On Wednesday, October 19, 2022 at 6:40:01 PM UTC-4, Richard Hachel wrote:
> Le 19/10/2022 à 23:40, rotchm a écrit :
> >> Si l'on prend x=(1/2)aTr², on peut alors poser a=2x/Tr².
> > And, if you declare that "a" is the proper acceleration, then the formula for x
> > as a function of
> > tau (your Tr here?) is not x=(1/2)aTr².
> No.
>
> YOU, you say that tau (or Tr) using here is not x=1/2aTr².
>
> And me, I said it is.

No it is not. You are very confused.
If 'a' is proper acceleration, then x=1/2aTr² is not true by definition.
So either you are using the wrong formula, or the wrong word for your 'a'.

As pointed out in my previous post, it's your 'a' that does not represent proper acceleration;
It is something else (with units of acceleration). Give it another name instead of proper acceleration. This is because, proper acceleration already has a very specific meaning in physics, and it is not x=1/2aTr² .

Le 20/10/2022 à 00:58, rotchm a écrit :

> Give it another name instead of proper acceleration.

And why?

>This is because, proper acceleration already has a very specific meaning in
>physics, and it is not x=1/2aTr² .

Yes, it is.

R.H.

On Wednesday, October 19, 2022 at 7:10:57 PM UTC-4, Richard Hachel wrote:
> Le 20/10/2022 à 00:58, rotchm a écrit :
>
> > Give it another name instead of proper acceleration.
> And why?

Already answered that many times. can't you read? or, comprehend what you read?
See below.

> >This is because, proper acceleration already has a very specific meaning in
> >physics, and it is not x=1/2aTr² .

That's why, DuH.

> Yes, it is.

Can you support your claim?
Where is your source that says that that is the definition or meaning ?

On Wednesday, October 19, 2022 at 1:58:20 PM UTC-7, Richard Hachel wrote:
> > In terms of the x,t coordinates, the velocity is defined as
> > dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1), which gives at x=3 and
> > x=3.1 the velocities v1=0.970620 and v2=0.972074. This is
> > what the phrase "velocity in terms of x,t" means. This is not
> > open to debate, it is the definition. These are the velocities
> > you would measure with standard rulers and clocks at rest
> > and inertially synchronized in the x,t frame. Agreed?
>
> No. The correct direct equation is...

Again, in terms of the x,t coordinate system, you agreed that the trajectory of the rocket is t = sqrt(x^2 + 2x/a), and by simple high school math the velocity is dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1). Therefore, at x=3 and x=3.1 the velocities are v1=0.970620 and v2=0.972074 respectively. Agreed? Anything else would logically contradict the relation t = sqrt(x^2 + 2x/a), which you have accepted. Understand?

Le 20/10/2022 à 01:43, Stan Fultoni a écrit :
> On Wednesday, October 19, 2022 at 1:58:20 PM UTC-7, Richard Hachel wrote:
>> > In terms of the x,t coordinates, the velocity is defined as
>> > dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1), which gives at x=3 and
>> > x=3.1 the velocities v1=0.970620 and v2=0.972074. This is
>> > what the phrase "velocity in terms of x,t" means. This is not
>> > open to debate, it is the definition. These are the velocities
>> > you would measure with standard rulers and clocks at rest
>> > and inertially synchronized in the x,t frame. Agreed?
>>
>> No. The correct direct equation is...
>
> Again, in terms of the x,t coordinate system, you agreed that the trajectory of
> the rocket is t = sqrt(x^2 + 2x/a), and by simple high school math the velocity is
> dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1). Therefore, at x=3 and x=3.1 the velocities are
> v1=0.970620 and v2=0.972074 respectively. Agreed? Anything else would logically
> contradict the relation t = sqrt(x^2 + 2x/a), which you have accepted.
> Understand?

On Wednesday, October 19, 2022 at 7:48:20 PM UTC-4, Richard Hachel wrote:
> Le 20/10/2022 à 01:43, Stan Fultoni a écrit :
> > On Wednesday, October 19, 2022 at 1:58:20 PM UTC-7, Richard Hachel wrote:
> >> > In terms of the x,t coordinates, the velocity is defined as
> >> > dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1), which gives at x=3 and
> >> > x=3.1 the velocities v1=0.970620 and v2=0.972074. This is
> >> > what the phrase "velocity in terms of x,t" means. This is not
> >> > open to debate, it is the definition. These are the velocities
> >> > you would measure with standard rulers and clocks at rest
> >> > and inertially synchronized in the x,t frame. Agreed?
> >>
> >> No. The correct direct equation is...
> >
> > Again, in terms of the x,t coordinate system, you agreed that the trajectory of
> > the rocket is t = sqrt(x^2 + 2x/a), and by simple high school math the velocity is
> > dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1). Therefore, at x=3 and x=3.1 the velocities are
> > v1=0.970620 and v2=0.972074 respectively. Agreed? Anything else would logically
> > contradict the relation t = sqrt(x^2 + 2x/a), which you have accepted..
> > Understand?

Replying without adding any content is useless. You are just spamming.
Did you agree that the trajectory of the rocket is t = sqrt(x^2 + 2x/a)?

Le 20/10/2022 à 01:43, Stan Fultoni a écrit :
> On Wednesday, October 19, 2022 at 1:58:20 PM UTC-7, Richard Hachel wrote:
>> > In terms of the x,t coordinates, the velocity is defined as
>> > dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1), which gives at x=3 and
>> > x=3.1 the velocities v1=0.970620 and v2=0.972074. This is
>> > what the phrase "velocity in terms of x,t" means. This is not
>> > open to debate, it is the definition. These are the velocities
>> > you would measure with standard rulers and clocks at rest
>> > and inertially synchronized in the x,t frame. Agreed?
>>
>> No. The correct direct equation is...
>
> Again, in terms of the x,t coordinate system, you agreed that the trajectory of
> the rocket is t = sqrt(x^2 + 2x/a),

I have told you a hundred times already that I completely agree with
relativistic physicists when they gave their equation
To=(x/c).sqrt(1+2c²/ax) equation that I have found myself, and that I
write as you see.

>and by simple high school math the velocity is dx/dt = a*sqrt(x^2 +
>2x/a)/(ax+1).

? ? ?

Vo=x/To

Vo=x/[(x/c).sqrt(1+2c²/ax)]

Vo/c=[1+c²/2ax]^-0.5

A l'école française, on m'apprenait comme ça.

De quelle nationalité es-tu Stan Fultoni?

> Therefore, at x=3 and x=3.1 the velocities are v1=0.970620

No.

>and v2=0.972074 respectively.

Nope.

Stan, please, I beg you to make an effort

N.B. Please, what is your nationality?

R.H.

Le 20/10/2022 à 01:57, rotchm a écrit :
> On Wednesday, October 19, 2022 at 7:48:20 PM UTC-4, Richard Hachel wrote:
>> Le 20/10/2022 à 01:43, Stan Fultoni a écrit :
>> > On Wednesday, October 19, 2022 at 1:58:20 PM UTC-7, Richard Hachel wrote:
>> >> > In terms of the x,t coordinates, the velocity is defined as
>> >> > dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1), which gives at x=3 and
>> >> > x=3.1 the velocities v1=0.970620 and v2=0.972074. This is
>> >> > what the phrase "velocity in terms of x,t" means. This is not
>> >> > open to debate, it is the definition. These are the velocities
>> >> > you would measure with standard rulers and clocks at rest
>> >> > and inertially synchronized in the x,t frame. Agreed?
>> >>
>> >> No. The correct direct equation is...
>> >
>> > Again, in terms of the x,t coordinate system, you agreed that the trajectory
>> of
>> > the rocket is t = sqrt(x^2 + 2x/a), and by simple high school math the
>> velocity is
>> > dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1). Therefore, at x=3 and x=3.1 the velocities
>> are
>> > v1=0.970620 and v2=0.972074 respectively. Agreed? Anything else would
>> logically
>> > contradict the relation t = sqrt(x^2 + 2x/a), which you have accepted.
>> > Understand?
>
> Replying without adding any content is useless. You are just spamming.
> Did you agree that the trajectory of the rocket is t = sqrt(x^2 + 2x/a)?

Oui, je l'ai dit.

Son équation est correcte.

C'est la même chose que ce que j'écris moi (avec ma notation
particulière).

Mais après, je ne sais pas ce qu'il fout.

Il fait n'importe quoi et trouve n'importe quoi...

R.H.

On Wednesday, October 19, 2022 at 8:11:32 PM UTC-4, Richard Hachel wrote:
> Le 20/10/2022 à 01:57, rotchm a écrit :
> > On Wednesday, October 19, 2022 at 7:48:20 PM UTC-4, Richard Hachel wrote:
> >> Le 20/10/2022 à 01:43, Stan Fultoni a écrit :

> >> > the rocket is t = sqrt(x^2 + 2x/a), and by simple high school math the
> >> > dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1). Therefore, at x=3 and x=3.1 the velocities
> >> > Understand?

> > Did you agree that the trajectory of the rocket is t = sqrt(x^2 + 2x/a)?
> Oui, je l'ai dit.

Why are you changing language? That does not help for the good flow of the conversation.

> C'est la même chose que ce que j'écris moi (avec ma notation particulière).
> Mais après, je ne sais pas ce qu'il fout.
> Il fait n'importe quoi et trouve n'importe quoi...

Are you saying that you can't do basic high school algebra and one derivative (calculus)?

velocity is defined by dx/dt right?
And since you agree that t = sqrt(x^2 + 2x/a), then can you show that dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1) ?
It's just algebra so surely you can show, it no?

On Wednesday, October 19, 2022 at 5:07:01 PM UTC-7, Richard Hachel wrote:
> > Again, in terms of the x,t coordinate system, you agreed that the trajectory of
> > the rocket is t = sqrt(x^2 + 2x/a),
>
> I have told you a hundred times already that I completely agree...

Yes, but you have also contradicted yourself a hundred times by denying the derivative of this.

> > and by simple high school math the velocity is
> > dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1).
>
> ? ? ?

Let's take tiny baby steps. You agree that the rocket's trajectory is t = sqrt(x^2 + 2x/a), and this means that at x1=3.0 we have t1=3.834504, and at x2=3.1 we have t2=3.937453. Now, in terms of these coordinates, the average velocity of the rocket on this segment is (x2 - x1) / (t2 - t1) = 0.971356.

This is what the word "velocity" means, i.e., the change in spatial position for a given change in time. Do you understand this?

On 10/18/2022 9:22 PM, Richard Hachel wrote:

> But strange thing, it is the theory which sometimes takes precedence
> over the experiment.

So you *are* a reality denier!

On Thursday, 20 October 2022 at 05:23:35 UTC+2, Volney wrote:
> On 10/18/2022 9:22 PM, Richard Hachel wrote:
>
> > But strange thing, it is the theory which sometimes takes precedence
> > over the experiment.
>
> So you *are* a reality denier!

Sorry, stupid Mike, you have no clue what the reality is.

Le 20/10/2022 à 05:23, Volney a écrit :
> On 10/18/2022 9:22 PM, Richard Hachel wrote:
>
>> But strange thing, it is the theory which sometimes takes precedence
>> over the experiment.
>
> So you *are* a reality denier!

No, clearly not.

R.H.

Le 20/10/2022 à 01:43, Stan Fultoni a écrit :

> Again, in terms of the x,t coordinate system, you agreed that the trajectory of
> the rocket is t = sqrt(x^2 + 2x/a),

Yes.

To=sqrt(1+2c²/ax)

I kept repeating it.

It is unavoidable. I really appreciate that you can understand and
approve of it.

>and by simple high school math the velocity is dx/dt = a*sqrt(x^2 +
>2x/a)/(ax+1).

This is where your error lies.

Your calculation is correct but based on physics which is wrong.

This equation is not good.

The good équation is :

Vo=x/To then Vo=x/[sqrt(1+2c²/ax)] then Vo/c=[(c²/2ax)+1]^-0.5

Not yours.

> Therefore, at x=3 and x=3.1 the velocities are v1=0.970620 and v2=0.972074
> respectively. Agreed? Anything else would logically contradict the relation t =
> sqrt(x^2 + 2x/a), which you have accepted. Understand?

No.

Therefore, at x=3 and x=3.1 the velocities are Vo1=0.78237c and
Vo2=0.78731c

Note : Vr1=5.02455c/4 for x=12/4=3

R.H.

On Thursday, October 20, 2022 at 7:24:41 AM UTC-4, Richard Hachel wrote:
> Le 20/10/2022 à 01:43, Stan Fultoni a écrit :
> > Again, in terms of the x,t coordinate system, you agreed that the trajectory of
> > the rocket is t = sqrt(x^2 + 2x/a),
> Yes.
> >and by simple high school math the velocity is dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1).
> This is where your error lies. Your calculation is correct but based on physics which is wrong.

The 'physics' will be wrong if it makes incorrect predictions.

> This equation is not good.

The equation comes from the definition of speed, dx/dt and this gives
a*sqrt(x^2 + 2x/a)/(ax+1). You agreed that this is the result of dx/dt.
Now, suffice to experimentally verify it to see if its good.
This exp (or runoffs) have been done and the formulas are confirmed.
Hence the physics (the model) remains good: it makes correct predictions.

> The good équation is :
> Vo=x/To then Vo=x/[sqrt(1+2c²/ax)] then Vo/c=[(c²/2ax)+1]^-0.5
> Not yours.

But 'ours' agrees with actual exps and your doesn't.
So you are a reality denier.

> > Therefore, at x=3 and x=3.1 the velocities are v1=0.970620 and v2=0.972074
> > respectively. Agreed? Anything else would logically contradict the relation t =
> > sqrt(x^2 + 2x/a), which you have accepted. Understand?
> No.
> Therefore, at x=3 and x=3.1 the velocities are Vo1=0.78237c and
> Vo2=0.78731c

If you plug the value x=3 in a*sqrt(x^2 + 2x/a)/(ax+1) what result do you get ?

From t = sqrt(x^2 + 2x/a), can you show that dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1) ?
Do answer this one because it will show us if you have a stable thinking process and
if you have the required logical and math skills to partake in these analyses.

> > Why are you changing language? That does not help for the good flow of the conversation.

No answer?

> > velocity is defined by dx/dt right?

No answer?

On Thursday, 20 October 2022 at 14:38:46 UTC+2, rotchm wrote:
> On Thursday, October 20, 2022 at 7:24:41 AM UTC-4, Richard Hachel wrote:
> > Le 20/10/2022 à 01:43, Stan Fultoni a écrit :
> > > Again, in terms of the x,t coordinate system, you agreed that the trajectory of
> > > the rocket is t = sqrt(x^2 + 2x/a),
> > Yes.
> > >and by simple high school math the velocity is dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1).
> > This is where your error lies. Your calculation is correct but based on physics which is wrong.
> The 'physics' will be wrong if it makes incorrect predictions.
> > This equation is not good.
> The equation comes from the definition of speed, dx/dt

Which is a common sense prejudice, refuted by your idiot
gurus with their "inflation" discovery.

On Thursday, October 20, 2022 at 8:47:55 AM UTC-4, maluw...@gmail.com wrote:
> On Thursday, 20 October 2022 at 14:38:46 UTC+2, rotchm wrote:

> > The equation comes from the definition of speed, dx/dt
> Which is a common sense prejudice,

So, a definition is 'prejudice' ? LOL. You are really a crockpot now.
Tell me, what is the definition of velocity or speed in our context?

> refuted by your idiot
> gurus with their "inflation" discovery.

A lie from The Fanatic trash you are.
We the gurus still use the definition of velocity as dx/dt, even in 'inflation discovery'.

But you are a crackpot that denies reality. This is why you are stuck in a situation you are; you
don't have the brains to progress nor to get out of your own troubles... You are a... crackpot.

On Wednesday, October 19, 2022 at 5:07:01 PM UTC-7, Richard Hachel wrote:
> > Again, in terms of the x,t coordinate system, you agreed that the trajectory of
> > the rocket is t = sqrt(x^2 + 2x/a), and by simple high school math the velocity is
> > dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1).
>
> ? ? ?

Let's take small steps. You agree that the rocket's trajectory is t = sqrt(x^2 + 2x/a), and this means that at x1=3.0 we have t1=3.834504, and at x2=3.1 we have t2=3.937453. Now, in terms of these coordinates, the average velocity of the rocket on this segment is (x2 - x1) / (t2 - t1) = 0.971356. This is what the word "velocity" means, i.e., the change in spatial position for a given change in time. Do you understand this?

n Thursday, 20 October 2022 at 14:55:01 UTC+2, rotchm wrote:
> On Thursday, October 20, 2022 at 8:47:55 AM UTC-4, maluw...@gmail.com wrote:
> > On Thursday, 20 October 2022 at 14:38:46 UTC+2, rotchm wrote:
>
> > > The equation comes from the definition of speed, dx/dt
> > Which is a common sense prejudice,
> So, a definition is 'prejudice' ? LOL. You are really a crockpot now.
> Tell me, what is the definition of velocity or speed in our context?

You have no, you're mumbling without any meaning,
like some idiots you are.

> > refuted by your idiot
> > gurus with their "inflation" discovery.
> A lie from The Fanatic trash you are.
> We the gurus still use the definition of velocity as dx/dt, even in 'inflation discovery'.

Oh, really? Quoting one of them:
"In an inflationary Universe, any two particles, beyond a tiny fraction of a second, will see the other one recede from them at speeds appearing to be faster-than-light. But the reason for this isn't because the particles themselves are moving, but rather because the space between them is expanding."

While it's a piece of mumble, of course - it's clear
that dx/dt >c here with "not themselves [particles]
are moving"

On 10/20/2022 7:05 AM, Richard Hachel wrote:
> Le 20/10/2022 à 05:23, Volney a écrit :
>> On 10/18/2022 9:22 PM, Richard Hachel wrote:
>>
>>> But strange thing, it is the theory which sometimes takes precedence
>>> over the experiment.
>>
>> So you *are* a reality denier!
>
> No, clearly not.
>
Stating that the theory sometimes takes precedence over experiment is
denying reality. Denying the reality of experimental results in favor of
some "theory" is denying reality. In this topic, many others have
pointed out the flaws in your claims. Denying the math of integration is
also denying reality.

Denying reality is delusional, and delusions are a symptom of various
forms of mental illnesses. Please seek a mental health professional.

On Thursday, October 20, 2022 at 9:33:52 AM UTC-4, maluw...@gmail.com wrote:
> n Thursday, 20 October 2022 at 14:55:01 UTC+2, rotchm wrote:

> > Tell me, what is the definition of velocity or speed in our context?
> You have no, you're mumbling without any meaning,

So you admit that you have no clue on what speed means.
You are nil in language as you are in physics and math. You don't belong here.
Have you tried rock collecting yet?

> While it's a piece of mumble, of course - it's clear
> that dx/dt >c here with "not themselves [particles]
> are moving"

Nope. You don't understand what that means. So what you don't understand what you read you can invent any nonsense you want in your demented head. I told you, stick to rock collecting.

rotchm wrote:

> On Thursday, October 20, 2022 at 9:33:52 AM UTC-4, maluw...@gmail.com
> wrote:
>> n Thursday, 20 October 2022 at 14:55:01 UTC+2, rotchm wrote:
>
>> > Tell me, what is the definition of velocity or speed in our context?
>> You have no, you're mumbling without any meaning,
>
> So you admit that you have no clue on what speed means.
> You are nil in language as you are in physics and math. You don't belong
> here.
> Have you tried rock collecting yet?

shut the fuck up, fool! *I_fuck_your_ass!*.

On Thursday, 20 October 2022 at 17:36:39 UTC+2, rotchm wrote:
> On Thursday, October 20, 2022 at 9:33:52 AM UTC-4, maluw...@gmail.com wrote:
> > n Thursday, 20 October 2022 at 14:55:01 UTC+2, rotchm wrote:
>
> > > Tell me, what is the definition of velocity or speed in our context?
> > You have no, you're mumbling without any meaning,
> So you admit that you have no clue on what speed means.

No, I don't admit. You're fabricating and lying, as
expected from relativistc scum in general and
from you especially.

> You are nil in language as you are in physics and math. You don't belong here.
> Have you tried rock collecting yet?
> > While it's a piece of mumble, of course - it's clear
> > that dx/dt >c here with "not themselves [particles]
> > are moving"
> Nope. You don't understand

Yes. You're denying blindly, like always.

On Wednesday, October 19, 2022 at 5:07:01 PM UTC-7, Richard Hachel wrote:
> > Again, in terms of the x,t coordinate system, you agreed that the trajectory of
> > the rocket is t = sqrt(x^2 + 2x/a), and by simple high school math the velocity is
> > dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1).
>
> ? ? ? [I still don't understand. Please explain it to me even more s . l . o .w . l . y.]

Okay, let's take small steps: You agree that the rocket's trajectory is t = sqrt(x^2 + 2x/a), and this means that at x1=3.0 we have t1=3.834504, and at x2=3.1 we have t2=3.937453. Now, in terms of these coordinates, the average velocity of the rocket on this segment is (x2 - x1) / (t2 - t1) = 0.971356. This is what "velocity in terms of these coordinates" means, i.e., the change in spatial position for a given change in time. Agreed?

Le 21/10/2022 à 01:23, Stan Fultoni a écrit :
> On Wednesday, October 19, 2022 at 5:07:01 PM UTC-7, Richard Hachel wrote:
>> > Again, in terms of the x,t coordinate system, you agreed that the trajectory
>> of
>> > the rocket is t = sqrt(x^2 + 2x/a), and by simple high school math the
>> velocity is
>> > dx/dt = a*sqrt(x^2 + 2x/a)/(ax+1).
>>
>> ? ? ? [I still don't understand. Please explain it to me even more s . l . o
>> .w . l . y.]
>
> Okay, let's take small steps: You agree that the rocket's trajectory is t =
> sqrt(x^2 + 2x/a), and this means that at x1=3.0 we have t1=3.834504, and at x2=3.1
> we have t2=3.937453. Now, in terms of these coordinates, the average velocity of
> the rocket on this segment is (x2 - x1) / (t2 - t1) = 0.971356. This is what
> "velocity in terms of these coordinates" means, i.e., the change in spatial
> position for a given change in time. Agreed?

Vous lisez les réponses que l'on vous fait?

R.H.