Hi Michael...

www.google.com "how to use google"

I appreciate your reply. What a wonderful and unique idea you have
discovered for the human race to try looking on Google for research
information!

I'm not sure if any human being has ever tried using
Google to perform research before, so maybe on your advise I'll be the
first!

It sounds like you may have a knack for gathering specific data
through Google. Can you please tell me a URL that you can find that
documents the internal pressure of a ping pong ball?

Is this a homework problem? What have you done toward solving the

Yes, this question is sort of homework, but not for school. I am the
Science Advisor for the third season of a very popular cable show that
proves or disproves urban stories through the use of scientific
experiments and blowing stuff up.

We are working on a story right now
for which I need this data.

Sarcasm aside

, of course I did scour the web and USENET with Google
looking for this data...actually for 2 days with no success...before
posting this inquiry as a last resort before conducting an experiment
myself. I also had our researchers telephone a ha'f dozen ping pong
ball manufacturers for the info with no results.

So, Michael, I am very curious if you yourself can actually use Google
to find the data on the PSI inside a standard ping pong ball.

Please
let me know what you find by posting your findings here, and I'll post
my results next week (Monday or Tuesday) after I conduct the tests.

I have a followup on this thread.

Today I conducted the soap bubble experiment as was suggested by Uncle
Al, and it resulted in absolutely no bubble when a hole was poked
through the wall poked with a small needle. I ran the test several
times, and even with the ball surface slightly below the water, and
saw no bubbles at all. Also, I confirmed that no water was sucked
into the hole in the ball, because the ball weighed the same after
dried off than it did before being poked under water (2.88g before and
after).

I want to tell you all how much I appreciate your suggestions.

Later this week we will be taking a sample of ping pong balls to a
local pressure chamber and determining the pressure at which they
implode. You'll probably see that on the **show** but I'll report
back with the numbers when I get them.

Finally, I would like to respond to the critisism directed towards me
for my questions. Even though I am the Science Advisor for the
**show**, I am certainly not an EXPERT in EVERY area of science. Few
people are. I am part of the **show** because I have GENERAL
knowledge in MANY areas of science, as well as skills in mechanical
design and fabrication, wide-topic research, and technical writing and
illustration.

When I am trying to research a topic for the **show** which is outside
of my area of knowledge, Instead of guessing or making up some BS to
sound smart, I seek the advice of those who know the subject, and then
I do the experiment.

What's wrong with that?

Now, if only someone could tell me how to get 300,000 ping pong balls
through customs. ;) Just kidding! They're on their way.

Thanks!

Hi Uncle Al...

This is a great idea for a method to find out if the ping pong ball

pressurized, and if so, to what degree.

By "the Ideal Gas Law" do you mean "the Perfect Gas Law"? That is
(from my Pocket Ref that I always have with me):

PV = nRT

where:

P = pressure in atmospheres (would be "1")
V = volume in liters (would be estimated from the soap bubble)
n = number of moles
R = gas constant (0.0821 liter-atms / K / mole)
T = temperature in K (would be 295 for room temperature)

So I would have:

V = n * R * 295

I don't really understand "n" and "R".

Let's say I do the pin ***** soap test, and I get a soap gas bubble
that is, say, 1.25" in diameter. The volume if it were a perfect
sphere would be 1.022 cubic inches, or 16.748 ml. How would I apply
my numbers to the Perfect Gas Law?

Out of curiosity, would I get different numbers if the ball

pressurized helium? (Not that it would be likely...it is probably
filled with mixed air or nitrogen.) This is probably where "n" and
"R" come into play.

Could I just figure out the percentage of compression from the

of the ping pong ball at 1 atm plus the volume of the bubble?

If the volume of the soap bubble was exactly the same as the

volume of the ping pong ball, then could I say that the ball was
pressurized to 2 atms or 14.696 PSI * 2 or 29.392 PSI?

Hi Uncle Al...

This is a great idea for a method to find out if the ping pong ball is
pressurized, and if so, to what degree.

By "the Ideal Gas Law" do you mean "the Perfect Gas Law"? That is
(from my Pocket Ref that I always have with me):

PV = nRT

where:

P = pressure in atmospheres (would be "1")
V = volume in liters (would be estimated from the soap bubble)
n = number of moles
R = gas constant (0.0821 liter-atms / K / mole)
T = temperature in K (would be 295 for room temperature)

So I would have:

V = n * R * 295

I don't really understand "n" and "R".

Let's say I do the pin ***** soap test, and I get a soap gas bubble
that is, say, 1.25" in diameter. The volume if it were a perfect
sphere would be 1.022 cubic inches, or 16.748 ml. How would I apply
my numbers to the Perfect Gas Law?

Out of curiosity, would I get different numbers if the ball contained
pressurized helium? (Not that it would be likely...it is probably
filled with mixed air or nitrogen.) This is probably where "n" and
"R" come into play.

Could I just figure out the percentage of compression from the volume
of the ping pong ball at 1 atm plus the volume of the bubble?

If the volume of the soap bubble was exactly the same as the internal
volume of the ping pong ball, then could I say that the ball was
pressurized to 2 atms or 14.696 PSI * 2 or 29.392 PSI?

This is a great idea! This is what USENET is for!

BTW, I'm glad you pointed out that the mechanical strength of the
shell is what I really need to look at. But wouldn't a higher
internal pressure help support a greater external pressure given an
equal shell strength?

We have a small pressure chamber at the studios (it was in a scene in
last season's show when it violently exploded while one of our hosts
was using it). I will fix the chamber seal and use it to try to
implode a ball and compare the failure pressure with the depth of
water that would produce that pressure (I already put together a table
of those figures down to 50 feet in fresh water).

Thanks, Uncle Al!

Uncle Al <UncleAl0@hate.spam.net> wrote in message

Ike wrote:

I have some questions about the standard 40mm ping pong ball:

1.) Is a ping pong ball filled with pressurized air, or air at normal
atmospheric pressure?

2.) How could I determine the internal pressure of a ping pong ball
through an experiment?

3.) Based on a given internal pressure from the determination above,
how deep could you submerge a ping pong ball in fresh water before it
would implode?

1) Wet surface of ping-pong ball with shampoo in water,
2) Pin *****.
3) See whether it sucks or blows. If it blows, measure volume of
bubble and work backwards with the Ideal Gas law.

Internal pressure does not determine implosion depth. Mechanical
failure of the shell does that. There can be a large metastable
region in-between.

Hi Uncle Al...

This is a great idea for a method to find out if the ping pong ball is
pressurized, and if so, to what degree.

By "the Ideal Gas Law" do you mean "the Perfect Gas Law"? That is
(from my Pocket Ref that I always have with me):

PV = nRT

where:

P = pressure in atmospheres (would be "1")
V = volume in liters (would be estimated from the soap bubble)
n = number of moles
R = gas constant (0.0821 liter-atms / K / mole)
T = temperature in K (would be 295 for room temperature)

So I would have:

V = n * R * 295

I don't really understand "n" and "R".

Let's say I do the pin ***** soap test, and I get a soap gas bubble
that is, say, 1.25" in diameter. The volume if it were a perfect
sphere would be 1.022 cubic inches, or 16.748 ml. How would I apply
my numbers to the Perfect Gas Law?

Out of curiosity, would I get different numbers if the ball contained
pressurized helium? (Not that it would be likely...it is probably
filled with mixed air or nitrogen.) This is probably where "n" and
"R" come into play.

Could I just figure out the percentage of compression from the volume
of the ping pong ball at 1 atm plus the volume of the bubble?

If the volume of the soap bubble was exactly the same as the internal
volume of the ping pong ball, then could I say that the ball was
pressurized to 2 atms or 14.696 PSI * 2 or 29.392 PSI?

This is a great idea! This is what USENET is for!

BTW, I'm glad you pointed out that the mechanical strength of the
shell is what I really need to look at. But wouldn't a higher
internal pressure help support a greater external pressure given an
equal shell strength?

We have a small pressure chamber at the studios (it was in a scene in
last season's show when it violently exploded while one of our hosts
was using it). I will fix the chamber seal and use it to try to
implode a ball and compare the failure pressure with the depth of
water that would produce that pressure (I already put together a table
of those figures down to 50 feet in fresh water).

Thanks, Uncle Al!

By "the Ideal Gas Law" do you mean "the Perfect Gas Law"? That is
(from my Pocket Ref that I always have with me):

PV = nRT

where:

P = pressure in atmospheres (would be "1")
V = volume in liters (would be estimated from the soap bubble)
n = number of moles
R = gas constant (0.0821 liter-atms / K / mole)
T = temperature in K (would be 295 for room temperature)

So I would have:

V = n * R * 295

I don't really understand "n" and "R".

Let's say I do the pin ***** soap test, and I get a soap gas bubble
that is, say, 1.25" in diameter. The volume if it were a perfect
sphere would be 1.022 cubic inches, or 16.748 ml. How would I apply
my numbers to the Perfect Gas Law?

Out of curiosity, would I get different numbers if the ball contained
pressurized helium?

(Not that it would be likely...it is probably
filled with mixed air or nitrogen.) This is probably where "n" and
"R" come into play.

Could I just figure out the percentage of compression from the volume
of the ping pong ball at 1 atm plus the volume of the bubble?

If the volume of the soap bubble was exactly the same as the internal
volume of the ping pong ball, then could I say that the ball was
pressurized to 2 atms or 14.696 PSI * 2 or 29.392 PSI?

This is a great idea! This is what USENET is for!

BTW, I'm glad you pointed out that the mechanical strength of the
shell is what I really need to look at. But wouldn't a higher
internal pressure help support a greater external pressure given an
equal shell strength?

We have a small pressure chamber at the studios (it was in a scene in
last season's show when it violently exploded while one of our hosts
was using it). I will fix the chamber seal and use it to try to
implode a ball and compare the failure pressure with the depth of
water that would produce that pressure (I already put together a table
of those figures down to 50 feet in fresh water).

On Wed, 19 May 2004 15:33:25 -0700, Uncle Al <UncleAl0@hate.spam.net>
wrote:

Ike wrote:

I have some questions about the standard 40mm ping pong ball:

1.) Is a ping pong ball filled with pressurized air, or air at normal
atmospheric pressure?

2.) How could I determine the internal pressure of a ping pong ball
through an experiment?

3.) Based on a given internal pressure from the determination above,
how deep could you submerge a ping pong ball in fresh water before it
would implode?

1) Wet surface of ping-pong ball with shampoo in water,
2) Pin *****.
3) See whether it sucks or blows. If it blows, measure volume of
bubble and work backwards with the Ideal Gas law.

Internal pressure does not determine implosion depth.

---
Beg to differ...

1. Starting at STP, take a ping-pong ball and internally pressurize it
until it explodes. Record that pressure and call it Pboom.

2. Again, starting at STP, take an identical ping-pong ball and
externally pressurize it until it implodes. Record that pressure
and call it Psquish.

3. Take a third identical ping-pong ball at STP and internally
pressurize it until (according to the measurement you made in 1.)
it's almost ready to blow up, and then seal it up.

4. Now, take that same ping-pong ball and start externally
pressurizing it. Before it can even _think_ about imploding it'll
have to go through the point where Psquish - Pboom
= 0, and then to crush it the new external pressure (let's call it
Psquishnew) will have to increase to the point where Psquishnew =
Psquish + Pboom, so internal pressure _does_ determine implosion
depth.

I agree with you that there'll be a region of uncertainty (the
so-called "metastable" region) where the collapse pressure will be
undefinable, but that window should be able to be narrowed by
controlling the chemical compositon of the ping-pong ball's shell, its
thickness, porosity, homogeneity and, probably most important, the
sphericity of its inner and outer walls and their concentricity.

You may have noticed that there are some inconsistencies in my
discussion with respect to temperature differentials and to small
pressure differentials tied to those temperature differences, but
that was intentional in that it's just a real pain in the ***** to get
into the nitty-gritty of it unless it's really, really, necessary. :-)

On Wed, 19 May 2004 15:33:25 -0700, Uncle Al <UncleAl0@hate.spam.net>
wrote:

Ike wrote:

I have some questions about the standard 40mm ping pong ball:

1.) Is a ping pong ball filled with pressurized air, or air at normal
atmospheric pressure?

2.) How could I determine the internal pressure of a ping pong ball
through an experiment?

3.) Based on a given internal pressure from the determination above,
how deep could you submerge a ping pong ball in fresh water before it
would implode?

1) Wet surface of ping-pong ball with shampoo in water,
2) Pin *****.
3) See whether it sucks or blows. If it blows, measure volume of
bubble and work backwards with the Ideal Gas law.

Internal pressure does not determine implosion depth.

---
Beg to differ...

1. Starting at STP, take a ping-pong ball and internally pressurize it
until it explodes. Record that pressure and call it Pboom.

2. Again, starting at STP, take an identical ping-pong ball and
externally pressurize it until it implodes. Record that pressure
and call it Psquish.

3. Take a third identical ping-pong ball at STP and internally
pressurize it until (according to the measurement you made in 1.)
it's almost ready to blow up, and then seal it up.

4. Now, take that same ping-pong ball and start externally
pressurizing it. Before it can even _think_ about imploding it'll
have to go through the point where Psquish - Pboom
= 0, and then to crush it the new external pressure (let's call it
Psquishnew) will have to increase to the point where Psquishnew =
Psquish + Pboom, so internal pressure _does_ determine implosion
depth.

I agree with you that there'll be a region of uncertainty (the
so-called "metastable" region) where the collapse pressure will be
undefinable, but that window should be able to be narrowed by
controlling the chemical compositon of the ping-pong ball's shell, its
thickness, porosity, homogeneity and, probably most important, the
sphericity of its inner and outer walls and their concentricity.

You may have noticed that there are some inconsistencies in my
discussion with respect to temperature differentials and to small
pressure differentials tied to those temperature differences, but
that was intentional in that it's just a real pain in the ***** to get
into the nitty-gritty of it unless it's really, really, necessary. :-)