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Surveyors and navigators ignore the gravimetric variations at theirOn 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
What a surprise, you're wrong again, AlwaysWrong. Not that it matters.On Wed, 16 Jun 2010 19:15:29 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 16 Jun 2010 16:15:35 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 17:59:49 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
No "scale" 'reports weight in Newtons'.
AlwaysWrong is *ALWAYS* wrong.
http://www.oldwillknottscales.com/ohaus/pull-type-scale-large.jpg
It is relative, and un-calibrated, idiot.
That's why it's marked in Newtons (and pounds). Why are you *ALWAYS* wrong,
AlwaysWrong?
Having a graduated scale does NOT connote being calibrated or even being
able to be calibrated.
However, a FORCE GAUGE is NOT a scale.On Wed, 16 Jun 2010 16:17:12 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 17:59:49 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
A force gauge 'reports' applied force in Newtons.
A force gauge *IS* a scale, AlwaysWrong.
No. A force gauge IS a gauge, and the analog models CARRIED a scale on
them that covered the range of operation they were made for.
WTF are you yammering on about. Talking about word salad, hold any
understanding.
It is still a GUAGE, however, and NOT a scale.
A scale *IS* a gauge. It gauges weight, AlwaysWrong. What a putz!
krw rears his retarded head again.Wrong, ALwaysWrong. A "pressure point" is something Spock grabs on the TeeVee
show.
Sure. Any change in the mass of my body (add oxygen, release moistureOn Wed, 16 Jun 2010 12:05:00 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 11:05:15 -0700 (PDT), George Herold
gherold@teachspin.com> wrote:
On Jun 16, 12:31 pm, 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?
2. Is this significant to the measurement?
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.
Extra credit, one more minute:
3. Is the position of the moon significant to the measurement?
John- Hide quoted text -
- Show quoted text -
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.
My thinking was...
Delta one mile out of 4000 is 1 part in 4000. Account for the r^2
thing and you get 1 part in 2000. My weight, or the accuracy of the
scale, will change a lot more than that in three hours, even if I
don't stop to pee. So the change in altitude is way down in the
measurement noise.
Jeez. This idiot is tossing things into this like 'his weight' change
due to evaporation or other retarded parameter.
Wrong. A fishing scale measures weight as calibrated at a given,Also called a SCALE, AlwaysWrong. These things are commonly known as "fishing
scales".
You are so goddamned retarded that you cannot even follow a thread. ItIs 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.
Laser: 1960On 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
Even among spring scales, there are no truly linear springs for theYou also forgot that said
difference makes for a different offset depending on the scale mechanism
utilized, since we are talking about weight measure and weight measuring
scales.
You're always wrong, AlwaysWrong.
A balance will read the same regardless of what sized spheroid it isA balance would yield the same mass reading at both locations.
You're always wrong, AlwaysWrong. ...and that wasn't the question, so you're
wrong again. Of course none of this surprises anyone, AlwaysWrong.
YOUR weight change does NOT "affect the scale", idiot. It affects thethat affect the scale,
I've got access to a huge map of gravitational anomalies-- If IOn 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
---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.
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.
You STILL assume a linear scale too. Another mistake.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.
A scale is a force gauge, AlwaysWrong.On Wed, 16 Jun 2010 19:17:25 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Wed, 16 Jun 2010 16:17:12 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 17:59:49 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
A force gauge 'reports' applied force in Newtons.
A force gauge *IS* a scale, AlwaysWrong.
No. A force gauge IS a gauge, and the analog models CARRIED a scale on
them that covered the range of operation they were made for.
WTF are you yammering on about. Talking about word salad, hold any
understanding.
It is still a GUAGE, however, and NOT a scale.
A scale *IS* a gauge. It gauges weight, AlwaysWrong. What a putz!
However, a FORCE GAUGE is NOT a scale.
You can find a length scale, too, AlwaysWrong. So what?You can FIND a WEIGHT SCALE. You can find a MASS SCALE
Irrelevant, AlwaysWrong.The weight scale measures the FORCE applied against a spring, and is
proportional to a mass reading, but is not exact.
If the force is G, it is a scale, AlwaysWrong. That doesn't change the *fact*A FORCE GAUGE measure the force applied against or pulled away from a
test node attached to the GAUGE.
No, AlwaysWrong, you won't.Maybe you'll eventually get it.
Just pointing out your imbecility, Nymbecile.On Wed, 16 Jun 2010 19:21:28 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
Wrong, ALwaysWrong. A "pressure point" is something Spock grabs on the TeeVee
show.
krw rears his retarded head again.
AlwaysWrong loves his strawmen.On Wed, 16 Jun 2010 19:21:28 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
Also called a SCALE, AlwaysWrong. These things are commonly known as "fishing
scales".
Wrong. A fishing scale measures weight as calibrated at a given,
specific altitude.
AlwaysWrong really is stupid, huh?!It reads specifically the force applied to the spring
within it and for weight measure it uses gravity as the attractor that it
was calibrated against. It cannot be used to accurately measure force
except in the axis perpendicular to the planet surface.
AlwaysWrong is always *so* wrong, and irrelevant.A true force gauge does not use gravity at all, and measures the force
applied to it in any plane, and includes features with which to zero the
range scale at the time the measurement is being taken.
More irrelevance from DimBulb.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
Irrelevant and wrong, AlwaysWrong.On Wed, 16 Jun 2010 19:29:53 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
You also forgot that said
difference makes for a different offset depending on the scale mechanism
utilized, since we are talking about weight measure and weight measuring
scales.
You're always wrong, AlwaysWrong.
Even among spring scales, there are no truly linear springs for the
purpose. That poses a problem for reading between two different scales
that were calibrated at one location and then read at another.
LOL, it shows how stupid you really are, AlwaysWrong.Shows just how little you know about it.
Wrong, and what does a "spheroid" have to do with anything, ALwaysWrong. YouA balance would yield the same mass reading at both locations.
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.
Irrelevant, as you, ALwaysWrong.Kilogram reference of one side, kilogram test mass on the other.
Regardles of location, the scale reads EXACTLY the same.
That's just so wrong in *so* many ways, AlwaysWrong.As usual, John Larkin, you are WRONG AGAIN.
Stop in Sacramento? No way!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 think it operates accurately by some other means?On Wed, 16 Jun 2010 18:48:05 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 19:21:28 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
Also called a SCALE, AlwaysWrong. These things are commonly known as "fishing
scales".
Wrong. A fishing scale measures weight as calibrated at a given,
specific altitude.
AlwaysWrong loves his strawmen.
You think that a fish scale will read properly sideways?It reads specifically the force applied to the spring
within it and for weight measure it uses gravity as the attractor that it
was calibrated against. It cannot be used to accurately measure force
except in the axis perpendicular to the planet surface.
AlwaysWrong really is stupid, huh?!
You are the one that said that a force gauge was a scale, and thenA true force gauge does not use gravity at all, and measures the force
applied to it in any plane, and includes features with which to zero the
range scale at the time the measurement is being taken.
AlwaysWrong is always *so* wrong, and irrelevant.