Someone is placed in a capsule of an electro-magnetic repulsion
accelerator in the weightless vacuum of space and accelerated along
its entire length. If the g force the person feels is 2g (32 ft/s-
squared x 2) during the entire time along the accelerator and the
accelerator is 60,000 miles long, how fast would the person be
traveling when they leave the accelerator and how long would their run
on the accelerator last? And if a machine part was fired off on the
same accelerator at 100g, why do I get the feeling that they be going
a lot faster than 50 times as fast the 2g-accelerated human and their
run taking a lot less than 1/50th the time? Or is my gut wrong and
they would only be going 50 times faster and their run would only be
1/50th the time?

If I understand g forces in the weightless of space, the person/object
has to at a constant rate of acceleration to so experience the desired
g force. The moment their velocity remains the same, they return to
their weightless existence. Thus when the person/machine-part finally
leaves the 60,000-mile long accelerator, they will return to a
weightless existence.

Thanks in advance!

Scott

: STJensen <RecreationalPoker@gmail.com>
: Someone is placed in a capsule of an electro-magnetic repulsion
: accelerator in the weightless vacuum of space and accelerated along
: its entire length. If the g force the person feels is 2g (32 ft/s-
: squared x 2) during the entire time along the accelerator and the
: accelerator is 60,000 miles long, how fast would the person be
: traveling when they leave the accelerator and how long would their run
: on the accelerator last? And if a machine part was fired off on the
: same accelerator at 100g, why do I get the feeling that they be going
: a lot faster than 50 times as fast the 2g-accelerated human and their
: run taking a lot less than 1/50th the time? Or is my gut wrong and
: they would only be going 50 times faster and their run would only be
: 1/50th the time?

Neither. Considerably *less* than 50 times as fast, and considerably
*more* than a 50th of the time.

Do the arithmetic yourself: t=sqrt(2d/a), and v=at. You can choose a
few numerical examples, or simplify sqrt(2d/a)/sqrt(2d/(50a)), which
of course turns out to be sqrt(1/50) for the time, and sqrt(50) for speed,
or, about seven times the speed.


Wayne Throop throopw@sheol.org http://sheol.org/throopw

STJensen wrote:

> Someone is placed in a capsule of an electro-magnetic repulsion
> accelerator in the weightless vacuum of space and accelerated along
> its entire length. If the g force the person feels is 2g (32 ft/s-
> squared x 2) during the entire time along the accelerator and the
> accelerator is 60,000 miles long, how fast would the person be
> traveling when they leave the accelerator and how long would their run
> on the accelerator last?

For straight-line, constant acceleration, starting from rest, with no
other forces acting, the equations are:

s = (1/2) a t^2

v = a t

where t is the elapsed time, v is the accumulated speed, s is the
accumulated distance, and a is the acceleration.

Solving one for t and substituting (since you're not concerned with the
total time), you get

v^2 = 2 a s.

Now plug in numbers and experiment.

> If I understand g forces in the weightless of space, the person/object
> has to at a constant rate of acceleration to so experience the desired
> g force. The moment their velocity remains the same, they return to
> their weightless existence. Thus when the person/machine-part finally
> leaves the 60,000-mile long accelerator, they will return to a
> weightless existence.

Yes.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
All delays are dangerous in war.
-- John Dryden, 1631-1700

Haven't done the sums but I would have thought you would have left Newton
far behind and would need to use special relativity


"STJensen" <RecreationalPoker@gmail.com> wrote in message
news:d5a10743-041e-47bf-9408-31309ddd9700@72g2000hsu.googlegroups.com...
> Someone is placed in a capsule of an electro-magnetic repulsion
> accelerator in the weightless vacuum of space and accelerated along
> its entire length. If the g force the person feels is 2g (32 ft/s-
> squared x 2) during the entire time along the accelerator and the
> accelerator is 60,000 miles long, how fast would the person be
> traveling when they leave the accelerator and how long would their run
> on the accelerator last? And if a machine part was fired off on the
> same accelerator at 100g, why do I get the feeling that they be going
> a lot faster than 50 times as fast the 2g-accelerated human and their
> run taking a lot less than 1/50th the time? Or is my gut wrong and
> they would only be going 50 times faster and their run would only be
> 1/50th the time?
>
> If I understand g forces in the weightless of space, the person/object
> has to at a constant rate of acceleration to so experience the desired
> g force. The moment their velocity remains the same, they return to
> their weightless existence. Thus when the person/machine-part finally
> leaves the 60,000-mile long accelerator, they will return to a
> weightless existence.
>
> Thanks in advance!
>
> Scott

: "Rod" <rodrodrodrod@hotmail.com>
: Haven't done the sums but I would have thought you would have left Newton
: far behind and would need to use special relativity

Not really. A hundred g in a 60k mile accelerator would take you less
than ten minutes. To reach significant fractions of lightspeed would
take you many hours.


Wayne Throop throopw@sheol.org http://sheol.org/throopw

Rod wrote:

> Haven't done the sums but I would have thought you would have left Newton
> far behind and would need to use special relativity

For 100 gee over 60 000 mi, the resulting speed is about 430 km/s.
That's awfully high, but a tiny fraction of that of light, which is 300
000 000 km/s. Special relativistic effects would be insignificant for
the accuracies we're talking about here.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
What would physics look like without gravitation?
-- Albert Einstein

"Erik Max Francis" <max@alcyone.com> wrote in message
news:I7SdnaoYWtaCDyfanZ2dnUVZ_vqpnZ2d@speakeasy.ne t...
> Rod wrote:
>
>> Haven't done the sums but I would have thought you would have left Newton
>> far behind and would need to use special relativity
>
> For 100 gee over 60 000 mi, the resulting speed is about 430 km/s. That's
> awfully high, but a tiny fraction of that of light, which is 300 000 000
> km/s. Special relativistic effects would be insignificant for the
> accuracies we're talking about here.


I should have done the sum;-)
Think first, post later.


>
> --
> Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
> San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
> What would physics look like without gravitation?
> -- Albert Einstein

throopw@sheol.org (Wayne Throop) writes:

> : "Rod" <rodrodrodrod@hotmail.com>
> : Haven't done the sums but I would have thought you would have left Newton
> :
> : far behind and would need to use special relativity
>
> Not really. A hundred g in a 60k mile accelerator would take you less
> than ten minutes. To reach significant fractions of lightspeed would
> take you many hours.

At 100 g ( 980 m/sec^2), to reach lightspeed (3*10^8 m/sec)
would take about 3.06*10^5 sec (ignoring relativity, of course)
or 85 hours. The accelerator would have to be about 4.59*10^13 m long,
or 307 AU, extending into the inner Oort cloud.
--
Robert Israel israel@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

: Robert Israel <israel@math.MyUniversitysInitials.ca>
: At 100 g ( 980 m/sec^2), to reach lightspeed (3*10^8 m/sec)
: would take about 3.06*10^5 sec (ignoring relativity, of course)
: or 85 hours. The accelerator would have to be about 4.59*10^13 m long,
: or 307 AU, extending into the inner Oort cloud.

Which oddly enough (unix "units" confirms for me), is about a hundredth
of a year. Which one would expect right off the bat if one happens
to remember that 1g is about 1c/yr. So, from my perspective, I would
say to myself, 100g gives you lightspeed in a hundredth of a year,
or three or four days. Arbitrarily saying a tenth lightspeed is below
what's needed for significant relativistic effects, that means it takes
you something like 8 hours at 100g before you get significant effects.
Very handwavy and imprecise, but gives you the general ballpark with
minimal calculation.

( Though doing the calculation to yield "bam, zoom, to the inner
Oord cloud, Alice" is nifty... )


"He wasn't an astronaut, he was a sitcom actor.
And he was only using space travel as a metaphor
for beating his wife."

--- Philip J. Fry


Wayne Throop throopw@sheol.org http://sheol.org/throopw

Wayne Throop wrote:
> : Robert Israel <israel@math.MyUniversitysInitials.ca>
> : At 100 g ( 980 m/sec^2), to reach lightspeed (3*10^8 m/sec)
> : would take about 3.06*10^5 sec (ignoring relativity, of course)
> : or 85 hours. The accelerator would have to be about 4.59*10^13 m long,
> : or 307 AU, extending into the inner Oort cloud.
>
> Which oddly enough (unix "units" confirms for me), is about a hundredth
> of a year. Which one would expect right off the bat if one happens
> to remember that 1g is about 1c/yr. So, from my perspective, I would
> say to myself, 100g gives you lightspeed in a hundredth of a year,
> or three or four days. Arbitrarily saying a tenth lightspeed is below
> what's needed for significant relativistic effects, that means it takes
> you something like 8 hours at 100g before you get significant effects.
> Very handwavy and imprecise, but gives you the general ballpark with
> minimal calculation.

For long trips at constant acceleration, a good fast approximation is

a*t = a*d + 1 = exp( a*s ) = gamma

for units where c = 1
a = shipboard acceleration
s = shipboard time
t = standard time
d = standard distance
gamma = ( 1 - v^2 )^(-1/2)

The correct, unapproximated equations are at
http://www.weburbia.com/physics/rocket.html
and elsewhere.

Jim Burns

[Oops++]

James Burns wrote:
> Wayne Throop wrote:
>
>> : Robert Israel <israel@math.MyUniversitysInitials.ca>
>> : At 100 g ( 980 m/sec^2), to reach lightspeed (3*10^8 m/sec)
>> : would take about 3.06*10^5 sec (ignoring relativity, of course) : or
>> 85 hours. The accelerator would have to be about 4.59*10^13 m long,
>> : or 307 AU, extending into the inner Oort cloud.
>>
>> Which oddly enough (unix "units" confirms for me), is about a hundredth
>> of a year. Which one would expect right off the bat if one happens
>> to remember that 1g is about 1c/yr. So, from my perspective, I would
>> say to myself, 100g gives you lightspeed in a hundredth of a year,
>> or three or four days. Arbitrarily saying a tenth lightspeed is below
>> what's needed for significant relativistic effects, that means it takes
>> you something like 8 hours at 100g before you get significant effects.
>> Very handwavy and imprecise, but gives you the general ballpark with
>> minimal calculation.
>
>
> For long trips at constant acceleration, a good fast approximation is
>
a*t = a*d + 1 = (1/2)exp( a*s ) = gamma

Not what I wrote earlier, which is
> a*t = a*d + 1 = exp( a*s ) = gamma
>
> for units where c = 1
> a = shipboard acceleration
> s = shipboard time
> t = standard time
> d = standard distance
> gamma = ( 1 - v^2 )^(-1/2)
>
> The correct, unapproximated equations are at
> http://www.weburbia.com/physics/rocket.html
> and elsewhere.

Which I just noticed can be put in nearly as compact a form:

gamma = cosh(a*T) = sqrt( (at)^2 + 1 ) = ad + 1

Jim Burns

"Robert Israel" <israel@math.MyUniversitysInitials.ca> wrote in message
news:rbisrael.20080219191254$0fdf@news.ks.uiuc.edu ...
>
> The accelerator would have to be about 4.59*10^13 m long,
> or 307 AU, extending into the inner Oort cloud.

And if you think that's too fantastic for anybody to propose with a straight
face, check out a book by Marshall Savage named "The Millennial Project".

--


Regards,
Mike Combs
----------------------------------------------------------------------
By all that you hold dear on this good Earth
I bid you stand, Men of the West!
Aragorn

On Feb 19, 1:23 am, Erik Max Francis <m...@alcyone.com> wrote:

> light, which is 300 000 000 km/s.

How's that again?

Knobby wrote:

> On Feb 19, 1:23 am, Erik Max Francis <m...@alcyone.com> wrote:
>
>> light, which is 300 000 000 km/s.
>
> How's that again?

Typo. 300 000 km/s.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
My reputation grows with every failure.
-- George Bernard Shaw

On Feb 19, 4:31 pm, Erik Max Francis <m...@alcyone.com> wrote:
> STJensen wrote:
> > Someone is placed in a capsule of an electro-magnetic repulsion
> > accelerator in the weightless vacuum of space and accelerated along
> > its entire length.  If the g force the person feels is 2g (32 ft/s-
> > squared x 2) during the entire time along the accelerator and the
> > accelerator is 60,000 miles long, how fast would the person be
> > traveling when they leave the accelerator and how long would their run
> > on the accelerator last?
>
> For straight-line, constant acceleration, starting from rest, with no
> other forces acting, the equations are:
>
>         s = (1/2) a t^2
>
>         v = a t
>
> where t is the elapsed time, v is the accumulated speed, s is the
> accumulated distance, and a is the acceleration.
>
> Solving one for t and substituting (since you're not concerned with the
> total time), you get
>
>         v^2 = 2 a s.
>
> Now plug in numbers and experiment.
>
> > If I understand g forces in the weightless of space, the person/object
> > has to at a constant rate of acceleration to so experience the desired
> > g force.  The moment their velocity remains the same, they return to
> > their weightless existence.  Thus when the person/machine-part finally
> > leaves the 60,000-mile long accelerator, they will return to a
> > weightless existence.
>
> Yes.
>
> --
> Erik Max Francis && m...@alcyone.com &&http://www.alcyone.com/max/
>   San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
>    All delays are dangerous in war.
>     -- John Dryden, 1631-1700

How do you 'square' time...?

On Feb 20, 7:17 am, thro...@sheol.org (Wayne Throop) wrote:
> : Robert Israel <isr...@math.MyUniversitysInitials.ca>
> : At 100 g ( 980 m/sec^2), to reach lightspeed (3*10^8 m/sec)
> : would take about 3.06*10^5 sec (ignoring relativity, of course)
> : or 85 hours.  The accelerator would have to be about 4.59*10^13 m long,
> : or 307 AU, extending into the inner Oort cloud.
>
> Which oddly enough (unix "units" confirms for me), is about a hundredth
> of a year.  Which one would expect right off the bat if one happens
> to remember that 1g is about 1c/yr.  So, from my perspective, I would
> say to myself, 100g gives you lightspeed in a hundredth of a year,
> or three or four days.  Arbitrarily saying a tenth lightspeed is below
> what's needed for significant relativistic effects, that means it takes
> you something like 8 hours at 100g before you get significant effects.
> Very handwavy and imprecise, but gives you the general ballpark with
> minimal calculation.
>
>  ( Though doing the calculation to yield "bam, zoom, to the inner
>    Oord cloud, Alice" is nifty... )
>
>     "He wasn't an astronaut, he was a sitcom actor.
>      And he was only using space travel as a metaphor
>      for beating his wife."
>
>            --- Philip J. Fry
>
> Wayne Throop   thro...@sheol.org  http://sheol.org/throopw

Where is the 'boundary' between Newton and Einstein?
What percentage of light 'speed'?

On Feb 20, 8:12 am, James Burns <burns...@osu.edu> wrote:
> Wayne Throop wrote:
> > : Robert Israel <isr...@math.MyUniversitysInitials.ca>
> > : At 100 g ( 980 m/sec^2), to reach lightspeed (3*10^8 m/sec)
> > : would take about 3.06*10^5 sec (ignoring relativity, of course)
> > : or 85 hours.  The accelerator would have to be about 4.59*10^13 m long,
> > : or 307 AU, extending into the inner Oort cloud.
>
> > Which oddly enough (unix "units" confirms for me), is about a hundredth
> > of a year.  Which one would expect right off the bat if one happens
> > to remember that 1g is about 1c/yr.  So, from my perspective, I would
> > say to myself, 100g gives you lightspeed in a hundredth of a year,
> > or three or four days.  Arbitrarily saying a tenth lightspeed is below
> > what's needed for significant relativistic effects, that means it takes
> > you something like 8 hours at 100g before you get significant effects.
> > Very handwavy and imprecise, but gives you the general ballpark with
> > minimal calculation.
>
> For long trips at constant acceleration, a good fast approximation is
>
>     a*t  =  a*d + 1  =  exp( a*s ) = gamma
>
> for units where c = 1
>     a = shipboard acceleration
>     s = shipboard time
>     t = standard time
>     d = standard distance
>     gamma = ( 1 - v^2 )^(-1/2)
>
> The correct, unapproximated equations are at
>    http://www.weburbia.com/physics/rocket.html
> and elsewhere.
>
> Jim Burns- Hide quoted text -
>
> - Show quoted text -

Regarding rocket.html
A photon is an 'object'.
It's mass=0 but it has momentum...
E=m*c^2 but a photon has energy...
Is this not incongruent?

finite guy wrote:

> How do you 'square' time...?

The same way you square anything: Multiply it by itself.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
Do we ask what profit the little bird hopes for in singing?
-- Johannes Kepler, 1571-1630

finite guy wrote:

> Where is the 'boundary' between Newton and Einstein?
> What percentage of light 'speed'?

It depends on what accuracy level you're going for. A good rule of
thumb is that 10% c is the point at which relativistic effects start to
creep into the third significant figure.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
Do we ask what profit the little bird hopes for in singing?
-- Johannes Kepler, 1571-1630

finite guy wrote:

> Regarding rocket.html
> A photon is an 'object'.
> It's mass=0 but it has momentum...
> E=m*c^2 but a photon has energy...
> Is this not incongruent?

Well, photons can't undergo constant acceleration due to rocket thrust,
so that's really relevant to the equations James was just talking about.

The answer to your question is that there is no problem with photons
have no mass but having momentum and energy because the most general
equation in special relativity relating energy, momentum, and mass is

E^2 = m^2 c^4 + p^2 c^2.

For a particle at rest (p = 0), this decays to E = m c^2, the more
familiar popularized Einstein equation. For a massless particle like a
photon (m = 0), this decays to E = p c, the equation relating a luxon's
energy to its momentum.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
Do we ask what profit the little bird hopes for in singing?
-- Johannes Kepler, 1571-1630

On Feb 22, 9:23 am, Erik Max Francis <m...@alcyone.com> wrote:
> finite guy wrote:
> > How do you 'square' time...?
>
> The same way you square anything:  Multiply it by itself.
>
> --
> Erik Max Francis && m...@alcyone.com &&http://www.alcyone.com/max/
>   San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
>    Do we ask what profit the little bird hopes for in singing?
>     -- Johannes Kepler, 1571-1630

Does that make a 'square'?

x of X axis times x of X axis is x^2 of X axis.

Or do you mean to introduce that which is not there... Y axis?
x*x is not x*y, is it?

On Feb 22, 9:26 am, Erik Max Francis <m...@alcyone.com> wrote:
> finite guy wrote:
> > Where is the 'boundary' between Newton and Einstein?
> > What percentage of light 'speed'?
>
> It depends on what accuracy level you're going for.  A good rule of
> thumb is that 10% c is the point at which relativistic effects start to
> creep into the third significant figure.
>
> --
> Erik Max Francis && m...@alcyone.com &&http://www.alcyone.com/max/
>   San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
>    Do we ask what profit the little bird hopes for in singing?
>     -- Johannes Kepler, 1571-1630

There are too many 'rules of thumb' used for such precise
calculations...
Shall we 'square' the thumb also? Or mutilple it by a finger?

On Feb 22, 9:29 am, Erik Max Francis <m...@alcyone.com> wrote:
> finite guy wrote:
> > Regarding rocket.html
> > A photon is an 'object'.
> > It's mass=0 but it has momentum...
> > E=m*c^2 but a photon has energy...
> > Is this not incongruent?
>
> Well, photons can't undergo constant acceleration due to rocket thrust,
> so that's really relevant to the equations James was just talking about.
>
> The answer to your question is that there is no problem with photons
> have no mass but having momentum and energy because the most general
> equation in special relativity relating energy, momentum, and mass is
>
>         E^2 = m^2 c^4 + p^2 c^2.
>
> For a particle at rest (p = 0), this decays to E = m c^2, the more
> familiar popularized Einstein equation.  For a massless particle like a
> photon (m = 0), this decays to E = p c, the equation relating a luxon's
> energy to its momentum.
>
> --
> Erik Max Francis && m...@alcyone.com &&http://www.alcyone.com/max/
>   San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
>    Do we ask what profit the little bird hopes for in singing?
>     -- Johannes Kepler, 1571-1630

A particle is 'at rest' at delta time = 0...
How do you successfully divide by zero?

In rec.arts.sf.science finite guy <adamlewis@amnet.net.au> wrote:
> On Feb 22, 9:26?am, Erik Max Francis <m...@alcyone.com> wrote:
>> finite guy wrote:
>> > Where is the 'boundary' between Newton and Einstein?
>> > What percentage of light 'speed'?
>>
>> It depends on what accuracy level you're going for. ?A good rule of
>> thumb is that 10% c is the point at which relativistic effects start to
>> creep into the third significant figure.
>
> There are too many 'rules of thumb' used for such precise
> calculations...
> Shall we 'square' the thumb also? Or mutilple it by a finger?

If you don't like rules of thumb then *you* need to be more precise.
Decide how accurate you want to be. This will then tell you where the
boundary lies.

For example, if I'm only interested in being within 10% of the true
answer, then I can use pure Newtonian dynamics up to, I believe, 80% or so
of light speed. But if I'm building a GPS system and need ten or so
decimal places of accuracy, I need to start thinking about relativity even
though my satellites are only doing a few miles per second.

--
Michael Ash
Rogue Amoeba Software

finite guy wrote:
> On Feb 22, 9:26 am, Erik Max Francis <m...@alcyone.com> wrote:
>> finite guy wrote:
>>> Where is the 'boundary' between Newton and Einstein?
>>> What percentage of light 'speed'?
>> It depends on what accuracy level you're going for. A good rule of
>> thumb is that 10% c is the point at which relativistic effects start to
>> creep into the third significant figure.
>
> There are too many 'rules of thumb' used for such precise
> calculations...

As I said, it depends on what level of accuracy you're going for. If
you want a more precise figure, you can solve for beta(gamma) yourself.
It comes out to roughly 10% c for third-significant-figure effects,
30% for second, and 70% for first.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
History is a set of lies agreed upon.
-- Napoleon Bonaparte

finite guy wrote:
> On Feb 22, 9:29 am, Erik Max Francis <m...@alcyone.com> wrote:
>> finite guy wrote:
>>> Regarding rocket.html
>>> A photon is an 'object'.
>>> It's mass=0 but it has momentum...
>>> E=m*c^2 but a photon has energy...
>>> Is this not incongruent?
>> Well, photons can't undergo constant acceleration due to rocket thrust,
>> so that's really relevant to the equations James was just talking about.
>>
>> The answer to your question is that there is no problem with photons
>> have no mass but having momentum and energy because the most general
>> equation in special relativity relating energy, momentum, and mass is
>>
>> E^2 = m^2 c^4 + p^2 c^2.
>>
>> For a particle at rest (p = 0), this decays to E = m c^2, the more
>> familiar popularized Einstein equation. For a massless particle like a
>> photon (m = 0), this decays to E = p c, the equation relating a luxon's
>> energy to its momentum.
>
> A particle is 'at rest' at delta time = 0...
> How do you successfully divide by zero?

No division by zero is taking place in anything I wrote, so I don't know
what you're talking about.

If you're asking how to consider instantaneously changing variables,
then the answer is differential calculus.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
History is a set of lies agreed upon.
-- Napoleon Bonaparte

finite guy wrote:
> On Feb 22, 9:23 am, Erik Max Francis <m...@alcyone.com> wrote:
>> finite guy wrote:
>>> How do you 'square' time...?
>> The same way you square anything: Multiply it by itself.
>
> Does that make a 'square'?
>
> x of X axis times x of X axis is x^2 of X axis.
>
> Or do you mean to introduce that which is not there... Y axis?
> x*x is not x*y, is it?

I have no idea what you're talking about. Plot y = x^2 on the x-y axis.
It contains a term that is squared, and isn't that hard to understand.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
History is a set of lies agreed upon.
-- Napoleon Bonaparte

On Feb 22, 10:11 am, Erik Max Francis <m...@alcyone.com> wrote:
> finite guy wrote:
> > On Feb 22, 9:26 am, Erik Max Francis <m...@alcyone.com> wrote:
> >> finite guy wrote:
> >>> Where is the 'boundary' between Newton and Einstein?
> >>> What percentage of light 'speed'?
> >> It depends on what accuracy level you're going for.  A good rule of
> >> thumb is that 10% c is the point at which relativistic effects start to
> >> creep into the third significant figure.
>
> > There are too many 'rules of thumb' used for such precise
> > calculations...
>
> As I said, it depends on what level of accuracy you're going for.  If
> you want a more precise figure, you can solve for beta(gamma) yourself.
>   It comes out to roughly 10% c for third-significant-figure effects,
> 30% for second, and 70% for first.
>
> --
> Erik Max Francis && m...@alcyone.com &&http://www.alcyone.com/max/
>   San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
>    History is a set of lies agreed upon.
>     -- Napoleon Bonaparte

Do you mean that the 'boundary' is subject to my decision...?

Roughly, is that a base-10 'third-significant-figure' or something
more precise? :-)

finite guy wrote:
> On Feb 22, 10:11 am, Erik Max Francis <m...@alcyone.com> wrote:
>> finite guy wrote:
>>> On Feb 22, 9:26 am, Erik Max Francis <m...@alcyone.com> wrote:
>>>> finite guy wrote:
>>>>> Where is the 'boundary' between Newton and Einstein?
>>>>> What percentage of light 'speed'?
>>>> It depends on what accuracy level you're going for. A good rule of
>>>> thumb is that 10% c is the point at which relativistic effects start to
>>>> creep into the third significant figure.
>>> There are too many 'rules of thumb' used for such precise
>>> calculations...
>> As I said, it depends on what level of accuracy you're going for. If
>> you want a more precise figure, you can solve for beta(gamma) yourself.
>> It comes out to roughly 10% c for third-significant-figure effects,
>> 30% for second, and 70% for first.
>
> Do you mean that the 'boundary' is subject to my decision...?

There is no boundary, just a continuous transition. So when you ask
what "the 'boundary'" is, the answer depends on what _you_ mean by the
boundary. To answer that, you have to specify quantitatively where you
think that relativistic effects become significant, and you haven't done
that.

However, even without that information I've given about the most
complete answer possible, so I don't see what your point is. Did you
want your question answered, or did you want to try to argue?

> Roughly, is that a base-10 'third-significant-figure' or something
> more precise? :-)

It is the standard meaning. I'm suspecting you're just trying to be
difficult at this point.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
There is no fate that cannot be surmounted by scorn.
-- Albert Camus

On Feb 22, 10:13 am, Erik Max Francis <m...@alcyone.com> wrote:
> finite guy wrote:
> > On Feb 22, 9:29 am, Erik Max Francis <m...@alcyone.com> wrote:
> >> finite guy wrote:
> >>> Regarding rocket.html
> >>> A photon is an 'object'.
> >>> It's mass=0 but it has momentum...
> >>> E=m*c^2 but a photon has energy...
> >>> Is this not incongruent?
> >> Well, photons can't undergo constant acceleration due to rocket thrust,
> >> so that's really relevant to the equations James was just talking about..
>
> >> The answer to your question is that there is no problem with photons
> >> have no mass but having momentum and energy because the most general
> >> equation in special relativity relating energy, momentum, and mass is
>
> >>         E^2 = m^2 c^4 + p^2 c^2.
>
> >> For a particle at rest (p = 0), this decays to E = m c^2, the more
> >> familiar popularized Einstein equation.  For a massless particle likea
> >> photon (m = 0), this decays to E = p c, the equation relating a luxon's
> >> energy to its momentum.
>
> > A particle is 'at rest' at delta time = 0...
> > How do you successfully divide by zero?
>
> No division by zero is taking place in anything I wrote, so I don't know
> what you're talking about.
>
> If you're asking how to consider instantaneously changing variables,
> then the answer is differential calculus.
>
> --
> Erik Max Francis && m...@alcyone.com &&http://www.alcyone.com/max/
>   San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
>    History is a set of lies agreed upon.
>     -- Napoleon Bonaparte- Hide quoted text -
>
> - Show quoted text -

"this decays to E = m c^2"
Is not the 'speed of light' in the form d/t ???
And squared is d^2/t^2.

So how do YOU 'square' time?

Perhaps it is 'lateral time' X 'axial time'.
Afterall they are two different axial 'times', aren't they?

On Feb 22, 10:13 am, Erik Max Francis <m...@alcyone.com> wrote:
> finite guy wrote:
> > On Feb 22, 9:29 am, Erik Max Francis <m...@alcyone.com> wrote:
> >> finite guy wrote:
> >>> Regarding rocket.html
> >>> A photon is an 'object'.
> >>> It's mass=0 but it has momentum...
> >>> E=m*c^2 but a photon has energy...
> >>> Is this not incongruent?
> >> Well, photons can't undergo constant acceleration due to rocket thrust,
> >> so that's really relevant to the equations James was just talking about..
>
> >> The answer to your question is that there is no problem with photons
> >> have no mass but having momentum and energy because the most general
> >> equation in special relativity relating energy, momentum, and mass is
>
> >>         E^2 = m^2 c^4 + p^2 c^2.
>
> >> For a particle at rest (p = 0), this decays to E = m c^2, the more
> >> familiar popularized Einstein equation.  For a massless particle likea
> >> photon (m = 0), this decays to E = p c, the equation relating a luxon's
> >> energy to its momentum.
>
> > A particle is 'at rest' at delta time = 0...
> > How do you successfully divide by zero?
>
> No division by zero is taking place in anything I wrote, so I don't know
> what you're talking about.
>
> If you're asking how to consider instantaneously changing variables,
> then the answer is differential calculus.
>
> --
> Erik Max Francis && m...@alcyone.com &&http://www.alcyone.com/max/
>   San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
>    History is a set of lies agreed upon.
>     -- Napoleon Bonaparte- Hide quoted text -
>
> - Show quoted text -

I'm asking you to consider how a photon moves from here to there.

On Feb 22, 11:42 am, Erik Max Francis <m...@alcyone.com> wrote:
> finite guy wrote:
> > On Feb 22, 10:11 am, Erik Max Francis <m...@alcyone.com> wrote:
> >> finite guy wrote:
> >>> On Feb 22, 9:26 am, Erik Max Francis <m...@alcyone.com> wrote:
> >>>> finite guy wrote:
> >>>>> Where is the 'boundary' between Newton and Einstein?
> >>>>> What percentage of light 'speed'?
> >>>> It depends on what accuracy level you're going for.  A good rule of
> >>>> thumb is that 10% c is the point at which relativistic effects start to
> >>>> creep into the third significant figure.
> >>> There are too many 'rules of thumb' used for such precise
> >>> calculations...
> >> As I said, it depends on what level of accuracy you're going for.  If
> >> you want a more precise figure, you can solve for beta(gamma) yourself.
> >>   It comes out to roughly 10% c for third-significant-figure effects,
> >> 30% for second, and 70% for first.
>
> > Do you mean that the 'boundary' is subject to my decision...?
>
> There is no boundary, just a continuous transition.  So when you ask
> what "the 'boundary'" is, the answer depends on what _you_ mean by the
> boundary.  To answer that, you have to specify quantitatively where you
> think that relativistic effects become significant, and you haven't done
> that.
>
> However, even without that information I've given about the most
> complete answer possible, so I don't see what your point is.  Did you
> want your question answered, or did you want to try to argue?
>
> > Roughly, is that a base-10 'third-significant-figure' or something
> > more precise?   :-)
>
> It is the standard meaning.  I'm suspecting you're just trying to be
> difficult at this point.
>
> --
> Erik Max Francis && m...@alcyone.com &&http://www.alcyone.com/max/
>   San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
>    There is no fate that cannot be surmounted by scorn.
>     -- Albert Camus- Hide quoted text -
>
> - Show quoted text -

Sorry, I'm not trying to be difficult.

I am alluding to the question: "What makes it so-called
significant...?"

finite guy wrote:

> "this decays to E = m c^2"
> Is not the 'speed of light' in the form d/t ???
> And squared is d^2/t^2.
>
> So how do YOU 'square' time?

You still square it just like any other quantity with units. The square
of a distance is an area. The square of a time is a unit we don't
commonly refer to with a separate name, but it comes up all the time in
physics. The square of a speed is, among other things, an energy density.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
There is no easy way from the Earth to the stars.
-- Seneca, ca. 1st C.

finite guy wrote:
> On Feb 22, 10:13 am, Erik Max Francis <m...@alcyone.com> wrote:
>> finite guy wrote:
>>> On Feb 22, 9:29 am, Erik Max Francis <m...@alcyone.com> wrote:
>>>> finite guy wrote:
>>>>> Regarding rocket.html
>>>>> A photon is an 'object'.
>>>>> It's mass=0 but it has momentum...
>>>>> E=m*c^2 but a photon has energy...
>>>>> Is this not incongruent?
>>>> Well, photons can't undergo constant acceleration due to rocket thrust,
>>>> so that's really relevant to the equations James was just talking about.
>>>> The answer to your question is that there is no problem with photons
>>>> have no mass but having momentum and energy because the most general
>>>> equation in special relativity relating energy, momentum, and mass is
>>>> E^2 = m^2 c^4 + p^2 c^2.
>>>> For a particle at rest (p = 0), this decays to E = m c^2, the more
>>>> familiar popularized Einstein equation. For a massless particle like a
>>>> photon (m = 0), this decays to E = p c, the equation relating a luxon's
>>>> energy to its momentum.
>>> A particle is 'at rest' at delta time = 0...
>>> How do you successfully divide by zero?
>> No division by zero is taking place in anything I wrote, so I don't know
>> what you're talking about.
>>
>> If you're asking how to consider instantaneously changing variables,
>> then the answer is differential calculus.
>
> I'm asking you to consider how a photon moves from here to there.

You were asking about particles at rest, which can't include photons, so
I don't know what you're asking.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
There is no easy way from the Earth to the stars.
-- Seneca, ca. 1st C.

finite guy wrote:
> On Feb 22, 11:42 am, Erik Max Francis <m...@alcyone.com> wrote:
>> finite guy wrote:
>>> On Feb 22, 10:11 am, Erik Max Francis <m...@alcyone.com> wrote:
>>>> finite guy wrote:
>>>>> On Feb 22, 9:26 am, Erik Max Francis <m...@alcyone.com> wrote:
>>>>>> finite guy wrote:
>>>>>>> Where is the 'boundary' between Newton and Einstein?
>>>>>>> What percentage of light 'speed'?
>>>>>> It depends on what accuracy level you're going for. A good rule of
>>>>>> thumb is that 10% c is the point at which relativistic effects start to
>>>>>> creep into the third significant figure.
>>>>> There are too many 'rules of thumb' used for such precise
>>>>> calculations...
>>>> As I said, it depends on what level of accuracy you're going for. If
>>>> you want a more precise figure, you can solve for beta(gamma) yourself.
>>>> It comes out to roughly 10% c for third-significant-figure effects,
>>>> 30% for second, and 70% for first.
>>> Do you mean that the 'boundary' is subject to my decision...?
>> There is no boundary, just a continuous transition. So when you ask
>> what "the 'boundary'" is, the answer depends on what _you_ mean by the
>> boundary. To answer that, you have to specify quantitatively where you
>> think that relativistic effects become significant, and you haven't done
>> that.
>>
>> However, even without that information I've given about the most
>> complete answer possible, so I don't see what your point is. Did you
>> want your question answered, or did you want to try to argue?
>>
>>> Roughly, is that a base-10 'third-significant-figure' or something
>>> more precise? :-)
>> It is the standard meaning. I'm suspecting you're just trying to be
>> difficult at this point.
>
> Sorry, I'm not trying to be difficult.
>
> I am alluding to the question: "What makes it so-called
> significant...?"

If you're asking what significant figures are, they're a way of
measuring accuracy, and yes, they're base 10. Something accurate to
three significant figures would be 1.64 -- it's not 1.63, and it's not
1.65. The more significant figures, the more accurate it is.

At speeds of about 10% c, relativistic effects will tend to start
creeping into that third significant figure, so they could change a 1.63
to a 1.64 or a 1.65.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
There is no easy way from the Earth to the stars.
-- Seneca, ca. 1st C.

On Feb 22, 12:10 pm, Erik Max Francis <m...@alcyone.com> wrote:
> finite guy wrote:
> > "this decays to E = m c^2"
> > Is not the 'speed of light' in the form d/t ???
> > And squared is d^2/t^2.
>
> > So how do YOU 'square' time?
>
> You still square it just like any other quantity with units.  The square
> of a distance is an area.  The square of a time is a unit we don't
> commonly refer to with a separate name, but it comes up all the time in
> physics.  The square of a speed is, among other things, an energy density.
>
> --
> Erik Max Francis && m...@alcyone.com &&http://www.alcyone.com/max/
>   San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
>    There is no easy way from the Earth to the stars.
>     -- Seneca, ca. 1st C.

That's cool but it doesn't really sddress the problem.

What is the 'PLANE' of time then?

If it comes up in physics all of the time,
why is the plane of time (lateral time X axial time) not commonly
understood?

It is there in the equation, E=m*d^2/t^2.

What is commonly mis-understood?
That there is only one axis of 'time'...

On Feb 22, 12:23 pm, Erik Max Francis <m...@alcyone.com> wrote:
> finite guy wrote:
> > On Feb 22, 10:13 am, Erik Max Francis <m...@alcyone.com> wrote:
> >> finite guy wrote:
> >>> On Feb 22, 9:29 am, Erik Max Francis <m...@alcyone.com> wrote:
> >>>> finite guy wrote:
> >>>>> Regarding rocket.html
> >>>>> A photon is an 'object'.
> >>>>> It's mass=0 but it has momentum...
> >>>>> E=m*c^2 but a photon has energy...
> >>>>> Is this not incongruent?
> >>>> Well, photons can't undergo constant acceleration due to rocket thrust,
> >>>> so that's really relevant to the equations James was just talking about.
> >>>> The answer to your question is that there is no problem with photons
> >>>> have no mass but having momentum and energy because the most general
> >>>> equation in special relativity relating energy, momentum, and mass is
> >>>>         E^2 = m^2 c^4 + p^2 c^2.
> >>>> For a particle at rest (p = 0), this decays to E = m c^2, the more
> >>>> familiar popularized Einstein equation.  For a massless particle like a
> >>>> photon (m = 0), this decays to E = p c, the equation relating a luxon's
> >>>> energy to its momentum.
> >>> A particle is 'at rest' at delta time = 0...
> >>> How do you successfully divide by zero?
> >> No division by zero is taking place in anything I wrote, so I don't know
> >> what you're talking about.
>
> >> If you're asking how to consider instantaneously changing variables,
> >> then the answer is differential calculus.
>
> > I'm asking you to consider how a photon moves from here to there.
>
> You were asking about particles at rest, which can't include photons, so
> I don't know what you're asking.
>
> --
> Erik Max Francis && m...@alcyone.com &&http://www.alcyone.com/max/
>   San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
>    There is no easy way from the Earth to the stars.
>     -- Seneca, ca. 1st C.- Hide quoted text -
>
> - Show quoted text -

You mean that you don't know how to correctly include photons in your
concept.
At delta time = zero, ALL things are static.
That is mathematically and conceptually obvious...

What is your commonly held misconception?
Perhaps it starts with thinking that the Pythagoras Theorem involves
'squares'.
x^2 + y^2 = e^2 does not show Pythagoras 'Squares'.
They are areas, as you point out...

Halve all the squares to become triangles and equate.
1/2*x^2 + 1/2*y^2 = 1/2*e^2.
1/2*(x^2 + y^2) = 1/2*(e^2).
x^2 + y^2 = e^2... looks the same for ANY shape...
Pythagoras Squares are a special case only.

So, just as in your physics equations, constants need to be involved.
Consider the more general application and
you find that 3D (or spatial' to us) is also a 2D plane.
Gosh, that's why E=m*c^2 is a planar equation -
not a 'spatial' one with space "extending to infinity".

Perhaps this is the beginning of the problem.
Think so...?

finite guy wrote:
> On Feb 22, 12:10 pm, Erik Max Francis <m...@alcyone.com> wrote:
>> finite guy wrote:
>>> "this decays to E = m c^2"
>>> Is not the 'speed of light' in the form d/t ???
>>> And squared is d^2/t^2.
>>> So how do YOU 'square' time?
>> You still square it just like any other quantity with units. The square
>> of a distance is an area. The square of a time is a unit we don't
>> commonly refer to with a separate name, but it comes up all the time in
>> physics. The square of a speed is, among other things, an energy density.
>
> That's cool but it doesn't really sddress the problem.
>
> What is the 'PLANE' of time then?
>
> If it comes up in physics all of the time,
> why is the plane of time (lateral time X axial time) not commonly
> understood?
>
> It is there in the equation, E=m*d^2/t^2.
>
> What is commonly mis-understood?
> That there is only one axis of 'time'...

Squaring a quantity does not imply multiple independent "dimensions" of
that quantity. This isn't a problem except in your visualization of
what's going on; just stop visualizing it like that.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
So take a look at yourself / What's this secret you can't tell me
-- Level 42

finite guy wrote:
> On Feb 22, 12:23 pm, Erik Max Francis <m...@alcyone.com> wrote:
>> You were asking about particles at rest, which can't include photons, so
>> I don't know what you're asking.
>
> You mean that you don't know how to correctly include photons in your
> concept.

You asked about particles at rest, which can't be photons, so of course
I didn't say anything about them. I didn't mention fruit flies, either,
because you didn't say anything about them.

"My concept" is describing basic special relativistic equations (or
basic mechanics, if you read my simpler response about calculus) which
are uncontroversial outside of crankdom. Hmm, am I starting to think
we've crossed over to the other side by someone crossposting to sci.math
.... ?

> At delta time = zero, ALL things are static.
> That is mathematically and conceptually obvious...

Time flows, so that hardly matters. If you freeze time, everything is
static. So what?

> What is your commonly held misconception?
> Perhaps it starts with thinking that the Pythagoras Theorem involves
> 'squares'.
> x^2 + y^2 = e^2 does not show Pythagoras 'Squares'.
> They are areas, as you point out...

I never said anything about the Pythagoras theorem and pointed no such
thing out. And I now see that this is a waste of time.

At first I thought you were just being difficult. Then I thought you
were just really, really confused and clueless, but sincere. Now I see
that you're just a sci.math-imported crank.

> Perhaps this is the beginning of the problem.
> Think so...?

Think I'm done.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
So take a look at yourself / What's this secret you can't tell me
-- Level 42

On Feb 22, 1:30 pm, Erik Max Francis <m...@alcyone.com> wrote:
> finite guy wrote:
> > On Feb 22, 12:23 pm, Erik Max Francis <m...@alcyone.com> wrote:
> >> You were asking about particles at rest, which can't include photons, so
> >> I don't know what you're asking.
>
> > You mean that you don't know how to correctly include photons in your
> > concept.
>
> You asked about particles at rest, which can't be photons, so of course
> I didn't say anything about them.  I didn't mention fruit flies, either,
> because you didn't say anything about them.
>
> "My concept" is describing basic special relativistic equations (or
> basic mechanics, if you read my simpler response about calculus) which
> are uncontroversial outside of crankdom.  Hmm, am I starting to think
> we've crossed over to the other side by someone crossposting to sci.math
> ... ?
>
> > At delta time = zero, ALL things are static.
> > That is mathematically and conceptually obvious...
>
> Time flows, so that hardly matters.  If you freeze time, everything is
> static.  So what?
>
> > What is your commonly held misconception?
> > Perhaps it starts with thinking that the Pythagoras Theorem involves
> > 'squares'.
> > x^2 + y^2 = e^2 does not show Pythagoras 'Squares'.
> > They are areas, as you point out...
>
> I never said anything about the Pythagoras theorem and pointed no such
> thing out.  And I now see that this is a waste of time.
>
> At first I thought you were just being difficult.  Then I thought you
> were just really, really confused and clueless, but sincere.  Now I see
> that you're just a sci.math-imported crank.
>
> > Perhaps this is the beginning of the problem.
> > Think so...?
>
> Think I'm done.
>
> --
> Erik Max Francis && m...@alcyone.com &&http://www.alcyone.com/max/
>   San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
>    So take a look at yourself / What's this secret you can't tell me
>     -- Level 42

I appreciate what you are trying to saying and I don't want to bicker.

You may not realise what you have said :-
"You asked about particles at rest, which CAN'T be photons..."
"If you freeze time, EVERYTHING is static..."

Why do you not include photons in "everything"?
Aren't they something...?
Aren't they also static at delta time = zero like everything... else?
Which one is it - static or not?
How far can a photon travel in zero time?

You didn't mention Pythagoras, I know.
But it is an underpinning of your thinking and mathematics and not
fully correct.
Please don't take this as 'crank' offensiveness -
your understanding is intelligent and educated but incongruent as
shown above.

The Pythagoras Theorem and Einstein E=m*c^2 are quite simply
in the same form and therefore relevant to the discussion.
If you cannot discuss it; it is not because of the mathematics
but because of your belief system which does not account for the plane
of time.

Please clarify if photons are part of everything or not.

Thanks.

finite guy skreiv:

> Where is the 'boundary' between Newton and Einstein?
> What percentage of light 'speed'?

Where is the boundary between being tall and being short ?

There is no boundary whatsoever. Mathemathically speaking Einstein is
-always- more correct than Newton.

At low speeds, the difference is -very- small, at large speeds the
difference can be very large.

Thus how long you can continue to use Newton depends ONLY on how
accurate an answer you need.


Eivind Kjørstad

finite guy wrote:

> You may not realise what you have said :-
> "You asked about particles at rest, which CAN'T be photons..."
> "If you freeze time, EVERYTHING is static..."
>
> Why do you not include photons in "everything"?
> Aren't they something...?
> Aren't they also static at delta time = zero like everything... else?
> Which one is it - static or not?
> How far can a photon travel in zero time?

Time _isn't_ frozen, so it's an academic question. If all particles
were really fruit flies, then you could ask questions about the
properties of those particle/fruit flies. However, since they aren't,
it's a pointless question to try to answer, or consider meaningful.

--
Erik Max Francis && max@alcyone.com && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
It is dangerous to be right in matters on which the established
authorities are wrong. -- Voltaire

On Feb 21, 11:30 pm, Erik Max Francis <m...@alcyone.com> wrote:

<snip>

> At first I thought you were just being difficult.  Then I thought you
> were just really, really confused and clueless, but sincere.  Now I see
> that you're just a sci.math-imported crank.

Erik, you should have stopped there. You're absolutely right - finite
guy is more than a crank, he's an *******.

Your discussions in this thread were very informative. Don't let that
******* get to you.

On Feb 23, 8:14 am, Erik Max Francis <m...@alcyone.com> wrote:
> finite guy wrote:
> > You may not realise what you have said :-
> > "You asked about particles at rest, which CAN'T be photons..."
> > "If you freeze time, EVERYTHING is static..."
>
> > Why do you not include photons in "everything"?
> > Aren't they something...?
> > Aren't they also static at delta time = zero like everything... else?
> > Which one is it - static or not?
> > How far can a photon travel in zero time?
>
> Time _isn't_ frozen, so it's an academic question.  If all particles
> were really fruit flies, then you could ask questions about the
> properties of those particle/fruit flies.  However, since they aren't,
> it's a pointless question to try to answer, or consider meaningful.
>
> --
> Erik Max Francis && m...@alcyone.com &&http://www.alcyone.com/max/
>   San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
>    It is dangerous to be right in matters on which the established
>     authorities are wrong. -- Voltaire

Folly.
Time is 'frozen' at delta time =0.

You deny meaning of the obvious.

On Feb 23, 9:01 am, Neilist <lattora...@gmail.com> wrote:
> On Feb 21, 11:30 pm, Erik Max Francis <m...@alcyone.com> wrote:
>
> <snip>
>
> > At first I thought you were just being difficult.  Then I thought you
> > were just really, really confused and clueless, but sincere.  Now I see
> > that you're just a sci.math-imported crank.
>
> Erik, you should have stopped there.  You're absolutely right - finite
> guy is more than a crank, he's an *******.
>
> Your discussions in this thread were very informative.  Don't let that
> ******* get to you.

You have an ******* - try wiping it, not me...

You just don't want to answer obvious fallacies in your double-think
understanding of reality.

On Feb 22, 11:42 am, Erik Max Francis <m...@alcyone.com> wrote:
> finite guy wrote:
> > On Feb 22, 10:11 am, Erik Max Francis <m...@alcyone.com> wrote:
> >> finite guy wrote:
> >>> On Feb 22, 9:26 am, Erik Max Francis <m...@alcyone.com> wrote:
> >>>> finite guy wrote:
> >>>>> Where is the 'boundary' between Newton and Einstein?
> >>>>> What percentage of light 'speed'?
> >>>> It depends on what accuracy level you're going for.  A good rule of
> >>>> thumb is that 10% c is the point at which relativistic effects start to
> >>>> creep into the third significant figure.
> >>> There are too many 'rules of thumb' used for such precise
> >>> calculations...
> >> As I said, it depends on what level of accuracy you're going for.  If
> >> you want a more precise figure, you can solve for beta(gamma) yourself.
> >>   It comes out to roughly 10% c for third-significant-figure effects,
> >> 30% for second, and 70% for first.
>
> > Do you mean that the 'boundary' is subject to my decision...?
>
> There is no boundary, just a continuous transition.  So when you ask
> what "the 'boundary'" is, the answer depends on what _you_ mean by the
> boundary.  To answer that, you have to specify quantitatively where you
> think that relativistic effects become significant, and you haven't done
> that.
>
> However, even without that information I've given about the most
> complete answer possible, so I don't see what your point is.  Did you
> want your question answered, or did you want to try to argue?
>
> > Roughly, is that a base-10 'third-significant-figure' or something
> > more precise?   :-)
>
> It is the standard meaning.  I'm suspecting you're just trying to be
> difficult at this point.
>
> --
> Erik Max Francis && m...@alcyone.com &&http://www.alcyone.com/max/
>   San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
>    There is no fate that cannot be surmounted by scorn.
>     -- Albert Camus- Hide quoted text -
>
> - Show quoted text -

Sorry for my phraseology.

My question was directed to you because you made a 'distinction'
between Newton and Einstein but no one seems to be able to quantify it
OR qualify it.
It is all wishy-washy about the 'level of accuracy' you choose.

Quite honestly, considering the accuracy of calculations and
experimentation, I would have thought that it was clearly defined when
to use one or the other.

I'm not 'trying to be difficult' but it seems that it is a difficult
question for you to answer clearly.

My apologies.

On Feb 23, 9:01 am, Neilist <lattora...@gmail.com> wrote:
> On Feb 21, 11:30 pm, Erik Max Francis <m...@alcyone.com> wrote:
>
> <snip>
>
> > At first I thought you were just being difficult.  Then I thought you
> > were just really, really confused and clueless, but sincere.  Now I see
> > that you're just a sci.math-imported crank.
>
> Erik, you should have stopped there.  You're absolutely right - finite
> guy is more than a crank, he's an *******.
>
> Your discussions in this thread were very informative.  Don't let that
> ******* get to you.

Yes, the discussions are informative but certainly not definitive.
Hey, choose your 'level of accuracy'.
Being an intelligent, upright guy with your head up your own ***, your
gun sight will probably point at your feet.
If you target this *******, you may only have accuracy to hit... your
own foot.

Certainly, like most people here, we can be asses.
Question is: 'Is it **** or fertiliser...?'

If you can't grow, it may be because you treat good fertiliser like
your same verbal diarrhea.

Honestly, although it can be frustrating to get inane replies, my
attitude is not one like yours. I say this 'tongue in cheek(s).'

: finite guy <adamlewis@amnet.net.au>
: My question was directed to you because you made a 'distinction'
: between Newton and Einstein but no one seems to be able to quantify it
: OR qualify it.

Several people have provided expressions for it.
The quantity is a matter of plugging in a velocity and doing some
arithmetic. The fact that you aren't satisfied does not constitute a
"fallacy" nor any "double-think" in either newtonian or einsteinian physics.

: Message-ID: <526f5a92-d2f8-4c60-8225-2011df5b294e@h25g2000hsf.googlegroups.com>
: Yes, the discussions are informative but certainly not definitive.

What, specifically, has not been defined? The meaning of "significant"?
Use any dictionary. The earliest post I see mentioning "significant"
said fair and square that the standard of significance is arbitrary, and
this arbitrary standard, used for purposes of discussion, was quantified.

Again, none of this constitutes a "fallacy" or any "double-think".


Wayne Throop throopw@sheol.org http://sheol.org/throopw

On 2008-02-25, finite guy <adamlewis@amnet.net.au> wrote:
> Quite honestly, considering the accuracy of calculations and
> experimentation, I would have thought that it was clearly defined
> when to use one or the other.

For any given limits of required precision, it is. That precision
will be determined by the application for which you are employing the
theoretical model, and upon any other simplifications you may already
have made.


- Tim

On Feb 25, 12:35 pm, thro...@sheol.org (Wayne Throop) wrote:
> : finite guy <adamle...@amnet.net.au>
> : My question was directed to you because you made a 'distinction'
> : between Newton and Einstein but no one seems to be able to quantify it
> : OR qualify it.
>
> Several people have provided expressions for it.
> The quantity is a matter of plugging in a velocity and doing some
> arithmetic.  The fact that you aren't satisfied does not constitute a
> "fallacy" nor any "double-think" in either newtonian or einsteinian physics.
>
> : Message-ID: <526f5a92-d2f8-4c60-8225-2011df5b2...@h25g2000hsf.googlegroups.com>
> : Yes, the discussions are informative but certainly not definitive.
>
> What, specifically, has not been defined?  The meaning of "significant"?
> Use any dictionary.  The earliest post I see mentioning "significant"
> said fair and square that the standard of significance is arbitrary, and
> this arbitrary standard, used for purposes of discussion, was quantified.
>
> Again, none of this constitutes a "fallacy" or any "double-think".
>
> Wayne Throop   thro...@sheol.org  http://sheol.org/throopw


Thanks for the polite and prety lucid response.

Why is the 'most accurate' method just not used all of the time?

Certainly, Newtonian and Einsteinian ideas are regarded somewhat
'differently'.
In what way are they really different.

You seem to have a reasonable and questioning attitude by your
website.
Would you, if possible, explain why E=mc^2 is a planar equation and
why the 'plane of time' in it is not conceptualised at all?

: finite guy <adamlewis@amnet.net.au>
: Why is the 'most accurate' method just not used all of the time?

Because it's often simpler to use newtonian expressions, and if you've
only measured to one part in a billion, or if you're only concerned with
modeling situations that you will only be able to verify to one part
in a billion, the extra effort is wasted. (The "one part in a billion"
is an arbitrary illustrative figure only.)

: Certainly, Newtonian and Einsteinian ideas are regarded somewhat
: 'differently'. In what way are they really different.

When you get right down to it, they differ in the coordinate transform
wrt which the laws of physics are invariant. (Phrasing it that way leads
to wondering why the coordinates matter... they don't, really, but it's
an easy way to see what's going on (or, it seems easy to me).) This in
turn leads to differing formulas for kinematics and dynamics, in particular
velocity "addition" (or actually, velocity transforms are mere addition),
and in expressions for momentum and kinetic energy, and other miscelania.
If you simplify the einsteinian expressions by taking the limit as velocity
approaches zero, they turn into the newtonian ones.

: Would you, if possible, explain why E=mc^2 is a planar equation and
: why the 'plane of time' in it is not conceptualised at all?

I'm afraid I don't understand why there *should* be a "plane" of time.
Certainly, just because a quantity is multiplied by itself doesn't
really imply there's a "plane" of time. If it did, then the newtonian
expression for distance covered in a given time at a given acceleration
(ie, the newtonian approximation of d=(1/2)at^2) would involve a "plane
of time", and it doesn't seem necessary.

Or, just consider what acceleration is: velocity per time. How much
do you speed up in a second. Since velocity is distance per time,
acceleration has units of disance per time squared. Again, this has
nothing to do with geometry, except perhaps in a *very* abstract sense.

Similarly, the d^2/t^2 units of "c^2" in E=mc^2, do not imply that there
are two different directions of time, or a geometrical plane. Now, to get
into "where does the c^2 come from anyways" would be a bit more complex.
In the abstract, it's a matter of units analysis. Energy has *units*
of mass times velocity squared, for reasons similar to the reason why
acceleratio has units of distance per time squared. It doesn't have to
do with there being a "plane" involved, it has to do with how you use
things with velocity terms in them to calculate the amount of energy.
Potential energy (in the newtonian approximation) is m*d*a, and kinetic
energy is (1/2)mv^2, for reasons that have nothing to do with "planes".
Instead it has to do with the same reason you get time squared when you
use time and acceleration to calculate distance.


Wayne Throop throopw@sheol.org http://sheol.org/throopw

On Feb 25, 1:40 pm, thro...@sheol.org (Wayne Throop) wrote:
> : finite guy <adamle...@amnet.net.au>
> : Why is the 'most accurate' method just not used all of the time?
>
> Because it's often simpler to use newtonian expressions, and if you've
> only measured to one part in a billion, or if you're only concerned with
> modeling situations that you will only be able to verify to one part
> in a billion, the extra effort is wasted.  (The "one part in a billion"
> is an arbitrary illustrative figure only.)
>
> : Certainly, Newtonian and Einsteinian ideas are regarded somewhat
> : 'differently'.  In what way are they really different.
>
> When you get right down to it, they differ in the coordinate transform
> wrt which the laws of physics are invariant.  (Phrasing it that way leads
> to wondering why the coordinates matter... they don't, really, but it's
> an easy way to see what's going on (or, it seems easy to me).)  This in
> turn leads to differing formulas for kinematics and dynamics, in particular
> velocity "addition" (or actually, velocity transforms are mere addition),
> and in expressions for momentum and kinetic energy, and other miscelania.
> If you simplify the einsteinian expressions by taking the limit as velocity
> approaches zero, they turn into the newtonian ones.
>
> : Would you, if possible, explain why E=mc^2 is a planar equation and
> : why the 'plane of time' in it is not conceptualised at all?
>
> I'm afraid I don't understand why there *should* be a "plane" of time.
> Certainly, just because a quantity is multiplied by itself doesn't
> really imply there's a "plane" of time.  If it did, then the newtonian
> expression for distance covered in a given time at a given acceleration
> (ie, the newtonian approximation of d=(1/2)at^2) would involve a "plane
> of time", and it doesn't seem necessary.
>
> Or, just consider what acceleration is: velocity per time.  How much
> do you speed up in a second.  Since velocity is distance per time,
> acceleration has units of disance per time squared.  Again, this has
> nothing to do with geometry, except perhaps in a *very* abstract sense.
>
> Similarly, the d^2/t^2 units of "c^2" in E=mc^2, do not imply that there
> are two different directions of time, or a geometrical plane.  Now, to get
> into "where does the c^2 come from anyways" would be a bit more complex.
> In the abstract, it's a matter of units analysis.  Energy has *units*
> of mass times velocity squared, for reasons similar to the reason why
> acceleratio has units of distance per time squared.  It doesn't have to
> do with there being a "plane" involved, it has to do with how you use
> things with velocity terms in them to calculate the amount of energy.
> Potential energy (in the newtonian approximation) is m*d*a, and kinetic
> energy is (1/2)mv^2, for reasons that have nothing to do with "planes".
> Instead it has to do with the same reason you get time squared when you
> use time and acceleration to calculate distance.
>
> Wayne Throop   thro...@sheol.org  http://sheol.org/throopw

Thanks again for a lucid response.
I will reply more to your thoughts later today.
Most assuredly though, there are 2 different directions of 'time'.
Lateral Time through the spatial plane and Axial Time from spatial
plane to spatial plane.

Even neglecting time momentarily,
why is E=mc^2 a purely planar equation for spatial (cubic)
reality...?
I am curious as to how this is commonly expressed?

Thanks.

: finite guy <adamlewis@amnet.net.au>
: Most assuredly though, there are 2 different directions of 'time'.
: Lateral Time through the spatial plane and Axial Time from spatial
: plane to spatial plane.

Sorry, I don't know what you mean by those terms.

: Even neglecting time momentarily, why is E=mc^2 a purely planar
: equation for spatial (cubic) reality...?

Sorry, I don't understand your obsession with calling a squared term "planar".
If it's a bad thing to have a squared term to define a quantity in 3d space,
then newtonian mechanics is just as guilty. Shrug.

: I am curious as to how this is commonly expressed?

You mean, how is E=mc^2 commonly expressed? Hrm. Well, it's actually
a simplification of the relativisitic expression for total energy; it's
what the expression gets simplified to if you set v=0.

But I expect that's not what you wanted to know, and wanted to know how
planes normally come into it. I must say, they most commonly don't.
It doesn't get expressed in that way. Hm. Possibly "integrating an area
under a curve" would come close to an answer to your question; but that's
area on a graph, not in the real world. To consider a simpler case to
illustrate this, the corresponding notion for deriving d=(1/2)at^2 is
to plot v vs t on a graph, and then the distance is represented by the
area under that line. The two-dimentional-ness of it is just because
v is a linear function of t.

Again, not a matter of the physics, but of a visual aid for the math.


Wayne Throop throopw@sheol.org http://sheol.org/throopw

On Feb 24, 9:57 pm, finite guy <adamle...@amnet.net.au> wrote:

<snip>

> You have an ******* - try wiping it, not me...

So, you admit to being an ******* that can be wiped.

If you can't stop spewing crap, wrap your head in a diaper.

> You just don't want to answer obvious fallacies in your double-think
> understanding of reality.

Blah blah blah, stop spewing, you crank Finite Guy.

On Feb 24, 9:55 pm, finite guy <adamle...@amnet.net.au> wrote:

<snip>

> Folly.
> Time is 'frozen' at delta time =0.
>
> You deny meaning of the obvious.


Finite Guy is a blithering idiot.

On Feb 24, 10:22 pm, finite guy <adamle...@amnet.net.au> wrote:

<snipped crap>

Blah blah blah, you Finite Guy are a waste of time.

On Feb 25, 2:09 am, finite guy <adamle...@amnet.net.au> wrote:
> Even neglecting time momentarily,
> why is E=mc^2 a purely planar equation for spatial (cubic)
> reality...?

It's a relationship between two scalar quantities.

What do you mean by being a "purely planar equation".
Surely you aren't saying the power of 2 makes it
planar, and everything involving 3-space should
have a power of 3 in it?

For instance F = ma is a three-dimensional
equation, relating two three-dimensional vector
quantities F and a. It is a short hand for the
three equations

Fx = m*ax
Fy = m*ay
Fz = m*az

and can also be written

(Fx, Fy, Fz) = m*(ax, ay, az)

If that isn't 3-D, what is? But nary a
power of 3 to be found.

- Randy

On Feb 26, 12:53 am, Randy Poe <poespam-t...@yahoo.com> wrote:
> On Feb 25, 2:09 am, finite guy <adamle...@amnet.net.au> wrote:
>
> > Even neglecting time momentarily,
> > why is E=mc^2 a purely planar equation for spatial (cubic)
> > reality...?
>
> It's a relationship between two scalar quantities.
>
> What do you mean by being a "purely planar equation".
> Surely you aren't saying the power of 2 makes it
> planar, and everything involving 3-space should
> have a power of 3 in it?
>
> For instance F = ma is a three-dimensional
> equation, relating two three-dimensional vector
> quantities F and a. It is a short hand for the
> three equations
>
>    Fx = m*ax
>    Fy = m*ay
>    Fz = m*az
>
> and can also be written
>
> (Fx, Fy, Fz) = m*(ax, ay, az)
>
> If that isn't 3-D, what is? But nary a
> power of 3 to be found.
>
>            - Randy

It isn't '3D/spatial'... actually.
I do understand what you mean but you neglect that Fermat points out
(and Wiles has sort of 'proved') that increasing rational/unitised
cubes do not exist.

That is one key to the understanding of the planar relationships of
'space' and time. x^2 + y^2 = e^2 DOES NOT WORK for n > 2.
The ususal suggestion is that you cannot make the e^3 but in actual
fact the x^3 and the y^3 are also impossible.

If you read the Fermats Last Theroem statement - he states that a cube
is a 4th power entity; which it is.
This alone implies a mathematical problem in going from the
traditional (1,1,1) to (2,2,2).

But, hey, that's just the mathematics - not the philosophical
underpinnings...

On Feb 25, 1:17 pm, Tim Little <t...@soprano.little-possums.net>
wrote:
> On 2008-02-25, finite guy <adamle...@amnet.net.au> wrote:
>
> > Quite honestly, considering the accuracy of calculations and
> > experimentation, I would have thought that it was clearly defined
> > when to use one or the other.
>
> For any given limits of required precision, it is.  That precision
> will be determined by the application for which you are employing the
> theoretical model, and upon any other simplifications you may already
> have made.
>
> - Tim

Tim, you are just restating the problem I think.

On Feb 22, 4:47 pm, Eivind Kjorstad <eivindor...@gmail.com> wrote:
> finite guy skreiv:
>
> > Where is the 'boundary' between Newton and Einstein?
> > What percentage of light 'speed'?
>
> Where is the boundary between being tall and being short ?
>
> There is no boundary whatsoever. Mathemathically speaking Einstein is
> -always- more correct than Newton.
>
> At low speeds, the difference is -very- small, at large speeds the
> difference can be very large.
>
> Thus how long you can continue to use Newton depends ONLY on how
> accurate an answer you need.
>
>         Eivind Kjørstad

Where is the boundary between 'slow' and 'fast' speeds...?

Anyway, ignoring Einstein, you have the same issue with a simple ol'
static triangle...
x^2 + y^2 = e^2, x=1, y=1 e=root2.
What will be your arbitary degree of calculation of root2?

Wait, if we use a triangle (pretty basic), we ARE using Einstein...
hah!!!

On Feb 26, 12:04 am, Neilist <lattora...@gmail.com> wrote:
> On Feb 24, 10:22 pm, finite guy <adamle...@amnet.net.au> wrote:
>
> <snipped crap>
>
> Blah blah blah, you Finite Guy are a waste of time.

A waste of time is planed to see...

On Feb 26, 12:04 am, Neilist <lattora...@gmail.com> wrote:
> On Feb 24, 10:22 pm, finite guy <adamle...@amnet.net.au> wrote:
>
> <snipped crap>
>
> Blah blah blah, you Finite Guy are a waste of time.

You have a cutting wit...
truely impressive.

Well, maybe half a cutting wit...

Or, a very dull edge to your cutting wit.

In rec.arts.sf.science bibinm7@gmail.com wrote:
> On Feb 25, 1:17?pm, Tim Little <t...@soprano.little-possums.net>
> wrote:
>> On 2008-02-25, finite guy <adamle...@amnet.net.au> wrote:
>>
>> > Quite honestly, considering the accuracy of calculations and
>> > experimentation, I would have thought that it was clearly defined
>> > when to use one or the other.
>>
>> For any given limits of required precision, it is. ?That precision
>> will be determined by the application for which you are employing the
>> theoretical model, and upon any other simplifications you may already
>> have made.
>
> Tim, you are just restating the problem I think.

If the problem is that you don't know how much accuracy you need, then you
have a very big problem which can't be solved on usenet.

--
Michael Ash
Rogue Amoeba Software

On Feb 26, 9:10 am, bibi...@gmail.com wrote:
> On Feb 26, 12:04 am, Neilist <lattora...@gmail.com> wrote:
>
> > On Feb 24, 10:22 pm, finite guy <adamle...@amnet.net.au> wrote:
>
> > <snipped crap>
>
> > Blah blah blah, you Finite Guy are a waste of time.
>
> A waste of time is planed to see...

Sorry guys, 'bibi' was 'finite guy'.
Someone else was logged in to Google email and I didn't notice that.

On Feb 26, 9:58 am, Michael Ash <m...@mikeash.com> wrote:
> In rec.arts.sf.science bibi...@gmail.com wrote:
> > On Feb 25, 1:17?pm, Tim Little <t...@soprano.little-possums.net>
> > wrote:
> >> On 2008-02-25, finite guy <adamle...@amnet.net.au> wrote:
>
> >> > Quite honestly, considering the accuracy of calculations and
> >> > experimentation, I would have thought that it was clearly defined
> >> > when to use one or the other.
>
> >> For any given limits of required precision, it is. ?That precision
> >> will be determined by the application for which you are employing the
> >> theoretical model, and upon any other simplifications you may already
> >> have made.
>
> > Tim, you are just restating the problem I think.
>
> If the problem is that you don't know how much accuracy you need, then you
> have a very big problem which can't be solved on usenet.
>
> --
> Michael Ash
> Rogue Amoeba Software

According to everyones opinions stated here... how big is 'very
big'...? :-)

In rec.arts.sf.science finite guy <adamlewis@amnet.net.au> wrote:
> On Feb 26, 9:58?am, Michael Ash <m...@mikeash.com> wrote:
>> In rec.arts.sf.science bibi...@gmail.com wrote:
>> > On Feb 25, 1:17?pm, Tim Little <t...@soprano.little-possums.net>
>> > wrote:
>> >> On 2008-02-25, finite guy <adamle...@amnet.net.au> wrote:
>>
>> >> > Quite honestly, considering the accuracy of calculations and
>> >> > experimentation, I would have thought that it was clearly defined
>> >> > when to use one or the other.
>>
>> >> For any given limits of required precision, it is. ?That precision
>> >> will be determined by the application for which you are employing the
>> >> theoretical model, and upon any other simplifications you may already
>> >> have made.
>>
>> > Tim, you are just restating the problem I think.
>>
>> If the problem is that you don't know how much accuracy you need, then you
>> have a very big problem which can't be solved on usenet.
>
> According to everyones opinions stated here... how big is 'very
> big'...? :-)

You are not asking the right question. This is like asking someone what
color paint you should use, or how large of a table you should buy.
Necessary accuracy is related to application and context. You can't just
waltz in and demand a universal answer. It will depend on what *you* are
doing with it and how much accuracy *you* *personally* want or need.

--
Michael Ash
Rogue Amoeba Software

On Feb 26, 3:04 pm, Michael Ash <m...@mikeash.com> wrote:
> In rec.arts.sf.science finite guy <adamle...@amnet.net.au> wrote:
>
>
>
>
>
> > On Feb 26, 9:58?am, Michael Ash <m...@mikeash.com> wrote:
> >> In rec.arts.sf.science bibi...@gmail.com wrote:
> >> > On Feb 25, 1:17?pm, Tim Little <t...@soprano.little-possums.net>
> >> > wrote:
> >> >> On 2008-02-25, finite guy <adamle...@amnet.net.au> wrote:
>
> >> >> > Quite honestly, considering the accuracy of calculations and
> >> >> > experimentation, I would have thought that it was clearly defined
> >> >> > when to use one or the other.
>
> >> >> For any given limits of required precision, it is. ?That precision
> >> >> will be determined by the application for which you are employing the
> >> >> theoretical model, and upon any other simplifications you may already
> >> >> have made.
>
> >> > Tim, you are just restating the problem I think.
>
> >> If the problem is that you don't know how much accuracy you need, then you
> >> have a very big problem which can't be solved on usenet.
>
> > According to everyones opinions stated here... how big is 'very
> > big'...?   :-)
>
> You are not asking the right question. This is like asking someone what
> color paint you should use, or how large of a table you should buy.
> Necessary accuracy is related to application and context. You can't just
> waltz in and demand a universal answer. It will depend on what *you* are
> doing with it and how much accuracy *you* *personally* want or need.
>
> --
> Michael Ash
> Rogue Amoeba Software- Hide quoted text -
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> - Show quoted text -

I understand.
You may think that this applies to Einstein, true, but it also applies
to your simple, basic triangle...
The 'precision' that you mention relates also to the determination of
root2.

Root2 is FINITELY EXACT, is it not?
But we cannot even finitely measure it...
Our finite ability in and perception of absolute reality is the
problem - not the absoluteness of reality itself.
Our inability to measure even root2 is a commentary on us, not
root2...

Precision is a somewhat minor point compared to concept.
Concept is built upon philosophy.
Your philosophy, presuming you agree with most of the ideas at this
site, is errant.

(I digress.)
Finite is finite. Finite can never be infinite in ANY way.
This means that real numbers, which are finite entities individually,
are NOT infinite in total quantity.
Yet, I presume that you believe that the quantity of real numbers is
infinite - if you are an acceptable mathematician.

Error is error; whether it is accepted or not.
Truth is truth; likewise.
How do we know error from truth...? Ay, there's the rub.
But you do know s o m e truths, so not mix them so much with
conceptual errors.
That is some of the issue to which I elude.