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

[George Herold]

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.

 

[Spehro Pefhany]

On Wed, 16 Jun 2010 22:03:46 -0400, Spehro Pefhany
<speffSNIP@interlogDOTyou.knowwhat> wrote:

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

On Wed, 16 Jun 2010 16:54:42 -0700 (PDT), Richard Henry
pomerado@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.

Can't relate exactly to the locations, but it looks like the anomalies
amount to no more than +/- 50mGal.

The problem with interpreting those numbers is understanding what
those anomalies are relative to- underlying model is the geoid plus
corrections such as the Bouguer gradient correction (which assumes
rock beneath the measuring point). A little much to delve into right
now, I'll see if I can bend the ear of a real geophysist some time..

 

[George Herold]

On Jun 16, 11:45 pm, Archimedes' Lever
<OneBigLe...@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 20:01:14 -0700 (PDT), George Herold





gher...@teachspin.com> wrote:
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

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?

  The answer is no.- Hide quoted text -

- Show quoted text -

It's amazing! (Hey if he is so perfectly wrong could we use him to
pick stocks on the market.)

Hey A. Lever, should I buy or sell BP stock?

George H.

 

[Archimedes' Lever]

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

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.


Small, but measureable.

http://en.wikipedia.org/wiki/File:GRACE_globe_animation.gif

 

[Archimedes' Lever]

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.

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

 

[Archimedes' Lever]

On Thu, 17 Jun 2010 10:51:20 -0400, Spehro Pefhany
<speffSNIP@interlogDOTyou.knowwhat> wrote:

On Wed, 16 Jun 2010 22:03:46 -0400, Spehro Pefhany
speffSNIP@interlogDOTyou.knowwhat> wrote:

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

On Wed, 16 Jun 2010 16:54:42 -0700 (PDT), Richard Henry
pomerado@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.

Can't relate exactly to the locations, but it looks like the anomalies
amount to no more than +/- 50mGal.

The problem with interpreting those numbers is understanding what
those anomalies are relative to- underlying model is the geoid plus
corrections such as the Bouguer gradient correction (which assumes
rock beneath the measuring point). A little much to delve into right
now, I'll see if I can bend the ear of a real geophysist some time..

http://en.wikipedia.org/wiki/Petr_Van%C3%AD%C4%8Dek

 

[George Herold]

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.

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.
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.)
3.)...

George H.

 

[John Larkin]

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

 

[George Herold]

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.

 

[John Larkin]

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.

John

 

[Archimedes' Lever]

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

I think that you still don't get it.

 

[Archimedes' Lever]

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.

John

Yes, the laws between large separated bodies is normalized and has no
geoidal variations.

Down here on the surface, however, there are differences. Differences
that Newton probably had no knowledge or concept of.

 

[John Larkin]

On Thu, 17 Jun 2010 08:50:01 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> 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.

John

Yes, the laws between large separated bodies is normalized and has no
geoidal variations.

What does that mean?

Down here on the surface, however, there are differences. Differences
that Newton probably had no knowledge or concept of.

Uh, excuse me, the law of gravitation is universal. The only changes
since Newton has been the addition of relativistic effects and the
finite speed of gravity.

John

 

[VWWall]

Spehro Pefhany wrote:

On Wed, 16 Jun 2010 22:03:46 -0400, Spehro Pefhany
speffSNIP@interlogDOTyou.knowwhat> wrote:

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.

Can't relate exactly to the locations, but it looks like the anomalies
amount to no more than +/- 50mGal.

The problem with interpreting those numbers is understanding what
those anomalies are relative to- underlying model is the geoid plus
corrections such as the Bouguer gradient correction (which assumes
rock beneath the measuring point). A little much to delve into right
now, I'll see if I can bend the ear of a real geophysist some time..

In the mid 1960s, I was involved with the Initial Defense Communication

Satellite Program, which launched communications satellites, eight at a
time. These were is a slightly sub-synchronous orbit and had no
"station keeping" ability. The measurement of their orbits gave some
interesting information about gravitational variation.

Of course the orbital data had to be corrected for the sun/moon
influence. One of the outcomes was the fact that the assumed position
of some of the tracking stations had to be corrected.

The existing GPS system requires very accurate knowledge of the orbital
positions of the orbiting satellites as well as precise time.

I find it interesting that we find so much information about our Earth
by observing happenings in space.

--
Virg Wall, P.E.

 

[Archimedes' Lever]

On Thu, 17 Jun 2010 08:58:12 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

Down here on the surface, however, there are differences. Differences
that Newton probably had no knowledge or concept of.

Uh, excuse me, the law of gravitation is universal. The only changes
since Newton has been the addition of relativistic effects and the
finite speed of gravity.

John

Are you saying that Newton was aware of geoidal variations?

He was not, and obviously neither are you.

 

[Archimedes' Lever]

On Thu, 17 Jun 2010 09:08:10 -0700 (PDT), George Herold
<gherold@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.

 

[Archimedes' Lever]

On Thu, 17 Jun 2010 09:35:57 -0700, VWWall <vwall@large.invalid> wrote:

I find it interesting that we find so much information about our Earth
by observing happenings in space.

But we will never learn anything from observing the huge, vast, empty
space between John Larkin's ears.

 

[John Larkin]

On Wed, 16 Jun 2010 22:08:32 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Tue, 15 Jun 2010 14:36:50 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 15 Jun 2010 13:48:18 -0700 (PDT), Rich Grise on Google groups
richardgrise@yahoo.com> wrote:

On Jun 14, 10:09 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Mon, 14 Jun 2010 09:38:18 -0700, Archimedes' Lever

 It is an applied force, regardless of how you attack my grammatical
error in describing it.

Torque is not force. The units are different.

Well, mass isn't weight either, but people use them interchangeably.

"Weight" isn't clear.

---
Nonsense.

Weight is the product of mass and the force exerted on it by
gravitational acceleration.
---

Often it means mass.

---
More nonsense.

It never means mass, since mass is simply the measure of inertial
resistance.
---

Mass (kg) and force
(newtons) are formal SI things; "weight" isn't.

---
Even more nonsense.

Weight is derived from the relationship between mass and force, both
of which are prime, so your disingenuous attempt at bastardizing
weight by denying it other than prime status, for your own ends is, at
the very least, despicable.

But then, that's how you work, isn't it?

I work in SI units. Mass is kilograms, force is newtons, and I don't
have to worry about what "weight" might mean to various people.

John

 

[John Larkin]

On Wed, 16 Jun 2010 21:58:44 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Wed, 16 Jun 2010 16:22:13 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 17:58:51 -0500, John Fields
jfields@austininstruments.com> wrote:

On Wed, 16 Jun 2010 14:06:39 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 16:44:17 -0400, Spehro Pefhany
speffSNIP@interlogDOTyou.knowwhat> wrote:

On Wed, 16 Jun 2010 13:01:06 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 12:54:58 -0700 (PDT), George Herold
gherold@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!


Well, you can say that about most anything.

But for g, sure.

2 * 2 = 4

1.1 * 1.1 = 1.21

1.01 * 1.01 = 1.0201

1.001 * 1.001 = 1.002001

and like that.

John

Vertical gravity gradient is around 3000 Eotvos (3E-6 sec^-2) at sea
level, and it's not very sensitive to altitude. 6.4 k feet is about
2km, so around 0.06% less acceleration. Less than a slice of
cheesecake. That's assuming a non-rotating uniform spherical earth
with no atmosphere.

Do we need to take the delta of your buoyancy into account at this
level of precision? Assume, say, 2.5 or 3 cubic feet for your volume.
;-) I'm guessing yes.

About 2.4, by my calcs.

---
Seems small for someone with such a fat head.

You dislike me, so you want to prove me wrong at every opportunity.

---
No, I don't.

I'm not interested in _proving_ you wrong, all I'm interested in is
showing you your errors and letting you prove to yourself that you
did, in fact, make an error.

Words, words. "Error" means bad numbers.

The reason I dislike you is because you're disingenuous and because
you won't admit to error when it's pointed out to you even if it's as
plain as the nose on your face that you made an error.

Again, a heap of words, arguing about definitions. Most every time
you've actually done a simulation, or presented real numbers (like the
recent newton/kg thing) you've been wrong. Why don't you check your
work?

John

 

[Archimedes' Lever]

On Thu, 17 Jun 2010 10:33:10 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 22:08:32 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 15 Jun 2010 14:36:50 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 15 Jun 2010 13:48:18 -0700 (PDT), Rich Grise on Google groups
richardgrise@yahoo.com> wrote:

On Jun 14, 10:09 am, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Mon, 14 Jun 2010 09:38:18 -0700, Archimedes' Lever

 It is an applied force, regardless of how you attack my grammatical
error in describing it.

Torque is not force. The units are different.

Well, mass isn't weight either, but people use them interchangeably.

"Weight" isn't clear.

---
Nonsense.

Weight is the product of mass and the force exerted on it by
gravitational acceleration.
---

Often it means mass.

---
More nonsense.

It never means mass, since mass is simply the measure of inertial
resistance.
---

Mass (kg) and force
(newtons) are formal SI things; "weight" isn't.

---
Even more nonsense.

Weight is derived from the relationship between mass and force, both
of which are prime, so your disingenuous attempt at bastardizing
weight by denying it other than prime status, for your own ends is, at
the very least, despicable.

But then, that's how you work, isn't it?

I work in SI units. Mass is kilograms, force is newtons, and I don't
have to worry about what "weight" might mean to various people.

John

You do when you declare your spring tension based bathroom scale to be
measuring mass simply because of the unit utilized on the scale.

The printed 'scale' on a scale does not determine what it measures. The
type of scale does.