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

On Wed, 16 Jun 2010 20:39:43 -0700, "JosephKK"<quiettechblue@yahoo.com>
wrote:

Though it is not something a bathroom scale can reliably detect, serious
inertial navigation systems contain up-to-date geodal height maps of the
whole planet. Hint, they have since the 1960s. For more, give "geodal
height" to one or more search engines.
Click to expand...
Simply go to the link that has been posted into the thread several
times already:

http://en.wikipedia.org/wiki/Earth%27s_gravity

The term is geoidal

geoid:

http://en.wikipedia.org/wiki/File:GRACE_globe_animation.gif
 
On Wed, 16 Jun 2010 19:14:37 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 21:07:21 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

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 it operates accurately by some other means?
Click to expand...
Your strawmen don't operate, AlwaysWrong. You continue to gather straw,
though.

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 think that a fish scale will read properly sideways?
Click to expand...
Measuring gravity, no. You'd probably think that your magic scale does,
though.

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.

AlwaysWrong is always *so* wrong, and irrelevant.

You are the one that said that a force gauge was a scale, and then
pointed at a fish scale as an example of a force gauge, which it is not.
Click to expand...
No, a scale is a force gauge. You're the one that insists that they're
different things. Your strawmen *are* irrelevant.

In other words, you are still an idiot, and have not proven me to be
wrong ever at any time you ever claimed that I was.
Click to expand...
I don't have to prove you wrong. You *are* AlwaysWrong and everyone here
knows it.

You are 100% pathetic, and could not present an argument if your
fingernails were being pulled out. You are sub-human, at best.
Click to expand...
Time to put away mommy's computer. You have an hour to play in her hamper
before it's beddy-by time.
 
On Wed, 16 Jun 2010 23:05:15 -0500, "krw@att.bizzzzzzzzzzzz"
<krw@att.bizzzzzzzzzzzz> wrote:

You think that a fish scale will read properly sideways?

Measuring gravity, no. You'd probably think that your magic scale does,
though.
Click to expand...
A fish scale is not for measuring gravity, idiot. Nor did I ever
mention even once in this thread, the measuring of gravity.
 
On Wed, 16 Jun 2010 23:05:15 -0500, "krw@att.bizzzzzzzzzzzz"
<krw@att.bizzzzzzzzzzzz> wrote:

No, a scale is a force gauge.
Click to expand...
That is what I said, you retarded fuck. YOU are the asswipe that said
that a fish scale was a force gauge, you retarded piece of shit.

You're the one that insists that they're
different things. Your strawmen *are* irrelevant.
Click to expand...

Show me where I made any such statement, you lying, pathetic piece of
shit.
 
On Wed, 16 Jun 2010 20:54:45 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 20:05:09 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 19:32:18 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 19:17:38 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

A zillion web sites say less, and use the
1/r^2 equation to demonstrate it, but that equation works if you're in
a balloon, not sitting on solid rock.

You are an idiot.

It depends on how spikey the mountain is.

No, it does not.

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

What does that cartoon have to say about the top-of-mountain issue?

John

Read it.

From about 400 km down to sea level there is a value given for both.

Linearize and compute.
Click to expand...
But the gap there is air. What if an object is above sea level, but
the space from it to the center of the earth is solid rock?

Consider: If I were 10 miles above sea level in a balloon, g would be
less than at sea level. But if you then add 10 miles to the radius of
earth below me, g would increase. Bigger planets have higher surface g
than small ones, because the mass of the planet goes as r^3, and g
goes as 1/r^2, so mass wins.

So whether g is higher or lower on a mountaintop, as compared to sea
level, must depend on how spikey the mountain is. If it's a skinny
column, it's not much better than air, so g falls as 1/r^2. If it so
slopey that it covers the entire planet, g increases as r.

I just can't estimate the intermediate cases, like for real mountains.
Calculus.

John
 
On Wed, 16 Jun 2010 19:39:39 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Jun 16, 10:04 pm, Archimedes' Lever
OneBigLe...@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 18:42:40 -0700 (PDT), George Herold





gher...@teachspin.com> wrote:
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 STILL assume a linear scale too.  Another mistake.- Hide quoted text -

- Show quoted text -

Dang, with changes at the 10^-3 level second order effect is at
10^-6. Do you know about Taylor expansions? Gravity is one of the
hardest things to measure. I'm not sure we know the value of big G to
a part in 10^6.
Click to expand...
Doesn't look like we're even close to 1:10^6:

http://www.search.com/reference/Gravitational_constant#Recent_measurement

"In the January 5, 2007 issue of Science (page 74), the report "Atom
Interferometer Measurement of the Newtonian Constant of Gravity" (J. B.
Fixler, G. T. Foster, J. M. McGuirk, and M. A. Kasevich) describes a new
measurement of the gravitational constant. According to the abstract: "Here,
we report a value of G = 6.693 x 10–11 cubic meters per kilogram second
squared, with a standard error of the mean of ą0.027 x 10–11 and a
systematic error of ą0.021 x 10–11 cubic meters per kilogram second
squared.".["
 
On Mon, 14 Jun 2010 22:30:21 +0000 (UTC), Tim Watts <tw@dionic.net>
wrote:

On Mon, 14 Jun 2010 15:17:41 -0700, VWWall <vwall@large.invalid> wibbled:

Tim Watts wrote:
On Mon, 14 Jun 2010 07:31:13 -0700, StickThatInYourPipeAndSmokeIt
Zarathustra@thusspoke.org> wibbled:

At least we are not measuring things by 'curling stones' or the
like.

It is still possible AFAIK here to go to a small time brewery and buy a
firkin (8 gallons) of beer. Or a barrel (4 firkins). If you're areal
pissartist, you'd probably want a hogshead, butt or tun though.

And the answer to: "Are you getting any lately?" has always been given
in furlongs per fortnight!

Incidentally, that brings up a question:

When you americans talk of "a butt load of <blah>" - does that derive
from the "butt" as in 144 gallons? I always mentally associated it with
butt=ass - but the former makes more sense...
Click to expand...
That sounds like a meaning shift due to forgetting larger units of
measure in place of more familiar ones (55 gal barrel).
 
On Wed, 16 Jun 2010 19:15:04 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 21:08:31 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

On Wed, 16 Jun 2010 18:50:06 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 19:26:58 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

Is that true in the "near field"? Integrate the volume of the earth over the
square of the distances and I'd expect to see that not all points affect
gravity the same. IOW, I'd expect the apparent "center" of the Earth to be
somewhat closer than it really is.

You are so goddamned retarded that you cannot even follow a thread. It
was posted three times already, dumbass.

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

More irrelevance from DimBulb.


It is absolutely relevant.
Click to expand...
You're full of shit. ...but you like it that way.
 
On Wed, 16 Jun 2010 19:39:39 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

Gravity is one of the
hardest things to measure. I'm not sure we know the value of big G to
a part in 10^6.

George H.
Click to expand...

http://en.wikipedia.org/wiki/Petr_Van%C3%AD%C4%8Dek
 
On Wed, 16 Jun 2010 19:26:58 -0500, "krw@att.bizzzzzzzzzzzz"
<krw@att.bizzzzzzzzzzzz> wrote:

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

On Wed, 16 Jun 2010 15:36:13 -0700 (PDT), Richard Henry
pomerado@hotmail.com> wrote:

On Jun 16, 12:53 pm, George Herold <gher...@teachspin.com> wrote:
On Jun 16, 2:31 pm, Archimedes' Lever <OneBigLe...@InfiniteSeries.Org
wrote:

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

gher...@teachspin.com> wrote:
Cool, I have to scribble numbers on the paper though.  6400 feet is
about 2000m, the Earth is about 6E6 m in radius, Since we only want a
small change I can ignore the r^2 stuff and just multiple the ratio by
2.  something like 4 parts out of 6,000.  much smaller than the
divisions on your scale.

  No you cannot.  What makes you think that G decreases (or
increases)linearly?

Big G doesn't change at all.  Little g (the force of gravity on the
Earths surface) will go as 1/r^2.  For small changes in r the change
is approximately linear... first term in the taylor expansion if you
want to think of it that way.  And it does go as 2*delta-R/Rearth

Not exactly. The mass of the Earth is not actually concentrated at a
single point.

Outside of a uniform spherical mass, g does behave as if all the mass
were concentrated at the center. The earth is non-homogenous, but
close enough. The short answer to my little problem is that the change
is about 1 part in 2000, too small to matter in the context of the
other measurement uncertainties.

Is that true in the "near field"? Integrate the volume of the earth over the
square of the distances and I'd expect to see that not all points affect
gravity the same. IOW, I'd expect the apparent "center" of the Earth to be
somewhat closer than it really is.
Click to expand...
If the density of a sphere is uniform, and the observer is anywhere
outside of the sphere, gravity acts the same as if all the mass were
concentreted in the center. Newton said that.

Of course, earth is denser deep inside, so it's not exactly the same.
But the mass is concentrated towards the center, all that iron, so it
must be close.

John
 
On Wed, 16 Jun 2010 20:46:44 -0700,
"JosephKK"<quiettechblue@yahoo.com> wrote:

On Wed, 16 Jun 2010 10:29:12 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Tue, 15 Jun 2010 23:34:50 -0700,
"JosephKK"<quiettechblue@yahoo.com> wrote:

On Tue, 15 Jun 2010 00:31:35 -0500, John Fields
jfields@austininstruments.com> wrote:

On Mon, 14 Jun 2010 08:25:57 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Mon, 14 Jun 2010 07:23:14 -0700, Archimedes' Lever
OneBigLever@InfiniteSeries.Org> wrote:

On Mon, 14 Jun 2010 07:19:37 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

Fluid of course. Few people ever measure force. And most liquids used
in everydat life have a s.g. near 1, so an ounce of tabasco is
unambiguous.

Hundreds, even thousands of folks measure force every day, and many of
those use ounces in their scales of measure. Many use Newtons.


Of course hundreds, maybe even thousands of people measure force every
day. But there are 300 million people in the USA. Most people never
measure force; they do measure weight, or mass actually.

---
Since weight is mass multiplied by the acceleration of gravity and
most people use scales instead of beam balances and calibrated
reference masses to do the measurement, they measure weight, not mass.

http://en.wikipedia.org/wiki/Weighing_scale

Balances compare the gravitationally induced forces of test masses versus
reference masses. Please note that this assumes that the gravitational
field is reasonably uniform in terms of area included in the balance the
masses to be compared and the distances between them. And many scales
measure deflection of (more or less) well documented structures deflected
by the forces in an assumed uniform gravitational field. Compare the
usefulness of balance versus scales in any accelerated (including
rotated) frame of reference.

Words, words, words. Do the puzzler.

John

I am not here to play with your "puzzlers" JL. Post electronics with
correct circuits, values, and explanations instead please.
Click to expand...
After you.

John
 
On Wed, 16 Jun 2010 21:10:41 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

But the gap there is air. What if an object is above sea level, but
the space from it to the center of the earth is solid rock?
Click to expand...
You really did not look at the image very closely, did you, Johnny?

Try looking at the Geoid image instead, and if that still has you
scratching your head, look up what this guy did with his life:

http://en.wikipedia.org/wiki/Petr_Van%C3%AD%C4%8Dek
 
On Wed, 16 Jun 2010 19:29:11 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 21:11:39 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:


You're always wrong, AlwaysWrong. ...and that wasn't the question, so you're
wrong again. Of course none of this surprises anyone, AlwaysWrong.

A balance will read the same regardless of what sized spheroid it is
placed onto.

Wrong, and what does a "spheroid" have to do with anything, ALwaysWrong. You
out of straw again?

The Earth is a spheroid, idiot. Spheroids have gravity.
Click to expand...
So do cubeoids.

A spring scale's reading is gravity dependent.
Click to expand...
And mass dependent. So?

It will read differently here than on the Moon, a smaller spheroid.
Click to expand...
It'll read differently on an egg, too.

A balance will not. It will read the same in both places.
Click to expand...
You assume a uniform gravity field. Raise the G high enough or dimensions
small enough and it won't.

You lose, again.
Click to expand...
You're *always* wrong, no exceptions, AlwaysWrong.
 
On Wed, 16 Jun 2010 19:53:55 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:

On Jun 16, 10:20 pm, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:
On Wed, 16 Jun 2010 18:51:38 -0700, Archimedes' Lever





OneBigLe...@InfiniteSeries.Org> wrote:
On Wed, 16 Jun 2010 17:28:39 -0700, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 16:44:44 -0700, Archimedes' Lever
OneBigLe...@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 16:41:11 -0700, John Larkin
jjlar...@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 09:36:11 +0300, Paul Keinanen <keina...@sci.fi
wrote:

On Tue, 15 Jun 2010 19:58:20 -0500, "k...@att.bizzzzzzzzzzzz"
k...@att.bizzzzzzzzzzzz> wrote:

Then, by your "logic", "millimeter" is an Imperial term since
1mm = 0.03937"

No, because the inch is defined as being 25.4mm.  The metric measure is not a
derivative of the English.

In actuality, what makes the carat a metric term is that the weight of
gemstones is measured using the metric system and described in metric
units.

Imperial units are defined using the metric system.  Does that mean that the
US uses the metric system?

I people are so allergic about the term "metric", why not go directly
to the primary definitions ?

The meter was previously defined as 1,650,763.73 krypton-86
wavelengths, thus 1 inch = 41,929.398,742 wavelengths.

I wonder if they actually counted the 1,650,763.73 fringes. I sure
hope that did it twice.

I'm impressed that the krypton line is narrow enough to have a meter
of coherence length. I think the measurement was made pre-laser.

John

 Yes, John.  Krypton based atomic clocks were around before the advent
of the laser in 1960.

http://en.wikipedia.org/wiki/Atomic_clock#History

Looks like you're wrong again. Why do you say stuff like this when you
could check google, like everybody else?

John

 Laser:  1960

Cesium atomic clock:  1955

 You lose. again.

Krypton atomic clock?

John- Hide quoted text -

- Show quoted text -

That's how all supermen tell time.
Click to expand...
Well, that's how they tell what time it is at home.
 
On Wed, 16 Jun 2010 21:14:15 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

Of course, earth is denser deep inside, so it's not exactly the same.
But the mass is concentrated towards the center, all that iron, so it
must be close.

John
Click to expand...
LOOK at the damned image, you stubborn fuckhead! It tells you exactly
what the differing densities are with respect to a homogenous 'norm'
reference.

You are the most dense person in the room.
 
On Wed, 16 Jun 2010 21:07:23 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 23:05:15 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

You think that a fish scale will read properly sideways?

Measuring gravity, no. You'd probably think that your magic scale does,
though.

A fish scale is not for measuring gravity, idiot.
Click to expand...
With a known mass, it'll do in a pinch, AlwaysWrong.

Nor did I ever
mention even once in this thread, the measuring of gravity.
Click to expand...
You want to measure weights. It's in there. You really should try to get
*something* right in your life. Mommy would be so pleased.
 
On Wed, 16 Jun 2010 21:08:46 -0700, Archimedes' Lever
<OneBigLever@InfiniteSeries.Org> wrote:

On Wed, 16 Jun 2010 23:05:15 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:


No, a scale is a force gauge.

That is what I said, you retarded fuck. YOU are the asswipe that said
that a fish scale was a force gauge, you retarded piece of shit.

You're the one that insists that they're
different things. Your strawmen *are* irrelevant.


Show me where I made any such statement, you lying, pathetic piece of
shit.
Click to expand...
AlwaysWrong, is *so* wrong. ...a liar, too.
 
On Wed, 16 Jun 2010 23:15:55 -0500, "krw@att.bizzzzzzzzzzzz"
<krw@att.bizzzzzzzzzzzz> wrote:

A balance will not. It will read the same in both places.

You assume a uniform gravity field. Raise the G high enough or dimensions
small enough and it won't.
Click to expand...

Wrong. The balance measures force applied against two arms. It is not
dependent on the gravity field it is in. One kg on left, one kg on right.
Same reading everywhere. Same force applied to balance point.

Same measure on Charon as on Jupiter. The spring scale, placed on IO,
however, will vary so widely and so much that you could probably watch it
as the moon spins and Jupiter rises. More proof for Johnny that the Moon
does have an effect on a spring scale, but not on a balance.

Pretty much basic physics, Williams.
 
On Wed, 16 Jun 2010 21:14:15 -0700, John Larkin
<jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:

On Wed, 16 Jun 2010 19:26:58 -0500, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:

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

On Wed, 16 Jun 2010 15:36:13 -0700 (PDT), Richard Henry
pomerado@hotmail.com> wrote:

On Jun 16, 12:53 pm, George Herold <gher...@teachspin.com> wrote:
On Jun 16, 2:31 pm, Archimedes' Lever <OneBigLe...@InfiniteSeries.Org
wrote:

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

gher...@teachspin.com> wrote:
Cool, I have to scribble numbers on the paper though.  6400 feet is
about 2000m, the Earth is about 6E6 m in radius, Since we only want a
small change I can ignore the r^2 stuff and just multiple the ratio by
2.  something like 4 parts out of 6,000.  much smaller than the
divisions on your scale.

  No you cannot.  What makes you think that G decreases (or
increases)linearly?

Big G doesn't change at all.  Little g (the force of gravity on the
Earths surface) will go as 1/r^2.  For small changes in r the change
is approximately linear... first term in the taylor expansion if you
want to think of it that way.  And it does go as 2*delta-R/Rearth

Not exactly. The mass of the Earth is not actually concentrated at a
single point.

Outside of a uniform spherical mass, g does behave as if all the mass
were concentrated at the center. The earth is non-homogenous, but
close enough. The short answer to my little problem is that the change
is about 1 part in 2000, too small to matter in the context of the
other measurement uncertainties.

Is that true in the "near field"? Integrate the volume of the earth over the
square of the distances and I'd expect to see that not all points affect
gravity the same. IOW, I'd expect the apparent "center" of the Earth to be
somewhat closer than it really is.

If the density of a sphere is uniform, and the observer is anywhere
outside of the sphere, gravity acts the same as if all the mass were
concentreted in the center. Newton said that.
Click to expand...
Certainly inside the sphere that's true, so in the limiting case it must be
too. It still seems that for the case outside the sphere, if you integrate
the contribution from all points those closer to the object should contribute
more.

Of course, earth is denser deep inside, so it's not exactly the same.
But the mass is concentrated towards the center, all that iron, so it
must be close.
Click to expand...
Sure, only DimBulb was hung up on that.
 
On Wed, 16 Jun 2010 21:12:03 -0700, "JosephKK"<quiettechblue@yahoo.com> wrote:

On Mon, 14 Jun 2010 22:30:21 +0000 (UTC), Tim Watts <tw@dionic.net
wrote:

On Mon, 14 Jun 2010 15:17:41 -0700, VWWall <vwall@large.invalid> wibbled:

Tim Watts wrote:
On Mon, 14 Jun 2010 07:31:13 -0700, StickThatInYourPipeAndSmokeIt
Zarathustra@thusspoke.org> wibbled:

At least we are not measuring things by 'curling stones' or the
like.

It is still possible AFAIK here to go to a small time brewery and buy a
firkin (8 gallons) of beer. Or a barrel (4 firkins). If you're areal
pissartist, you'd probably want a hogshead, butt or tun though.

And the answer to: "Are you getting any lately?" has always been given
in furlongs per fortnight!

Incidentally, that brings up a question:

When you americans talk of "a butt load of <blah>" - does that derive
from the "butt" as in 144 gallons? I always mentally associated it with
butt=ass - but the former makes more sense...

That sounds like a meaning shift due to forgetting larger units of
measure in place of more familiar ones (55 gal barrel).
Click to expand...
You mean like the 40gal oil barrel?
 
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