(4m/s^2)/2s = ?
Dennis B
Does anyone know how to solve the following problem:
(4m/s^2)/2s = ?
Key:
m = meters
s = seconds
I believe the correct solution is 2m/s. Is this correct?
-Dennis
Sam Wormley
Sam Wormley
Randy Poe
No. It's 2 m/s^3. See the results of the google calculator Sam
Wormley linked to.
Now a question for you: What is the symbolic expression for
the product of x and y, which tells you to multiply x by y
(so for instance if x = 4 and y = 2, this expression has the
value 4*2 = 8)?
- Randy
Dennis B
m/s^3 is a unit of velocity, equivalent to m/s.
-Dennis
Randy Poe
Dennis B wrote:
> m/s^3 is a unit of velocity, equivalent to m/s.
Incorrect. s^3 and s are not the same thing, any more
than meters and cubic meters are the same thing.
- Randy
Sam Wormley
>
> m/s^3 is a unit of velocity, equivalent to m/s.
>
> -Dennis
>
m/s^3 is a unit of jerk--the derivative of acceleration
with respect to time, commonly specified for components
in military aircraft, ordinance and ballistic missiles.
Sam Wormley
>
> m/s^3 is a unit of velocity, equivalent to m/s.
> -Dennis
>
T Wake
"Dennis B" <Uto...@kaxy.com> wrote in message
news:1152658145.3...@p79g2000cwp.googlegroups.com...
Where on Earth do you get that from?
Bill Snyder
Proctology.
--
Bill Snyder [This space unintentionally left blank.]
T Wake
"Bill Snyder" <bsn...@airmail.net> wrote in message
news:qnd8b2h4mmlv1f78e...@4ax.com...
Seems about right to me.
Dennis B
a = v/t
v = at = v/t * t = (m/s)/s * s = m/s^3
-Dennis B
Dennis B
a = v/t
Sam Wormley
>
> a = v/t
>
> v = at = v/t * t = (m/s)/s * s = m/s^3
>
> -Dennis B
>
T Wake
So you think that meters per second means the same thing as meters per
second cubed? That doesn't seem odd to you?
Do you not think there may be an error in your working..?
Do you think the / and * symbols mean the same thing? Try this with a=10m
so^-2 and t=10s.
You end up with 100m so^-1 not 100 ms^3.
Why is that?
Dennis B
v = at
a = v/t = at/t = at^2
That the t's in at/t aren't cancelled doesn't seem odd to you?
> Do you not think there may be an error in your working..?
>
Only if there is an error in a = v/t = at/t = at^2
-Dennis B
Randy Poe
Dennis B wrote:
> v = at
> a = v/t = at/t = at^2
>
> That the t's in at/t aren't cancelled doesn't seem odd to you?
It is odd. at/t = a. It doesn't equal at^2
You still haven't answered the question yet as to whether
you think at is equal to the value of a multiplied by
the value of t.
> > Do you not think there may be an error in your working..?
> >
>
> Only if there is an error in a = v/t = at/t = at^2
There is, it's in that last step. at/t = at^2 is wrong.
- Randy
Randy Poe
Dennis B wrote:
> v = at
> a = v/t = at/t = at^2
You still haven't answered the question yet as to whether
you think at is equal to the value of a multiplied by
the value of t.
- Randy
T Wake
at/t is not the same as at^2
at/t = a.
Why are you saying they dont cancel?
>> Do you not think there may be an error in your working..?
>>
>
> Only if there is an error in a = v/t = at/t = at^2
There sure is. at/t is not at^2. at*t is at^2. See the difference?
Dennis B
I understand that the t's do cancel. Yet, it seems to ME that the
definitions of a=v/t and v=a/t are circular. Which is madness. Such
madness may be the source of my confusion
-Dennis B
Sam Wormley
> I understand that the t's do cancel. Yet, it seems to ME that the
> definitions of a=v/t and v=a/t are circular. Which is madness. Such
> madness may be the source of my confusion
>
> -Dennis B
>
a = v/t
v = at (not what you posted, which is wrong)
You best learn some basic algebra, followed by calculus and
differental equations. Actually a = dv/dt .
T Wake
Well, they are a bit. v=at, a=v/t. They are only circular if you already
know all the data. If you know a and t for example you can work out v. If v
didn't equal at you couldn't. There is no mileage in trying to read more
into the fact that v=(v/t)t.
That doesn't make it circular as such, it just allows you to do the
calculations.
Art Deco
at/t != at^2, you dunderhead. Go back to grade school.
>
>> Do you not think there may be an error in your working..?
>>
>
>Only if there is an error in a = v/t = at/t = at^2
There is -- the error is in your brain.
>
>-Dennis B
>
--
COOSN-266-06-39716
Official Associate AFA-B Vote Rustler
Official Overseer of Kooks and Saucerheads in alt.astronomy
Co-Winner, alt.(f)lame Worst Flame War, December 2005
Official "Usenet psychopath and born-again LLPOF minion",
as designated by Brad Guth
"Get a rope! Say did you see on the National news about a
'Cross burning' in Arkansas? The black guy was scared
shitless."
-- "Honest" John the crackah without a brain
Art Deco
The real definitions are:
a = v' = x''
or
2
dv d x
a = -- = ---
dt 2
dt
Randy Poe
Dennis B wrote:
> I understand that the t's do cancel. Yet, it seems to ME that the
> definitions of a=v/t and v=a/t are circular.
There is no definition of v = a/t.
> Which is madness.
Possibly. I haven't decided whether it's that or a joke.
- Randy
Randy Poe
Dennis B wrote:
> [snip]
krp
I see it too, I hate when they do these simple simple equations,and try
to use them as a means of intimidating others.
Like any tard in a community college couldn't get it...
Dennis B
Randy Poe wrote:
> Dennis B wrote:
>
> > I understand that the t's do cancel. Yet, it seems to ME that the
> > definitions of a=v/t and v=a/t are circular.
>
> There is no definition of v = a/t.
>
That was probably a "typo". I believe I meant at. Yet, considering my
argument is that a/t = at...I'm not sure that ther is a difference.
> > Which is madness.
>
> Possibly. I haven't decided whether it's that or a joke.
>
> - Randy
It's not a joke.
-Dennis B
Dennis B
at = a * t
-Dennis
Dennis B
Randy Poe wrote:
> Dennis B wrote:
> > v = at
> > a = v/t = at/t = at^2
> >
> > That the t's in at/t aren't cancelled doesn't seem odd to you?
>
> It is odd. at/t = a. It doesn't equal at^2
>
Yet, by cancelling the t's one destroys the meaning of "a". What is
"a"? It is v/t = at/t.
> You still haven't answered the question yet as to whether
> you think at is equal to the value of a multiplied by
> the value of t.
>
> > > Do you not think there may be an error in your working..?
> > >
> >
> > Only if there is an error in a = v/t = at/t = at^2
>
> There is, it's in that last step. at/t = at^2 is wrong.
>
> - Randy
Yet, the same reasoning is ultimately used to convert v/t (which equals
m/s/s) into vt (m/s^2). Why do you not argue against the logic of that?
v/t = at/t and vt = at^2. Therefore, if v/t = vt then at/t = at^2.
Perhaps you argue with the logic of my math because if you agreed with
it, you would be indirectly agreeing with the equation I have proposed,
E = W = F = Mass?
-Dennis
Cranks Reply
Dennis B wrote:
> Randy Poe wrote:
> > Dennis B wrote:
> >
> > > I understand that the t's do cancel. Yet, it seems to ME that the
> > > definitions of a=v/t and v=a/t are circular.
> >
> > There is no definition of v = a/t.
> >
>
> That was probably a "typo". I believe I meant at. Yet, considering my
> argument is that a/t = at...I'm not sure that ther is a difference.
there is. your argument is all fucked up.
> > > Which is madness.
> >
> > Possibly. I haven't decided whether it's that or a joke.
> >
> > - Randy
>
> It's not a joke.
it sure frigging is.
Dennis B
The following equations prove that (m/s^3) is a unit of velocity
equivalent to (m/s):
d = vt
v = d/t = (v)t/t = (m/s)s/s = m/s^3
-Dennis B
Cranks Reply
you are fucking insane.
CWatters
>
> v = at = v/t * t = (m/s)/s * s
>= m/s^3
No is isn't.
(m/s)/s * s = ((m/s)/s) * s = m/s
Randy Poe
Dennis B wrote:
> Randy Poe wrote:
> > Dennis B wrote:
> > > [snip]
> >
> > You still haven't answered the question yet as to whether
> > you think at is equal to the value of a multiplied by
> > the value of t.
> >
>
So if a=4 and t=2, is the value of at = 4*2 = 8?
- Randy
Randy Poe
Dennis B wrote:
> Randy Poe wrote:
> >
>
Only by you.
> to convert v/t (which equals m/s/s)
Wrong. A statement made only by you.
> into vt (m/s^2).
Wrong. v/t is not "converted to vt". Two different things.
Wrong. The units of vt are not m/s^2.
> Why do you not argue against the logic of that?
I do.
> v/t = at/t
Yes. And at/t = a.
> and vt = at^2.
Yes.
> Therefore, if v/t = vt
It doesn't.
> then at/t = at^2.
It doesn't.
> Perhaps you argue with the logic of my math because if you agreed with
> it, you would be indirectly agreeing with the equation I have proposed,
> E = W = F = Mass?
I argue with the "logic" because it's wrong.
You agree that at means a multiplied by t, right? So if a=4 and t=2,
is at = 4*2 = 8?
- Randy
T Wake
"Art Deco" <er...@netcabal.com> wrote in message
news:110720061926036683%er...@netcabal.com...
I just want to make every one aware that your message was not actually a
reply to mine.....
:-)
T Wake
>
>
> Randy Poe wrote:
>> Dennis B wrote:
>> > v = at
>> > a = v/t = at/t = at^2
>> >
>> > That the t's in at/t aren't cancelled doesn't seem odd to you?
>>
>> It is odd. at/t = a. It doesn't equal at^2
>>
>
> Yet, by cancelling the t's one destroys the meaning of "a". What is
> "a"? It is v/t = at/t.
Nonsense. You are getting hung up on idiocy.
a is a. The fact it can be calculated by v/t doesnt change anything.
>> You still haven't answered the question yet as to whether
>> you think at is equal to the value of a multiplied by
>> the value of t.
>>
>> > > Do you not think there may be an error in your working..?
>> > >
>> >
>> > Only if there is an error in a = v/t = at/t = at^2
>>
>> There is, it's in that last step. at/t = at^2 is wrong.
>>
>> - Randy
>
> Yet, the same reasoning is ultimately used to convert v/t (which equals
> m/s/s) into vt (m/s^2). Why do you not argue against the logic of that?
Because you are the only person I have seen who has tried to convert v/t
into vt in this manner.
vt is not m/s^2. vt is in m. v is m s^-1. t is s. If you multiply m s^-1 by
s you get left with m.
> v/t = at/t and vt = at^2. Therefore, if v/t = vt then at/t = at^2.
Strawman if ever I saw one. Only you say v/t=vt.
Normal people would say a=v/t. v=at. t=v/a. You have some fixation with
simply swapping / and *.
> Perhaps you argue with the logic of my math because if you agreed with
> it, you would be indirectly agreeing with the equation I have proposed,
> E = W = F = Mass?
No. Your mathematics has no logic to argue with.
PD
Dennis B wrote:
> Does anyone know how to solve the following problem:
>
> (4m/s^2)/2s = ?
>
> Key:
>
> m = meters
> s = seconds
>
> I believe the correct solution is 2m/s. Is this correct?
>
Let's start from the beginning.
v = (delta-d) / (delta-t)
In SI units both sides have units of m/s
Some people shorten this further to write in algebraic shorthand
v = d/t
but when you write this you should really remember that the terms on
the right hand side are changes in those values, not the values
themselves.
Other forms of this same equation
d = v * t
In SI units both sides have units m. On the rhs, (m/s) * s happens to
be the same as m
t = d / v
In SI units both sides have units s. On the rhs, m / (m/s) happens to
be the same as s
Moving on,
a = (delta-v) / (delta-t)
which well shorten further with the same caveat to write
a = v /t
In SI units both sides have units m/s/s or m/s^2 (those mean the same
thing).
Other forms of this same equation
v = a * t
In SI units both sides have units m/s. On the rhs, (m/s^2) * s happens
to be the same as m/s
t = v / a
In SI units both sides have units s. On the rhs (m/s) / (m/s^2) happens
to be the same as s
Now, if you can follow all this and reproduce the SI units of
everything I've outlined here, you may be in shape to try other things.
If you can't, then you need some extended remedial practice in basic
algebra before continuing.
PD
Art Deco
Correct; sorry, my bad. In my haste I forgot who I was replying to.
dave hillstrom
Dennis B
I believe the correct solution would be 8m/2s. Or would it be 8m/2s^3?
-Dennis B
Dennis B
I believe the correct solution would be 8m/2s. Or would it be 8m/2s^3?
-Dennis B
Randy Poe
So you believe 4*2 = 8/2?
Let's back up.
Do you agree that a*t means the value of a times the
value of t?
If I say a=4, what is the value of a?
If I say t=2, what is the value of t?
If I say a=4 and t=2, what is the value of a times the
value of t?
- Randy
Phineas T Puddleduck
Dennis B <Uto...@kaxy.com> wrote:
This is ELEMENTARY algebra, multiplication and division that you've
managed to get consistently wrong. Surely you are not this seriously
bad at basic math?
--
Relf's Law? -+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
"Bullshit repeated to the limit of infinity asymptotically approaches
the odour of roses."
Corollary -+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
łIt approaches the asymptote faster, the more Śpseduosą you throw in
your formulas.˛
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
łGravity is one of the four fundamental interactions. The classical
theory of gravity - Einstein's general relativity - is the subject
of this book.˛ : Hartle/ Gravity pg 1
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Jaffa cakes. Sweet delicious orange jaffa goodness, and an abject lesson
why parroting information from the web will not teach you cosmology.
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Dennis B
Randy Poe wrote:
> Dennis B wrote:
> > Randy Poe wrote:
> > > Dennis B wrote:
> > > > Randy Poe wrote:
> > > > > Dennis B wrote:
> > > > > > [snip]
> > > > >
> > > > > You still haven't answered the question yet as to whether
> > > > > you think at is equal to the value of a multiplied by
> > > > > the value of t.
> > > > >
> > > >
> > > > at = a * t
> > >
> > > So if a=4 and t=2, is the value of at = 4*2 = 8?
> > >
> > > - Randy
> >
> > I believe the correct solution would be 8m/2s. Or would it be 8m/2s^3?
>
> So you believe 4*2 = 8/2?
>
> Let's back up.
>
> Do you agree that a*t means the value of a times the
> value of t?
>
No. a doesn't have a "value" represented by a single symbol as you've
specified.
> If I say a=4, what is the value of a?
>
That is meaningless.
> If I say t=2, what is the value of t?
>
That is also meaningless non-sense.
> If I say a=4 and t=2, what is the value of a times the
> value of t?
>
at = a * t = (m/s^2) * (s) = m/s^3 = m/s
-Dennis
Phineas T Puddleduck
Dennis B <Uto...@kaxy.com> wrote:
>
> > If I say a=4, what is the value of a?
> >
>
> That is meaningless.
>
> > If I say t=2, what is the value of t?
> >
>
> That is also meaningless non-sense.
Well there goes algebra
>
> > If I say a=4 and t=2, what is the value of a times the
> > value of t?
> >
>
> at = a * t = (m/s^2) * (s) = m/s^3 = m/s
at = at = ms^-2 * s = ms^-2
Your middle step is nonsense.
I now cannot seriously beleive that someone over the age of 12 is
capable of butchering algebra in such a fashion. The only conclusion I
can come to is that you are taking the piss. As it stands, your
proclamations on physics are worthless if you are consistently failing
to understand literally the first steps of algebra.
If you seriously believe at = a/t then the kindest word I can use to
describe you is deluded.
Randy Poe
Phineas T Puddleduck wrote:
> In article <1152757499.1...@35g2000cwc.googlegroups.com>,
> Dennis B <Uto...@kaxy.com> wrote:
> I now cannot seriously beleive that someone over the age of 12 is
> capable of butchering algebra in such a fashion. The only conclusion I
> can come to is that you are taking the piss.
I agree. Dennis seems to be using two different personalities.
> As it stands, your
> proclamations on physics are worthless
I agree with that too.
- Randy
Dennis B
Phineas T Puddleduck wrote:
> In article <1152757499.1...@35g2000cwc.googlegroups.com>,
> Dennis B <Uto...@kaxy.com> wrote:
>
> >
> > > If I say a=4, what is the value of a?
> > >
> >
> > That is meaningless.
> >
> > > If I say t=2, what is the value of t?
> > >
> >
> > That is also meaningless non-sense.
>
> Well there goes algebra
>
> >
> > > If I say a=4 and t=2, what is the value of a times the
> > > value of t?
> > >
> >
> > at = a * t = (m/s^2) * (s) = m/s^3 = m/s
>
> at = at = ms^-2 * s = ms^-2
>
at = velocity
The units of velocity are m/s (and perhaps m/s^3). The product of (a *
t) is a velocity and should be units of velocity (m/s or possibly
m/s^3) and not acceleration (m/s^2). According to *your* math, the
product is acceleration (m/s^2)?
-Dennis
Dennis B
Randy Poe wrote:
> Phineas T Puddleduck wrote:
> > In article <1152757499.1...@35g2000cwc.googlegroups.com>,
> > Dennis B <Uto...@kaxy.com> wrote:
> [it doesn't matter]
> > I now cannot seriously beleive that someone over the age of 12 is
> > capable of butchering algebra in such a fashion. The only conclusion I
> > can come to is that you are taking the piss.
>
> I agree. Dennis seems to be using two different personalities.
>
Perhaps you say this because I provided a different solution to the
dilema of how to solve (a * t), wherein a=4m/s^2 and t=2s, than I had
the day before? That is because I am still learning. I am only a
student. Nevertheless, I have accomplished more with my limited
abilities in the past few weeks than many have accomplished over their
entire lives. If you re-arrange the last solution I provided, so as to
conform to v/t = vt and at = a/t...I believe you will find it is a step
in the right direction.
> > As it stands, your
> > proclamations on physics are worthless
>
> I agree with that too.
>
> - Randy
It is YOUR proclamations concerning physics which are worthless! Do you
actually believe the gravitational force (and the resulting
acceleration) between two masses is dependent only upon the greater of
the two interacting masses? Does not GMm/r^2 teach you differently? The
force between the moon and Earth is much greater than the force between
the Earth and a 20 ton asteroid. Thus, if the moon were not orbiting
the Earth...it would accelerate at a greater velocity (per unit of
time) than the 20 ton asteroid would. Perhaps this explains why the
human ape has yet to evolve beyond the use of the primitive wheel for
transportation and conquer the kingdom of the heavens.
-Dennis B
Ben Newsam
>at = a * t = (m/s^2) * (s) = m/s^3 = m/s
Breathtaking. <Falls off chair>
Phineas T Puddleduck
Dennis B <Uto...@kaxy.com> wrote:
> It is YOUR proclamations concerning physics which are worthless! Do you
> actually believe the gravitational force (and the resulting
> acceleration) between two masses is dependent only upon the greater of
> the two interacting masses?
Says the man who thinks, force, energy and momentum are the same thing.
> Does not GMm/r^2 teach you differently? The
> force between the moon and Earth is much greater than the force between
> the Earth and a 20 ton asteroid. Thus, if the moon were not orbiting
> the Earth...it would accelerate at a greater velocity (per unit of
> time) than the 20 ton asteroid would. Perhaps this explains why the
> human ape has yet to evolve beyond the use of the primitive wheel for
> transportation and conquer the kingdom of the heavens.
Considering your inability to do algebra, particularly simple division
and multiplication, then I think we are justified in not expecting you
to lead us to a new physics paradigm.
Phineas T Puddleduck
Dennis B <Uto...@kaxy.com> wrote:
NOPE it should be ms^-1 where it is obvious from the above your madness
is catching, it is my cockup
at = at = ms^-2 * s = ms^-1
Randy Poe
Dennis B wrote:
> Randy Poe wrote:
> > Phineas T Puddleduck wrote:
> > > In article <1152757499.1...@35g2000cwc.googlegroups.com>,
> > > Dennis B <Uto...@kaxy.com> wrote:
> > [it doesn't matter]
> > > I now cannot seriously beleive that someone over the age of 12 is
> > > capable of butchering algebra in such a fashion. The only conclusion I
> > > can come to is that you are taking the piss.
> >
> > I agree. Dennis seems to be using two different personalities.
> >
>
We have the persona (A) carrying on a more or less coherent,
if misinformed, conversation about particle physics in other
threads.
And we have this persona (B) who won't be pinned down on
the meaning of xy or x/y, and pretends to be puzzled by
trying to figure out xy if x=4 and y=2.
It is B who makes statements about revolutionizing physics.
- Randy
T Wake
Don't get ahead of yourself. No units were mentioned here.
For days now you have been claiming xy = x/y. This is not the case.
if x=4 and y=2, then xy = 8 and x/y = 2. They are not the same. Can you see
that?
T Wake
In what realm of madness do you get m s^-3 = m s^-1?
Randy Poe
Dennis B wrote:
> Randy Poe wrote:
> > Phineas T Puddleduck wrote:
> > > As it stands, your
> > > proclamations on physics are worthless
> >
> > I agree with that too.
> >
> actually believe the gravitational force (and the resulting
> acceleration) between two masses is dependent only upon the greater of
> the two interacting masses?
No. But then I didn't say so (about force), did I?
Acceleration is a more complicated story if the smaller
mass is not negligible compared to the larger, as you
have correctly pointed out in some other messages.
- Randy
T Wake
>
>
> Randy Poe wrote:
>> Phineas T Puddleduck wrote:
>> > In article <1152757499.1...@35g2000cwc.googlegroups.com>,
>> > Dennis B <Uto...@kaxy.com> wrote:
>> [it doesn't matter]
>> > I now cannot seriously beleive that someone over the age of 12 is
>> > capable of butchering algebra in such a fashion. The only conclusion I
>> > can come to is that you are taking the piss.
>>
>> I agree. Dennis seems to be using two different personalities.
>>
>
> Perhaps you say this because I provided a different solution to the
> dilema of how to solve (a * t), wherein a=4m/s^2 and t=2s, than I had
> the day before? That is because I am still learning. I am only a
> student.
No, you are not learning. You have an "end solution" and you are trying to
corrupt things to make it valid.
Your basic premise that xy=x/y is flawed. The fact that you try to confuse
matters by being as foolish with the SI units as you are with the values is
irrelevant.
Your whole argument is self defeating.
> Nevertheless, I have accomplished more with my limited
> abilities in the past few weeks than many have accomplished over their
> entire lives. If you re-arrange the last solution I provided, so as to
> conform to v/t = vt and at = a/t...I believe you will find it is a step
> in the right direction.
No solution you have provided shows v/t = vt. Seriously. You need to go back
to some very, very basic mathematical principles which were proven around
the time of Alexander the Great.
(now I may make an error here, it is hot and I have been at a barbeque all
day, and drunk far too much to be correcting nonsense on the internet - if I
am wrong some one sane will correct it)
Your argument is based on reducing the divisor to one and then tricking
things by the properties of the number 1 - in that anything divided by 1 is
the same as multiplied by 1.
The problem is you are reducing an equation incorrectly. Lets try a worked
example, where the v=8 and t=4.
v/t = 8/4, you reduce it by dividing by 4 (t) giving 2/1, which you then
invert to make 2*1 = vt.
However, what you have done is v/t = (v/t)/(t/t) [This does not change it to
vt] - do you see how this is a pointless stage? It still cancels to v/t and
not vt.
The whole matter gets worse when you add in the SI units....... You think
s^-3 is the same as s^-1. That in itself speaks volumes.
>> > As it stands, your
>> > proclamations on physics are worthless
>>
>> I agree with that too.
>>
>> - Randy
>
> It is YOUR proclamations concerning physics which are worthless! Do you
> actually believe the gravitational force (and the resulting
> acceleration) between two masses is dependent only upon the greater of
> the two interacting masses? Does not GMm/r^2 teach you differently? The
> force between the moon and Earth is much greater than the force between
> the Earth and a 20 ton asteroid. Thus, if the moon were not orbiting
> the Earth...it would accelerate at a greater velocity (per unit of
> time) than the 20 ton asteroid would. Perhaps this explains why the
> human ape has yet to evolve beyond the use of the primitive wheel for
> transportation and conquer the kingdom of the heavens.
At last the k00kfr0th comes out. I was wondering how long you could remain
sane and lucid for.
T Wake
Argh, the madness hurts my eyes!
a = (number)m s^-2
t = (number) s
a*t = m s^-1
a/t = m s^-3
They are not even close to being the same.
T Wake
"Phineas T Puddleduck" <phineasp...@googlemail.com_NOSPAM> wrote in
message news:130720061133548255%phineasp...@googlemail.com_NOSPAM...
> In article <1152759135....@h48g2000cwc.googlegroups.com>,
> Dennis B <Uto...@kaxy.com> wrote:
>
>> > at = at = ms^-2 * s = ms^-2
>> >
>>
>> at = velocity
>>
>> The units of velocity are m/s (and perhaps m/s^3). The product of (a *
>> t) is a velocity and should be units of velocity (m/s or possibly
>> m/s^3) and not acceleration (m/s^2). According to *your* math, the
>> product is acceleration (m/s^2)?
>>
>> -Dennis
>
> NOPE it should be ms^-1 where it is obvious from the above your madness
> is catching, it is my cockup
>
> at = at = ms^-2 * s = ms^-1
>
I have caught myself a few times, after reading the drivel it is hard not to
get all word blind and mix up the terms.
Madness is catching and can be spread by the internet...
T Wake
Madness. How do you get 4*2 = 8/2 ?????
T Wake
Its OK, I forgive you :-) You were correct, there are errors in Dennis B's
brain! :-)
PD
Dennis B wrote:
>
> It is YOUR proclamations concerning physics which are worthless! Do you
> actually believe the gravitational force (and the resulting
> acceleration) between two masses is dependent only upon the greater of
> the two interacting masses? Does not GMm/r^2 teach you differently? The
> force between the moon and Earth is much greater than the force between
> the Earth and a 20 ton asteroid. Thus, if the moon were not orbiting
> the Earth...it would accelerate at a greater velocity (per unit of
> time) than the 20 ton asteroid would. Perhaps this explains why the
> human ape has yet to evolve beyond the use of the primitive wheel for
> transportation and conquer the kingdom of the heavens.
>
> -Dennis B
The answer is not the same for force and acceleration. Forget the
GMm/r^2 for a second. Just consider a piano and a golf ball, both held
20 feet above the ground. The force of gravitational attraction between
the Earth and the piano is a *whole lot* more than the force of
gravitational attraction between the Earth and the golf ball. That much
is unquestioned. (We call this force of attraction near the surface of
the earth "weight".) But now release both the piano and the golf ball,
and they both accelerate towards the ground at *identical* rates. Now
how can the force be so much different, but the acceleration be the
same?
Think about that for a minute and get back to me.
PD
Dennis B
at = instantaneous velocity
a/t = average velocity
-Dennis B
Phineas T Puddleduck
Dennis B <Uto...@kaxy.com> wrote:
>
> at = instantaneous velocity
>
> a/t = average velocity
>
> -Dennis B
at = ms^-2 * s = ms^-1 <= units of velocity
a/t= ms^-2 / s = ms^-3 <= different units entirely
Where the hell do you get the idea that s^-3 = s^-1?
Dennis B
Phineas T Puddleduck wrote:
> In article <1152837442....@m73g2000cwd.googlegroups.com>,
> Dennis B <Uto...@kaxy.com> wrote:
>
> >
> > at = instantaneous velocity
> >
> > a/t = average velocity
> >
> > -Dennis B
>
> at = ms^-2 * s = ms^-1 <= units of velocity
> a/t= ms^-2 / s = ms^-3 <= different units entirely
>
> Where the hell do you get the idea that s^-3 = s^-1?
>
d = vt
v = d/t = m/s = vt/t = d/t^2/t = d/t^3 = m/s^3
-Dennis B
Phineas T Puddleduck
Dennis B <Uto...@kaxy.com> wrote:
> Phineas T Puddleduck wrote:
> > In article <1152837442....@m73g2000cwd.googlegroups.com>,
> > Dennis B <Uto...@kaxy.com> wrote:
> >
> > >
> > > at = instantaneous velocity
> > >
> > > a/t = average velocity
> > >
> > > -Dennis B
> >
> > at = ms^-2 * s = ms^-1 <= units of velocity
> > a/t= ms^-2 / s = ms^-3 <= different units entirely
> >
> > Where the hell do you get the idea that s^-3 = s^-1?
> >
>
> d = vt
> v = d/t = m/s = vt/t = d/t^2/t = d/t^3 = m/s^3
>
> -Dennis B
>
Your algebra is absolutely appalling. I've known children do better
then this. If this is the basis of your "physics myths" then the world
has absolutely nothing to worry about.
v = d/t = m/s
vt/t = v
vt/t ms^-1 * s /s = ms^-1 = m/s
This is a joke right? Surely, to paraphrase John McEnroe, you cannot be
serious?
http://www.mathpower.com/tutorial.htm
For God's sake Dennis you should stop posting this. It makes you look
an idiot - unless that is your intention.
Even you can see that units of s^-3 are NOT EQUAL to s^-1
Dennis B
I'm just calling it as I see it. My math skills are admittedly
atrophied at the present time. Over the past few days I have been
strengthening my skills and I can see everyone else's point of view
more clearly. Yet, I still believe there is something of value to be
found in the formulas and strategies I am using. I will let you know
where I stand once I feel that my math skills are fully developed.
-Dennis B
Phineas T Puddleduck
Dennis B <Uto...@kaxy.com> wrote:
They're not atorphied, they are non existent. Your algebra is such that
anything bulit on top of it is instantly suspect. You have no
strategies save a form of dimensional dyslexia, and there is nothing of
value in torturing poor innocent quantities such as velocity and
acceleration to further a delusion that force, energy and momentum are
the same quantity.
If you are serious about learning physics, then sign off usenet and go
to an educational establishment. If you want to break the rules, you
first have to learn what the rules are.
Dennis B
Although the standard solution would be 8m/s...it is important to
remember that this is the instantaneous velocity (v) after 2 seconds of
acceleration (at a rate of 4m/s^2). Therefore it is not wrong to say
that the velocity is 8m/2s. Of course, the average velocity (a-v) after
two seconds would be 4m/s (presuming constant acceleration)...which I
believe can be found by the equation I've proposed: a/t=v. After only
one second, the instantaneous velocity would be 4m/s and the average
velocity would be 2m/s. I will admit that I may be wrong about a/t=v.
Yet, if I am right...a method must be implemented to convert the
average-v into the instantaneous-v and vice verse so that the numbers
truly are identical. Nevertheless, even in the absence of such a
method, at=a/t would remain true (if it is true) because
instantaneous-v and average-v are interdependent.
-Dennis B
Cranks Reply
you fricking retard. your math skills are in the toilet. there is no
value to your asinine swaps between /and*.
you need to learn basics before you rewrite the text books.
Phineas T Puddleduck
Dennis B <Uto...@kaxy.com> wrote:
> a method must be implemented to convert the
> average-v into the instantaneous-v and vice verse so that the numbers
> truly are identical. Nevertheless, even in the absence of such a
> method, at=a/t would remain true (if it is true) because
> instantaneous-v and average-v are interdependent.
>
> -Dennis B
at = a/t?
Surely even you can see the problems there?
T Wake
Not relevant.
You are implying that at = a/t. Which also implies that any value time
another value is the same as the values divided. This is nonsense.
I will ask again - if x=4 and y=2, then xy = 8 and x/y = 2. They are not
T Wake
Amazing.
Did you go to school?
T Wake
>
>
> T Wake wrote:
>> "Dennis B" <Uto...@kaxy.com> wrote in message
>> news:1152756971....@m79g2000cwm.googlegroups.com...
>> >
>> >
>> > Randy Poe wrote:
>> >> Dennis B wrote:
>> >> > Randy Poe wrote:
>> >> > > Dennis B wrote:
>> >> > > > [snip]
>> >> > >
>> >> > > You still haven't answered the question yet as to whether
>> >> > > you think at is equal to the value of a multiplied by
>> >> > > the value of t.
>> >> > >
>> >> >
>> >> > at = a * t
>> >>
>> >> So if a=4 and t=2, is the value of at = 4*2 = 8?
>> >>
>> >> - Randy
>> >
>> > I believe the correct solution would be 8m/2s. Or would it be 8m/2s^3?
>>
>> Madness. How do you get 4*2 = 8/2 ?????
>
> Although the standard solution would be 8m/s...
No. The standard solution would be 8.
I have not used any units of measure.
Forget what a and t mean in this context.
If you were correct an at = a/t then 16x8 would be 2.
If you cant see the madness then give up. If you can see the madness then
stop trolling.
> it is important to
> remember that this is the instantaneous velocity (v) after 2 seconds of
> acceleration (at a rate of 4m/s^2). Therefore it is not wrong to say
> that the velocity is 8m/2s. Of course, the average velocity (a-v) after
> two seconds would be 4m/s (presuming constant acceleration)...which I
> believe can be found by the equation I've proposed: a/t=v. After only
> one second, the instantaneous velocity would be 4m/s and the average
> velocity would be 2m/s. I will admit that I may be wrong about a/t=v.
> Yet, if I am right...a method must be implemented to convert the
> average-v into the instantaneous-v and vice verse so that the numbers
> truly are identical. Nevertheless, even in the absence of such a
> method, at=a/t would remain true (if it is true)
It isnt.
PD
Whoops. Rusty algebra.
v = d/t (units m/s) = vt/t = (d/t)*t/t (not d/t^2/t)
Just to be clear, (d/t)*t =/= d/(t*t).
PD
>
> -Dennis B
Dennis B
Perhaps you have concluded that the greater the mass of an object, the
greater the force required to move that mass at the same rate of
acceleration as a smaller mass? This would seem to solve the question
of why a greater mass doesn't accelerate faster, yet...I'm not so sure
that there isn't a minute difference between the acceleration of two
differing masses. In fact, I can cite several facts which appear to
prove that there should be a difference, even though it may be
virtually undetectable. First, there is the example of how a mass
"falling" towards the (Earth's) Moon is accelerated at only 1/6 the
rate it would if "falling" towards the Earth, which elucidates the fact
that the strength of the mutual gravitational force between two masses
affects the rate of acceleration. Then there is the fact that the rate
of acceleration between the Earth and a mass equal to the Moon (not in
orbit) should be greater than a 20 ton asteroid which started off at an
equal distance to the moon-like mass on the opposite side of the
Earth...unless one invokes inertial "frame drag" so that the
accelerated Earth effectively pulls harder on the smaller mass (and
"pushes" slightly on the greater mass) so as to offset the differences
which would otherwise manifest. Alas, there is the phenomena of
"gravitational lensing" in which the trajectory of photons are altered
by gravity. The degree to which the photon trajectory is altered is
dependent upon the frequency (and perhaps the amplitude) of the
photons. Thus, gravitational lensing is actually refraction and proves
Newton's intuition that a prism refracts light due to the gravitational
interaction of the matter comprising the prism with the light passing
through the prism. The obvious explanation for gravitational lensing is
that the photons are being "accelerated" by the gravitational field.
The greater the frequency, the greater the energy of the photon, and
therefore the greater the gravitationally induced acceleration (due to
the equivalence of mass and energy...E = mass times velocity (of light)
squared). If the velocity of light is truly constant, regardless of
frequency, then the acceleration is merely a change of direction.
Photons of higher frequency (supposedly) have greater energy and are
therefore "heavier", so are they subject to a greater rate of
gravitationally induced "acceleration"? Or is it possible that the
speed of light is NOT truly constant and that the different photon
frequencies are created by different velocities? Perhaps both a
difference of photon energy and velocity accounts for the difference of
trajectory between photons of different frequency as a result of
gravitational effects? Whatever the underlying reality may be, the
undeniable fact is that photon trajectories are affected by gravity. In
my opinion, the phenomena of gravitational lensing reveals a method of
detecting and measuring quantum gravitational effects and may answer
the question of whether the gravitational constant is truly
constant...or non-linear. Perhaps in close proximity to nuclei, G
becomes more powerful than the electro-weak force. Perhaps the strong
nuclear force is actually the gravitational force?
-Dennis B
Dennis B
I believe you are correct. These fractal equations are rather
disorienting. Thank you for pointing out the error.
-Dennis B
T Wake
>
>
> PD wrote:
>> Dennis B wrote:
>> > Phineas T Puddleduck wrote:
>> > > In article <1152837442....@m73g2000cwd.googlegroups.com>,
>> > > Dennis B <Uto...@kaxy.com> wrote:
>> > >
>> > > >
>> > > > at = instantaneous velocity
>> > > >
>> > > > a/t = average velocity
>> > > >
>> > > > -Dennis B
>> > >
>> > > at = ms^-2 * s = ms^-1 <= units of velocity
>> > > a/t= ms^-2 / s = ms^-3 <= different units entirely
>> > >
>> > > Where the hell do you get the idea that s^-3 = s^-1?
>> > >
>> >
>> > d = vt
>> > v = d/t = m/s = vt/t = d/t^2/t = d/t^3 = m/s^3
>>
>> Whoops. Rusty algebra.
>>
>> v = d/t (units m/s) = vt/t = (d/t)*t/t (not d/t^2/t)
>>
>> Just to be clear, (d/t)*t =/= d/(t*t).
>>
>> PD
>>
>> >
>> > -Dennis B
>
> I believe you are correct.
Good call. PD is correct in about 98% of his posts.
You may find life easier if you stop trying to prove something which is
wrong.
PD
Actually, I didn't conclude it. Newton did about 350 years ago.
This statement is algebraically condensed in the following shorthand
notation.
F = m*a
You may have heard of it.
The rest of what follows below does not seem to address this directly.
Was it the result of drinking a whole pot of coffee?
PD
PD
For the record, I'm really embarassed about the 2%, to the point where
it seems a whole lot more to me.
T Wake
"PD" <TheDrap...@gmail.com> wrote in message
news:1152905462.8...@b28g2000cwb.googlegroups.com...
:-) To be honest, I thought I was being excessive.... I am only talking
about your posts here - you may be wrong 99% of the time at home. (I know I
seem to be).
PD
As a general rule, if you start off with a quantity that has certain
units and then, through a series of algebraic equalities, end up with a
quantity with different units, then you can be sure that you've made an
algebraic error someplace. This skill is inculcated into freshman
chemistry and physics students who frequently make errors of this kind.
Repeat: If this happens, you've discovered an error, not new physics.
This is what several folks here have, each in their own way, been
trying to point out to you.
PD
cnctut
> As a general rule, if you start off with a quantity that has certain
> units and then, through a series of algebraic equalities, end up with a
> quantity with different units, then you can be sure that you've made an
> algebraic error someplace. This skill is inculcated into freshman
> chemistry and physics students who frequently make errors of this kind.
>
> Repeat: If this happens, you've discovered an error, not new physics.
> This is what several folks here have, each in their own way, been
> trying to point out to you.
>
> PD
Tut writes:
PD,
Now do it with a sliderule and tell me where the decimal point
is--along 'with' the proper units. ;-))
VR
Tut
Randy Poe
If you are going to use F = GMm/r^2 as a starting point, then
everything else you wrote below is nonsense. For instance,
if you want to start with assuming F = GMm/r^2, but
then have a = GM/r^2 depend on m, that is nonsense.
Are you hypothesizing a modification of F = GMm/r^2?
GR says that F = GMm/r^2 is very slightly inaccurate, but
the breakdown is only in extreme cases. Even the well-known
error in the precession of the orbit of Mercury, as close as it
is to the sun, is measured in arc-seconds per CENTURY.
- Randy
PD
cnctut wrote:
>
> PD,
>
> Now do it with a sliderule and tell me where the decimal point
> is--along 'with' the proper units. ;-))
>
I still have a slide rule or two and I think I remember how to use it.
And you're right, there are three skills involved:
- Using the slide rule to calculate the mantissa
- Using addition and subtraction to calculate the exponent
- Using fraction multiplication to calculate the units
The real beauty of the slide rule is that it kept everything to 3 sig
figs. :>)
PD
Phineas T Puddleduck
Randy Poe <poespa...@yahoo.com> wrote:
> If you are going to use F = GMm/r^2 as a starting point, then
> everything else you wrote below is nonsense. For instance,
> if you want to start with assuming F = GMm/r^2, but
> then have a = GM/r^2 depend on m, that is nonsense.
>
> Are you hypothesizing a modification of F = GMm/r^2?
>
> GR says that F = GMm/r^2 is very slightly inaccurate, but
> the breakdown is only in extreme cases. Even the well-known
> error in the precession of the orbit of Mercury, as close as it
> is to the sun, is measured in arc-seconds per CENTURY.
>
> - Randy
42 approximately. Douglas Adams would be proud ;-)
Dennis B
Yes, but the question is whether the greater force of a greater mass
perhaps leads to a greater acceleration...even though the difference
may be undetectable over the distances and times typically measured. In
my opinion, the answer already exists in the phenomena of gravitational
lensing.
> The rest of what follows below does not seem to address this directly.
> Was it the result of drinking a whole pot of coffee?
>
> PD
>
No. I haven't drinken any coffee for several days and do not intend to
for quite some time, if ever. I only drink coffee for medical reasons
and I've found an alternative which works more reliably...without the
undesirable side-effects.
-Dennis B
Dennis B
How do YOU explain gravitational lensing?
-Dennis B
Randy Poe
As one of the things predicted by the GR picture of gravity.
It's not a Newtonian effect.
A non sequitur, as usual.
- Randy
Dennis B
Yet, how can you disregard "m"? Gravitational force is the result of a
coupling of the gravitational fields of two masses, not one. The
g-force is dependent upon both masses and not just one. If the moon
were ever dislodged from it's orbit by a comet, asteroid, or other
celestial body so as to descend towards the Earth, would it not
accelerate at a greater rate towards the Earth than a 20-ton
asteroid???
> Are you hypothesizing a modification of F = GMm/r^2?
>
No. I am attempting to elucidate what I believe to be it's true
meaning.
> GR says that F = GMm/r^2 is very slightly inaccurate, but
> the breakdown is only in extreme cases. Even the well-known
> error in the precession of the orbit of Mercury, as close as it
> is to the sun, is measured in arc-seconds per CENTURY.
>
> - Randy
Such differences may be insignificant for most purposes, yet may be
very important in evaluating phenomena such as gravitational lensing
and quantum gravity.
-Dennis B
Randy Poe
If I say that F = GMm/r^2, I'm not disregarding m.
But since F = m a = m(GM/r^2), then clearly a = GM/r^2.
> Gravitational force is the result of a
> coupling of the gravitational fields of two masses, not one.
Indeed. F = (GM/r^2)m. It depends on m.
> > Are you hypothesizing a modification of F = GMm/r^2?
> >
>
> No. I am attempting to elucidate what I believe to be it's true
> meaning.
F = (GM/r^2)*m and F = a*m.
So if you aren't trying to revise that m*a = m*(GM/r^2), then
are you trying to revise F = m*a?
- Randy
Phineas T Puddleduck
Dennis B <Uto...@kaxy.com> wrote:
> > GR says that F = GMm/r^2 is very slightly inaccurate, but
> > the breakdown is only in extreme cases. Even the well-known
> > error in the precession of the orbit of Mercury, as close as it
> > is to the sun, is measured in arc-seconds per CENTURY.
> >
> > - Randy
>
> How do YOU explain gravitational lensing?
>
> -Dennis B
Light movement in null geodesics through curved spacetime.
Phineas T Puddleduck
Randy Poe <poespa...@yahoo.com> wrote:
> > How do YOU explain gravitational lensing?
>
> As one of the things predicted by the GR picture of gravity.
>
> It's not a Newtonian effect.
>
> A non sequitur, as usual.
>
> - Randy
Actually there is a formula for a Newtonian derivation of light
deflection, it is half the value of the GR one.
Randy Poe
Phineas T Puddleduck wrote:
> In article <1152936791....@p79g2000cwp.googlegroups.com>,
> Randy Poe <poespa...@yahoo.com> wrote:
>
> > > How do YOU explain gravitational lensing?
> >
> > As one of the things predicted by the GR picture of gravity.
> >
> > It's not a Newtonian effect.
> >
> > A non sequitur, as usual.
> >
> > - Randy
>
> Actually there is a formula for a Newtonian derivation of light
> deflection, it is half the value of the GR one.
I've heard of it. I consider it bogus. Basically you consider
the deflection of a nonzero mass and take the limit as m->0.
The basic problem is that "f(m) at m=0" and "lim f(m) as m->0" mean
two different things. Newtonian gravity will affect a nonzero mass
which is arbitrarily small, but the effect of Newtonian gravity
on a mass of exactly zero, is exactly zero.
- Randy
Dennis B
Phineas T Puddleduck wrote:
> In article <1152934082....@m73g2000cwd.googlegroups.com>,
> Dennis B <Uto...@kaxy.com> wrote:
>
> > > GR says that F = GMm/r^2 is very slightly inaccurate, but
> > > the breakdown is only in extreme cases. Even the well-known
> > > error in the precession of the orbit of Mercury, as close as it
> > > is to the sun, is measured in arc-seconds per CENTURY.
> > >
> > > - Randy
> >
> > How do YOU explain gravitational lensing?
> >
> > -Dennis B
>
> Light movement in null geodesics through curved spacetime.
>
The acceleration of a mass by gravity is explained as the the result of
curved space-time. There is no difference.
-Dennis B
Dennis B
Would you agree that v/t=vt?
v = at
a = v/t = at/t = at * t = v * t = at^2 = vt
Thus:
v/t=vt
Hence...
d = at^2/2 = vt/2
Of course, d=vt/2 is the calculation for computing the distance
traversed when accelerating. If velocity is constant, then d=vt.
Correct? It seems to me that one could use the average-v to compute
distance. For example:
d = average-v * t
If velocity is constant, then average-v = instantaneous-v. Correct?
Therefore, d=(average-v * t) could be used both when velocity is
constant AND when accelerating.
-Dennis B
Dennis B
Please explain your math. The mass corresponding to m in (ma) would be
both masses in my opinion. They both accelerate. Is this not what
GMm/r^2 truly reveals? I will conceded that since no inertial
acceleration forces are experienced in gravitationally induced
acceleration, and therefore...if you are falling towards the Earth due
to gravity, then it may appear to you that it is the Earth which is
accelerating toward you. As opposed to you accelerating toward the
Earth. Yet, the fact is that BOTH M and m are accelerating.
> > Gravitational force is the result of a
> > coupling of the gravitational fields of two masses, not one.
>
> Indeed. F = (GM/r^2)m. It depends on m.
>
It depends upon both M AND m.
> > > Are you hypothesizing a modification of F = GMm/r^2?
> > >
> >
> > No. I am attempting to elucidate what I believe to be it's true
> > meaning.
>
> F = (GM/r^2)*m and F = a*m.
>
> So if you aren't trying to revise that m*a = m*(GM/r^2), then
> are you trying to revise F = m*a?
>
> - Randy
I am not attempting to revise either. I am saying that
F=m*a=m*(GMm/r^2)=M*(Gm/r^2)...!!! BOTH M and m are accelerating!!!
Dennis B
Randy Poe
No, none of us would agree to that.
Example: v=5, t=5.
v/t = 1
vt = 25.
Not equal.
- Randy