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

[George Herold]

On Jun 16, 3:05 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 11:05:15 -0700 (PDT), George Herold





gher...@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.



Moon?  forget it... though I'll have to think a bit to put some sort
of number on the force...

240,000 miles is 60 earth radii. G falls as r^2, so 3600. And the moon
is small and light. Again, way down in the noise.

People who can't do this should sell shoes for a living.

John- Hide quoted text -

- Show quoted text -

Yeah I just don't have numbers for the earth moon distance and such in
my head.... 6 x 10^6 meters is in my head... (I don't know why? I
TAed freshman physics many moons ago... I'm afraid I learned a lot
more than the students.)

George H.

I was thinking I might be able to 'get there' by knowing the period...
But anyway.

 

[Archimedes' Lever]

On Wed, 16 Jun 2010 14:45:56 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 16:20:52 -0500, John Fields
jfields@austininstruments.com> wrote:

---
I don't think he likes anything but showing off and trying to make
himself look smart by trying to make others look stupid.

I was trying to figure out who here actually understands a little
physics. Now I know.

John

It is obvious that you are in fact, quite clueless.

 

[George Herold]

On Jun 16, 4:01 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.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!

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- Hide quoted text -

- Show quoted text -

Exactly, you put the same factor of two in your calculation.

George h.

 

[John Larkin]

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.

John

 

[John Fields]

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.

 

[krw@att.bizzzzzzzzzzzz]

On Wed, 16 Jun 2010 09:50:31 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 06:29:56 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

Does your scale report your weight in newtons?

John

No "scale" 'reports weight in Newtons'.

AlwaysWrong is *ALWAYS* wrong.

http://www.oldwillknottscales.com/ohaus/pull-type-scale-large.jpg

A force gauge 'reports' applied force in Newtons.

A force gauge *IS* a scale, AlwaysWrong.

Said force
application must be pushed against or pulled away from the calibrated
node of the measuring device.

Nodes are not calibrated, AlwaysWrong. Springs are calibrated.

Said force gauge is NOT a "scale".

You're *always* wrong, ALwaysWrong. Why is that?

 

[John Larkin]

On Wed, 16 Jun 2010 15:13:07 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 09:31:57 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:



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?

To start with, NONE of these three "questions" calls for any "number"
to be the answer.

The first one sure did.

But for you, I'll say that the amount is less than a 0.01% difference.

It's around 1 part in 2000, about 0.05%. But I bet you looked it up
somewhere.

John

 

[John Fields]

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!

---
Not if you take at least 3 samples.

 

[John Larkin]

On Wed, 16 Jun 2010 17:59:49 -0500, "krw@att.bizzzzzzzzzzzz"
<krw@att.bizzzzzzzzzzzz> wrote:

On Wed, 16 Jun 2010 09:50:31 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 06:29:56 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

Does your scale report your weight in newtons?

John

No "scale" 'reports weight in Newtons'.

AlwaysWrong is *ALWAYS* wrong.

http://www.oldwillknottscales.com/ohaus/pull-type-scale-large.jpg


A force gauge 'reports' applied force in Newtons.

A force gauge *IS* a scale, AlwaysWrong.

Said force
application must be pushed against or pulled away from the calibrated
node of the measuring device.

Nodes are not calibrated, AlwaysWrong. Springs are calibrated.

Said force gauge is NOT a "scale".

You're *always* wrong, ALwaysWrong. Why is that?

Must be some really strange mental model of reality.

John

 

[Archimedes' Lever]

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

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

You never reached your goal, nor did you meet it yourself.

And your assessments of others aptitude have about as much credence as
Casey Anthony's mother telling folks she had nothing to do with her
death.

 

[Archimedes' Lever]

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.

 

[Archimedes' Lever]

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.

It is still a GUAGE, however, and NOT a scale.

 

[Archimedes' Lever]

On Wed, 16 Jun 2010 17:59:49 -0500, "krw@att.bizzzzzzzzzzzz"
<krw@att.bizzzzzzzzzzzz> wrote:

Said force
application must be pushed against or pulled away from the calibrated
node of the measuring device.

Nodes are not calibrated, AlwaysWrong. Springs are calibrated.

You do not understand physics terms. I'll use a lay term that might
allow you to grasp it. "pressure point".

The gauge is calibrated to read the force or pressure applied against
the test node or applied to it such that it pulls the test node away from
the gauge instrument.

I have one at work that I use to test RF shielding honeycomb
installation integrity with using a Teflon test node that was supplied by
the gauge manufacturer and referred to in their manual.

So much for proving again just how little you know about it.

Also, this force gauge does not use a spring.

 

[John Larkin]

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.

John

 

[Archimedes' Lever]

On Wed, 16 Jun 2010 16:03:22 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

It's around 1 part in 2000, about 0.05%. But I bet you looked it up
somewhere.

John

You'd be wrong. Both about the figure you gave, and about mine being
wrong, and about how I derived it.

 

[krw@att.bizzzzzzzzzzzz]

On Wed, 16 Jun 2010 09:31:57 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 07:45:34 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 15 Jun 2010 19:42:04 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 15 Jun 2010 13:44:39 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 15 Jun 2010 07:00:03 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.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



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. ;-)

 

[Archimedes' Lever]

On Wed, 16 Jun 2010 18:23:16 -0500, "krw@att.bizzzzzzzzzzzz"
<krw@att.bizzzzzzzzzzzz> wrote:

On Wed, 16 Jun 2010 09:31:57 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 07:45:34 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 15 Jun 2010 19:42:04 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 15 Jun 2010 13:44:39 -0500, John Fields
jfields@austininstruments.com> wrote:

On Tue, 15 Jun 2010 07:00:03 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.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



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. ;-)

You forgot about the difference in gravity. 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.

A balance would yield the same mass reading at both locations.

 

[krw@att.bizzzzzzzzzzzz]

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 ?

Besides the point, but why increase complexity and cost if it's not needed?
The imperial system works just fine for 99% of the people. For the other 1%,
it's simple enough to do the conversion. The metric system isn't magic. Even
the metric system has a bunch of numbers that have to be remembered.

The meter was previously defined as 1,650,763.73 krypton-86

See.

wavelengths, thus 1 inch = 41,929.398,742 wavelengths.

Currently 1 m is defined as the distance the light propagates in
1/299,792,458 seconds. Light propagates 1 inch in
254/2,997,924,580,000 seconds, which can further be reduced to
127/1,498,962,290,000 seconds.

Sure, but I have a hard time reading inverse seconds on my tape measure.
....and counting that high.

 

[krw@att.bizzzzzzzzzzzz]

On Wed, 16 Jun 2010 07:35:02 -0700, Fred Abse <excretatauris@invalid.invalid>
wrote:

On Mon, 14 Jun 2010 22:01:51 +0000, Tim Watts wrote:

SI rules for science, imperial rules for everyday living. Except
carpentry - I hate fractions!

You would rather multiply a piece of timber by 0.5, than cut it in half?

;-)

If you multiply the timer you'll tend to make it too long, which is easier to
correct for than cutting it too short. "I cut that board three times and it's
*still* too short."

 

[krw@att.bizzzzzzzzzzzz]

On Tue, 15 Jun 2010 21:12:41 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Tue, 15 Jun 2010 22:25:01 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Tue, 15 Jun 2010 18:28:34 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Tue, 15 Jun 2010 20:14:30 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

IOW, you don't know. We know you're always wrong, AlwaysWrong. You don't
have to prove it with every post.


I have framed 50 more Joe Plumber El Cheapo pre-fab houses than you
have and ten more full custom rich fucker houses than you. I also have
done some drywall, and you are again, wrong, as usual.

What a dummy, DimBulb. YOU are AlwaysWrong, not me! A liar, too, but that's
been obvious, too, for years.


Are you saying that I never worked as a carpenter, framing houses?

I'm saying that you're a liar, ALwaysWrong. You claim to have done everything
conceivable, yet are *always* wrong, about *everything*. One way or another,
you're a damned liar.