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

[John Larkin]

On Wed, 16 Jun 2010 11:31:53 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

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

 

[Archimedes' Lever]

On Wed, 16 Jun 2010 11:55:29 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 11:31:53 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

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

It does vary, just not linearly.

 

[John Larkin]

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.

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

 

[John Larkin]

On Wed, 16 Jun 2010 11:08:09 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 10:25:36 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 10:02:17 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

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

1. About how much has my measured weight changed due to the change of
G with altitude?

Depends on the scale, and its calibrated range of operation, and the
atmospheric pressure.

The scale I gave you a picture of (link) shows you exactly what the
difference is in G.

Time's up. You blew it.

John


Sad thing is that you have obviously blown something else that most
real men possess. Something that it is now clear that you'll never
recover from. In that area, you are simply irrecoverably 'hard wired
dumb'.

Quit flailing. Real man can do math.

John

 

[Archimedes' Lever]

On Wed, 16 Jun 2010 11:55:29 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 11:31:53 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

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

An apple dropped from 120,000 ft and read as it passes through 119,000
ft will fall ever so slightly slower than an apple dropped from 1000 ft
above the dead sea shore.

For the purpose of the claim, if you wish, you can assume the falls
occur inside tubes containing no gas (vacuum).

 

[Archimedes' Lever]

On 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.

John, you are one lame fucktard.

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.

No, it isn't, idiot.

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

Better get on Monster. Beat feet, dumbfuck.

 

[John Larkin]

On Wed, 16 Jun 2010 12:10:14 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 11:55:29 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 11:31:53 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

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


An apple dropped from 120,000 ft and read as it passes through 119,000
ft will fall ever so slightly slower than an apple dropped from 1000 ft
above the dead sea shore.

How much slower?

John

For the purpose of the claim, if you wish, you can assume the falls
occur inside tubes containing no gas (vacuum).

 

[ihis groE]

Yes folks, the US will never go metric because we stuck our foot in
our mouths.

Why the U.S. should use the metric system.. See link below

http://www.wimp.com/metricsystem/

 

[John Fields]

On Wed, 16 Jun 2010 10:40:57 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 12:23:29 -0500, John Fields
jfields@austininstruments.com> 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?

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?

---
That's too much like work; I decline.

Around our shop, we do stuff like this all the time, calculate corner
frequencies or power dissipations or noise levels or loop stabilities,
in our heads, in seconds, standing in a corridor or at a whiteboard.
Rough calculations like this should be second nature to engineers.

---
Oh, well...
---

See Williams, 1991, p 295, or google lightning empiricism

---
That's too much like work; I decline.
---

There's a Feynman story somewhere about this, too.

---
And???

Johm Fields

 

[John Larkin]

On Wed, 16 Jun 2010 14:31:44 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Wed, 16 Jun 2010 10:40:57 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 12:23:29 -0500, John Fields
jfields@austininstruments.com> 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?

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?

---
That's too much like work; I decline.

Around our shop, we do stuff like this all the time, calculate corner
frequencies or power dissipations or noise levels or loop stabilities,
in our heads, in seconds, standing in a corridor or at a whiteboard.
Rough calculations like this should be second nature to engineers.

---
Oh, well...
---

See Williams, 1991, p 295, or google lightning empiricism

---
That's too much like work; I decline.
---

There's a Feynman story somewhere about this, too.

---
And???

Johm Fields

To paraphrase, from memory: Some guys at some science facility
challenged everyone to come up with a problem that they couldn't solve
in a minute, within 5%. Feynman thought about it for a second and said
"the tangent of ten to the 40th power."

John

 

[John Larkin]

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

 

[George Herold]

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.

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

George H.

Opps just thinking... is my minute up? The r^2 is on the bottom I
think I screwed up the factor of two... (now I do have to write
things.)

G.

 

[Spehro Pefhany]

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.

 

[John Larkin]

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.

John

 

[John Fields]

On Wed, 16 Jun 2010 11:06:27 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 12:23:29 -0500, John Fields
jfields@austininstruments.com> wrote:

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

That's too much like work; I decline.

I did it. Just the way he said, and I even did it before I read his
rules.

He may not like the answer though.

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

 

[John Larkin]

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

On Wed, 16 Jun 2010 11:06:27 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 12:23:29 -0500, John Fields
jfields@austininstruments.com> wrote:

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

That's too much like work; I decline.

I did it. Just the way he said, and I even did it before I read his
rules.

He may not like the answer though.

---
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

 

[George Herold]

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

George H.

 

[George Herold]

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!

George H.

 

[Archimedes' Lever]

On Wed, 16 Jun 2010 12:53:26 -0700 (PDT), George Herold
<gherold@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

George H.

http://en.wikipedia.org/wiki/File:Earth-G-force.png

 

[Archimedes' Lever]

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.

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

2. Is this significant to the measurement?

That depends on what you are acquiring the measurement for. Again, no
numbers needed.

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?

Yes, but also only by a small amount. The position of the sun also
plays in. And again, no number needed, nor are YOU even intelligent
enough to apply them correctly.