Fiv (EMF?)

[Strange]

1 hour ago, Capiert said:

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

Of course it doesn't.

1 hour ago, Capiert said:

According to (my) experiment(s)

Presumably you have don them wrong.

If you put as much care into your experiments as you do into your writing, then they are worthless.

58 minutes ago, John Cuthber said:

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

Ditto.

[swansont]

7 hours ago, Capiert said:

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

I'm pointing out that you are wrong in your conclusions.

7 hours ago, Capiert said:

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

No, it doesn't.

7 hours ago, Capiert said:

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

It's been pointed out a number of times that you can't use the power equation in the experiment you describe.

7 hours ago, Capiert said:

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).

No, you get the wrong answer because you aren't doing it correctly, and keep repeating your errors instead of trying to understand the physics.

7 hours ago, Capiert said:

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

You can't, since I*V has units of Watts, and taking the square root does not get you to Newtons. Force is not the square root of power.

7 hours ago, Capiert said:

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?

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).

If there is no motion there is no work. That's not a limitation of the formula, it's a consequence of it. 

7 hours ago, Capiert said:

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).

I however, am looking for a more universal formula instead

  of those (limited) restrictions.

restrictions (=limits)

Which you can't have, since you don't get to tell nature how to behave.

7 hours ago, Capiert said:

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

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

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).

Power is the wrong application here, as has been explained several times. There is no power being dissipated from a mechanical standpoint.

7 hours ago, Capiert said:

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

 between the electrical versus mechanical watt values.

So at least 1 of them is doubtful.

What's doubtful is your mastery of the topic. But instead of learning the physics and asking questions, you claim it's wrong.

7 hours ago, Capiert said:

  Fiv=(I*V)^0.5

Still not a valid equation, no matter how many times you use it.

 

6 hours ago, John Cuthber said:

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

Agree, and also it would be helpful to stick to one topic, instead of trying to argue a dozen different scenarios in one thread.

[Capiert]

On ‎2018‎ ‎08‎ ‎20 at 1:48 AM, swansont said:

I'm pointing out that you are wrong in your conclusions.

No, it doesn't.

What I mean is:

The electrical "watt" (definition), conflicts (severely) with the (mechanical) angle value, by an exponential amount.

I have to root the electrical power

 to get a linear (identical) correlation

 with the (meter's displacement, deflection) angle.

(Can you understand that?

Or did I say it wrong?)

Quote

It's been pointed out a number of times that you can't use the power equation in the experiment you describe.

How can you explain the excellent correlation

 between deflection angle,

 versus rooted power? (Luck, I guess?)

Your formulas do NOT do that;
 but mine do(es).

 

(I'll take it for granted

 you have (correctly) understood:

 the number( value)s agree;

 but not the units,

 if you have ever done the (=that=my) experiment).

(Normally, according to you,

 that ((excellent) correlation) is NOT suppose to be possible.)

 

But I used neither mechanical power

 nor (mechanical) energy

 to get that (mechanical angle) correlation.

Quote

No, you get the wrong answer because you aren't doing it correctly, and keep repeating your errors instead of trying to understand the physics.

How do you know if I get the wrong answer or NOT?

Did you do the experiment?

(It's NOT difficult.)

What is the correct answer?

(E.g. Do you have an example?)

 

I suspect it is very difficult for you to accept

 that I might have discovered something

 you do not understand yet.?

Quote

You can't, since I*V has units of Watts, and taking the square root does not get you to Newtons.

That's the problem.

Quote

Force is not the square root of power.

Doing the math & the experiment

 gives me every indication

 to the contrary of your last statement.

Thus I must conclude

 there is a units conflict

 based on a physic's (doubtful) derivation (standard)

 (that nobody else has bothered with, yet;

 or else you have ignored them (=those people) too).

You physicists are ignoring something important.

Quote

If there is no motion there is no work.

Yes. I think that's clear.

The work definition

 is useless to explain the power lost

 once the needle stops

 at steady state.

(I.e. Electrical Power is being lost

 by producing mechanical force,

 & heat.)

Thus work is useless for explaining

 conservation of energy

 for the meter.

Quote

That's not a limitation of the formula, it's a consequence of it. 

(It's) A consequenting limitation.

Its limitation is a consequence of its non_thorough design.

1 condition (i.e. the static condition) does NOT work in that formula for accounting energy.

It never was intended to (either).

Thus (the formula is) useless for static cases.

That sure looks like a limitation to me.

The work formula (energy, definition)

 was never designed for no motion.

(& they never wanted it to do that, either.)

Why then do cosmologists use energy (instead of momentum)?

Quote

Which you can't have, since you don't get to tell nature how to behave.

Fewer limits are NOT (necessarily) no limits.

Again:

I am looking for a more universal formula instead

 (i.e. such as momentum)

 (to explain the losses);

 without being limited

 by NOT being able to account for zero speed (examples)

 (as in the work definition).

E.g. My 1 kg levitation thread.

(But I suspect you know my intentions already (by now).?)

Quote

Power is the wrong application here, as has been explained several times.

You're right! Rooted (electrical) power is correct, instead.

Quote

There is no power being dissipated from a mechanical standpoint.

Unless it's friction?

 

That's a funny=strange statement there.

Mechanical dissipation?

What's that?

I thought dissipation was used (as "random" (scattering, equalizing)) for heat (energy);

 & mechanical was used for work. ?

I guess you mean 100% mechanical (power) has zero heat (dissipated).

It's either mechanical motion produced, or else heat

 from electricity.

Or a mixture.

"There is no (electrical heat) power being dissipated from (wrt) a mechanical standpoint."

?

Quote

What's doubtful is your mastery of the topic. But instead of learning the physics and asking questions, you claim it's wrong.

Still not a valid equation, no matter how many times you use it.

Then how should that force formula Fiv=(I*V)^0.5

 be stated relative

 the spring's angles (displacement)?

Does it need a correction constant, for the (crazy) units

 (which to me seem redundant)?

Quote

Agree, and also it would be helpful to stick to one topic, instead of trying to argue a dozen different scenarios in one thread.

I guess I'll have to start a new thread for elastic collision.

Edited August 24, 2018 by Capiert

[Strange]

10 minutes ago, Capiert said:

What I mean is:

The electrical "watt" (definition), conflicts (severely) with the (mechanical) angle value, by an exponential amount.

It doesn’t matter how often you repeat that it still isn’t true. 

[Capiert]

36 minutes ago, Strange said:

It doesn’t matter how often you repeat that it still isn’t true. 

Then, please state to me the meter's deflection angle wrt electrical power (correctly).

I can vary the (meter's electrical) power (via voltage) to get different angles.

I.e. What is your relation?

E.g. values &/or formula.

Edited August 24, 2018 by Capiert

[John Cuthber]

As I said earlier, in principle, I can make a current meter out of superconducting materials.

If I put it in a circuit with a battery and a bulb then a current will flow through it and the meter will turn through some angle.

But the voltage across the meter will be zero- because it has zero resistance.

So the power dissipated in the meter will be zero.

 

If I use two bulbs in parallel then there will be twice the current.

And the voltage across the meter will be zero and the power dissipated in the meter will still be zero- even though the deflection will be twice what it was.

You need to distinguish the power lost in the meter from the power delivered to the load.

 

[swansont]

20 hours ago, Capiert said:

What I mean is:

The electrical "watt" (definition), conflicts (severely) with the (mechanical) angle value, by an exponential amount.

I have to root the electrical power

 to get a linear (identical) correlation

 with the (meter's displacement, deflection) angle.

(Can you understand that?

Or did I say it wrong?)

It is difficult to understand you for a variety of reasons.

The square root of power is not force. You are using force power, and work as if they were the same thing, or they have certain relationships between them, and none of that is true.

20 hours ago, Capiert said:

How can you explain the excellent correlation

 between deflection angle,

 versus rooted power? (Luck, I guess?)

I have zero confidence in your ability to do an experiment and accurately report what is going on. You don't use standard (understandable) terminology correctly, and you clearly don't understand critical aspects of physics.

So I don't really know what has excellent correlation with each other.

20 hours ago, Capiert said:

Your formulas do NOT do that;

Since you are misusing "my" formulas, it's really hard to take this as a serious objection.

20 hours ago, Capiert said:

(I'll take it for granted

 you have (correctly) understood:

 the number( value)s agree;

 but not the units,

If the units don't agree, the fact that the numbers do doesn't mean very much.

 

[Capiert]

On 24 August 2018 at 9:45 PM, John Cuthber said:

As I said earlier, in principle, I can make a current meter out of superconducting materials.

I'd love to see your meter, John!

Please show it too me.

What did it cost to build?

Did the bureau of standards have to calibrate it many times?

I didn't know you had made so many.

Quote

If I put it in a circuit with a battery and a bulb then a current will flow through it and the meter will turn through some angle.

Yes, probably maximum deflection for any current,

 thus it can't be used to measure power.

Quote

But the voltage across the meter will be zero- because it has zero resistance.

How do you know if it's exactly zero?

E.g. 300 years ago everybody thought light (speed) was instant (e.g. infinity, takes no (=zero) time), including Descartes;

 till Roemer came along, noticing  a "slight" delay. (I.e. no longer infinite speed.)

How do I know something like a slight resistance does NOT exist in super conductors?

I mean the next 300 years is a long time for me to wait.

Quote

So the power dissipated in the meter will be zero.

Yes (maybe?) although the so called meter tells you infinite (=maximum deflection).

Quote

If I use two bulbs in parallel then there will be twice the current.

And still the infinite reading.

Quote

And the voltage across the meter will be zero and the power dissipated in the meter will still be zero- even though the deflection will be twice what it was.

I see, twice of maximum deflection (e.g. twice infinity).

Very good.

Quote

You need to distinguish the power lost in the meter from the power delivered to the load.

I know you are trying to help,

 but I would appreciate if you would stay on topic.

This thread is about a standard copper wire D'Arsenval meter

 that anyone can aford

 & test.

(Please use my superconductivity thread instead, for your imaginary cool perspective discussion.)

Edited August 26, 2018 by Capiert

[John Cuthber]

38 minutes ago, Capiert said:

I'd love to see your meter, John!

Please show it too me.

What did it cost to build?

Did the bureau of standards have to calibrate it many times?

I didn't know you had made so many.

You need to find out what "in principle" means.

39 minutes ago, Capiert said:

Yes, probably maximum deflection for any current,

No.
Why would you think that?

50 minutes ago, Capiert said:

How do I know something like a slight resistance does NOT exist in super conductors?

You don't, but the people who understand how superconductors work do kow this.

51 minutes ago, Capiert said:

Yes (maybe?) although the so called meter tells you infinite (=maximum deflection).

No

Where do you gt that silly idea?

Apart from anything else, if that were the case, I' not have used it as an example.

52 minutes ago, Capiert said:

but I would appreciate if you would stay on topic.

It is the same topic- it' s just that you don't understand it.

You are looking at the power lost in the meter- due to imperfect design and muddling it with the power delivered to the load.

They are different things.

I can show that by a thought experiment where you use a meter with zero resistance.
Unfortunately, you think you are too clever to learn- so you won't listen.

 

55 minutes ago, Capiert said:

And still the infinite reading.

No

55 minutes ago, Capiert said:

I see, twice of maximum deflection (e.g. twice infinity).

Very good

No, you don't see.

You  don't understand that the current is limited by the bulb.

[Capiert]

16 minutes ago, John Cuthber said:

You need to find out what "in principle" means.

No.
Why would you think that?

Because of the extreme magnetic field that "kicks" in at the superconducting threshold.

[John Cuthber]

Just now, Capiert said:

Because of the extreme magnetic field that "kicks" in at the superconducting threshold.

That's obviously something you have made up.

Apart from anything else strong magnetic fields "quench" superconductivity.

A superconducting galvanometer would behave just the same as an ordinary one- but with zero resistance.

[Capiert]

1 hour ago, John Cuthber said:

That's obviously something you have made up.

Apart from anything else strong magnetic fields "quench" superconductivity.

A superconducting galvanometer would behave just the same as an ordinary one- but with zero resistance.

Why hasn't anyone built 1 yet?
(It's a fascinating subject.)

That means low power (current) applications would function well.?
How well do 2 repelling superconducting coils function together? Strictly inductive (no resistive losses)? No quenching?

I guess weak magnetic fields do NOT quench the magnetic effect. They do NOT erase (or demagnetize) the magnetized field.?

P.S. I see super conductivity as electromagnetism. The magnetic field around the conductor confirms the (electric) current (flowing).
As you stated above with zero resistance, then that would mean zero heat production.
 & nothing more, no bizarre (super) magnetic effects, e.g. supermagnetics.?

Simply missing resistance.

 

[John Cuthber]

15 minutes ago, Capiert said:

. I see super conductivity as electromagnetism.

That may explain some of your mistakes.

 

15 minutes ago, Capiert said:

Why hasn't anyone built 1 yet?

 

2 hours ago, John Cuthber said:

A superconducting galvanometer would behave just the same as an ordinary one- but with zero resistance.

So why bother?

[Capiert]

But (electrical) resistance is important for limiting current flow.
How should I see non_limited current flow?
Surely it (=the non_limited electric current, electrons) will NOT accelerate because there is no (extra) force accounted,
 thus no cause for that acceleration.
Will it go (=flow) near or at the speed of light?

33 minutes ago, John Cuthber said:

That may explain some of your mistakes.

 

 

So why bother?

To build 1 (=superconducting galvanometer), to verify your hypothesis
that it will function precisely with no problems, nor (new) anomalies.
Just to be sure.

Your scientists (were absolutely sure, when they) wrongly predicted the proton gyromagnetic ratio,
 & had to insert a correction factor g

https://en.wikipedia.org/wiki/G-factor_(physics

later, after they measured.

Alone a factor of 2 is 100% error (=totally wrong).
1=perfect, zero error.

Electromagnetism will always have (future) surprises (for you).
I.e. Those correction factors still exist, & are being used.

Edited August 26, 2018 by Capiert

[swansont]

27 minutes ago, Capiert said:

But (electrical) resistance is important for limiting current flow.
How should I see non_limited current flow?

Why would the flow in a circuit be non-limited? The ammeter isn't the only component of the circuit.

27 minutes ago, Capiert said:

Will it go (=flow) near or at the speed of light?

I hesitate to ask, but why would it?

27 minutes ago, Capiert said:

To build 1 (=superconducting galvanometer), to verify your hypothesis
that it will function precisely with no problems, nor (new) anomalies.
Just to be sure.

Your scientists (were absolutely sure, when they) wrongly predicted the proton gyromagnetic ratio,
 & had to insert a correction factor g

https://en.wikipedia.org/wiki/G-factor_(physics

later, after they measured.

Alone a factor of 2 is 100% error (=totally wrong).
1=perfect, zero error.

Electromagnetism will always have (future) surprises (for you).

Yes QM was a surprise. How is this relevant? or is  it just the fallacious reasoning of "scientists were wrong about one thing, therefore they are wrong about everything"?

I mean, if you had some actual physics to point to in your complaint, that would be one thing. But instead you are making physics up, and then complaining that it's wrong. Most people who are unfamiliar with a topic, find that their intuition doesn't match accepted physics, assume their understanding is at fault. Some small fraction, which apparently includes you, has the chutzpah to assume that physics is at fault. Where does such confidence come from? 

[Capiert]

1 hour ago, swansont said:

Why would the flow in a circuit be non-limited?

Resistance is the limiting factor (against), it hinders current flow thus decreasing current.

Yes, I find it difficult to imagine infinite flow in a circuit too, not to mention in super conductors as well.

Quote

The ammeter isn't the only component of the circuit.

Yes so I can't see how John will get his super galvanometer to work properly the way he expects.
The devil is in the details, e.g. connectors.

Quote

I hesitate to ask, but why would it (Will it go (=flow) near or at the speed of light?)?

Resistance is the only thing that would limit a(n applied) voltage to accelerate electrons.
Voltage_drop (=electric_potential_difference) is the driving force (so to speak).
But constant force, with no opposition means constantly accelerating (although that acceleration diminishes (=decreases)).

Quote

Yes QM was a surprise. How is this relevant?

Fiv is electromagnetic. (& mechanical too, if I may say).
If you get it right, (then) that will work down to atoms (size).

Quote

or is  it just the fallacious reasoning of "scientists were wrong about one thing, therefore they are wrong about everything"?

I doubt that you are wrong about everything,
 but I have noticed a few things need corrections & improvements.

Quote

I mean, if you had some actual physics to point to in your complaint, that would be one thing.

I think I've given you enough hints, to doubt (e.g g_factor, cosmologist's dark energy).
Your accuracy at research & experimental precision far surpasses my ability & resources.
But I'm NOT the only 1 that complains.
Surely, you must be able to recognize from the sheer number of complaints
 that something is NOT completely in order,
 even if they can NOT describe things adequately, at all.

Quote

But instead you are making physics up, and then complaining that it's wrong.

I'm trying to explain things the way I see them,
 in an attempt to get "near" to the problems.
I'm NOT perfect.

Quote

Most people who are unfamiliar with a topic, find that their intuition doesn't match accepted physics, assume their understanding is at fault. Some small fraction, which apparently includes you, has the chutzpah to assume that physics is at fault. Where does such confidence come from? 

Trial & error.
BBC documentaries (unveiling the errors).
The bits & pieces add up.
Finding out what (does &) does not work well (for me).
If it doesn't work the way you say then I have to try it on my own.
(From) mom, =m*v.
Algebra.

A bit of psychology, how people react & behave.
Attempting (alternative) perspectives, til they fail.

P.S. My chutzpah requested a D'Arsenval meter's "angle versus electrical power" formula (&/or values) from your team
 (to improve my way of thinking),
 which still has NOT been delivered yet.

My chutzpah comes from
 the search for trying to find something reliable to depend on,
 because I've been disappointed, mislead & deceived too often.

Edited August 26, 2018 by Capiert

[Capiert]

On 24 August 2018 at 9:45 PM, John Cuthber said:

As I said earlier, in principle, I can make a current meter out of superconducting materials.

If I put it in a circuit with a battery and a bulb then a current will flow through it and the meter will turn through some angle.

But the voltage across the meter will be zero- because it has zero resistance.

So the power dissipated in the meter will be zero.

 

If I use two bulbs in parallel then there will be twice the current.

And the voltage across the meter will be zero and the power dissipated in the meter will still be zero- even though the deflection will be twice what it was.

You need to distinguish the power lost in the meter from the power delivered to the load.

I have to admire you at sidetracking this thread to the real cause, skipping all the intermediates.

Are you suggesting the (electro)magnetic force is purely inductive (repulsion)? E.g. Reactive.

 

Electromagnetism

 (closed circuit (circle) integral of B*dl=uo*I, Oersted's law https://en.m.wikipedia.org/wiki/Oersted's_law)

 is the magnetic field B produced around a wire

 (of length l, by electron current flow I, magnetic permeability in free space uo).

 

(Although, the length element's dl integral, looks like it's on the wrong side of the formula for me;

otherwise B is already unitized as a per "standard something?";

 we'd have to state how B is defined.)

 

Exclusively:

The mechanical force on the D'Arsenval meter's spring (from the torque's radius)

Fiv=I*(R^0.5)

 depends on the current I flowing thru the coil,

 & the circuit's (e.g. coil's) rooted_resistance (R^0.5).

We would have to insert Oersted law as current

 I=closed integral of B*dl/uo

 & use the light bulbs rooted_resistance

 to obtain the meter's deflection force Fiv, correctly.

E.g. Super conductor galvanometer.

 

As the NPL National Physics Lab stated (way back when),

 voltage is not trackable, instead current is.

 

I have used Piv=I*V because it's most commonly recognized, as easy,

 although its voltage V is not quite the real truth.

A D'Arsenval voltmeter uses an ampmeter with extra resistance to display what we call voltage.

 

(High impedance (voltage) measurements are a no_no=distortion (of the truth), only implying the real current.

Although I do NOT doubt their accuracy.)

Edited September 3, 2018 by Capiert

[Strange]

4 hours ago, Capiert said:

Are you suggesting the (electro)magnetic force is purely inductive (repulsion)? E.g. Reactive.

The words inductive, repulsive and reactive are not synonyms so this question is meaningless. (Apart from being a strawman.)

[Capiert]

 

5 hours ago, Strange said:

The words inductive, repulsive and reactive are not synonyms so this question is meaningless.

I mean, (to narrow in) are you suggesting the (repelling) magnetic force field

 (which I assume is repelling, something like a Lenz's law (inductive effect) on an atomic scale)

 is (kind of) caused by (something like) the inductive (current) reactance (part) of the coil?

I.e. Not the resistive (heating) part of the coil('s impedance equation, circuit. E.g. The inductance L).

 

I could have asked so (as 3 possibilities, which is closer=nearer?):

Are you suggesting the (electro)magnetic force is purely inductive? repulsive? or Reactive?

E.g. Whatever is needed to picture this thing (=effect) better, without resistance.

5 hours ago, Strange said:

(Apart from being a strawman.)

Otherwise I would have had to consider John's virtual (superconductor, zero (instead of minimal) resistance) galvanometer

 as the strawman (highjack)

 evading the copper (resistance) D'Arsenval (meter) question.

But I don't want to do that (because his proposal is so interesting, now).

Edited September 3, 2018 by Capiert

[swansont]

9 hours ago, Capiert said:

 

Exclusively:

The mechanical force on the D'Arsenval meter's spring (from the torque's radius)

Fiv=I*(R^0.5)

 

THAT IS NOT THE FORCE

R has units of kg m ² s ^-3 · A ^-2

I has units of amps, so in your equation that will cancel, leaving you with units of (kg m²/s³)^(0.5)

Force has units of kg m/s² 

You have the wrong units, thus, the equation must be incorrect.

[Capiert]

2 hours ago, swansont said:

THAT IS NOT THE FORCE

R has units of kg m ² s ^-3 · A ^-2

How do we know what units R has exactly?

George Ohm used R as a(n arbitrary) proportionality constant,

 just to "fit" his observations, V=I*(R).

https://en.m.wikipedia.org/wiki/Ohm's_law#History

In other words, didn't you guys just make it (=R's units) up, to fit best?

(E.g. without knowing exactly what a volt is,

 from the jolts Cavendish got.)

I mean, I could do the same non_sense (to pacify your needs)

 & multiply by a(n arbritrary) proportionality constant k3=kiv=1 [N/((V*A)^0.5]

 (NOT [N/(W^0.5)] in which the mechanical watt W is)

 giving

 Fiv=I*(R^0.5)*k3

 to explain the jolts Cavendish received on himself

 from electrification when touching the charged Leyden jar capacitors(' salted tube water).

But that still does NOT tell us what rooted_resistance is

 (aside from [(V/A)^0.5]).

Quote

I has units of amps, so in your equation that will cancel, leaving you with units of (kg m²/s³)^(0.5)

Force has units of kg m/s² 

Why do you write "amps" small case with the units having their short form as large (capital A);

 but force (is written with the 1st letter) large "Force" (although also large acronymn F)?

Is there any consistency in what you do (there)?

I mean sometimes you (would) write current (1st letter not captialized) in the middle of a sentence.

(If I'm not mistaken.?)

If the concepts are not people's names, then they are 1st letter capital

(exactly opposite of what is taught in school);

 & a person's name (place or thing) is 1st character capital.

Is there any method to (all) that madness?

Quote

You have the wrong units, thus, the equation must be incorrect.

The fact that my proportionality constant's value is exactly 1 (with above mentioned units),

 does NOT indicate I am (now) wrong.

Edited September 3, 2018 by Capiert

[Capiert]

P.S. When distinguishing between mechanical (watt) & (ridiculous) electrical watt.

From what I see (experimentally), (obviously, & intuitively) current's [Ampere] multiplied by rooted_resistance's [rooted_ohm] should give (the) force [Newton(s)].
That has "nothing" to do with your math finesse.
(You guys have fixed it so it won't!). So..

How was the electrical watt "derived" (by the Committee)?
 

Edited September 3, 2018 by Capiert

[Phi for All]

!

Moderator Note

Capiert, we entertain speculative ideas with the hopes that members with ideas outside the mainstream may pursue them within the methodology we know works best. If you refuse to allow members to correct your mistaken physics, if you refuse to take on board reasonable arguments against your concepts in order to either strengthen or discard them, then you aren't doing science. In addition, because of your refusal to learn, you keep repeating the same ignorant ideas, which means you're soapboxing, trying to lecture/preach to people who can easily see where you keep going wrong, and have been more than willing to tell you about it. 

Discussing anything with anyone who isn't listening is unrewarding and pointless. Please don't start anymore threads here at SFN unless you're willing to be swayed by superior evidence.