Bicycling gifts

On Sun, 10 Jan 2021 13:18:17 -0500, Frank Krygowski
wrote:

On 1/10/2021 3:33 AM, Jeff Liebermann wrote:
On Sun, 10 Jan 2021 06:53:40 +0000 (UTC), Ralph Barone
wrote:

If I had to calibrate a pressure gauge, I’d consider building a manometer
with different diameter tubing on both sides. Then with some water and some
math, you can calibrate from first principles.

Well... the column of water would need to weigh the same as the
maximum gauge pressure. Basically, the air pressure in the tire will
need to lift a column of water that produces the same pressure.

Water weighs 8.33 lbs/gallon. To make the math easy, we start with a
tube that has an area of 1 sq-inch at each end. That's about:
1 sq-in = Pi * r^2
r = (1 / 3.14)^0.5 = 0.564 in = 1.12 in diameter
For 100 psi full scale, the volume of water would be:
100 lbs / 8.33 lbs/gallon = 12 gallons
1 gallon of water is 231 cubic inches. For a 1.12 in diameter tube,
the necessary column height would be:
h = vol / (Pi * r^2) = 231 / (3.14 * 0.564^2) = 231 inches = 19 ft
Yes, it might work, but would rather huge and messy to build.
Dissimilar tube diameters and a denser liquid (mercury) would help,
but it would still be huge and require 100 lbs of water. Radically
dissimilar diameter tubes will certainly work, if I wanted a small
water pond in my shop. No matter what shape I build the open end tube
of the manometer, the water still needs to weigh at least 100 lbs and
occupy 12 gallons. Mercury is 13.5 times as dense as water and
therefore will require 13.5 times less liquid or about:
12 gallons / 13.5 = 0.89 gallons of mercury
I have some mercury, but not that much:
http://www.learnbydestroying.com/jeffl/crud/mercury.jpg
Also, stuffing a Presta or Schrader valve into a rubber stopper, which
is then pounded into the open end of the glass manometer tube, with
100 lbs of water behind it, does not seem like something I would want
in my shop. Ummm... do you have any better less dangerous ideas?

Note: It's past midnight here. My math ability is normally terrible
and deteriorates rapidly with lack of sleep. Please check my numbers.

I'll give you some slack because of your sleepiness; but sorry, you're
way, way off.

Don't be sorry. I hate being wrong, but I always appreciate
corrections.

There was more going on than lack of sleep. On Friday (01/10/2021), I
was working at a friends house on his hot tube, computer, vinegar deck
wash, etc. when I started getting severe stomach cramps. Food
poisoning was the best guess. I managed to drive home without killing
anyone and rolled into bed. After throwing up a few times, I started
feeling better. The next day, I sorta drifted around the house doing
nothing. I was reading on Usenet but not posting. By midnight, I
wasn't thinking very clearly and decided to answer some of the email
and postings. Trying to work through the manometer problem was
probably a bad idea. Not waiting an hour and checking my assumptions
and math just made it worse. My apologies for the mess.

It's not math so much as units of measurement, starting
with your statement that "the column of water would need to weigh the
same as the maximum gauge pressure." That doesn't work, because weight
and pressure are fundamentally different quantities that can't be equated.

That's why I setup the example using 1 sq-in area for the column. That
should have made the weight of the column in lbs be numerically the
same as the pressure in lbs/sq-in. Now, I'm not so sure that works.
Where I really screwed up was getting my units of measure mixed up,

I suspect that I have a problem understanding how a manometer works. I
still don't understand it. I'll do some more reading. (Today, I get
to rebuild the carburetors for 2 chainsaws). Thank again for the
corrections.

The relationship you need is this: Pressure = weight density x depth.
So the necessary depth (or height of a manometer column) is the desired
pressure divided by the weight density of the fluid.

Water's weight density is about 62.4 lb/ft^3 or 0.0361 lb/in^3.
100 psi divided by 0.0361 lb/1n^3 gives 2770 inches or 231 feet. [Hmm...
coincidentally, the number of cubic inches in a gallon. Weird.]

Anyway, a water manometer would need to be 231 feet tall to give 100
psi. It's not practical. A mercury manometer would be about 16.7 feet tall.

And BTW, the idea of different diameters doesn't work at all. Pressure =
density x height has no "area" term in it. Consider a tank of liquid
with a thin transparent sight tube on its outside to show the depth
within the tank. The pressure at the bottom of the sight tube (maybe
1/4" ID) is the same as the pressure just inside the tank at that level,
even though the tank's area is much, much greater.

--
Jeff Liebermann
PO Box 272 http://www.LearnByDestroying.com
Ben Lomond CA 95005-0272
Skype: JeffLiebermann AE6KS 831-336-2558

Frank Krygowski writes:

On 1/10/2021 3:33 AM, Jeff Liebermann wrote:
On Sun, 10 Jan 2021 06:53:40 +0000 (UTC), Ralph Barone
wrote:

If I had to calibrate a pressure gauge, I’d consider building a manometer
with different diameter tubing on both sides. Then with some water and some
math, you can calibrate from first principles.

Well... the column of water would need to weigh the same as the
maximum gauge pressure. Basically, the air pressure in the tire will
need to lift a column of water that produces the same pressure.

Water weighs 8.33 lbs/gallon. To make the math easy, we start with a
tube that has an area of 1 sq-inch at each end. That's about:
1 sq-in = Pi * r^2
r = (1 / 3.14)^0.5 = 0.564 in = 1.12 in diameter
For 100 psi full scale, the volume of water would be:
100 lbs / 8.33 lbs/gallon = 12 gallons
1 gallon of water is 231 cubic inches. For a 1.12 in diameter tube,
the necessary column height would be:
h = vol / (Pi * r^2) = 231 / (3.14 * 0.564^2) = 231 inches = 19 ft
Yes, it might work, but would rather huge and messy to build.
Dissimilar tube diameters and a denser liquid (mercury) would help,
but it would still be huge and require 100 lbs of water. Radically
dissimilar diameter tubes will certainly work, if I wanted a small
water pond in my shop. No matter what shape I build the open end tube
of the manometer, the water still needs to weigh at least 100 lbs and
occupy 12 gallons. Mercury is 13.5 times as dense as water and
therefore will require 13.5 times less liquid or about:
12 gallons / 13.5 = 0.89 gallons of mercury
I have some mercury, but not that much:
http://www.learnbydestroying.com/jeffl/crud/mercury.jpg
Also, stuffing a Presta or Schrader valve into a rubber stopper, which
is then pounded into the open end of the glass manometer tube, with
100 lbs of water behind it, does not seem like something I would want
in my shop. Ummm... do you have any better less dangerous ideas?

Note: It's past midnight here. My math ability is normally terrible
and deteriorates rapidly with lack of sleep. Please check my numbers.

I'll give you some slack because of your sleepiness; but sorry, you're
way, way off. It's not math so much as units of measurement, starting
with your statement that "the column of water would need to weigh the
same as the maximum gauge pressure." That doesn't work, because weight
and pressure are fundamentally different quantities that can't be
equated.

The relationship you need is this: Pressure = weight density x depth.
So the necessary depth (or height of a manometer column) is the
desired pressure divided by the weight density of the fluid.

Water's weight density is about 62.4 lb/ft^3 or 0.0361 lb/in^3.
100 psi divided by 0.0361 lb/1n^3 gives 2770 inches or 231
feet. [Hmm... coincidentally, the number of cubic inches in a
gallon. Weird.]

Anyway, a water manometer would need to be 231 feet tall to give 100
psi. It's not practical. A mercury manometer would be about 16.7 feet
tall.

And BTW, the idea of different diameters doesn't work at all. Pressure
= density x height has no "area" term in it. Consider a tank of liquid
with a thin transparent sight tube on its outside to show the depth
within the tank. The pressure at the bottom of the sight tube (maybe
1/4" ID) is the same as the pressure just inside the tank at that
level, even though the tank's area is much, much greater.

I'm not at all sure on that last statement. The example of the sight
gauge is not relevant, because the pressure inside the tank is the same
as that at the top of the sight gauge. There are devices called "well
manometers" that, by design, use different areas for the tube exposed to
the measured pressure than they do for the one exposed to the reference
pressure. Typically the tube for the measured pressure is larger, which
makes the height variations in the other tube larger than they would be
for a bog standard manometer.

A mercury barometer is normally a well-type manometer: there is an
enlarged reservoir exposed to atmospheric pressure, so that the effect
of the falling level on that side is small enough that the scale on the
other side in millimeters is close to a measurement in real millimeters.
If both sides were the same size then the "millimeters" on the scale
would actually be only half a millimeter long.

On 1/14/2021 4:20 PM, Radey Shouman wrote:
Frank Krygowski writes:

On 1/10/2021 3:33 AM, Jeff Liebermann wrote:
On Sun, 10 Jan 2021 06:53:40 +0000 (UTC), Ralph Barone
wrote:

If I had to calibrate a pressure gauge, I’d consider building a manometer
with different diameter tubing on both sides. Then with some water and some
math, you can calibrate from first principles.

Well... the column of water would need to weigh the same as the
maximum gauge pressure. Basically, the air pressure in the tire will
need to lift a column of water that produces the same pressure.

Water weighs 8.33 lbs/gallon. To make the math easy, we start with a
tube that has an area of 1 sq-inch at each end. That's about:
1 sq-in = Pi * r^2
r = (1 / 3.14)^0.5 = 0.564 in = 1.12 in diameter
For 100 psi full scale, the volume of water would be:
100 lbs / 8.33 lbs/gallon = 12 gallons
1 gallon of water is 231 cubic inches. For a 1.12 in diameter tube,
the necessary column height would be:
h = vol / (Pi * r^2) = 231 / (3.14 * 0.564^2) = 231 inches = 19 ft
Yes, it might work, but would rather huge and messy to build.
Dissimilar tube diameters and a denser liquid (mercury) would help,
but it would still be huge and require 100 lbs of water. Radically
dissimilar diameter tubes will certainly work, if I wanted a small
water pond in my shop. No matter what shape I build the open end tube
of the manometer, the water still needs to weigh at least 100 lbs and
occupy 12 gallons. Mercury is 13.5 times as dense as water and
therefore will require 13.5 times less liquid or about:
12 gallons / 13.5 = 0.89 gallons of mercury
I have some mercury, but not that much:
http://www.learnbydestroying.com/jeffl/crud/mercury.jpg
Also, stuffing a Presta or Schrader valve into a rubber stopper, which
is then pounded into the open end of the glass manometer tube, with
100 lbs of water behind it, does not seem like something I would want
in my shop. Ummm... do you have any better less dangerous ideas?

Note: It's past midnight here. My math ability is normally terrible
and deteriorates rapidly with lack of sleep. Please check my numbers.

I'll give you some slack because of your sleepiness; but sorry, you're
way, way off. It's not math so much as units of measurement, starting
with your statement that "the column of water would need to weigh the
same as the maximum gauge pressure." That doesn't work, because weight
and pressure are fundamentally different quantities that can't be
equated.

The relationship you need is this: Pressure = weight density x depth.
So the necessary depth (or height of a manometer column) is the
desired pressure divided by the weight density of the fluid.

Water's weight density is about 62.4 lb/ft^3 or 0.0361 lb/in^3.
100 psi divided by 0.0361 lb/1n^3 gives 2770 inches or 231
feet. [Hmm... coincidentally, the number of cubic inches in a
gallon. Weird.]

Anyway, a water manometer would need to be 231 feet tall to give 100
psi. It's not practical. A mercury manometer would be about 16.7 feet
tall.

And BTW, the idea of different diameters doesn't work at all. Pressure
= density x height has no "area" term in it. Consider a tank of liquid
with a thin transparent sight tube on its outside to show the depth
within the tank. The pressure at the bottom of the sight tube (maybe
1/4" ID) is the same as the pressure just inside the tank at that
level, even though the tank's area is much, much greater.

I'm not at all sure on that last statement. The example of the sight
gauge is not relevant, because the pressure inside the tank is the same
as that at the top of the sight gauge. There are devices called "well
manometers" that, by design, use different areas for the tube exposed to
the measured pressure than they do for the one exposed to the reference
pressure. Typically the tube for the measured pressure is larger, which
makes the height variations in the other tube larger than they would be
for a bog standard manometer.

I'm familiar with well manometers. But you seem to be misstating their
benefit. They allow for more convenient reading at a certain level of
sensitivity, but it's still true that the height difference in the legs
depends only on the pressure difference. Area doesn't come into play,
except to shift (almost) all the height change to one leg.

In a standard u-tube manometer, the area of the "reading" leg (where a
scale is typically attached) is the same as the area of the sensing leg
(attached to the pressure being measured). When pressure in the sensing
leg increases, its level drops and the level in the reading leg rises by
the same amount.

This is fine, but again, the pressure measurement is indicated by the
_difference_ in height of the two. So to read that height difference,
you have two choices: 1) Use a moveable scale, so for each new
measurement you have to lower the "zero" point to the liquid level in
the sensing leg; or 2) use a fixed scale against the reading leg, with
marks half as far apart as they'd normally be, since the height
difference is split between the two legs.

The problems are these: Scheme #1 is inconvenient; and scheme #2 is half
as sensitive.

A well manometer has a much greater area in the sensing leg. (If it's 10
times the diameter, which is usually easy, it's 100 times the area.)
That means the level in the well will drop much, much less - 1/100 of
the U-tube's drop. That's usually negligible, and it allows using a
fixed scale, because 99% of the depth difference is occurring against
that scale. But the pressure difference is still proportional to the
height difference, and nothing else.

Even more sensitive are inclined well manometers. Their reading leg runs
at a shallow angle. The liquid level travels farther along that slope to
generate the same height difference, so their scale is amplified and
smaller pressure changes are visible. But it's still height difference
that actually matters.

A mercury barometer is normally a well-type manometer: there is an
enlarged reservoir exposed to atmospheric pressure, so that the effect
of the falling level on that side is small enough that the scale on the
other side in millimeters is close to a measurement in real millimeters.
If both sides were the same size then the "millimeters" on the scale
would actually be only half a millimeter long.

Right, as I described above. We agree.

I once had use of a super-precise mercury barometer. It featured a
movable scale with a tiny pointer to touch the surface of the mercury
pool and establish "zero." It also had a vernier scale for precisely
measuring height. Note, though, that if that barometer were modified so
its well was even larger - say, twice the diameter - nothing else would
need to be changed.

--
- Frank Krygowski

Frank Krygowski writes:

On 1/14/2021 4:20 PM, Radey Shouman wrote:
Frank Krygowski writes:

On 1/10/2021 3:33 AM, Jeff Liebermann wrote:
On Sun, 10 Jan 2021 06:53:40 +0000 (UTC), Ralph Barone
wrote:

If I had to calibrate a pressure gauge, I’d consider building a manometer
with different diameter tubing on both sides. Then with some water and some
math, you can calibrate from first principles.

Well... the column of water would need to weigh the same as the
maximum gauge pressure. Basically, the air pressure in the tire will
need to lift a column of water that produces the same pressure.

Water weighs 8.33 lbs/gallon. To make the math easy, we start with a
tube that has an area of 1 sq-inch at each end. That's about:
1 sq-in = Pi * r^2
r = (1 / 3.14)^0.5 = 0.564 in = 1.12 in diameter
For 100 psi full scale, the volume of water would be:
100 lbs / 8.33 lbs/gallon = 12 gallons
1 gallon of water is 231 cubic inches. For a 1.12 in diameter tube,
the necessary column height would be:
h = vol / (Pi * r^2) = 231 / (3.14 * 0.564^2) = 231 inches = 19 ft
Yes, it might work, but would rather huge and messy to build.
Dissimilar tube diameters and a denser liquid (mercury) would help,
but it would still be huge and require 100 lbs of water. Radically
dissimilar diameter tubes will certainly work, if I wanted a small
water pond in my shop. No matter what shape I build the open end tube
of the manometer, the water still needs to weigh at least 100 lbs and
occupy 12 gallons. Mercury is 13.5 times as dense as water and
therefore will require 13.5 times less liquid or about:
12 gallons / 13.5 = 0.89 gallons of mercury
I have some mercury, but not that much:
http://www.learnbydestroying.com/jeffl/crud/mercury.jpg
Also, stuffing a Presta or Schrader valve into a rubber stopper, which
is then pounded into the open end of the glass manometer tube, with
100 lbs of water behind it, does not seem like something I would want
in my shop. Ummm... do you have any better less dangerous ideas?

Note: It's past midnight here. My math ability is normally terrible
and deteriorates rapidly with lack of sleep. Please check my numbers.

I'll give you some slack because of your sleepiness; but sorry, you're
way, way off. It's not math so much as units of measurement, starting
with your statement that "the column of water would need to weigh the
same as the maximum gauge pressure." That doesn't work, because weight
and pressure are fundamentally different quantities that can't be
equated.

The relationship you need is this: Pressure = weight density x depth.
So the necessary depth (or height of a manometer column) is the
desired pressure divided by the weight density of the fluid.

Water's weight density is about 62.4 lb/ft^3 or 0.0361 lb/in^3.
100 psi divided by 0.0361 lb/1n^3 gives 2770 inches or 231
feet. [Hmm... coincidentally, the number of cubic inches in a
gallon. Weird.]

Anyway, a water manometer would need to be 231 feet tall to give 100
psi. It's not practical. A mercury manometer would be about 16.7 feet
tall.

And BTW, the idea of different diameters doesn't work at all. Pressure
= density x height has no "area" term in it. Consider a tank of liquid
with a thin transparent sight tube on its outside to show the depth
within the tank. The pressure at the bottom of the sight tube (maybe
1/4" ID) is the same as the pressure just inside the tank at that
level, even though the tank's area is much, much greater.

I'm not at all sure on that last statement. The example of the sight
gauge is not relevant, because the pressure inside the tank is the same
as that at the top of the sight gauge. There are devices called "well
manometers" that, by design, use different areas for the tube exposed to
the measured pressure than they do for the one exposed to the reference
pressure. Typically the tube for the measured pressure is larger, which
makes the height variations in the other tube larger than they would be
for a bog standard manometer.

I'm familiar with well manometers. But you seem to be misstating their
benefit. They allow for more convenient reading at a certain level of
sensitivity, but it's still true that the height difference in the
legs depends only on the pressure difference. Area doesn't come into
play, except to shift (almost) all the height change to one leg.

That's true.

In a standard u-tube manometer, the area of the "reading" leg (where a
scale is typically attached) is the same as the area of the sensing
leg (attached to the pressure being measured). When pressure in the
sensing leg increases, its level drops and the level in the reading
leg rises by the same amount.

This is fine, but again, the pressure measurement is indicated by the
_difference_ in height of the two. So to read that height difference,
you have two choices: 1) Use a moveable scale, so for each new
measurement you have to lower the "zero" point to the liquid level in
the sensing leg; or 2) use a fixed scale against the reading leg, with
marks half as far apart as they'd normally be, since the height
difference is split between the two legs.

I agree that in a not-too-idealized world it's the difference in heights
that is proportional to the difference in pressure. The height
difference really depends not on the density of the manometer fluid, but
on the difference in density between the manometer fluid and whatever it
displaces. If that fluid is different in the two legs (eg air on one
side and almost nothing on the other) then I *think* that differences in
the area between them could change the height difference.

The problems are these: Scheme #1 is inconvenient; and scheme #2 is
half as sensitive.

Scheme #1 is also less accurate, because two readings with error have to
be subtracted, and the errors come close to adding.

A well manometer has a much greater area in the sensing leg. (If it's
10 times the diameter, which is usually easy, it's 100 times the
area.) That means the level in the well will drop much, much less -
1/100 of the U-tube's drop. That's usually negligible, and it allows
using a fixed scale, because 99% of the depth difference is occurring
against that scale. But the pressure difference is still proportional
to the height difference, and nothing else.

Even more sensitive are inclined well manometers. Their reading leg
runs at a shallow angle. The liquid level travels farther along that
slope to generate the same height difference, so their scale is
amplified and smaller pressure changes are visible. But it's still
height difference that actually matters.

A mercury barometer is normally a well-type manometer: there is an
enlarged reservoir exposed to atmospheric pressure, so that the effect
of the falling level on that side is small enough that the scale on the
other side in millimeters is close to a measurement in real millimeters.
If both sides were the same size then the "millimeters" on the scale
would actually be only half a millimeter long.

Right, as I described above. We agree.

I once had use of a super-precise mercury barometer. It featured a
movable scale with a tiny pointer to touch the surface of the mercury
pool and establish "zero." It also had a vernier scale for precisely
measuring height. Note, though, that if that barometer were modified
so its well was even larger - say, twice the diameter - nothing else
would need to be changed.

If you actually measure the difference that's true. Garden variety
manometers, like the one my old doctor used for blood pressure, are
just read on one leg.

On Thursday, January 14, 2021 at 6:24:34 PM UTC-8, Radey Shouman wrote:
Frank Krygowski writes:

On 1/14/2021 4:20 PM, Radey Shouman wrote:
Frank Krygowski writes:

On 1/10/2021 3:33 AM, Jeff Liebermann wrote:
On Sun, 10 Jan 2021 06:53:40 +0000 (UTC), Ralph Barone
wrote:

If I had to calibrate a pressure gauge, I’d consider building a manometer
with different diameter tubing on both sides. Then with some water and some
math, you can calibrate from first principles.

Well... the column of water would need to weigh the same as the
maximum gauge pressure. Basically, the air pressure in the tire will
need to lift a column of water that produces the same pressure.

Water weighs 8.33 lbs/gallon. To make the math easy, we start with a
tube that has an area of 1 sq-inch at each end. That's about:
1 sq-in = Pi * r^2
r = (1 / 3.14)^0.5 = 0.564 in = 1.12 in diameter
For 100 psi full scale, the volume of water would be:
100 lbs / 8.33 lbs/gallon = 12 gallons
1 gallon of water is 231 cubic inches. For a 1.12 in diameter tube,
the necessary column height would be:
h = vol / (Pi * r^2) = 231 / (3.14 * 0.564^2) = 231 inches = 19 ft
Yes, it might work, but would rather huge and messy to build.
Dissimilar tube diameters and a denser liquid (mercury) would help,
but it would still be huge and require 100 lbs of water. Radically
dissimilar diameter tubes will certainly work, if I wanted a small
water pond in my shop. No matter what shape I build the open end tube
of the manometer, the water still needs to weigh at least 100 lbs and
occupy 12 gallons. Mercury is 13.5 times as dense as water and
therefore will require 13.5 times less liquid or about:
12 gallons / 13.5 = 0.89 gallons of mercury
I have some mercury, but not that much:
http://www.learnbydestroying.com/jeffl/crud/mercury.jpg
Also, stuffing a Presta or Schrader valve into a rubber stopper, which
is then pounded into the open end of the glass manometer tube, with
100 lbs of water behind it, does not seem like something I would want
in my shop. Ummm... do you have any better less dangerous ideas?

Note: It's past midnight here. My math ability is normally terrible
and deteriorates rapidly with lack of sleep. Please check my numbers..

I'll give you some slack because of your sleepiness; but sorry, you're
way, way off. It's not math so much as units of measurement, starting
with your statement that "the column of water would need to weigh the
same as the maximum gauge pressure." That doesn't work, because weight
and pressure are fundamentally different quantities that can't be
equated.

The relationship you need is this: Pressure = weight density x depth.
So the necessary depth (or height of a manometer column) is the
desired pressure divided by the weight density of the fluid.

Water's weight density is about 62.4 lb/ft^3 or 0.0361 lb/in^3.
100 psi divided by 0.0361 lb/1n^3 gives 2770 inches or 231
feet. [Hmm... coincidentally, the number of cubic inches in a
gallon. Weird.]

Anyway, a water manometer would need to be 231 feet tall to give 100
psi. It's not practical. A mercury manometer would be about 16.7 feet
tall.

And BTW, the idea of different diameters doesn't work at all. Pressure
= density x height has no "area" term in it. Consider a tank of liquid
with a thin transparent sight tube on its outside to show the depth
within the tank. The pressure at the bottom of the sight tube (maybe
1/4" ID) is the same as the pressure just inside the tank at that
level, even though the tank's area is much, much greater.

I'm not at all sure on that last statement. The example of the sight
gauge is not relevant, because the pressure inside the tank is the same
as that at the top of the sight gauge. There are devices called "well
manometers" that, by design, use different areas for the tube exposed to
the measured pressure than they do for the one exposed to the reference
pressure. Typically the tube for the measured pressure is larger, which
makes the height variations in the other tube larger than they would be
for a bog standard manometer.

I'm familiar with well manometers. But you seem to be misstating their
benefit. They allow for more convenient reading at a certain level of
sensitivity, but it's still true that the height difference in the
legs depends only on the pressure difference. Area doesn't come into
play, except to shift (almost) all the height change to one leg.
That's true.
In a standard u-tube manometer, the area of the "reading" leg (where a
scale is typically attached) is the same as the area of the sensing
leg (attached to the pressure being measured). When pressure in the
sensing leg increases, its level drops and the level in the reading
leg rises by the same amount.

This is fine, but again, the pressure measurement is indicated by the
_difference_ in height of the two. So to read that height difference,
you have two choices: 1) Use a moveable scale, so for each new
measurement you have to lower the "zero" point to the liquid level in
the sensing leg; or 2) use a fixed scale against the reading leg, with
marks half as far apart as they'd normally be, since the height
difference is split between the two legs.
I agree that in a not-too-idealized world it's the difference in heights
that is proportional to the difference in pressure. The height
difference really depends not on the density of the manometer fluid, but
on the difference in density between the manometer fluid and whatever it
displaces. If that fluid is different in the two legs (eg air on one
side and almost nothing on the other) then I *think* that differences in
the area between them could change the height difference.
The problems are these: Scheme #1 is inconvenient; and scheme #2 is
half as sensitive.
Scheme #1 is also less accurate, because two readings with error have to
be subtracted, and the errors come close to adding.
A well manometer has a much greater area in the sensing leg. (If it's
10 times the diameter, which is usually easy, it's 100 times the
area.) That means the level in the well will drop much, much less -
1/100 of the U-tube's drop. That's usually negligible, and it allows
using a fixed scale, because 99% of the depth difference is occurring
against that scale. But the pressure difference is still proportional
to the height difference, and nothing else.

Even more sensitive are inclined well manometers. Their reading leg
runs at a shallow angle. The liquid level travels farther along that
slope to generate the same height difference, so their scale is
amplified and smaller pressure changes are visible. But it's still
height difference that actually matters.

A mercury barometer is normally a well-type manometer: there is an
enlarged reservoir exposed to atmospheric pressure, so that the effect
of the falling level on that side is small enough that the scale on the
other side in millimeters is close to a measurement in real millimeters.
If both sides were the same size then the "millimeters" on the scale
would actually be only half a millimeter long.

Right, as I described above. We agree.

I once had use of a super-precise mercury barometer. It featured a
movable scale with a tiny pointer to touch the surface of the mercury
pool and establish "zero." It also had a vernier scale for precisely
measuring height. Note, though, that if that barometer were modified
so its well was even larger - say, twice the diameter - nothing else
would need to be changed.
If you actually measure the difference that's true. Garden variety
manometers, like the one my old doctor used for blood pressure, are
just read on one leg.
I guess I'm just losing this discussion. Pressure is measured in Newtons per square meter, so area is the most important point in a manometer all other things being equal.

On 1/15/2021 3:04 PM, Tom Kunich wrote:
On Thursday, January 14, 2021 at 6:24:34 PM UTC-8, Radey Shouman wrote:
Frank Krygowski writes:

On 1/14/2021 4:20 PM, Radey Shouman wrote:
Frank Krygowski writes:

On 1/10/2021 3:33 AM, Jeff Liebermann wrote:
On Sun, 10 Jan 2021 06:53:40 +0000 (UTC), Ralph Barone
wrote:

If I had to calibrate a pressure gauge, I’d consider building a manometer
with different diameter tubing on both sides. Then with some water and some
math, you can calibrate from first principles.

Well... the column of water would need to weigh the same as the
maximum gauge pressure. Basically, the air pressure in the tire will
need to lift a column of water that produces the same pressure.

Water weighs 8.33 lbs/gallon. To make the math easy, we start with a
tube that has an area of 1 sq-inch at each end. That's about:
1 sq-in = Pi * r^2
r = (1 / 3.14)^0.5 = 0.564 in = 1.12 in diameter
For 100 psi full scale, the volume of water would be:
100 lbs / 8.33 lbs/gallon = 12 gallons
1 gallon of water is 231 cubic inches. For a 1.12 in diameter tube,
the necessary column height would be:
h = vol / (Pi * r^2) = 231 / (3.14 * 0.564^2) = 231 inches = 19 ft
Yes, it might work, but would rather huge and messy to build.
Dissimilar tube diameters and a denser liquid (mercury) would help,
but it would still be huge and require 100 lbs of water. Radically
dissimilar diameter tubes will certainly work, if I wanted a small
water pond in my shop. No matter what shape I build the open end tube
of the manometer, the water still needs to weigh at least 100 lbs and
occupy 12 gallons. Mercury is 13.5 times as dense as water and
therefore will require 13.5 times less liquid or about:
12 gallons / 13.5 = 0.89 gallons of mercury
I have some mercury, but not that much:
http://www.learnbydestroying.com/jeffl/crud/mercury.jpg
Also, stuffing a Presta or Schrader valve into a rubber stopper, which
is then pounded into the open end of the glass manometer tube, with
100 lbs of water behind it, does not seem like something I would want
in my shop. Ummm... do you have any better less dangerous ideas?

Note: It's past midnight here. My math ability is normally terrible
and deteriorates rapidly with lack of sleep. Please check my numbers.

I'll give you some slack because of your sleepiness; but sorry, you're
way, way off. It's not math so much as units of measurement, starting
with your statement that "the column of water would need to weigh the
same as the maximum gauge pressure." That doesn't work, because weight
and pressure are fundamentally different quantities that can't be
equated.

The relationship you need is this: Pressure = weight density x depth.
So the necessary depth (or height of a manometer column) is the
desired pressure divided by the weight density of the fluid.

Water's weight density is about 62.4 lb/ft^3 or 0.0361 lb/in^3.
100 psi divided by 0.0361 lb/1n^3 gives 2770 inches or 231
feet. [Hmm... coincidentally, the number of cubic inches in a
gallon. Weird.]

Anyway, a water manometer would need to be 231 feet tall to give 100
psi. It's not practical. A mercury manometer would be about 16.7 feet
tall.

And BTW, the idea of different diameters doesn't work at all. Pressure
= density x height has no "area" term in it. Consider a tank of liquid
with a thin transparent sight tube on its outside to show the depth
within the tank. The pressure at the bottom of the sight tube (maybe
1/4" ID) is the same as the pressure just inside the tank at that
level, even though the tank's area is much, much greater.

I'm not at all sure on that last statement. The example of the sight
gauge is not relevant, because the pressure inside the tank is the same
as that at the top of the sight gauge. There are devices called "well
manometers" that, by design, use different areas for the tube exposed to
the measured pressure than they do for the one exposed to the reference
pressure. Typically the tube for the measured pressure is larger, which
makes the height variations in the other tube larger than they would be
for a bog standard manometer.

I'm familiar with well manometers. But you seem to be misstating their
benefit. They allow for more convenient reading at a certain level of
sensitivity, but it's still true that the height difference in the
legs depends only on the pressure difference. Area doesn't come into
play, except to shift (almost) all the height change to one leg.
That's true.
In a standard u-tube manometer, the area of the "reading" leg (where a
scale is typically attached) is the same as the area of the sensing
leg (attached to the pressure being measured). When pressure in the
sensing leg increases, its level drops and the level in the reading
leg rises by the same amount.

This is fine, but again, the pressure measurement is indicated by the
_difference_ in height of the two. So to read that height difference,
you have two choices: 1) Use a moveable scale, so for each new
measurement you have to lower the "zero" point to the liquid level in
the sensing leg; or 2) use a fixed scale against the reading leg, with
marks half as far apart as they'd normally be, since the height
difference is split between the two legs.
I agree that in a not-too-idealized world it's the difference in heights
that is proportional to the difference in pressure. The height
difference really depends not on the density of the manometer fluid, but
on the difference in density between the manometer fluid and whatever it
displaces. If that fluid is different in the two legs (eg air on one
side and almost nothing on the other) then I *think* that differences in
the area between them could change the height difference.
The problems are these: Scheme #1 is inconvenient; and scheme #2 is
half as sensitive.
Scheme #1 is also less accurate, because two readings with error have to
be subtracted, and the errors come close to adding.
A well manometer has a much greater area in the sensing leg. (If it's
10 times the diameter, which is usually easy, it's 100 times the
area.) That means the level in the well will drop much, much less -
1/100 of the U-tube's drop. That's usually negligible, and it allows
using a fixed scale, because 99% of the depth difference is occurring
against that scale. But the pressure difference is still proportional
to the height difference, and nothing else.

Even more sensitive are inclined well manometers. Their reading leg
runs at a shallow angle. The liquid level travels farther along that
slope to generate the same height difference, so their scale is
amplified and smaller pressure changes are visible. But it's still
height difference that actually matters.

A mercury barometer is normally a well-type manometer: there is an
enlarged reservoir exposed to atmospheric pressure, so that the effect
of the falling level on that side is small enough that the scale on the
other side in millimeters is close to a measurement in real millimeters.
If both sides were the same size then the "millimeters" on the scale
would actually be only half a millimeter long.

Right, as I described above. We agree.

I once had use of a super-precise mercury barometer. It featured a
movable scale with a tiny pointer to touch the surface of the mercury
pool and establish "zero." It also had a vernier scale for precisely
measuring height. Note, though, that if that barometer were modified
so its well was even larger - say, twice the diameter - nothing else
would need to be changed.
If you actually measure the difference that's true. Garden variety
manometers, like the one my old doctor used for blood pressure, are
just read on one leg.
I guess I'm just losing this discussion. Pressure is measured in Newtons per square meter, so area is the most important point in a manometer all other things being equal.


SMH


--
- Frank Krygowski

On Friday, January 15, 2021 at 1:34:35 PM UTC-8, Frank Krygowski wrote:
On 1/15/2021 3:04 PM, Tom Kunich wrote:
On Thursday, January 14, 2021 at 6:24:34 PM UTC-8, Radey Shouman wrote:
Frank Krygowski writes:

On 1/14/2021 4:20 PM, Radey Shouman wrote:
Frank Krygowski writes:

On 1/10/2021 3:33 AM, Jeff Liebermann wrote:
On Sun, 10 Jan 2021 06:53:40 +0000 (UTC), Ralph Barone
wrote:

If I had to calibrate a pressure gauge, I’d consider building a manometer
with different diameter tubing on both sides. Then with some water and some
math, you can calibrate from first principles.

Well... the column of water would need to weigh the same as the
maximum gauge pressure. Basically, the air pressure in the tire will
need to lift a column of water that produces the same pressure.

Water weighs 8.33 lbs/gallon. To make the math easy, we start with a
tube that has an area of 1 sq-inch at each end. That's about:
1 sq-in = Pi * r^2
r = (1 / 3.14)^0.5 = 0.564 in = 1.12 in diameter
For 100 psi full scale, the volume of water would be:
100 lbs / 8.33 lbs/gallon = 12 gallons
1 gallon of water is 231 cubic inches. For a 1.12 in diameter tube,
the necessary column height would be:
h = vol / (Pi * r^2) = 231 / (3.14 * 0.564^2) = 231 inches = 19 ft
Yes, it might work, but would rather huge and messy to build.
Dissimilar tube diameters and a denser liquid (mercury) would help,
but it would still be huge and require 100 lbs of water. Radically
dissimilar diameter tubes will certainly work, if I wanted a small
water pond in my shop. No matter what shape I build the open end tube
of the manometer, the water still needs to weigh at least 100 lbs and
occupy 12 gallons. Mercury is 13.5 times as dense as water and
therefore will require 13.5 times less liquid or about:
12 gallons / 13.5 = 0.89 gallons of mercury
I have some mercury, but not that much:
http://www.learnbydestroying.com/jeffl/crud/mercury.jpg
Also, stuffing a Presta or Schrader valve into a rubber stopper, which
is then pounded into the open end of the glass manometer tube, with
100 lbs of water behind it, does not seem like something I would want
in my shop. Ummm... do you have any better less dangerous ideas?

Note: It's past midnight here. My math ability is normally terrible
and deteriorates rapidly with lack of sleep. Please check my numbers.

I'll give you some slack because of your sleepiness; but sorry, you're
way, way off. It's not math so much as units of measurement, starting
with your statement that "the column of water would need to weigh the
same as the maximum gauge pressure." That doesn't work, because weight
and pressure are fundamentally different quantities that can't be
equated.

The relationship you need is this: Pressure = weight density x depth.
So the necessary depth (or height of a manometer column) is the
desired pressure divided by the weight density of the fluid.

Water's weight density is about 62.4 lb/ft^3 or 0.0361 lb/in^3.
100 psi divided by 0.0361 lb/1n^3 gives 2770 inches or 231
feet. [Hmm... coincidentally, the number of cubic inches in a
gallon. Weird.]

Anyway, a water manometer would need to be 231 feet tall to give 100
psi. It's not practical. A mercury manometer would be about 16.7 feet
tall.

And BTW, the idea of different diameters doesn't work at all. Pressure
= density x height has no "area" term in it. Consider a tank of liquid
with a thin transparent sight tube on its outside to show the depth
within the tank. The pressure at the bottom of the sight tube (maybe
1/4" ID) is the same as the pressure just inside the tank at that
level, even though the tank's area is much, much greater.

I'm not at all sure on that last statement. The example of the sight
gauge is not relevant, because the pressure inside the tank is the same
as that at the top of the sight gauge. There are devices called "well
manometers" that, by design, use different areas for the tube exposed to
the measured pressure than they do for the one exposed to the reference
pressure. Typically the tube for the measured pressure is larger, which
makes the height variations in the other tube larger than they would be
for a bog standard manometer.

I'm familiar with well manometers. But you seem to be misstating their
benefit. They allow for more convenient reading at a certain level of
sensitivity, but it's still true that the height difference in the
legs depends only on the pressure difference. Area doesn't come into
play, except to shift (almost) all the height change to one leg.
That's true.
In a standard u-tube manometer, the area of the "reading" leg (where a
scale is typically attached) is the same as the area of the sensing
leg (attached to the pressure being measured). When pressure in the
sensing leg increases, its level drops and the level in the reading
leg rises by the same amount.

This is fine, but again, the pressure measurement is indicated by the
_difference_ in height of the two. So to read that height difference,
you have two choices: 1) Use a moveable scale, so for each new
measurement you have to lower the "zero" point to the liquid level in
the sensing leg; or 2) use a fixed scale against the reading leg, with
marks half as far apart as they'd normally be, since the height
difference is split between the two legs.
I agree that in a not-too-idealized world it's the difference in heights
that is proportional to the difference in pressure. The height
difference really depends not on the density of the manometer fluid, but
on the difference in density between the manometer fluid and whatever it
displaces. If that fluid is different in the two legs (eg air on one
side and almost nothing on the other) then I *think* that differences in
the area between them could change the height difference.
The problems are these: Scheme #1 is inconvenient; and scheme #2 is
half as sensitive.
Scheme #1 is also less accurate, because two readings with error have to
be subtracted, and the errors come close to adding.
A well manometer has a much greater area in the sensing leg. (If it's
10 times the diameter, which is usually easy, it's 100 times the
area.) That means the level in the well will drop much, much less -
1/100 of the U-tube's drop. That's usually negligible, and it allows
using a fixed scale, because 99% of the depth difference is occurring
against that scale. But the pressure difference is still proportional
to the height difference, and nothing else.

Even more sensitive are inclined well manometers. Their reading leg
runs at a shallow angle. The liquid level travels farther along that
slope to generate the same height difference, so their scale is
amplified and smaller pressure changes are visible. But it's still
height difference that actually matters.

A mercury barometer is normally a well-type manometer: there is an
enlarged reservoir exposed to atmospheric pressure, so that the effect
of the falling level on that side is small enough that the scale on the
other side in millimeters is close to a measurement in real millimeters.
If both sides were the same size then the "millimeters" on the scale
would actually be only half a millimeter long.

Right, as I described above. We agree.

I once had use of a super-precise mercury barometer. It featured a
movable scale with a tiny pointer to touch the surface of the mercury
pool and establish "zero." It also had a vernier scale for precisely
measuring height. Note, though, that if that barometer were modified
so its well was even larger - say, twice the diameter - nothing else
would need to be changed.
If you actually measure the difference that's true. Garden variety
manometers, like the one my old doctor used for blood pressure, are
just read on one leg.
I guess I'm just losing this discussion. Pressure is measured in Newtons per square meter, so area is the most important point in a manometer all other things being equal.

SMH

About what Frank? You are supposed to be a teacher and appear to be doing a terrible job of explaining this issue of a manometer not needing to be calculated using area.

On 1/15/2021 4:57 PM, Tom Kunich wrote:
On Friday, January 15, 2021 at 1:34:35 PM UTC-8, Frank Krygowski wrote:
On 1/15/2021 3:04 PM, Tom Kunich wrote:
On Thursday, January 14, 2021 at 6:24:34 PM UTC-8, Radey Shouman wrote:
Frank Krygowski writes:

On 1/14/2021 4:20 PM, Radey Shouman wrote:
Frank Krygowski writes:

On 1/10/2021 3:33 AM, Jeff Liebermann wrote:
On Sun, 10 Jan 2021 06:53:40 +0000 (UTC), Ralph Barone
wrote:

If I had to calibrate a pressure gauge, I’d consider building a manometer
with different diameter tubing on both sides. Then with some water and some
math, you can calibrate from first principles.

Well... the column of water would need to weigh the same as the
maximum gauge pressure. Basically, the air pressure in the tire will
need to lift a column of water that produces the same pressure.

Water weighs 8.33 lbs/gallon. To make the math easy, we start with a
tube that has an area of 1 sq-inch at each end. That's about:
1 sq-in = Pi * r^2
r = (1 / 3.14)^0.5 = 0.564 in = 1.12 in diameter
For 100 psi full scale, the volume of water would be:
100 lbs / 8.33 lbs/gallon = 12 gallons
1 gallon of water is 231 cubic inches. For a 1.12 in diameter tube,
the necessary column height would be:
h = vol / (Pi * r^2) = 231 / (3.14 * 0.564^2) = 231 inches = 19 ft
Yes, it might work, but would rather huge and messy to build.
Dissimilar tube diameters and a denser liquid (mercury) would help,
but it would still be huge and require 100 lbs of water. Radically
dissimilar diameter tubes will certainly work, if I wanted a small
water pond in my shop. No matter what shape I build the open end tube
of the manometer, the water still needs to weigh at least 100 lbs and
occupy 12 gallons. Mercury is 13.5 times as dense as water and
therefore will require 13.5 times less liquid or about:
12 gallons / 13.5 = 0.89 gallons of mercury
I have some mercury, but not that much:
http://www.learnbydestroying.com/jeffl/crud/mercury.jpg
Also, stuffing a Presta or Schrader valve into a rubber stopper, which
is then pounded into the open end of the glass manometer tube, with
100 lbs of water behind it, does not seem like something I would want
in my shop. Ummm... do you have any better less dangerous ideas?

Note: It's past midnight here. My math ability is normally terrible
and deteriorates rapidly with lack of sleep. Please check my numbers.

I'll give you some slack because of your sleepiness; but sorry, you're
way, way off. It's not math so much as units of measurement, starting
with your statement that "the column of water would need to weigh the
same as the maximum gauge pressure." That doesn't work, because weight
and pressure are fundamentally different quantities that can't be
equated.

The relationship you need is this: Pressure = weight density x depth.
So the necessary depth (or height of a manometer column) is the
desired pressure divided by the weight density of the fluid.

Water's weight density is about 62.4 lb/ft^3 or 0.0361 lb/in^3.
100 psi divided by 0.0361 lb/1n^3 gives 2770 inches or 231
feet. [Hmm... coincidentally, the number of cubic inches in a
gallon. Weird.]

Anyway, a water manometer would need to be 231 feet tall to give 100
psi. It's not practical. A mercury manometer would be about 16.7 feet
tall.

And BTW, the idea of different diameters doesn't work at all. Pressure
= density x height has no "area" term in it. Consider a tank of liquid
with a thin transparent sight tube on its outside to show the depth
within the tank. The pressure at the bottom of the sight tube (maybe
1/4" ID) is the same as the pressure just inside the tank at that
level, even though the tank's area is much, much greater.

I'm not at all sure on that last statement. The example of the sight
gauge is not relevant, because the pressure inside the tank is the same
as that at the top of the sight gauge. There are devices called "well
manometers" that, by design, use different areas for the tube exposed to
the measured pressure than they do for the one exposed to the reference
pressure. Typically the tube for the measured pressure is larger, which
makes the height variations in the other tube larger than they would be
for a bog standard manometer.

I'm familiar with well manometers. But you seem to be misstating their
benefit. They allow for more convenient reading at a certain level of
sensitivity, but it's still true that the height difference in the
legs depends only on the pressure difference. Area doesn't come into
play, except to shift (almost) all the height change to one leg.
That's true.
In a standard u-tube manometer, the area of the "reading" leg (where a
scale is typically attached) is the same as the area of the sensing
leg (attached to the pressure being measured). When pressure in the
sensing leg increases, its level drops and the level in the reading
leg rises by the same amount.

This is fine, but again, the pressure measurement is indicated by the
_difference_ in height of the two. So to read that height difference,
you have two choices: 1) Use a moveable scale, so for each new
measurement you have to lower the "zero" point to the liquid level in
the sensing leg; or 2) use a fixed scale against the reading leg, with
marks half as far apart as they'd normally be, since the height
difference is split between the two legs.
I agree that in a not-too-idealized world it's the difference in heights
that is proportional to the difference in pressure. The height
difference really depends not on the density of the manometer fluid, but
on the difference in density between the manometer fluid and whatever it
displaces. If that fluid is different in the two legs (eg air on one
side and almost nothing on the other) then I *think* that differences in
the area between them could change the height difference.
The problems are these: Scheme #1 is inconvenient; and scheme #2 is
half as sensitive.
Scheme #1 is also less accurate, because two readings with error have to
be subtracted, and the errors come close to adding.
A well manometer has a much greater area in the sensing leg. (If it's
10 times the diameter, which is usually easy, it's 100 times the
area.) That means the level in the well will drop much, much less -
1/100 of the U-tube's drop. That's usually negligible, and it allows
using a fixed scale, because 99% of the depth difference is occurring
against that scale. But the pressure difference is still proportional
to the height difference, and nothing else.

Even more sensitive are inclined well manometers. Their reading leg
runs at a shallow angle. The liquid level travels farther along that
slope to generate the same height difference, so their scale is
amplified and smaller pressure changes are visible. But it's still
height difference that actually matters.

A mercury barometer is normally a well-type manometer: there is an
enlarged reservoir exposed to atmospheric pressure, so that the effect
of the falling level on that side is small enough that the scale on the
other side in millimeters is close to a measurement in real millimeters.
If both sides were the same size then the "millimeters" on the scale
would actually be only half a millimeter long.

Right, as I described above. We agree.

I once had use of a super-precise mercury barometer. It featured a
movable scale with a tiny pointer to touch the surface of the mercury
pool and establish "zero." It also had a vernier scale for precisely
measuring height. Note, though, that if that barometer were modified
so its well was even larger - say, twice the diameter - nothing else
would need to be changed.
If you actually measure the difference that's true. Garden variety
manometers, like the one my old doctor used for blood pressure, are
just read on one leg.
I guess I'm just losing this discussion. Pressure is measured in Newtons per square meter, so area is the most important point in a manometer all other things being equal.

SMH

About what Frank? You are supposed to be a teacher and appear to be doing a terrible job of explaining this issue of a manometer not needing to be calculated using area.

I'd say instead that you're doing a terrible job of understanding it.

Tom, you don't have the qualifications to enter as a freshman into the
program I ran. And despite your overconfidence, you don't have the
background in physics to understand this manometer issue.

Go to a library, take out a book on Fluid Mechanics and read all the
chapters on Fluid Statics. Then come back with specific questions.

--
- Frank Krygowski

On Fri, 15 Jan 2021 13:57:21 -0800, Tom Kunich scribed:

On Friday, January 15, 2021 at 1:34:35 PM UTC-8, Frank Krygowski wrote:
On 1/15/2021 3:04 PM, Tom Kunich wrote:

If you actually measure the difference that's true. Garden variety
manometers, like the one my old doctor used for blood pressure, are
just read on one leg.
I guess I'm just losing this discussion. Pressure is measured in
Newtons per square meter, so area is the most important point in a
manometer all other things being equal.

SMH

About what Frank? You are supposed to be a teacher and appear to be
doing a terrible job of explaining this issue of a manometer not needing
to be calculated using area.

Did you say you had problems with blood pressure?
That can be solved by going to a doctor with a smaller manometer.
Simple enough.

On Fri, 15 Jan 2021 18:30:56 -0500, Frank Krygowski
wrote:

On 1/15/2021 4:57 PM, Tom Kunich wrote:
On Friday, January 15, 2021 at 1:34:35 PM UTC-8, Frank Krygowski wrote:
On 1/15/2021 3:04 PM, Tom Kunich wrote:
On Thursday, January 14, 2021 at 6:24:34 PM UTC-8, Radey Shouman wrote:
Frank Krygowski writes:

On 1/14/2021 4:20 PM, Radey Shouman wrote:
Frank Krygowski writes:

On 1/10/2021 3:33 AM, Jeff Liebermann wrote:
On Sun, 10 Jan 2021 06:53:40 +0000 (UTC), Ralph Barone
wrote:

If I had to calibrate a pressure gauge, I’d consider building a manometer
with different diameter tubing on both sides. Then with some water and some
math, you can calibrate from first principles.

Well... the column of water would need to weigh the same as the
maximum gauge pressure. Basically, the air pressure in the tire will
need to lift a column of water that produces the same pressure.

Water weighs 8.33 lbs/gallon. To make the math easy, we start with a
tube that has an area of 1 sq-inch at each end. That's about:
1 sq-in = Pi * r^2
r = (1 / 3.14)^0.5 = 0.564 in = 1.12 in diameter
For 100 psi full scale, the volume of water would be:
100 lbs / 8.33 lbs/gallon = 12 gallons
1 gallon of water is 231 cubic inches. For a 1.12 in diameter tube,
the necessary column height would be:
h = vol / (Pi * r^2) = 231 / (3.14 * 0.564^2) = 231 inches = 19 ft
Yes, it might work, but would rather huge and messy to build.
Dissimilar tube diameters and a denser liquid (mercury) would help,
but it would still be huge and require 100 lbs of water. Radically
dissimilar diameter tubes will certainly work, if I wanted a small
water pond in my shop. No matter what shape I build the open end tube
of the manometer, the water still needs to weigh at least 100 lbs and
occupy 12 gallons. Mercury is 13.5 times as dense as water and
therefore will require 13.5 times less liquid or about:
12 gallons / 13.5 = 0.89 gallons of mercury
I have some mercury, but not that much:
http://www.learnbydestroying.com/jeffl/crud/mercury.jpg
Also, stuffing a Presta or Schrader valve into a rubber stopper, which
is then pounded into the open end of the glass manometer tube, with
100 lbs of water behind it, does not seem like something I would want
in my shop. Ummm... do you have any better less dangerous ideas?

Note: It's past midnight here. My math ability is normally terrible
and deteriorates rapidly with lack of sleep. Please check my numbers.

I'll give you some slack because of your sleepiness; but sorry, you're
way, way off. It's not math so much as units of measurement, starting
with your statement that "the column of water would need to weigh the
same as the maximum gauge pressure." That doesn't work, because weight
and pressure are fundamentally different quantities that can't be
equated.

The relationship you need is this: Pressure = weight density x depth.
So the necessary depth (or height of a manometer column) is the
desired pressure divided by the weight density of the fluid.

Water's weight density is about 62.4 lb/ft^3 or 0.0361 lb/in^3.
100 psi divided by 0.0361 lb/1n^3 gives 2770 inches or 231
feet. [Hmm... coincidentally, the number of cubic inches in a
gallon. Weird.]

Anyway, a water manometer would need to be 231 feet tall to give 100
psi. It's not practical. A mercury manometer would be about 16.7 feet
tall.

And BTW, the idea of different diameters doesn't work at all. Pressure
= density x height has no "area" term in it. Consider a tank of liquid
with a thin transparent sight tube on its outside to show the depth
within the tank. The pressure at the bottom of the sight tube (maybe
1/4" ID) is the same as the pressure just inside the tank at that
level, even though the tank's area is much, much greater.

I'm not at all sure on that last statement. The example of the sight
gauge is not relevant, because the pressure inside the tank is the same
as that at the top of the sight gauge. There are devices called "well
manometers" that, by design, use different areas for the tube exposed to
the measured pressure than they do for the one exposed to the reference
pressure. Typically the tube for the measured pressure is larger, which
makes the height variations in the other tube larger than they would be
for a bog standard manometer.

I'm familiar with well manometers. But you seem to be misstating their
benefit. They allow for more convenient reading at a certain level of
sensitivity, but it's still true that the height difference in the
legs depends only on the pressure difference. Area doesn't come into
play, except to shift (almost) all the height change to one leg.
That's true.
In a standard u-tube manometer, the area of the "reading" leg (where a
scale is typically attached) is the same as the area of the sensing
leg (attached to the pressure being measured). When pressure in the
sensing leg increases, its level drops and the level in the reading
leg rises by the same amount.

This is fine, but again, the pressure measurement is indicated by the
_difference_ in height of the two. So to read that height difference,
you have two choices: 1) Use a moveable scale, so for each new
measurement you have to lower the "zero" point to the liquid level in
the sensing leg; or 2) use a fixed scale against the reading leg, with
marks half as far apart as they'd normally be, since the height
difference is split between the two legs.
I agree that in a not-too-idealized world it's the difference in heights
that is proportional to the difference in pressure. The height
difference really depends not on the density of the manometer fluid, but
on the difference in density between the manometer fluid and whatever it
displaces. If that fluid is different in the two legs (eg air on one
side and almost nothing on the other) then I *think* that differences in
the area between them could change the height difference.
The problems are these: Scheme #1 is inconvenient; and scheme #2 is
half as sensitive.
Scheme #1 is also less accurate, because two readings with error have to
be subtracted, and the errors come close to adding.
A well manometer has a much greater area in the sensing leg. (If it's
10 times the diameter, which is usually easy, it's 100 times the
area.) That means the level in the well will drop much, much less -
1/100 of the U-tube's drop. That's usually negligible, and it allows
using a fixed scale, because 99% of the depth difference is occurring
against that scale. But the pressure difference is still proportional
to the height difference, and nothing else.

Even more sensitive are inclined well manometers. Their reading leg
runs at a shallow angle. The liquid level travels farther along that
slope to generate the same height difference, so their scale is
amplified and smaller pressure changes are visible. But it's still
height difference that actually matters.

A mercury barometer is normally a well-type manometer: there is an
enlarged reservoir exposed to atmospheric pressure, so that the effect
of the falling level on that side is small enough that the scale on the
other side in millimeters is close to a measurement in real millimeters.
If both sides were the same size then the "millimeters" on the scale
would actually be only half a millimeter long.

Right, as I described above. We agree.

I once had use of a super-precise mercury barometer. It featured a
movable scale with a tiny pointer to touch the surface of the mercury
pool and establish "zero." It also had a vernier scale for precisely
measuring height. Note, though, that if that barometer were modified
so its well was even larger - say, twice the diameter - nothing else
would need to be changed.
If you actually measure the difference that's true. Garden variety
manometers, like the one my old doctor used for blood pressure, are
just read on one leg.
I guess I'm just losing this discussion. Pressure is measured in Newtons per square meter, so area is the most important point in a manometer all other things being equal.

SMH

About what Frank? You are supposed to be a teacher and appear to be doing a terrible job of explaining this issue of a manometer not needing to be calculated using area.

I'd say instead that you're doing a terrible job of understanding it.

Tom, you don't have the qualifications to enter as a freshman into the
program I ran. And despite your overconfidence, you don't have the
background in physics to understand this manometer issue.

Go to a library, take out a book on Fluid Mechanics and read all the
chapters on Fluid Statics. Then come back with specific questions.

Tom don't need no steenking books... He's on facebook!
--
Cheers,

John B.