| Topic: |
Science > Physics |
| User: |
"Math1723" |
| Date: |
23 May 2007 04:26:32 PM |
| Object: |
Re: Infinitesimal Arithmetic |
I don't follow. What is r and what are these "infinitessimals"
you are talking about?
Lester believes that for a circle with radius r, that dr/dt <> 0.
Actually centripetal dr/dt>0.
I'm still not following. The term "dr/dt" is calculus notation,
referring to the derivative with respect to t of some function r. It
is perhaps better written as: d/dt r(t). Yet, I do not understand
from the previous posts this function r you are referring to.
One of the main problems is that none of you can explain how
circular rotation is mechanized in the context of constant
tangential velocity and centripetal acceleration.
What do you mean by "mechanized"? And is r(t) supposed to be the
rotational velocity?
.
|
|
| User: "Lester Zick" |
|
| Title: Re: Infinitesimal Arithmetic |
23 May 2007 06:14:31 PM |
|
|
On 23 May 2007 14:26:32 -0700, Math1723 <anonym1723@aol.com> wrote:
I don't follow. What is r and what are these "infinitessimals"
you are talking about?
Lester believes that for a circle with radius r, that dr/dt <> 0.
Actually centripetal dr/dt>0.
I'm still not following. The term "dr/dt" is calculus notation,
referring to the derivative with respect to t of some function r. It
is perhaps better written as: d/dt r(t). Yet, I do not understand
from the previous posts this function r you are referring to.
What function r? r is a finite radius of rotation in the presence of
linear velocity v=dr/dt and transverse acceleration dv/dt. I've seen
different notations used to describe the effect but v=dr/dt is the one
I'm familiar with.
One of the main problems is that none of you can explain how
circular rotation is mechanized in the context of constant
tangential velocity and centripetal acceleration.
What do you mean by "mechanized"? And is r(t) supposed to be the
rotational velocity?
r is simply the radius of rotation in the presence of linear velocity
v=dr/dt and transverse acceleration dv/dt to produce curvilinear
forms. As to what I mean by "mechanized" I suggest you read the brief
root post to my thread "Epistemology 401: Tautological Mechanics" if
you want a taste of mechanics in boolean contexts.
~v~~
.
|
|
|
|
| User: "Randy Poe" |
|
| Title: Re: Infinitesimal Arithmetic |
23 May 2007 05:09:48 PM |
|
|
On May 23, 5:26 pm, Math1723 <anonym1...@aol.com> wrote:
I don't follow. What is r and what are these "infinitessimals"
you are talking about?
Lester believes that for a circle with radius r, that dr/dt <> 0.
Actually centripetal dr/dt>0.
I'm still not following. The term "dr/dt" is calculus notation,
referring to the derivative with respect to t of some function r. It
is perhaps better written as: d/dt r(t). Yet, I do not understand
from the previous posts this function r you are referring to.
Lester is referring to an old argument about centripetal motion,
a particle moving in a circle of radius r, i.e. r(t) = constant,
theta(t) = w*t.
Lester believes that nevertheless dr/dt is nonzero in this
case.
One of the main problems is that none of you can explain how
circular rotation is mechanized in the context of constant
tangential velocity and centripetal acceleration.
What do you mean by "mechanized"? And is r(t) supposed to be the
rotational velocity?
Nope. Radius in constant-radius motion. No idea what "mechanized"
might mean here.
- Randy
.
|
|
|
| User: "Math1723" |
|
| Title: Re: Infinitesimal Arithmetic |
23 May 2007 05:23:26 PM |
|
|
I'm still not following. The term "dr/dt" is calculus notation,
referring to the derivative with respect to t of some function
r. It is perhaps better written as: d/dt r(t). Yet, I do not
understand from the previous posts this function r you are
referring to.
Lester is referring to an old argument about centripetal motion,
a particle moving in a circle of radius r, i.e. r(t) = constant,
theta(t) = w*t.
Lester believes that nevertheless dr/dt is nonzero in this
case.
If r(t) is constant, then dr/dt = r'(t) = 0. This is basic Freshman
Calculus. Why does he think a constant function is going to have a
non-zero derivative?
One of the main problems is that none of you can explain how
circular rotation is mechanized in the context of constant
tangential velocity and centripetal acceleration.
What do you mean by "mechanized"? And is r(t) supposed to be
the rotational velocity?
Nope. Radius in constant-radius motion. No idea what "mechanized"
might mean here.
Hmmm...sounds like another potential entry for http://www.crank.net/maths.html
.. ;-)
.
|
|
|
| User: "" |
|
| Title: Re: Infinitesimal Arithmetic |
23 May 2007 07:39:31 PM |
|
|
In sci.math Math1723 <anonym1723@aol.com> wrote:
I'm still not following. The term "dr/dt" is calculus notation,
referring to the derivative with respect to t of some function
r. It is perhaps better written as: d/dt r(t). Yet, I do not
understand from the previous posts this function r you are
referring to.
Lester is referring to an old argument about centripetal motion,
a particle moving in a circle of radius r, i.e. r(t) = constant,
theta(t) = w*t.
Lester believes that nevertheless dr/dt is nonzero in this
case.
If r(t) is constant, then dr/dt = r'(t) = 0. This is basic Freshman
Calculus. Why does he think a constant function is going to have a
non-zero derivative?
Lester does not know or understand Calculus. He just copies
symbols and phrases that he has seen others use, with no understanding
of what the symbols mean.
Stephen
.
|
|
|
| User: "Lester Zick" |
|
| Title: Re: Infinitesimal Arithmetic |
24 May 2007 01:07:26 PM |
|
|
On Thu, 24 May 2007 00:39:31 +0000 (UTC), wrote:
In sci.math Math1723 <anonym1723@aol.com> wrote:
I'm still not following. The term "dr/dt" is calculus notation,
referring to the derivative with respect to t of some function
r. It is perhaps better written as: d/dt r(t). Yet, I do not
understand from the previous posts this function r you are
referring to.
Lester is referring to an old argument about centripetal motion,
a particle moving in a circle of radius r, i.e. r(t) = constant,
theta(t) = w*t.
Lester believes that nevertheless dr/dt is nonzero in this
case.
If r(t) is constant, then dr/dt = r'(t) = 0. This is basic Freshman
Calculus. Why does he think a constant function is going to have a
non-zero derivative?
Lester does not know or understand Calculus. He just copies
symbols and phrases that he has seen others use, with no understanding
of what the symbols mean.
Which puts me a little up on you, Stephen. Or perhaps you'd care to
explain exactly how curvilinear motion is mechanized in the absence of
linear v=dr/dt, tangential dr, transverse acceleration a=dv/dt, finite
transverse v=dr/dt, and transverse dr as the definite infinitesimal
integral of v=dr/dt between 0 and dt. I mean just so we can move on to
more difficult topics like analysis of Planck's constant, particle
spin, and such. Otherwise if we have nothing to do but hold your lazy
stupid hand for the rest of eternity we aren't going to make much
progress. It's time to shape up or ship out, Stephen,
~v~~
.
|
|
|
|
|
| User: "Lester Zick" |
|
| Title: Re: Infinitesimal Arithmetic |
23 May 2007 06:42:28 PM |
|
|
On 23 May 2007 15:23:26 -0700, Math1723 <anonym1723@aol.com> wrote:
I'm still not following. The term "dr/dt" is calculus notation,
referring to the derivative with respect to t of some function
r. It is perhaps better written as: d/dt r(t). Yet, I do not
understand from the previous posts this function r you are
referring to.
Lester is referring to an old argument about centripetal motion,
a particle moving in a circle of radius r, i.e. r(t) = constant,
theta(t) = w*t.
Lester believes that nevertheless dr/dt is nonzero in this
case.
If r(t) is constant, then dr/dt = r'(t) = 0. This is basic Freshman
Calculus. Why does he think a constant function is going to have a
non-zero derivative?
Not exactly, sport. See, like the other morons around here you're
arguing backwards. You say r is constant; consequently, dr/dt=0. That
isn't correct. What you know and what is true is that r is finitely
constant. r could still be and is infinitesimally variable despite
being finitely constant which is the point of this whole exercise.
On the other hand if you can explain how curvilinear motion can be
mechanized without finite centripetal v=dr/dt and infinitesimal dr
mechanized through the definite infinitesimal integral of centripetal
acceleration between 0 and dt and the definite infinitesimal integral
of finite centripetal v=dr/dt between 0 and dt I shall definitely go
gently into this good night without a whimper and you shall never hear
from me again (What never? Well hardly ever!) because the last time I
checked tangential velocities only combine vectorially with other
velocities and not with centripetal acceleration and a rotation of r
requires tangential dr as a result of the definite infinitesimal
integral of tangential v=dr/dt between 0 and dt to combine with a
centripetal dr as the definite infinitesimal integral of centripetal
v=dr/dt between 0 and dt. I realize this is a lot of words but do the
best you can unlike the rest who appear pretty much stuck on stupid.
One of the main problems is that none of you can explain how
circular rotation is mechanized in the context of constant
tangential velocity and centripetal acceleration.
What do you mean by "mechanized"? And is r(t) supposed to be
the rotational velocity?
Nope. Radius in constant-radius motion. No idea what "mechanized"
might mean here.
Hmmm...sounds like another potential entry for http://www.crank.net/maths.html
. ;-)
Oh yes by all means. The problem is that these turkeys are too
fascinated by the truth to excise me from their must-read lists. The
fact is I would be amazed if I'm not already on the crackpot list.
~v~~
.
|
|
|
| User: "Math1723" |
|
| Title: Re: Infinitesimal Arithmetic |
23 May 2007 09:26:56 PM |
|
|
Not exactly, sport. See, like the other morons around here
you're arguing backwards. You say r is constant;
consequently, dr/dt=0. That isn't correct. What you know
and what is true is that r is finitely constant. r could
still be and is infinitesimally variable despite being
finitely constant which is the point of this whole
exercise.
Okay, let me understand this then. You are saying that this r(t) is
in fact not a constant. When you say that r(t) is "finitely
constant", is what you mean that the standard part of r(t) is constant
but the non-standard part varies? if so, you are correct that r'(t)
would not necessarily be 0.
Do you have some formula for r(t)? This would go a long way to
solving your problem.
On the other hand if you can explain how curvilinear motion
can be mechanized without finite centripetal v=dr/dt and
infinitesimal dr mechanized through the definite...
<snip>
Whoah, whoah. Slow down. Please define what you mean by
"mechanize". I cannot understand the rest of this if I don't know
what you mean by this term.
...infinitesimal integral of centripetal acceleration
between 0 and dt and the definite infinitesimal integral of
finite centripetal v=dr/dt between 0 and dt I shall...
You appear to be using the names "dr" and "dt" as if they are actual
variables. They are not. The term "dr/dt" is simply short-hand
notation for "the first derivative of r with respect to t". "dr" and
"dt" have no meaning outside that context.
You might consider picking up a text on Freshman Calculus, where these
terms are defined. You may find it enlightening for your pursuits.
.
|
|
|
| User: "Lester Zick" |
|
| Title: Re: Infinitesimal Arithmetic |
24 May 2007 01:45:58 PM |
|
|
On 23 May 2007 19:26:56 -0700, Math1723 <anonym1723@aol.com> wrote:
Not exactly, sport. See, like the other morons around here
you're arguing backwards. You say r is constant;
consequently, dr/dt=0. That isn't correct. What you know
and what is true is that r is finitely constant. r could
still be and is infinitesimally variable despite being
finitely constant which is the point of this whole
exercise.
Okay, let me understand this then. You are saying that this r(t) is
in fact not a constant. When you say that r(t) is "finitely
constant", is what you mean that the standard part of r(t) is constant
but the non-standard part varies? if so, you are correct that r'(t)
would not necessarily be 0.
No *****? Gee, sport, thanks. Next time think before you speak.
Do you have some formula for r(t)? This would go a long way to
solving your problem.
My problem??? What is this "my problem" thingie? Mechanics has the
problem. I'm just trying to solve it. You don't like my solution? Then
please be so good as to solve it for us by explaining the mechanics of
circular rotation in some other terms.
On the other hand if you can explain how curvilinear motion
can be mechanized without finite centripetal v=dr/dt and
infinitesimal dr mechanized through the definite...
<snip>
Whoah, whoah. Slow down. Please define what you mean by
"mechanize". I cannot understand the rest of this if I don't know
what you mean by this term.
You know, dude, like "making it happen". Now Stringfellow is of the
opinion tying a piece of string to something and whirling it around
his head is a sufficient mechanical explanation for circular rotation
however I'm less confident that's how god mechanizes particle rotation
or even rotation in ordinary macroscopic bodies. Personally I find
definite integration of linear v=dr/dt and transverse acceleration
a=dv/dt and finite transverse v=dr/dt to be a sufficient mechanical
explanation but if you have something more exotic in mind such as
magic little "circular rotatons" instead please feel free to offer it.
The term "mechanics" or "mechanical" in general refers to a single
predicate compounded in terms of itself which explains how things
happen and what is true of everything in fully reduced exhaustive
terms.
...infinitesimal integral of centripetal acceleration
between 0 and dt and the definite infinitesimal integral of
finite centripetal v=dr/dt between 0 and dt I shall...
You appear to be using the names "dr" and "dt" as if they are actual
variables. They are not. The term "dr/dt" is simply short-hand
notation for "the first derivative of r with respect to t". "dr" and
"dt" have no meaning outside that context.
Gee that's just swell. Or maybe they have meaning as results of
definite infinitesimal integration between 0 and dt? Or are we
speaking before we think again?
You might consider picking up a text on Freshman Calculus, where these
terms are defined. You may find it enlightening for your pursuits.
But apparently not for yours.
~v~~
.
|
|
|
| User: "Bob Cain" |
|
| Title: Re: Infinitesimal Arithmetic |
24 May 2007 05:41:18 PM |
|
|
Lester Zick wrote:
On 23 May 2007 19:26:56 -0700, Math1723 <anonym1723@aol.com> wrote:
Okay, let me understand this then. You are saying that this r(t) is
in fact not a constant. When you say that r(t) is "finitely
constant", is what you mean that the standard part of r(t) is constant
but the non-standard part varies? if so, you are correct that r'(t)
would not necessarily be 0.
No *****? Gee, sport, thanks. Next time think before you speak.
This twee dance around what math1723 asked is what Zick considers wit.
He offers no more substance than that ever. Attempting any
substantial reply to him or requesting anything of substance from him
is as rewarding as conversing with a brick.
Zick is all and only about the pretty dance he sees himself doing as
viewed on the back of his own eyelids. The rest of us sadly see a
brick repeatedly attempting a pirouette.
This is why most of us have come to speak of him rather than to him.
Bob
--
"Things should be described as simply as possible, but no simpler."
A. Einstein
.
|
|
|
| User: "Lester Zick" |
|
| Title: Re: Infinitesimal Arithmetic |
24 May 2007 06:19:22 PM |
|
|
On Thu, 24 May 2007 15:41:18 -0700, Bob Cain
<arcane@arcanemethods.com> wrote:
Lester Zick wrote:
On 23 May 2007 19:26:56 -0700, Math1723 <anonym1723@aol.com> wrote:
Okay, let me understand this then. You are saying that this r(t) is
in fact not a constant. When you say that r(t) is "finitely
constant", is what you mean that the standard part of r(t) is constant
but the non-standard part varies? if so, you are correct that r'(t)
would not necessarily be 0.
No *****? Gee, sport, thanks. Next time think before you speak.
This twee dance around what math1723 asked is what Zick considers wit.
He offers no more substance than that ever. Attempting any
substantial reply to him or requesting anything of substance from him
is as rewarding as conversing with a brick.
Or conversing with Stringfellow. After all I never suggested circular
rotation might be effected with a brick.
Zick is all and only about the pretty dance he sees himself doing as
viewed on the back of his own eyelids. The rest of us sadly see a
brick repeatedly attempting a pirouette.
Weeeeell now, Stringfellow, I defy you to cite any instance where I
have suggested circular rotation might be effected with brick, your
psychological insights to the contrary notwithstanding.
This is why most of us have come to speak of him rather than to him.
With reverance no doubt.
~v~~
.
|
|
|
|
|
|
|
|
| User: "PD" |
|
| Title: Re: Infinitesimal Arithmetic |
24 May 2007 07:08:51 AM |
|
|
On May 23, 5:23 pm, Math1723 <anonym1...@aol.com> wrote:
I'm still not following. The term "dr/dt" is calculus notation,
referring to the derivative with respect to t of some function
r. It is perhaps better written as: d/dt r(t). Yet, I do not
understand from the previous posts this function r you are
referring to.
Lester is referring to an old argument about centripetal motion,
a particle moving in a circle of radius r, i.e. r(t) = constant,
theta(t) = w*t.
Lester believes that nevertheless dr/dt is nonzero in this
case.
If r(t) is constant, then dr/dt = r'(t) = 0. This is basic Freshman
Calculus. Why does he think a constant function is going to have a
non-zero derivative?
Because Lester hasn't the vaguest acquaintance with either freshman
calculus or freshman physics. He does, however, enjoy making noises
similar to those heard in freshman calculus or freshman physics. Doing
so makes him feel pretty.
PD
One of the main problems is that none of you can explain how
circular rotation is mechanized in the context of constant
tangential velocity and centripetal acceleration.
What do you mean by "mechanized"? And is r(t) supposed to be
the rotational velocity?
Nope. Radius in constant-radius motion. No idea what "mechanized"
might mean here.
Hmmm...sounds like another potential entry forhttp://www.crank.net/maths.html
. ;-)
.
|
|
|
| User: "Lester Zick" |
|
| Title: Re: Infinitesimal Arithmetic |
24 May 2007 12:50:01 PM |
|
|
On 24 May 2007 05:08:51 -0700, PD <TheDraperFamily@gmail.com> wrote:
Because Lester hasn't the vaguest acquaintance with either freshman
calculus or freshman physics. He does, however, enjoy making noises
similar to those heard in freshman calculus or freshman physics. Doing
so makes him feel pretty.
Just as it makes you feel bitchy.
~v~~
.
|
|
|
| User: "Don Stockbauer" |
|
| Title: Re: Infinitesimal Arithmetic |
24 May 2007 12:55:29 PM |
|
|
Lord. Someone who disbelieves in empiricism. Getting one's hands
dirty is how knowledge is advanced as opposed to endless usenet
discussions.
.
|
|
|
| User: "Lester Zick" |
|
| Title: Re: Infinitesimal Arithmetic |
24 May 2007 06:14:04 PM |
|
|
On 24 May 2007 10:55:29 -0700, Don Stockbauer
<donstockbauer@hotmail.com> wrote:
Lord. Someone who disbelieves in empiricism. Getting one's hands
dirty is how knowledge is advanced as opposed to endless usenet
discussions.
Well to be more precise, Don, getting ones hands dirty is how dirty
knowledge is advanced as opposed to endless usenet knowledge which may
or may not be dirty but will certainly be dirty if acquired with dirty
hands.
~v~~
.
|
|
|
|
|
|
|
| User: "Randy Poe" |
|
| Title: Re: Infinitesimal Arithmetic |
23 May 2007 06:18:10 PM |
|
|
On May 23, 6:23 pm, Math1723 <anonym1...@aol.com> wrote:
I'm still not following. The term "dr/dt" is calculus notation,
referring to the derivative with respect to t of some function
r. It is perhaps better written as: d/dt r(t). Yet, I do not
understand from the previous posts this function r you are
referring to.
Lester is referring to an old argument about centripetal motion,
a particle moving in a circle of radius r, i.e. r(t) = constant,
theta(t) = w*t.
Lester believes that nevertheless dr/dt is nonzero in this
case.
If r(t) is constant, then dr/dt = r'(t) = 0. This is basic Freshman
Calculus. Why does he think a constant function is going to have a
non-zero derivative?
I see you're stuck in the old, pre-Zickian mode of mathematics
and physics.
It's hard enough keeping track of *what* Lester believes.
Figuring out *why* is beyond the capabilities of science.
As for the what, you need Lester to explain in his own words.
Here are a couple of threads to wander through:
http://groups.google.com/group/sci.math/browse_thread/thread/3fa33a57803725ca/214b8afe53522e42
http://groups.google.com/group/sci.physics/browse_thread/thread/2d8b72593cdd8adb/03830c611619c704
- Randy
.
|
|
|
| User: "Lester Zick" |
|
| Title: Re: Infinitesimal Arithmetic |
24 May 2007 12:30:41 PM |
|
|
On 23 May 2007 16:18:10 -0700, Randy Poe <poespam-trap@yahoo.com>
wrote:
On May 23, 6:23 pm, Math1723 <anonym1...@aol.com> wrote:
I'm still not following. The term "dr/dt" is calculus notation,
referring to the derivative with respect to t of some function
r. It is perhaps better written as: d/dt r(t). Yet, I do not
understand from the previous posts this function r you are
referring to.
Lester is referring to an old argument about centripetal motion,
a particle moving in a circle of radius r, i.e. r(t) = constant,
theta(t) = w*t.
Lester believes that nevertheless dr/dt is nonzero in this
case.
If r(t) is constant, then dr/dt = r'(t) = 0. This is basic Freshman
Calculus. Why does he think a constant function is going to have a
non-zero derivative?
I see you're stuck in the old, pre-Zickian mode of mathematics
and physics.
Most empirics are. There are those who appreciate truth and those who
appreciate modern math and empiricism instead.
It's hard enough keeping track of *what* Lester believes.
Figuring out *why* is beyond the capabilities of science.
Science??? Shirley you jest. Obviously you mean beyond the
capabilities of those too lazy or stupid to do much beyond the
assumption of truth instead of science.
As for the what, you need Lester to explain in his own words.
Ah much appreciated, Randy. There's nothing quite so refreshing as a
trip down the memory lane of truth.
Here are a couple of threads to wander through:
http://groups.google.com/group/sci.math/browse_thread/thread/3fa33a57803725ca/214b8afe53522e42
http://groups.google.com/group/sci.physics/browse_thread/thread/2d8b72593cdd8adb/03830c611619c704
- Randy
~v~~
.
|
|
|
|
|
|
| User: "Lester Zick" |
|
| Title: Re: Infinitesimal Arithmetic |
23 May 2007 06:20:26 PM |
|
|
On 23 May 2007 15:09:48 -0700, Randy Poe <poespam-trap@yahoo.com>
wrote:
On May 23, 5:26 pm, Math1723 <anonym1...@aol.com> wrote:
I don't follow. What is r and what are these "infinitessimals"
you are talking about?
Lester believes that for a circle with radius r, that dr/dt <> 0.
Actually centripetal dr/dt>0.
I'm still not following. The term "dr/dt" is calculus notation,
referring to the derivative with respect to t of some function r. It
is perhaps better written as: d/dt r(t). Yet, I do not understand
from the previous posts this function r you are referring to.
Lester is referring to an old argument about centripetal motion,
a particle moving in a circle of radius r, i.e. r(t) = constant,
theta(t) = w*t.
Lester believes that nevertheless dr/dt is nonzero in this
case.
Has to be, Randy, unless you or the other solons around here can
explain how constant finite centripetal acceleration but zero finite
centripetal velocity in combination with finite tangential velocity
produce curvilinear motion. I mean we're all ears. Don't keep us in
suspense. Last time I checked velocities only combined vectorially
with other velocities. But maybe the rules have been repealed by
modern math.
One of the main problems is that none of you can explain how
circular rotation is mechanized in the context of constant
tangential velocity and centripetal acceleration.
What do you mean by "mechanized"? And is r(t) supposed to be the
rotational velocity?
Nope. Radius in constant-radius motion. No idea what "mechanized"
might mean here.
One doesn't generally have ideas with respect to what one cannot even
be bothered to examine.
~v~~
.
|
|
|
|
|
|
Related Articles |
|
|
