Le 23/11/2022 à 01:03, Stan Fultoni a écrit :
>> These very simple equations for accelerated repositories are:
>> To = x.sqrt(1 + 2/ax)
>> Vo = [1 + 1/2ax]^(-1/2)
>
> You contradict yourself, because you claimed that Vo=x/To, but then your first
> equation implies Vo = [1 + 2/(ax)]^(-1/2), whereas your second equation is Vo = [1
> + 1/(2ax)]^(-1/2), which implies 1/2 = 2. Of course, you failed to inser
> parentheses so your expressions are ambiguous (not to mention senseless), but this
> just illustrtates one absurdity compatible with what you typed. Agreed?

There is no contradiction, only misunderstanding.

I SAY:
Vo/c=[1+1/2ax]^(-1/2)

I am talking about the instantaneous speed when passing through x.

Which gives Vr=sqrt(2ax)

These are instantaneous speeds at the crossing point.

When I write Vo/c=[1+2c²/ax]^(-1/2), it is the average observable
velocity between and x.

It is in this case the same equation as
Vo/c=(x/c)/To=(x/c)/[(x/c).sqrt(1+2c²/ax)]

R.H.

Le 23/11/2022 à 05:18, JanPB a écrit :

> Whatever. Get real.
>
> --
> Jan

For 36 years I have been trying to be real, and to show what the reality
of relativistic things is.

I don't want to do fantasy, but reality.

Here, the reality is not scientific, but human. This is terrible dogma.

This dogma I compare it to the Muslim religion, which believes that
Muhammad came from Mecca to Jerusalem on a winged white horse.

There are fanatics who are ready to kill for it.

Come and tell me, that there is no paradox in RR, and that if I find any
it means that I understand pmal (while on the contrary never anyone in the
history of humanity no one explained things better than me, and reduced
all the paradoxes (the real ones) to nothing, it's ridiculous.

It is you who are ridiculous by arrogance and fanaticism.

But how to convince yourself of fanaticism?

I tried to show you that the paradox becomes total if we don't do as I say
by using the notion of apparent speeds.

But 95% of the people who answer me (and I'm optimistic maybe) don't even
know what "apparent speed" is.

I swear it's true.

So how do we do?

R.H.

On Wednesday, November 23, 2022 at 4:10:40 AM UTC-8, Richard Hachel wrote:
> > Place two identically constructed clocks beside the road, 1 km apart, and
> > synchronize them by shooting identical bullets from identical guns at rest at the
> > mid point between them, setting each clocks to 12:00 noon when the respective
> > bullet arrives. Now, a bicycle moves along the road, and it passes one clock when
> > it reads 1:00pm, and it passes the next clock when it reads 1:06 pm. What is the
> > bicycle's average speed between those clocks?
>
> You don't understand at all what I'm trying to tell you.

It would help me to understand if you would answer the question. What is the average velocity of the bicycle during that interval in terms of those coordinates?

Please answer the question.

> > - Example 1: One twin is stationary at the origin for 2 hours, and the other
> > twin moves at -5 mph for 1 hour and then abruptly accelerates and moves at +5 mph
> > for 1 hour. Do the twins have the same elapsed proper time?
> >
> > - Example 2: One twin moves at 10 mph for 2 hours, and the other twin moves at 5
> > mph for 1 hour, and then abruptly accelerates and moves at 15 mph for 1 hour. Do
> > the twins have the same elapsed proper time?
>
> In your Tau Ceti example...

Again, it would help me understand if you would answer the questions.

Please answer the questions.

>>> These very simple equations for accelerated repositories are:
>>> To = x.sqrt(1 + 2/ax)
>>> Vo = [1 + 1/2ax]^(-1/2)
>>
>> You contradict yourself, because you claimed that Vo=x/To, but then your first
>> equation implies Vo = [1 + 2/(ax)]^(-1/2), whereas your second equation is Vo = [1
>> + 1/(2ax)]^(-1/2), which implies 1/2 = 2. Of course, you failed to inser
>> parentheses so your expressions are ambiguous (not to mention senseless), but this
>> just illustrtates one absurdity compatible with what you typed. Agreed?
>
> I SAY: Vo/c=[1+1/2ax]^(-1/2) [is] the instantaneous velocity...
> [but] when I write Vo/c = [1+2c²/ax]^(-1/2), it is the average velocity
> between [the origin and] x.

Again, you contradict yourself. You are saying the symbol "Vo" represents both the instantaneous velocity of the trajectory at x and the average velocity over the interval from the origin to x. Those values are different, and you are using the same symbol to refer to both of them, so you are contradicting yourself.

Also, you are contradicting yourself still further, because even if you give those expressions different names, such as Vo and Vo_ave[0,x], we can directly calculate the average value of Vo from your equation over the range from the origin to x and it does not equal your exporession for Vo_ave[0,x]. So, your beliefs imply 1=0, which is absurd.

Le 23/11/2022 à 14:08, Stan Fultoni a écrit :
>> > Place two identically constructed clocks beside the road, 1 km apart, and
>> > synchronize them by shooting identical bullets from identical guns at rest at
>> the
>> > mid point between them, setting each clocks to 12:00 noon when the respective
>> > bullet arrives. Now, a bicycle moves along the road, and it passes one clock
>> when
>> > it reads 1:00pm, and it passes the next clock when it reads 1:06 pm. What is
>> the
>> > bicycle's average speed between those clocks?
>>

It takes six minutes to cover 1 km.

Its average speed is therefore v=x/t,
i.e. 1000m/360s.

But we are not here in a relativistic problem.

In a relativistic problem, things get complicated because there are two
new factors.

Anisochrony and dyschronotropy.

Anisochrony (the start and finish clocks are not the same, and cannot be
absolutely synchronized).

Dyschronotropia (the relativistic cyclist's watch
will actually turn slower than that of the supporter on the sidewalk for
this supporter) and the travel time will not be the same for both.

R.H.

On Wednesday, November 23, 2022 at 7:21:59 AM UTC-5, Richard Hachel wrote:

> And that we cannot logically do T2-T1 without falling into a major
> blunder.
> Why do you think we can do it?

Because it is the *definition* (of time interval).
When people discuss or argue about something, they must agree upon the meaning of the words used.
In physic, delta_t or t2-t1 or ' time interval' is defined as t2 - t1, t1 even if they are taken from different clocks.
This is the *definition*. Do you understand this?

> Always, always, always relativists make the same mistake.

No, only you've make the mistake, the mistake being that you do not know the meanings of the words you use.

On Wednesday, November 23, 2022 at 7:34:30 AM UTC-5, Richard Hachel wrote:

> When I write Vo/c=[1+2c²/ax]^(-1/2), it is the average observable
> velocity between and x.

Between *what* two points?
You said "between and x" which makes no sense.

And if x = 0, what is the average velocity according to your formula?

> It is in this case the same equation as
> Vo/c=(x/c)/To=(x/c)/[(x/c).sqrt(1+2c²/ax)]

So if x = 0, what is the average velocity?

On Wednesday, November 23, 2022 at 5:27:32 AM UTC-8, Richard Hachel wrote:
> > Place two identically constructed clocks beside the road, 1 km apart, and
> > synchronize them by shooting identical bullets from identical guns at rest at the
> > mid point between them, setting each clocks to 12:00 noon when the respective
> > bullet arrives. Now, a bicycle moves along the road, and it passes one clock when
> > it reads 1:00pm, and it passes the next clock when it reads 1:06 pm. What is the
> > bicycle's average speed between those clocks?
>
> Its average speed is therefore v=x/t,

No, the first clock is at x1=37005 meters and the second clock is at x2=38005 meters. Also, the values of t (if we set t=0 at synchronization noon) as the bicycle passes are t1=3600 seconds and t2=3960 seconds. So, the average velocity of the bicycle is v_ave[1,2] = (x2-x1)/(t2-t1). Agreed?

> i.e. 1000m/360s.

That's the correct numerical value, but that is not equal to x/t. For example, x1/t1 = 10.279 meters/sec, but the average speed is actually 2.777 meters/sec. That's why it makes no sense to say v=x/t, you must say v_ave[1,2] = (x2-x1)/(t2-t1). Agreed?

> But we are not here in a relativistic problem.

Right, your misunderstandings are much more fundamental. You deny the most fundamental facts until I spend days forcing you to admit them. Now we can finally talk about velocities in a rational way. The average velocity of the bicycle in terms of those coordinates is 2.777 meters/sec, and for a smaller and smaller interval around a given event the average velocity approaches the instantaneous at that event. Agreed?

On Wednesday, November 23, 2022 at 8:27:32 AM UTC-5, Richard Hachel wrote:
> Le 23/11/2022 à 14:08, Stan Fultoni a écrit :
> >> > Place two identically constructed clocks beside the road, 1 km apart, and
> >> > synchronize them by shooting identical bullets from identical guns at rest at
> >> the
> >> > mid point between them, setting each clocks to 12:00 noon when the respective
> >> > bullet arrives. Now, a bicycle moves along the road, and it passes one clock
> >> when it reads 1:00pm, and it passes the next clock when it reads 1:06 pm. What is
> >> the bicycle's average speed between those clocks?

> It takes six minutes to cover 1 km.

How did you calculate this?

> Its average speed is therefore v=x/t,

No. Its (x2-x1)/(t2-t1). Learn to use the appropriate language and notations, Elsie will not be understood.

> But we are not here in a relativistic problem.

So?

> Anisochrony (the start and finish clocks are not the same, and cannot be
> absolutely synchronized).

And neither in Newtonian Mechanics.
So what are you using two different clocks to calculate the
average speed of the bike?

> Dyschronotropia (the relativistic cyclist's watch
> will actually turn slower

Irrelevant. His own watch, what he had for breakfast, the clothes that he's wearing, has nothing to do with the definition of average speed.
His average speed is (x2-x1)/(t2-t1) just like in NM.

> this supporter) and the travel time will not be the same for both.

The question was clear: what is the average speed of the cyclist (wrt the ground frame, the 'supporter') ?

Le 23/11/2022 à 14:54, rotchm a écrit :
> Because it is the *definition* (of time interval).
> When people discuss or argue about something, they must agree upon the meaning
> of the words used.
> In physic, delta_t or t2-t1 or ' time interval' is defined as t2 - t1, t1 even
> if they are taken from different clocks.
> This is the *definition*. Do you understand this?
>
>> Always, always, always relativists make the same mistake.
>
> No, only you've make the mistake, the mistake being that you do not know the
> meanings of the words you use.

It is imperative, when talking about science, to use clear concepts.

Let's put a watch in A, and a watch in B.

Watch A is three quarters of an hour behind B.

In one hour (on his watch) a cyclist travels 20 km between A and B.

Data are collected starting at 3:15 p.m.

Arrival in B (on watch B) 4:15 p.m.

Let's say it took an hour.

We will not understand that the watches were out of step.

This is what happens in the theory of relativity.

I've always been amazed that men didn't understand this.

Their dreadful "present time plan" ideology has driven them all insane.

R.H.

Le 23/11/2022 à 14:58, rotchm a écrit :
> On Wednesday, November 23, 2022 at 7:34:30 AM UTC-5, Richard Hachel wrote:
>
>> When I write Vo/c=[1+2c²/ax]^(-1/2), it is the average observable
>> velocity between and x.

Between 0 and x.

> Between *what* two points?
> You said "between and x" which makes no sense.
>
> And if x = 0, what is the average velocity according to your formula?
>
>> It is in this case the same equation as
>> Vo/c=(x/c)/To=(x/c)/[(x/c).sqrt(1+2c²/ax)]
>
> So if x = 0, what is the average velocity?

If x=0 : no move.

LOL.

R.H.

Le 23/11/2022 à 15:05, rotchm a écrit :

> His average speed is (x2-x1)/(t2-t1) just like in NM.

We are in a different world where the effects of relativity must be
applied.

And there, we can no longer do, v+u=v+u, nor t2-t1=t2-t1.

This is what you don't WANT to understand.

It is a human problem, I remind you, "we don't WANT to understand".

I don't know how long it took to understand that v+u is not equal to v+u
in relativity.

But to understand that t2-t1, in accelerated repositories, it doesn't make
t2-t1, it will take at least 50 to 75 years.

In short, when Stan puts v=(x2-x1)/(To2-To1) in relativity, it doesn't
work.

On the other hand, it would work if it posed:
Vr=(x2-x1)/(Tr2-Tr1)

There, yes.

Then he then transposed by:
v=Vr/sqrt(1+Vr²/c²)

But as apparently, what I say, he doesn't care...

R.H.

On Wednesday, November 23, 2022 at 6:23:45 AM UTC-8, Richard Hachel wrote:
> We are in a different world where the effects of relativity must be
> applied.

Everything we've been discussing about velocities in terms of coordinate systems applies equally well in all situations. We do not set aside rationality or sanity when we discuss relativity. The average velocity is always (x2-x1)/(t2-t1).

> And there, we can no longer do, v+u=v+u, nor t2-t1=t2-t1.

That is false. We covered this before, remeber? 0.5+0.5=1, it does not equal 0.8. According to special relativity, we have v+u = v+u. Relativity did not overthrow arithmetic. You are conflating addition with composition. Your belief that t2-t1 does not equal t2-t1 is utterly insane. Surely you can see this?

On Wednesday, November 23, 2022 at 9:12:17 AM UTC-5, Richard Hachel wrote:
> Le 23/11/2022 à 14:54, rotchm a écrit :
> > Because it is the *definition* (of time interval).
> > When people discuss or argue about something, they must agree upon the meaning
> > of the words used.
> > In physic, delta_t or t2-t1 or ' time interval' is defined as t2 - t1, t1 even
> > if they are taken from different clocks.
> > This is the *definition*. Do you understand this?

No answer?

> It is imperative, when talking about science, to use clear concepts.

And this is what people here do, except for you.
We asked you for your answers, clear definitions, and you refuse to answer.
So, why do you refuse to use clear Concepts?

> Let's put a watch in A, and a watch in B.

"in" ?
If a and b are points, you cannot put a clock in them.
Or, are A and B reference frames?
You need to use Clear Concepts, you need to write clearly. Start over.

< rest of post snip until you address the above >

On Wednesday, November 23, 2022 at 9:14:07 AM UTC-5, Richard Hachel wrote:
> Le 23/11/2022 à 14:58, rotchm a écrit :
> > On Wednesday, November 23, 2022 at 7:34:30 AM UTC-5, Richard Hachel wrote:
> >
> >> When I write Vo/c=[1+2c²/ax]^(-1/2), it is the average observable
> >> velocity between and x.
> Between 0 and x.
> > Between *what* two points?
> > You said "between and x" which makes no sense.

No answer?

> > And if x = 0, what is the average velocity according to your formula?
> >> It is in this case the same equation as
> >> Vo/c=(x/c)/To=(x/c)/[(x/c).sqrt(1+2c²/ax)]
> >
> > So if x = 0, what is the average velocity?
> If x=0 : no move.

So if someone leaves from x = 0, goes away and comes back to x=0, then you declare that he did not move?
Good one!

On Wednesday, November 23, 2022 at 9:23:45 AM UTC-5, Richard Hachel wrote:
> Le 23/11/2022 à 15:05, rotchm a écrit :
>
> > His average speed is (x2-x1)/(t2-t1) just like in NM.
> We are in a different world where the effects of relativity must be
> applied.
>
> And there, we can no longer do, v+u=v+u, nor t2-t1=t2-t1.

This is your hand waving claims.
The fact of the matter is, that the definition of a time interval is t2-t1.
This is true in physics, in Newtonian mechanics, in relativistic physics.
It's all the same definition.

> This is what you don't WANT to understand.

It has nothing to do about understanding but about applying a definition.

> I don't know how long it took to understand that v+u is not equal to v+u
> in relativity.

In Newtonian mechanics and in relativity, v+u is still v+u.

> But to understand that t2-t1, in accelerated repositories,

You mean accelerated *frames* ?

> it doesn't make t2-t1, it will take at least 50 to 75 years.

t2 - t1 equals t2 - t1.

> In short, when Stan puts v=(x2-x1)/(To2-To1) in relativity, it doesn't
> work.

The definition of average speed is v_avg = (x2-x1)/(t2-t1).
Why are those parameters are the space-time coordinates of the two events in question as described by the given inertial frame.
Agree?

On Wednesday, 23 November 2022 at 16:39:40 UTC+1, rotchm wrote:

> The definition of average speed is v_avg = (x2-x1)/(t2-t1).

Let's take a single dimensional space - inflating.
Suppose 2 objects are immobile (their speed
is 0) in such space, Is their x2-x1 const, poor
halfbrain?

Le 23/11/2022 à 16:39, rotchm a écrit :

> The definition of average speed is v_avg = (x2-x1)/(t2-t1).
> Why are those parameters are the space-time coordinates of the two events in
> question as described by the given inertial frame.
> Agree?

Absolutly not.

R.H.

Le 23/11/2022 à 16:39, rotchm a écrit :

> The definition of average speed is v_avg = (x2-x1)/(t2-t1).
> Why are those parameters are the space-time coordinates of the two events in
> question as described by the given inertial frame.
> Agree?

Absolutly not.

For this to be consistent, the measurements must be made in the same frame
of reference, but in relativity, contrary to popular belief, this is not
possible.

Take the case of measuring the height of a house; and we ask two workers
to measure it to paint the facade.

The workers agree to say "me, I measure the bottom of the house, up to the
top of this window, and you, from the top of the window, up to the
cornice. We will do the addition".

One measures in meters, the other in inches.

It is not consistent without conversion for logical result.

The same goes for Stan's blunder, which subtracts a value taken with one
watch from a value taken with another.

The problem is that the blunder is so fine that no one sees that there is
a blunder.

The clock which measures the time which goes from 0 to 3 ly, is ideally
placed at x=1.5ly.

The clock that measures time, which goes from 0 to 3.1ly is ideally placed
at 1.55ly.

They are in the same frame, but, with Hachel, they are not in the same
spatio-temporal frame of reference.

The measured times cannot be subtracted like this.

This will lead to a Vo(avg)=(x2-x1)/(T2? ? ?-T1? ? ?) value that is too
high.

I've been telling you for weeks, we can't do just anything, subtractions,
additions, integrations.

Relativity is not Newtonian kinetics.

R.H.

On Wednesday, November 23, 2022 at 11:08:37 AM UTC-5, maluw...@gmail.com wrote:
> On Wednesday, 23 November 2022 at 16:39:40 UTC+1, rotchm wrote:
>
> > The definition of average speed is v_avg = (x2-x1)/(t2-t1).
> Let's take a single dimensional space - inflating.
> Suppose 2 objects are immobile (their speed
> is 0) in such space, Is their x2-x1 const,...?

If the coordinates x1 & x2 continue to have their same value, then yes.
Dont forget, you have *coordinated* your space, inflating or not.

Take a "stretchable " of 12 inches. Stretch it to 14 inches.
The endpoint is still labeled 12 and the begining 0.
The difference in those values is still 12-0 = 12 , with units inches.
IOW, in the frame of the ruler, its length is still 12 inches.
In the frame of "me", my hand, the ground etc, the length is now 14 inches.
Very simple concept!

On Wednesday, 23 November 2022 at 18:44:50 UTC+1, rotchm wrote:
> On Wednesday, November 23, 2022 at 11:08:37 AM UTC-5, maluw...@gmail.com wrote:
> > On Wednesday, 23 November 2022 at 16:39:40 UTC+1, rotchm wrote:
> >
> > > The definition of average speed is v_avg = (x2-x1)/(t2-t1).
> > Let's take a single dimensional space - inflating.
> > Suppose 2 objects are immobile (their speed
> > is 0) in such space, Is their x2-x1 const,...?
>
>
> If the coordinates x1 & x2 continue to have their same value, then yes.
> Dont forget, you have *coordinated* your space, inflating or not.
>
> Take a "stretchable " of 12 inches. Stretch it to 14 inches.
> The endpoint is still labeled 12 and the begining 0.
> The difference in those values is still 12-0 = 12 , with units inches.

Well, a logic of a ralativistic fanatic can't surprise
me anymore, but it's still funny.

On Wednesday, November 23, 2022 at 11:33:48 AM UTC-5, Richard Hachel wrote:
> Le 23/11/2022 à 16:39, rotchm a écrit :
> > The definition of average speed is v_avg = (x2-x1)/(t2-t1).
> > Why are those parameters are the space-time coordinates of the two events in
> > question as described by the given inertial frame.
> > Agree?
> Absolutly not.

:/

Should have read "Those parameters are the space-time coordinates of the two events in
question as described by the given inertial frame. Agree?"

The "Why are" should not be there. (voice typing).

So you are saying that if there is an event at x=0 and t=0, labeled as (0,0), this is not its
coordinates (as pre-defined by our notations & conventions) ?
Ditto for the event (x=1,t=6) ?

> For this to be consistent, the measurements must be made in the same frame
> of reference,

Correct. And that is what we/physicists are doing here in these scenarios.

> but in relativity, contrary to popular belief, this is not possible.

Its not a question of relativity. Its a question of definition.
Coordinate your space (meaning, to coordinate positions and time) as prescribed.
The coordinates of an event is then *defined* as the value of the location (the mark on the ruler, say) and
the value of the clock there.

Tis is the *definition*, agree?
Note that we are discussing the meaning, the definition of position & time, the definition of coordinate.
This has nothing to do with NM or SR.

> Take the case of measuring the height of a house; and we ask two workers
> to measure it to paint the facade.
> The workers agree to say "me, I measure the bottom of the house, up to the
> top of this window, and you, from the top of the window, up to the
> cornice. We will do the addition".

Exactly. They must agree *beforehand* what is meant by "to measure...".

> One measures in meters, the other in inches.

So they have not agreed on the meaning (the operational definition) of 'to measure'.
So they cant proceed sensibly. They can each use their definition. for themselves.
If they want then to compare their results with someone else definition, they need to translate their results to the other's.
This 'translation' is the 'transformation'.

> It is not consistent without conversion for logical result.
>
> The same goes for Stan's blunder, which subtracts a value taken with one
> watch from a value taken with another.

Thats not a blunder. Its the *definition* of 'time interval' for that observer (reference frame).

> The problem is that the blunder is so fine that no one sees that there is
> a blunder.

You mean that you are seeing blunders where there are none.
You are saying that you do not know the meaning of the words/concepts used in physics.

> The clock which measures the time which goes from 0 to 3 ly, is ideally
> placed at x=1.5ly.

The definition is that you take the value of the clock at x=0 and the clock at x=3.

> This will lead to a Vo(avg)=(x2-x1)/(T2? ? ?-T1? ? ?) value that is too
> high.

Irrelevant. The defintion of average speed is (x2-x1)/(t2-t1), where the variables are
the *values* (of the ruler and clocks) located at the events. This is the definition, agree?

> I've been telling you for weeks, we can't do just anything, subtractions,
> additions, integrations.

Correct. What we can do is apply the definition, the rules that have been agreed upon.
And that is what NM, SR, physics does.

Le 23/11/2022 à 19:03, rotchm a écrit :

> Irrelevant. The defintion of average speed is (x2-x1)/(t2-t1), where the
> variables are
> the *values* (of the ruler and clocks) located at the events. This is the
> definition, agree?

Je vous ai dit plusieurs fois que non.

Cette égalité ne veut rien dire en cas de référentiels accélérés
dans lesquels une horloge mesure un temps,
(de 0 à 3.1ly), une autre un autre temps, car c'est une AUTRE montre
(qui mesure de 0 à 3.1ly).

Le fait de faire ensuite une soustraction biaiseuse, c'est débile.

Je vous l'ai expliqué, vous n'êtes pas capables de le comprendre.

R.H.

On Wednesday, November 23, 2022 at 2:04:06 PM UTC-5, Richard Hachel wrote:
> Le 23/11/2022 à 19:03, rotchm a écrit :
>
> > Irrelevant. The defintion of average speed is (x2-x1)/(t2-t1), where the
> > variables are
> > the *values* (of the ruler and clocks) located at the events. This is the
> > definition, agree?
> Je vous ai dit plusieurs fois que non.
>
> Cette égalité

Can you state that in the language we were using, so that I may understand better?

The language we were using (English) was understood by us both.
Now you changed language. Why?

On Wednesday, November 23, 2022 at 10:33:41 AM UTC-5, rotchm wrote:
> On Wednesday, November 23, 2022 at 9:14:07 AM UTC-5, Richard Hachel wrote:

> > >> Vo/c=(x/c)/To=(x/c)/[(x/c).sqrt(1+2c²/ax)]
> > >
> > > So if x = 0, what is the average velocity?
> > If x=0 : no move.
> So if someone leaves from x = 0, goes away and comes back to x=0, then you declare that he did not move?

No answer?

On Wednesday, November 23, 2022 at 11:04:06 AM UTC-8, Richard Hachel wrote:
[evasions]

Please answer the questions:

Example 1: One twin is stationary at the origin for 2 hours, and the other twin moves at -5 mph for 1 hour and then abruptly accelerates and moves at +5 mph for 1 hour. Do the twins have the same elapsed proper time?

Example 2: One twin moves at 10 mph for 2 hours, and the other twin moves at 5 mph for 1 hour, and then abruptly accelerates and moves at 15 mph for 1 hour. Do the twins have the same elapsed proper time?

> I SAY: Vo/c=[1+1/2ax]^(-1/2) [is] the instantaneous velocity...
> [but] when I write Vo/c = [1+2c²/ax]^(-1/2), it is the average velocity
> between [the origin and] x.

Again, you contradict yourself. You're saying the symbol "Vo" represents both the instantaneous velocity of the trajectory at x and the average velocity over the interval from the origin to x. Those values are different, and you are using the same symbol to refer to both of them, so you are contradicting yourself.

Moreover, you are contradicting yourself still further, because even if you give those expressions different names, such as Vo and Vo_ave[0,x], we can directly calculate the average value of Vo from your equation over the range from the origin to x and it does not equal your exporession for Vo_ave[0,x]. So, your beliefs imply 1=0, which is absurd. Right?