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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.
I guess the question is whether g at the top of a mountain is greaterOn 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.
Krypton atomic clock?On Wed, 16 Jun 2010 17:28:39 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 16:44:44 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 16:41:11 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 09:36:11 +0300, Paul Keinanen <keinanen@sci.fi
wrote:
On Tue, 15 Jun 2010 19:58:20 -0500, "krw@att.bizzzzzzzzzzzz"
krw@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.
The Earth is a spheroid, idiot. Spheroids have gravity.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 fact that you say "the question is" tells us that you do not know.I guess the question is whether g at the top of a mountain is greater
or less than g at sea level.
You are an idiot.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.
No, it does not.It depends on how spikey the mountain is.
Krypton length standard.Krypton atomic clock?
John
---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.
---But for Pete's sake, check your work before you post, and quit
lecturing on subjects you don't understand.
---It does your mission no
good when you post stuff that's flat wrong.
---That's just plain good engineering: check your work.
What does that cartoon have to say about the top-of-mountain issue?On Wed, 16 Jun 2010 19:17:38 -0700, John Larkin
jjlarkin@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
A genuine quote:On Wed, 16 Jun 2010 19:20:38 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
Krypton atomic clock?
John
Krypton length standard.
- AlwaysWrong, 2010Yes, John. Krypton based atomic clocks were around before the advent
of the laser in 1960.
---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.
---Often it means mass.
---Mass (kg) and force
(newtons) are formal SI things; "weight" isn't.
I was doing a 'back of the envelope' calculation. The bouancy effectOn 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.- Hide quoted text -
- Show quoted text -
Slick. You could digitize the photodiode signal and curve-fit the heckOn 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.
Though it is not something a bathroom scale can reliably detect, seriousOn 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 something is small enough, we engineers just write it off. A quick,
rough calculation is usually enough to decide if that's safe.
2kM is 4000 is small enough for me to write off by inspection. A pit stop on
the way swamps these numbers.
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.
Some have, some haven't. ;-)
What? if the change is a part in 1,000, then the second orderOn 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 -
The answer is no.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?
I am not here to play with your "puzzlers" JL. Post electronics withOn Tue, 15 Jun 2010 23:34:50 -0700,
"JosephKK"<quiettechblue@yahoo.com> wrote:
On Tue, 15 Jun 2010 00:31:35 -0500, John Fields
jfields@austininstruments.com> wrote:
On Mon, 14 Jun 2010 08:25:57 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Mon, 14 Jun 2010 07:23:14 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:
On Mon, 14 Jun 2010 07:19:37 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
Fluid of course. Few people ever measure force. And most liquids used
in everydat life have a s.g. near 1, so an ounce of tabasco is
unambiguous.
Hundreds, even thousands of folks measure force every day, and many of
those use ounces in their scales of measure. Many use Newtons.
Of course hundreds, maybe even thousands of people measure force every
day. But there are 300 million people in the USA. Most people never
measure force; they do measure weight, or mass actually.
---
Since weight is mass multiplied by the acceleration of gravity and
most people use scales instead of beam balances and calibrated
reference masses to do the measurement, they measure weight, not mass.
http://en.wikipedia.org/wiki/Weighing_scale
Balances compare the gravitationally induced forces of test masses versus
reference masses. Please note that this assumes that the gravitational
field is reasonably uniform in terms of area included in the balance the
masses to be compared and the distances between them. And many scales
measure deflection of (more or less) well documented structures deflected
by the forces in an assumed uniform gravitational field. Compare the
usefulness of balance versus scales in any accelerated (including
rotated) frame of reference.
Words, words, words. Do the puzzler.
John
Cool, that's less than 1/2 the gravity effect.. but still veryOn Wed, 16 Jun 2010 09:31:57 -0700, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 07:45:34 -0500, John Fields
jfie...@austininstruments.com> wrote:
On Tue, 15 Jun 2010 19:42:04 -0700, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 15 Jun 2010 13:44:39 -0500, John Fields
jfie...@austininstruments.com> wrote:
On Tue, 15 Jun 2010 07:00:03 -0700, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Tue, 15 Jun 2010 00:31:35 -0500, John Fields
jfie...@austininstruments.com> wrote:
On Mon, 14 Jun 2010 08:25:57 -0700, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Mon, 14 Jun 2010 07:23:14 -0700, Archimedes' Lever
OneBigLe...@InfiniteSeries.Org> wrote:
On Mon, 14 Jun 2010 07:19:37 -0700, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
Fluid of course. Few people ever measure force. And most liquids used
in everydat life have a s.g. near 1, so an ounce of tabasco is
unambiguous.
Hundreds, even thousands of folks measure force every day, and many of
those use ounces in their scales of measure. Many use Newtons.
Of course hundreds, maybe even thousands of people measure force every
day. But there are 300 million people in the USA. Most people never
measure force; they do measure weight, or mass actually.
---
Since weight is mass multiplied by the acceleration of gravity and
most people use scales instead of beam balances and calibrated
reference masses to do the measurement, they measure weight, not mass.
http://en.wikipedia.org/wiki/Weighing_scale
Most people in the world use SI units, and they weigh things in
kilograms. A kg is a unit of mass.
Whether they use springs or balance beams or load cells, the reported
result is mass. kg, not newtons.
---
Sorry, but no.
The result of the measurement is caused by a force acting on a mass,
the product of which is called a "newton" if the mass is 1kg and the
force is the attraction due to gravity, 9.8m/s .
Entirely wrong:
http://en.wikipedia.org/wiki/Newton_%28unit%29
---
Yup.
I got the mass wrong, (it should be about 102 grams) but the fact
still remains that what a scale does is measure weight, not mass.
OK, today's puzzler:
Suppose I weigh myself at home, using my ordinary spring-based
bathroom scale. Home is 365 feet above sea level. Now I drive to
Truckee; it takes about 3 hours if I push it, 80+ MPH except for the
speed trap at Clipper Gap. When I arrive I use the same scale to weigh
myself, now at 6400 feet. Latitude is about the same.
1. About how much has my measured weight changed due to the change of
G with altitude?
A body will more at 6400ft than at 365feet, assuming no pit stops or coffee
breaks. Assuming a body weighing 200lbs and is mostly made up of water,
that's about three cubic feet, so you're displacing about 3cu. ft. of air.. Air
is about .075 lbs/ft^3 at MSL and about .06lbs per cubic foot at 6000ft, for a
difference of .015 lbs/ft*3. You'll be .045 lbs heavier at 6400 ft. Assuming
of course that you didn't stop for coffee, a potty break, or sweat too hard
(or breathe) getting there.
Now, gravity...
2. Is this significant to the measurement?
Define "significant". No, because you likely did breathe.
Rules: you have one minute to deliver an answer. Use no paper, pencils
or equivalent, calculators, computers, books, or any external
assistance or references of any kind. Keep your eyes closed. Do it
entirely in your head.
Oh, I did have to look up the density of air.
Extra credit, one more minute:
3. Is the position of the moon significant to the measurement?
Nope. No air on the moon. ;-)- Hide quoted text -
- Show quoted text -
Read it.On Wed, 16 Jun 2010 19:32:18 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 19:17:38 -0700, John Larkin
jjlarkin@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
What does that cartoon have to say about the top-of-mountain issue?
John
A genuine dope. Yes, you are.A genuine