# Capiert

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Classical physics
(no calculus)
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Classical physics

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1. ## Power?

Well done! But if the proportionality to distance is a (correct) mass_squared; instead of (only mass) NOT squared, then you have a definite difference that can NOT be pushed_away (=denied!) (& hid under the carpet). The numbers (resulting) for mass_squared versus only mass (NOT squared). Btw. That ("squared" versus NOT_squared (mass)) is math; NOT physics. Math_Physics at the most. Quite right. But with what (=which?) method you choose, to describe, determines your outccome. Choose momentum & then you are probably (quite) right. Choose Energy & then it can go wrong when mass changes in the collision. Choose mass*Energy & then you might get it right (again), when dealing with changing mass. The choice is (really) yours (NOT mine). (I'( woul)d call that bias.) ..What? (Please explain yourself. "NOT" what?) Quite right. Mass_squared does NOT exist in either of those (mom or Energy) equations, (& because they are inferior, e.g. limited). Mass_squared is a new concept from Ewert (1996) & it does seem to work right=correctly, e.g. better, such as in problems where the mass changes. & the cool (=neat) thing about it (=mass_squared) is m*E, mass*Energy (m*E) is VERY similar to (only) Energy so you hardly (=barely) need to change a thing. It'( i)s very easy. Simply multiply mass by Energy & you might (=should) get the correct answers. That'( i)s over_simplified of course. What sort of problem am I making up? I see definite problems with Energy that can NOT be over_come (otherwise). Dark_Energy is a real eye_sore. But the so_called energy loss in an inelastic (=NON_elastic) collision is most prevalent (=obvious). & I am NOT making that 1 up. YOU are! The Energy NEVER left the system. & I can tell you where it is. I (wrongly) said that you "lost" -1 [J]; but that is wrong only because you guys condition me into your same bad habits. (& I fell for it, as naive as I am.) You haveN'T lost a thing! Take a look at (conservation of momentum, COM) mom1+mom2=mom3. That'( i)s a perfectly balanced equation. Left_side (is mom1+mom2) is "before" the collision & right_side (is mom3) is "after" the collision. Square both sides & it is still balanced (=perfect)! (mom1+mom2)2=mom32. But that produces mom12+mom22+2*mom1*mom2=mom32. That +2*mom1*mom2 belongs to "before" the collision (=left_side of the equation); NOT "after" the collision (right_side of the equation). Again: That +2*mom1*mom2 belongs to the "before"_ side of the collision equation; NOT the "after"_side of the collision equation. Putting the 2*mom1*mom2 on the "after" side of the collision equation means it "needs" to be NEGATIVE. I.e. -2*mom1*mom2. NON_elastic collisions might seem to have lost energy on the (heavier) result; but they have NOT lost a thing (in that sense); you need only observe the math to correctly determine what has happened. & (thus) you physicists (have) preach(ed) an untrue gospel (=NONSENSE!) Sorry! But more SORRY for your (misled) followers.
2. ## Power?

(amount (of mass)) But it (that mass) is distributed differently (into the (mass) terms) so that it is only (1 mass, now e.g.) together. & that ((different) togetherness) (now) has a different affect on the math answer. are now 1 total mass m3=m1+m2 which is now larger than either 1 of its initial components (e.g. either: m1; or m2) & If mass is squared, then its parts produce different (math, number, value) results compared to its (=the total mass) sum squared. m32=(m1+m2)2 m32=m12+m22+2*m1*m2 <---But look at that! m32#m12+m22 Mass squared does NOT simply add only the squared terms; because the (NON_squared term) +2*m1*m2 is missing! I hope you get my drift. (What I am "trying" to tell you.) Such naive math (just) can NOT work, (sometimes).

8. ## Power?

Undoubtably average_power. I was NOT aware that James Watt used instantaneous power. He used how fast a (180 lb (force) ~82 kg (mass)) "horse" walking (continuously) a circle, using yards (=3 feet=0.9144 m) & seconds (time), could wheel up water from a well against (the horse's) gravity's weight (806 N). ~737 Watt (rounded to 740 W for 75 kg). 746 W would have needed 75.59 kg (~167 lb) at 1m/s. I question that math was any great deed of calculus. Watt determined that a horse could turn a mill wheel 144 times in an hour (or 2.4 times a minute).[6] The wheel was 12 feet (3.7 m) in radius; therefore, the horse travelled 2.4 × 2π × 12 feet in one minute. Watt judged that the horse could pull with a force of 180 pounds-force (800 N). So: Watt defined and calculated the horsepower as 32,572 ft⋅lbf/min, which was rounded to an even 33,000 ft⋅lbf/min.[7] Watt determined that a pony could lift an average 220 lbf (0.98 kN) 100 ft (30 m) per minute over a four-hour working shift.[8] Watt then judged a horse was 50% more powerful than a pony and thus arrived at the 33,000 ft⋅lbf/min figure. https://en.wikipedia.org/wiki/Horsepower Then the "speed" should be constant. But the real (more probing) question is from where or what "speed" (reference) are you using to measure the "speed" you are observing. (Obviously, your own speed is the reference, there.)
9. ## Power?

Continued, (from my previous) That speed_change is the acceleration e.g. to or from a collision, but once that collision (duration e.g. t<1 , stops=) finishes the(n that) acceleration stops; so that friction decelerates the ball (further) to a (slow, much longer duration, e.g. t>1) stop. But, an Earthquake (e.g. an abrupt change in the Earth_plates' (rotational) speed) could shake the (table &) balls (in)to motion. Done. How would you describe (your) KE? (..without problems). That was mine (=my description). Identical speeds does NOT mean zero KE. It only means zero KE between the 2 objects that have the same speed. Constant speed does NOT mean zero KE. In fact any speed (helps) indicate a KE. (E.g. If you (also) know the mass(es), too.) But any (observed) speed(_difference) v is a (speed_)difference v, i.e. that uses a reference (initial_)speed vi (as basis), to observe the (final_)speed vf=vi+v. (But) we do NOT see the (final_)speed vf because it has the (initial_)speed vi in it, which we exclude. E.g. The Earth's_rotation_speed vi~1000 [m/s] eastwards remains invisible (to us) because everything (static) around us is (also) moving at (approx.) that same speed. How can we solve your problem? P.S. When Earthquakes happen 1 of the problems is the ((tectonic) plates') rotation slows down & then continues to speed up; repeatedly. That kind of (kinetic) Energy can bring down skyscrapers (crashing), (it's so powerful!) (m*PE=m*KE) m*m*a*d=m*m*(vf2-vi2) only equates the accelerated distance with the speed change of a mass; NOTHING more. a*d=((vf2)-(vi2)).
10. ## Power?

Why is (your) initial_speed vi (=vref) NOT zero? You have said the 10 kg mass is "moving" at (vf=vi+v=0+v=) 10 m/s, so what is the reference that you used?
11. ## Power?

Yuk! What a messterpiece. I'm trying to simplify (to algebra) so (even) a 5 year old can understand (Trump mentality, noted (by reporters) for its advantages), but it explodes & gets scattered into so much complexity (some interesting though), (really avoiding the question, with (fake) substitutes) (unfortunately demonstrating how little was understood, in some cases). When the cat'( i)s away the mouse will play. To get back on track. On 6/3/2022 at 2:50 AM, joigus said: What's the speculation here? Capiert: The (basic) question was whether the (Jame's Watt's) Power_equation's (forced) speed is a speed_difference v=vf-vi (No?, but most commonly claimed?) or the average_speed va=d/t (Yes? New). The algebra seems to confirm (the later, & itsself), (Unfortunately opening the question why the calculus did NOT do the job (right, for that question), which I DON'T really want to discuss (calculus yet). Maybe for later.) (I guess) I was a little sloppy because gravity's weight force was intended only as an example of force; NOT all forces in general as Swansont put it. Here is an example of using a specific equation and trying to apply it in general. Power applies to more than the situation of an object falling under gravity, so one cannot make a general claim that force is weight, since there are other forces. If you use weight, then the equation will only apply to a falling object, and also only if you can assume that this is a constant force. Should (better be) Power P=F*d/t is the Force F=m*a, multiplied by the distance d per time t. The average_speed va=d/t is the distance d per time t E.g. Let the (linear) acceleration a=g be gravitational free_fall's acceleration, & the force F=Wt be the Weight Wt=m*g, & the distance d=h height (fallen). P=Wt*h/t. I did NOT say that (an object moving at a constant velocity has zero KE); you did. If initial & final speeds are the same then the KE (between them) is zero. E.g. For 2 objects=bodies. E.g. For 1 object=body the (same, single) mass has NOT be accelerated (e.g. faster) from its initial speed; or e.g. it has NOT been accelerated from its final speed. Take your pick. I (still) do NOT see your problem. E.g. Same speeds (for same mass(es)) means NO kinetic_energy(_difference) KE, but KE is a difference (KEd=KEf-KEi) of (the 2) KE's; although I think Swansont will want to correct me there. No it does NOT. The equation uses 2 (different) speeds even if 1 of them (speeds) is allowed to be zero. A difference of KE's means that you are subtracting 2 KE's. So your components are KE's. E.g. KEd=m*(vf2/2)-m*(vi2/2). KEf=m*(vf2/2). KEi=m*(vi2/2). vf=vi+v vi=vf-v v=vd=vf-vi. Please show me the error (if so). As far as I know that math (syntax) works (for me). Maybe you have a task, example where I can use it? My concept (syntax) of KE (=KEd=KEf-KEi) is (a (KE) difference, &) similar to yours, but (differs in that it) is (also) extrapolated thru to (both) the initial_KE KEi=KEf-KE & the final_KE KEf=KEi+KE. (But Swansont commented=identified that is delta_KE.) What I'm saying is, the initial_speed vi is [often] invisible (for same speed objects); but NOT zero! Even if objects=bodies seem static; they are (really) moving! Same speed objects can NOT accelerate each other (because they do NOT collide with each other). Again PS: Very interesting. It looks like you did NOT get it (=the (2) perspective(s)). How can anything moving, NOT have KE? If vf = vi (then that) means you can NOT see the(ir) motion. (..when compared). The 2 objects (seem to) stand still, (perhaps) with a constant (separation) distance. E.g. 2 billiard balls on a (billiard) table (although the world=Earth is turning). Our reference(_speed) vi could be the Earth's rotation speed e.g. ruffly ~1000 [m/s] eastwards. The (2) balls are NOT going to do anything (e.g. (they are NOT going to) collide) with each other because they stay still (wrt the billiard table, on Earth); but (both (balls)) are rotating with the Earth_speed vi~1000 [m/s] eastwards (i.e. wrt the Earth)! Their KE would NOT become obvious until they collide with something else moving at a completely different speed. E.g. a cue hitting a ball. ((Please) allow me to exaggerate.) E.g. If the cue travelled ~-1000 [m/s] e.g. westwards (to hit the ball) to compensate against the Earth's rotational speed (so that the cue would seem like zero speed wrt the Earth's center). KE, (e.g. the speed_(energy)_change) can only be [acquired=] "received from", or else "transferred to" another mass('s motion).
12. ## Power?

Yes. I agree that there is NO such thing as conservation of Energy (COE); although conservation of mass*Energy m*E might exist. COE is a FAKE, because it often FAILS, although NOBODY has the guts to kick it OUT; they all still (perhaps naively) defend it (as traditional BRAINWASHING, e.g. NOT to upset things=tradition). Conservation of ENERGY does NOT deserve to be mentioned except for its (crude) history e.g. development of Physics. (That is:) Wasted time, considering my (just=immediately) previous comment. But what do you mean by extend the time frame? Distance d (x,y,z), d=P2(x2,y2,z2)-P1(x1,y1,z1), is the difference between 2 position(s points): e.g. P2(x2,y2,z2) minus P1(x1,y1,z1) ; (where P1(x1,y1,z1) is the origin(al starting point) which determines the positive direction). e.g. x=x2-x1 y=y2-y1 z=z2-z1. Please explain. (e.g. "contrary" to "self-referential" frame). (What is that?) (E.g. A Point source?) (Is ego supreme?) Everything exists "in" the universe; NOT OUTSIDE of it (=the universe). Everything is "a" part of the universe (=connected); NOT "apart" !. (E.g. I am in the universe; NOT independent from the universe). (Please give me a clue (as) to help understand you better.) But if I understand you correctly then the "initial" kinetic_energy KEi might help fix=remedy things; for the total=final kinetic_energy KEf=KEi+KE. E.g. Kinetic_Energy KE=KEf-KEi is the difference between final (kinetic_energy KEf) & initial (kinetic_energy KEi) similar to speed(_differrence vd vd=v) v=vf-vi where the subscripts are: final f & initial i. ("you can tease out the difference between") I'( a)m sorry (but) you have lost me (there). I can NOT imagine any speed without direction. "I DON'T know where we'( a)re going Captain; but (at warp_speed) we are getting there in a he(ck) of a hurry!"-Scotty. Direction (e.g. angle) is also relative. E.g. to a line. E.g. 2 points (P1 & P2, each an x,y,z); but the 1st (point) must be established from the 2nd, to complete that relation (for positve, (versus negative) direction) i.e. relativity. I will assume (with (the word) map) you are implying x,y,z (e.g. wrt to some other reference (of similar structure, e.g. Ref, e.g Frame)). I (can still) consider a speed(_difference) v (=vd), to be (composed of) components vx, vy, & vz. (But) I do NOT use instantaneous_speed; I use average_speed va=d/t, instead. I(' am sorry, I) DON'T follow (you). My (speed_difference) v is a mixture of x,y,z speeds. Where is the problem? (I suspect you are adding unneeded complexity; where NONE is needed. Am I right or wrong? If wrong please explain.)
13. ## Power?

We are assuming they are moving on parallel lines, in the same direction. Can (=May) I pacify that argument(?) by saying: (I acknowledge) an "initial" Kinetic_Energy KEi=m*vdi*vai where the intial_speed vi=vi-0 is (simply) extrapolated from a predacesser speed_difference with its maximum_speed being (only) that excluded initial_speed vdi=vi-0 (whatever zero=0 speed should be e.g. relative to something else('s motion_speed)). vai=(0+vi)/2 (analogy (similar to)) va=(vf+vi)/2). e.g. minus zero, where zero is that (next, smaller) reference. That is all done so because KE is (already) relative to its (own) initial_speed vi. Thus the kinetic_energies can be added (sequentially).
14. ## Power?

Please explain that problem. Kinetic_Energy is (already) relative (with respect) to the initial_speed vi. (-c<)vi<c can be (almost) anything less than light's_speed (+/-)c. That means the initial_speed vi is excluded in that (amount of) "kinetic_energy('s)" change of speed. E.g. 2 masses moving at the same speed (wrt each other), have NO speed_difference v=vf-vi (wrt each other), thus a constant distance is maintained (=kept) between them. Only if a speed_difference v exists between them can they affect each other in e.g. a collision, to change the other's speed e.g. via Newton's 3rd law, (equal & opposite) reaction, Repulsion. E.g. Assuming an (EM_)Field wrt (a decreasing) distance, to ((elastically) repulsively) bounce. The catch there is electric_repulsion is inversely proportional to the radial_distance "squared"! That is NO longer a linear relation; but instead exponential (wrt distance)!