| Topic: |
Science > Physics |
| User: |
"Donald G. Shead" |
| Date: |
22 Jul 2004 10:38:13 AM |
| Object: |
Are all changes in velocity acceleration |
I'd have to say yes; even though a force is not exerted on the
observer, but on what is being observed, the acceleration observed
required an accelerating force:
Average acceleration is (defined as) the rate of change in velocity [a
= (vt-vi)/t], and instantaneous acceleration is defined as "something
else".
Average acceleration is proposed to be the result of an external
force, and is found to be proportional to the force exerted on and/or
by a body's mass, and/or inertia.
So in any event any observed acceleration is either a forced
acceleration of the observer, and/or a forced acceleration of whatever
is observed.
That should pretty well explain the rest of Newton' s first law; that
a body continues to remain at rest, or move in a straight line with
constant velocity.
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| User: "Sam Wormley" |
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| Title: Re: Are all changes in velocity acceleration |
22 Jul 2004 03:01:51 PM |
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"Donald G. Shead" wrote:
I'd have to say yes; even though a force is not exerted on the
observer, but on what is being observed, the acceleration observed
required an accelerating force:
See: http://scienceworld.wolfram.com/physics/Acceleration.html
http://scienceworld.wolfram.com/physics/Force.html
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| User: "Uncle Al" |
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| Title: Re: Are all changes in velocity acceleration |
22 Jul 2004 11:11:02 AM |
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"Donald G. Shead" wrote:
[snip]
Nothing.
Fucking imbecile Dumb Donny *****.
http://arXiv.org/abs/gr-qc/0407022
Geometric basis of inertial frames
Senile drooling psychotic idiot.
http://www.apa.org/journals/psp/psp7761121.html
http://insti.physics.sunysb.edu/~siegel/quack.html
<http://www.firehead.org/~jessh/film/kubrick/Kubrick-Psycho.html>
<http://www.naturalchild.com/elliott_barker/prisons.html>
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
.
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| User: "Paul Cardinale" |
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| Title: Re: Are all changes in velocity acceleration |
23 Jul 2004 08:40:12 AM |
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(Donald G. Shead) wrote in message news:<48402bae.0407220738.4c8bc0ce@posting.google.com>...
I'd have to say yes; even though a force is not exerted on the
observer, but on what is being observed, the acceleration observed
required an accelerating force:
Average acceleration is (defined as) the rate of change in velocity [a
= (vt-vi)/t], and instantaneous acceleration is defined as "something
else".
Idiot. a = dv/dt. v = ds/dt.
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| User: "Donald G. Shead" |
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| Title: Re: Are all changes in velocity acceleration |
24 Jul 2004 07:49:02 AM |
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(Paul Cardinale) wrote in message news:<64050551.0407230540.5dd0ac34@posting.google.com>...
dcshead@charter.net (Donald G. Shead) wrote in message news:<48402bae.0407220738.4c8bc0ce@posting.google.com>...
I'd have to say yes; even though a force is not exerted on the
observer, but on what is being observed, the acceleration observed
required an accelerating force:
Average acceleration is (defined as) the rate of change in velocity [a
= (vt-vi)/t], and instantaneous acceleration is defined as "something
else".
Idiot. a = dv/dt. v = ds/dt.
Do you know Paul that dv and dt are just short rectangular coordinates
plotted on velocity-time graphs, and that ds and dt are just those
rectangular coordinates plotted on a distance time graph?
They are used to find the slopes v/t and s/t at points located along
the time coordinate.
Couldn't the slopes at points along the time coordinate be found more
easily and more accurately with something like [Galileo's] equation [s
= at^2/2, and a = 2s/t^2]? Rather than with the calculus? I think so.
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| User: "Paul Cardinale" |
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| Title: Re: Are all changes in velocity acceleration |
27 Jul 2004 09:23:53 AM |
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(Donald G. Shead) wrote in message news:<48402bae.0407240449.39de1d0b@posting.google.com>...
pcardinale@volcanomail.com (Paul Cardinale) wrote in message news:<64050551.0407230540.5dd0ac34@posting.google.com>...
(Donald G. Shead) wrote in message news:<48402bae.0407220738.4c8bc0ce@posting.google.com>...
I'd have to say yes; even though a force is not exerted on the
observer, but on what is being observed, the acceleration observed
required an accelerating force:
Average acceleration is (defined as) the rate of change in velocity [a
= (vt-vi)/t], and instantaneous acceleration is defined as "something
else".
Idiot. a = dv/dt. v = ds/dt.
Do you know Paul that dv and dt are just short rectangular coordinates
plotted on velocity-time graphs, and that ds and dt are just those
rectangular coordinates plotted on a distance time graph?
Idiot. dv, ds, and dt are not coordinates. They are differentials.
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| User: "Donald G. Shead" |
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| Title: Re: Are all changes in velocity acceleration |
27 Jul 2004 05:45:24 PM |
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(Paul Cardinale) wrote
Cut<
Idiot. dv, ds, and dt are not coordinates. They are differentials.
No they are two sides of a right triangle; where the third side
approximates a tangent to the point between its ends; which point is
that at which we wish to know the value of the slope: The shorter the
coordinate sides the closer the third side's accuracy is; which will
be on the money when the sides become infinitesimal.
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| User: "Harry Conover" |
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| Title: Re: Are all changes in velocity acceleration |
27 Jul 2004 04:12:43 PM |
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(Paul Cardinale) wrote in message news:<64050551.0407270623.78147bb5@posting.google.com>...
dcshead@charter.net (Donald G. Shead) wrote in message news:<48402bae.0407240449.39de1d0b@posting.google.com>...
(Paul Cardinale) wrote in message news:<64050551.0407230540.5dd0ac34@posting.google.com>...
dcshead@charter.net (Donald G. Shead) wrote in message news:<48402bae.0407220738.4c8bc0ce@posting.google.com>...
I'd have to say yes; even though a force is not exerted on the
observer, but on what is being observed, the acceleration observed
required an accelerating force:
Average acceleration is (defined as) the rate of change in velocity [a
= (vt-vi)/t], and instantaneous acceleration is defined as "something
else".
Idiot. a = dv/dt. v = ds/dt.
Do you know Paul that dv and dt are just short rectangular coordinates
plotted on velocity-time graphs, and that ds and dt are just those
rectangular coordinates plotted on a distance time graph?
Idiot. dv, ds, and dt are not coordinates. They are differentials.
I really don't believe that the poster was an idiot, simply a
highschool kid getting his first introduction to physics and math and
not quite yet grasping it.
Kids today are somewhat slow in grasping initial or fundamental
concepts, but when these concepts really sink in, wow and stand back!
I know that I must have been a 'dumb *****', but it took me at least a
month to grasp the true meaning of a derivative and a differential.
Part of the problem was the way that it was being taught at that time,
by highschool teachers who had no comprehension whatsoever of the
subject that they were attempting to teach.
Fortunately, when I entered college it was a different breed of
instructors, whe actually knew their subject but didn't give a 'rat's
*****' about their students. Sink or swim was the philosophy of these
guys...who I learned to admire far more than the well intended but
incompetent teachers that I encountered in highschool.
Arguably the most feared dude in my university was Dr. Tartler, whose
infamious legend lives on to this very day at Drexel. He actually
caused his own son to flunk out of the university by failing him in a
Calc II course. Tartler was of strict Tutonic upbringing, and would
begin to immediately lecture and write on the blackboards right at the
stroke of the clock. The end of his lecture was punctuated precisely
at the end of the class period, at which point he had filled all the
blackboards surrounding the room with his illustrations, then just as
promptly as he had arrived walked out of the room. I personally
learned more math from this guy than from any other teacher that I
ever encountered! That strict ***** gave me all 'A' grades for three
math courses, so I'm not complaining! :-)
Let's give the poster benefit of the doubt and simply hope that he is
a beginner.
Harry C.
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| User: "tadchem" |
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| Title: Re: Are all changes in velocity acceleration |
28 Jul 2004 03:46:42 PM |
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"Harry Conover" <hhc314@yahoo.com> wrote in message
news:7ce4e226.0407271312.13d54243@posting.google.com...
<snip>
I really don't believe that the poster was an idiot, simply a
highschool kid getting his first introduction to physics and math and
not quite yet grasping it.
This poster is a well known troll hereabouts who calls himself "Donald G.
Shead" and is better known by many less flattering names.
sHead (pronounced"s*** head") does not believe in calculus, and is evidently
unable to grasp the concept of mathematical operators such as differentials
and integrals.
He claims to be a retired "designer of bridges" in his '70s - too old to
learn new tricks.
Tom Davidson
Richmond, VA
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| User: "Richard Henry" |
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| Title: Re: Are all changes in velocity acceleration |
27 Jul 2004 04:48:28 PM |
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"Harry Conover" <hhc314@yahoo.com> wrote in message
news:7ce4e226.0407271312.13d54243@posting.google.com...
Arguably the most feared dude in my university was Dr. Tartler, whose
infamious legend lives on to this very day at Drexel. He actually
caused his own son to flunk out of the university by failing him in a
Calc II course. Tartler was of strict Tutonic upbringing, and would
begin to immediately lecture and write on the blackboards right at the
stroke of the clock. The end of his lecture was punctuated precisely
at the end of the class period, at which point he had filled all the
blackboards surrounding the room with his illustrations, then just as
promptly as he had arrived walked out of the room. I personally
learned more math from this guy than from any other teacher that I
ever encountered! That strict ***** gave me all 'A' grades for three
math courses, so I'm not complaining! :-)
I took an University Extension class in Neual Networks at UCSD from Dr. Bart
Kosko. who was at the time best-known as Dr. Lofti Zedah's protege in the
field of Fuzzy Logic. The classes were conducted once a week for three
hours. Usually by the end of the first hour, Dr. Kosko had covered every
writing surface. He would open a can of Diet Pepsi, and we had until he
finished it to catch up on our notes. He would then erase every board and
start again.
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| User: "" |
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| Title: Re: Are all changes in velocity acceleration |
27 Jul 2004 04:31:44 PM |
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In sci.physics Harry Conover <hhc314@yahoo.com> wrote:
I really don't believe that the poster was an idiot, simply a
highschool kid getting his first introduction to physics and math and
not quite yet grasping it.
<snip story>
Let's give the poster benefit of the doubt and simply hope that he is
a beginner.
You must be new.
He's a middle aged idiot.
See http://www.hyperdeath.co.uk/spaceman/
or see http://www.crank.net/ where he makes it several times.
--
Jim Pennino
Remove -spam-sux to reply.
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| User: "Marcus Wellpoth" |
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| Title: Re: Are all changes in velocity acceleration |
25 Jul 2004 11:54:54 AM |
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Donald G. Shead wrote:
pcardinale@volcanomail.com (Paul Cardinale) wrote in message
news:<64050551.0407230540.5dd0ac34@posting.google.com>...
dcshead@charter.net (Donald G. Shead) wrote in message
news:<48402bae.0407220738.4c8bc0ce@posting.google.com>...
I'd have to say yes; even though a force is not exerted on the
observer, but on what is being observed, the acceleration observed
required an accelerating force:
Average acceleration is (defined as) the rate of change in velocity [a
= (vt-vi)/t], and instantaneous acceleration is defined as "something
else".
Idiot. a = dv/dt. v = ds/dt.
Do you know Paul that dv and dt are just short rectangular coordinates
plotted on velocity-time graphs, and that ds and dt are just those
rectangular coordinates plotted on a distance time graph?
They are used to find the slopes v/t and s/t at points located along
the time coordinate.
Couldn't the slopes at points along the time coordinate be found more
easily and more accurately with something like [Galileo's] equation [s
= at^2/2, and a = 2s/t^2]? Rather than with the calculus? I think so.
----------------------------------------------------------------------------------
Would you please integrate d^2 s/dt^2 = const. and see what you get.
If you are unable to integrate the above properly you should simply give up
posting on this list.
mw
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| User: "Donald G. Shead" |
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| Title: Re: Are all changes in velocity acceleration |
24 Jul 2004 07:58:59 PM |
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Marcus Wellpoth <wellpoth@gmx.de> wrote in message news:<410294d4$0$19728$9b4e6d93@newsread2.arcor-online.net>...
Donald G. Shead wrote:
pcardinale@volcanomail.com (Paul Cardinale) wrote in message
news:<64050551.0407230540.5dd0ac34@posting.google.com>...
dcshead@charter.net (Donald G. Shead) wrote in message
news:<48402bae.0407220738.4c8bc0ce@posting.google.com>...
I'd have to say yes; even though a force is not exerted on the
observer, but on what is being observed, the acceleration observed
required an accelerating force:
Average acceleration is (defined as) the rate of change in velocity [a
= (vt-vi)/t], and instantaneous acceleration is defined as "something
else".
Idiot. a = dv/dt. v = ds/dt.
Do you know Paul that dv and dt are just short rectangular coordinates
plotted on velocity-time graphs, and that ds and dt are just those
rectangular coordinates plotted on a distance time graph?
They are used to find the slopes v/t and s/t at points located along
the time coordinate.
Couldn't the slopes at points along the time coordinate be found more
easily and more accurately with something like [Galileo's] equation [s
= at^2/2, and a = 2s/t^2]? Rather than with the calculus? I think so.
----------------------------------------------------------------------------------
Would you please integrate d^2 s/dt^2 = const. and see what you get.
For d^2 s/dt^2; I get ds/t^2, just by canceling the extra d, and it's
sure not a constant: What do _you_ get by integrating?
If you are unable to integrate the above properly you should simply give up
posting on this list.
mw
Maybe it's you who should hangup on long island;^!
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| User: "The Ghost In The Machine" |
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| Title: Re: Are all changes in velocity acceleration |
25 Jul 2004 03:00:53 PM |
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In sci.math, Donald G. Shead
<dcshead@charter.net>
wrote
on 24 Jul 2004 17:58:59 -0700
<48402bae.0407241658.622db73c@posting.google.com>:
Marcus Wellpoth <wellpoth@gmx.de> wrote in message news:<410294d4$0$19728$9b4e6d93@newsread2.arcor-online.net>...
Donald G. Shead wrote:
pcardinale@volcanomail.com (Paul Cardinale) wrote in message
news:<64050551.0407230540.5dd0ac34@posting.google.com>...
dcshead@charter.net (Donald G. Shead) wrote in message
news:<48402bae.0407220738.4c8bc0ce@posting.google.com>...
I'd have to say yes; even though a force is not exerted on the
observer, but on what is being observed, the acceleration observed
required an accelerating force:
Average acceleration is (defined as) the rate of change in velocity [a
= (vt-vi)/t], and instantaneous acceleration is defined as "something
else".
Idiot. a = dv/dt. v = ds/dt.
Do you know Paul that dv and dt are just short rectangular coordinates
plotted on velocity-time graphs, and that ds and dt are just those
rectangular coordinates plotted on a distance time graph?
They are used to find the slopes v/t and s/t at points located along
the time coordinate.
Couldn't the slopes at points along the time coordinate be found more
easily and more accurately with something like [Galileo's] equation [s
= at^2/2, and a = 2s/t^2]? Rather than with the calculus? I think so.
----------------------------------------------------------------------------------
Would you please integrate d^2 s/dt^2 = const. and see what you get.
For d^2 s/dt^2; I get ds/t^2, just by canceling the extra d, and it's
sure not a constant: What do _you_ get by integrating?
Do you know how to properly use Leibnitz notation?
(This sort of abuse may explain the current (?) usage of D_2(f)
instead. Or was it D^(2)(f)? ASCII is a pain. One other usage:
s' = ds/dt, s" = d^s/dt^2.)
Anyway...no, you can't cancel the d's in this context. However,
here's a thought for you.
dy/dx is a corruption/mutation/extension of
\{Delta}y/\{Delta}x, where \{Delta} is the Greek letter
Delta, which looks like a triangle, base on the baseline,
apex pointing up. What this *means* is:
lim (\{Delta}x -> 0) ( (s(x + \{Delta}x) - s(x)) / \{Delta}x)
or, if you prefer,
lim (h -> 0) ( (s(x + h) - s(x)) / h )
which is the way I've seen it in at least one book, and
very straightforward for straight ASCII, as opposed to TeX.
I'm not entirely certain how one gets d^2s / dt^2, but that
is probably an abbreviation for (d/dt)(ds/dt). Now 'd/dt'
in this context is not a fraction, but an *operator*
-- differentiation with respect to t. In this case,
d^2s/dt^2 = s" in some notations, where
s" = lim (h' -> 0) (s'(x + h') - s'(x) / h')
and s' = ds/dt = lim (h -> 0) ( (s(x + h) - s(x)) / h ) .
Where physics comes in isn't too hard to understand;
in order to get from a point x to a point x + vt in time
t, one has to be traveling with an average velocity v.
(v(t) can be uniform, but often isn't.) It turns out
that in order to get from a velocity v to a velocity v +
at in time t, that one has to accelerate with an average
acceleration a. (Again, a(t) can be uniform, but often
isn't, although for areas near the Earth -- e.g., thrown
balls -- one usually takes a = g to be constant.)
Now, does this illuminate any light bulbs? How about running
a charge through a wire, or moving a magnet? Hint hint...
As for integrating d^s/dt^2 dt from 0 to t, one gets
(ds/dt)(t=t) - (ds/dt)(t=0)
as one should. Or one can simply integrate using a formalism, and
get ds/dt + K, where K is an arbitrary integration constant.
Now, to me this is extremely clear, straightforward entry-level
college physics. (I'd say high school but apparently high
schools don't quite get around to teaching integral calculus,
and barely manage differential calculus.)
If you are unable to integrate the above properly you should simply give up
posting on this list.
mw
Maybe it's you who should hangup on long island;^!
--
#191,
It's still legal to go .sigless.
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| User: "Marcus Wellpoth" |
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| Title: Re: Are all changes in velocity acceleration |
25 Jul 2004 03:42:46 PM |
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The Ghost In The Machine wrote:
[..]
I'm not entirely certain how one gets d^2s / dt^2, but that
is probably an abbreviation for (d/dt)(ds/dt). Now 'd/dt'
in this context is not a fraction, but an *operator*
-- differentiation with respect to t. In this case,
d^2s/dt^2 = s" in some notations, where
s" = lim (h' -> 0) (s'(x + h') - s'(x) / h')
and s' = ds/dt = lim (h -> 0) ( (s(x + h) - s(x)) / h ) .
--------------------------------------------------------------------------------
I'm sorry but d^2 x/d t^2 is the standart notation in my textbooks on
physics. It's used by Wolfgang Nolting, Thorsten Fließbach and Arnold
Sommerfeld. And since the first edition of Sommerfelds "Vorlesungen über
Theoretische Physik" was published back in 1942 i assume this notation to
be around a long time. But i must admit that the notation is not very ASCII
friendly.
--------------------------------------------------------------------------------
Where physics comes in isn't too hard to understand;
in order to get from a point x to a point x + vt in time
t, one has to be traveling with an average velocity v.
(v(t) can be uniform, but often isn't.) It turns out
that in order to get from a velocity v to a velocity v +
at in time t, that one has to accelerate with an average
acceleration a. (Again, a(t) can be uniform, but often
isn't, although for areas near the Earth -- e.g., thrown
balls -- one usually takes a = g to be constant.)
Now, does this illuminate any light bulbs? How about running
a charge through a wire, or moving a magnet? Hint hint...
As for integrating d^s/dt^2 dt from 0 to t, one gets
(ds/dt)(t=t) - (ds/dt)(t=0)
as one should. Or one can simply integrate using a formalism, and
get ds/dt + K, where K is an arbitrary integration constant.
Now, to me this is extremely clear, straightforward entry-level
college physics. (I'd say high school but apparently high
schools don't quite get around to teaching integral calculus,
and barely manage differential calculus.)
----------------------------------------------------------------------------------
I wanted it to be on this level just to see where Mr Sheads deficits begin.
Unfortunately i don't know of a good textbook on calculus written in
English.
mw
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| User: "Eric Gisse" |
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| Title: Re: Are all changes in velocity acceleration |
25 Jul 2004 04:22:48 PM |
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On Sun, 25 Jul 2004 22:42:46 +0200, Marcus Wellpoth <wellpoth@gmx.de>
wrote:
The Ghost In The Machine wrote:
[..]
I'm not entirely certain how one gets d^2s / dt^2, but that
is probably an abbreviation for (d/dt)(ds/dt). Now 'd/dt'
in this context is not a fraction, but an *operator*
-- differentiation with respect to t. In this case,
d^2s/dt^2 = s" in some notations, where
s" = lim (h' -> 0) (s'(x + h') - s'(x) / h')
and s' = ds/dt = lim (h -> 0) ( (s(x + h) - s(x)) / h ) .
--------------------------------------------------------------------------------
I'm sorry but d^2 x/d t^2 is the standart notation in my textbooks on
physics. It's used by Wolfgang Nolting, Thorsten Fließbach and Arnold
Sommerfeld. And since the first edition of Sommerfelds "Vorlesungen über
Theoretische Physik" was published back in 1942 i assume this notation to
be around a long time. But i must admit that the notation is not very ASCII
friendly.
d^2x / dt^2 = [2 dots above x to denote double differentiation, and
x``. All are equivlant.
--------------------------------------------------------------------------------
Where physics comes in isn't too hard to understand;
in order to get from a point x to a point x + vt in time
t, one has to be traveling with an average velocity v.
(v(t) can be uniform, but often isn't.) It turns out
that in order to get from a velocity v to a velocity v +
at in time t, that one has to accelerate with an average
acceleration a. (Again, a(t) can be uniform, but often
isn't, although for areas near the Earth -- e.g., thrown
balls -- one usually takes a = g to be constant.)
Now, does this illuminate any light bulbs? How about running
a charge through a wire, or moving a magnet? Hint hint...
As for integrating d^s/dt^2 dt from 0 to t, one gets
(ds/dt)(t=t) - (ds/dt)(t=0)
as one should. Or one can simply integrate using a formalism, and
get ds/dt + K, where K is an arbitrary integration constant.
Now, to me this is extremely clear, straightforward entry-level
college physics. (I'd say high school but apparently high
schools don't quite get around to teaching integral calculus,
and barely manage differential calculus.)
----------------------------------------------------------------------------------
I wanted it to be on this level just to see where Mr Sheads deficits begin.
Unfortunately i don't know of a good textbook on calculus written in
English.
mw
I like the one I use. Though I don't use others, it suits my purposes
well enough to not make me look for another. It is starting to get
worn at the edges :D
Thomas' Calculus - 10th Edition. ISBN 0-201-44141-1
Coveres everything from basic calculus to vector calculus, with a
smattering of things like geometry and linear algebra in the last
third of the book.
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| User: "Donald G. Shead" |
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| Title: Re: Are all changes in velocity acceleration |
26 Jul 2004 08:09:27 AM |
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The Ghost In The Machine <ewill@aurigae.athghost7038suus.net> wrote
CUT<
The point of this thread was to show that all changes in velocity are
NOT due to acceleration: Changes in velocity may be due to force
acting on an observer, who would then be accelerated; or on what the
observer is observing so that the observer is not accelerated.
You know dualisim and viewpoints and all that stuff.
As always, the discussion has turned to what Mr. Shead can and cannot
do or be: For better or worse Mr. Shead did design some bridges, and
none have caved in yet. Why can't you people stick to discussing the
thread?
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| User: "Goran Jakupovic" |
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| Title: Re: Are all changes in velocity acceleration |
26 Jul 2004 08:54:28 AM |
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"Donald G. Shead" <dcshead@charter.net> wrote in message
news:48402bae.0407260509.25459fa3@posting.google.com...
The Ghost In The Machine <ewill@aurigae.athghost7038suus.net> wrote
CUT<
The point of this thread was to show that all changes in velocity are
NOT due to acceleration: Changes in velocity may be due to force
acting on an observer, who would then be accelerated; or on what the
observer is observing so that the observer is not accelerated.
No Mr. Shead if there is change of velocity then there must be
acceleration which is not equal to zero because acceleration is first
derivative of velocity. If you try to learn some calculus you may have
ability to understand this.
You know dualisim and viewpoints and all that stuff.
As always, the discussion has turned to what Mr. Shead can and cannot
do or be: For better or worse Mr. Shead did design some bridges, and
none have caved in yet.
You still haven't provided us with references to those bridges haven't you?
Why can't you people stick to discussing the
thread?
But they are discussing the thread, it's you who repeatedly fail to
understand what are they trying to explain to you.
.
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| User: "The Ghost In The Machine" |
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| Title: Re: Are all changes in velocity acceleration |
26 Jul 2004 11:01:11 AM |
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In sci.math, Donald G. Shead
<dcshead@charter.net>
wrote
on 26 Jul 2004 06:09:27 -0700
<48402bae.0407260509.25459fa3@posting.google.com>:
The Ghost In The Machine <ewill@aurigae.athghost7038suus.net> wrote
CUT<
The point of this thread was to show that all changes in velocity are
NOT due to acceleration: Changes in velocity may be due to force
acting on an observer, who would then be accelerated; or on what the
observer is observing so that the observer is not accelerated.
You know dualisim and viewpoints and all that stuff.
As always, the discussion has turned to what Mr. Shead can and cannot
do or be: For better or worse Mr. Shead did design some bridges, and
none have caved in yet. Why can't you people stick to discussing the
thread?
So some changes in velocity relative to an observer could
be because the observer is being accelerated?
Possible, and indeed that would have an effect on the
observation. Clocks on GPS satellites in particular are
deliberately constructed to run slowly, as the observer
is in a distorted space field -- on the Earth's surface
standing thereon in 1g acceleration. Normally we don't
think of this as accelerated, as the Earth is pressing
back with a force equivalent to the force we experience
as gravity. (Think of a trampoline, which distends, for
example, to compensate for increased weight -- not mass --
placed in its center, usually someone who wants to jump
up and down.)
As for designing some bridges -- I would be curious as
to whether a accelerating car would affect the structural
components thereof, and by how much -- especially if the
car is still nowhere near the bridge...
--
#191,
It's still legal to go .sigless.
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| User: "Eric Gisse" |
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| Title: Re: Are all changes in velocity acceleration |
25 Jul 2004 04:31:40 PM |
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On 24 Jul 2004 17:58:59 -0700, (Donald G. Shead)
wrote:
Marcus Wellpoth <wellpoth@gmx.de> wrote in message news:<410294d4$0$19728$9b4e6d93@newsread2.arcor-online.net>...
Donald G. Shead wrote:
pcardinale@volcanomail.com (Paul Cardinale) wrote in message
news:<64050551.0407230540.5dd0ac34@posting.google.com>...
(Donald G. Shead) wrote in message
news:<48402bae.0407220738.4c8bc0ce@posting.google.com>...
I'd have to say yes; even though a force is not exerted on the
observer, but on what is being observed, the acceleration observed
required an accelerating force:
Average acceleration is (defined as) the rate of change in velocity [a
= (vt-vi)/t], and instantaneous acceleration is defined as "something
else".
Idiot. a = dv/dt. v = ds/dt.
Do you know Paul that dv and dt are just short rectangular coordinates
plotted on velocity-time graphs, and that ds and dt are just those
rectangular coordinates plotted on a distance time graph?
Meaningless babble.
They are used to find the slopes v/t and s/t at points located along
the time coordinate.
Couldn't the slopes at points along the time coordinate be found more
easily and more accurately with something like [Galileo's] equation [s
= at^2/2, and a = 2s/t^2]? Rather than with the calculus? I think so.
----------------------------------------------------------------------------------
Would you please integrate d^2 s/dt^2 = const. and see what you get.
For d^2 s/dt^2; I get ds/t^2, just by canceling the extra d, and it's
sure not a constant: What do _you_ get by integrating?
Hopeless idiot!
If you are unable to integrate the above properly you should simply give up
posting on this list.
mw
Maybe it's you who should hangup on long island;^!
No Dumb Donny, you can't do basic calculus.
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| User: "tadchem" |
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| Title: Re: Are all changes in velocity acceleration |
27 Jul 2004 04:02:14 PM |
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"Eric Gisse" <fsegg@uaf.edu> wrote in message
news:i598g0tf6bc3t1jfg122rgmjaluva0gcbi@4ax.com...
On 24 Jul 2004 17:58:59 -0700, (Donald G. Shead)
wrote:
<snip>
For d^2 s/dt^2; I get ds/t^2, just by canceling the extra d, and it's
sure not a constant: What do _you_ get by integrating?
Hopeless idiot!
That explains a lot, though. sHead never learned the difference between a
variable and an operator.
Tom Davidson
Richmond, VA
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| User: "David Bandel" |
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| Title: Re: Are all changes in velocity acceleration |
22 Jul 2004 07:32:55 PM |
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(Donald G. Shead) wrote in message news:<48402bae.0407220738.4c8bc0ce@posting.google.com>...
I'd have to say yes; even though a force is not exerted on the
observer, but on what is being observed, the acceleration observed
required an accelerating force:
Average acceleration is (defined as) the rate of change in velocity [a
= (vt-vi)/t], and instantaneous acceleration is defined as "something
else".
Average acceleration is proposed to be the result of an external
force, and is found to be proportional to the force exerted on and/or
by a body's mass, and/or inertia.
So in any event any observed acceleration is either a forced
acceleration of the observer, and/or a forced acceleration of whatever
is observed.
That should pretty well explain the rest of Newton' s first law; that
a body continues to remain at rest, or move in a straight line with
constant velocity.
so first you say yes.. then you deviate to a completely unrelated
topic about observers. why did you send me an email bragging about
your rights to being a homosexual pedophile?
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| User: "Eric Gisse" |
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| Title: Re: Are all changes in velocity acceleration |
22 Jul 2004 09:37:20 PM |
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On 22 Jul 2004 08:38:13 -0700, (Donald G. Shead)
wrote:
Yes.
Christ, learn some calculus.
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| User: "The Ghost In The Machine" |
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| Title: Re: Are all changes in velocity acceleration |
23 Jul 2004 03:01:39 AM |
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In sci.math, Eric Gisse
<fsegg@uaf.edu>
wrote
on Thu, 22 Jul 2004 18:37:20 -0800
<heu0g0hasc35ojltg6c15i429rhqugec5q@4ax.com>:
On 22 Jul 2004 08:38:13 -0700, (Donald G. Shead)
wrote:
Yes.
Christ, learn some calculus.
An interesting command, since Jesus pre-dated Newton and Leibnitz
by over a millennium....
:-)
--
#191,
It's still legal to go .sigless.
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| User: "Alex Hunsley" |
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| Title: Re: Are all changes in velocity acceleration |
23 Jul 2004 10:18:45 AM |
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The Ghost In The Machine wrote:
In sci.math, Eric Gisse
<fsegg@uaf.edu>
wrote
on Thu, 22 Jul 2004 18:37:20 -0800
<heu0g0hasc35ojltg6c15i429rhqugec5q@4ax.com>:
On 22 Jul 2004 08:38:13 -0700, (Donald G. Shead)
wrote:
Yes.
Christ, learn some calculus.
An interesting command, since Jesus pre-dated Newton and Leibnitz
by over a millennium....
:-)
Calculate the miracle volume if a point-source saviour, moving at 5 furlongs a
second, encounters a incontinent crippled man who is executing brownian motion
......
[5 marks]
etc. etc.
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| User: "MorituriMax" |
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| Title: Re: Are all changes in velocity acceleration |
23 Jul 2004 11:03:33 AM |
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Alex Hunsley wrote:
Calculate the miracle volume if a point-source saviour, moving at 5 furlongs a
second, encounters a incontinent crippled man who is executing brownian motion
.....
[5 marks]
etc. etc.
Ah dam man, warn me next time.. I got pepsi everywhere.... ROFL..
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| User: "Alex Hunsley" |
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| Title: Re: Are all changes in velocity acceleration |
05 Aug 2004 08:54:51 AM |
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MorituriMax wrote:
Alex Hunsley wrote:
Calculate the miracle volume if a point-source saviour, moving at 5 furlongs a
second, encounters a incontinent crippled man who is executing brownian motion
.....
[5 marks]
etc. etc.
Ah dam man, warn me next time.. I got pepsi everywhere.... ROFL..
oops! hope you got that pepsi cleaned up.
I think the ultimate purpose of usenet is the ability to make people far away
eject fluid/croutons from their nose.
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| User: "Creighton Hogg" |
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| Title: Re: Are all changes in velocity acceleration |
23 Jul 2004 10:34:48 AM |
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On Fri, 23 Jul 2004, Alex Hunsley wrote:
The Ghost In The Machine wrote:
In sci.math, Eric Gisse
<fsegg@uaf.edu>
wrote
on Thu, 22 Jul 2004 18:37:20 -0800
<heu0g0hasc35ojltg6c15i429rhqugec5q@4ax.com>:
On 22 Jul 2004 08:38:13 -0700, (Donald G. Shead)
wrote:
Yes.
Christ, learn some calculus.
An interesting command, since Jesus pre-dated Newton and Leibnitz
by over a millennium....
:-)
Calculate the miracle volume if a point-source saviour, moving at 5 furlongs a
second, encounters a incontinent crippled man who is executing brownian motion
.....
[5 marks]
I'm sorry but I find it somewhat sacreligious to assume a spherical Jesus.
:P
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| User: "MorituriMax" |
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| Title: Re: Are all changes in velocity acceleration |
23 Jul 2004 11:10:38 AM |
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Creighton Hogg wrote:
Calculate the miracle volume if a point-source saviour, moving at 5 furlongs
a
second, encounters a incontinent crippled man who is executing brownian
motion
.....
[5 marks]
I'm sorry but I find it somewhat sacreligious to assume a spherical Jesus.
Ahh, no problem then.. Jesus is a point source, his *area of effect volume* is
spherical.
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| User: "The Ghost In The Machine" |
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| Title: Re: Are all changes in velocity acceleration |
24 Jul 2004 11:01:13 AM |
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In sci.physics, Creighton Hogg
<wchogg@hep.wisc.edu>
wrote
on Fri, 23 Jul 2004 10:34:48 -0500
<Pine.LNX.4.44.0407231034150.3660-100000@erodium.hep.wisc.edu>:
On Fri, 23 Jul 2004, Alex Hunsley wrote:
The Ghost In The Machine wrote:
In sci.math, Eric Gisse
<fsegg@uaf.edu>
wrote
on Thu, 22 Jul 2004 18:37:20 -0800
<heu0g0hasc35ojltg6c15i429rhqugec5q@4ax.com>:
On 22 Jul 2004 08:38:13 -0700, (Donald G. Shead)
wrote:
Yes.
Christ, learn some calculus.
An interesting command, since Jesus pre-dated Newton and Leibnitz
by over a millennium....
:-)
Calculate the miracle volume if a point-source saviour, moving at 5 furlongs a
second, encounters a incontinent crippled man who is executing brownian motion
.....
[5 marks]
I'm sorry but I find it somewhat sacreligious to assume a spherical Jesus.
:P
As opposed to a plastic one? :-)
--
#191,
It's still legal to go .sigless.
.
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| User: "MorituriMax" |
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| Title: Re: Are all changes in velocity acceleration |
23 Jul 2004 12:22:45 AM |
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Yes... or deceleration.
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