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

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 -
Click to expand...
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.

George H.
 
On Wed, 16 Jun 2010 21:21:29 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

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

John

LOOK at the damned image, you stubborn fuckhead! It tells you exactly
what the differing densities are with respect to a homogenous 'norm'
reference.
Click to expand...
It's a cartoon. Nothing is exact.

John
 
On Mon, 14 Jun 2010 18:04:39 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Mon, 14 Jun 2010 17:57:06 -0700, VWWall <vwall@large.invalid
wrote:

John Larkin wrote:
On Mon, 14 Jun 2010 22:05:21 +0000 (UTC), Tim Watts <tw@dionic.net
wrote:

On Mon, 14 Jun 2010 07:31:13 -0700, StickThatInYourPipeAndSmokeIt
Zarathustra@thusspoke.org> wibbled:

At least we are not measuring things by 'curling stones' or the like.
It is still possible AFAIK here to go to a small time brewery and buy a
firkin (8 gallons) of beer. Or a barrel (4 firkins). If you're areal
pissartist, you'd probably want a hogshead, butt or tun though.

Bulk beer here comes in kegs and, for quiet get-togethers, half-kegs.

A keg is 15 gallons. Or maybe 10.


See: http://www.cockeyed.com/inside/keg/keg.html

We once did a bar-top taste test of Michelob vs Bud. Nobody could tell
the difference. All these A-B rice beers give me a headache.

John
Click to expand...
I don't understand how you cannot tell the difference. Setup proper
double blind tests and i bet i can get it right between the two of them
over 90 % of the time.
 
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 -
Click to expand...
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.
 
On Jun 16, 10:20 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 18:51:38 -0700, Archimedes' Lever





OneBigLe...@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 17:28:39 -0700, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 16:44:44 -0700, Archimedes' Lever
OneBigLe...@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 16:41:11 -0700, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 09:36:11 +0300, Paul Keinanen <keina...@sci.fi
wrote:

On Tue, 15 Jun 2010 19:58:20 -0500, "k...@att.bizzzzzzzzzzzz"
k...@att.bizzzzzzzzzzzz> wrote:

Then, by your "logic", "millimeter" is an Imperial term since
1mm = 0.03937"

No, because the inch is defined as being 25.4mm.  The metric measure is not a
derivative of the English.

In actuality, what makes the carat a metric term is that the weight of
gemstones is measured using the metric system and described in metric
units.

Imperial units are defined using the metric system.  Does that mean that the
US uses the metric system?

I people are so allergic about the term "metric", why not go directly
to the primary definitions ?

The meter was previously defined as 1,650,763.73 krypton-86
wavelengths, thus 1 inch = 41,929.398,742 wavelengths.

I wonder if they actually counted the 1,650,763.73 fringes. I sure
hope that did it twice.

I'm impressed that the krypton line is narrow enough to have a meter
of coherence length. I think the measurement was made pre-laser.

John

 Yes, John.  Krypton based atomic clocks were around before the advent
of the laser in 1960.

http://en.wikipedia.org/wiki/Atomic_clock#History

Looks like you're wrong again. Why do you say stuff like this when you
could check google, like everybody else?

John

 Laser:  1960

Cesium atomic clock:  1955

 You lose. again.

Krypton atomic clock?

John- Hide quoted text -

- Show quoted text -
Click to expand...
That's how all supermen tell time.

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

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.

John
 
On Jun 16, 10:32 pm, Archimedes' Lever
<OneBigLe...@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 19:17:38 -0700, John Larkin

jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
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.

  You are an idiot.



It depends on how spikey the mountain is.

  No, it does not.

http://en.wikipedia.org/wiki/File:Earth-G-force.png
Click to expand...
How 'bout if I buried a buried a 10 meter diameter sphere of gold
right under John's house. Would that change the local g force?

If the answer is yes, can you estimate how much?

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

You want to measure weights.
Click to expand...
No, I do not. Johnny does.
 
"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.
Click to expand...

Not even in his hooker heels.


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

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.
Click to expand...

His fig cookies are good, too! ;-)


--
Anyone wanting to run for any political office in the US should have to
have a DD214, and a honorable discharge.
 
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.
Click to expand...
You retards still want to average everything. Too bad reality
disagrees.

There are SEVERAL density changes, you stupid fucktards.
 
On Wed, 16 Jun 2010 21:42:25 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 21:21:29 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

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

John

LOOK at the damned image, you stubborn fuckhead! It tells you exactly
what the differing densities are with respect to a homogenous 'norm'
reference.


It's a cartoon.
Click to expand...

You can't get anything right. It is not a cartoon. You now show us
that you have no clue what that is either.

Nothing is exact.
Click to expand...
Certainly not you... ever.
 
On Wed, 16 Jun 2010 21:46:21 -0700, "JosephKK"<quiettechblue@yahoo.com>
wrote:

On Mon, 14 Jun 2010 18:04:39 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Mon, 14 Jun 2010 17:57:06 -0700, VWWall <vwall@large.invalid
wrote:

John Larkin wrote:
On Mon, 14 Jun 2010 22:05:21 +0000 (UTC), Tim Watts <tw@dionic.net
wrote:

On Mon, 14 Jun 2010 07:31:13 -0700, StickThatInYourPipeAndSmokeIt
Zarathustra@thusspoke.org> wibbled:

At least we are not measuring things by 'curling stones' or the like.
It is still possible AFAIK here to go to a small time brewery and buy a
firkin (8 gallons) of beer. Or a barrel (4 firkins). If you're areal
pissartist, you'd probably want a hogshead, butt or tun though.

Bulk beer here comes in kegs and, for quiet get-togethers, half-kegs.

A keg is 15 gallons. Or maybe 10.


See: http://www.cockeyed.com/inside/keg/keg.html

We once did a bar-top taste test of Michelob vs Bud. Nobody could tell
the difference. All these A-B rice beers give me a headache.

John

I don't understand how you cannot tell the difference. Setup proper
double blind tests and i bet i can get it right between the two of them
over 90 % of the time.
Click to expand...
Any idiots that cannot tell rice from barley should all get together
and do a purple tennis shoe kool-aid party together. Take your kids with
you. and their kids too, if any.
 
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.
Click to expand...
Bullshit. That would be a black hole.
 
On Jun 16, 9:30 pm, "k...@att.bizzzzzzzzzzzz"
<k...@att.bizzzzzzzzzzzz> wrote:
On Wed, 16 Jun 2010 21:12:03 -0700, "JosephKK"<quiettechb...@yahoo.com> wrote:
On Mon, 14 Jun 2010 22:30:21 +0000 (UTC), Tim Watts <t...@dionic.net
wrote:

On Mon, 14 Jun 2010 15:17:41 -0700, VWWall <vw...@large.invalid> wibbled:

Tim Watts wrote:
On Mon, 14 Jun 2010 07:31:13 -0700, StickThatInYourPipeAndSmokeIt
Zarathus...@thusspoke.org> wibbled:

  At least we are not measuring things by 'curling stones' or the
  like.

It is still possible AFAIK here to go to a small time brewery and buy a
firkin (8 gallons) of beer. Or a barrel (4 firkins). If you're  areal
pissartist, you'd probably want a hogshead, butt or tun though.

And the answer to: "Are you getting any lately?" has always been given
in furlongs per fortnight!

Incidentally, that brings up a question:

When you americans talk of "a butt load of <blah>" - does that derive
from the "butt" as in 144 gallons? I always mentally associated it with
butt=ass - but the former makes more sense...

That sounds like a meaning shift due to forgetting larger units of
measure in place of more familiar ones (55 gal barrel).

You mean like the 40gal oil barrel?
Click to expand...
42 gallons.
 
On Wed, 16 Jun 2010 09:44:56 -0700 Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote in Message id:
<6ivh161qv9sfp326srpqo1aet9mt7bcikj@4ax.com>:

On Wed, 16 Jun 2010 03:58:23 -0700 (PDT), Proteus IIV
proteusiiv@gmail.com> wrote:

I AM PROTEUS

Snipped retarded, Usenet abusing post, made by a google groups
subscriber agreement abusing dumbass.

One would think that some little retard on a Mac would be able to find
and use the caps lock key.

One would also think that most folks that have seen your pathetic crap
up in the engineering group would be offended by seeing you stalking
folks down in this group.

One would think that they would file a battery of complaints against
your pathetic ass, akin to the one I just filed on you.
Click to expand...
Tsk, tsk... Looks like sig material!

From: Archimedes' Lever <OneBigLever@InfiniteSeries.Org>
Message-ID: <cd0i16p041nsps2onevsi0r0d8hoanpe1v@4ax.com>
"Netkkkop wanna be dumbfucks should be in EVERYONE's kill file.
You epitomize someone that should be ignored."

"One would think that they would file a battery of complaints against
your pathetic ass, akin to the one I just filed on you."
Netkkkop wanna be dumbfuck Archimedes' Lever steps on his crank in
Message-ID: <6ivh161qv9sfp326srpqo1aet9mt7bcikj@4ax.com>

One would have to also think that google groups would have a modicum of
civility and would shitcan your pathetic, forum abusing ass.

Continue your abuse and stalking behavior and you will find yourself in
a world of legal shit, boy.
Click to expand...
Oh FFS, grow up and wipe the snot off your nose, AlwaysWrong.
 
On Thu, 17 Jun 2010 06:24:51 -0400, JW <none@dev.null> wrote:

Oh FFS,
Click to expand...

JW, eat shit and die. Fuck off and die.

Hell, you retarded fuck... just die already!
You goddamned immature, retarded bastard.
Your mother should be in prison for not flushing you.
 
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?
Click to expand...
---
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.
 
On Wed, 16 Jun 2010 19:29:11 -0700, Archimedes' Lever wrote:

On Wed, 16 Jun 2010 21:11:39 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:


You're always wrong, AlwaysWrong. ...and that wasn't the question, so you're
wrong again. Of course none of this surprises anyone, AlwaysWrong.

A balance will read the same regardless of what sized spheroid it is
placed onto.

Wrong, and what does a "spheroid" have to do with anything, ALwaysWrong. You
out of straw again?

The Earth is a spheroid, idiot. Spheroids have gravity.
Click to expand...

ALL masses have gravity.


A spring scale's reading is gravity dependent.

It will read differently here than on the Moon, a smaller spheroid.
Click to expand...
A smaller mass.

A balance will not. It will read the same in both places.
Click to expand...
Only if the sample and mass standard both have the same density, See my
earlier post regarding buoyancy. The moon has negligible atmosphere.


--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)
 
On Wed, 16 Jun 2010 16:29:27 -0700, Archimedes' Lever wrote:

A balance would yield the same mass reading at both locations.
Click to expand...
The locations are at different elevations, hence air density will be different.
Buoyancy effects need to be considered.

"If a substance is weighed in air, and found to be balanced by mass
standards ("weights") of value M, the true mass of the substance is

M + Mda [(1/d) - (1/ds)]

where d is the density of the substance, ds is the density of the mass
standards, and da is the density of the air."

(Kaye & Laby, "Tables of physical and chemical constants", Fifteenth
Edition.)

--
"For a successful technology, reality must take precedence
over public relations, for nature cannot be fooled."
(Richard Feynman)
 
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