Capiert Posted May 24 Share Posted May 24 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 Link to comment Share on other sites More sharing options...

swansont Posted May 24 Share Posted May 24 55 minutes ago, Capiert said: 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. Missing from what? This not an equation for g. It comes from a different equation, rearranged, which you have not shared.(i.e. g is typically not an unknown in kinematics problems) IOW, you are skipping some necessary steps in the discussion. Link to comment Share on other sites More sharing options...

Capiert Posted May 24 Author Share Posted May 24 (edited) 16 minutes ago, swansont said: Missing from what? From g. You obviously disagree, otherwise you would NOT have asked. Quote This not an equation for g. Why NOT? It works. Quote It comes from a different equation, rearranged, which you have not shared. (As far as I know) I have shared the equation: the fallen height (taken from a graph), h=vi*t+g*(t^2)/2. Quote (i.e. g is typically not an unknown in kinematics problems) 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)? Edited May 24 by Capiert Link to comment Share on other sites More sharing options...

exchemist Posted May 24 Share Posted May 24 (edited) 23 minutes ago, Capiert said: 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)? I'm not quite sure what this is about but regarding the effect on g of the Earth's rotation, this once came up elsewhere and I worked out that even at the equator, the reduction in apparent g due to the Earth's rotation was only about 0.3%. It went as follows: - The radius of the Earth is ~4000miles, so its circumference, say at the equator is 2 x π x 4000 which comes to ~25,000miles. - The tangential speed at the equator is therefore 25000miles/day, which is close to 1000mph. - The centripetal acceleration needed to keep an object moving with tangential velocity v, in a circle of radius r, is v²/r. So, for an object at the equator, the apparent centrifugal acceleration it experiences, counter to the acceleration of gravity, will be 1,000,000/4000 = 250 miles/hr². What we need to know is how this acceleration compares with g, the acceleration due to gravity, which is about 10m/sec². To do that, we need to get this result into the same units as g is quoted in:- 1 mile is ~1600m. And 1hr is 3600 seconds. So 250 miles/hr² becomes 250 x 1600/(3600)² = 25 x 16/(360 x 36) = 25 x 4/(360 x 9) = 100/3240 = ~ 0.03 m/sec². So the centrifugal force at the equator, as a proportion of the force of gravity, is of the order of 0.03/10 =0.003, or 0.3%. At higher latitudes it will be less, presumably by a factor of the cosine of the latitude, falling to zero at the poles. Edited May 24 by exchemist Link to comment Share on other sites More sharing options...

swansont Posted May 24 Share Posted May 24 1 hour ago, Capiert said: (As far as I know) I have shared the equation: the fallen height (taken from a graph), h=vi*t+g*(t^2)/2. 1. No, you did not post this equation 2. vi*t is clearly present in it. So what is missing? Link to comment Share on other sites More sharing options...

Capiert Posted May 24 Author Share Posted May 24 1 hour ago, swansont said: 1. No, you did not post this equation 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. 1 hour ago, swansont said: 2. vi*t is clearly present in it. Good! (for that product). But now for the quotient -2*vi/t in g (alone). 1 hour ago, swansont said: So what is missing? 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. Link to comment Share on other sites More sharing options...

swansont Posted May 25 Share Posted May 25 28 minutes ago, Capiert said: 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. Sorry, you didn’t post it before the other equation. I only read the first section, because of the nonsense value. Plus I would’ve skipped the strikethrough text anyway. 28 minutes ago, Capiert said: 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). Clearly this is bollocks. h=vi*t+g*(t^2)/2 (or an equivalent equation) is standard fare in any physics textbook that presents kinematics equations. equation 3 https://www.khanacademy.org/science/physics/one-dimensional-motion/kinematic-formulas/a/what-are-the-kinematic-formulas top left of the “big 4” https://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations 3rd equation https://www.pasco.com/products/guides/kinematic-equations and so on... Your complaint is entirely fabricated. (you can also notice that nobody is presenting this as an equation for acceleration) Link to comment Share on other sites More sharing options...

Capiert Posted May 25 Author Share Posted May 25 11 hours ago, swansont said: Swansont: Sorry, you didn’t post it before the other equation. Sorry, that'( i)s NOT true! It's in other threads before this 1 Is half a year too early for you? 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 11 hours ago, swansont said: I only read the first section, because of the nonsense value. 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. 11 hours ago, swansont said: Plus I would’ve skipped the strikethrough text anyway. 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. 11 hours ago, swansont said: 11 hours ago, Capiert said: 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). Swansont: Clearly this is bollocks. h=vi*t+g*(t^2)/2 (or an equivalent equation) is standard fare in any physics textbook that presents kinematics equations. Yes, we (all) know that. 11 hours ago, swansont said: equation 3 https://www.khanacademy.org/science/physics/one-dimensional-motion/kinematic-formulas/a/what-are-the-kinematic-formulas top left of the “big 4” https://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations 3rd equation https://www.pasco.com/products/guides/kinematic-equations and so on... Your complaint is entirely fabricated. (you can also notice that nobody is presenting this as an equation for acceleration) 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). 11 hours ago, swansont said: 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]. -2 Link to comment Share on other sites More sharing options...

swansont Posted May 25 Share Posted May 25 6 minutes ago, Capiert said: 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. Don’t injure yourself patting yourself on the back. Rearranging the equation is algebra, so any high-schooler should be able to manage that. I am focused solely on your claim that vi is missing from the equation, when that’s clearly not true. 11 minutes ago, Capiert said: Capiert: (Kinetic_Energy is already relative, 2020 11 23 16:28) If you’re going to reference an earlier thread, you should post an actual link, rather than give a time/date. 17 minutes ago, Capiert said: 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).!) You should learn calculus. a = dv/dt Integrate twice, applying boundary conditions, and you get that result (as would many thousands of people who can do the math) That you can’t do it in no way implies that the equation is wrong. Link to comment Share on other sites More sharing options...

Capiert Posted May 25 Author Share Posted May 25 3 minutes ago, swansont said: Don’t injure yourself patting yourself on the back. I'll do my best! Thanks. (I got a good chuckle (out of that).) 3 minutes ago, swansont said: Rearranging the equation is algebra, so any high-schooler should be able to manage that. Yes! That is my intention. You obviously have NOT read this thread's file. It's NOT complicated! It's easy. 3 minutes ago, swansont said: I am focused solely on your claim that vi is missing from the equation, when that’s clearly not true. Please explain. vi included: is the general equation. But without vi: is a limited (specific example) formula, that can NOT work for all cases. 3 minutes ago, swansont said: If you’re going to reference an earlier thread, you should post an actual link, rather than give a time/date. Locked threads DON'T allow quoting. Maybe (other_options) sharing can help linking, a bit.? Link to comment Share on other sites More sharing options...

swansont Posted May 25 Share Posted May 25 4 hours ago, Capiert said: You obviously have NOT read this thread's file. You should know that requiring this is not in compliance with the rules. 4 hours ago, Capiert said: Please explain. vi included: is the general equation. But without vi: is a limited (specific example) formula, that can NOT work for all cases. Yes. That’s what you do - write down the general equation and eliminate the terms that are zero in a particular problem. In the other thread I expressly said dropping from rest, making vi = 0. There’s no reason to continue to include it for that problem. There’s no implication that it applies to all cases once you have imposed such restrictions 4 hours ago, Capiert said: Locked threads DON'T allow quoting. Locked threads also aren’t supposed to be resurrected Link to comment Share on other sites More sharing options...

Capiert Posted June 29 Author Share Posted June 29 (edited) On 5/25/2021 at 6:42 PM, Capiert said: Quote I am focused solely on your claim that vi is missing from the equation, when that’s clearly not true. Please explain. vi included: is the general equation. But without vi: is a limited (specific example) formula, that can NOT work for all cases. On 5/25/2021 at 7:55 PM, swansont said: Yes. That’s what you do - write down the general equation and eliminate the terms that are zero in a particular problem. In the other thread I expressly said dropping from rest, making vi = 0. There’s no reason to continue to include it for that problem. There’s no implication that it applies to all cases once you have imposed such restrictions 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. Edited June 29 by Capiert An example. Link to comment Share on other sites More sharing options...

swansont Posted June 29 Share Posted June 29 2 hours ago, Capiert said: What I mean is: (vi was missing from the "general" equation, (NOT (just a) formula.) I provided three links that shows this to be in error. Have you provided any examples of it being "hidden"? Quote 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. Usually it's derived with few constraints to it, and no initial conditions. e.g. start with a = dv/dt, which is the definition of acceleration (always true) Add in the condition that acceleration is constant, and integrate the equation and you get v = v_{0} +at Apply the definition of velocity, v = dx/dt Integrate again and you get s = v_{0}t + 1/2 at^{2} The initial velocity is there (as it is in so many textbooks, if you'd only bother to look), and the only restriction is that you must have a constant acceleration Quote vi remains an invisible (hidden (excluded)) term Only if you forget to include it, but that's on you Link to comment Share on other sites More sharing options...

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