Balloon science 101

Balloons are biodegradeable. The stress that occurs to the balloon when it's inflated speeds this process, which begins almost immediately. Exposure to sunlight quickens the process, but a combination of oxygen and ozone will attack natural rubber even in the dark.
- the Balloon Council
Latex is anything but simple. Here are several material science discussions about balloons, ranging from the highly technical, to the easy to understand.

Stupid Human Tricks

Here's a real winner for "Stupid Pet Tricks"; Get some liquid nitrogen and drop a pet balloon dog into it. The liquid nitrogen is so cold that it will condense all gaseous oxygen and nitrogen in the balloon, causing the animal to shrivel up. If you then carefully remove it and set it on the table, it will re-inflate in front of your eyes as it warms up, and the twists will stay intact.

Or take a bell jar connected to a vacuum pump. Inside the bell jar put a round balloon. Turn on the pump. The demonstration shows the balloon inflating as the vacuum pump worked. Next make a balloon dog for the demonstration. This works really well - the students can't wait to see the dog blown up. The whole laboratory filled with cheers when it popped!!

Latex hysteresis effects

T. Myers showed me one of the demos he does at his workshops. He inflated a jewel tone 260 and:
bent it over on itself one way,
then straightened it,
then bent it over on itself the other way at the same point,
then straightened it.
He then drew my attention to the darker band and the slight variation in balloon diameter which remained at the location of the bend. (actually, you get this band from twisting and slowly releasing the twist in a balloon too.)

Tom writes: "In the workshop I'm making a point about keeping the balloon as strong as possible so you can weaken it to help control its shape. The more of a difference in strength between the inside of a curve and the outside of a curve the better the curve will hold its shape. So I bend a balloon to make an angle and them back again to straight. The balloon gets weakened in one spot, all the way around. I would expect that spot to be more relaxed and make a bulge. That's not the case. It gets slightly narrower.... Yes, I think the color change at that spot is due to (a change in wall) thickness."

Tom said that the balloon wall was thicker at that band.

I believe the dark band is due to a wall thickness variation caused by hysteresis in the cyclic stress/strain response of the latex. If you slowly twist a balloon in torsion in front of a light, you can watch the dark section of balloon develop. A torsion stress state in the wall of a thin tube can be resolved into perpendicular tensile and compressive stresses by a construct called "Mohr's circle." You can see this physically when you wring a wet towel - there you can squeeze someone's fingers in the folds (compression) or rip the towel perpendicular to the folds (tension). Anyway, when we make a twist, we keep applying torque until the resolved compressive stress in the balloon wall exceeds the buckling limit. If you twist very slowly, you'll notice that the balloon darkens in the location where the first buckle subsequently forms. This makes sense because we expect the latex to be thickening there. In between buckles is the portion of the balloon carrying the resolved tension and that is lighter in color, again as you'd expect. Completed twists are themselves very dark, because there is a fair amount of latex in compression in there. In fact, you can suppress the buckling and darkening by pulling axially on the balloon as you apply the torque. And, you can get rid of a resulting dark band by pulling on the balloon. I think all these observations support the "_thickness variation_ caused by hysteresis in the cyclic stress/strain response" reasoning.

The band could have been the result of damage (crazing, etc.) caused by the high local stress/strain at that point, except that the band disappears upon applying a tensile stress so it can't be.

Put something in and you get something out. Here, as a result of applying stress we get "Strain". Strain is the engineering quantity proportional to the deflection or "stretch" that occurs when you apply a stress (remember that stress is proportional to force) to anything. When you inflate a 260, the diameter and length each increase by 500 to 600 % (we say this is 500 to 600% "hoop strain" and "axial strain," respectively), and then you reach a point where it gets very difficult to blow up any further. If you continue to inflate it further, it will burst. If we graphed the stress vs strain (think of it as force vs stretch) for latex we would get a sigmoidal (S-shaped) plot like the following:

 S   |                           (*) < == ultimate strength,
 T   |                            {      or burst strength
 R   |                            |
 E   |                            }
 S   |                           /
 S   |                         ,"
     |                       .'
     |               ._ - '"
     |     , - ~ '""
     |  ,~
   0 +-------------------------------->
             S T R A I N

Well, when you stretch latex to a point below it's ultimate strength and then slowly release it, the latex doesn't retrace its "stretch curve." Instead it relaxes along a new curve.

 S   |                           
 T   |
 R   |                            |
 E   |                            }
 S   |                           /|
 S   |     stretch             ," /
     |      curve            .'  "
     |               ._ - '"   .'
     |     , - ~ '""        _-"  relax
     |  ,~           , - ~'"     curve
     |,'    , - ~ '" 
-----+--_-" -------------------------->
             S T R A I N

Note that the "zero" of the relax curve (the point where the relax curve crosses the Strain axis, where the Stress is zero) does not occur at zero strain! Instead, it occurs at some positive strain - this is the permanent stretch you see after blowing up and then deflating a balloon.

If you've read the chapter on how balloons pop, you know that inflating a balloon stresses the latex to a certain level. Draw a horizontal (constant stress) line through our graph.

 S   |                           
 T   |
 R   |                            |
 E   |                            }
 S   |                           /|
 S   |     stretch             ," /
     |      curve            .'  "
-----|--------------------'"---.'--------constant stress line
     |     , - ~ '""        _-"  relax
     |  ,~           , - ~'"     curve
     |,'    , - ~ '" 
-----+--_-" -------------------------- >
             S T R A I N

Note that this constant stress line crosses both our stretch curve AND the relax curve. Thus, if I asked you "how much has the balloon stretched upon inflation and twisting?" (what is the strain at a particular stress level?), you could give me 2 answers! This means that sections of a balloon at the same stress can have 2 different values of strain! All our advanced twisting and shaping tricks depend on this property!

Note also that to get back to the initial size (zero strain) we have to apply a compressive (push) stress!

If we had some way to apply one complete stretch-relax-compress cycle, we might be able to get a closed loop called a "hysteresis loop"

 S   |
 T   |
 R   |                            |
 E   |                            }
 S   |                           /|
 S   |     stretch             ," /
     |      curve            .'  "
     |               ._ - '"   .'
     |     , - ~ '""        _-"  relax
     |  ,~           , - ~'"     curve
     |,'    , - ~ '" 
--- ,+--_-" -------------------------- >
   { ,~'
   |/        S T R A I N
   { | 
   | |

Many processes and behaviors found in nature trace out hysteresis loops when they are graphed on appropriate axes (because most things are not ideally reversible - they are functions of the path taken and not just their final state). Here, the area enclosed by a hysteresis loop is representative of the energy lost in the process of stretching-relaxing-compressing the latex. Where does the energy go? Well do this experiment: Take a balloon in both hands so that you have about two inches of unsupported balloon between your hands. Press the unsupported section of balloon lengthwise against your lips. Then move it away from your face and completely stretch and relax the unsupported section 10 times, as quickly as you can. Immediately after the tenth time, press the unsupported section of balloon lengthwise against your lips again, and you will notice that its temperature has increased. The energy wasn't really lost; rather it was converted into heat.

Entropy and the second law of thermodynamics as applied to latex

The first law of thermodynamics says that the change in internal energy (dE) is equal to the change in heat absorbed (dH) or released plus the work done on the system (dW). dE = dH + dW.

The second law of thermodynamics defines a quantity called "entropy" which is a measure of the randomness of a system. A highly ordered system (like toys in a toybox in a child's clean room) has low entropy. A random system (like toys spread randomly all around a child's room) has high entropy. Suffice it to say that in natural processes, entropy stays constant or increases.

The second law of thermodynamics says that for a reversible process, the change in heat absorbed (dH) is equal to the Temperature (T) times the change in the entropy (dS). dH = T * dS

An adiabatic process is one in which no heat is transferred to/from the surroundings. For an adiabatic process, dH = 0. The first law then tells us that the work done on the system is converted entirely to internal (stored) energy.

After a little calculus, the second law tells us that for a reversible adiabatic process, T * S is a constant. The product of two variables equal to a constant is the equation of a hyperbola, where when one variable increases, the other must decrease.

If you've read the chapter on how balloons pop, you know that latex has a structure composed of many coiled-up, intertwined, chain-like molecules. Since the chains prefer a random, curled configuration, their initial degree of order is low and their entropy is high. However, when a tensile load is applied, the entropy decreases as the chains become straightened and aligned.

What does all this mean? Let's do an experiment.

Take a balloon in both hands so that you have about two inches of unsupported balloon between your hands. Press the unsupported section of balloon lengthwise against your lips. Keeping the balloon pressed against your lips, stretch the unsupported section as quickly as you can and hold it. The balloon heats up!

Stretching the balloon quickly allows us to call the process adiabatic because there is no time for heat to be transferred to the surroundings. The first law tells us that all the work we've done stretching the balloon has gone directly into internal stored energy in the balloon. The second law tells us that if the entropy decreased, the temperature has to increase!

Now, keeping the balloon stretched, remove it from your lips. Hold it stretched for 30 seconds, so that it cools back down to room temperature. Then press it back against your lips and relax the unsupported section section as quickly as you can. (Don't punch yourself in the nose!) The balloon gets cold!

Relaxing the balloon quickly allows us to call the process adiabatic because there is no time for heat to be transferred from the surroundings. The first law tells us that all the internal stored energy in the balloon was converted to work done on us as we relaxed the balloon. The second law tells us that if the entropy increased, the temperature has to decrease!

Effect of water and ozone on balloons

Tom writes:
Tyen the Magic Mime of LA called me twice with an interesting problem. He is convinced that 260Q's don't hold air as long as they used to. He's been making poodles and marking the dates. They only last him a couple of days.

Adrienne writes:
I have noticed that the animals I leave lying around the house are dying faster than usual. I have a Winnie the P**H given to me by Dave Lewis, and it died before I even had a chance to disect it. They seem to be lasting about 4 days before they really shrink up and are gone. It doesn't effect the quality of the balloons we are making for people, but I have had to stop telling people that they would last for weeks if they keep them in a cool place out of sunlight.

I haven't noticed a change in the time that Qualatex balloons stay inflated, but then I haven't looked for one either. Are the balloons thinner? I don't know. Did they change the formula? Here's what Tim Vlamis of Pioneer wrote two months ago:

Date: Thu, 21 Dec 1995 18:20:53 -0500
Subject: Re: Balloon quality query

In response to the question of whether or not we have changed 
our "formulas" I can give both a short and a long answer. The 
short answer is "no". We know that our business relies solely 
on making the best quality balloon (our prices tend to be higher 
than others, so we *have* to be better) and we would not endanger 
our business (or yours) by "cheapening" our manufacturing processes 
and formulas.

Making balloons is in many ways like baking a cake. We work with 
hundreds (if not thousands) of separate raw materials and variables
in the manufacturing process. Our main ingrediant (latex) is a 
natural product with all the vagaries of other natural raw materials.
In some senses, our formulas are like recipes, only far more 
complicated and larger in scale. Each "formula" for each color 
(over 50 in the Qualatex range) or size or shape (etc.) is slightly 
different. Depending on other factors including changes in raw 
material supply, environmental changes (temperature, pressure, 
humidity), what else is running on that manufacturing line, color 
sequence, etc., we may "tweak" our manufacturing formulas/process 
to maximize the quality of the product. So it isn't strictly fair 
to say that we don't change, in fact, we are changing all the time.
Balloons are special in that they are organic, they are alive. They 
change depending on the circumstances they are in. Most of us have 
tried to use 260Qs that have sat in a car during the summer for a 
few weeks and have experienced this!  :-)


What else could cause balloons to lose air? Degradation? Am I the only one who noticed that Tyen the Magic Mime lives in LA, and Adrienne V. lives right next to LA? LA, smog, ozone, hmmmm... It is well known that ozone attacks latex.

In their retailer kit, the Balloon Council (of which Pioneer's Dan Flynn is Chairman) literature says "Balloons are biodegradeable. The stress that occurs to the balloon when it's inflated speeds this process which begins almost immediately. Exposure to sunlight quickens the process, but a combination of oxygen and ozone will attack natural rubber even in the dark. Deterioration can be seen clearly only a few hours after a balloon is inflated, as it begins to oxidize or frost;"

But then LA is also next to the Pacific ocean. Ocean, water, hummidity, hmmmm...

The Qualatex book "Design" says that the chalkiness or "oxidation" which develops on latex balloons which have been left out for a few hours or days is due to humidity and the "quality" of the ambient air. Does "quality" mean ozone level? Is there any correlation between humidity (what "Design" states as the cause) and ozone levels? (well, I know that in a thunderstorm, humidity and ozone levels are both high... )

Above, Tim Vlamis wrote that humidity affects balloons during manufacture. Others have written:

The "oxidation" is caused by ozone in the air. I think you'll routinely see cautions about ozone degrading rubber in most materials texts.

Warm temps, heat, sun will cause oxidizing to start. Once started, it won't stop.

Here in the New Orleans area we're 3 ft below sea level and 85-90% humidity is normal. Latex doesn't take long to start breaking down in these conditions. Sometimes I even have to Hi-float air-filled arches and columns when they aren't in good A/C. You really know the humidity is high when you have to high float 260's in a hospital arrangement (not an easy task).

A couple of years ago someone wrote that twisters with years of experience say that humidity makes a big difference in the feel, resilience, and workability of balloons. It was also posted that round balloons can even be inflated to much larger sizes on a cool foggy day than in any other weather. This example was given: you can inflate a standard 16" round party balloon to 22" diameter on a 65 degree foggy day when the humidity is around 80-85%.

Four months ago we had a discussion on here about the workability of pump versus mouth inflated balloons. Dave Beedy wrote:

Date: 27 Oct 95 11:30:52 EDT
From: DAVE BEEDY < >
Subject: Pump vs. Mouth

Earlier Lorna wrote (in part):
> When a balloon is pump inflated there seems to be a lot more air in
> the balloon; the latex is tighter, and therefore less space for
> twisting.

I normally mouth inflate, but have a pump that I use on ocasion. I had
never really noticed the difference in balloons inflated one way or the
other. After I read Lorna's posting I got my pump out and inflated both
ways. Lorna is right!!! The pumped balloon are tighter and of a larger
diameter than the mouth inflated.

Good observation, Lorna!!! This can have impact on balloon twisting
instuctions and a lot more. Maybe some of the "M.I.T." types on the list
can tell us why there is a difference.

Dave "Buttons" Beedy

and James Dewitt replied:

Date: Sat, 28 Oct 95 01:44:48 -0500
From: (GS-7 James C. DeWitt)
Subject: Re: Pump vs. Mouth

Dave and Lorna,
     When you use a pump, you are putting room temperature air into
the balloon.  The air stays the same pressure and doesn't contract and
the pressure stays the same inside.  When you mouth inflate, you are
putting air that is around 98 degrees (probably higher since it is
from deep inside your body, Doc?) into the balloon.  The balloon is
surrounded by cooler, room temperature air, therefore it cools quickly
and contracts, which decreases the pressure inside the balloon.  And
that's basically why mouth inflated balloons are softer and thinner
than pumped balloons.

James Dewitt

Yes, this is part (probably most) of the answer, but the air you exhale is also loaded with more water vapor than the air you pump. I never responded to Dave and James' discussion four months ago, but about six months ago I posted the following question to several chemistry and materials science newsgroups on Usenet:

From: (Mark Balzer)
Newsgroups: sci.materials,sci.chem,sci.polymers,sci.eng.chem
Subject: Help with humidity effects on latex rubber?
Date: Sep 95

 I recently read in a manual published by a balloon manufacturer
 that the chalkiness or "oxidation" which develops on latex balloons
 which have been left out for a few hours or days is due to humidity
 and the "quality" of the ambient air.   I have also heard that humid
 conditions during balloon manufacture can even affect product quality.

 Balloon workers with  years of experience say that humidity makes a
 big difference in the feel, resilience, and workability of balloons.
 There is also a difference in the feel of balloons blown up by mouth
 and by pump - blown balloons contain warm moist air from your lungs
 while pumped balloons are filled with drier ambient air.

 Balloons can even be inflated to much larger sizes on a cool foggy day
 than in any other weather.  If you ever wish to impress someone,
 you can inflate a standard 16" round party balloon to 22" diameter
 on a 65 degree foggy day when the humidity is around 80-85%.

 Does anyone know why any of these effects occur?
 Are there any mechanisms that would explain them?
 I would really appreciate any explanations or references I could look up
 to find an answer.

and I got the following responses, the first one from a chemistry professor here at the U. of Illinois:

From: (Rich Masel)
Newsgroups: sci.materials,sci.chem,sci.polymers,sci.eng.chem
Subject: Re: Help with humidity effects on latex rubber?
Date: 7 Sep 95 16:00:21 GMT (Mark Balzer) writes:
> I recently read in a manual published by a balloon manufacturer
> that the chalkiness or "oxidation" which develops on latex balloons
> which have been left out for a few hours or days is due to humidity
> and the "quality" of the ambient air.   I have also heard that humid
> conditions during balloon manufacture can even affect product quality.

I have some related observations which may help explain this.  My company
makes latex rubber bands for orthodontics.  We usually saturate the rubber
bands with water, so their elastic properties are constant when the rubber
bands are put in a patients mouth.  However, if we dry the rubber bands
the color of the rubber bands changes from a milky white to brown.  The
elastic constant also increases.  The process can be reversed by soaking
the rubber bands in water.  (the rubber bands swell!!).

Our observations then are that water is slightly soluble in the rubber,
and the presence of the water acts like a filler to reduce the elastic
constant of the rubber.

At first sight one would not think that water would be soluble in
rubber.  However, natural rubber contains about 2% of water soluble
components (surfactant and protein).

Rich Masel (

From: (Kirk Mueller)
Newsgroups: sci.materials,sci.chem,sci.polymers,sci.eng.chem
Subject: Re: Help with humidity effects on latex rubber?
Date: Tue, 12 Sep 1995 17:46:05 -0700
Organization: Hughes Aircraft Co., RCS

   Latex rubber, like many materials, absorbs moisture.  The moisture
changes the structure of the rubber making it more 'rubbery' and
stretchable, in the short term.  It usually increases the material volume
and undoubtedly increases the balloon's thickness here.  In the long term
the balloon's life is shortened by moisture exposure.  However most of us
don't care to have a balloon last more than a couple days anyway.
   For some compounds, including latex rubber, the moisture causes
irreversible chemical reactions, what we call lack of 'hydrolytic
stability' in the Defense industry.  (We are required to use
hydrolytically stable materials for Defense as I can't count on dry air.)
The reaction byproducts may be what you're seeing on the balloon.  The
white residue might also just be mold release or something similar
however.  Usually moisture reaction products degrade the original
compound.  I suspect the balloons loose their stretch and fail from
cumulative damage.
   You should consult text books on Fick's Law for the rate of moisture
ingress and egress.  I'm sure you'll find it takes only a few minutes at
most for a typical balloon to equilibrate with the surrounding atmosphere.

   A short term moisture effect test:  Try filling a balloon with moist
air from your lungs (it should stretch oversize) and then leaving it out
in the desert (out of the sun to avoid heating effects) to dry out for an
hour or so.  Once the moisture leaves the rubber it should pop because it
now can't stretch as far.
   A long term moisture effect test:   Soak some balloons in water for
about a week.  Keep some more as dry as possible for the same time
period.  Blow all balloons up (it doesn't matter if it's humid, mouth, air
or not as long as all use the same air) and measure the maximum diameter
(before popping).  The wet, degraded balloons shouldn't blow up as large
(if they blow up at all).
Kirk Mueller
Hughes Aircraft Co., Radar and Communications Sector
El Segundo, CA
---  All comments are strictly my own. ---

From: (david rogers)
Subject: Re: Help with humidity effects on latex rubber?
Date: Tue, 12 Sep 95 00:39:53 PST

I carried our research funded by the Science Council of British Columbia
in Vancouver Canada in 1992-93 to extrude an Improved Porous Elastic
Irrigation Pipe made from TDP Tire (Rubber) Derived Product.

The key mechanism involved at the core of the research was that of the
permeability of rubbers to water.

Key references were:

(1) "Water and Rubber do mix" by D.C. Edwards. Chemtech October 1986

(2) "Predicting Water Diffusivity in Elastomers" by T.M. Aminabhavi,
     R.W. Thomas and P.E. Cassidy Polymer Eng. and SCi., Dec. 1984,
     Vol. 24, No. 18.

(3) "Surface Enthalphy and Entropy and The Physio-Chemical Nature
     of Hydrophobic and Hydrophilic Interactions" Journ. Dispersion
     Sci. and Technology, 12(3&4), 273-287 (1991)

The key factor in the admission of water molecules into a rubber matrix
are the prescence of polar nucleating sites ( typically salts ) around
which form pools of water molecules. The driving force is both
osmotic pressure ( vapour pressure ) and epectrophoresis ( Polar forces ).

D.C. Edwards work was such a breakthrough it helped explain the basis
for many phenomena previously not fully understood associated with the
work I was investigating.

Dr. Patrick Cassidy was working on the use of starches in bioderadable
polyethylene films for use as agricultural mulches in China when I last
spoke to him some 20 months ago RE: Ref. (2) above.

Ref.(3) is an example of an avenue for investigation I explored RE:
engineering a solution to controlling water egress into rubbers i.e.
using polar compounds.


I went to the library and got copies all the references which David mentioned above. The first one is a very interesting paper, and I will post a translation it into layman's terms when I get a chance.

I'm sorry I can't answer the original question, but I hope I provided some insight into what may be behind it.

MB 7/6/96
MB 1/4/97

The following material has been saved from posts on the mailing lists. Rather than keeping it hidden away, it has been temporarily placed here until the guide editors get a chance to move it to its proper location in this chapter. Feel free to make use of it.
We have been experiencing some balloon discoloration.  This has occured
mainly on white(doesn't matter what size) and turns parts of the balloon an
ugly dark yellow.  Can anyone tell what is happening or if we are doing
something wrong with storage? We keep all balloons in cool dark storage.
My husband smokes, has anyone else had this problem because of smoke filled

>We have been selling bouquets, etc. to our local taverns and supper
>clubs for sports games.  Our latex balloons barely make it the length of
>the game.  Highfloat is used and they are filled with helium but within a
>short time of being delivered they cloud up (I can understand that) and
>visibly shrink.  Any ideas for us?  We have gone to using only mylar in the

>I just got a request for some balloons to be picked up here in Denver, 5,280'
>and then taken up to Evergreen which is at 7,500'. Some other twister told
>this customer that the balloons would pop during the elevation change. I doubt
>that that amount of elevation change would make alot of difference, but don't

From a "US Standard Atmosphere" table, I interpolated the air pressures and
temperatures at the two altitudes you listed:

height       temp       pressure
5280         4.5 C      0.824 atm
7500         0.1 C      0.760 atm

Running it through the equation relating pressure, absolute temp and volume
for a fixed mass of an ideal gas (ideal gas behavior is an excellent
assumption for both air and helium at atmospheric pressure and temp.)

 P1 * T2     V2
--------- = ----
 P2 * T1     V1

0.824 * (273.15 + 0.10)
---------------------- = 1.067
0.760 * (273.15 + 4.50)

Assuming the balloons are outside and the air temp and pressure is per the
US Std. Atmosphere table, their volumes will increase by almost 7%.

Until it is fully inflated, the diameter of a round balloon is free to
change.  11" round latex balloons at 5280' will become 11 1/4" balloons at
7500' (assuming spherical balloons, this is more than a 2% increase in
diameter, since diameter scales as the cube root of volume for a sphere.
This also assumes that the air pressure variation inside the balloon is
negligible for small diameter changes.)

To a first approximation, the diameter of a 260 balloon is not free to
change; only the length of a not-fully-inflated 260 is free to change.
Therefore, 260's that have been inflated (but not twisted) leaving some
length of uninflated nipple will behave differently.  In that case, the
change in altitude will cause the length of the 2" diameter bubble to
increase by almost 7%.

260's that have been inflated and twisted into tight bubbles will... hmmmm,
well...  the bubbles will get tighter.  How much tighter requires a lot
more time than I'm willing to devote to this problem. (there are highly
nonlinear effects at work in the simultaneously occurring interactions
between the geometry changes, the stress-strain properties of latex under
high biaxial strains, the pressure/volume/temp relationships of the gases,
etc.  Even time is involved when you include gas diffusion rates, stress
relaxation effects, etc.

Assuming the balloons are inside and room temps are the same, their volumes
will increase by about 8%.  11" round latex balloons at 5280' will become
11.3" balloons at 7500' (almost 3% larger in diameter).  The length of the
2" diameter bubble (on 260's that have been inflated but not twisted,
leaving some length of uninflated nipple) will increase by almost 8%.

>I figured I could just make them somewhat "soft", by a little extra
>squeezing while twisting.

Good idea.


Balloon Wrap Oxidation

The oxidation of balloons generally occurs on the outside surface of a 
balloon.  The whitening of a diamond clear balloon causes a "foggy" 
effect. On colored balloons this "foggy" effect causes a "satin" effect.  
Actually the term foggy is a misnomer.  The balloon is not full of fog 
- "water vapor", it just looks that way.  Actually the outside of the 
balloon is degrading due to natural "photodegradation" or oxidation.  
On an atomic level - free ions are attaching themselves to the natural 
"stable" atoms of the balloon.

The main culprit here is low altitude ozone.  O-2 is pure oxygen - 
very stable atom.  0-3 is Ozone and it is a very unstable atom.  Ozone 
occurs in larger quantities during a high pressure system "watch the 
newscast" and also the associated high temperatures.  During the 
winter, ozone is produced by heating systems.  Place an unprotected 
balloon by a hot furnace duct for a day and you will see what I 
mean.  Computers also produce ozone, e.g. not too many balloon 
decorators hang around in a CompUSA store with their balloons.

Hospitals are known to use air filtration systems that utilize ozone to 
"cleanse the air".  Balloons do not fare well in this type of 
environment. Beauty salons can also have many airborne chemicals 
which can accelerate oxidation.

The following is Grolier's explanation concerning the oxidizing power 
of natural ozone.

ozone --------------------------------  Ozone is a form of OXYGEN in 
which three atoms combine to form a molecule, instead of the usual 
diatomic form of the element. It is a blue gas with a pungent odor, 
noticeable when the gas is formed by an electrical discharge (like 
lightning). Ozone is a powerful "oxidizing agent" and an effective 
antiseptic and bleaching agent;  in high concentrations it is a severe 
irritant.  The atmosphere's ozone layer protects life against harmful 
solar radiation (see OZONE LAYER), but ozone produced in the lower 
atmosphere by industry and automobile exhaust is a pollutant. It 
damages crops and may be indirectly linked to some breathing 
disorders. Bibliography: Fishman, J., Kalish, B., Global Smog (1990).


How do we protect the 18" balloon wraps?

liquid solutions to the problem are easily available.  STP 'Son-of-a-
Gun' is one solution - 'Balloon Shine' is another.  Both of these 
products create a liquid barrier on the outside of the balloon which 
will ward off those pesky oxidation atoms.  But both of these 
products dry out in a day or two and mother nature will soon oxidize 
the balloon in streaks or patches as it dries.

The only long lasting barrier is plastic.  This can be found with 
balloon bags.  Place a balloon in a plastic bag, seal it with a twist tie 
and then place a balloon on your window sill.  In just two days you 
will clearly see the difference.   Balloon bags are available in ultra-
clear or typical extruded grade plastic.

A unique "plastic bag" that is on the market today is a product called 
'Mr. Clear'.  It is a sleeve of crystal clear shrink wrap plastic.  It can 
be placed on a 18-inch Upside-down balloon, twist tie - and then you use 
a heat gun or strong hair dryer to slowly shrink the plastic to a form 
fitting barrier.  The advantage to this method is that the plastic is 
barely discernible from a distance.  This gives the customers a great 
view of the actual balloon wrapped product.  The product has it's 
drawbacks, cost, time and wrinkles when exposed to cold 
temperatures, but it does the job of protecting the balloon very well.  
Invented and produced at Incredible Balloon this product is always 
in high demand.

For the most part balloon wraps are an impulse item.  Here today 
exploded tomorrow, but when you need a balloon to last a long time 
there are a few choices.

You might as well get adjusted to the oxidation that occurs on 
outdoor installations.  I have found that using the STP's etc. has 
hastened the demise of the balloon itself, particularly in hot weather.  
Oxidation will even occur on some indoor installations if the humidity 
overcomes the airconditioning.  Some of the elements that affect your 
oxidation are:         Blowing the balloon up closer to its designed size. 
This retards the oxidation  process a little.
The darker the color the more heat it obsorbs, this hastens the 
Heat, humidty and sun are killers together.
Cool weather and sun is not to bad
Cold weather has been pretty good for us
Freezing weather, if you blow were it is displayed, with air or Helium 
the same temp as the enviroment, has been amazingly good.

We have not tried this product outside, but inside we rejuvinate our 
displays  with a spray from Design Master called Floral Master, 
formerly called glitter glue.

Does anyone know how an Air Purifier affects Latex Balloons?

We put one of these machines in our retail store and have noticed 
that any flat balloons not in a bag faded very fast.  And our air-filled 
latex balloon displays only lasted 2-3 days.

If the air purifier is a HEPA filtration unit, it should have  zero 
impact. If it is a unit that generates ozone in addition to simply 
moving air through a multistage filter, you will experience 
deterioration of your balloons. (ozone chemically attacks natural 

if you have one that is using high voltage to generatge ozone, it will 
damage balloons very quickly.  I learned it the hard way...  A balloon 
arch lasted less than four hours near one.

If the Air Purifier emits ozone, which many models do, the ozone will 
speed up the oxidation of the latex balloons. Many of the "high end"  
models will cause quite a high level of ozone in the air, especially if  
it is an enclosed room with poor ventilation. This might be the cause 
of  the shortened life span of your latex balloons.

In a nut shell, if the machine generates ozone, which mine does, it 
will break down the latex balloons faster than if there was no air 
purifer.  It does the same thing as if you pur balloons outside, but 
much faster. I  will make sure my flat balloons are in bags to prevent 
the problem.  The air machine is worth it since the man in the next 
door business smokes like a champ and the smoke comes through the 
walls.  We hate smoke! Karen


On the Mad Scientist Network, Everett Rubel writes:

Humidity in the air makes the air lighter, less dense.
The presence of water vapor (humidity) in the air lowers the average mass of 
the molecules that are in the air. This is because a molecule of water only 
masses about 18 atomic mass units, while air is about 29 AMU, which is an 
between the molecules of nitrogen (28) and oxygen (32).  Airplane pilots 
know that when the humidity is high, it takes longer for them to lift off a 
runway because the air is less dense.  This is also why low air pressure, (a 
falling barometer), means a storm is on the way, because of all the moist 
air of a storm being lighter than plain old dry air.
BTW, this is why helium-filled balloons don't float very long in humid 

Bobbie writes:
>When I got to the shop, all but about 10
>of the helium balloons where on the floor...
>Knowing that it is really hot and
>humid outside,  I only inflated the 11" to 9".    

That sounds like your problem right there.

You are dealing with two effects here: humidity and temperature.
You need to deal with each variable separately.

Humidity makes the air less dense.  Therefore, the buoyancy effect that you 
are counting on to float your balloons, decreases.  The ways to fight this 

  1) maximize buoyancy 
      * inflate to the largest possible diameter (see the chapter in 
       the Guide on "how balloons float" for an explanation of why).
      * inflate with pure helium (not a helium/air mix),
      * knot tightly to prevent helium leakage. 
  2) minimize balloon weight
      * cut off any extra nozzle beyond the knot
      * don't use extra hi-float,
      * use thin, narrow ribbon.

The presence of water vapor (humidity) in the air lowers the average mass of 
any given volume of air. This is because equal volumes of water vapor and dry 
air have a mass (or weight) ratio of 18 to 29.  Airplane pilots know that when 
the humidity is high, it takes longer for them to lift off a runway because 
the air is less dense.  This is also why low air pressure, (a falling 
barometer), means a storm is on the way, because of all the moist air of a 
storm being lighter than plain old dry air.

Let's say it was 70F (21.1C) inside and 100F (37.8C) outside.  Then inflating 
to 9" inside would only result in a 9.17" balloon outside.  If you wanted an 
11" balloon outside (for maximum float time on a humid day), you should have 
inflated to 10.75" inside.

For the temperature correction, just combine Charles' Law from high school 
chemistry with a little geometry. (the following equations must be viewed in a 
non-proportional font like "courier")  Charles' Law says:

  Initial Volume       Initial (absolute) temperature
 ----------------  =  --------------------------------
   Final Volume         Final (absolute) temperature

Assuming spherical balloons, volume is proportional to the diameter cubed.
Also, absolute temperature in degrees Rankine = degrees Fahrenheit + 460.
Substituting and rearranging gives:

                       [                         (460 + final temp in F)   ]
Final dia = cube root  [ (initial diameter)^3 x -------------------------- ]
                       [                         (460 + initial temp in F) ]

>I went back a couple hours later to make sure the
>balloons where alright, since it was really hot today.  Blue balloons were
>popped everywhere.  I don;t know why.  4 colors, and only the blues where
>popped.   Can anyone tell me what I did wrong.  

Were the balloons in the sun or in the shade?  
Were the other three colors lighter or more transparent than the blue?  

Dark balloons in the sun absorb more of the sun's energy, get hotter and 
expand more  (why sonny, when I was young, those blue balloons got so hot, you 
could fry an egg on 'em!)  plus the heat simultaneously weakens the latex. 

ps - It would be nice to have some actual temperature vs time data: it 
wouldn't be hard to place a bunch of different color balloons in the sun and 
chart the air temperature inside each one, plus the air temp in the shade as a 
function of time.  It would make for a nice magazine article.  Hmmmmm...

buy a few bags of asst balloons,do some tests,make some columns,topiary 
balls,air filled,helium filled,see how they handle being outside,check what 
with dark vs light colours,differently sized balloons,air vs helium
filled,duplet tied garlands vs string of pearl helium arches etc.You will
then gain alot of confidence when you know what to expect.

Most distributers sell a balloon art practice kit which usually runs
under $15, and is the best learning tool I've ever encountered.  It can teach
you spiral arches, quads, sixes, string of pearls, stuffed balloons, air-
filled, helium-filled.  It can help you learn float times and how to use hi-
float.  You can increase your production speed, discover the right heights for
your centerpieces - Oh! there's nothing this kit can't do!  It's called: "Buy
a bag of balloons and experiment with them"  Also available by the case!  It's
the way to get to Carnegie Hall - "Practice! Practice! Practice!"

I have found that latex balloons in columns, after they get wet and then dry, 
can 'fuse' together.  When there is any wind or movement, the friction can pop 

> I need the formula to determine the volume of various sizes of balloons
> to determine how much helium I need for my projects. I have a short
> chart, but it does not give me enough information. It includes the
> following equation:
> 4/3 x r3
> -----------   = Volume in cubic feet
>   1728
> My question is, does the "r" mean radius? And is the radius cubed or
> multiplied by 3? 

As you've already mentioned, using that formula you will get different
results from what your chart gives (assuming the chart is correct).  I'm
just going to answer this briefly now.  We have had discussions on this in
the past.  I found a couple of messages mentioning this in the list archive
(, click on "decorating"), but not the one I
was looking for, so I'll leave all of you to hunt for it if you'd like.

The r in the formula is the radius, and it is being cubed.  there is an
important element missing however.  The volume of a sphere is 

     4/3 x pi x r^3.

Without multiplying by pi (which is roughly 3.14), you're not going to come
close at all.  Now, I say close because you're only going to get an
approximation.  Balloons aren't perfectly spherical.

The last thing to note is that your original formula divides the volume by
1728.  1728 is the number of cubic inches in one cubic foot (12^3).  If you
plug in a value for r in inches, you will need to divide by 1728 to get
cubic feet.

Ex: To determine volume of a sphere with an 11 inch diameter (radius is
half of the diameter) -
	4/3 x pi x r^3 = 4/3 x pi x 5.5^3 

which is roughly 697 cubic inches, or .4 cubic feet.

Don B. writes:
>We have found it helpful to blow up all the balloons with air to their
>maximum size and then deflating them. Usually a 16" can easily go to near
>17" depending on the color. Now you have prestretched all the balloons so
>when you undersize with helium they should be more reliable. By preinflating
>you now know that they should withstand the effects heat enlarging.
>This prestretching can be done far in advance of the actual event.

Thanks for sharing this advice.  I'd like to ask if you could provide any 
specifics about your statement in the last line.  

The "How Balloons Pop" chapter of the Guide to Balloons and Ballooning talks 
about the "viscoelastic" nature of latex responsible for its time-dependant 
properties, but it seems to me that the the majority of the prestretching 
effect would only last several days before the balloons recovered their 
initial shape and lost the prestretching benefits. (Of course, this time 
period should depend on the storage temperature, with lower temperatures 
extending the recovery times).

Since I've never prestretched and stored, I was wondering if you could comment 
on what timeframe you had in mind when you wrote "far in advance," and whether 
or not you prestretch and refrigerate when preparing them far in advance?  

Finally, have you or anyone else actually done a direct comparison test in 
which you noticed a significant difference in the lives of prestretched vs 
virgin balloons placed outdoors in the sun?  I noticed that you only say they 
"_should_ be more reliable."  Can you quantify the improvement or put any 
numbers to it?  While I'm certain that the prestretch cycle improves the 
reliability of a balloon used indoors (like, say in an SDS application where 
prestretching is recommended), it just occurred to me that truly high outdoor 
temperatures would speed the recovery and tend to obviate the benefit of 

If you aren't sure what I am getting at, try this:  take three identical 
balloons.  Inflate two to maximum size (prestretch), hold each for a minute 
and then deflate.  Compare the sizes of the (deflated) prestretched and virgin 
balloons.  Now hold one of the deflated prestretched balloons in front of a 
hair dryer (1 foot from the nozzle, the air coming out of my hair dryer is 106 
F) and watch it quickly shrink back to original size.  Left at room temp, the 
other prestretched balloon will slowly shrink back.


A long-loading, slow-running but interesting applet of balloon and gas
models is seen at (used as educational experiment aid):