# How can we slow down g?

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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))

..13:39

I was a little puzzled

that I couldn’t make that conclusion

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

e.g. why it confused me at 1st;

& 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 1st 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.

Edited by Capiert
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None of that changes g, and making something smaller does not change its sign.

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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.

Quote

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).

Edited by Capiert
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13 minutes ago, Capiert said:

I get the idea those changes are altering the polarity (sign).

Then your "idea" is clearly wrong.

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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?

Edited by Capiert
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11 minutes ago, Capiert said:

What's wrong?

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

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How you're able to come up with chaniging the font colors without noticing the nonsense your're posting is beyond me.

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10 minutes ago, Strange said:

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

Do you think the penny will drop one day?

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10 hours ago, Capiert said:

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

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.

10 hours ago, Capiert said:

in stunts.

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.

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.

10 hours ago, Capiert said:

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).

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.

9 hours ago, Capiert said:

Zero polarity is equilibrium, there.

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

What's wrong?

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

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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.)

Quote

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).

Edited by Capiert
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4 hours ago, Capiert said:

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

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

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

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6 hours ago, Capiert said:

Nobody is arguing with you there.

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

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

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.

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.)

Direction matters as well  with velocity.

Edited by Endy0816
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11 hours ago, Capiert said:

Nobody is arguing with you there.

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

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

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

11 hours ago, Capiert said:

Never.

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

You don't know which one, though.

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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.)

Quote

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

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.

Edited by Capiert
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10 minutes ago, Capiert said:

then the orbiting object is neither falling, nor rising.

No

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

11 minutes ago, Capiert said:

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

11 minutes ago, Capiert said:

Maybe that is why you did not understand?)

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

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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

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).

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3 minutes ago, Capiert said:

because the height is constant

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?

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.

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.

Edited by Strange
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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)?

Quote

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?

Quote

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

Yes, sometimes.

(I'm not perfect.)

Edited by Capiert
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1 minute ago, Capiert said:

but that doesn't make sense to me

What a surprise.

1 minute ago, Capiert said:

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

No.

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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

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.)

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

Edited by Capiert
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11 minutes ago, Capiert said:

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

They are entirely figments of your fevered imagination.

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20 minutes ago, Strange said:

What a surprise.

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

No.

Why not?

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12 minutes ago, Capiert said:

How then is it falling?

Do you really expect to understand a different explanation?

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

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

Just now, Capiert said:

Why not?

Because it isn't orbiting. Sheesh. I think I will have to put you on ignore. I am worried that this level of stupidity might be catching.

And didn't you mean to write:

Quote

w*H--y!

N

((o)]]

t

>>>?

In the vain hope that ONE of these might get through your thick skull, here are a few more explanations of what an orbit is:

Please read these before continuing. And I mean read them and not just look at them and go "Oh they must be wrong b'cos I am such a frigging genius and know those dumb scientists can't even add up."

And an interactive web page where you can try it yourself:

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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.

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

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

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13 minutes ago, Capiert said:

I explained it was NOT falling in radially coordinates;

but you insist on using cartisian coordinates with them radial

I didn't use ANY coordinates.

Are you deliberately lying or are you just incapable of understanding what you read?

14 minutes ago, Capiert said:

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

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

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