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.

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.

 

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.

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.)

?

Have you ever considered

 superconductivity might be a frozen magnetism (effect)

 in the cooling fluid,

 around a (super)conductor (catalyst)?

 E.g. The cooling fluid has frozen magnetism

 around near the superconductor surface.

That (surface) acts as a catalyst

 or seed starter for magnetism (domains).

The effect is external to the conductor, not internal;

 & it is (very strongly) magnetic

 on the outside of the conductor.

The superconductor's surface acts as a catalyst

 to kickstart the magnetic fluid's field,

 dealing with frozen magnetism

 in a cold fluid,

 requiring different temperatures for various superconducting metal's surfaces.

E.g. Normal non_magnetic materials

 such as fluids

 can be magnetized at lower temperatures

 with the appropriate catalyst surface

 & current thru the catalyst.

It's a magnetic conduction

 outside of the wire (in the fluid)

 (similar to a high frequency skin effect;

 & larger conduction path around the wire

 having less impedance

 due to more conduction volume

 thru the fluid, around the conductor).

 

Default answer is: no.

 

I suggest that

 because when the temperature is raised,

 the apparatus explodes.

7 hours ago, Strange said:

That is what the CR key is for. Adding blank lines (the clue is in the name).

?
CR=
Carriage =buggy, horse & carriage, old_fashioned time westerns.
&
Return.
Where does blank_line come into place there?

Quote

 If you don't want so many blank lines, then don't press the CR key so often. 

And here is another, related, tip: How about NOT pressing CR in the middle of a line as well ! That might make your posts look slightly less idiotic, even if it doesn't improve the content.

Thanks for the tips.
But it looks like I need to use: Shift+CR, instead of CR.

I.e. (I have to do more (keys), to get less (spacing)!

Very logical! ?

It still doesn't work right using a tablet. (As you can see.)

1 hour ago, swansont said:

Which stems from conservation of energy,

 in your mind (maybe?). I.e.

Which can stem also from conservation of energy.

(It might stem from coe (in your thinking); but it does NOT have to (in mine)!)

Where (from what I said)
 did you get the idea that
 linear acceleration
 & speed change
 are suppose to be (according to you) exclusively from energy?

I only stated linear acceleration (equivalent)
 & speed change (difference).

I said nothing about energy (there)!
You did!

You are using your mind's memory
 & trying to project it onto what I said
 when I said nothing of the sort.

I do not mind,
 if you attempt to recognize things in your (own energy) terms.
But please avoid the distortion, & explain what you mean.

E.g.
Do you mean
 that linear acceleration (also) is not possible
 with (just only, some) Newton's force F=m*a
 & some speed_difference v=vf-vi? (a=v/t).

Or do you mean Hooke's law F=-x*k?

Quote

which you are not allowed to assume.

I can't follow you, (there, yet).

I'm studying motion
 (from point A to point B).
It's that simple.

I don't see the error
 (nor a potential error).

Please clarify.

Quote

It certainly doesn't come from conservation of momentum.

I never said it (=linear acceleration a=v/t; & speed_difference v=vf-vi) did, either.

Perhaps that is why (normal) people find it difficult
 to discuss with scientists?

Please help. (Signed -Troubled. You lost me at the bakery.)

21 hours ago, Capiert said:

Assuming a linear acceleration equivalent,
 each phase ((elastic) compression & decompression)
 have the same amount of speed change vd1.

I've assumed an equivalent,
 & said (=set) it (=acceleration, for a speed_difference) to be linear,
 so that I can use it (=half the speed_difference) twice
 as 2 halves.

To define a (virtual) maximum_compression('s)
 point_in_time.

I.e. A virtual maximum
 of elastic force,
 where both (elastic) forces are (maximum,) opposite & equal.

We can not know all the details (of acceleration & elasticity);
 but we can equate (virtual speeds) to them, with an equivalent.

 

 

P.S. Caution

Error (to the Software (Bug) team):

While editing,
 marking text,
 & then pressing the backspace (key)
 (=attempting to delete the text),
 ejects me out from the thread
 onto another page (somewhere else).

 

30 minutes ago, swansont said:

I have taken the liberty of removing all of your text modifications, which take poor formatting and make it even harder to decipher (I mean, you say you can't type a proper way because of a crappy keyboard, but you can add colored text? Stop making things harder to understand. Just stop.)

My development history is very complicated

 with various syntaxes & attempts

 (I often can NOT decide which is best),

 I tried to give you a good cut

 otherwise you would have seen ~5x too much.

Last minute (insertions) changes, were added to clarify.

(I doubted your chances to understand (well) without (them).)

Quote

Another shortcoming is your use of non-standard terminology and symbols. I can't decipher what you are trying to do.

If you tell me what you do understand

 (put things in your own words),

 maybe we can take it from there.?

Quote

For example, this. You don't explain what the primed variables are, though it seems they are after the collision. But you have a variable u. What is this mystery variable? And where does this equation come from? 

" common (=mutual) non_elastic speed u´=(mom1+mom2)/(m1+m2); "

 

u' is the (equivalent, united) speed (velocity) after a non_elastic collision.

mom1+mom2=momt, total momentum momt=

before=after

m1*v1+m2*v2=(m1+m2)*u´, (non_elastic).

Yes (you are right) prime ´ is (for) after the collision.

(I assumed you would recognize them.)

  11 hours ago, Capiert said:

If I multiplied force by time, instead

 I would get momentum,

 which I would prefer (& trust more).

 

Swansont suggested (from my Fiv thread)

OK. Solve some problems involving thermodynamics with conservation of momentum. Go ahead. Try it. Or even a block sliding down an inclined plane with friction. Or an elastic collision, without involving energy.

That, too, has limits on practicality.

 

Here goes.

A few assumptions are necessary.

 

Collision takes time:

 

While colliding

 the 2 balls are:

 in contact with each other

 for a specific duration time (amount);

 & (thus) decelerating

 (to wrt a (united) non_elastic common_speed u´=(mom1+mom2)/(m1+m2)

 (& 1 specific point in time),

 that they reach=attain;

 & (then (they)) continue decellerating)

 til they disconnect.

 

After disconnection:

 No further acceleration occurs

 (because there is no affecting cause (e.g. elastic force, e.g. Hooke's force Fs=-x*k)

 when NOT (elastically) in contact, any more

 (=contact is missing).

 

(Then) at that (disconnected) point in time afterwards,

 the speed( velocitie)s are final.

v1´=2*u´-v1

v2´=2*u´-v2.

---

What do we know?

 

To summarize:

 

Collision takes time:

The 2 balls connect,

 indent (compressing, & decelerating),

 til effectively stopping

 wrt to a common=mutual non_elastic speed u´=(mom1+mom2)/(m1+m2);

 & then continue to decelerate while undeforming

 til they disconnect.

Assuming a linear acceleration equivalent,

 each phase ((elastic) compression & decompression)

 have the same amount of speed change vd1.

Thus the total speed change (for mass1) is 2*vd1=v1´-v1;

 & half the total (speed_changed (difference), for mass1)

 vd1=(v1’-v1)/2

 is an average.

 

In other words:

 mass1’s (initial) speed (velocity) v1

 (e.g. moving from left to right (is positive polarity)),

 decelerates

 to the (non_elastic) mutual (masses’) speed (velocity) u´

 ((unfortunately) here indicated as the addition

 of its (negative) (speed_difference) value vd1 ((that is) lost))

 v1+vd1=u´ (compression);

 & mass1 continues decelerating (from the mutual (masses’) speed u´)

 ((with) the other (negative) (speed_difference) value vd1 ((that is) lost))

 u´+vd1=v1´ (decompression, =decompressed)

 til its (=mass1’s) final_speed v1´.

 

 

So I’ve split the deceleration (e.g. of mass1)

 into 2,

 by using the

 (doubling method (syntax))

 (2)*vd1,

 (of the speed_difference vd1;

 instead of simply after-before=final-initial speeds)

 so I can easily (=conveniently) pinpoint

 the non_elastic (collision’s) time_point equivalent.

 

E.g. For identical masses (when we let m1=m2)

 we can (equivalently) observe if & when the speeds (velocities) reverse.

 

So let vd1=(v1´-v1)/2 be half of mass1’s (decelerated)
 (final minus initial) speed (velocities’) difference.

 

Rewritten:

u´=v1+vd1 (compression result), &

u´=v1´-vd1 (uncompression), add both equations gives

2*u´=v1+v1´, /2=divide both sides by 2, gives

u´=(v1+v1´)/2

Again

vd1=u´-v1, &

vd1=v1´-u´.

Significant is (that)

 mass1’s final speed (velocity) is

v1´=2*u´-v1, for after the elastic collision,

 & using the mutual (united) non_elastic speed (velocity)

 u´=(mom1+mom2)/(m1+m2)

 for the total mass mt=m1+m2 (or mass_sum ms=mt),

 where

 mass1’s initial momentum is

 mom1=m1*v1,

 & mass2’s initial momentum is

 mom2=m2*v2.

 

The same can be done for mass2:

u´=v2+vd2 (compression result), &

u´=v2´-vd2 (decompression), add both equations gives

2*u´=v2+v2´, /2=divide both sides by 2, gives

u´=(v2+v2´)/2

Again

vd2=u´-v2, &

vd2=v2´-u´.

Significant is (that)

 mass2’s final speed (velocity) is

v2´=2*u´-v2, for after the elastic collision,

 & using the mutual (united) non_elastic speed (velocity)

 u´=(mom1+mom2)/(m1+m2),

 where

 mass1’s initial momentum is

 mom1=m1*v1,

 & mass2’s initial momentum is

 mom2=m2*v2.

 

Thus mass1’s final speed (velocity) (after elastic collision) is

v1´=2*((mom1+mom2)/(m1+m2))-v1, or

v1´=2*((m1*v1+m2*v2)/(m1+m2))-v1;

&

mass2’s final speed (velocity) (after elastic collision) is

v2´=2*((mom1+mom2)/(m1+m2))-v2, or

v2´=2*((m1*v1+m2*v2)/(m1+m2))-v2.

 

Is that something like what you want Swansont?

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.

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.

I.e. I can not confirm what does not exist, namely a change in r, wrt time,

 either as speed or acceleration.

But instead, (x,y) cartesian coordinates are perfect (for the job),

 I can confirm (fall of an orbit) with using all 3 (r, x, & y) parameters in 2D;

 or else in x,y,z, all 4 (r^2=x^2 + y^2 + z^2) parameters in 3D.

1 hour ago, Capiert said:

Sorry I forgot the rooting

That should read:

 

I'm content with the (circular) orbital period

T=2*Pi*((R/g)^0.5)

 is (like) the pendulum period

T=2*Pi*((L/g)^0.5)

 the pendulum's length L=R the orbit's radius;

 & the circumferential_speed vc=cir/T=2*Pi*R/T.

 

21 minutes ago, swansont said:

So you've failed basic physics. No surprise. 

I'm exclusively discussing vertically. i.e. radius r.

21 minutes ago, swansont said:

Objects with no acceleration mover in a straight line (Newton's first law). Are you familiar with Newton's laws of motion? They're kind of important.

No bout adout it in x & y.

21 minutes ago, swansont said:

It didn't crash — it missed us. And keeps missing us, because of the way it's moving. But it's accelerating toward us.

Quite right.

Again x,y,z.

To summarize,

 I don't find a delta r wrt time;

 I only find delta x

 & delta y

 wrt time.

You'll need a different strategy

 to convince otherwise

 for the radial coordinate system.

I cannot lie to myself

 & believe it.

Sorry it won't work until you guys start making sense

 on that point,

 so it's better to drop the radial theme

 til then.

47 minutes ago, Strange said:

So, you didn’t read the explanations in the links I provided. Lazy as well as arrogant. 

Sorry, I had to reinstall the adobe flash player. But you're right.

Am I so arrogant: to try new things, not known to you?

Is that being arrogant?

Quote

Of course not. If you understood it, you wouldn’t be making such ludicrous assertions. 

I can't say I liked the explanation.

It's a (little) deceptive (=misleading) picture, making use of falling on 1 (e.g. left) side,

 & (but) ascending on the other (e.g. right side).

 

So the spacecraft just falls all the way around the Earth, never hitting the surface. The curve of the spacecraft's path is about the same as the curve of Earth's surface.

So astronauts orbiting Earth aren't really weightless, they are just falling . . . and falling, and falling!

 

That doesn't make (complete) sense.

It only exploits the earthly falling mentality,

 such as falling (=starting to fall),

 & a roller coaster looping.

E.g. with gravity always working at the bottom of the picture,

 even below the earth.

Why should there be (even more) gravity below (under) the earth in that picture?

 

I'm content with the (circular) orbital period

T=2*Pi*(R/g)

 is (like) the pendulum period

T=2*Pi*(L/g),

 the pendulum's length L=R the orbit's radius;

 & the circumferential_speed vc=cir/T=2*Pi*R/T.

 

That doesn't work for the earth to moon distance d,

 in which I must conclude

 the barycenter radii would have to be in line,

 as to create maybe

 Re+d=Rm

 or

 d=Rm-Re.

& the only way to decern

which (Re or Rm) had the larger radius

would be to notice a wooble (magnitude or amount) wrt the stars (positions, after a sidereal day 23h 56m 4s).

10 minutes ago, Strange said:

I know exactly what I am doing: Trying to explain schoolboy physics to someone who is too arrogant to learn anything.

(What is falling?)

Any object in orbit.

Why doesn't its height change then

 if it is suppose to be falling?

I don't see any fall involved.

Quote

It doesn't matter what coordinate system you use. It is still falling.

I'm sorry that makes no sense.

Quote

The fact that you think changing the coordinates makes a difference is another illustration of how profoundly ignorant your are.

Thanks, profoundly.

Quote

Its not my choice. I haven't even reported it to the mods (your immaturity and ignorance is mildly amusing).

Thanks, I guess we have to leave the (radial) theme.

53 minutes ago, Strange said:

I didn't use ANY coordinates.

If you don't know (=recognize, exactly) what you are doing

 then I can't help you much

 until you recognize it yourself.

That might take you some time. Good luck.

I can only hint,

 & hope you get it someday.

If the radius stays the same,

 what is falling? Answer: only x & y's change (but that's cartesian).

r alone demonstrates no change, thus no acceleration (of the height).

Got it?

Quote

Are you deliberately lying

No. Not that I know.

Quote

or are you just incapable of understanding what you read?

Maybe. If it makes nonsense.

Please consider:

radius r=((x^2)+(y^2))^0.5, is (a constant or variable) legal for cartesian x,y coordinates;

 but x & y are not necessarily legal (as in reverse) for radial coordinates.

Either you're discussing (only) radial (which does NOT include x,y because they are named Cartesian)

 or else Cartesian (which can include the radius).

I guess your professors weren't very fussy.

Quote

All I can see is you making a hash of ... well, everything.

I see your education has made you incapable of distinguishing some things.

Let's drop (the radial theme), you don't get.

Please leave the thread open.

The significant (speculative) part has been said

 (but you're still against it).

2 minutes ago, Strange said:

Do you really expect to understand a different explanation? 

No, it was a rhetorical question.

I explained it was NOT falling in radially coordinates;

 but you insist on using cartisian coordinates with them radial

 to cheat the perspective.

2 minutes ago, Strange said:

You are incapable of learning. Full stop. There is nothing more to say. 

I question whether you can see the perspective mixture

 created for you in your institutions.

2 minutes ago, Strange said:

You might as well ask the mods to close this thread as all you are going to do is repeat the same nonsensical gibberish.

We can drop the (radial) theme, instead of the thread,

 if you are not willing to notice the hash in coordinates.

No sense if you're NOT ready for it yet.

20 minutes ago, Strange said:

What a surprise.

Does that mean a helicopter, hovering at constant height is falling (too)?

No. 

Why not?

26 minutes ago, Strange said:

Because the Earth is curved and the orbiting object is falling towards it. I mean, really. Sheesh.

Your ignorance is absolutely staggering. Unbelievable. I know people who propose their own "personal theories" are usually relatively uneducated but you don't know ANYTHING. You know NOTHING. I cannot believe how little you know. How did you manage to learn so little in school?

Yes I must admit it was difficult

(getting stuffed with too much info, that conflicts).

Quote

http://www.astronautix.com/n/newtonsorbitalcannon.html

Puhlease try and learn the sort of basic physics that schoolboys know before going round trying to say that all of modern physics is wrong. Your attitude is just delusional. No one will take you seriously when you base your ludicrous assertions on total ignorance.

I suppose learning is hard work and making stuff up is easy, but it is just pathetic to watch. Please stop it: You are embarrassing yourself and disappointing the rest of us.

Sorry, but the paradoxes stand out (to me).

Quote

Newton 1687. 

the first and best explanation of what an orbit is. An object in orbit is weightless not because 'it is beyond the earth's gravity' but because it is in 'free-fall' - just like a skydiver. The difference is that it has enough horizontal speed never to hit the ground.

How then is it falling? It does not fall (wrt radial coordinates;

instead (it falls) only wrt cartisian x,y,z coordinates, because it curves.)

Your (radial) perspective is inconsistent.

Radius r, & height h do NOT change when weightless.

26 minutes ago, John Cuthber said:

No

An object in orbit is always falling. It accelerates towards the thing it's orbiting round.

Sorry John,

 but that doesn't make sense to me

 from that perspective.

I haven't (really) noticed when the moon fell down onto the earth, & crashed.

Significantly the moon earth distance varies a bit,

 but radially to say "fall" as in "down",

 isn't that relatively out of the question in radial coordinates?

Does that mean a helicopter, hovering at constant height is falling (too)?

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It doesn't matter what coordinate system you use, it's still falling.

I'm talking about the observed phenomena of decent to the earth's surface.

Are clouds (that are hanging in the air) falling?

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Did it occur to you that it may not be our understanding that is wrong?

Yes, sometimes.

(I'm not perfect.)

On ‎2018‎ ‎08‎ ‎17 at 3:49 PM, swansont said:

That's even worse, since in a circular orbit (unchanging r) there must be an acceleration.

Radially there is no fall acceleration (neither) up (n)or down

 because the height is constant

 (as part of the radius);

 although, acceleration occurs in both x,y coordinates.

(I hope that's said ok, now.)

On ‎2018‎ ‎08‎ ‎17 at 3:49 PM, swansont said:

You don't know which one (=watch is wrong, when 2 exist), though. 

Naturally.

That's usually determined by which 1 is different of 2 from the 3rd (or else 3 watches). 

On ‎2018‎ ‎08‎ ‎17 at 8:12 AM, Strange said:

That makes no sense, even if we remove the random punctuation.

(For an orbit, I said zero (vertical) fall acceleration was equilibrium.)

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If it is orbiting then it is constantly accelerating and falling. 

(I suppose your are talking about x,y coordinates, instead of my radial coordinates?

Maybe that is why you did not understand?)

While orbiting, & the radial height distance is constant,

 then the orbiting object is (radially) neither falling, nor rising.

Just now, Strange said:

It is hidden so how can we know?

If the device is disassembled

 & no hidden mechanism is found

 then your assumed guess is useless.

Just now, Strange said:

Why not contact the person who created the video, politely point out that you know they are a fraud

Why should I do something like that

 & lie, doing something that I am not convinced of?

Just now, Strange said:

and ask if they would be willing to explain how it was done.

Your idea is perhaps not new,

 but nobody has succeeded

 (to report the exact mechanism,

 as far as I can see).

Just now, Strange said:

Good luck.

Thanks, but if it is so easy as you say,

 then surely you will have more success than me.

Sorry, that has (also) not answered my question.

(I've stated how it functions

 if the exact mechanism is not otherwise declared.)

On ‎2018‎ ‎08‎ ‎17 at 3:56 PM, swansont said:

Phineas Taylor

Thank you.

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A hidden mechanism. A cheat.

Sorry, that still didn't answer my question.

I.e. What is the mechanism?

Independent if hidden or not.

On ‎2018‎ ‎08‎ ‎17 at 4:14 PM, Strange said:

The fake perpetual motion machine isn't backed up by anything except a fake video.

I didn't ask whether it was fake or not.

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Magic. Or invisible pink unicorns. 

Sorry, that also didn't answer my question.

On ‎2018‎ ‎08‎ ‎17 at 4:38 PM, koti said:

What I don’t understand is people looking at obviously useless ways of finding energy when there are virtually infinite (at least for practical reasons) sources of energy like solar and hydrogen right at our fingertips to use.

Yes, it seems the people at this forum do not understand (my question).

Sorry, that did not answer my question.