| Topic: |
Science > Physics |
| User: |
"Don1" |
| Date: |
14 Sep 2005 07:39:42 PM |
| Object: |
Free fall |
Don1 writes:
The formula for finding the velocity of free fall is: v = 2s/t; where s
is the distance a body falls, and t is the time during which it falls.
The formula for finding the distance a body free falls is
d=vi+(a/2)t^2: Let's look at the progress of a body falling freely from
rest (vi=0) after 4 successive intervals of time:
During second 1, starting from rest, an object falls at 32'/sec^2; so
that the distance it falls is d=vi + (a/2)t^2=16 ft. and it will
attain a velocity of 2s/t=32 ft/sec.
During second 2, an object falls at 32'/sec^2; so that the distance it
falls is d=vi + (a/2)t^2= 64 ft. and it will attain a velocity of
2s/t=2x64 ft/2 sec = 64ft/sec.
During second 3, an object falls at 32'/sec^2; so that the distance it
falls is d=vi + (a/2)t^2= 144 ft. and it will attain a velocity of
2s/t=2x144 ft/3 sec = 96 ft/sec.
During second 4, an object falls at 32'/sec^2; so that the distance it
falls is d=vi + (a/2)t^2= 256 ft. and it will attain a velocity of
2s/t=2x 256 ft/4 sec = 128 ft/sec.
Using these simple formulas it is possible to find the velocity and
distance a body free falls during any period of time.
Don
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| User: "odin" |
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| Title: Re: Free fall |
14 Sep 2005 08:17:36 PM |
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The formula for finding the velocity of free fall is: v = 2s/t; where s
is the distance a body falls, and t is the time during which it falls.
Is the moon in free fall? If yes, how does that work with your equation
above. If not, then what the heck is it doing up there?
The formula for finding the distance a body free falls is
d=vi+(a/2)t^2: Let's look at the progress of a body falling freely from
rest (vi=0) after 4 successive intervals of time:
You do not take into account the vector nature of the problem. What if
velocity and force are not parallel?
You do not take into account the fact that mass is not constant, but depends
on velocity (relativistic).
You do not take into account the fact that the force is not constant, but
depends on displacement (universal gravitation)
Also, even given all these simplifying assumptions, you have an error...
you have:
d = vi + (a/2)*t^2
you should have:
d = vi*t + (a/2)*t^2
where d is the distance measured from when t=0. Not the distance measured
for a given time interval as you describe...
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| User: "Don1" |
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| Title: Re: Free fall |
14 Sep 2005 09:20:19 PM |
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odin wrote:
The formula for finding the velocity of free fall is: v = 2s/t; where s
is the distance a body falls, and t is the time during which it falls.
Is the moon in free fall? If yes, how does that work with your equation
above. If not, then what the heck is it doing up there?
It's falling, but missing, and passing by: Only to fall back again, and
again.
The formula for finding the distance a body free falls is
d=vi+(a/2)t^2: Let's look at the progress of a body falling freely from
rest (vi=0) after 4 successive intervals of time:
You do not take into account the vector nature of the problem. What if
velocity and force are not parallel?
Velocity is relative, and the force is centripetal.
You do not take into account the fact that mass is not constant, but depends
on velocity (relativistic).
Says you.
You do not take into account the fact that the force is not constant, but
depends on displacement (universal gravitation)
The gravitation is in the direction of the free fall, and is its cause.
Also, even given all these simplifying assumptions, you have an error...
you have:
d = vi + (a/2)*t^2
you should have:
d = vi*t + (a/2)*t^2
No vi is the point of beginning; the rate of the initial _inertial_
displacement l: It is equal to l/t; for vi = 0, it is 0/t.
where d is the distance measured from when t=0. Not the distance measured
for a given time interval as you describe...
Yes, d= the total distance from wherever it starts.
Don
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| User: "odin" |
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| Title: Re: Free fall |
15 Sep 2005 12:57:49 AM |
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The formula for finding the velocity of free fall is: v = 2s/t; where s
is the distance a body falls, and t is the time during which it falls.
Is the moon in free fall? If yes, how does that work with your equation
above. If not, then what the heck is it doing up there?
It's falling, but missing, and passing by: Only to fall back again, and
again.
So does the moon obey your equation or not? If not, why not?
You do not take into account the fact that the force is not constant, but
depends on displacement (universal gravitation)
The gravitation is in the direction of the free fall, and is its cause.
But gravitational field strength is not constant, is it?
Also, even given all these simplifying assumptions, you have an error...
you have:
d = vi + (a/2)*t^2
you should have:
d = vi*t + (a/2)*t^2
No vi is the point of beginning; the rate of the initial _inertial_
displacement l: It is equal to l/t; for vi = 0, it is 0/t.
Nope. The variable vi is the initial velocity. If you do not multiply it by
t, you will fail in your dimensional analysis.
If vi = 10 miles/hour, and t = 2 hours, assuming for now that a = 0 (like in
deep space), doesn't it make sense that after t, the displacement would be
vi*t or 10 miles/hour * 2 hours to give you 20 miles?
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| User: "Eric Gisse" |
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| Title: Re: Free fall |
15 Sep 2005 01:39:30 AM |
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odin wrote:
[snip]
Please, please, please stop validating his existance. The zero sum game
of trying to teach Don anything needs to stop.
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| User: "Timothy Little" |
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| Title: Re: Free fall |
15 Sep 2005 04:43:50 AM |
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Eric Gisse wrote:
Please, please, please stop validating his existance. The zero sum
game of trying to teach Don anything needs to stop.
Zero-sum? And here was I thinking it was negative-sum.
- Tim
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| User: "odin" |
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| Title: Re: Free fall |
15 Sep 2005 09:48:46 AM |
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Eric Gisse wrote:
Please, please, please stop validating his existance. The zero sum
game of trying to teach Don anything needs to stop.
Zero-sum? And here was I thinking it was negative-sum.
That is more like it. A negative sum game. There are no winners. Kind of
like abusing drugs......
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| User: "tadchem" |
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| Title: Re: Free fall |
15 Sep 2005 03:55:27 PM |
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"odin" <ragnarok@yahoo.com> wrote in message
news:k8-dnaAukKFNFLTeRVn-gQ@whidbeytel.com...
Eric Gisse wrote:
Please, please, please stop validating his existance. The zero sum
game of trying to teach Don anything needs to stop.
Zero-sum? And here was I thinking it was negative-sum.
That is more like it. A negative sum game. There are no winners. Kind of
like abusing drugs......
Just say 'No' to sHead.
Tom Davidson
Richmond, VA
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| User: "" |
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| Title: Re: Free fall |
15 Sep 2005 10:22:38 AM |
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In article <slrndiigim.to4.tim-usenet@soprano.little-possums.net>, Timothy Little <tim-usenet@little-possums.net> writes:
Eric Gisse wrote:
Please, please, please stop validating his existance. The zero sum
game of trying to teach Don anything needs to stop.
Zero-sum? And here was I thinking it was negative-sum.
More to the point, it is not fixed-sum.
John Briggs
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| User: "" |
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| Title: Re: Free fall |
15 Sep 2005 10:21:46 AM |
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In article <1126766370.621443.199350@o13g2000cwo.googlegroups.com>, "Eric Gisse" <jowr.pi@gmail.com> writes:
odin wrote:
[snip]
Please, please, please stop validating his existance. The zero sum game
of trying to teach Don anything needs to stop.
Please don't use the term "zero sum game" when it does not apply.
If it were zero sum then your advice would be worthless.
John Briggs
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| User: "" |
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| Title: Re: Free fall |
15 Sep 2005 08:23:36 PM |
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wrote:
In article <1126766370.621443.199350@o13g2000cwo.googlegroups.com>, "Eric Gisse" <jowr.pi@gmail.com> writes:
odin wrote:
[snip]
Please, please, please stop validating his existance. The zero sum game
of trying to teach Don anything needs to stop.
Please don't use the term "zero sum game" when it does not apply.
If it were zero sum then your advice would be worthless.
John Briggs
Watch for the next dumb donny post.
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| User: "Shmuel Seymour J. Metz" |
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| Title: Re: Free fall |
16 Sep 2005 07:38:15 AM |
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In <LyVCKjt$mKVu@eisner.encompasserve.org>, on 09/15/2005
at 10:21 AM, said:
Please don't use the term "zero sum game" when it does not apply.
It does apply. A non-zero sum game can be converted to a zero sum game
by adding an additional player.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>
Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply to
domain Patriot dot net user shmuel+news to contact me. Do not
reply to
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| User: "odin" |
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| Title: Re: Free fall |
16 Sep 2005 03:31:16 PM |
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Please don't use the term "zero sum game" when it does not apply.
It does apply. A non-zero sum game can be converted to a zero sum game
by adding an additional player.
Nope. When you add a player to a game, you change the game entirely. The new
game may become zero sum but the old game remains non-zero sum.
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| User: "Sam Wormley" |
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| Title: Re: Free fall |
14 Sep 2005 09:52:01 PM |
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Don1 wrote:
Don1 writes:
The formula for finding the velocity of free fall is....
No you have to start with the differential equation F = ma, and
this time acceleration is not constant.... it is quite a complicated
calculation. Newton gave us all the tools we need.
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| User: "Herman Trivilino" |
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| Title: Re: Free fall |
14 Sep 2005 10:57:19 PM |
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"Sam Wormley" <swormley1@mchsi.com> wrote ...
No you have to start with the differential equation F = ma, and
this time acceleration is not constant.... it is quite a complicated
calculation. Newton gave us all the tools we need.
And did all the calculations.
And did all that over 200 years ago.
And Don is still catching up.
Trying to understand what Galileo did over 300 years ago.
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| User: "G=EMC^2 Glazier" |
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| Title: Re: Free fall |
15 Sep 2005 08:01:47 AM |
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Hi Sam Free fall or force free motion has meaning only by comparison
with other objects. Bert
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| User: "Randy Poe" |
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| Title: Re: Free fall |
14 Sep 2005 07:49:04 PM |
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Don1 wrote:
Don1 writes:
The formula for finding the velocity of free fall is: v = 2s/t; where s
is the distance a body falls, and t is the time during which it falls.
The formula for finding the distance a body free falls is
d=vi+(a/2)t^2: Let's look at the progress of a body falling freely from
rest (vi=0) after 4 successive intervals of time:
During second 1, starting from rest, an object falls at 32'/sec^2; so
that the distance it falls is d=vi + (a/2)t^2=16 ft.
if vi = 0, not otherwise.
and it will
attain a velocity of 2s/t=32 ft/sec.
During second 2, an object falls at 32'/sec^2; so that the distance it
falls is d=vi + (a/2)t^2= 64 ft. and it will attain a velocity of
2s/t=2x64 ft/2 sec = 64ft/sec.
64 feet is the total distance it has fallen since t = 0. The
distance it traveled in second 2 was 64 - 16 = 48 feet.
During second 3, an object falls at 32'/sec^2; so that the distance it
falls is d=vi + (a/2)t^2= 144 ft.
Again, that's the total distance it has fallen (if vi=0).
The distance just during second 3 is 144 - 64 = 80 feet.
and it will attain a velocity of
2s/t=2x144 ft/3 sec = 96 ft/sec.
During second 4, an object falls at 32'/sec^2; so that the distance it
falls is d=vi + (a/2)t^2= 256 ft. and it will attain a velocity of
2s/t=2x 256 ft/4 sec = 128 ft/sec.
Using these simple formulas it is possible to find the velocity and
distance a body free falls during any period of time.
Yes, if correctly applied. This is the most correct post I've seen
from you but there is still a slight misinterpretation of the
meaning of the formulas.
- Randy
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| User: "Don1" |
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| Title: Re: Free fall |
15 Sep 2005 07:21:06 AM |
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Randy Poe wrote:
Don1 wrote:
Snip<
Using these simple formulas it is possible to find the velocity and
distance a body free falls during any period of time.
Yes, if correctly applied. This is the most correct post I've seen
from you but there is still a slight misinterpretation of the
meaning of the formulas.
- Randy
These formulas are just adaptations of those found in most basic
physics texts: With further adaptions they will apply to all kinds of
motion. You, or someone who will make the effort, and has the
fortitude, could be the one to do it. I'm running out of time and
energy now but have made substantial inroads which as time and energy
permit, I'll be continuing to persue.
Don
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| User: "Herman Trivilino" |
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| Title: Re: Free fall |
15 Sep 2005 10:05:12 PM |
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"Don1" <dcshead@charter.net> wrote ...
These formulas are just adaptations of those found in most basic
physics texts: With further adaptions they will apply to all kinds of
motion. You, or someone who will make the effort, and has the
fortitude, could be the one to do it. I'm running out of time and
energy now but have made substantial inroads which as time and energy
permit, I'll be continuing to persue.
With no intent to diminish your efforts, haven't you really been a student
of the work others have done? These substantial inroads are admirable, but
what makes you think others haven't gone before you? Are you asking that
their work be duplicated, or just studied more thoroughly?
Based on my studies, I've learned a lot about how to apply these formulas to
"other kinds of motion". And so have a lot of your other correspondents.
We've tried to share that knowledge with you, but I have to say, it's been
met with considerable resistance on your part.
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| User: "odin" |
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| Title: Re: Free fall |
15 Sep 2005 09:47:08 AM |
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These formulas are just adaptations of those found in most basic
physics texts.
Nope. These formulas are wrong. Go look at some physics texts. You are
forgetting to multiply the initial velocity by time to get displacement.
With further adaptions they will apply to all kinds of
motion. You, or someone who will make the effort, and has the
fortitude, could be the one to do it.
Ummm. Nobody needs to make the effort. It is done already. Newton's
Principia covered what you are trying to do and much more. And unlike you,
it was correct. And it was published in 1687. That was 318 years ago.
I'm running out of time and
energy now but have made substantial inroads which as time and energy
permit, I'll be continuing to persue.
Pleas run out of time and energy as soon as possible. You have made no
substantial inroads into anything. No inroads of any kind. You are a
pathetic fool.
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| User: "Randy Poe" |
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| Title: Re: Free fall |
15 Sep 2005 11:06:40 AM |
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Don1 wrote:
Randy Poe wrote:
Don1 wrote:
Snip<
Using these simple formulas it is possible to find the velocity and
distance a body free falls during any period of time.
Yes, if correctly applied. This is the most correct post I've seen
from you but there is still a slight misinterpretation of the
meaning of the formulas.
These formulas are just adaptations of those found in most basic
physics texts:
"Adapted" in that they were changed from correct to incorrect
(I missed your error in forgetting to put a "t" after "vi").
However, if you fix that error, then much of what you
said is not far from correct, although you are misinterpreting
the meaning of d.
It is not the distance traveled in the t-th second. It
is the distance traveled since t=0, and t need not be
an integer. Correctly interpreted, t can be any real
number.
With further adaptions they will apply to all kinds of
motion.
The unadapted equations already do that. Specifically:
F = ma
a = rate of change of velocity
v = rate of change of displacement.
Those three have served many a simulator or game designer
to give good, realistic motion.
- Randy
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| User: "Don1" |
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| Title: Re: Free fall |
15 Sep 2005 08:49:04 PM |
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Randy Poe wrote:
Don1 wrote:
These formulas are just adaptations of those found in most basic
physics texts:
"Adapted" in that they were changed from correct to incorrect
(I missed your error in forgetting to put a "t" after "vi").
That wasn't an error Randy: There is no t after vi because vi is a
_ratio_ of length l, to t. The formula in the textbooks is wrong: It
should be s=(v/i)xt + (a/2)t^2: Dividing all terms by t we should get
s/t=l/t + (a/2)t
However, if you fix that error, then much of what you
said is not far from correct, although you are misinterpreting
the meaning of d.
Look in your basic physics text: Although they probably use s instead
of d; the equation is given as s=(vi)t + at^2; No wonder it only works
when vi=0'/sec.
Don
- Randy
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| User: "odin" |
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| Title: Re: Free fall |
16 Sep 2005 12:53:03 AM |
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Look in your basic physics text: Although they probably use s instead
of d; the equation is given as s=(vi)t + at^2; No wonder it only works
when vi=0'/sec.
The equation s=(vi)t + at^2 is not found in any textbook unless it is a
typo. It should be s=(vi)t + (a/2)t^2. If you knew any calculus, you would
understand that. And no, it does not only work for when vi is zero.
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| User: "Don1" |
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| Title: Re: Free fall |
16 Sep 2005 07:37:44 AM |
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odin wrote:
Snip<
The equation s=(vi)t + at^2 is not found in any textbook unless it is a
typo. It should be s=(vi)t + (a/2)t^2.
The equation s=(vi)t + (a/2)t^2 is not found in any textbook that I
know of: The parentheses are mine! I put them around vi, because vi is
a distance l, divided by a unit of time; so that the equation should be
s=(l/t)t + (a/2)t^2.
The best part is that dividing all terms by t gives s/t=l/t + (a/2)t.
Try that in your peace-pipe.
Don
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| User: "" |
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| Title: [TROLL] Re: free fall |
16 Sep 2005 10:16:13 AM |
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In article <1126874264.435268.303620@g49g2000cwa.googlegroups.com>, "Don1" <dcshead@charter.net> writes:
odin wrote:
Snip<
The equation s=(vi)t + at^2 is not found in any textbook unless it is a
typo. It should be s=(vi)t + (a/2)t^2.
The equation s=(vi)t + (a/2)t^2 is not found in any textbook that I
know of: The parentheses are mine! I put them around vi, because vi is
a distance l, divided by a unit of time; so that the equation should be
s=(l/t)t + (a/2)t^2.
*sigh*. Responding to Don.
Note that the t in parentheses is a "unit of time" -- the time interval
during which a distance l was traversed. The t outside the parentheses
is the elapsed time during which a distance s has been covered.
The two t's are entirely separate, distinct and unrelated.
Real physicists, mathematicians and people operating with more than half
a brain try not to use the same symbol twice in the same equation with
incompatible meanings.
The best part is that dividing all terms by t gives s/t=l/t + (a/2)t.
Try that in your peace-pipe.
Dividing by t when there are two distinct t's is the fallacy of
equivocation. It is an error.
It also flies in the face of Don's previous erroneous assertion that
"terms" inside of parentheses are indivisible.
John Briggs
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| User: "Randy Poe" |
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| Title: Re: Free fall |
16 Sep 2005 07:50:38 AM |
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Don1 wrote:
Randy Poe wrote:
Don1 wrote:
These formulas are just adaptations of those found in most basic
physics texts:
"Adapted" in that they were changed from correct to incorrect
(I missed your error in forgetting to put a "t" after "vi").
That wasn't an error Randy: There is no t after vi because vi is a
_ratio_ of length l, to t. The formula in the textbooks is wrong: It
should be s=(v/i)xt + (a/2)t^2:
Um...
(1) That *is* the formula in textbooks, and
(2) You just put the t in after vi, after telling me
there isn't one. Though you made another change, from vi for
"initial velocity" to v/i for...?
Try your version s = vi + at^2/2 for a = 0, constant velocity
non-accelerated motion. Do you think s = vi correctly
describes the motion?
- Randy
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| User: "odin" |
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| Title: Re: Free fall |
16 Sep 2005 08:08:44 AM |
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Um...
What I am curious about is that Don1 thinks the textbooks are wrong on this
stuff. Yet his equation seems to be different and inconsistent every time he
posts a message about it. God only knows how vi became v/i for no good
reason in this thread. And God only knows what parenthesis actually mean in
his own twisted version of elementary algebra. God only knows why he eschews
calculus, vectors, quantum physics, relativity, etc. But in any case, I
wonder how he explains that everyone from Ford to NASA uses the appropriate
equations as they are described in the textbooks, and they all design
devices that actually work. Sometimes he is a throwback to Newton's time,
but loaded with errors. Sometimes he is a throwback to ancient Greek times.
His math never gets beyond the level of a failing grade seven student. He
never gets into the 19th century let alone the last one or this one.
Well, if these equations that have been in constant use for over 300 years
are wrong, then what the hell does Don1 think is going on? Well, I think I
know... Don1 does not believe in what he says. He has a keen interest in
annoying others and bringing humiliation on himself. For whatever reasons.
Of course, that is just my operating theory. He could also just simply be an
idiot.
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| User: "Herman Trivilino" |
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| Title: Re: Free fall |
17 Sep 2005 09:16:22 AM |
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"odin" <ragnarok@yahoo.com> wrote ...
What I am curious about is that Don1 thinks the textbooks are wrong on
this stuff. Yet his equation seems to be different and inconsistent every
time he posts a message about it.
You are using different criteria for establishing "wrong". To a physicist,
in this context, "wrong" means "doesn't match what's observed".
To Don, "wrong" means "doesn't satisfy my expectation of the the way it
ought to be".
We argue with Don, pointing out that his formulae and his meanings aren't
consistent with what's observed. He doesn't know what that means so he
ignores it, and continues on, claiming that we are wrong because the stuff
we tell him doesn't match his notion of the way things ought to be.
God only knows how vi became v/i for no good reason in this thread.
It doesn't mater to him because vi and v/i have the same meaning to him.
And God only knows what parenthesis actually mean in his own twisted
version of elementary algebra.
He's not doing algebra, he's expressing his notion of the way things ought
to be.
God only knows why he eschews calculus, vectors, quantum physics,
relativity, etc.
They aren't the way things ought to be.
But in any case, I wonder how he explains that everyone from Ford to NASA
uses the appropriate equations as they are described in the textbooks, and
they all design devices that actually work.
He dosn't see a connection between his worldview of the way things ought to
be and the worldview of a physicist who is modelling the way things actually
are.
Sometimes he is a throwback to Newton's time, but loaded with errors.
Sometimes he is a throwback to ancient Greek times. His math never gets
beyond the level of a failing grade seven student. He never gets into the
19th century let alone the last one or this one.
It took Western civilization centuries to sort out the notion that Nature
doesn't necessarily behave the way humans think it ought to behave, and that
one must actually observe Nature if one is to produce a successful model of
the way Nature actually behaves.
Well, if these equations that have been in constant use for over 300 years
are wrong, then what the hell does Don1 think is going on?
He thinks that physicists have a worldview that is wrong, according to his
notion of what it means to be wrong. It makes no difference that equations
are a successful model. That fact has no value whatever in his worldview.
Well, I think I know... Don1 does not believe in what he says. He has a
keen interest in annoying others and bringing humiliation on himself. For
whatever reasons. Of course, that is just my operating theory. He could
also just simply be an idiot.
All of what you've said in the abopve paragraph is probably true. True, but
not relevant.
I've said this before. Don may sound like an idiot, and he may look like an
idiot, but don't let that fool you. He really IS an idiot.
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| User: "" |
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| Title: Re: Free fall |
18 Sep 2005 02:36:13 PM |
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In article <1126966735_3339@spool6-east.superfeed.net>, "Herman Trivilino" <physhead@kingwoodREMOVECAPScable.com> writes:
"odin" <ragnarok@yahoo.com> wrote ...
What I am curious about is that Don1 thinks the textbooks are wrong on
this stuff. Yet his equation seems to be different and inconsistent every
time he posts a message about it.
You are using different criteria for establishing "wrong". To a physicist,
in this context, "wrong" means "doesn't match what's observed".
To Don, "wrong" means "doesn't satisfy my expectation of the the way it
ought to be".
And this, in a nutshell, sums up the disconnect between scientists and
cranks, as evidenced on this ng. We may use the same words but we
don't speak the same language.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "Bob Cain" |
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| Title: Re: Free fall |
16 Sep 2005 04:29:27 PM |
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odin wrote:
Um...
What I am curious about is that Don1 thinks the textbooks are wrong on this
stuff. Yet his equation seems to be different and inconsistent every time he
posts a message about it. God only knows how vi became v/i for no good
reason in this thread. And God only knows what parenthesis actually mean in
his own twisted version of elementary algebra. God only knows why he eschews
calculus, vectors, quantum physics, relativity, etc.
I know and I'm not god. It's doubletalk and all the
responses to it give him great fun. He couldn't do it so
well if he didn't actually know what's going on. The man's
an artist and this forum has been his canvas for years.
Bob
--
"Things should be described as simply as possible, but no
simpler."
A. Einstein
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| User: "" |
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| Title: Re: Free fall |
16 Sep 2005 08:52:41 PM |
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I know and I'm not god. It's doubletalk and all the
responses to it give him great fun. He couldn't do it so
well if he didn't actually know what's going on. The man's
an artist and this forum has been his canvas for years.
Bob
***********
Now who would do that? Reply sensibly one time, sarcastically the
next, yet harbor underlying knowledge which allows him to pull it all
off? Maybe someone who's given up on any sensible dialectic here and
just wants to point out how the noncooperation of a few ruin it for
all, the same thing which could still bring Humanity down at this late
stage despite all the progress we've made. Have a nice day.
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