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

[John Larkin]

On Thu, 17 Jun 2010 15:20:12 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Thu, 17 Jun 2010 12:20:39 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Thu, 17 Jun 2010 11:26:07 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:

On Jun 17, 1:06 pm, Archimedes' Lever <OneBigLe...@InfiniteSeries.Org
wrote:
On Thu, 17 Jun 2010 09:08:10 -0700 (PDT), George Herold

gher...@teachspin.com> wrote:
we can treat it as linear
and get the right number.

  No.  You get NEAR the right number. You no longer get a "reliable"
reading, regardless of the number the instrument reports.

I'm done with you.

Go pester some other news group.

George H.

I suppose at some point it becomes obvious that some posters aren't
just nasty ignorant SOBs, they are in fact mentally deficient for one
reason or another. If we met them in person, it would be obvious. On
usenet, it's harder to tell.

But time wounds all heels, so eventually your true colors (or lack of
them) surface.

And the more time you spend angry, the more your immune system will
stay revved up, and the faster you'll age. Really.

John

 

[John Fields]

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:

On Thu, 17 Jun 2010 09:26:30 -0500, John Fields
jfields@austininstruments.com> wrote:

On Wed, 16 Jun 2010 18:42:40 -0700 (PDT), George Herold
gherold@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?

---
Probably not, but that's not what I had in mind.

Here: Let's say that we have an exquisitely sensitive gravimeter
located on the surface of the earth, somewhere, and that we measure
and record the stregth of the gravitational field (g1) there.

Then, let's locate the gravimeter one meter above its starting
position and take another reading (g2) there.

Now, lets subtract g2 from g1 and call the difference d1.

Finally, let's take another reading (g3) 2 meters above the starting
point, subtract g3 from g2 and call that difference d2.

Voila! We now notice that since d2 isn't equal to d1, the rate of
change of the strength of the gravitational field with respect to
distance isn't constant, so it isn't linear.

Now, using only two readings, we can't really infer much about the
field, but by using three or more we can, and we can make the
distances between those points smaller and smaller until the
gravimeter runs into its noise floor.

With superconducting gravimeters being capable of measuring down to
1 nanogal, and the Earth's gradient being around 3.1 ľgal/meter, one
would think that any nonlinearity would be able to be easily seen with
3100 measuring points available per meter.

The fly in the ointment is that the gradient decreases from 3.083
ľgal/m at sea level to 2.638 ľgal/m at 340 km, which is a change in
gradient of 345 ľgal/340km, or 1.014 ľgal/km.

That means that even with the best gravimeters available, the
measurement points would have to be at least 1 km apart to detect the
nonlinearity.

Tethered balloon?

No need, and the good news is that somebody's already measured the
gradient, found it changed with equal increments of altitude, and
therefore proved it's not linear.



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˛

?

 

[Archimedes' Lever]

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

I know what mass means, bacause it's an SI unit. Weight means whatever
you care to define it to mean.

But, it is NOT, nor has it ever been, what you have been defining it to
mean.

 

[Archimedes' Lever]

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.

 

[Archimedes' Lever]

On Thu, 17 Jun 2010 13:28:55 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

Dang, you beat me to it. My copy is at home.

John

But you failed to grasp any of it.

 

[Archimedes' Lever]

On Thu, 17 Jun 2010 13:34:31 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

And the more time you spend angry, the more your immune system will
stay revved up, and the faster you'll age. Really.

John

Your problem is that you think that your assessments of anger in others
are accurate.

You fail at so many things, what makes you think that your
psychological assessments have ANY merit what-so-fucking-ever?

If your claims of anger in others are accurate, then our claims about
errors from you, and your inability to handle those events are also
accurate.

What is good for the goose is good for the gander, John.

And your heart attack will come soon enough, and the stressed you
induced upon yourself to cause it will not be apparent to you until that
moment.

Your demeanor foments a dislike for you. I am on that bandwagon, and I
dislike you more than the average Joe here.

Why? Because your pathetic immaturity tells me that you never got that
ass kicking that you needed so badly when you were younger. Instead, you
are hard wired asshole. It will get you soon enough.

That sting you feel is not pride, boy. It is shame.

 

[Archimedes' Lever]

On Thu, 17 Jun 2010 14:06:35 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

A big gooey oil slick spoiling our beautiful
threads.

This is why YOUR immature, retarded ass should be ignored by all.

You just cannot keep from poking others.

Shame. In the old west, I would poke you back, and you wouldn't be
getting back up. Then, I'd go have a drink with the local sheriff, and
he'd thank me for getting rid of yet another asshole in town..

 

[Sum Ting Wong]

On Thu, 17 Jun 2010 14:06:35 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Jun 17, 3:20 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Thu, 17 Jun 2010 11:26:07 -0700 (PDT), George Herold





gher...@teachspin.com> wrote:
On Jun 17, 1:06 pm, Archimedes' Lever <OneBigLe...@InfiniteSeries.Org
wrote:
On Thu, 17 Jun 2010 09:08:10 -0700 (PDT), George Herold

gher...@teachspin.com> wrote:
we can treat it as linear
and get the right number.

No. You get NEAR the right number. You no longer get a "reliable"
reading, regardless of the number the instrument reports.

I'm done with you.

Go pester some other news group.

George H.

I suppose at some point it becomes obvious that some posters aren't
just nasty ignorant SOBs, they are in fact mentally deficient for one
reason or another. If we met them in person, it would be obvious. On
usenet, it's harder to tell.

John- Hide quoted text -

- Show quoted text -

Oh I like discussing things with people, And honestly I'm thrilled
when I'm wrong, 'cause it means that I've learned something. But A*
Lever seems like a lost cause. I tend not to read any threads when I
see his name attached... and lately it seems like he's all over
everything on SED. A big gooey oil slick spoiling our beautiful
threads.

George H.

* A does not stand for Archimedes.

If it was "ass" lever, as you infer, you would already be gone from the
group, because my lever would have already pried your retarded ass OUT of
the group.

After all, that is what an ASS lever is for, ass.

 

[Archimedes' Lever]

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:

On Thu, 17 Jun 2010 09:26:30 -0500, John Fields
jfields@austininstruments.com> wrote:

On Wed, 16 Jun 2010 18:42:40 -0700 (PDT), George Herold
gherold@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?

---
Probably not, but that's not what I had in mind.

Here: Let's say that we have an exquisitely sensitive gravimeter
located on the surface of the earth, somewhere, and that we measure
and record the stregth of the gravitational field (g1) there.

Then, let's locate the gravimeter one meter above its starting
position and take another reading (g2) there.

Now, lets subtract g2 from g1 and call the difference d1.

Finally, let's take another reading (g3) 2 meters above the starting
point, subtract g3 from g2 and call that difference d2.

Voila! We now notice that since d2 isn't equal to d1, the rate of
change of the strength of the gravitational field with respect to
distance isn't constant, so it isn't linear.

Now, using only two readings, we can't really infer much about the
field, but by using three or more we can, and we can make the
distances between those points smaller and smaller until the
gravimeter runs into its noise floor.

With superconducting gravimeters being capable of measuring down to
1 nanogal, and the Earth's gradient being around 3.1 ľgal/meter, one
would think that any nonlinearity would be able to be easily seen with
3100 measuring points available per meter.

The fly in the ointment is that the gradient decreases from 3.083
ľgal/m at sea level to 2.638 ľgal/m at 340 km, which is a change in
gradient of 345 ľgal/340km, or 1.014 ľgal/km.

That means that even with the best gravimeters available, the
measurement points would have to be at least 1 km apart to detect the
nonlinearity.

Tethered balloon?

No need, and the good news is that somebody's already measured the
gradient, found it changed with equal increments of altitude, and
therefore proved it's not linear.



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˛

?
Good job.

 

[Sum Ting Wong]

On Thu, 17 Jun 2010 14:06:35 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Jun 17, 3:20 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Thu, 17 Jun 2010 11:26:07 -0700 (PDT), George Herold





gher...@teachspin.com> wrote:
On Jun 17, 1:06 pm, Archimedes' Lever <OneBigLe...@InfiniteSeries.Org
wrote:
On Thu, 17 Jun 2010 09:08:10 -0700 (PDT), George Herold

gher...@teachspin.com> wrote:
we can treat it as linear
and get the right number.

No. You get NEAR the right number. You no longer get a "reliable"
reading, regardless of the number the instrument reports.

I'm done with you.

Go pester some other news group.

George H.

I suppose at some point it becomes obvious that some posters aren't
just nasty ignorant SOBs, they are in fact mentally deficient for one
reason or another. If we met them in person, it would be obvious. On
usenet, it's harder to tell.

John- Hide quoted text -

- Show quoted text -

Oh I like discussing things with people, And honestly I'm thrilled
when I'm wrong, 'cause it means that I've learned something.

Yet you never once addressed anything I ever stated. Is that fear,
asshole? Or sociocentristic horseshit?

But A*
Lever seems like a lost cause. I tend not to read any threads when I
see his name attached...

Then your claim about liking to discuss things was an outright lie,
fucktard.

and lately it seems like he's all over
everything on SED.

I have been on SED for over a decade. Likely longer than you.

 

[Spehro Pefhany]

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:

On Thu, 17 Jun 2010 09:26:30 -0500, John Fields
jfields@austininstruments.com> wrote:

On Wed, 16 Jun 2010 18:42:40 -0700 (PDT), George Herold
gherold@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?

---
Probably not, but that's not what I had in mind.

Here: Let's say that we have an exquisitely sensitive gravimeter
located on the surface of the earth, somewhere, and that we measure
and record the stregth of the gravitational field (g1) there.

Then, let's locate the gravimeter one meter above its starting
position and take another reading (g2) there.

Now, lets subtract g2 from g1 and call the difference d1.

Finally, let's take another reading (g3) 2 meters above the starting
point, subtract g3 from g2 and call that difference d2.

Voila! We now notice that since d2 isn't equal to d1, the rate of
change of the strength of the gravitational field with respect to
distance isn't constant, so it isn't linear.

Now, using only two readings, we can't really infer much about the
field, but by using three or more we can, and we can make the
distances between those points smaller and smaller until the
gravimeter runs into its noise floor.

With superconducting gravimeters being capable of measuring down to
1 nanogal, and the Earth's gradient being around 3.1 ľgal/meter, one
would think that any nonlinearity would be able to be easily seen with
3100 measuring points available per meter.

The fly in the ointment is that the gradient decreases from 3.083
ľgal/m at sea level to 2.638 ľgal/m at 340 km, which is a change in
gradient of 345 ľgal/340km, or 1.014 ľgal/km.

That means that even with the best gravimeters available, the
measurement points would have to be at least 1 km apart to detect the
nonlinearity.

Tethered balloon?

No need, and the good news is that somebody's already measured the
gradient, found it changed with equal increments of altitude, and
therefore proved it's not linear.



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˛

?

You're arguing over whether Newton used r or d to represent the
distance between the centres of two spheres in his ca. 1687 text?

I'll let you learned gentlemen consult your original Latin texts and
apply erudite analysis, I'm afraid this is getting a little too deep
for me.

 

[John Fields]

On Thu, 17 Jun 2010 18:10:45 -0400, Spehro Pefhany
<speffSNIP@interlogDOTyou.knowwhat> 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˛

?


You're arguing over whether Newton used r or d to represent the
distance between the centres of two spheres in his ca. 1687 text?

I'll let you learned gentlemen consult your original Latin texts and
apply erudite analysis, I'm afraid this is getting a little too deep
for me.

---
Funny!

It's the radius of either of two intersecting circles drawn from the
centers of the spheres with the centers of the spheres lying on the
circumferences of the intersecting circles.

 

[krw@att.bizzzzzzzzzzzz]

On Wed, 16 Jun 2010 22:05:56 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 23:22:42 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:


You want to measure weights.

No, I do not. Johnny does.

You certainly do go ramble on, unintelligibly, about something you don't want,
AlwaysWrong.

 

[krw@att.bizzzzzzzzzzzz]

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...

  "In the January 5, 2007 issue of Science (page 74), the report "Atom
  Interferometer Measurement of the Newtonian Constant of Gravity" (J. B.
  Fixler, G. T. Foster, J. M. McGuirk, and M. A. Kasevich) describes a new
  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 ą0.027 x 10–11 and a
  systematic error of ą0.021 x 10–11 cubic meters per kilogram second
  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.

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?

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.)

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

3.)...

Pretty much.

 

[krw@att.bizzzzzzzzzzzz]

On Thu, 17 Jun 2010 21:33:10 +0200, Jeroen Belleman <jeroen@nospam.please>
wrote:

On 06/17/2010 06:12 AM, krw@att.bizzzzzzzzzzzz wrote:
On Wed, 16 Jun 2010 19:15:04 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 21:08:31 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Wed, 16 Jun 2010 18:50:06 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 19:26:58 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

Is that true in the "near field"? Integrate the volume of the earth over the
square of the distances and I'd expect to see that not all points affect
gravity the same. IOW, I'd expect the apparent "center" of the Earth to be
somewhat closer than it really is.

You are so goddamned retarded that you cannot even follow a thread. It
was posted three times already, dumbass.

http://en.wikipedia.org/wiki/File:Earth-G-force.png

More irrelevance from DimBulb.


It is absolutely relevant.

You're full of shit. ...but you like it that way.

Don't wrestle a pig. You'll get covered in mud and the pig likes it.

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

 

[krw@att.bizzzzzzzzzzzz]

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

On Wed, 16 Jun 2010 23:29:21 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Wed, 16 Jun 2010 21:14:15 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 19:26:58 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Wed, 16 Jun 2010 15:58:30 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 15:36:13 -0700 (PDT), Richard Henry
pomerado@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.

Is that true in the "near field"? Integrate the volume of the earth over the
square of the distances and I'd expect to see that not all points affect
gravity the same. IOW, I'd expect the apparent "center" of the Earth to be
somewhat closer than it really is.

If the density of a sphere is uniform, and the observer is anywhere
outside of the sphere, gravity acts the same as if all the mass were
concentreted in the center. Newton said that.

Certainly inside the sphere that's true, so in the limiting case it must be
too. It still seems that for the case outside the sphere, if you integrate
the contribution from all points those closer to the object should contribute
more.

Turns out that, seen from outside a uniform-density sphere, the sphere
acts as if all its mass was concentrated at the center.

I hear you, just trying to get my head around the calculus.

An observer inside a uniform sphere, at distance r from the center,
observes the sphere's gravitation as if he were on the surface of a
sphere of radius r. In other words, everything farther away from the
center than the observer sort of vanishes.

Newton figured both of those out. Smart guy.

No question. IIRC, he had to invent calculus first.

 

[krw@att.bizzzzzzzzzzzz]

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.

 

[krw@att.bizzzzzzzzzzzz]

On Wed, 16 Jun 2010 22:24:42 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 23:29:21 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:


Of course, earth is denser deep inside, so it's not exactly the same.
But the mass is concentrated towards the center, all that iron, so it
must be close.

Sure, only DimBulb was hung up on that.

You retards still want to average everything. Too bad reality
disagrees.

There are SEVERAL density changes, you stupid fucktards.

Irrelevant, as always. Go back to mommy's hamper, AlwaysWrong.

 

[krw@att.bizzzzzzzzzzzz]

On Wed, 16 Jun 2010 21:26:19 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 23:15:55 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

A balance will not. It will read the same in both places.

You assume a uniform gravity field. Raise the G high enough or dimensions
small enough and it won't.


Wrong. The balance measures force applied against two arms. It is not
dependent on the gravity field it is in. One kg on left, one kg on right.
Same reading everywhere. Same force applied to balance point.

You're illiterate, as always. It certainly does depend on a uniform gravity
field.

Same measure on Charon as on Jupiter. The spring scale, placed on IO,
however, will vary so widely and so much that you could probably watch it
as the moon spins and Jupiter rises. More proof for Johnny that the Moon
does have an effect on a spring scale, but not on a balance.

Try reading, AlwaysWrong.

Pretty much basic physics, Williams.

Pretty much primary school literacy, AlwaysWrong.

 

[krw@att.bizzzzzzzzzzzz]

On Thu, 17 Jun 2010 08:01:44 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Thu, 17 Jun 2010 07:50:12 -0700, Fred Abse
excretatauris@invalid.invalid> wrote:

Only if the sample and mass standard both have the same density, See my
earlier post regarding buoyancy. The moon has negligible atmosphere.



Only on a spheroid that also has an atmosphere of gas.

s/gas/fluid, AlwaysWrong.

Practically immeasurable. Especially compared to a spring scale and the
error that they carry.

Always wrong.