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Everything posted by Capiert

  1. 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.
  2. 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.?
  3. 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].
  4. 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.
  5. 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)?
  6. 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
  7. Capiert


    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.
  8. 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.
  9. 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 .
  10. 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).
  11. 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
  12. I'm glad you see it too. Thanks.
  13. It does NOT have to be a car. Surely simpler contraptions exist to test. Confirmation. Yes (& reliability). I'm interested in at least 1 formula I could rely on. From there I could adapt & fine_tune to a general formula if possible. How many things did Newton (have to) go thru til he could (finally) settle on F=m*a? "The proof is in the pudding." I'm surprised this non_linear acceleration (formula, stuff) is so rare (for something (that is) (supposed to be) so universal). Do you tell students to learn (useless) linear_acceleration in your universities because it's NOT universal? (Surely) I hope NOT. But that's what's taught. Free_fall's ("linear" acceleration) is pretty common throughout the whole universe. So linear acceleration is NOT so uncommon. Even if you don't dare to call it universal. I'm just curious about how non_linear (acceleration) is proportioned to linear_acceleration, & the methods of confirmation. Is my curiousity justified, in a dark energy dilemma?
  14. I want a reliable formula that I can equate to for comparisons. I can make up anything I want, but it doesn't mean it's accurate. Your team surely has better experience there, for real measurements. (I'm just guessing.)
  15. (Yes) but please give me a formula example so I can get feel for the number_values.
  16. Then I suppose your syntax v is my syntax vf. Is that true? Please elaborate, if not. Just attempting to be thorough(ly defined), which is NOT what I could claim everybody does. Capiert: I thought that (linear acceleration) was understood, that everybody knows that. It's clearly stated in the 3rd line (at the intro). Surely you have overseen my definition.? I gave you a specific example: the linear acceleration. But now you have made me curious. Please tell me more, e.g. an example (or more), for further (kinds of acceleration).
  17. Hi Swansont I'm sorry but I can't follow you there. Isn't speed (always) relative (to another speed), thus a difference (in speed)? Or are you insisting on the (non_linear, complicated) Fitzgerald_Lorentz transform (e.g. to watch cars' speed, v=100 m/s (wrt) on earth))? E.g. 2 cars each with the same speed as the other (wrt earth), appear at rest (=0 speed) (wrt each other (car)). Or are you suggesting I should multiply the basic (speed) unit [1 m/s] by the number (value, e.g. 100) as a product (instead of (subtract for a) difference). Which I don't find such a bad idea, to get rid of (some of) the math problems (complexity). I'd just have to get used to it (=the multiplication technique), instead. I don't (really) want complexity, I want simplicity (& solutions), instead (but NOT at the cost of precision, as (severe) errors). I thought that was understood, that everybody knows that. But I guess some things must be said anyway (just to be sure). The equation is simply derived from the (linear) 1st order of acceleration. High school physics. (Non_linear acceleration does make me curious, though.) I doubt that I would stubble into as many (of your physics) pitfalls if I used identical syntax. My alternative syntax allows me to compare your results. But (I suspect) you don't recognize the drawbacks. I don't prefer the (foreign) greek alphabet. "It's all greek to me!?" (Not English.) (The cool thing about physics is how it simplifies, but the nasty thing is how it encrypts (into only 1 possibility, when others are possible). Physics can be done in any language, with or without another language. So those languages, or rules, conventions are NOT physics. They are something else instead. I think the biggest confusion you physicists must (eventually) deal with is (dark) energy because it is NOT Newtonian. You ignore the initial_speed vi too much, setting it to zero. Relativity clearly indicates motion wrt other frames. Could it be an (invisible) common initial_speed vi=(KE/mom)-(v/2) (which you deny) is at least partly responsible for your (unknown) dark energy (ERROR)? (Not to mention speed's (incorrect) exponential (=non_linear) proportionality to mass. Something KE (definitely) defies. How can you possibly brush_off that (severe) incompatibility so lightly? I thought you were reasonable people. To do things right you would need mass_squared (instead of only mass), in energy equations. But that would be momentum_squared, instead of your (errorful) energy=non_sense.) I don't know what else to call it (energy as error) because it (dark energy) makes no_sense according to your (standard) college educations. Or does it? Astronomers are complaining to you on a cosmic scale, while little old me attempts to deal with (your energy math (incompatibility) problems) on a small earthly scale.
  18. Period versus 1/(vc), frequency versus vc ? I find it very confusing that rotation_speed vc=2*Pi*r/t looks (very similar, &) proportional to frequency f (cycles per second; or (1) cycle (e.g. circle) per time (in units of seconds, or less than a second)); & inverted_rotation_speed 1/(vc)=t/(2*Pi*r) (the time "per" cycle_or_circle_(circumference)) looks proportional to Period T=1/f which is inverted_frequency! (Thus units are suppose to be inverted.) So (confusing) that I must start with what we know. In fact I am so confused, that I know nothing except the rotation_speed vc=2*Pi*r/t (which is obvious), & the frequency f=1/T inverse_period relation (definition). The rest seems (to me) like non_sense! Thus I can derive everything (I need) from what we know: The circumferential_speed vc=cir(cle)/t(ime) vc=2*Pi*r/t. ~f. That looks like cycles (=circles) "per" second =[cps]=[c/s]. (Units in square brackets.) I.e. That looks (very much) like frequency! (Although you guys are only interested in the inverse(=1_divided_by)_time 1/t for the "number" value of f (without units, =excluding units [c/s]). We also know the "inverse" circumferential_speed 1/(vc)=t(ime)/cir(cle) 1/(vc)=t/(2*Pi*r). ~T. That looks like the time_(in seconds)_"per"_cycle(=circle) =[spc]=[s/c]. I.e. That looks (very much) like the Period! (Although you guys are only interested in time t for the "number" value of T (without units, =excluding units [s/c]), because you have defined frequency f=1/T as inverse_period. So the time ("for" a cycle) is t=(2*Pi*r)/(vc) time=(circle (or cycle)) "per" rotation_speed). That is the (number) value for what you guys call period (T). Your units are seconds (because all circles cancel). It's inverse 1/t=vc/(2*Pi*r) inverted_time=rotation_speed "per" (1) circle (or cycle), gives the number (value) for what you (guys & gals) call frequency (f). In other words, stripped of the cycle(s). You now have units [Hz] Herz (=hurts! Ouch!)
  19. study & discuss their (great?) work, energy WE=m*a*d using linear_acceleration a (=v/t, =2*((vi/t)+(d/(t*t)), =F/m) & the force(d) distance d (=va*t) of a mass m (=F/a). That (linear accelerated) force is F=m*a, where the average (accelerated) speed (velocity) va (=(vi+vf)/2), & speed (velocity, difference) is v=vf-vi, for final_speed vf, minus initial_speed vi. ((Even) although it would also be possible to use factoring (an initial unit_speed of 1 m/s), instead). The (moving) kinetic_energy is KE=m*v*va, (pronounced key), or KE=m*((v^2)/2)+v*vi) using initial_speed vi, or KE=m*((v^2)/2)-v*vf) with (respect to) final_speed vf. The potential_energy PE=m*g*h (pronounced pea, as in pee) is a mass m g accelerated (fallen, multiplied by) height h=d distance. PE=Wt*h is (the force F=) weight Wt (=m*g) multiplied by height h. (No distance fallen_able, is no potential_energy.) Equating (both energies) PE=KE (pronounced peek) m*g*h=m*v*va without mass (divided from both sides) is g*h=v*va is only linear_acceleration a*d=v*va a*d=(vf-vi)*((vi+vf)/2) of the distance d, a*d=((vf^2)-(vi^2))/2, *2 a*d*2=(vf^2)-(vi^2), (pronounced add too). That's standard mechanics. (Anything wrong there?) Swapped sides (vf^2)-(vi^2)=a*d*2, +(vi^2) (vf^2)=(vi^2)+a*d*2, ^0.5 vf=((vi^2)+a*d*2)^0.5 is the final_speed (velocity) (of linear_acceleration). The speed difference (velocity) v=vf-vi, vf=((a*d*2+(vi^2))^0.5) v=((a*d*2+(vi^2))^0.5)-vi is final_speed (velocity) vf, minus the initial_speed (velocity) vi. Momentum('s impulse) mom=m*v is mass m multiplied by (the accelerated) speed difference (velocity) v. mom=m*(((a*d*2+(vi^2))^0.5)-vi) mom=m*((a*d*2+(vi^2))^0.5)-m*vi is the final_momentum momf=m*vf (pronounced mumf, as in eating fast), minus the the initial_momentum momi=m*vi (pronounced mommy). The average momentum moma=m*va (pronounced mama) is the mass m multiplied by the average (accelerated) speed (velocity) va=(vi+va)/2. Any questions? Since KE & mom only use mass & speeds, & all energy can be equated to KE, then it seems imaginable to equate all energies into impulse(s) of mom=F*t. Mom=E/va. Mother nature (pronounce Eva (the German Eve); or else Elva from James Cameron's Avatar).
  20. I'm sorry, I did not make myself clear enough. I hope you do not make random answers when you answer. (=wide) (which they were) 'any' speed except c, but its true Michelson's sketch is an exageration. Was it v=c/10? I thought the point was (for) the M&M experiment of 1888, fig 2. Yes, so why aren't we (still) talking about the real experiment? Yes, any reasonable speed. Does that mean trial & error? (E.g. decide (=arbitrate) from random values.) E.g. It's (=the mirror has) got to be wide (=big) enough.? They were 5 cm wide (& made of metal). Quite the contrary, (I'm interested in specific details). If the mirror is made too small, wouldn't a small (& coherent enough) light beam get cutoff? E.g. Wouldn't a (vertically) 90 degree incident ray (either) partially (or never) hit a too small mirror b that (mirror b) is moving away (to the right) at v? How can you change 90 degree x,y,z orthogonality, by changing reference systems? E.g. Formula?
  21. I'm sorry that did not answer my questions, e.g. y/n's.
  22. In which case, wasn't that angle decided by the draftsperson (that was) sketching? (because in that case a new angle would be needed for every different speed v ?); or else if the upper mirror is long enough to be hit at 90 degrees, & reflection occurs in a wavefront, wouldn't the diagonal beam, where it's twice Michealson's (drafted) reflected angle, hit the semi_mirror in perfect sync with the horizontal path? i.e. identical delays for both (vertical & horizontal) paths? (I mean the (reflected) angle in question is <1 degree (difference, wrt incidence at 90 degrees) for the intended speed v (~c/1000). Surely a slur distortion, or front might account for that interception (fringe intensity) at the semi_mirror.?)
  23. "optimism" denied. Because I had not recognized you were impling WE=F*d is elastic; instead of stored momentum might be the (2*a*d)^0.5 part of the momentum mom=m*(((vi^2)+2*d*a)^0.5)-vi.? Sorry, (it's vague, general) typo, reminder. That comment is (awkwardly) in the wrong place. Is that a bit better? So momentum is the unlieing (basis) principle for collision.? For all collisions (non_elastic, partialy elastic, & totally elastic)? Which means we need to integrate then? But does that work well for (all) non_linear accelerations? I think it is zero wrt to my (immediate) surroundings (on earth), but I know I'm moving fast (eastwards) as the earth rotates thru the day. Yes, but our (so_called static) immediate surroundings are a deception. You know the earth is rotating, with no motion you are claiming identical speed, an inherent motion (i.e. speed) must exist (if we automatically imply (identical) speed). The real question is which (moving) reference to choose, to quantify (how fast). No motion is (truely) absurd. All things are moving (in the universe. Thus inherent motion exists.) Or isn't it? (I doubt that you can convince me otherwise. If you're clever enough you might. I don't know the outcome. But as the wording stands, it makes no sense otherwise (to me). & it's Newtonian.) Then nothing is moving.? That's not a conceptual failure, it's a fact; or else bad wording.? Everything in this universe moves although we can not say an independent speed wrt no reference. Wrt light we are moving at -c. I do not see an error. Einstein said there is no preferred reference, they all work well. You're right, they do not produce a difference, so that is the marvelous advantage when studying (=observing) collisions; until we get down to atoms & sub_atomic particles where the music changes (becoming significant). Then it is a non_elastic collision; or else a partially elastic collision. Yes. I did not convert after deriving (probably assuming (the momentum energy relation mom=E/va is so obvious) I could, later; & forgot that it (the existing work_energy formula) still was energy). It would matter for your challenge, if I had a momentum equivalent for your (physicist's) WE. Or wouldn't?
  24. I guess you mean work energy WE=F*d, so I'm stuck with (the fact) that the defomational distance d done by an applied force F is (defined by you physicists) as an energy concept (instead of a momentum concept). Would you please give me a hint what could go wrong? That might help me change my ways for the improvement. Thanks. I would hope so. I search for confirmation, from several aspects, even though I can not deal with all aspects, that are beyond me. The origin, of the big bang, seems to me the most relevant (for an underlieing inherent momentum). Other than that, the speed of light. Those 2, & the earth's are most relevant to me. I'll assume though, that the big bang had an origin (x,y,z=0,0,0) position. ? Everything moves in the universe, was my statement. Choosing the BBT origin, was my attempt at prescribing a simple reference, to describe that motion easily. e.g. generally. Quite right, & the discussion was about momentum storage (instead of KE=elastic). I.e. 1st, (so_called) static momentum must be established in order to discuss momentum storage. To do that we need motion, but if we see no motion then we must deny momentum. A terrible dilemma. But if we know everything moves anyway, (& static is only identical speed, wrt the (same) reference) then the problem is solved. What is the inherent speed of matter? Is it light's speed c? Ignorance is a continuing process. It repeatedly pops up. But you are right, we can approximate, for what is significant in most cases. Ok. But my question was (really more like) is the earth moving? I.e. apparent static (=at rest) is really moving. There is not (really) such a thing as not moving. The earth's rotation is 1 thing, but our speed thru the milkyway galaxy is another! & even there you will say that galatic_speed is not significant, till something happens. Ok. It's a simple explaination (analogy), to say it briefly. But they will deform. Or won't they (even though that deformation might be too small to measure)? Work energy WE=F*d. Now I recognize, that you have been using that WE formula, as your premise, because it is (previously) defined in physics (as your rights, to catalog so). I had no idea (how you were thinking) before that. Don't you mean KE=WE elastic, instead? E.g. KE=WE=KE'. Prime ' is for after the collision. Yes, it seems so. I did not recognize that the WE equation('s distance d) was (strictly, only in) an energy equation, after I had used it for my derivations. Mistakes happen. Please forgive. (Humble, humble). Momentum wise I would need (some kind of) a rooted_distance, instead. (Or am I still wrong, on that 1, too?)
  25. Then center of the big bang explosion as the reference instead. Well then I guess we can give up. You knew it (excluding your assumption of James Watt's horsepower history).
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