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Capiert

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About Capiert

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    Classical physics
    (no calculus)
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    Classical physics

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  1. I'm glad you see it too. Thanks.
  2. 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?
  3. 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.)
  4. (Yes) but please give me a formula example so I can get feel for the number_values.
  5. 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).
  6. 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.
  7. 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!)
  8. 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).
  9. 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?
  10. I'm sorry that did not answer my questions, e.g. y/n's.
  11. 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.?)
  12. "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?
  13. 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?)
  14. 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).
  15. Your 1st sentence saying "It depends" gave an inkling of hope, that neither yes nor no were completely right. Thus I concluded a yes possibility also exists. I hope now you do. Ok. What I usually write as F=mom/t (so I can make the connection for myself, & don't forget.) F=m*v/t. Is my notation wrong? Force will change a mass's speed (velocity). But I still see the acceleration a=F/m of a mass m as the observable. Momentum changed is like a bank account. What goes into the mass m (as a speed change differencee v=vf-vi) can come out. If the speed was increased, then decreasing the speed can return the mass to its original speed. How we invoke & do the (speed) change is another story. But I think I'm off topic there, momentum_squared is the better concept instead of momentum & kinetic energy, there. We have no momentum wrt to each other, true. But I don't think we can say that for any other reference, because we are moving thru the universe (no doubt). I mean surely there is an underlieing inherent momentum for every mass, based on let us say the universe's center (from where you (=physicists) say the big bang happened). Ok. That makes things easier. i.e. less to think about. I guess according to the big bang (hypothesis, since I doubt we'll ever have a time machine or viewer to the past to prove it), everything is suppose to be flying away from the universe's center & being also gravitationally attracted (according to you physicists). That looks like a radial acceleration to me, e.g. a non_balanced force (setup) in ruffly 1 direction out from the core (center). Does that mean the earth on which the balance is fixed is not turning. Surely not.? Stationary is a static (d)illusion. No motion does not exist in the universe, everything is moving. Stationary (or static, at rest) only means both (the object (e.g. mass), & its reference sytem) are moving at the "same" speed. To say they do not move (at all) is absurd. Or do you disagree? Math or instrument machine? Ok. I said the balls deform like a spring, in order to decellerate (elastically, during the collision). That deformation needs time, (even if it's very quick, & the 2 balls are very hard, like steel or glass. Both materials still have (hard) elastic properties). I hope you understand me now.? (It seems you (also) did not register squared_momentum.? Have I said something wrong? Are you still trying to figure out what that means (because you have not commented on it so I assumed you are not sure). Please ask that I can try to fill in the blanks. Sometimes the mind automatically blocks things out too fast to notice.)
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