Capiert

1 hour ago, John Cuthber said:

If that was true then the people who make power stations would have noticed.

I suspect they didn't because they (attempt to) use constant voltage(s) V.

Piv^0.5=((V^2)/R)^0.5=V/(R^0.5)

=I*(R^0.5).

Metering (for billing) is indicating current flowing.

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If your writing style was more readable I might bother to look for other problems in  your post.

 

On ‎2018‎ ‎08‎ ‎18 at 3:07 PM, swansont said:

The units are fine. You appear to simply not understand what's going on.

Unfortunately (I suspect) you might have missed my point?

The electrical "watt" (definition), conflicts (severely) with the mechanical value in (=by) an exponential amount. I.e.

According to (my) experiment(s), using Hooke's spring law, equating forces, for the same (=identical) angular displacement.

I can NOT verify the committee's (electrical) definition (because it is way off!). I do not get the correct (electrical) values (when compared to mechanical values) using their assumption (=definition).

I do get correct (electrical & mechanical, corelation) values using my formula Fiv=(I*V)^0.5, (=F=ma=Wt=m*g) instead.

That is why the(ir) (electrical) units are crazy!

(According to my formula derivations (the (electrical) units for), current multiplied by voltage should be (mechanical) watt_squared, instead of (only) watt units.)

& I don't know how to explain that (discrepancy, problem) to you otherwise. Maybe you can help?

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And the answer to your question is that no, the equations are fine, when used within the constraints of how they may be applied. Which can possibly be understood by studying some physics.

I think you have made your point,

 that (our earthly) work energy WE=F*d (concept)

 is a very limited (=restricted) basis

 for studying the total cosmic system (accurately)

 (because it lacks the static case,

 & can only document moving things).

It's useful for studying some (e.g. moving) things,

 but not everything (e.g. static wrt to our own (earth's) speed),

 thus useless in those cases).

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You should use them when the conditions allow their use.

I however, am looking for a more universal formula instead

  of those (limited) restrictions.

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The physics definition is reliable, if you understand the physics

restrictions (=limits)

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.

(That sounds (a bit) like using a crutch, as a sports car, for the (human) race.)

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If not, all bets are off and you will likely end up with nonsense.

(I think) I've made some sense out of the chaos.

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Displacement angle is not the same as displacement distance.

Quite right.

But do you have a problem with (using) a (D'Arsenval_meter's)  spiral_spring (versus linear_spring)

 & seeing the (for me, obvious) correlation, & (my self_allowed) substitution?

Either: weights or electrical rooted_power produce identical deflection angles

 (confirming Hooke's law F=-x*k, angularly F=-theta*k).

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The equations are very reliable.

As you can (or will) see, there are discrepancies

 between the electrical versus mechanical watt values.

So at least 1 of them is doubtful.

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The problem lies with the person using them, not understanding how to use them.

Quite true & I propose the committee's personal concerning the electrical concerns

 might need to brush up a bit on the facts.

To my person, I am always learning.

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Then it's like a neophyte using a power tool. Unfortunate things can happen..

I agree, as we have seen. e.g. my errata thread.

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Energy does not have a direction.

I would have said, (heat) energy('s randomness) could have a net direction of all directions,

 thus an exclusive single direction

 is NOT discernible.

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It can, however, have a gradient. Heat is energy transfer owing to a temperature gradient.

Yes.

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Yes. You have to use the equations that apply to the problem, rather than using ones which don't. If you need to account for thermodynamic events, you must apply thermodynamics. If you don't then simple mechanics will work for such a problem, e.g. you could simply be accounting for work done by a frictional force, rather than having to worry about heat and temperature. It depends on the problem. And, of course, if you have current and resistance, etc., you will need to use electrical equations. But if these effects are absent, you would not.

It depends on the parameters of the problem.

I tend to agree with you (improvising where possible).

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That energy is conserved can be derived. It is one of Noether's theorems.

Could you (please) summarize that for me that I could comprehend it better (that it would stay better in my memory)?

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Then derive it.

OK. It's a repeat (from above).

Newton's force

Fn=m*a
Wt=m*g

Hooke's (spring) force

Fs=-x*k

then angularly

Fa=-theta*k2

electrical (rooted power's) force

Fiv=(I*V)^0.5

Equate Newton's force (gravitationally as weight)

 with the electrical force

 because the angular displacements are identical gives

Wt=Fiv

m*g=(I*V)^0.5

Should I go further?

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It is true that one cannot always experimentally measure all of the energy in a system, so conservation of energy is not always useful to apply, even if it is conserved. In an inelastic collision, for example.

If I may say(?),

 during an elastic collision

 there is a brief period (=duration of time)

 when the collision is non_elastic.

?

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One cannot easily account for the kinetic energy that is converted into other forms. (i.e. collisions don't tend to conveniently take place inside of calorimeters).

What do you mean (there) by conveniently?

Are you saying, many collisions are very complicated?

I see that KE fails to account for everything, & that some of the other (so_called converted energy) forms might be (unjustifiable) guesses=assumptions.

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OK. Solve some problems involving thermodynamics with conservation of momentum.

I don't see the problem with putting things on a momentum basis,

 but I don't know the thermodynamics you are looking for.

(Could you please tell me the formulas you want & mean?)

Momentum is

mom=m*v

mom=m*((vi^2+2*h*g)^0.5)-vi)

with

speed_difference v=((vi^2+2*h*g)^0.5)-vi)

initial speed (velocity) vi, &

mass m.

The only problem there

 is sequentially dealing with negatives

 under the root sign, correctly,

 when squaring both sides.

Please give me a simple thermodynamic problem

 (to get started),

 that needs the momentum conversion.

If you give me enough hints

 maybe I can convert (for you)?

I do not know if I can,

 because I don't know enough of your thermodynamics (syntax & concepts you exclusively use)

 to evaluate if I can.

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Go ahead. Try it. Or even a block sliding down an inclined plane with friction.

I know (next to) nothing about them (inclined planes & friction).

If you help get me started (with the energy description setup),

 maybe I could (also) convert that to momentum

 for you?

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Or an elastic collision, without involving energy.

For (starting with) a non_elastic collision

The total momentum of the system is (also)

mom3^2=mom1+mom2, ^2=square both sides

mom3^2=(mom1+mom2)^2, gives

mom3^2=mom1^2+2*mom1*mom2+mom2^2, rearranged that'( i)s

mom3^2=(mom1^2)+(mom2^2)+(2*mom1*mom2).

So you (can) see there is some continuity.

Assume mom4^2=-2*mom1*mom2.

If mom1 & mom2 are the initial momentums,

 then mom3 & mom4 are the final momentums;

 where initial mass1 m1=m3 mass3 final,

 & initial mass2 m2=m4 mass4 final.

I.e. Odd postscript numbers are for the same mass.

& even postscript numbers are for the same mass.

Please choose some values,

 & we can (try to) extract some answers.

Before=After collision

(mom1^2)+(mom2^2)=(mom3^2)+(mom^4^2).

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Fortunately you are not the arbiter of acceptance in physics.

EM waves have no mass.

If a wave has momentum (e.g. impulse)

 then I would say

 it also has a very small mass m (=mom/c) value

 based on the momentum formula

 (& wave_speed c=v),

 only as a number.

I mean I can equate.

7 hours ago, swansont said:

When the units don't match, yes, it is difficult to believe.

I believe you! The units are crazy!

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You only suspect?

Yes. The answer depends on how things are defined

 (versus how they are derived=self_defined, instead);

 & which basis of reference(d definitions should be used

 for trouble_shooting).

I question both of James Watts

 definitions: mechanical_power P_W=F*va;

 & work_energy WE=F*d

 (as possible culprits

 of the problems).

Thus, I should isolate both

 & NOT_use (=avoid) them

 as my (starting) basis.

Instead I should choose

 another definition

 (that is more reliable)

 on which to rely (on)

 just (in case) to be careful

 & see what gets derived (out,

 from that (other or new) usage).

In my case I have used

 the 2 (simple) quantities:

 (Newton's) force;

 & displacement (angle) distance.

E.g. If a units problem occurs,

 then there is always the question:

 where (=at which end, or perspective)

 should we tackle the problem (1st)?

What could be corrupt?

 & what could be used, instead

 that is (more) reliable?

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If I send  current through a resistor, there is electric power

heat (=random mechanical energy, in 6 directions: +/-x,y,z)

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but no mechanical work

moved mainly in only 1 direction (.e.g. angle).

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.

If I push a cart, it's mechanical and no electrical.

But if the cart is braked, or too heavy

 when you push,

 then it will (also) NOT move

 (thus no mechanical work,

 even if you are using a bull_dozer

 e.g. to push a(n extremely large) tree, or mountain of granite

 that does not budge (a bit),

 although your gasoline gets wasted

 trying all day

 till the gas tank is empty

 (although paid, for that chemical energy).

James Watt's (2) definitions (P & WE)

 do NOT account

 for what does NOT get moved!

How can you rely

 on that deficit

 when dealing with

 (a declaration

 of) conservation of energy?

It's ridiculous!

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So, how can there be one equation?

Via a (self_defining) derivation.

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Curious you should ask, since you introduced the topic. Electrical power, P=IV, mechanical P=Fv.

Yes. Those told me how you are looking at things,

 so I can orientate better

 as to finding the problem

 (in the communication gap).

You do NOT see any problem with energy

 so you trust it completely,

 & automatically use it (instinctively);

 while I do the opposite

 avoiding energy as much as possible

 with the suspicion

 that it is corrupt

 & unreliable

 (in some cases).

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Multiply by time to get the energy.

If I multiplied force by time, instead

 I would get momentum,

 which I would prefer (& trust more).

I'd trust Newton's force & momentum;

 over Leibnitz's energy, & James Watt's power.

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As I think we discussed, there is no work being done while there is no motion.

Yes, that exclusion (of motion) in (James) Watt's definition of work (energy) & power

 is the (deficit) problem.

That deficit won't (always) allow a total account

 for registering energy conservation.

Intuitively, we know we can loose energy

 without being able to account for it (=the lost energy)

 with Watt's method. E.g. bull dozer against a mountain all day getting nothing done except getting warm=hot.

Here I (hope I) have shown why.

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You have too many examples to know which one you are referring to, and you tend to leave out sufficient detail to have a useful discussion.

I hope that (above) helps better.

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I don't trust that you have done a sufficiently careful experiment,

I do not brag about any of my experimental results (precision),

 they are only done to orientate me,

 for what tendancy is to be expected

 (based on my algebra).

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since you don't seem to understand that there is only mechanical work done while the needle is moving.

As I have said that's the weakness in (James) Watt's definition.

His definition is unacceptable (for me).

It does NOT fit in well enough

 with Newton's math

 (for me).

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And since there is no motion once the needle stops, and thus no work, how can you possibly think that the mechanical work(or power) equation can be applied?

I suspect you are right there.

I must drop the usage of mechanical work (energy)

 & attempt to exclusively use (Newtonian) momentum (& force);

 instead of (work) energy (& James Watt's mechanical power).

Otherwise I'm only asking for future problems

 with such unreliable definition as those from James Watt.

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EM wave motion is not mechanical.

Please explain a bit.

Isn't mechanical, a motional (=moving) mass study (analysis, e.g. measurements).

Don't waves have momentum mom=m*v?

4 hours ago, swansont said:

If they are not actually equal, then the equation will be wrong, and any conclusions you draw will be invalid.

But are they equal?

Yes. They are equal.

(Is that too difficult to believe?)

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IOW, is it possible to do both mechanical work and electrical work? 

(I suspect the answer (to that question) is yes.)

What do you mean (exactly) by electrical work? (E.g What formula are you referring to?).

I know I can achieve the same amount of spring displacement (angle distance),

 with either: weights; or electricity (as rooted power).

Their affect is identical=the_same (to the meter's spring, Hooke's law F=-k*x).

Similar forces, produce similar displacements,

 thus I achieve identical results

 with 2 different methods.

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You tried to tell us that one equation should work for all situations.

Yes, I believe that is true, I have been convinced it is possible.

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How is that an observation?

In that it contradicts your statement

 that 2 equations must (always) be used (separately), instead.

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4 equations is not 1 equation,

True.

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and none of the four deals with mechanical systems.

If you insist waves' motion is not mechanical.?)

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So there are even more than that.

(Most likely, at lot for me to still do (if possible, at all?).)

 

I stated Maxwell versus Heaviside as a tendancy (analogy)

 (towards simpification,

 from a non_PhD (Heaviside)

 to their wonder (e.g. at least Maxwell));

 but Oliver's uncle was Kirchhoff

 whose law pertains to the meter's functioning, as well.

(If that's the connection (relavance) you are searching for, from my statement,

 as to why I ever mentioned something like that, at all.?)

Naturally my (own, (formula) compression) ratio is not 20:4,

 but instead only 2:1.

19 hours ago, swansont said:

Yes, it can, but the issue here is that when you actually use an equals sign, it means you are saying they are equal. It does not mean proportionality.

What happens when I declare they are equal (as well, just to present my dilemma)?

I mean I'm assuming you can understand my problem.

(I "can" equate both (as mechanical) forces.)

How should I proceede (acceptably)?

I have applied rooted_power to a D'Arsenval meter;

 & I have weighted the needle with weights (without electrical power)

 & found (calculating the torques' force)

 that they (both: electrical (rooted_power) force & weight force) are "directly" proportional.

I claim no great precision in my own measurements,

 you are the experts there.

But the effect is obvious (to me, experimentally; & mathematically).

I recommend you to (also) investigate (to convince) yourselves (also), if possible.

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Not when you do things correctly.

No, what is wrong is the way you are understanding this.

I do not see the error you imply. (Enlighten me.)

(My problem is more, getting "you" to understand my perspective & observations.

It's often easier to explain to a simple person on the street, than it is to scientists.)

I hope the above can explain my perspective.

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We are describing how nature behaves.

Me too.

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We don't get to tell nature how to behave.

I have tried to tell "you" what I observed.

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So if it takes more than one equation to do the former, then that's how it is. Nature is a tad complicated.

(Prof) Maxwell started with 20 equations.

The telegraph operator Heaviside compacted them (20) into only 4.

Amazing things are possible. Especially simplification.

(Nature often uses similar blueprints. There is often a similar general scheme (to be found (in nature)).)

17 hours ago, swansont said:

Here's the thing: you need to be precise, and use the proper definitions in order to be understood. If you want to change the acceleration of an object, that's what you say. You don't talk about changing g.

If you have two objects on a pulley, changing the mass could change the direction of the acceleration. It's all described by the equations.

Since you tend to use radial coordinates, no. r is always positive, and v is always positive. You can talk about the sign of the change in either of these, which could be positive or negative.

Polarity isn't the correct terminology here. And neither is fall vs ascension. Having zero acceleration does not mean that the object is stationary,

Nobody is arguing with you there.

(For an orbit (=moving, =not stationary)), I said zero (fall) acceleration was equilibrium.)

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and acceleration downward does not mean that the object is moving downward.

In my equation, it does. (Doesn't it?)

On 16 August 2018 at 2:13 AM, StringJunky said:

Do you think the penny will drop one day?

Never.

Capiert says: a person with 2 watches can know when 1 is wrong.

A person with 3 might tell which is wrong.

Even if all 3, or less.

On 16 August 2018 at 2:02 AM, Strange said:

I guess (a) you don't know what you are talking about and (b) you are unable to express yourself clearly.

Wrong (guess): I don't know what "you" are talking about, there.

(I.e. You said, my idea of (a2's fall acceleration) polarity getting changed

(via mass, or orbit speed change)

was wrong;

 after I asked what was wrong, giving you a + & - hint.)p

I guess "you" did not understand me.

(I do make mistakes.)

On 16 August 2018 at 1:33 AM, Capiert said:

I get the idea those (orbit speed) changes are altering the (fall acceleration's) polarity (sign, away from zero, when they happen).

 

On 6 June 2018 at 5:15 AM, Phi for All said:

There was no filmed demonstration, presumably because it didn't work. We had a thread some time ago on perpetual motion and he was mentioned. Somewhere on the web there's some correspondence where Costa admits his current machine failed, but swore a bigger one would work, only he wasn't being allowed to build it (which didn't make sense to me).

I like the crowbar bit. Very PT Barnum.

PT=?

Sorry, rumours don't interest me either

 (unless they can be backed up, & traced).

I.e. That still did NOT answer my (debunk) question:

What makes that "10 Tons" constantly turn?

2 hours ago, Strange said:

As always, the problem is with your lack of knowledge of physics.

(I feel) Resignation

((but) accepting your criticism

(as your good intent)).

2 hours ago, Strange said:

Nonsense.

Let's try again:

(James) Watt defined ((horse_)power):

P=F*va

with the (average) speed va,
horses could pull weights Wt=m*g=F.
(I'm sorry you missed your history lesson (if you did?);
 or was I too brief (& that's why you did not understand (me))?).

I'll assume the later.

I suspect a specific quotient (instead) might have fit Newtonian (physics) structure better (tendencial). 

2 hours ago, Strange said:

https://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf#page=52

Yes, mechanical versus electrical definitions clash.

It would be easier (avoiding conflicts)
 to maintain the electrical units Ampere*Voltage as maybe A*V;
 instead of (re)naming them Watts W,

 (unlike) as the committee did.

The unit Watt has unfortunately been double defined
(e.g. electrically & mechanically),
 & that is more than double trouble!

Theoretically, the 1st definition receives priority (in honour=respect)
 to be maintained,
 while the 2nd (named usage) must (=should) be stopped.

2 hours ago, Strange said:

This is not a "facit". 

Watt's definition does NOT give us the whole picture,
 & that is 1 aspect (point of view)
 indicating its weakness
 (in this argument).

How would you describe that point?

2 hours ago, Strange said:

Ignorant nonsense.

Your cosmologists are still trying to make sense of energy,
 because they report their stupidity not knowing what "dark energy" is.

(The exception can define some rules.)

Your Nasa article below confirms
they don't know "what" they are talking about
 (if you'll excuse the pun).

2 hours ago, Strange said:

Nonsense.

" It turns out that roughly 68% of the universe is dark energy." (https://science.nasa.gov/astrophysics/focus-areas/what-is-dark-energy)

Add the +27% dark matter (related to energy by E=m*(c^2))
 & that adds to 95% dark (universe).

2 hours ago, Strange said:

Nonsense.

So the expansion of the universe has not been slowing due to gravity, as (scientists   wished) everyone thought, it has been accelerating. No one expected this (although I did, (but) why don’t they mention me?), no one knew how to explain it (but I did (with gravitational push*). Is that quote suppose to be scientific propaganda=brainwashing?). But something was causing it.

 

*I predicted Hubble’s constant starts small, increases, & (then) tapers off as we progress outwards from the (universe’s) center.

Scientists originally did NOT know if it was a straight line, curved: up; or down (=taper off).

 

Energy is Leibnitz’s (math) physics (NOT Newtonian).

Momentum is Newtonian physics.

 

Maybe you should (also) deal with the real McCoy?

 

 

Theorists still don't know what the correct explanation is..

Their solution is called “dark energy”.

 

More is unknown than is known. We know how much dark energy there is because we know how it affects the universe's expansion. Other than that, it is a complete mystery.

 

So they are pleading their (own) stupidity,

 & attempting to put everyone in the same boat.

 

roughly 68% of the universe is dark energy. Dark matter makes up about 27%.

 

Together that (& mass as energy) constitutes 95% total (of the universe).

2 affects, both dark=unknown why or what.

 

Other than that, the article (is friendly &) explains well. Thanks.

 

Another explanation for how space acquires energy comes from the quantum theory of matter. In this theory, "empty space" is actually full of temporary ("virtual") particles that continually form and then disappear. But when physicists tried to calculate how much energy this would give empty space, the answer came out wrong - wrong by a lot. The number came out 10¹²⁰ times too big. That's a 1 with 120 zeros after it. It's hard to get an answer that bad. So the mystery continues.

 

At least they are honest, (thus) making themselves credible.

Now if Swansont says (or continues to say) Physics has NO problems;
 & that only I have the problems,

 then I guess he was not (really) aware of that 1.

 

The article confirms you guys need help,

 because you’ve got big problems

 you have NOT solved.

If I believed you 10 years ago,

 it’s all wrong today.

How do I know what you now declare,

 is not wrong,

 & will not be declared outdated, again?

Who had expected a 180° change, in views?

(Some) TV documentaries are (now) suggesting physics may need to be newly redefined.

 

(& just guess who has been working on that?)

 

A last possibility is that Einstein's theory of gravity is not correct.

But if it does turn out that a new theory of gravity is needed,
 what kind of theory would it be?

I don’t want to get (too) “pushy” (t)here.

 

Btw the book on the (book)shelf (mentioned earlier)

 is being pushed up (& maintained) at that (bookshelf) height,

 by the bookshelf (itself).

The maximum pressure is on the lowest side (of the book).

 

2 hours ago, Strange said:

Nonsense. You are saying that physical systems will behave differently with changes in time and space (Noether's Theorem).

I'm sorry I am not familiar with him.
Or is that an acronym for no_ether?

2 hours ago, Strange said:

This is not what we observe. We can also measure conservation of energy and momentum directly.

Would you mind showing me how.
I can't make it always work mathematically.
How can you?

2 hours ago, Strange said:

So all the evidence shows that you are wrong. 

?

2 hours ago, Strange said:

You are the only one getting that.

Your Nasa paper (above), also indicates the non_sense.

I can't take the "credit"
 for being the only 1
 that gets non_sense from the physics theory.

4 minutes ago, Strange said:

Then your "idea" is clearly wrong.

Zero polarity is equilibrium, there.

(E.g. Neither fall (-), nor ascension (+).)

What's wrong?

9 hours ago, swansont said:

None of that changes g,

True, g is defined as "free" fall,

 & being bound to a rope is not (being) free.

But I assume you understood my intention

 to discuss altering our fall acceleration a2

 in stunts.

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and making something smaller does not change its sign.

In that equation,

 changing "something" like 1 of those masses (i.e. m2) by making it smaller (wrt the other mass m1),

 will change the polarity (sign) of its (fall_acceleration) a2.

 

There is a sidekick, however to that theme when considering orbits:

Faster orbits will increase the (orbit's existing) radius;

 while slower orbit speed will decrease the orbit radius

 to a new equilibrium.

I get the idea those changes are altering the polarity (sign, away from zero, when they happen).

1 hour ago, Strange said:

Well, I'm relieved it isn't something more serious. Why don't you replace it.

I'm not rich like you. It's in a laptop that overheats while in internet (& hangs in slow motion);

 & I don't know how to get the hackers out, either.

I'm attached to it.

1 hour ago, Capiert said:

 

Quote

If the needle isn't moving, no mechanical work is being done.

That's Watt's definition.

I challenge it,

 because it (still) costs.

Watt defined with the speed horses could pull weights.

But if the weights were too heavy, they wouldn't budge (=move, a bit);

 & the horses would get exhausted,

 wasting their (bio_chemical) energy,

 on getting "NO WORK" done.

Facit: Watt's definition does NOT give us the whole picture.

& your cosmologists are still trying to make sense of energy,

 because they report their stupidity

 not knowing what "dark energy" is (as much as 95%?, in the universe).

(A few years back it was only ~75%?)

I'm trying to tell you the culprit is energy's definition.

ENERGY DOES NOT FIT with momentum.

You can NOT serve both conservation "laws?" at the same time.

Somewhere along the line you are going to get hash=chaos=trash=garbage non_sense (if you do)! 

1 minute ago, Strange said:

Is there something wrong with the keyboard on your computer?

Yes.

On ‎2018‎ ‎07‎ ‎14 at 8:06 PM, swansont said:

You have written these as an equality, but say it's a proportionality.

An (exact) equality (=equation) can state a proportionality,

 but a ruff idea proportionality

 cannot always state an equality.

 

On ‎2018‎ ‎07‎ ‎14 at 8:06 PM, swansont said:

(The units are wrong for it being an equality).

That'( i)s my (1, major) complaint.

The units totally conflict,

 although the concepts (excluding the units)

 are proportional.

Something is very wrong (with the definitions)!

On ‎2018‎ ‎07‎ ‎14 at 8:06 PM, swansont said:

Why should mechanical power and electromagnetic power have the same equation?

Simply because we are dealing with displacive force.

That is the link, the basis, for equating.

That can be translated into pressure (multiplied by area).

Let me ask it so:

 wouldn't it be better (=an advantage, as less bother, simplification)

 if electromechanical power had the same (=only 1) equation?

On ‎2018‎ ‎07‎ ‎14 at 8:06 PM, swansont said:

(plus, one is a proportionality)

Stating 1 as a proportionality

 was the only way I could state the problem

 (without getting into too much difficulty, & conflicts).

On ‎2018‎ ‎07‎ ‎14 at 8:06 PM, swansont said:

If the needle isn't moving, no mechanical work is being done.

That's Watt's definition.

I challenge it,

 because it (still) costs.

12 minutes ago, Strange said:

I *ca&n't! U n d e ... rst (an)yth(i)n[g] u w-r-i-t-e 

Thanks for your emogies.

On ‎2018‎ ‎07‎ ‎14 at 5:46 PM, John Cuthber said:

A shelf does not need power to hold up a book.

Atomic repulsion (power?) is what keeps the atoms away from each other.

But you'( a)re right,

 the bookshelf doesn't have an electrical bill

 (after it was made; & transported into place.

Those costs (seem to) fall away).

Quote

In principle,

=(in) theory (=where'( i)s the practice?)

Quote

it's possible to make a meter

 Are you sure you are talking about a D'Arsenval meter here? I was not aware he had superconductors in his time.

Quote

entirely from superconductors so the needle would stay in place without the supply of any power.

Please show me 1, (that) you mean.

But I'd hate to pay for that power bill (for making those superconductors (& that they) work =stay cold & magnetized).

 

 

Here'( i)s a spin_off (=sidetrack))

 of Ewert's pulley experiment.

The driving (=pulling) weight difference (force)

 (m2-m1)*g=(m1+m2)*a2

 (of 2 different masses, stringed over a pully)

 must accelerate (to drive, all=)

 the ((total) whole=100%, =sum of)

 mass m3=m2+m1

 (because the rest (=not_different=non_difference)

 is (equal=) equated_out (as the same, or identical)).

 a2=g*(m2-m1)/(m1+m2).

 &

 a1=-a2, so

 a1=-g*(m2-m1)/(m1+m2)

 or swapping the mass difference

 a1=g*(m1-m2)/(m1+m2)

 

Thus we have the eccentric equation(s)

 & mass2('s excess, wrt mass1) determines

 its (own) acceleration_direction (polarity).

Note: g=-9.8 m/(s^2)

 is negative,

 thus a large(r) mass2 (than mass1)

 will naturally have a negative acceleration2=a2 (down(wards)).

 

I suspect (=find) that is a remarkable formula,

 (i.e. + & -, =eccentric);

 & quite good for some TV stunts,

 e.g. the 3 Musketeers

 jumping on a rope, pullyed with a weight

 similar to the person

 to slow the descent(‘s acceleration).

E.g. To get more done in the scene,

 like in slow motion.

 

E.g.

How can we slow down g

 (& make it smaller).

 

The final trick

 would be to make the (negative) g

 so small,

 that it becomes positive,

 & will gently levitate

 to rise a person.

Assuming the person is m2,

 then m1 must be(come) larger (than m2).

Identical (masses) m1=m2 is (simply) levitation.

Please not(ic)e, that slowed (=smaller) acceleration a1=-a2

 is (still) “acceleration”

 if (=when) the pulley “friction” is cancelled out!

So a constant speed is NOT expected

 for that trick (=stunt).

 

..13:29

 

E.g. Star Wars: falls; & ascenssions.

 (Blue (or green) box, video overlay effects.)

Very slow acceleration,

 can be made to (optically) look like expansion (=expanding, against the background)

 or shrinking (in the opposite (slow) acceleration).

 

Over_simplified math

 just won’t do (the same (effect))

 in our heads(‘ imagination).-Disclaimer.

..13:39

 

I was a little puzzled

 that I couldn’t make that conclusion

 in (only) 1 (=the 1^st) session (=sitting)

 e.g. why it confused me at 1^st;

 & that I had to let my intuition take over

 to finish the job.

Lucky enough that it did.

E.g. My intuition gave me the task in the 1^st place,

 as a curiosity (=peculiarity).

If it didn’t then somebody else did, e.g. ET’s etc (moma=mother nature)?

I can’t blame it (=success) on my ego,

 because that (=ego) failed.

22 minutes ago, John Cuthber said:

Once the needle has swung into place, all the power delivered to the meter is dissipated as heat so, if you only look at mechanical energy, it doesn't seem to be conserved.

Isn't it (=the meter) dissipating heat whether it (=the needle) swings or not, while electricity flows?

I.e. It (=the needle's deflection) "costs" (continuous) power (=E/t, energy E, per time t),

 not just a once_only energy to keep that needle in place (beyond zero).

Isn't there something wrong with James Watt's (mechanical power) definition?

E.g. It's not universal. It does NOT include the static (equilibrium) case.

The electromagnetic force of a D'Arsenval (spring_loaded) meter

 Fiv=P^0.5

 is proportional to the rooted electrical_power

 Piv=I*V

 (small postscripts used)

 for current I

 voltage V,

 resistance R.

 

Isn't that in conflict to James Watt's definition of mechanical_power (caution: small postscript m used)

 Pm=F*va

 (not to be confused, with the m)

 for (Newton's) force F=m*a

 mass m

 acceleration a,

 average_speed (velocity) va=(vi+vf)/2

 initial_speed (velocity) vi

 final_speed (velocity) vf.

 

(I mean:)

Both forces

 Fiv=F

 produce mechanical displacement

 that can be equated.

---

(Sorry for the sloppy syntax, but yours confuses me too.)

Re_done that would be:

 

The electromagnetic force of a D'Arsenval (spring_loaded) meter

 F_I*V=sqrt(P_I*V)

 is proportional to the rooted electrical_power

 P_I*V=I*V

 (small postscripts used)

 for current I

 voltage V,

 resistance R.

 

Isn't that in conflict to James Watt's definition of mechanical_power (caution: small postscript m used)

 P_m=F*v_a

 (not to be confused)

 for (Newton's) force F=m*a

 mass m

 acceleration a,

 average_velocity v_a=(v_i+v_f)/2

 initial_velocity v_i

 final_velocity v_f.

 

(I mean:)

Both forces

 F_I*V=F

 produce mechanical displacement

 that can be equated.

7 minutes ago, Phi for All said:

It was my understanding that the wheel stalls when this is attempted. Otherwise, Mr Costa would be demonstrating THAT.

I don't know. I didn't see that demonstration you mentioned. Do you have any link with time point?

Maybe I should have said, much less than 1 kW?

Thanks for all the interesting input.

The objective here was to find the arguments against the gravity push concept

 & give you all a chance to change my mind.

But it looks like I have more questions

 needed to help me figure out what you are saying.

I chose the lounge because I thought it was an informal (casual) peaceful (non_hostile) place

 to discuss anything

 (in a scientific matter).

 

Sensei, what charge does a neutron have?

How are neutrons, accelerated, or deflected in their particle accelerators?

Does the earth have a (variable) net charge on its surface?

How then can we say the atoms are (completely) neutral

 when we can not measure all things?

E.g. position & momentum.

 

Swansont, aren't coordinate systems convertable?

It does not matter what coordinate system is used. Einstein.

Your selection of coordinate system might make the numbers easier for you.

(But) Orbits occur, for any coordinate system (you want to use).

I did NOT disagree that Cavendish demonstrated a pull,

 even if you insist (only) radially.

(E.g. most efficient.)

Then,

 nothing (=no equatorial pull) should be expected (by Cavendish's experiment)

 if (the 2nd mass is slightly) offset from the equatorial plane?

 

Strange, are atoms neutral?

How can that be?

Most of them are tied up (=bound [up]) in molecules, (& are)

 victims of electrostatic & magnetic fields.

A lot is happening, more than meets the eye.

 

How do you know matter is not expanding,

 if all matter around you

 (including your instruments)

 is also expanding?

I mean, how can you exclude that possibility?

If the medium (which you call space*time) is flexible, why can it not "expand" & compress?

(I didn't make the rules. Bending is none other than a deformation.

But wrt what? Another volume (reference)?)

 

Push is from electrostatic repulsion.

(Please don't ask me how that works.

You (all would) state: like charges repel.

 but have NOT explained (me) why (they repel).)

(Similar to water waves produced

 from dropping a stone

 into a pail of water,

 so the amplitude of the waves decreases

 further away from the source:)

If each atom's volume (as wave) increased ~a millionth of its volume pro second,

 then adding those (volume increases=) parts, for large objects

 gives noticeable numbers

 e.g. for planets' "surface" motion (& thus push)?

Are you with me? (=Can you understand?)

That (volume expansion) concept affects throughout all matter (=stuff)

 simply because matter is a wave.

The concept that the universe is running down (=dissipating) energy is well known.

The energy density is decreasing,

 due to the volume expansion

 (if the amount of energy is constant).

But (e.g.) drop something

 from a very high height.

The object will be destroyed

 when it hits the (earth's) surface

 (if it does not bounce).

E.g.

The air's pressure is being maintained

 against a vacuum

 (from outer space).

E.g.

The earth has a changing climate

 (although mostly due to the sun light).

E.g.?

I doubt that the earth rotates

 (against friction)

 due to gravity('s push, disapation).

But why doesn't the earth slow down to a stop?

Why do galaxies's (arms) curve?

Everything is moving,

 & getting faster,

 although we don't always notice it (=the acceleration).

The details of gravity are probably quite hairy,

 but what do you mean a push

 does not explain gravity correctly?

Please explain.

You probably have some specific examples in mind

 that you can share.

 

You're on.

(Unlike you) I see (vertical) gravity as a push.

(But that's putting the cart before the horse, so to speak.)

 

(I see that) there are (really) 2 types of gravity: vertical & horizontal.

It's rather obvious (that) the Cavendish experiment is a horizontal pull.

I interpret that (pull) as vertically moving (e.g. accelerating) charges of atoms

 are electromagnetically attracted together.

But has it ever occurred to you why you can NOT shield against gravity vertically?

Perhaps because that might only be the acceleration of matter moving upwards.

Thus any other reasoning is missing (as nonsense).

(E.g. because there is nothing there to shield against.?)

Hubble provided us with an interesting perspective:

 The universe appears to be expanding.

& if so why do NOT matter waves in that (weak) vacuum also expand?

(E.g. osmosis: moving from higher concentration to lower).

 

Maybe you can convince me why vertical gravity should NOT be a push?

 

Btw Asmov's Fantastic Voyage, of shrinking things

 for a period

 would be possible

 if we could (find a way to) restrict matter's expansion.

 

1 hour ago, John Cuthber said:

The body is generally quite good at converting chemicals.

Starch, for example, is quite rapidly degraded in the stomach to form glucose which is then absorbed.

About a quarter or a half of  the fructose is converted to glucose- the rest is metabolised via different pathways.

 

What is in beer that makes people fat is, essentially, calories.

It doesn't matter much if those calories are in the form of sugar or alcohol; if you eat more energy than you use, the body stores the excess by making fat.

So decreasing the carbs (=carbohydrates) is a good strategy, to start with?

That would mean fermenting more (CO2) gas than alcohol,

 & toasting (or burning) your bread to burn off energy.

But what can we do with butter (fat)?

I thought the trend nowadays

 was good vs bad bacteria.

The bad 1's make people thick or fat

 because they are so efficient

 using the food molecules.

Quote

Incidentally, a linguistic point; when talking about people in English, fat is not the same as thick.

Fat means they weigh too much.

Thick is usually taken to mean "thick headed" i.e. stupid.

So, yes, beer makes people thick- but not in the way you meant.

(Yes) ok, so what sort of (grain?) molecule is responsible for that: (stupidity)? alcohol?;

but fat? Malt sugar?, or is there some other specific chemical compound?

Or should I be asking,

 what molecules are missing

 to prevent that process?

PS I've also noticed 2 different distilled (liquer) alcohols.

1 caused a mental collapse, (e.g. 1 particular gin (others not), so a sugar was probably involved);

 while most rums allow maintaining consciousness to the same percentage intake.

What could be responsible for the 2 different affects? Chemical tampering?;

Surely not age, because many (old) scotches are reknown for their quality.

What determines the alcohol's efficiency?

 

5 minutes ago, John Cuthber said:

You have the wrong idea.

Please, explain (a bit).

I mean not as well as others.

Don't some starches or sugars go straight to fat (production).

What is in beer that makes people thick (=fat)?