J
John Larkin
Guest
On Wed, 16 Jun 2010 11:31:53 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:
John
<OneBigLever@InfiniteSeries.Org> wrote:
I doesn't!
John
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A
Archimedes' Lever
Guest
On Wed, 16 Jun 2010 11:55:29 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
<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.
J
John Larkin
Guest
On Wed, 16 Jun 2010 11:05:15 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:
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.
is small and light. Again, way down in the noise.
People who can't do this should sell shoes for a living.
John
<gherold@teachspin.com> wrote:
My thinking was...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.
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.
240,000 miles is 60 earth radii. G falls as r^2, so 3600. And the moonMoon? forget it... though I'll have to think a bit to put some sort
of number on the force...
is small and light. Again, way down in the noise.
People who can't do this should sell shoes for a living.
John
J
John Larkin
Guest
On Wed, 16 Jun 2010 11:08:09 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:
John
<OneBigLever@InfiniteSeries.Org> wrote:
Quit flailing. Real man can do math.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'.
John
A
Archimedes' Lever
Guest
On Wed, 16 Jun 2010 11:55:29 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
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).
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
An apple dropped from 120,000 ft and read as it passes through 119,000On 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
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).
A
Archimedes' Lever
Guest
On Wed, 16 Jun 2010 12:05:00 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
due to evaporation or other retarded parameter.
John, you are one lame fucktard.
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
Jeez. This idiot is tossing things into this like 'his weight' changeOn 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.
due to evaporation or other retarded parameter.
John, you are one lame fucktard.
No, it isn't, idiot.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.
Better get on Monster. Beat feet, dumbfuck.People who can't do this should sell shoes for a living.
J
John Larkin
Guest
On Wed, 16 Jun 2010 12:10:14 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:
John
<OneBigLever@InfiniteSeries.Org> wrote:
How much slower?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.
John
For the purpose of the claim, if you wish, you can assume the falls
occur inside tubes containing no gas (vacuum).
I
ihis groE
Guest
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/
J
John Fields
Guest
On Wed, 16 Jun 2010 10:40:57 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
Oh, well...
---
That's too much like work; I decline.
---
And???
Johm Fields
<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
J
John Larkin
Guest
On Wed, 16 Jun 2010 14:31:44 -0500, John Fields
<jfields@austininstruments.com> wrote:
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
<jfields@austininstruments.com> wrote:
To paraphrase, from memory: Some guys at some science facilityOn 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
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
J
John Larkin
Guest
On Wed, 16 Jun 2010 12:54:58 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:
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
<gherold@teachspin.com> wrote:
Well, you can say that about most anything.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!
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
G
George Herold
Guest
On Jun 16, 12:31 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
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.
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
Cool, I have to scribble numbers on the paper though. 6400 feet isOn 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 -
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.
S
Spehro Pefhany
Guest
On Wed, 16 Jun 2010 13:01:06 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
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.
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
Vertical gravity gradient is around 3000 Eotvos (3E-6 sec^-2) at seaOn 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
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.
J
John Larkin
Guest
On Wed, 16 Jun 2010 16:44:17 -0400, Spehro Pefhany
<speffSNIP@interlogDOTyou.knowwhat> wrote:
John
<speffSNIP@interlogDOTyou.knowwhat> wrote:
About 2.4, by my calcs.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
J
John Fields
Guest
On Wed, 16 Jun 2010 11:06:27 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> 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.
<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.
J
John Larkin
Guest
On Wed, 16 Jun 2010 16:20:52 -0500, John Fields
<jfields@austininstruments.com> wrote:
physics. Now I know.
John
<jfields@austininstruments.com> wrote:
I was trying to figure out who here actually understands a littleOn 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.
physics. Now I know.
John
G
George Herold
Guest
On Jun 16, 2:31 pm, Archimedes' Lever <OneBigLe...@InfiniteSeries.Org>
wrote:
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.
wrote:
Big G doesn't change at all. Little g (the force of gravity on theOn 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?
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.
G
George Herold
Guest
On Jun 16, 2:55 pm, John Larkin
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
George H.
<jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
for small enough changes it is linear!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
George H.
A
Archimedes' Lever
Guest
On Wed, 16 Jun 2010 12:53:26 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:
<gherold@teachspin.com> wrote:
http://en.wikipedia.org/wiki/File:Earth-G-force.pngOn 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.
A
Archimedes' Lever
Guest
On Wed, 16 Jun 2010 09:31:57 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
to be the answer.
But for you, I'll say that the amount is less than a 0.01% difference.
numbers needed.
plays in. And again, no number needed, nor are YOU even intelligent
enough to apply them correctly.
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
To start with, NONE of these three "questions" calls for any "number"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 be the answer.
But for you, I'll say that the amount is less than a 0.01% difference.
That depends on what you are acquiring the measurement for. Again, no2. Is this significant to the measurement?
numbers needed.
Yes, but also only by a small amount. The position of the sun alsoRules: 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?
plays in. And again, no number needed, nor are YOU even intelligent
enough to apply them correctly.