OT: Why the US will never go metric....

[Michael A. Terrell]

"krw@att.bizzzzzzzzzzzz" wrote:

On Thu, 17 Jun 2010 17:57:21 -0700, The Great Attractor
SuperM@ssiveBlackHoleAtTheCenterOfTheMilkyWayGalaxy.org> wrote:

On Thu, 17 Jun 2010 20:45:02 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


"krw@att.bizzzzzzzzzzzz" wrote:

On Wed, 16 Jun 2010 22:28:30 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 21:54:23 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

the sphere
acts as if all its mass was concentrated at the center.

Bullshit. That would be a black hole.

Always wrong.


The black hole is between Dimmie's ears. We have conformation of
this, by the fact that nothing intelligent ever escapes his head.


That is why none of you hang around. If you were intelligent, you
would already be inside my head. Since you are an utter retard, you have
no attraction to my head. You lose... Again.

He's got you hooked, why bother reeling you in?

Very true. The sandy tar balls washing up on the beaches brings more
than you could get for a dead barn cat.


--
Anyone wanting to run for any political office in the US should have to
have a DD214, and a honorable discharge.

 

[Michael A. Terrell]

"krw@att.bizzzzzzzzzzzz" wrote:

On Thu, 17 Jun 2010 20:46:47 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


"krw@att.bizzzzzzzzzzzz" wrote:

On Thu, 17 Jun 2010 01:08:10 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


"krw@att.bizzzzzzzzzzzz" wrote:

On Wed, 16 Jun 2010 16:50:38 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 18:43:32 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Wed, 16 Jun 2010 01:39:05 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


Tim Watts wrote:

On Mon, 14 Jun 2010 22:32:09 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wibbled:

My Vermont house, other than the living and family rooms (cathedral
ceilings) had 7' 2" ceilings; definitely not standard.

This first floor of this house has 9' ceilings and the two bedrooms
upstairs 8', with the great room 18', and higher. ;-)

You should try my village, which dates back to 1066 - in fact the Battle
of Hastings was fought and shamefully lost (especially when you visit the
field and see the massive tactical advantage Harold had), 3 miles down
the road in a town called "Battle" (hmm) and not actually in Hastings
which is rather further down the road.

I digress...

Ceilings you can brush your head on and 5' front doors or less on some of
the old timber framed houses.


You sublet from a Leprechaun?

Someone with DimBulb's stature.

I am tall enough to beat knots onto the top of your skull, boy.

Not.


Not even in his hooker heels.

Not even the ones he got out of mommy's hamper?


That's where he hides them, so his 'daddy' doesn't spank him for
dressing like a hooker.

Not possible. That would imply that he was a product of sexual reproduction.
Ewwww. That's just *wrong*.

Read it again. 'daddy' implies something else.


--
Anyone wanting to run for any political office in the US should have to
have a DD214, and a honorable discharge.

 

[Archimedes' Lever]

On Thu, 17 Jun 2010 21:22:54 -0400, "Michael A. Terrell"
<mike.terrell@earthlink.net> wrote:

"krw@att.bizzzzzzzzzzzz" wrote:

Michael A. Terrell wrote:

"krw@att.bizzzzzzzzzzzz" wrote:

Close, but AlwaysWrong has a fetish for feces. He's really into it.


Too much, apparently. They fired him from his job of mucking out
horse stalls.

The horses were embarrassed.


Horses know that there is no such thing as an amateur proctologist.

I find it interesting that when I call one of you retarded fucks a
"shithead", you spew ten posts about YOUR fantasies about what you claim
I have a problem with.

You stupid fucks gotz issues. None of which carry any viable
information for man's use. In other words, all you retarded fucks are is
shit.

 

[Archimedes' Lever]

On Thu, 17 Jun 2010 21:26:06 -0400, "Michael A. Terrell"
<mike.terrell@earthlink.net> wrote:

"krw@att.bizzzzzzzzzzzz" wrote:

On Thu, 17 Jun 2010 20:46:47 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


"krw@att.bizzzzzzzzzzzz" wrote:

On Thu, 17 Jun 2010 01:08:10 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


"krw@att.bizzzzzzzzzzzz" wrote:

On Wed, 16 Jun 2010 16:50:38 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 18:43:32 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Wed, 16 Jun 2010 01:39:05 -0400, "Michael A. Terrell"
mike.terrell@earthlink.net> wrote:


Tim Watts wrote:

On Mon, 14 Jun 2010 22:32:09 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wibbled:

My Vermont house, other than the living and family rooms (cathedral
ceilings) had 7' 2" ceilings; definitely not standard.

This first floor of this house has 9' ceilings and the two bedrooms
upstairs 8', with the great room 18', and higher. ;-)

You should try my village, which dates back to 1066 - in fact the Battle
of Hastings was fought and shamefully lost (especially when you visit the
field and see the massive tactical advantage Harold had), 3 miles down
the road in a town called "Battle" (hmm) and not actually in Hastings
which is rather further down the road.

I digress...

Ceilings you can brush your head on and 5' front doors or less on some of
the old timber framed houses.


You sublet from a Leprechaun?

Someone with DimBulb's stature.

I am tall enough to beat knots onto the top of your skull, boy.

Not.


Not even in his hooker heels.

Not even the ones he got out of mommy's hamper?


That's where he hides them, so his 'daddy' doesn't spank him for
dressing like a hooker.

Not possible. That would imply that he was a product of sexual reproduction.
Ewwww. That's just *wrong*.


Read it again. 'daddy' implies something else.

More proof that it is you and the krw utter retard that have the
biggest issues here.

Neither of you qualify as men. Hell, neither of you qualify as a
member of the human species.

 

[JosephKK]

On Thu, 17 Jun 2010 08:40:07 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 17 Jun 2010 06:41:25 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Jun 17, 8:50 am, George Herold <gher...@teachspin.com> wrote:
On Jun 16, 11:38 pm, John Larkin





jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 19:52:34 -0700 (PDT), George Herold

gher...@teachspin.com> wrote:
On Jun 16, 10:17 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 22:03:46 -0400, Spehro Pefhany

speffS...@interlogDOTyou.knowwhat> wrote:
On Wed, 16 Jun 2010 17:23:13 -0700, the renowned John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 16:54:42 -0700 (PDT), Richard Henry
pomer...@hotmail.com> wrote:

On Jun 16, 3:58 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 15:36:13 -0700 (PDT), Richard Henry

pomer...@hotmail.com> wrote:
On Jun 16, 12:53 pm, George Herold <gher...@teachspin.com> wrote:
On Jun 16, 2:31 pm, Archimedes' Lever <OneBigLe...@InfiniteSeries.Org
wrote:

On Wed, 16 Jun 2010 11:05:15 -0700 (PDT), George Herold

gher...@teachspin.com> wrote:
Cool, I have to scribble numbers on the paper though. 6400 feet is
about 2000m, the Earth is about 6E6 m in radius, Since we only want a
small change I can ignore the r^2 stuff and just multiple the ratio by
2. something like 4 parts out of 6,000. much smaller than the
divisions on your scale.

No you cannot. What makes you think that G decreases (or
increases)linearly?

Big G doesn't change at all. Little g (the force of gravity on the
Earths surface) will go as 1/r^2. For small changes in r the change
is approximately linear... first term in the taylor expansion if you
want to think of it that way. And it does go as 2*delta-R/Rearth

Not exactly.  The mass of the Earth is not actually concentrated at a
single point.

Outside of a uniform spherical mass, g does behave as if all the mass
were concentrated at the center. The earth is non-homogenous, but
close enough. The short answer to my little problem is that the change
is about 1 part in 2000, too small to matter in the context of the
other measurement uncertainties.

If something is small enough, we engineers just write it off. A quick,
rough calculation is usually enough to decide if that's safe.

After all the profound word salads and hand-waving about forces and
masses and weights, it was fun to see if the lecturers could do a
simple high-school physics exercize.

John

Surveyors and navigators ignore the gravimetric variations at their
own peril.

Actually, delta-g may be less than 1:2000. After all, Truckee isn't
floating on air, it's sitting on rock. It's sort of, not quite, like
being on a planet that's one mile bigger in radius.

This is one I *can't* do in my head.

John

I've got access to a huge map of gravitational anomalies-- If I
remember, I'll take a gander at Truckee vs. San Francisco next time I
get a chance.

I guess the question is whether g at the top of a mountain is greater
or less than g at sea level. A zillion web sites say less, and use the
1/r^2 equation to demonstrate it, but that equation works if you're in
a balloon, not sitting on solid rock.

It depends on how spikey the mountain is.

John- Hide quoted text -

- Show quoted text -

Mountains are the light pieces of earth floating on the magma
underneath.  Hey if we want to calculate the effect of the mountain...
well I'm a physicist... approximate the mountain as a sphere

I heard of this new'ish' gravimeter where they drop a corner cube
reflector as one arm of an interferometer and count fringes as it
drops... At least that's how I think it works.

George h.

Slick. You could digitize the photodiode signal and curve-fit the heck
out of it.

John- Hide quoted text -

- Show quoted text -

Yeah, I'll see if I can find a link.  (I saw this at an APS trade
show.)  I think the primary use is for miners mapping the local g
field to figure out where to dig.

George H.- Hide quoted text -

- Show quoted text -

http://principles.ou.edu/grav_ex/absolute.html

Pretty cool.

George H.

Neat. But at 350 kG, it must disturb the g field it's trying to
measure!

John

Oh come on JL. What you normally spout is tantamount to saying leverage
that distortion to your advantage.

 

[krw@att.bizzzzzzzzzzzz]

On Thu, 17 Jun 2010 18:02:17 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 19:59:22 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Thu, 17 Jun 2010 16:55:04 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 18:41:16 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Thu, 17 Jun 2010 16:36:54 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 17 Jun 2010 14:46:50 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 13:26:12 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

What matters is that 1 N of force accelerates 1 kg of mass by 1
m/sec^2.

Your problem is that you actually inferred that those numbers carried
through to any location on the planet, until you were proven (or shown)
that you were wrong. They do not. You lose... again.

Those numbers work anywhere in the solar system.

...by definition. ;-)

Except that he left part of the definition out. That being "in a
vacuum".

AlwaysWrong again lives up to his name.

So, said force against said mass in a slurry of pudding yields the same
rate? Nope.

AlwaysWrong, is *always* wrong.

Same thing goes for air, dipshit.

Always.

 

[krw@att.bizzzzzzzzzzzz]

On Thu, 17 Jun 2010 18:03:08 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 20:00:58 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Thu, 17 Jun 2010 16:59:41 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 16:46:27 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 17 Jun 2010 16:23:27 -0500, John Fields
jfields@austininstruments.com> wrote:

---
OK, but isn't it written:

G m1 m2
Fg = ---------
r˛

?

Some people might, but they have to define what they mean by "r" or
"d" or whatever they use. In Russian or Greek or Arabic or Japanese,
they might use different symbols. Apparently Newton used the Latin
text word for "distance", and expressed the force relation in prose.
Modern mathematical notation wasn't used in his day. He invented
calculus, but lots of people came along later and reformulated the way
it looks today. Same thing happened with Maxwell's equations: he
described them, but others put then into our modern forms.

m1
F=(G ------)m2
r2


Same thing.

Wrong again, AlwaysWrong. You really are a zero.

Dumbfuck, it is the same. Do you not understand transposition?

*ALWAYS* wrong. You mean "communicative" (you're wrong), but you'd still be
*WRONG*, AlwaysWrong. You really are innumerate. You can't even do *simple*
algebra.

 

[krw@att.bizzzzzzzzzzzz]

On Thu, 17 Jun 2010 21:22:54 -0400, "Michael A. Terrell"
<mike.terrell@earthlink.net> wrote:

"krw@att.bizzzzzzzzzzzz" wrote:

Michael A. Terrell wrote:

"krw@att.bizzzzzzzzzzzz" wrote:

Close, but AlwaysWrong has a fetish for feces. He's really into it.


Too much, apparently. They fired him from his job of mucking out
horse stalls.

The horses were embarrassed.


Horses know that there is no such thing as an amateur proctologist.

They knew it wasn't DimBulb's finger, too.

 

[Archimedes' Lever]

On Thu, 17 Jun 2010 22:31:20 -0500, "krw@att.bizzzzzzzzzzzz"
<krw@att.bizzzzzzzzzzzz> wrote:

On Thu, 17 Jun 2010 18:03:08 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 20:00:58 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Thu, 17 Jun 2010 16:59:41 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 16:46:27 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 17 Jun 2010 16:23:27 -0500, John Fields
jfields@austininstruments.com> wrote:

---
OK, but isn't it written:

G m1 m2
Fg = ---------
r˛

?

Some people might, but they have to define what they mean by "r" or
"d" or whatever they use. In Russian or Greek or Arabic or Japanese,
they might use different symbols. Apparently Newton used the Latin
text word for "distance", and expressed the force relation in prose.
Modern mathematical notation wasn't used in his day. He invented
calculus, but lots of people came along later and reformulated the way
it looks today. Same thing happened with Maxwell's equations: he
described them, but others put then into our modern forms.

m1
F=(G ------)m2
r˛


Same thing.

Wrong again, AlwaysWrong. You really are a zero.

Dumbfuck, it is the same. Do you not understand transposition?

*ALWAYS* wrong. You mean "communicative" (you're wrong), but you'd still be
*WRONG*, AlwaysWrong. You really are innumerate. You can't even do *simple*
algebra.

You're an idiot and it was quoted... from principia, idiot.

You want it all, so you can put your foot further into your mouth?

The equations are the same.

m1m2 m1
F = G ------ = (G ----) m2
r˛ r˛

They are the same, idiot.

 

[George Herold]

krw@att.bizzzzzzzzzzzz wrote:

On Thu, 17 Jun 2010 06:12:37 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Jun 17, 12:11�am, "k...@att.bizzzzzzzzzzzz"
k...@att.bizzzzzzzzzzzz> wrote:
On Wed, 16 Jun 2010 19:39:39 -0700 (PDT), George Herold





gher...@teachspin.com> wrote:
On Jun 16, 10:04�pm, Archimedes' Lever
OneBigLe...@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 18:42:40 -0700 (PDT), George Herold

gher...@teachspin.com> wrote:
On Jun 16, 7:11�pm, John Fields <jfie...@austininstruments..com> wrote:
On Wed, 16 Jun 2010 12:54:58 -0700 (PDT), George Herold

gher...@teachspin.com> wrote:
On Jun 16, 2:55�pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 11:31:53 -0700, Archimedes' Lever

OneBigLe...@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 11:05:15 -0700 (PDT), George Herold
gher...@teachspin.com> wrote:

Cool, I have to scribble numbers on the paper though. �6400 feet is
about 2000m, the Earth is about 6E6 m in radius, Since we only want a
small change I can ignore the r^2 stuff and just multiple the ratio by
2. �something like 4 parts out of 6,000. �much smaller than the
divisions on your scale.

�No you cannot. �What makes you think that G decreases (or
increases)linearly?

I doesn't!

John

for small enough changes it is linear!

---
Not if you take at least 3 samples.- Hide quoted text -

- Show quoted text -

What? �if the change is a part in 1,000, then the second order
correction is at the one out of 10^6 level. John L's going to stop 1/2
between SF and Truckee and weigh himself with six figure acurracy?

George H.

� You STILL assume a linear scale too. �Another mistake.- Hide quoted text -

- Show quoted text -

Dang, with changes at the 10^-3 level second order effect is at
10^-6. �Do you know about Taylor expansions? �Gravity is one of the
hardest things to measure. �I'm not sure we know the value of big G to
a part in 10^6.

Doesn't look like we're even close to 1:10^6:

http://www.search.com/reference/Gravitational_constant#Recent_measure....

ďż˝ "In the January 5, 2007 issue of Science (page 74), the report "Atom
ďż˝ Interferometer Measurement of the Newtonian Constant of Gravity" (J. B.
ďż˝ Fixler, G. T. Foster, J. M. McGuirk, and M. A. Kasevich) describes a new
ďż˝ measurement of the gravitational constant. According to the abstract: "Here,
� we report a value of G = 6.693 x 10�11 cubic meters per kilogram second
� squared, with a standard error of the mean of �0.027 x 10�11 and a
� systematic error of �0.021 x 10�11 cubic meters per kilogram second
ďż˝ squared.".["- Hide quoted text -

- Show quoted text -

Thanks, looks like a part in 10^4 or so. I'm not sure if that's the
best that's been done to date. I know people are working hard on
this.

Yeah, but I doubt they've bettered it by two orders of magnitude in three
years. It really is a tough nut.

Yeah some silicon sphere and trying to count the number of atoms.

We've got ideas for an improved Cavendish type balance, but I'm don't
think we'll ever try building it. Too many other more profitable
things to do.

1.) do an AC not DC measurement.

How do you vary G?

Oh, There is nothing really new in any of my ideas, Others have
thought of them much earlier. (Faller for one.) You use a torsional
oscillator and look at how the period changes when you move masses
closer or farther away.

2.) let the test mass move through the gravity mass, (like the problem
of dropping rocks down a hole that goes through the center of the
earth.)

Well there has to be a hole through the big mass and maybe also a slot
to let the oscillator 'arms' swing through.

...and we think BP has problems! ;-)

3.)...

Pretty much.

Oh I've got lots of other ideas I'd like to try. Here's my favorite
'crazy'. Float the whole movable section in a pail of water and over
top the edges of the pail (with out spilling any) so that the surface
tension keeps the whole floating part of the apparatus centered in the
pail.

George H.

 

[JosephKK]

On Wed, 16 Jun 2010 21:14:59 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 20:46:44 -0700,
"JosephKK"<quiettechblue@yahoo.com> wrote:

On Wed, 16 Jun 2010 10:29:12 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 15 Jun 2010 23:34:50 -0700,
"JosephKK"<quiettechblue@yahoo.com> wrote:

On Tue, 15 Jun 2010 00:31:35 -0500, John Fields
jfields@austininstruments.com> wrote:

On Mon, 14 Jun 2010 08:25:57 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Mon, 14 Jun 2010 07:23:14 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Mon, 14 Jun 2010 07:19:37 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

Fluid of course. Few people ever measure force. And most liquids used
in everydat life have a s.g. near 1, so an ounce of tabasco is
unambiguous.

Hundreds, even thousands of folks measure force every day, and many of
those use ounces in their scales of measure. Many use Newtons.


Of course hundreds, maybe even thousands of people measure force every
day. But there are 300 million people in the USA. Most people never
measure force; they do measure weight, or mass actually.

---
Since weight is mass multiplied by the acceleration of gravity and
most people use scales instead of beam balances and calibrated
reference masses to do the measurement, they measure weight, not mass.

http://en.wikipedia.org/wiki/Weighing_scale

Balances compare the gravitationally induced forces of test masses versus
reference masses. Please note that this assumes that the gravitational
field is reasonably uniform in terms of area included in the balance the
masses to be compared and the distances between them. And many scales
measure deflection of (more or less) well documented structures deflected
by the forces in an assumed uniform gravitational field. Compare the
usefulness of balance versus scales in any accelerated (including
rotated) frame of reference.

Words, words, words. Do the puzzler.

John

I am not here to play with your "puzzlers" JL. Post electronics with
correct circuits, values, and explanations instead please.

After you.

John

Oh great leader, i cannot let you lead from behind. Just the same, i
will post something obvious someday. Hold your breath if you wish.

 

[krw@att.bizzzzzzzzzzzz]

On Thu, 17 Jun 2010 20:38:53 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 22:31:20 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Thu, 17 Jun 2010 18:03:08 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 20:00:58 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Thu, 17 Jun 2010 16:59:41 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 16:46:27 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 17 Jun 2010 16:23:27 -0500, John Fields
jfields@austininstruments.com> wrote:

---
OK, but isn't it written:

G m1 m2
Fg = ---------
r˛

?

Some people might, but they have to define what they mean by "r" or
"d" or whatever they use. In Russian or Greek or Arabic or Japanese,
they might use different symbols. Apparently Newton used the Latin
text word for "distance", and expressed the force relation in prose.
Modern mathematical notation wasn't used in his day. He invented
calculus, but lots of people came along later and reformulated the way
it looks today. Same thing happened with Maxwell's equations: he
described them, but others put then into our modern forms.

m1
F=(G ------)m2
r˛


Same thing.

Wrong again, AlwaysWrong. You really are a zero.

Dumbfuck, it is the same. Do you not understand transposition?

*ALWAYS* wrong. You mean "communicative" (you're wrong), but you'd still be
*WRONG*, AlwaysWrong. You really are innumerate. You can't even do *simple*
algebra.

You're an idiot and it was quoted... from principia, idiot.

Wrong.

You want it all, so you can put your foot further into your mouth?

Not my foot, AlwaysWrong.

The equations are the same.

m1m2 m1
F = G ------ = (G ----) m2
r˛ r˛

That is *not* what you wrote, AlwaysWrong. ...and it's not "transposition",
so again you're wrong, as always.

They are the same, idiot.

Wrong, always.

 

[Archimedes' Lever]

On Thu, 17 Jun 2010 22:54:58 -0500, "krw@att.bizzzzzzzzzzzz"
<krw@att.bizzzzzzzzzzzz> wrote:

On Thu, 17 Jun 2010 20:38:53 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 22:31:20 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Thu, 17 Jun 2010 18:03:08 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 20:00:58 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Thu, 17 Jun 2010 16:59:41 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 16:46:27 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 17 Jun 2010 16:23:27 -0500, John Fields
jfields@austininstruments.com> wrote:

---
OK, but isn't it written:

G m1 m2
Fg = ---------
r˛

?

Some people might, but they have to define what they mean by "r" or
"d" or whatever they use. In Russian or Greek or Arabic or Japanese,
they might use different symbols. Apparently Newton used the Latin
text word for "distance", and expressed the force relation in prose.
Modern mathematical notation wasn't used in his day. He invented
calculus, but lots of people came along later and reformulated the way
it looks today. Same thing happened with Maxwell's equations: he
described them, but others put then into our modern forms.

m1
F=(G ------)m2
r˛


Same thing.

Wrong again, AlwaysWrong. You really are a zero.

Dumbfuck, it is the same. Do you not understand transposition?

*ALWAYS* wrong. You mean "communicative" (you're wrong), but you'd still be
*WRONG*, AlwaysWrong. You really are innumerate. You can't even do *simple*
algebra.

You're an idiot and it was quoted... from principia, idiot.

Wrong.

You want it all, so you can put your foot further into your mouth?

Not my foot, AlwaysWrong.

The equations are the same.

m1m2 m1
F = G ------ = (G ----) m2
r˛ r˛

That is *not* what you wrote,

You obviously cannot read. I said that I would put up the entire
equation to show you that they are the same.

I did not expect you to have enough brains to see it though.

AlwaysWrong. ...and it's not "transposition",
so again you're wrong, as always.

You wouldn't know.

They are the same, idiot.

Wrong, always.

Yes, you are, Williams.

 

[John Larkin]

On Thu, 17 Jun 2010 20:22:35 -0700,
"JosephKK"<quiettechblue@yahoo.com> wrote:

On Thu, 17 Jun 2010 08:40:07 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 17 Jun 2010 06:41:25 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Jun 17, 8:50 am, George Herold <gher...@teachspin.com> wrote:
On Jun 16, 11:38 pm, John Larkin





jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 19:52:34 -0700 (PDT), George Herold

gher...@teachspin.com> wrote:
On Jun 16, 10:17 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 22:03:46 -0400, Spehro Pefhany

speffS...@interlogDOTyou.knowwhat> wrote:
On Wed, 16 Jun 2010 17:23:13 -0700, the renowned John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 16:54:42 -0700 (PDT), Richard Henry
pomer...@hotmail.com> wrote:

On Jun 16, 3:58 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 15:36:13 -0700 (PDT), Richard Henry

pomer...@hotmail.com> wrote:
On Jun 16, 12:53 pm, George Herold <gher...@teachspin.com> wrote:
On Jun 16, 2:31 pm, Archimedes' Lever <OneBigLe...@InfiniteSeries.Org
wrote:

On Wed, 16 Jun 2010 11:05:15 -0700 (PDT), George Herold

gher...@teachspin.com> wrote:
Cool, I have to scribble numbers on the paper though. 6400 feet is
about 2000m, the Earth is about 6E6 m in radius, Since we only want a
small change I can ignore the r^2 stuff and just multiple the ratio by
2. something like 4 parts out of 6,000. much smaller than the
divisions on your scale.

No you cannot. What makes you think that G decreases (or
increases)linearly?

Big G doesn't change at all. Little g (the force of gravity on the
Earths surface) will go as 1/r^2. For small changes in r the change
is approximately linear... first term in the taylor expansion if you
want to think of it that way. And it does go as 2*delta-R/Rearth

Not exactly.  The mass of the Earth is not actually concentrated at a
single point.

Outside of a uniform spherical mass, g does behave as if all the mass
were concentrated at the center. The earth is non-homogenous, but
close enough. The short answer to my little problem is that the change
is about 1 part in 2000, too small to matter in the context of the
other measurement uncertainties.

If something is small enough, we engineers just write it off. A quick,
rough calculation is usually enough to decide if that's safe.

After all the profound word salads and hand-waving about forces and
masses and weights, it was fun to see if the lecturers could do a
simple high-school physics exercize.

John

Surveyors and navigators ignore the gravimetric variations at their
own peril.

Actually, delta-g may be less than 1:2000. After all, Truckee isn't
floating on air, it's sitting on rock. It's sort of, not quite, like
being on a planet that's one mile bigger in radius.

This is one I *can't* do in my head.

John

I've got access to a huge map of gravitational anomalies-- If I
remember, I'll take a gander at Truckee vs. San Francisco next time I
get a chance.

I guess the question is whether g at the top of a mountain is greater
or less than g at sea level. A zillion web sites say less, and use the
1/r^2 equation to demonstrate it, but that equation works if you're in
a balloon, not sitting on solid rock.

It depends on how spikey the mountain is.

John- Hide quoted text -

- Show quoted text -

Mountains are the light pieces of earth floating on the magma
underneath.  Hey if we want to calculate the effect of the mountain...
well I'm a physicist... approximate the mountain as a sphere

I heard of this new'ish' gravimeter where they drop a corner cube
reflector as one arm of an interferometer and count fringes as it
drops... At least that's how I think it works.

George h.

Slick. You could digitize the photodiode signal and curve-fit the heck
out of it.

John- Hide quoted text -

- Show quoted text -

Yeah, I'll see if I can find a link.  (I saw this at an APS trade
show.)  I think the primary use is for miners mapping the local g
field to figure out where to dig.

George H.- Hide quoted text -

- Show quoted text -

http://principles.ou.edu/grav_ex/absolute.html

Pretty cool.

George H.

Neat. But at 350 kG, it must disturb the g field it's trying to
measure!

John

Oh come on JL. What you normally spout is tantamount to saying leverage
that distortion to your advantage.

If that sentence made any sense, I'd know if you were trying to insult
me or not.

John

 

[John Larkin]

On Thu, 17 Jun 2010 20:49:11 -0700,
"JosephKK"<quiettechblue@yahoo.com> wrote:

On Wed, 16 Jun 2010 21:14:59 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 20:46:44 -0700,
"JosephKK"<quiettechblue@yahoo.com> wrote:

On Wed, 16 Jun 2010 10:29:12 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 15 Jun 2010 23:34:50 -0700,
"JosephKK"<quiettechblue@yahoo.com> wrote:

On Tue, 15 Jun 2010 00:31:35 -0500, John Fields
jfields@austininstruments.com> wrote:

On Mon, 14 Jun 2010 08:25:57 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Mon, 14 Jun 2010 07:23:14 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Mon, 14 Jun 2010 07:19:37 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

Fluid of course. Few people ever measure force. And most liquids used
in everydat life have a s.g. near 1, so an ounce of tabasco is
unambiguous.

Hundreds, even thousands of folks measure force every day, and many of
those use ounces in their scales of measure. Many use Newtons.


Of course hundreds, maybe even thousands of people measure force every
day. But there are 300 million people in the USA. Most people never
measure force; they do measure weight, or mass actually.

---
Since weight is mass multiplied by the acceleration of gravity and
most people use scales instead of beam balances and calibrated
reference masses to do the measurement, they measure weight, not mass.

http://en.wikipedia.org/wiki/Weighing_scale

Balances compare the gravitationally induced forces of test masses versus
reference masses. Please note that this assumes that the gravitational
field is reasonably uniform in terms of area included in the balance the
masses to be compared and the distances between them. And many scales
measure deflection of (more or less) well documented structures deflected
by the forces in an assumed uniform gravitational field. Compare the
usefulness of balance versus scales in any accelerated (including
rotated) frame of reference.

Words, words, words. Do the puzzler.

John

I am not here to play with your "puzzlers" JL. Post electronics with
correct circuits, values, and explanations instead please.

After you.

John

Oh great leader, i cannot let you lead from behind. Just the same, i
will post something obvious someday. Hold your breath if you wish.

Post circuits, or whine about circuits. We all have our functions in
life.

John

 

[krw@att.bizzzzzzzzzzzz]

On Thu, 17 Jun 2010 21:23:38 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 22:54:58 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Thu, 17 Jun 2010 20:38:53 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 22:31:20 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Thu, 17 Jun 2010 18:03:08 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 20:00:58 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Thu, 17 Jun 2010 16:59:41 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 16:46:27 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 17 Jun 2010 16:23:27 -0500, John Fields
jfields@austininstruments.com> wrote:

---
OK, but isn't it written:

G m1 m2
Fg = ---------
r˛

?

Some people might, but they have to define what they mean by "r" or
"d" or whatever they use. In Russian or Greek or Arabic or Japanese,
they might use different symbols. Apparently Newton used the Latin
text word for "distance", and expressed the force relation in prose.
Modern mathematical notation wasn't used in his day. He invented
calculus, but lots of people came along later and reformulated the way
it looks today. Same thing happened with Maxwell's equations: he
described them, but others put then into our modern forms.

m1
F=(G ------)m2
r˛


Same thing.

Wrong again, AlwaysWrong. You really are a zero.

Dumbfuck, it is the same. Do you not understand transposition?

*ALWAYS* wrong. You mean "communicative" (you're wrong), but you'd still be
*WRONG*, AlwaysWrong. You really are innumerate. You can't even do *simple*
algebra.

You're an idiot and it was quoted... from principia, idiot.

Wrong.

You want it all, so you can put your foot further into your mouth?

Not my foot, AlwaysWrong.

The equations are the same.

m1m2 m1
F = G ------ = (G ----) m2
r˛ r˛

That is *not* what you wrote,

You obviously cannot read. I said that I would put up the entire
equation to show you that they are the same.

You're illiterate as well as innumerate. That is obvious.

I did not expect you to have enough brains to see it though.

I wouldn't expect you to be able to read what I write (you are dumber than a
rock), but not being able to read what *you* write is really stupid.

AlwaysWrong. ...and it's not "transposition",
so again you're wrong, as always.

You wouldn't know.

I know you're always wrong, AlwaysWrong. You never let us down on that one.

They are the same, idiot.

Wrong, always.

Yes, you are, Williams.

Wrong again, AlwaysWrong.

 

[Greegor]

Archie > Most of us that have a modicum of
Archie > sense DO use it, so you are the one
Archie > that it decades behind.
Archie > I have been using it for decades.

When measuring your junk you use millimeters?

 

[Archimedes' Lever]

On Thu, 17 Jun 2010 23:13:25 -0700 (PDT), Greegor <greegor47@gmail.com>
wrote:

Archie > Most of us that have a modicum of
Archie > sense DO use it, so you are the one
Archie > that it decades behind.
Archie > I have been using it for decades.

When measuring your junk you use millimeters?

If you do not know how to quote in Usenet, you do not deserve a
response, you forum invading little bitch!

Now, IF you fucking EVER learn about this forum which you have invaded,
and you learn about POSTING CONVENTIONS, you might deserve a response.

I am not going to go hunting up whatever post you think you responded
to though. Get the fuck out of Usenet until you learn about it, you
little wussified invader.

 

[Archimedes' Lever]

On Thu, 17 Jun 2010 23:13:25 -0700 (PDT), Greegor <greegor47@gmail.com>
wrote:

When measuring your junk you use millimeters?

You are pathetic, and no longer deserve to be responded to.

 

[John Fields]

On Thu, 17 Jun 2010 16:46:27 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 17 Jun 2010 16:23:27 -0500, John Fields
jfields@austininstruments.com> wrote:

On Thu, 17 Jun 2010 15:32:07 -0400, Spehro Pefhany
speffSNIP@interlogDOTyou.knowwhat> wrote:

On Thu, 17 Jun 2010 13:20:41 -0500, John Fields
jfields@austininstruments.com> wrote:

On Thu, 17 Jun 2010 08:45:55 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

Isaac Newton demonstrated that gravitational attraction goes as 1/d^2.
It explains the orbits of the planets.

---
Yes, but I think he said 1/r^2.

In two spheres mutually gravitating each towards the other if the
matter in places on all sides round about and equi-distant from the
centres is similar, the weight of either sphere towards the other will
be reciprocally as the square of the distance between their centres.

-- Principia Book 1 "De motu corporum" by Newton
(translated into English by Motte, 1848)

---
OK, but isn't it written:

G m1 m2
Fg = ---------
r˛

?

Some people might, but they have to define what they mean by "r" or
"d" or whatever they use. In Russian or Greek or Arabic or Japanese,
they might use different symbols. Apparently Newton used the Latin
text word for "distance", and expressed the force relation in prose.
Modern mathematical notation wasn't used in his day. He invented
calculus, but lots of people came along later and reformulated the way
it looks today. Same thing happened with Maxwell's equations: he
described them, but others put then into our modern forms.

Good book about that: The Maxwellians, by Bruce Hunt.