((or a hat) (like a magician)))? (That's ridiculous.) Why should the rest_mass be in your relativistic equations "twice"? E.g. Originally (inherently); & then again (randomly, by you) because you thought you missed (=forgot) it, before. I DON'T need to add rest_mass to my KE calculations because it's already there. You however, think you do (need to do that). The rest_frame is when the initial_speed vi=0 [m/s]. I'( have) a bundle of them some read several times. But they DON'T solve my problems (=paradoxes). They (=those books) only "help" me solve them. Those answers are NOT in the books. In fact the books are (sometimes) misleading. Too much NONSENSE makes my brain shut off. (I CAN'T tolerate it.) So I have to take my time & unravel the puzzle. I have to try (new) alternatives; NOT (old) failures. All your textbooks only leed to dark_energy (=NONSENSE, errorful calculations). I'm sorry but I think you are behind the times judging me so. I'm searching for a recalibration to bring physics up to date, instead of the scattered mess it is in now. You guys (your team) is either going to help me, or not.

Sorry, but I only live here (on Earth); I'm going to no other planet. So Earth was my best example. va is relative to the initial_speed vi which for this example is the earth's(_speed, which could be anything (reasonable)). So if we are traveling at the same speed as the Earth then that initial_speed vi=0 is zero for at rest wrt on Earth. I agree. I only used it (=Earth) as an (easy, simple) example to "try" & get the message across. The Work_energy's element F multiplied by the distance element dx. How do we know that acceleration a=x/(t^2)? If the momentum mom=m*v & mom^2=(m*v)^2. I DON'T see the coherence. WE=F*d. I would prefer to say the mass*Energy m*E is conserved, instead. NOT the energy. Capiert said: Which should be KE=m*(((v^2)/2)+v*vi), instead. Algebra. The average_speed is va=(vi+vf)/2. *2 2*va=vi+vf, swap sides vi+vf=2*va, -vi vf=2*va-vi. The speed_difference is v=vf-vi, swap sides vf-vi=v, +vi vf=v+vi. Both vf's can be equated also vf=vf v+vi=2*va-vi, -vi v=2*va-2*vi v=2*(va-vi), /2 2*v=(va-vi), +vi 2*v+vi=va, swap sides va=2*v+vi. Or expanding the kinetic_energy KE=m*((vf^2)-(vi^2))/2, ((vf^2)-(vi^2))=(vf-vi)*(vf+vi) KE=m*(vf-vi)*((vf+vi)/2), (v=vf-vi & thus) vf=v+vi (Notice: KE=m*v*va, for v=(vf-vi) & va=(vi+vf)/2 but continue from above for below) KE=m*(v+vi-vi)*((v+vi+vi)/2), vi-vi=0 & vi+vi=vi*2 KE=m*(v)*((v+vi*2)/2), KE=m*(v*v+v*vi*2)/2, expand KE=m*((v*v/2)+(v*vi)). I derived (=equated) that above (for you also further above) via substitution, & I see no error in my equations, thus I must conclude they are correct. They are simple, algebra. Mechanics: Gravity's (free_fall, linear_acceleration a=g) fallen_height h=hf-hi=vi*t+g*t*t/2 for time t & initial_speed vi gives final_speed vf=((vi^2)+g*h*2)^0.5 Capiert said: Does a (virtual) moving_frame need (to have) mass? I'( wi)ll assume, (your answer is) no. (=constant_speed) Sorry, my mistake I used a (virtual, constant_speeed=inertial) perspective, instead of non_inertial=accelerating frame. Thanks for clearing that. Capiert said: Galilean_relativity NEVER worked (right) because it did NOT limit speed to a maximum. So here, in this example of (simple) classical_relativity I have limited the speed (maximum) to c. It looks questionable (=doubtful) to me. The author tried to get rid of it (e.g. avoid it) in 1920 chapter 22. He didn't recommend it (anymore); NOR did the Nobel committee give him a prize for Relativity. (Why would they NOT if it were really important for Physics?) I guess they also had their doubts. Capiert said: I see no problems & get astounding results (for constant_speeds). It's values are different from yours by a factor of -1/2. How do you know that? How do you know if SR is correct? E.g. When its author discouraged its usage, 1920 chapter 22. Capiert said: What do you mean by "inertial"_frame? Capiert said: 1_stone (1905) calculated the photons mass m=KE/(c^2). What do you mean by that? Newton believed photons were particles. That means small (=less) mass; NOT NO_mass. That'( energy i)s what he used for a photon's mass in 1905. What do you mean by "No"? Are you trying to say E#m*(c^2). Mass is NOT (a kind of) energy? To me it seems rather obvious what things are, with an equation (=equality). I do NOT understand why you shirk (~avoid) from using math for physics, to get to the bottom of things. What prevents the (mass versus energy) connection? Light's_speed squared c^2? I.e. Mass & energy are NOT identical. If you are implying (=saying) NO_mass, then how do you know that? Or are you implying photos are (sort of like) energy, but not yet mass? How can you possibly have energy, without mass? Mass is only a construct, a coefficient (=factor). Capiert said: =? (light? or speed?) If you mean speed, I was only using the earth's speed as an (easy) example. Sorry. I suspect yes. & I can convert to your SR answers with the factor -2; & visa versa from SR values to mine with the factor -1/2. Answer: the final_speed is vf=((vi^2)+g*h*2)^0.5 .

If v=vf-vi & KE=m*((vf^2)-(vi^2))/2 & ((vf^2)-(vi^2))/2=(vf-vi)*(vf+vi)/2, & va=(vi+vf))/2 then why is NOT ((vf^2)-(vi^2))/2=v*va? & How can KE possibly be m*v*v/2? When should be KE=m*(((v^2)/2)+v*vi), instead. Does a (virtual) moving_frame need (to have) mass? I did NOT declare an inertial frame, I used a (virtual, non_inertial) perspective. (I do NOT have to touch a(ny) thing.) I simply stated a speed_difference u=c-v between the maximum possible speed c & the moving_point's speed v. u is only the (complementary_) speed needed to add to v (in order) to be(come) c. That is simple algebra. No magic. Galilean_relativity NEVER worked (right) because it did NOT limit speed to a maximum. So here, in this example of (simple) classical_relativity I have limited the speed (maximum) to c. I see no problems & get astounding results (for constant_speeds). It's values are different from yours by a factor of -1/2. What do you mean by "inertial"_frame? 1_stone (1905) calculated the photons mass m=KE/(c^2). =? (light? or speed?) If you mean speed, I was only using the earth's speed as an example. Considering my results, I don't (believe) I need your SR form. Perhaps you can convince me otherwise? What I wanted to say is: Please notice (If I let) c=v+u Why NOT? I can't comment your deletions. My point is "virtual". Standard definition: no size. & moving (at constant speed). The (so_called) Rest_mass is inherent (=included) in (my) KE's opposite_perspective (-KE'); & (thus it) does NOT have to be added extra (by haphazard random guessing).

wrt the initial_speed vi; & relativistic if you (only) swap (=reverse) its perspective from wrt earth's minimum_speed 0 to wrt earth's maximum_speed c. Reversing the Kinetic_energy (perspective) KE=m*v*va (wrt Earth’s_speed), *(-1)' gives -KE'=m'*(-)*v'*va' (wrt light’s_speed) which is already relativistic. Please notice if I let c=v+u, u=c-v, u=-v', (prime_symbol ‘ is wrt light’s_speed) then the (negative) speed_difference (wrt light’s_speed), is -v'=-(vf'-vi')=vi'-vf' & the average_(accelerated)_speed (wrt light’s_speed), is va'=(vi'+vf')/2 for initial_speed vi' (=-0'=v_min, wrt light’s_speed, or v_max=c wrt Earth’s_speed) & final_speed vf' (wrt light’s_speed). -KE'=m'*(vi'-vf')*((vi'+vf')/2), combine brackets -KE'=m'*((vi'^2)-(vf'^2))/2, let vi’=-0’=c -KE'=m'*((vi'^2)-(vf'^2))/2, bring c^2 out from the brackets (c^2)/(c^2)=1/1=1 -KE'=m'*(c^2)*(1-(vf'^2)/(c^2))/2, let gamma’^2=(1-(vf'^2)/(c^2)) -KE'=m'*(c^2)*gamma’*gamma’/2. That equation has (=contains) "half" of (DePretto's 1903, vis_viva), Energy E=m*(c^2) & 2 (Fitzgerald_Lorentz, relativistic_contraction similar) coefficients (named) gamma_primed gamma’=(1-(vf’^2)/(c^2))^0.5 where the rest_mass m=m' is (the same, =identical) constant(_variable) for either perspective. I.e. Conservation of mass com (wrt Earth’s_speed) m=m' (wrt light’s_speed). Speeds are the variables (instead). 2020_11_23_1010_KE_is_already_relative__Preliminary__2020 11 23 1419 PS Wi_(stripped).pdf 2020_11_23_1010_KE_is_already_relative__Preliminary__2020 11 23 1356 PS Wi__with old _Excel_ formula text_(mix).pdf

I'm glad you see it too. Thanks.