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  1. Capiert


    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).
  2. Capiert


    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).
  3. Capiert


    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).
  4. Capiert


    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].
  5. Capiert


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


    Are number(_value)s multiplied by the unit? (Surely NOT added.) 1 * meter=1 [m].
  7. 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.
  8. 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.?
  9. 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].
  10. 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.
  11. 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)?
  12. 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
  13. 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.
  14. 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.
  15. 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 .
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