Jim Pennino wrote:

>>>> coma separated decimal, with an arbitrary number of decimals, are
>>>> decimal numbers. Float or double, etc. You are sticking you fingers
>>>> in the shit, instead of catching the point in context of the
>>>> discussion. You are useless.
>>>
>>> Gibberish, try again in English.
>>
>> you suck dicks in many languages, I can see. Which part you fucking
>> stupid can't undrestand? We are here to help.
>
> We?
> Maybe you all should get together and attend an English class or two.

he said, he fucks your mother in she ass, in front of you, and pays
nothing. But you may take pictures, if you wish.

rotchm wrote:

> On Friday, November 4, 2022 at 8:46:08 PM UTC-4, Jim Pennino wrote:
>> In sci.physics Stefan Russo <ft...@urstsoto.ee> wrote:
>
>> Why is you English so pathetic and full of absurd phraseology?
>
> So as to sucker you into stroking him.
> It worked. This well known troll got you!

here's another imbecile in fucking english. Lol. Not even him undrestand
what he wants to say. This forum is assholes populated. See you in Kiev.

Le 05/11/2022 à 00:50, Stan Fultoni a écrit :
> On Friday, November 4, 2022 at 2:42:27 PM UTC-7, Richard Hachel wrote:
>> For the trajectory defined by t = sqrt(x^2 + 2x/a), the elapsed coordinate
>> time between t=3 and t=3.1 is not 0.10295 but 0.1075 .
>
> You are mistaken. At x=3.0 the time coordinate is t=3.834504, and at x=3.1 the
> time coordinate is t=3.937453, so we have Δx = 0.1 and Δt = 0.102949.
>
> To isolate your mistake, please tell me the value of t at x=3.0 on the
> trajectory t = sqrt(x^2 + 2x/a). Assume a=1.052.

There is no mistake, there is no need to search.

I repeat it, thanking you for your kindness, your competence and your
help: there is NO mistake.

As for the time measurements at x=3 and at x=3.1ly, I have exactly the
same as you.

I have :
To1=3.834504 and To2=3.937453

So far, there is no error.

And there will be no error either if you set Vo=Δx/ΔTo

I repeat: the error does not come from there.

It comes from the fact that you are using a geometry different from mine.

The two geometries giving different results, both mathematically correct,
but physically incompatible.

One of the two geometries being erroneous.

So when you do the subtraction between To2 and To1,
YOU, you find Δt=0.10295

And me, I find Δt=0.10751

Your subtraction is incorrect, in the sense that, for example, 0.8c-0.5c
is not 0.3c but 0.5c.

Here, it's the same, you can't subtract "observable" values ​​like
that, whereas if you work in real values ​​(Tr2 and Tr1) everything is
correct and without any calculation difficulty.

This is why you find Vo values ​​that are much too high for all the
instantaneous observable velocities.

I don't have this problem.

As for the formula to be used, once understood, it is very simple and
should be taught, too (thank you Richard Hachel) to all physicists in the
world.

v/c=[1+c²/2ax]^(-1/2)

You then obtain an instantaneous speed of Vo1=0.92910c
and Vo2=0.93116c

R.H.

Jim Pennino wrote:

>>> Gibberish, try again in English.
>>
>> which one, fucking stoopid. You don't undrestand shit anyway, you
>> uneducated and unskilled pile of shit. Engilsh is for pigs. It's
>> Biblical. You don't know you are a pig?? I have a planet saying that you
>> ARE a pig. Lol.
>>
> What a childish, illiterate moron.

try pulling his finger, then.

On Friday, November 4, 2022 at 9:24:27 AM UTC-4, Stefan Russo wrote:
> J. J. Lodder wrote:
>
> >> >> never mix decimal point numbers in lemmas and theories. Use 1/2, not
> >> >> 0.5
> >> >
> >> > If you want.
> >> > The equation will be no less exact. v/c=[1+c?/2ax]^(-1/2)
> >>
> >> that's the point, fractions ARE exact. 0.(3) is not exact, but 1/3 IS
> >> exact.
> >
> > If you want 0.(3) to mean 0.333... , so 0.3 repeating, then it is exact..
> > 0.333... is just another notation for 1/3,
> you knownothing idiot, how can it be exact, when it NEVER converges to
> 1/3?? You never passed an Numerical_Analysis exam, you repeatedly
> impertinent idiot.

Apparently you fail to understand limit theory. This is the standard argument explaining the equality of 0.(3) and 1/3. To reject it find the logical error: First, 10* 0.(3) is just 3.(3), ok? Then 10*0.(3) - 0.(3) = 9*0.(3) = 3.0, simple subtraction, ok? Since 9 does not equal 0, divide both sides of the equation by it, 1*0.(3) = 0.(3) = 3.0/9 = 1/3. Ok? 0.(3) = 1/3. If you can't find a logical flaw, your assertion that they are different is illogical. You apparently don't understand the meaning of converge in the context of limit. It is incorrect to say that the repeating decimal "NEVER converges" when it DOES converge, at INFINITY. Any truncated representation is obviously finite and fails the test for convergence, because the actual point of convergence is still infinitely far away. The rules for infinity are not the same as the rules for finite.

On Saturday, November 5, 2022 at 4:38:31 AM UTC-7, Richard Hachel wrote:
> >> For the trajectory defined by t = sqrt(x^2 + 2x/a), the elapsed coordinate
> >> time between t=3 and t=3.1 is not 0.10295 but 0.1075 .
> >
> > You are mistaken. At x=3.0 the time coordinate is t=3.834504, and at x=3.1 the
> > time coordinate is t=3.937453, so we have Δx = 0.1 and Δt = 0.102949.
> >
> > To isolate your mistake, please tell me the value of t at x=3.0 on the
> > trajectory t = sqrt(x^2 + 2x/a). Assume a=1.052.
>
> At x1=3 and x2=3.1 I have : t1=3.834504 and t2=3.937453

Good. So the coordinate time at x=3 is 3.834504 and the coordinate time at x=3.1 is 3.937453, so the change in coordinate time along the trajectory from x=3 to x=3.1 is 0.102949.

> So when you do the subtraction between t2 and t1 you find Δt=0.102949

Yes, and it isn't just me. Every sane human agrees that 5-2=3 and 7-1=6 and 3.937453 - 3.834504 = 0.102949.

> Your subtraction is incorrect, in the sense that, for example, 0.8c-0.5c
> is not 0.3c but 0.5c.

No, the correct result is 0.8c - 0.5c = 0.3c. This was explained to you before. Again, special relativity did not overthrow arithmetic or sanity or rational thought. The change in t from t=3.834504 to t=3.937453 is 0..102949. The denial of this is equivalent to claiming 1=0. It is absurd.. It is just as absurd as claiming 0.8 - 0.5 = 0.5, or claiming 8 - 5 = 5, or claiming 8=10.

> For the trajectory t=sqrt(x^2 + 2x/a) we have dx/dt = [1+c²/2ax]^(-1/2)

That is false. Given that t=sqrt(x^2 + 2x/a) we have dx/dt = sqrt[1 - 1/(1+ax)^2]. We covered this before. Again, your beliefs imply 1=0, so they are absurd.

Remember, 0.8c - 0.5c = 0.3c. Agreed?

Le 05/11/2022 à 15:33, Stan Fultoni a écrit :
> On Saturday, November 5, 2022 at 4:38:31 AM UTC-7, Richard Hachel wrote:
>> >> For the trajectory defined by t = sqrt(x^2 + 2x/a), the elapsed coordinate
>> >> time between t=3 and t=3.1 is not 0.10295 but 0.1075 .
>> >
>> > You are mistaken. At x=3.0 the time coordinate is t=3.834504, and at x=3.1 the
>>
>> > time coordinate is t=3.937453, so we have Δx = 0.1 and Δt = 0.102949.
>> >
>> > To isolate your mistake, please tell me the value of t at x=3.0 on the
>> > trajectory t = sqrt(x^2 + 2x/a). Assume a=1.052.
>>
>> At x1=3 and x2=3.1 I have : t1=3.834504 and t2=3.937453
>
> Good. So the coordinate time at x=3 is 3.834504 and the coordinate time at
> x=3.1 is 3.937453, so the change in coordinate time along the trajectory from x=3
> to x=3.1 is 0.102949.
>
>> So when you do the subtraction between t2 and t1 you find Δt=0.102949
>
> Yes, and it isn't just me. Every sane human agrees that 5-2=3 and 7-1=6 and
> 3.937453 - 3.834504 = 0.102949.
>
>> Your subtraction is incorrect, in the sense that, for example, 0.8c-0.5c
>> is not 0.3c but 0.5c.
>
> No, the correct result is 0.8c - 0.5c = 0.3c. This was explained to you before.
> Again, special relativity did not overthrow arithmetic or sanity or rational
> thought. The change in t from t=3.834504 to t=3.937453 is 0.102949. The denial
> of this is equivalent to claiming 1=0. It is absurd. It is just as absurd as
> claiming 0.8 - 0.5 = 0.5, or claiming 8 - 5 = 5, or claiming 8=10.
>
>> For the trajectory t=sqrt(x^2 + 2x/a) we have dx/dt = [1+c²/2ax]^(-1/2)
>
> That is false. Given that t=sqrt(x^2 + 2x/a) we have dx/dt = sqrt[1 -
> 1/(1+ax)^2]. We covered this before. Again, your beliefs imply 1=0, so they are
> absurd.
>
> Remember, 0.8c - 0.5c = 0.3c. Agreed?

Faites comme vous voulez.

Moi, je fatigue.

R.H.

Il giorno venerdì 4 novembre 2022 alle 13:39:39 UTC+1 Stefan Russo ha scritto:
> Richard Hachel wrote:
>
> >>> We simply set v/c=[1+c²/2ax]^-0.5
> >>
> >> Hopeless. PointedEars
> >
> > Don't like my equation?
> never mix decimal point numbers in lemmas and theories. Use 1/2, not 0.5

Ismael Stabile pretended :
> FromTheRafters wrote:
>
>> on 11/4/2022, Stefan Russo supposed :
>>> *I_fuck_your_ass*, fool!
>>>
>>> https://www.mathsisfun.com/decimals.html
>>> https://www.splashlearn.com/math-vocabulary/decimals/decimal
>>
>> Yep, ten, just as I said. You are confused about the decimal system and
>> the fact that a decimal radix point is often called a decimal point.
>
> no, he is right. Read the fucking papers. It's about *decimals* or
> *decimal_numbers*, kindergarten stuff. And NOT *base_ten*, as you write,
> you fucking idiot.

What does the 'dec' in decimal mean?

Ismael Stabile wrote on 11/5/2022 :
> FromTheRafters wrote:
>
>> Jim Pennino expressed precisely :
>>> If you mean a number that has a fractional part, which requires a
>>> decimal point, there are N+1 symbols where N=the number base.
>>
>> Wrong, base ten has ten symbols (0,1,2,3,4,5,6,7,8,9) and base sixteen
>> has sixteen symbols (0,1,2,3,4,5,6,7,8,9,a,b,c,d,e,f) not your N+1 crap.
>> The radix point makes it a radix system, not necessarily a decimal
>> system.
>
> you repulsive ignorants, was about decimal numbers, not *base_ten*. lol

Ha ha ha. Decimal numbers *are* base ten.

On Saturday, November 5, 2022 at 8:23:48 AM UTC-7, Richard Hachel wrote:
> >> >> For the trajectory defined by t = sqrt(x^2 + 2x/a), the elapsed coordinate
> >> >> time between t=3 and t=3.1 is not 0.10295 but 0.1075 .
> >> >
> >> > You are mistaken. At x=3.0 the time coordinate is t=3.834504, and at x=3.1 the
> >>
> >> > time coordinate is t=3.937453, so we have Δx = 0.1 and Δt = 0.102949.
> >> >
> >> > To isolate your mistake, please tell me the value of t at x=3.0 on the
> >> > trajectory t = sqrt(x^2 + 2x/a). Assume a=1.052.
> >>
> >> At x1=3 and x2=3.1 I have : t1=3.834504 and t2=3.937453
> >
> > Good. So the coordinate time at x=3 is 3.834504 and the coordinate time at
> > x=3.1 is 3.937453, so the change in coordinate time along the trajectory from x=3
> > to x=3.1 is 0.102949.
> >
> >> So when you do the subtraction between t2 and t1 you find Δt=0.102949
> >
> > Yes, and it isn't just me. Every sane human agrees that 5-2=3 and 7-1=6 and
> > 3.937453 - 3.834504 = 0.102949.
> >
> >> Your subtraction is incorrect, in the sense that, for example, 0.8c-0.5c
> >> is not 0.3c but 0.5c.
> >
> > No, the correct result is 0.8c - 0.5c = 0.3c. This was explained to you before.
> > Again, special relativity did not overthrow arithmetic or sanity or rational
> > thought. The change in t from t=3.834504 to t=3.937453 is 0.102949. The denial
> > of this is equivalent to claiming 1=0. It is absurd. It is just as absurd as
> > claiming 0.8 - 0.5 = 0.5, or claiming 8 - 5 = 5, or claiming 8=10..
> >
> >> For the trajectory t=sqrt(x^2 + 2x/a) we have dx/dt = [1+c²/2ax]^(-1/2)
> >
> > That is false. Given that t=sqrt(x^2 + 2x/a) we have dx/dt = sqrt[1 -
> > 1/(1+ax)^2]. We covered this before. Again, your beliefs imply 1=0, so they are
> > absurd.
> >
> > Remember, 0.8c - 0.5c = 0.3c. Agreed?
>
> Faites comme vous voulez. Moi, je fatigue.

You see? When your errors are explained, you never provide a rational response, you simply make excuses and run away. That's why you never learn. We've clearly shown that your beliefs imply 1=0, so your beliefs are absurd.

Tom Capizzi wrote:

>> > If you want 0.(3) to mean 0.333... , so 0.3 repeating, then it is
>> > exact.
>> > 0.333... is just another notation for 1/3,
>> you knownothing idiot, how can it be exact, when it NEVER converges to
>> 1/3?? You never passed an Numerical_Analysis exam, you repeatedly
>> impertinent idiot.
>
> Apparently you fail to understand limit theory. This is the standard

nonsense, but you fail to understand the equality and convergence theory,
which are primordial.

FromTheRafters wrote:

>>> Wrong, base ten has ten symbols (0,1,2,3,4,5,6,7,8,9) and base sixteen
>>> has sixteen symbols (0,1,2,3,4,5,6,7,8,9,a,b,c,d,e,f) not your N+1
>>> crap.
>>> The radix point makes it a radix system, not necessarily a decimal
>>> system.
>>
>> you repulsive ignorants, was about decimal numbers, not *base_ten*. lol
>
> Ha ha ha. Decimal numbers *are* base ten.

so what, you talk terminology of different things. Now go back to sleep.
And do not fart. It's disgusting for long time, depending on your
circumstance.

Richard Hachel wrote:

> Le 04/11/2022 à 08:40, Thomas 'PointedEars' Lahn a écrit :
>> Richard Hachel wrote:
>>> How to know the exact speed of a uniformly accelerated body according to
>>> its position on a trajectory?
>>>
>>> We simply set v/c=[1+c²/2ax]^-0.5
>> Hopeless.
>
> Don't like my equation?

You still don’t get it. It’s not a matter of taste, but correctness.
Your equation is just wrong: It does not describe the universe which
we live in.

<https://www.feynmanlectures.caltech.edu/fml.html#2>

PointedEars
--
Q: What happens when electrons lose their energy?
A: They get Bohr'ed.

(from: WolframAlpha)

On Sunday, 20 November 2022 at 02:51:40 UTC+1, Thomas 'PointedEars' Lahn wrote:
> Richard Hachel wrote:
>
> > Le 04/11/2022 à 08:40, Thomas 'PointedEars' Lahn a écrit :
> >> Richard Hachel wrote:
> >>> How to know the exact speed of a uniformly accelerated body according to
> >>> its position on a trajectory?
> >>>
> >>> We simply set v/c=[1+c²/2ax]^-0.5
> >> Hopeless.
> >
> > Don't like my equation?
> You still don’t get it. It’s not a matter of taste, but correctness.
> Your equation is just wrong: It does not describe the universe which
> we live in.

It doesn't describe the gedankenwelt where you live,
but it does describe the gedankenwelt where he lives.

Le 20/11/2022 à 02:51, Thomas 'PointedEars' Lahn a écrit :
> Richard Hachel wrote:
>
>> Le 04/11/2022 à 08:40, Thomas 'PointedEars' Lahn a écrit :

>>>> We simply set v/c=[1+c²/2ax]^-0.5

> Your equation is just wrong

Oh no....

Sniffff...

> PointedEars

R.H.