Capiert

Way back (long ago) & (I) was confused why the lamp didn't light at all. I didn't measure the voltage (with a digital meter) like you. Your answer helps a lot with the fotos. Because back then the lamp didn't light at all so I thought years later maybe a diode effect happened. But you're right, most of the flashlights (I had used) back then were 2 cells (occationally 3). Studiots fotos are very convincing (=helpful). Yes. I used to play around with the batteries. Good tip. I'll keep it in mind. Right now I only find LED flashlights, also often with 3 cells but they only work 1 way because of the LEDiode. Thanks everybody. You've helped a lot & got me thinking (right (again), hopefully back on track). If I get a chance someday I can try to measure.

Yes well done. Yes, sorry, you are diplomatic enough the website formally is still awkward for me. Confirmation of him being a dentist came late(r). I.e. His answer (12:10 "Yes it is embarrassing that I a dentist ..") was not ignored. If you already know then you are bullying and harassing me. I misinterpretted, (the same) as he did.

If I interpret what Dr. Turner('s math model) is implying, he's saying (in a different perspective) that if mass radiates space, then that additional space will be added (whether negative space might also be possible?, is not stated, because only addition is mentioned) to the existing space thus warping it (=the previous (space)) to a new total i.e. (thus) affecting the (total amount of) spacetime around the mass. Gravity is vertical, (not tangential (around like an orbit, "curve")). So I can't support Koti's arguement on that point. I'm surprised Studiot hit the nail on the head with tooth pulling, & he (Dr Turner) did admit being a dentist shortly after, so repeating the question (which Dr) seems inappropriate. Can we cool down a bit & try to understand him. He does seem to know a bit, just fill in the holes. (Einstein wrote Relativity, for the common man with a high school education, with the intent to entertain us with some joyful times, in the intro.) I mean, wouldn't it be helpful to (try to) be friendly. Everybody that lands here goes into high gear (spontaineously, informal) but nobody knows so much about physics as you all. All you have to do is ask the details & they might notice the pitfalls. Appologies in advance.

Voltage addition? If 1 (1.5 V) battery of a 3 battery flashlight is wrong (=the opposite (polarity)) (then) the total voltage is zero, not 1.5 V so the lamp (filament) will NOT go on (=won't light) when the switch is closed (=turned on). Kirchhoff's voltage law adds voltage( drop)s on loads, but seems to collapse when dealing with (battery) sources of mixed polarity. (=I find that (unexpectedly) rather (stunningly) peculiar, the math fails!) Can anyone help me to (simply) understand what's happening, (e.g. with like charges repel, unlike attract) why the voltages do NOT add anymore (for mixed polarity batteries)? E.g. total voltage (of batteries) adds for same polarities Vt=V1+V2+V3 4.5 V = 1.5 V + 1.5 V + 1.5 V, & -4.5 V = -1.5 V - 1.5 V - 1.5 V, but (makes no sense with mixed polarities) 0 V # 1.5 V + 1.5 V - 1.5 V, & 0 V # 1.5 V - 1.5 V + 1.5 V, & 0 V # -1.5 V + 1.5 V + 1.5 V, & 0 V # -1.5 V - 1.5 V + 1.5 V, & 0 V # -1.5 V + 1.5 V - 1.5 V, & 0 V # 1.5 V - 1.5 V - 1.5 V.

I'm a believer! (=Yes it must be truely amazing, for the learning amount & different ways.) (Mind boggling!) Penrose, wasn't he the 1 that said the past does NOT exist, it's a stack of nows? (Oh we'( wi)ll have to start a new thread, if this (=these fascinating discusions) keeps up, & the scalar time?) Btw Isn't the basic problem with vectors, dealing with "even" negative_multiples? Even multiples of a dimension x^(2*n) for n=0, 1, 2, 3, .. are all scalars. (Negative) "Polarity is lost" (= not tracked, anymore). (So the polarity stays positive, =never goes negative). Odd multiples of a dimension (e.g. x^1, x^3, x^5, .. are (all) vectors &) have complex (=complicated, sometimes wierd) multiplication rules, (in order) to track ((negative) polarity). Thus no (single) general formula (rule) for both (vectors & scalars) ? E.g. Vector notation would be prefered for both (vectors & scalars) because it's most (cap)able (to deal with negatives(' tracking)). (But dread to think,) what happens when we root a scalar, that was based on x^2? (E.g. The other( scalar)s, with even exponents larger than 2 would NOT be a problem because their scalar status would NOT change.) Yes, that (SI standards) must be the best way to cut thru the confusion.

Does anybody have a link to Gravesande's (original) experiment &/or measurements? E.g. Brass ball weights dropped vertically from different heights into clay (=mud) that produces dents of various area (diameter) sizes, & depths.

Moved.

(I'm) Also glad to help. Your archives must be immense.

Yes. I always wondered (for years) why Lorentz (1904, Electromagnetic phenomena in a system moving with any velocity smaller than that of light, in German, from das Relativitaetsprinzip "1913" (=not otherwise, thus found here) https://archive.org/details/dasrelativittsp00minkgoog) would go into such detail of all 3 dimensions, to tackle the (analysis) problem(s) for the first 19/20 terms of GR (i.e. that was very admirable, to get into the nitty gritty e.g. every step of the way, when nothing else worked (right) (but I couldn't follow it (=his breakdown method) then(, why)); & (I also had wondered (for years) why a gyroscope had the ability to (180 degree) rotate & transfer a(n asymetrical, =1 sided) force, or acceleration (such as g) (for precession) at slow speeds (e.g. hanging 90 degrees, horizontal; away from the vertical rotating axis, at 0 degrees). Now I know (why). The (rotating) acceleration, is "non"_linear for the 2 rotating coordinates (e.g. x, & y; when z is vertical); (although gravitional acceleration g=-9.8 m/(s^2) is linear). The asymetric (rotational) acceleration (wrt time) causes the (momentum, & acceleration) transfer to another dimension (of the 3D's). The acceleration of 1 (rotating) coordinate (when added) is different for every step of time. Yes. Thank you, the article; your do's & don'ts; & tips are very helpful at getting me started. Just the right amount (of guidence: not too exhaustingly (long &) boring; nor too little to grasp) with torque example, to get the (=my mind's) ball rolling (& make it click (for me)). Enough to get me curious enough to ask why (in view of the paradoxes (to solve)). Thanks both (of you).

Yes, but I suppose there are (at least) 2 different possibilities: Elastic (e.g. bounced, having more acceleration (damage?)), & (absorbed) non_elastic collisions (e.g. like Gravesande's experiment, 1D).

I want to agree with you there, but I'( woul)d have to say, (theoretical) physics has to be fit to the observations (=experimental results). E.g. Dark (=unknown) energy, needs some filing (=rework, fitting).

That means: (KE is) the amount of work(_energy) to (move=) accelerate a mass m, from initial_velocity vi=0 m/s to the final_velocity vf at the time of impact. But that sounds like force F=m*a (instead). I suspect the (forced, travelled) distance d=va*t is a clever way to involve the amount of time t into the equation, as d=t*va (for momentum mom=F*t). Work_energy is WE=F*d, d=t*va WE=F*t*va, mom=F*t WE=mom*va. (Momentum is in the core of the (work) energy equation.)

KE=p * v_a (p & v_a are multiplied together; NOT added) So the answer to my question: (Can we know the (simple=) linear velocity (components) of a rotating object?) is no! We do NOT have linear velocity formulas for rotating objects' x & y components.

Moved.

I can't guarantee physics, they way you want, on your terms. There is no error correction mechanism, for your faulty software which adds errors into the syntax, as well as the syntax changes which the new software demands. Maybe you need better programmers to get (all) the bugs out. I can't deal with everything. As for Lazy. (Buff.) My mind shuts down when too many problems exist. I'm happy to have been able to communicate til this far. I think your software bugs are screwing the people up. As for got it, you now know what my pdf's va is all about. Are there any errors? If not then a major problem (still) needs to be solved. "A rose by any other name smells just as sweet." That's interesting: Newton's physics; Leibnitz's (calculus) syntax. Wasn't Newton's inertia renamed momentum? What a mix. Funny that Newton didn't use it. Heaven (on earth). ? But I'm no fan of your mixed up conventions. (e.g. Conventional flow versus electron flow). I don't worship & praise your syntax as the best solution, like you do. (Your syntax conventions are more like my downfall because it goes so (=too) fast. My brain latches (=locks_on) skipping the problem too quick.) (I also have my preferences, like any human being does. That you want to dictate them implies a monarchial desire.) I'll admit it is neat & fun to have a single character syntax, e.g. a minimum (syntax), but it doesn't help me enough to deal with (infected) computers to check the math, & (assist) error (reduction techniques) control. You've got more resources than me, are richer, & expect everybody on the same level. My PC constantly overheats, & stops. The times have changed since the midages. I'm looking for basic problems, but you all are convinced they do not exist (=no problems exist); your methods are perfect; & (it's) the only way to do them. Btw You still have not answered my question: what is the linear x component speed of a rotating object.

That's where you & I differ. You see force as a cause, & I see it (at least partly) as an effect, an observation. I am not completely decided on that as to whether force is a mechanism or not. Sometimes you say rediculous things. I can explore anything I want (without definitions, = definitionless). Many people use physics, but not all people. I use other syntax, because yours is not adequate enough for me (& my model testing). There are not enough non_foreign characters in the english alphabet. I want to test things independently (=from different perspectives), (with the ability to (test)) every step of the way. If its foreign (for you & me) then there are better chances of (me) not falling into the same booby traps you do. That aside, I'm not convinced of everything in (theoretical) physics & I would like to make (partial) comparisons to sort it out & identify (where) the problems are. Sorry, (if you want whorship, you've got the wrong person) I'm not going to fall for that 1. If I do everything exactly like you, I will fall into the snags that you do & never learn a different perspective (than yours) to solve the problems because I'm stuck in them. (It's a lousey job, but somebody has got to do it.) Nature has variation (& creation). "Take me to your leader (earthling!)"-ET. Pun aside, I'll (try to) work on it. Ruder was to kick the indians out, & now declare it (=what was stolen, is) yours. -Squatter's rights.

mom=KE/va. (Sorry, va. Bold type for vector.) Swansont: "Dividing by a vector is not defined." By the look of that equation, it sure looks (to me) like high time that (the) defining should begin. E.g. At least derived.

Stated (simply=) linearly, that (component's velocity, e.g. vx) would (then) be .. ? (please continue). Swansont: "I have no idea why you think I would not include the other body." Swansont: Momentum is conserved when the net external force on an object or system is zero. Capiert: Meaning no net force (F=m*a, Newton's 2nd law), or balanced forces (F1+F2=0, Newton's 3rd law reaction (is zero)). Effectively, no acceration a=F/m ((of) Newton's 2nd law). Swansont: That's not what Newton's third law says. Action-reaction force pairs always have equal magnitudes, even when the net force is not zero. But they act on different objects, and are never both in an equation for net force. Capiert: You've defined the net force as on a single object because (from your experience) it's useless on a force pair system. I did not have that restriction, because I (still) explore (whether useless or not). But I only mentioned Newton's 3rd law to (further) indicate that when a single object gets accelerated (e.g. Newton's 2nd law), (then) it "depends" on a 2nd object, too. We often view only half the story from our (exclusive) limited perspective. I only wanted to hint, more was going_on (=involved) than the single particle (mass).

Which parts do you mean. I often don't (recognize=) understand your writing style (without direct examples). (Perhaps it's too careful for my informal attitude? An example often helps to cut thru the communication gap.) Maybe you are mentioning what I already know, in other words; or I've (obviously) said things wrong. ? Isn't the total reacted (=reacting) force of 2 collided bodies (as a system) zero? Although you wouldn't include the other body, I could (just to evaluate the system('s net total)). Or what do you mean exactly? (Please put me on track.) ? mom=KE/va. (Sorry, va. Bold type for vector.) Otherwise, I'm sorry (I can't follow you), that formula is too simple to be declared incorrect. Anybody could see its connection (if they wanted to). Thus, m3*E3*(v3/va3)=m1*E1*(v1/va1)+mom1*mom2+m2*E2*(v2/va2), m3=m1+m2. But, wasn't KE derived (wrt PE) from (only) linear acceleration? How (then) can you imply otherwise. (va is the average "linear" accelerated speed.) I thought everyone knew KE is a "linear accelerated" energy (concept, equivalence), for its final resulting speed. I have never seen non_linear acceleration in the textbooks. I guess things get too complicated for most (people) when non_linear (e.g. exponential). ? Newton's centrifugal acceleration (equivalence) ac=(vc^2)/r, for (tangential) circular_speed vc=cir/T=2*Pi*r/T, (cycle's) Period T, & radius r. But, circular rotation (at constant_speed) is non_linear acceleration of each x & y component. Can we (simply) know the velocity of a single component? (I.e. The displacement is obvious (with Pythagoras's r^2=y^2 + y^2), but velocity?)

The only difference I see is (negative) polarity (ability) (which I attribute to (180 degrees) angle): 1 (=vector) has it; the other (=energy) does NOT. Differences usually (always) matter. How can you be (so) sure? Sorry for my sloppy style, I was using Newton's 3rd as a starting point (concept, theme) to explain further. True. No arguement there. True. ? ? I know they (mom & E) are different, & I know they are related (to each other, & (by) how (much, they are)): KE=mom*va mom=KE/va. (That last 1 is rather remarkable, don't you think?)

Because they are related by the same units: mass & speed (as a factor, e,g. multiplied). m*v is common to both concepts: E(nergy) & mom. That is so obvious (it rolls me over). Meaning no net force (F=m*a, Newton's 2nd law), or balanced forces (F1+F2=0, Newton's 3rd law reaction (is zero)). Effectively, no acceration a=F/m ((of) Newton's 2nd law). I.e. no changes of mom(entum) =constant mom(entum), (=conserved). (Newton's 1st law: constant speed of a mass, mom=m*v.) Ojects in motion [tend to] stay in motion when NOT acted_on (=NOT accelerated). wrt distance (d). Work's energy WE=F*d ? E=KE=PE=WE=EE=... E3#E1+E2. (See below.) ? wrt time (t). mom=F*t. ? 2*m3*KE3*(v3/va3)=2*m1*KE1*(v1/va1)+2*mom1*mom2+2*m2*KE2*(v2/va2), m3=m1+m2. ? ?

Thanks. You've done an awful lot of preparation. (I've got (way too) much to tackle. ..as you said.) "did you even understand this as per the scalar modelling formula under the first link? " No, I don't think it registered (well enough). But my major interest is in Gravesande's experiment & the energy discrepancies for a(n energy versus momentum) calibration, (preferably exclusively with algebra, because I don't trust calculus). I don't see why we need, tensors, calculus, or relativity, when (serious (algebraic)) math problems are already obvious. (& the others are not my specialty, at all, anyway.) You know a vast amount of details, to point me in the right direction. Put simply, momentum & KE deal with only 2 concepts (at sub_light speeds): mass & velocity. That means if we know those 2 then we (always) know momentum & KE & know they don't agree. Just dump the info in an excel table (spreadsheet) & it won't lie that there is a (big) problem. & it's simply algebra.

I think we're a bit off topic there. Gravesande dropped balls into clay. I.e. (dropped) brass balls were (passively, linearly) accelerated (with constant acceleration) from rest (velocity) vi=0 m/s to a final (fall) velocity vf (obtainable) without any calculus. Your bullet was accelerated in the gun before travelling to strike with impact. I don't know how it's acceleration was during the explosion, & it's too complicated for me to calculate (that) as such. (Anyway: ) If I have a 1/2 in that (constant, linear acceleration) equation it is often simply due to an average using the triangle's area A=b*h/2, b=base, h=height, rule, when plotted graphically. That's simple enough for me to understand. I also know something will fall g=-9.8 m/(s^2), meaning (starting at rest vi=0 m/s) after 1 second, it will have fallen (the height) h=-4.9 m (e.g. faster). From that, & using standard mechanics, displacement (distance) (s=vi*t + a*t*t/2) I can derive the rest of the free fall equation (at least confirm it as) h=vi*t+g*t*t/2, where g=-9.8 m/(s^2), g=v/t, so v=g*t, v=vf-vi, & h=hf-hi, (postscripts) f=final, & i=initial for (duration) time t. Their (complicated?) math does NOT show me the connection between the 2 conservation laws (com & coe) which my math shows me does NOT exist. Maybe you know a simpler strategy?

If the bullet recoils (=bounces, backwards) then that (reversed) speed also plays a role (for the transfer). -Newton's 3rd law. (I suspect) the average (accelerated) velocity (only) simplifies the syntax (formula), (as an alternative). E.g. an equivalent, or substitution. (That impact) Energy has the ability to accelerate mass. I guess it only helps me: with constant linear acceleration g versus fallen height h. e.g. va=h/t.

No I did NOT. I (really) provided KE=mass x velocity x average_(accelerated)_velocity (as Imatfaal pointed_out =indicated, a bit, was needed). Which is a good (standard) method to trouble_shoot. which I did (have) No it's NOT because "you" are using 2 different syntaxes at the same time for multiplication, instead of being consistent with only 1. ke does NOT mean k x e in your syntax, does it? Why then should va mean v x a? Well at least now we know where that 1 went wrong. It's an honest error (typo). It's just convient (to simplify the formula, compacter; & make some sense of it (=interpret it, for me)), that's all. A helpful (equality) tool? The math tells me what it is. It looks easier (to me).