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FAQ's
How long will balloons float?
12" latex balloons will float for 15-20 hours or 3-5 days when treated with Hi-Float. Mylar balloons will float for up to one week.
What is Hi-Float?
Hi-Float is a liquid sealant used to coat the interior surface of latex balloons and reduces the speed at which helium escapes through the latex skin. It can increase the float time by several days. Hi-Float is non-toxic and water soluble.
Do you deliver the same day?
Yes, with a 2-3 Hour Notice.
Can you deliver to hospitals?
Yes. Some hospitals accept only Mylar balloons, some hospitals do not accept balloons and?deliveries must be completed by 4 pm.
When will my balloons arrive?
Your balloons will arrive within an hour of requested time (sometimes, we get caught in traffic!)
If I blow up balloons by mouth, will they float?
No, only balloons inflated with helium will float.
Do you do balloon drops or releases?
Yes.
Do you accept credit cards?
Yes.
Can you come visit the event location?
Yes. Just call and make an appointment.
Will table centerpieces obstruct the view of the person across from me?
No. They are designed for socializing and will float above eye-level.
Can I have my balloons delivered a day before my event?
Yes, but latex balloons will have to be treated with Hi-Float at 50 cents per balloon. Mylar will be fine overnight.
Are your balloons on strings?
They float on 6 foot ribbons or longer.
Can my bouquet be personalized?
Yes.
Can I get balloons in any color?
Yes.
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Why do balloons float? Can you fill a balloon part way with helium, make an animal out of it and have it float? Definitely in water. In air... well, that depends. This is an application of "Archimedes' Principle" which states that a body immersed in a fluid is buoyed up by a force equal to the weight of the fluid displaced. If an object immersed in a fluid is heavier than the fluid it displaces, it will sink to the bottom, and if lighter than the fluid it displaces, it will rise. In our case, the "body" is a Helium-filled 260, and the "fluid" is air. If we use the "Ideal Gas Law" we can quickly find the mass of the Helium or air in a 260. The "Ideal Gas Law says that m = PVM/RT where: m = mass of the gas in grams P = pressure inside the balloon, in atmospheres V = volume in liters M = Molecular weight of the gas R = a constant = 0.082 T = Absolute temperature, Kelvin At sea-level, the air pressure is 1 atmosphere and say we are at room temperature. Then P = 1 and T = 293K. Then approximate a fully inflated 260 as a 1.75" dia cylinder that is 50" long. Since the pressure in the balloon is only slightly greater than the pressure outside the balloon, let's call them the same for now (both equal to 1 atmosphere). The volume is pi x 0.875" x 0.875" x 50" long = 120.2 in^3 or: 1 liter V = 120.2 in^3 x --------- = 2.0 liters. 61 in^3 We next look up the molecular weight of the gasses: M = 4.00 for Helium, M = 28.97 for the mixture of gasses we call "air" and plug all these numbers into the Ideal Gas Law to find: 1 x 2.0 x 4.00 the mass of Helium in a 260 = m = ---------------- = 0.33 grams 0.082 x 293 1 x 2.0 x 28.97 the mass of air displaced = m = ----------------- = 2.41 grams 0.082 x 293OK, we're almost done. The mass of a 260 is 1.6884 grams (it's nice to have a laboratory balance that weighs to a ten-thousandth of a gram... ), but let's just call it 1.70 grams since we're all friends here. Now add that to the 0.33 grams of Helium inside the balloon for a total weight: The total mass of the "body" (balloon + Helium) is 2.03 grams The mass of the fluid displaced (air) is 2.41 grams An object's mass doesn't change, but it's "weight" depends on the force of gravity. Converting these masses to equivalent weights on earth, the Helium-filled balloon weighs 0.0045 pounds and the displaced air pushes up with a weight of 0.0053 pounds. This results in a net buoyant force or "lift" of 0.0007 pounds on a Helium-filled 260, so it should float (but just barely!) !!! So, let's see if any of this works in the real world: PFFFFffffffffffffffttt! PFFFFffffffffffffffttt! PFFFFffffffffffffffttt! PFFFFffffffffffffffttt! PFFFFffffffffffffffttt! PFFFFffffffffffffffttt! Squeak, squeek, SQUEEEEEAAAK, SQUeak, squEAK, squeak... Ta-dah! Well, I now have a parasol floating above my head, so yes, fully inflated 260 sculptures with a minimal number of twists will indeed float (just not for very long...). This would be good for a quick demo, if nothing else. The Helium diffuses out rapidly, and there is just not enough reserve lift to keep it aloft as the 260's start to shrink. HOWEVER, Freda asked "Can you fill a balloon *part way* with helium and make an animal out of it and have it float?" Yes, you could make an animal that will float for a short period of time if you inflate the balloon *most of the way*, leaving only a short nipple. Neglecting the lift (but not the weight) for the uninflated nipple, the math says that for neutral buoyancy (the "just" floating condition where the weight equals the lift): 0.33 g 2.41 g 1.7 g + ------------ x (inflated length) = ------------ x (inflated length) 50" inflated 50" inflatedSolving this equation for the "inflated length" term gives the minimum inflated length for buoyancy = 41". So if you could fill a 260 up to 1.75" diameter and make a figure that had a total bubble length of at least 41", it should hang in the air for a while. How do you get more than a few minutes of flying time out of a Helium-filled 260? inflate all the way and don't twist it much. inflate to as large a diameter as you can (don't burp it). cut off the rolled nozzle after you tie it (shed those milligrams!). To get as large a diameter as possible for # 2, try this trick: put a paperclip across the nozzle of the uninflated 260 and drop it into some boiling water for a while. Take the balloon out of the water, remove the clip (which has kept any water from getting inside the balloon) and *immediately* inflate it with helium without even drying the outside. You'll see that water has a huge effect on the properties of latex. What is the effect of the latex compressing the helium? You can fill a balloon, then you can fill it some more before it pops. Assuming the balloon is not getting as much bigger as it had been during inflation, the helium is more compressed.
How do balloons pop?
Can static electricity pop a balloon?
Ever get an hour or so into a gig and then BAM! BAM! BAM! - balloons start to pop one after another? When many balloons start to spontaneously explode, static electricity is often the culprit. If you work in the same general area for a prolonged period of time and that area happens to be carpeted, and/or in low humidity (you're inside a heated room on a cold, winter day) there's a really good chance for static build up. Static electricity sparks will cause your balloons to pop. So, here's what you can do:
- Get yourself a can of "Static Guard," a product to prevent static buildup on clothing, and spray profusely whenever you feel like you're a victim of static cling. Anti-Static agents for textiles quite generally work by increasing the electrical conductivity of the fabric. In this way, electrical charge is prevented from accumulating on the fabric, and cannot cause discharges. The conductivity is improved by attracting humidity from the air. Ingredients of antistatic sprays thus contain humidity-loving (hygroscopic) substances, e.g. salts. Alberto-Culver Co., the original manufacturer of Static Guard, obtained a patent in 1976 on an antistatic spray:TI Antistatic spray IN Dasher, George F.; Fiebig, August E., Jr. PA Alberto - Culver Co., USA SO U.S., 5 pp. CODEN: USXXAM PI US 4129505 781212 AI US 76-725361 760922 DT Patent LA English AB Antistatic, sprayable (pumper or aerosol) solns. for clothing comprise dimethylbis(tallow alkyl)ammonium chloride softener 1-1.4%, H2O 0.15-5%, NH4OAc 0.2%, and ethanol [64-17-5] 93-8%.
Among the ingredients, both the alkylammonium chloride and the NH4OAc are hygroscopic compounds. - The effect from "Static Guard" only lasts a few hours and it smells badly. Try 3M "Scotchguard for Furniture." It has wonderful anti-static properties.
- I explained the problems with static that balloon people experience to an aerospace Static Control Engineer. He told me a spritzer bottle of water will do the same trick as Static Guard without odor or contamination problems.
- On dry, windy days, we keep a cool air humidifier going in our shop.
- I've used Static Guard on the inside of plastic release bags with success.
- I've encountered this static twice before; both in carpeted rooms, and one of those times in a room with lots of copiers and computers.
- In regards to helium balloons in a computer store - don't do it! We tried to do single arches in a computer retail store and every balloon popped. At one point 5 balloons popped at the same time! It has to do with static. It has nothing to do with the balloon size - we've even done it in a computer training center in an office tower - same problem.
- Recently I did a job in a mall conference room with new carpet and a TV (which was off). I don't recall breaking as many balloons in my entire 14 years of being in business. So I went to the drug store in the mall and bought Static Guard, sprayed it on the balloons before I blew them up and also sprayed it on the bouquets. The problem stopped. I have used the same exact balloons for other jobs and had no problems. The only thing I could figure was the new carpet and/or the TV.
- On one job my balloons were doing fine and then started popping like mad. Then I realized that the hair on my arms were stading on end. Static, big time. Now I keep a "Bounce" dryer sheet in one of my apron pockets. When static builds up, I take it out and rub my hands on it - works for me!
- Buy a static gun (under $10) at a record store - They takes the static away from record surfaces. I've used the gun on my bag of balloons and it seems to help. Every once in a while I squeeze the gun at the plastic bag and it takes the charge away.
- If you are standing in one spot twisting and your balloons start to pop, move over a couple feet. Perhaps try grounding yourself.
- Attach a computer static mat around your pump and touch that before you touch the balloon.
- Don't use a pump! Inflate by mouth. The air in the balloon will be more humid.
- To prevent popping, try underinflating or working with a soft balloon.
- Use baby oil or glycerin on my hands. Glycerin will make them squeak a LOT more and baby oil needs time to soak in but both work well. A hand lotion that contains glycerin is also good.
- Every time I go to my local hospital with bouquets, they start popping. Once, while delivering to the emergency room, and back behind the doors, they started! One at a time- bang! bang! Out of a dozen latex balloons in the arrangement, I had 3 left! I went back to the studio to make another arrangement. This time I underinflated a bit, and I was able to make the delivery.
- We find that when we hi-float and preinflate, so the balloons are dry when we get on site, they will take a lot more abuse regarding handling and static.
- I've asked some engineers what could cause the popping; they told me it most likely has to do with ions in the air that are charged in such a way that they react with the balloon. I (we) don't believe it to be just static. How many of us have ever made a balloon stick to our hair with static electricity? One engineer recommended moving the air around with a fan to help keep the problem at bay.
Why do balloons pop around an adhered point when subjected to movement (pulling rubbing etc.)?
First create an adhered point by cleaning all the talc/cornstarch from a balloon, inflating it, and rubbing it against itself with a lot of force. Or by making a lock-twist. Or tie a knot in the nozzle and roll the ever-tightening knot towards the nozzle.
If you press your clean palm against a clean table, press down and then try to slide, you will get a jerky, stop-and-go motion called "stick-slip motion" or "stiction" (sticking-friction). If you look very carefully at the adhered point on the balloon when subjected to movement, you will get the same thing - the rubber welds to itself (sticks) and then tears (as it slides) because of the low shear and tensile strength of the latex. You can actually see the tearing debris accumulate when you do this (wear goggles!) Once this mechanism tears a hole in the balloon, the shape of the hole, stress in the wall of the balloon, toughness, and thickness of the latex all come into play in determining how the balloon responds. This kind of thing is studied in a field called fracture mechanics - the study of crack formation and growth.
Look at a balloon after it has popped. See that straight edge in the rubber that looks as if it was cut by a razor blade? That is the fracture surface along which the crack ran at the speed of sound in the latex. You can trace it right back to the origin of the fracture (the original tear). A familiar example are the cracks you get in glass windows when you hit them with a rock - you can tell where the rock hit, can't you? Because of the stress distribution present in an inflated 260Q, I'd expect the crack to run substantially more longitudinally than circumferentially.
Probably a better example are the cracks that can form in pressurized pipelines. When the pipelines are welded together, a crack can form and run for miles in a few seconds. After a few costly failures like this, pipeline designers now employ crack arrestors (bolted flange joints) every so often in pipelines to limit the maximum crack length (a crack can pass through a weld but can't pass through a bolted joint where the pipes are separated by a gasket in between the flanges).
Now let's talk about stress:
Normal Stress = Force / Area. Applying equal and opposite pulls of 1 pound, axially along to a 1" by 1" square bar results in a tensile stress of 1 pound per square inch (psi) at points far away from where the forces are applied. Pressure is a compressive stress, the same as you'd get by pushing on the above bar.
A 260 or 350 is what engineers would call a "thin-walled cylindrical pressure vessel with capped ends." The stress distribution near the ends can be very complex and requires methods of analysis that you learn in classes on "plates, shells and membranes" (where you even find solutions for stress distributions in toroidal shells... like a Geo donut!)
The stress distribution a few diameters from the ends is very simple though, and is derived in the most elementary textbooks. Skipping the derivation (...ha!... I remember having to derive it off the top of my head for the chief engineer who interviewed me 8 years ago for the job I used to have) the answer is:
axial (along the length) stress = Pr / 2t (tensile)
hoop (like if it was wearing a belt) stress = Pr / t (tensile)
radial (through the wall) stress = negligible
where:
P = pressure r = radius (1/2 the balloon diameter) t = balloon wall thickness
Notice this important point: the hoop stress is _always_ twice the axial stress for a 260, or a 350, or a 524.
Note, the above applies only to long skinnies. Using the thin wall pressure vessel approximations, the stress in a truly spherical balloon would be "equal biaxial tension plane stress"; Pr / 2t in each of the perpendicular "axial" and "hoop" directions (placed in quotes here because there is no one true pair of specific axial or hoop directions in a sphere because of the spherical symmetry) and again negligible radial stress.
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 | T | { R | | E | } S | / S | ," | .' | ._ - '" | , - ~ '"" | ,~ |,' 0 +--------------------------------> 0 S T R A I N
The slope of a line drawn tangent to the curve is the "elastic modulus" or stiffness (first derivative for those of you who remember calculus). The slope of a horizontal line is zero, and increases with angle (measured ccw like on a protractor) until it becomes infinite for a vertical line. Notice that the stiffness becomes very large at high strains. The reason for this unique behavior of latex is due to the way its molecular structure changes with stress.
Latex is a polymer, and specifically, a type of polymer called an elastomer (a rubber). It is made up of spaghetti-like molecules that are all coiled up and intertwined like... a plate of cooked spaghetti. Each molecule attracts the neighboring molecules with weak bonds called Van der Waals forces, much like the starch that sticks individual spaghetti strands together if you don't make it right. When you start to stress the latex (put 2 forks into the spaghetti and move them in opposite directions), the molecules uncoil and start to straighten out. As more and more molecules straighten out, the latex gets harder and harder to stretch (read: more and more stiff). It also gets easier to see through since light can pass in-between the straightened molecules, rather than getting lost in the tangled jungle of unstressed molecules (this is NOT an effect due to wall thinning). Once the molecules are stretched out straight, further deflection would require stretching the atomic bonds that make up the backbone of the molecule. This is very hard to do, so before that point the Van der Waals forces (starch) give out and the mostly straight molecules (spaghetti) slide past each other, initiating cracks.
Some 260 have a very clear running crack and some break straight across. Why?
In materials with uniform properties in all directions, cracks like to extend perpendicular to the maximum applied tensile stress. In a 260, we have two tensile stresses which are perpendicular to each other. But remember that hoop stress is always twice the magnitude of the axial stress. Since these are "principal stresses" (no shear stresses), it turns out that the hoop stress will always be the maximum stress. In an ideal world then, cracks will start at a flaw, then turn so that they run perpendicular to the maximum stress which means they really want to run axially. However, this isn't an ideal world, and latex does not have uniform properties in all directions when inflated....
Are the molecules arranged differently by the different stresses as the latex is stretched?
Yes. when you stretch an uninflated 260 like a rubber band, you straighten out and line up the molecules in the direction that you are pulling. When you inflate a round balloon, the molecules are stretched equally in all directions tangential to the balloon wall because the stress in all tangential directions is the same. But when you inflate a 260, twice as many molecules (probably twice, I can't really count 'em...) uncoil in the hoop direction as in the axial direction because of the 2:1 ratio of the stresses. Now to answer your question of why some 260's break straight across, think of this. If there are twice as many molecules lined up with their strong direction in the hoop direction, it shouldn't be that hard for a crack to overcome the weak Van Der Waals forces that hold those molecules against each other, right?
Upon inflation, balloons get easier to see through primarily because light can pass in-between the straightened, oriented molecules, rather than getting lost in the tangled jungle of unstressed molecules (this is NOT an effect due to wall thinning). The wall does get thinner upon inflation, but that's not the point here.
Somewhere, find a heavy piece of clear latex just the right thickness so that when you stress it to the same stress level present in an inflated balloon, it is exactly as thick as the wall of an UNinflated clear balloon.
You will be able to see through the stressed heavy latex sheet much better than through the wall of the UNinflated clear balloon.
Does the action of dipping line up molecules?
No. It is the stressing of the latex upon inflation that lines up (orients) the molecules.
As an aside, the plastic grocery bags that you get at the supermarket are made from oriented polyethylene (oriented by stretching during the rolling of the polyethylene sheets). They are very strong in the vertical (load bearing) direction because that is the way the molecules run, and they are rather weak in the sideways direction (play with one and see!).
For a definite example of orientation effects in latex, try pulling on a 260 as hard as you can as it is inflated. When you pull it really hard it is very difficult to inflate by mouth. With a pump you can inflate the 260 about the diameter of a 130. This is the way Roger Seigel (?I think?) gets a great looking elephant trunk in one of his books. A balloon inflated this way will take an incredible amount of abuse - much more than a standard inflated 260.
Additional stretching or pinching of the balloon will further stress, strain and orient the molecules. But there's more to it than that, due to the competing processes of stress relaxation and creep.
Try this experiment which will demonstrate some of the viscoelastic (time dependent) properties of rubber: hammer three nails into the wall. a) Stretch a fresh rubber band over two of the nails. Now since the nails can't move, the amount of strain (stretch) in this rubber band will not change with time. b) Place a fresh rubber band on the third nail, and hang a weight on the rubber band. Now since the weight can't change, the amount of stress (force) in this rubber band will not change with time.
Over time, the tensile stress in the rubber band in (a) will decrease as the molecules slide past each other. This is called "stress relaxation". Over time, the amount of stretch of the rubber band in (b) will increase as the molecules slide past each other. This is called "creep".
Creep and stress relaxation cause uninflated balloons to be larger after an inflation/deflation cycle. You can take advantage of the strengthening that can be achieved by reorienting the molecules, the decreased stress from stress relaxation, and the increased stretch from creep if you pre-inflate the balloon all the way, wait a few seconds, and then deflate it "severely". Because of creep and stress relaxation, a given balloon diameter can now be achieved at a lower latex stress level than for the inflated-directly-to-size condition, and lower stress makes the balloon more resistant to popping. That's what is recommended for pushing balloons into those new SDS metal grid frames used for creating balloon walls in large scale balloon decorating. While all of the above applies to 260's and 350's, in practice it should be limited to "burping" them, because inflating 260's and 350's all the way first will make them unworkable for twisting.
On a side note, vulcanizing "cross-links" some of the polymer molecules (chains). Vulcanization can be thought of as spot welding the spaghetti strands to their neighbors every so often, but leaving them free to wiggle in between the spot welds. The more vulcanization, the closer the spot welds become, and the harder/stiffer/stronger/less-flexible the latex becomes.
Is vulcanization a fancy name for cooking?
Yes and no.
Here's an interesting history of rubber from "Tinkers and Genius, the Story of the Yankee Inventors, by Edmund Fuller, Hastings House Publishers, NY, 1955."
..."India rubber"... Its original name was Caoutchouc (pronounced something like koochook). It was widely known as "gum elastic" but had come to be called "rubber" because its earliest recorded use (other than as balls to play with) by white men who fetched it from South America, was as an eraser. The "India" crept in as a joint reference to the South American Indians who gathered it and to the West Indies which became a trading channel for it.
...Around 1834... The India rubber trade was the next thing to being dead.... The plagued India rubber either melted and ran in the summer or petrified in the winter.... Scores of people were experimenting with the rubber problem.... As for Charles (Goodyear).... He recognized this as God's chosen work for him. Nothing would stop him.... The discovery of the sought-for secret came in 1839.... he was boiling rubber and sulphur on the kitchen stove, trying to make the curing process permeate it. A blob fell on the hot stove top and hardened. It was what came to be called "vulcanized."
His work was not finished. How much sulphur? How much dry heat? How long for the process? These things had to be worked out experimentally. But Charles had it, he genuinely had it.... In success, he was at the extreme of bankruptcy.... For five years more he wandered in poverty around New England, working out the process, begging facilities, seeking a backer... It wasn't until 1844 that he got a patent.
Are there all levels of vulcanization from runny to hard?
from "Chem One by Trublood, Waser and Knobler, McGraw-Hill, 1980"
The vulcanization of rubber, by heating it with sulphur, which converts the rubber from a soft, gummy material into a product of varying hardness depending on the amount of sulphur used, involves the creation of cross-links that consist of -S-S- groups
H H HCH H H | | | | THIS IS THE REPEATING UNIT IN ONE ... - C - C = C - C - ... POLYISOPRENE (NATURAL RUBBER) MOLECULE | | | H H H H H HCH H H | | | | ... - C - C - C - C - ... ONE POLYISOPRENE (NATURAL RUBBER) MOLECULE | | | | H H S H | VULCANIZED TO ANOTHER H S H H | | | | ... - C - C - C - C - ... POLYISOPRENE (NATURAL RUBBER) MOLECULE | | | | H H HCH H H
(Polyisoprene is the major constituent of natural rubber)
From Introduction to Material Science for Engineers, Shackelford, MacMillan Publishing Co, NY, 1985
The extent of cross-linking is controlled by the amount of sulfur addition. This permits control of the rubber behavior from a gummy material to a tough, elastic one and finally, a hard, brittle product as the sulfur content is increased.
Now Mark again: Oxygen is chemically very similar to sulfur, and can replace sulfur for cross-linking polyisoprene. They may take advantage of this fact in the balloon manufacturing process, or it may be what causes balloons to go bad from air exposure. I don't know for certain which - I'm a mechanical engineer, not a polymer chemist (though I play one on the net... :-)
Would not the amount of vulcanization make a great deal of difference in the balance of the long force and the side force of a 260?
No, the 2:1 balance of the hoop and axial stresses is a function of the pressure vessel (balloon) geometry, not the material. It's 2:1 for steel too.
There must be a best amount of vulcanization for a 260 to make the best balance of forces for a 260.
Well, there certainly must be a best amount of vulcanization for a 260 in order to make it best for twisting. It's probably determined by trial and error, then written down and kept in the company vault.
How would a 260 made from under cooked latex be different than a 260 that was made from over cooked latex?
The undercooked 260 would be red and bloody in the middle, and the overcooked 260 would taste smoky and burnt.
I wonder who is responsible for the cooking. Does Qualatex get the latex pre-cooked?
No. The "cooking" is done after the dipping.
In the Qualatex-published book "Design" by Gary Wells, they state that: "Qualatex balloons are made from 100% latex. No fillers or substitutes are used."
From Introduction to Material Science for Engineers, Shackelford, MacMillan Publishing Co, NY, 1985
A filler is added to strengthen a polymer primarily by restricting chain mobility. ("chain" is short for "polymer chain" or molecule) It provides dimensional stability and reduced cost.... Roughly one third of the typical automotive tire is a filler (carbon black).
Design also states: "Pioneer compounds its own latex and blends its own inks and dyes".
From Introduction to Material Science for Engineers, Shackelford, MacMillan Publishing Co, NY, 1985
Dyes are soluble organic colorants that can provide transparent colors... A pigment is an insoluble colored material added in powdered form.
monty writes:
> What about the thermal properties of latex? I still don't have any idea > why small deviations in temperature have such a massive effect other than > perhaps the fact that it is derived from TREE SAP. Which of course as we > all know "flows" slower in cold. That was the only theory I had to explain > the slow-mo spread and large pieces that never contracted in my outside > clean-up.
From Introduction to Material Science for Engineers Shackelford, MacMillan Publishing Co, NY, 1985
Polymer properties vary tremendously with temperature. To demonstrate the temperature dependence of polymer properties , the modulus of elasticity (stiffness) is typically plotted against Temperature.
At low temperatures (well below Tg), polymers behave like rigid solids (exhibiting a relatively constant, high stiffness). They deform elastically (like spring steel) and are quite brittle. In this temperature range, they are also referred to as "glassy".
^ | | : |_____: | :". | : ` | : \ S | g : } : T | l : { : I | a : ! : F | s : ! : : F | s : l \ : : N | y : e `.: _ - " ~-. E | : a " _ - " " :` S | : t : : \ S | : h : : i | : e : rubbery : | | : r : : viscous | : y : : | : : : | : : : +-----:----------:------------------------:--------> Tg Tm T E M P E R A T U R E
In the glass transition temperature range, Tg, the modulus (stiffness) drops precipitously and the mechanical behavior is termed "leathery". The polymer can be extensively deformed and slowly returns to its original shape upon stress removal.
Just above Tg, a "rubbery" plateau is observed. In this region, extensive deformation is possible with rapid spring back when the stress is removed. (Latex is an "elastomer" - a polymer with a predominant rubbery region. Note that the modulus of elasticity (stiffness) of rubbers INCREASES with temperature in the "rubbery" region.)
As the melting point Tm is approached, the modulus (stiffness) again drops precipitously as we enter the liquid-like "viscous" region where they behave like cake batter. The boundary between elastic and viscous behavior is known as the "glass transition temperature", Tg.
Images magazine says that after inflation, some vendors of round balloons will throw them in the dryer to get them back to original shape/size for reuse. The agitation and heat that the balloons get as they roll around inside the dryer is essential to restoring them to almost like-new condition. The dryer method works....but its safer to put them inside a terry cloth bag or old pillow slip first to avoid the possibility of melted latex on your dryer drum. Home dryers, even on low settings, can get quite hot! Use the lowest heat setting. Toss them for about 15 minutes and voila, the latex is shrunk back down to original size.
Viscoelastic materials like latex exhibit time dependent deformation behavior. Stretch a steel spring elastically (don't permanently deform it) and release it. You will observe that it instantly goes back to its original size. Now stretch a rubber band and release it. You will observe that the large initial contraction does not return it to its original size; instead it recovers its original size quite slowly. Likewise, inflate a balloon, and then deflate it. Again you will observe that it does not instantly return to its original size, but continues to shrink for quite some time. The rate of recovery can be increased by increasing the temperature. This is another great experiment that you can do with a hair dryer.
Increasing the temperature is what the mfrs. do when they "drum" the balloons in rotating industrial dryers to shrink them back to "like new" after they've been inflated for printing, etc. (Good quality printed balloons are inflated for printing - that's part of the reason they are so expensive). The drumming machines look like oversized stainless steel cement mixers, very similar to the coating drums used in the pharmaceutical industry, and work like the clothes dryer in your home.
There is another neat phenomenon called the "Gow-Joule Effect" wherein most elastomers (rubbers) contract when heated while stressed in tension (stretched). Quite the opposite of what you'd expect, but it's true as you can easily see for yourself. Hammer a nail in the wall, and put a rubber band on it. Hang a heavy weight from the rubber band and let the weight come to rest. Mark the spot on the wall where the weight is. As you gently heat the rubber band, like with a hair dryer (I used a propane torch, but then I like overkill...), you can watch the weight rise (until you start melting the rubber band, that is). The Gow-Joule Effect only works when tensile stress is present though.
Tom writes:
> The exact level of vulcanization may be a most important part of > making a 260.
Indeed. No doubt it's a crucial element of the total manufacturing process.
> The old Ashland balloons had a rubbery feel and seemed to decompose faster > than the Qualatex. The feel of a balloon is tricky because there are > different finishes but I wonder if they started with a different level of > vulcanization.
I expect that the exact process variables are proprietary trade secrets. This is schematically how vulcanizing affects the plot above (it raises and extends the stiffness curve at all temperatures):
^ | | : |_____:_ | :".`. vulcanized curve | : ` `. : / | : \ `. : / S | r : } `.: _ - " ~------. T | i : { ` _ - " " ` I | g : ! : F | i : ! : : F | d : l \ : : N | : e `.: _ - " ~-. E | : a " _ - " " :` S | : t : : \ S | : h : : i | : e : rubbery : | | : r : : viscous | : y : : | : : : | : : : +-----:----------:------------------------:--------> Tg Tm T E M P E R A T U R E
What can be inferred from how a balloon breaks?
Pretty much anything a Tarot Card can tell you if you're good at it. Give me a call at $3.99 a minute and I'll be happy to pop and then read some balloons for ya...
But seriously, fracture mechanics is a powerful tool for post-mortem analysis, and detailed examination of fractures and fracture surfaces can provide all sorts of information.
Monty writes:
> the strangest break line from a round balloon is the Sawtooth zig zag edge. > I suspect that it's from a pressure-overinflation rupture since that's > how some distributors recognize pretend "defectives" (mylar) that some > people try to return when they were actually overinflated.
In a round balloon the tensile stresses are the same in all directions tangential to the wall. Earlier we said cracks propagate perpendicular to the maximum tensile stress direction. Since all directions tangential to the wall are maximum tensile stress directions in a round balloon, a crack can run in any direction. What probably happens is that it runs until it is deflected into a new direction by a nub or other inclusion in the latex.
> I've also noticed that different balloon color "types" break differently. > Pearl Tones seem to "shred" (explode into dozens of TINY shards) which is > a pain on cleanup. > monty
The pearl-metallic sheen of "Pearl Tone" balloons derives from the same technique used to make the "pearl" paint colors used in custom automobile paint jobs: the addition of powdered mica or aluminum. The powder particles in the latex can be thought of as inclusions, and if the latex-particle bond is not as strong as the latex-latex bond it replaces, each powder particle can be thought of as being like... one perforation in the line of perforations between a check and its check stub. When you pull hard on a check and stub perpendicular to the line of perforations, a tear (another name for a crack) runs from perforation to perforation. Well, if you had a piece of paper that was full of perforations everywhere (like what is used in some brands of paper hand towels to make them feel soft), and you pulled on it equally in all directions in the plane of the sheet (like the stress distribution in a round balloon), nobody could predict how the cracks would "connect the perforations", and each time you tried it you'd get a different result.
Additionally, the initial popping of the balloon sends stress waves out through the already highly stressed wall of the balloon, much like the waves that spread out when you drop a rock in a puddle of water. The high stress from the superposition of the existing stress plus the stress wave stress just might be enough to initiate other cracks at sites that were teetering on the edge of bursting, leading to additional crack fronts.
Inflate a balloon of any size/shape. Tell (or bet) the audience you can stick a pin in the balloon, and it won't pop!
If you put a piece of cellophane tape on the surface of the balloon, and jab the pin through the tape into the balloon, it won't pop (at least not right away). The tape keeps that initial "crack" from developing. The air will leak around the pin, or through the hole if the pin is removed. After a while,, one or more cracks will work their way past the edge of the tape, and _then_ it will pop if there is still enough air in the balloon. (I used to use this effect as a sort of "time-delay fuse". The wider the tape, the longer you delay the pop because the tape helps carry the stress that the latex would normally have to carry alone.)
But this can even be done without the tape. You can put an oiled, polished steel needle through a special balloon without popping it. If you know the trick you know the needle goes through the thickest parts of the balloon; near the nozzle and opposite the nozzle. These are the last parts of the balloon to be stretched out completely as it inflates. Make sure that the needle is very sharp, and be sure to wipe a very thin film of Vaseline on the needle... as explained earlier, friction is deadly to latex and you want an initially clean opening, not a tear.
The "Needle thru Balloon" consists of a giant needle (about 16" long) threaded with a bit of yarn. The needle is lubricated with petroleum jelly or silicone grease, and is best stored in a hollow magic wand. The needle has a special conical taper at the tip to make penetration of the balloon more reliable - but a sharpened knitting needle will work in a pinch. Keep the needle very sharp and well lubricated. When it goes through, the Vaseline jelly will plug up the hole and the balloon won't deflate. Secondly, only blow up the balloon half way. It should be about the size of a cantaloupe. The trick should work - I only failed once in 32 years!
A magician writes: I perform the Needle Thru Balloon using clear 11 inch balloons. I usually use Vaseline to lubricate the needle. In an emergency I've even used Salad oil! I blow up an 11 inch balloon with four full blows and let a little out to keep it soft. I tie it, then squeeze the top of the balloon and I'm ready to go into my routine: I get a volunteer up from the audience and have him blow up one of the balloons while I blow up the other one. I ask the kid what would happen if I touched the balloon with the needle. Then I pop the first balloon. I then have the kid hold out his hand. I place the balloon in his hand and have him place his other hand on the top. I take out some ear muffs and put them on me. I realize the mistake and then place the muffs on him. Then I stick the needle through the top of the balloon and through the knot side. I then have the kid let go and then proceed with pulling the thread through. (I use yarn for the thread.) I hand the kid the balloon at the end of the routine and as he goes away, I pop the balloon and give him an animal instead. In my travels, I found different grades of clear balloons and had trouble finding the "spot" at the top of the balloon to go through. So I came up with a foolproof solution: Go to the theatrical make-up counter at your local clown or theater store and get a bottle of liquid latex. Brush some on the part of the uninflated balloon you want the needle to go through and let it dry (usually 5 - 15 minutes). The liquid latex creates the needed spot for the penetration! I have even brushed some on the sides of the balloon and performed a sideways penetration. ( I believe that this is the way Doug Henning did the penetration on one of his specials.)
Another approach is using an 18" clear round and inflating it to 11". It leaves so much barely stretched out rubber on the end that you can poke it like a pincushion if your needle is in good shape. Baby Vaseline in the needle wand (holder) coats the whole needle. Without a coating of lubricant the odds of the trick working go way down.
T.Myers recommends using an 18" clear balloon, blown up to about 11" size. This leaves lots of extra stretchiness to the balloon, and makes the trick much less likely to fail. Failure, as any of you who perform the trick know, is inevitable and happens almost 50% of the time using 11" balloons.
Hmmm, I've been using the "needle-through-balloon" balloons, maybe they are just regular balloons and the name has me psyched but I only rarely have a failure, I can only remember one in the last 10 or so that I've done. There is a technique to it that includes slightly pinching the end before putting in the needle to give yourself some slack. I also use the needle wand to keep my needle well lubed .
I've been using the 18" clear balloons in needle-thru-balloon for about six months now. My success with the 11" clear balloons was very spotty, and seemed to depend on the batch of balloons. The 18" clears have a very large neck that is a bit troublesome to tie, and the "thick spot" at the crown may be quite a bit off-center when the balloon is blown up to 11" size -- you have to look for it. Overall, I like the 18" clears, and have had no failures in performance. I have no plans to go back to the smaller ones.
I use 11" balloons. Of course, I do not inflate them to the full 11" size. I always leave them a little soft.
Ickle Pickle Products 808 Somerton Ridge Drive, St. Louis, MO 63141 (314)434-3630, sells both sizes of needle for the needle through the balloon trick. The small one can be stored in a clear tube. The small one will work with saliva as a lubricant, it doesn't need vasoline.
Has anyone mentioned the similarity between the mini-needle thru the balloon and a *hat pin*? Although real hat pins are rare in this part of the world, you can make your own easily. Go to a bead store, (Jewel-Art in our area), and ask for a hat-pin pin. I think they come in 5" - 8" sizes.You can even purchase a fancy bead to glue on the end. Choose carefully, you don't want to get *stuck* with a dull one.
Tom asks: is it better to first overinflate and then let air out to have less tension at the point of penetration or does it make more sense to work with the wall thickness and tension of a inflated-directly-to-80%-full balloon? What is the optimum look of the spaghetti to stick a needle through without creating a tear? The trick will work either way but it seems like one should be better than the other.
The problem is that you need more info to answer this: specifically, you need to know the fracture toughness (a property meaning "resistance to crack growth") and how it varies vs. strain. Well, you need 2 graphs - one for pre-stressed (oriented) latex, and one for virgin latex. Hmmm, actually, the fracture toughness will vary with angle relative to the molecular orientation direction... and with state of stress (equal or different hoop and axial components?)... and... what a mess!
So, what you really need to do is to find a balance point between all these competing mechanisms, (stress driving the crack, stress changing the fracture toughness of the material, orientation effects, etc...) and you'll need quantitative info (numbers) to do it, not just qualitative hand waving like everything written here. Plus, all of these properties are functions of temperature, so it really becomes complicated. Here's a report of how substantially the properties can change with temperature:
Monty writes:
Latex reacts oddly in the cold. In one instance, after cleaning up from a New Years party, I used a pin to dispose of some 16 inch balloons that were outside in 15 degree Fahrenheit weather. Instead of the usual physics, The latex refused to contract in real time and split into two large pieces. The topmost piece was so large that it floated about 12 feet away before the helium escaped from it's edges. In fact the latex never fully contracted until it was brought inside. It did not have High Float in it or anything else that would impede it.
It's probably a lot easier to just do a series of balloon/needle controlled experiments to figure it out. If it's any help, there's a recommendation for making balloons more resistant to popping when pushing them into those new SDS metal grid frames (used for creating balloon walls in large scale balloon decorating). They evidently tell you to pre-inflate the balloon all the way, wait a few seconds, and then deflate it "severely". Because of orientation, creep and stress relaxation, a given balloon diameter can now be achieved at a lower latex stress level than for the inflated-directly-to-size condition.
Balloons at High Altitudes - when do they pop?
Released helium balloons explode at a height of about 28,000 - 30,000 ft. 2 studies prove that. One by Don Burchette, inventor of Hi-Float and winner of the crystal award, and anotherfinanced by the Danish Department of Aviation and translated by Don Gebhard.
How far can a balloon travel before it bursts? According to Totex Corp. (one of the world's largest makers of weather balloons) the rate of ascent for a large balloons is 320 meters per minute or 17.5 ft. per second. A large balloon released at sea level would reach it's bursting height in 26 minutes. The rest is up to wind currents. Treb Heining has released over 1.4 million balloons at once. Millions are released each year in the US. The National Weather Service releases 50,000 five foot diameter balloons each year.
Heidi writes:
Why Do Balloons Go Bang?
The energy stored in the compressed air inside a balloon is not very large at all. Balloons create very little overpressure, apparently on the order of 5 or 6 mm of mercury when inflated to normal size. On inflation, the pressure must be higher as the rubber just starts to stretch because, from our stress equations above:
the modulus (stiffness) of the rubber is initially large, (it then drops off, to finally get VERY large with increasing strain) the balloon wall is initially thick, and the radius of the balloon is small. Pressure falls rapidly as the balloon grows in size. This follows from the stress/pressure relationship, and the stress/strain curve for latex.
There is a well-understood differential equation applying to soap bubbles relating surface tension, bubble shape and internal pressure. The surface tension can be thought of as a *constant* hoop and axial stress (NOT a function of strain, as in latex). Two soap bubbles inflated to about the same size and connected with a pipe form a system that is not stable. One soap bubble will always collapse and the other will inflate. The smaller bubble size requires a higher air pressure than the larger bubble; it tries to develop the higher pressure by shrinking, but since the bubbles are connected by a pipe, shrinking just forces the air into the larger bubble. As the bubble size difference increases, so does the pressure difference generated to drive the air flow. This speeds up the collapse of the small bubble. Now, remember that the volume of a spherical soap bubble is proportional to the cube of its diameter. Visually, the process *appears* to speed up even more, because even for a constant air flow rate through the pipe, the diameter of the small bubble will be decreasing at a much greater rate than the large bubble diameter will be increasing.
This can be demonstrated with balloons, but the size difference has to be rather noticeable before the process will begin. When it does begin, it can become rapid and it can suddenly halt. With balloons, this is a much more complex experiment than meets the eye because there are so many variables changing at once. The 500 - 600% strains make it a "large deflection" problem, in which we can't make any of the simplifying assumptions which we usually do. The geometry changes substantially, and latex displays highly nonlinear behavior.
The sudden halt even shows up in ONE balloon when you are using 260's. Partially inflate a 260 and what do you get? a large diameter, thin wall, high stress bubble with 500 - 600% strain, a small diameter, thick wall, low stress nipple with but a few % strain, and a transition region between them. Note that each of these two distinct sections contains the same pressure! How is this possible? It's possible because this large deflection problem in nonlinear elasticity (remember the sigmoidal stress-strain curve?) has more than one stable solution! Amazing if I do say so myself!
As balloons reach maximum expansion they get to a point where the latex runs out of stretch and gets stiff and resists further stretching. This is obvious in a fresh, overinflated balloon. It will become stiffer and get very rigid as all the latex molecules all become oriented in the tensile stress directions. This increase in stiffness will cause balloons, unlike soap bubbles, to increase in internal air pressure just before bursting.
The Big Bang Theory
Here is some info on balloon bangs, big and little.
While the air pressure inside the balloon does not contain much potential energy, the latex does store terrific potential energy as "elastic strain energy". The rapid release of the stored energy in the latex produces the resounding bang.
When a balloon bursts, the latex splits into various pieces as cracks develop. The speed of sound in latex is much higher than the speed of sound in air. The speed of the crack propagation through the latex approaches the speed of sound in the latex. Therefore, the velocity of the crack faces break the sound barrier in air and make a sonic boom. The latex then violently contracts. The ends of the latex contract so rapidly that they break the sound barrier. Just like the end of a bullwhip, and they make a shock wave. The more latex breaking the sound barrier, the bigger the bang. The faster the latex is going the bigger the bang. A few large very tight pieces of latex contracting will make a bigger bang.
This explains the following:
1. Very large weather balloons made with the very thin latex tend to go "foom" instead of bang. The latex does not develop the high degree of elastic tension needed to really accelerate when it rips apart.
2. A balloon that has been well stretched by severe inflation several times does not expire with as extreme a bang as an identical balloon blown up to the point of fatal overinflation without stopping. (Stress relaxation, creep and fatigue may all play a part here) Also it tends to rip into more pieces (because there's more strain energy to dissipate).
3. Even small balloons like nine inch rounds can produce a very big bang if they are strong high quality balloons and are blown up to the limit. They can develop fantastically high tensions and the latex develops very high speeds when it bursts. Of course a larger balloon blown up to a similar extreme tension all over would make an even bigger bang since more latex would be breaking the sound barrier when it burst.
Implications of the Big Bang Theory
This theory predicts that the very biggest bangs should be produced by:
1. Using high quality balloons made of latex that will stretch very far and is not stiff and limited in the amount of stretch possible.
2. Inflating balloons to the maximum possible extent in order to get the latex as stretched as possible. That is to say, just keep blowing and do not speed things up with a pin or other sharp object.
3. Stretching the balloon a minimum amount. One partial inflation will take away the initial stiffness of the latex and help insure that it will stretch further than a balloon that has not been annealed with one previous inflation. However, this should not be overdone.
4. Have the balloon clear of anything that will impede the latex when it does burst.
Environment and the Big Bang
Environment can have a major affect on the sound produced when even the most optimum balloon is properly inflated past its bursting point.
A padded environment with many complex shapes, an environment full of things to make many canceling reflections, will muffle the bang considerably.
A simple environment with few hard surfaces and a large volume will enhance the bang and produce great booming echoes.
Optimum places for inflating large strong balloons for maximum effect include:
Empty gymnasiums in the middle of the basketball court. Sometimes a fantastic rippling echo can return from the banks of seats.
Large dance studios, often used for aerobics classes, especially with mirrored walls. The mirrors enable one to see just how big the balloon is getting and make optimum sound reflective surfaces.
Concrete stairwells in large buildings, the larger and taller the better. Massive booms can be produced, especially if the balloon is overinflated one floor from the top or bottom of the stairwell. (like a giant organ pipe!)
MB 12/22/95
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.
At the end a needle through balloon routine where the balloon is busted (I name my balloons Sam) I thank my assistant for her dedication, but I thank Sam even more for his dedication! I liken it to the difference between dedication and commitment can be seen with a helping of ham and eggs. The chicken showed dedication but the pig showed commitment. -- Having too many balloons break makes you look bad. And it's very hard to keep that cheery attitude. It's happened to me and I get to talk to twisters right after they get home from a bad balloon day. I am aware of the frustration it can cause. Twisters have suggested I am responsible for their travel to and from the job, their damaged reputation, and the cost of that damage for the next couple of years. We have different ideas about who carries the responsibility for popping balloons. There are lots of possible reasons balloons break. From manufacturing to bad storage any number of things can happen. I know that all the problems are not due to storage. It is clear to me that every batch of balloons is different. Through the years, Qualatex has kept its standards higher than other companies. They are careful about checking their balloons. I check batches as they come through and my 260Q assorted stock turns over in 2 weeks but bad balloons occasionally make it to the customer. It's just a fact of life. You can take your chances and assume the balloons you just got are good or you can go to the trouble of testing them. The only way you can be sure the balloons you use in performance are good is to test each bag. You also need to buy them far enough in advance to get them replaced if necessary. Not testing them is kind of like not having car insurance. Chances are you will not have an accident. If you go through a lot of balloons, testing them sounds crazy. It's way too much work and expense. I'm in the same boat so what I do is spot check and pay attention to complaints. When I was cranking through lots of balloons on the street and they started popping, the first thing I'd do is slow down. If they kept popping I'd burp the balloons a little extra. If they kept popping I'd put that bag back in the cooler and use a different bag. Sometimes the bad bag worked better later in the day. I generally worked through them but if there were bags of that batch left, I'd get them replaced. Every order of balloons that I send out has a note to test your balloons and a quality report card. When you send in a card, we fax it right off to the manufacturer and I'll check my present stock for any specific complaints. I wish I could do more but until you inflate and twist a balloon, no one knows if it is going to pop. If you use enough balloons at some point you will get an unsatisfactory batch. When you do, let me know, maybe I can keep it from happening to someone else. T Myers -- You are the one the customer sees, the one who looks unprofessional when he/she gets frustrated and loses their temper over excessive popping. You can blame the balloons all you want to, and they may even be at fault, but that doesn't make much difference to the customers you serve. It doesn't stop the the ringing ears and whimpering kids and the angry parents. The manufacturer can't deal with that for you. That is yours to handle. Foremost would be a positive attitude. The more positive you stay, the better your chances are of coming through the trial, while still looking good. If you get negative, the people do too. If you start blaming the balloons & the manufacturer & just plain get into griping, you tend to set the tone for how everyone else is going to act. Try and keep your cool and stay as positive as you can. 3/17/99 Can you please tell me how big the openings are in a latex membrane? I write a question and answer column for newspapers in New Mexico. One of my readers asked why helium balloons deflate after a few days. I know the size of an helium atom but not the size of the hole it passes through. Can you please help me? Deflation of a helium-filled balloon by diffusion is a VERY complex, time and temperature dependant process. It's really not about openings in the way that air leaks out of a punctured tire. The Helium DIFFUSES through the latex by INTERSTITIAL DIFFUSION - essentially winding its way through a maze of narrow passageways between the tangled, long chain, polymer molecules. The force driving diffusion is essentially the same one driving osmosis... nature wants to even things out... to maximize randomness... to reach minimum entropy. Nature wants to spread that pressurized, high concentration of helium out into the atmosphere which is at a lower pressure and lower helium concentration. A one-dimensional form of Fick's law governs diffusion through membranes. Look it up in a college level materials textbook. It would take a fair bit of work (experiments, analysis) to determine the diffusion constants for this situation, and how they vary with temperature, not to mention the degree of vulcanizing. I would say that the diffusion constant will also show a significant dependence on the stress (and/or strain) state of the latex (which in turn changes with time due to stress relaxation). In an inflated balloon the initial strains in the latex are on the order of 500 to 600%, there is time-dependant stress relaxation occurring in the latex (which changes the internal pressure), plus the geometry, stress and strain states are all changing during deflation. This all means that balloons do NOT deflate at anywhere near a constant rate... it is a highly nonlinear process. BTW, experience has shown that in many cases the majority of the helium leaks out through loose knots. Tightening up the knots will often dramatically lengthen the float time of helium-filled balloons. Avoid those "Quick-fill" plastic plugs at all costs... they are also quick-leak :-) Also, use "helium quality" balloons - they are thicker than the typical balloon and have better quality control. Finally, float time can be extended 5 to 10X by using "Hi-Float" inside the balloon... it dries to form a polymer coating that is more impermeable to helium than latex. Of course, you just could do what I do: Slap a piece of tape on the balloon when no one is looking, and say "Hey kids, betcha can't do this. Here, everyone take a needle and try stabbing it through something." The only down-side to the tape "stick" is that this is a technique taught in boy-scout and cub-scout magic books. I've had more than one kid inform me that they learned how to do it in scouts so I let them inspect the balloon and when there is no tape you make major points (1) the needle in the balloon trick My favorite bit with this is one that I found in MUM (the magazine of the Socitey of American Magicians) a few years ago. I'm not sure about the exact reference. If I find time I'll look for it. Rather than doing a needle through balloon, do the power drill through balloon. I use a cordless drill with a very thick needle. Take a sharpie or other appropriate marker and paint a line around the needle to make it look like a drill bit. Fire up teh drill so they see it working. Then stop it at just the moment that you're about to insert it into the balloon. It passes in more soothly if it isn't rotating. If you draw a face on the balloon, there are a lot of story possibilities. The story line, depending on the audience, could range from a dentist's drill (this is the one from the magazine) to a lobotomy. With the right crowd, popping zits can also work.
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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 +-------------------------------->
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
From: TimosV@aol.com
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! :-)
Tim
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 < 74221.2015@compuserve.com >
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: jdewitt@ocdis01.tinker.af.mil (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: m-balzer@uiuc.edu (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: masel@aries.scs.uiuc.edu (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
m-balzer@ux5.cso.uiuc.edu (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 (r-masel@uiuc.edu)
From: kdmueller@ccgate.hac.com (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
(310)334-2586
Hughes Aircraft Co., Radar and Communications Sector
El Segundo, CA
--- All comments are strictly my own. ---
From: rogersda@direct.ca (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.
Sincerely,
David.
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.
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.
SCIENCE - OZONE
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
atmospheres?
>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
>bars.
P-V RELATIONSHIPS
>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
>know.
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.)
gives:
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.
--
OXIDATION - CARE - SCIENCE
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
oxidation.
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.
HOWPOP
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
rubber)
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
9/30/98
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
average
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
weather.
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
--------
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
are:
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.
TEMPERATURE
-----------
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
happens
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!"
POPPING
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
balloons.
> 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
(http://www.fooledya.com/balloon, 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
prestretching.
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.
Why do balloons float? Can you fill a balloon part way with helium, make an animal out of it and have it float? Definitely in water. In air... well, that depends. This is an application of "Archimedes' Principle" which states that a body immersed in a fluid is buoyed up by a force equal to the weight of the fluid displaced. If an object immersed in a fluid is heavier than the fluid it displaces, it will sink to the bottom, and if lighter than the fluid it displaces, it will rise. In our case, the "body" is a Helium-filled 260, and the "fluid" is air. If we use the "Ideal Gas Law" we can quickly find the mass of the Helium or air in a 260. The "Ideal Gas Law says that m = PVM/RT where: m = mass of the gas in grams P = pressure inside the balloon, in atmospheres V = volume in liters M = Molecular weight of the gas R = a constant = 0.082 T = Absolute temperature, Kelvin At sea-level, the air pressure is 1 atmosphere and say we are at room temperature. Then P = 1 and T = 293K. Then approximate a fully inflated 260 as a 1.75" dia cylinder that is 50" long. Since the pressure in the balloon is only slightly greater than the pressure outside the balloon, let's call them the same for now (both equal to 1 atmosphere). The volume is pi x 0.875" x 0.875" x 50" long = 120.2 in^3 or: 1 liter V = 120.2 in^3 x --------- = 2.0 liters. 61 in^3 We next look up the molecular weight of the gasses: M = 4.00 for Helium, M = 28.97 for the mixture of gasses we call "air" and plug all these numbers into the Ideal Gas Law to find: 1 x 2.0 x 4.00 the mass of Helium in a 260 = m = ---------------- = 0.33 grams 0.082 x 293 1 x 2.0 x 28.97 the mass of air displaced = m = ----------------- = 2.41 grams 0.082 x 293OK, we're almost done. The mass of a 260 is 1.6884 grams (it's nice to have a laboratory balance that weighs to a ten-thousandth of a gram... ), but let's just call it 1.70 grams since we're all friends here. Now add that to the 0.33 grams of Helium inside the balloon for a total weight: The total mass of the "body" (balloon + Helium) is 2.03 grams The mass of the fluid displaced (air) is 2.41 grams An object's mass doesn't change, but it's "weight" depends on the force of gravity. Converting these masses to equivalent weights on earth, the Helium-filled balloon weighs 0.0045 pounds and the displaced air pushes up with a weight of 0.0053 pounds. This results in a net buoyant force or "lift" of 0.0007 pounds on a Helium-filled 260, so it should float (but just barely!) !!! So, let's see if any of this works in the real world: PFFFFffffffffffffffttt! PFFFFffffffffffffffttt! PFFFFffffffffffffffttt! PFFFFffffffffffffffttt! PFFFFffffffffffffffttt! PFFFFffffffffffffffttt! Squeak, squeek, SQUEEEEEAAAK, SQUeak, squEAK, squeak... Ta-dah! Well, I now have a parasol floating above my head, so yes, fully inflated 260 sculptures with a minimal number of twists will indeed float (just not for very long...). This would be good for a quick demo, if nothing else. The Helium diffuses out rapidly, and there is just not enough reserve lift to keep it aloft as the 260's start to shrink. HOWEVER, Freda asked "Can you fill a balloon *part way* with helium and make an animal out of it and have it float?" Yes, you could make an animal that will float for a short period of time if you inflate the balloon *most of the way*, leaving only a short nipple. Neglecting the lift (but not the weight) for the uninflated nipple, the math says that for neutral buoyancy (the "just" floating condition where the weight equals the lift): 0.33 g 2.41 g 1.7 g + ------------ x (inflated length) = ------------ x (inflated length) 50" inflated 50" inflatedSolving this equation for the "inflated length" term gives the minimum inflated length for buoyancy = 41". So if you could fill a 260 up to 1.75" diameter and make a figure that had a total bubble length of at least 41", it should hang in the air for a while. How do you get more than a few minutes of flying time out of a Helium-filled 260? inflate all the way and don't twist it much. inflate to as large a diameter as you can (don't burp it). cut off the rolled nozzle after you tie it (shed those milligrams!). To get as large a diameter as possible for # 2, try this trick: put a paperclip across the nozzle of the uninflated 260 and drop it into some boiling water for a while. Take the balloon out of the water, remove the clip (which has kept any water from getting inside the balloon) and *immediately* inflate it with helium without even drying the outside. You'll see that water has a huge effect on the properties of latex. What is the effect of the latex compressing the helium? You can fill a balloon, then you can fill it some more before it pops. Assuming the balloon is not getting as much bigger as it had been during inflation, the helium is more compressed. I don't know what the exact pressures are because I don't have a pressure gage that measures fractions of a psi. I would have to use a manometer to measure the "head" (height that a water column is raised by the air pressure) in a 260 and then convert that to pressure. Is it heavier for having more helium inside? Yes. Consider how a tank full of compressed gas is heavier when full than when empty. Since we are talking about tenth-of-a-gram lift levels here, this can't be neglected for a 260. However, if the extra pressure causes the diameter to grow a bit, then we will most likely see an increase in lift. Where is the optimum lift on the inflation graph? The lift is directly proportional of the volume of helium in the balloon. The volume of helium is directly proportional to the length of the balloon. (A fully inflated 280 will have 33% more lift than a 260) The volume of helium is also directly proportional to the cross-sectional area of the 260. BUT, the cross-sectional area of the 260 is proportional to the square of the balloon diameter. So, the fastest way to increase the lift is to increase the diameter. That's why I suggested soaking the balloon in boiling water before inflating. The water is absorbed into the latex and reduces its elastic modulus, allowing it to stretch more. It will stretch more in the direction of maximum stress, the circumferential or "hoop" direction, and you'll see this as a diameter increase. (No, I don't know how much water is absorbed, and yes, the weight of the absorbed water does make the balloon heavier, and yes, there is probably an optimum amount of time to soak the balloon which maximizes the diameter increase to water-weight gain ratio... but that's a little deeper than I wanted to get into this). Inflating with heated air may also result in a diameter increase. Inflating & deflating a balloon several times will also increase it's diameter. You wouldn't have water weight gain to worry about either. It does tend to make the balloon thin-skinned & harder to work with, but the added volume of helium is well worth the effort. To answer your original question, to find the optimum lift as a function of helium pressure, you need a graph of the balloon diameter against internal pressure. Given those numbers and the formulas above, you could find the pressure for optimum lift. However, it might all be an exercise in futility because: the wall thickness variation may exceed the pressure/diameter variation, the balloon weight variation may exceed the few tengths of a gram lift we have, more pressure makes the helium diffuse out faster (but overall I'd expect it to last longer - just not twice as long for twice the pressure) the optimum may be at the burst point, which would vary for each balloon depending on it's surface and wall condition. Have you ever looked closely at an uninflated balloon? Each Qualatex 260 typically has a few particles embedded in the wall. In the paint industry, they call these specks of solid matter in a coat of paint or varnish "nibs". In the balloons, I am not certain what the nibs are made of - they may be hardened latex, dirt, lumps of coloring agent, etc. and it would take a little work to find out exactly what they are. Upon inflation, a thin-walled region develops around some of the nibs. The size of the nib in the wall seems to determine whether or not there is a thin area around it (as I'm certain the specific nib material would as well, due to the interfacial bond with the latex of the balloon). The larger nibs cause greater stress concentrations which create the thin areas around them (from elementary solid mechanics principles). I would say that the balloon would burst at these thin areas first, and that the variation in the burst point may outweigh the pressure/diameter variation. Are you saying that a twisted dog won't float because the gas makes it too heavy to float? No, just that the more gas you add, the heavier the balloon becomes. (It's the same as adding more water to a water balloon. It's true that air is a gas and water is a liquid; but they are both "fluids" in a technical sense and obey many of the same laws) Also I was saying that I don't know how much pressure a balloon will hold before it bursts, because I don't have the equipment to measure pressures that small. Part of Tom's question was: once the balloon is fully inflated, does adding more Helium increase the lift? Well, if the balloon doesn't get any bigger at all, the answer would be no. If the answer was that the balloon stretches a tiny bit bigger, then I'd have to know; did the tiny increase in size add enough lift to offset the extra weight of helium added. I don't know the answer because latex properties are highly nonlinear, the stress state is complicated, and you really can't calculate an answer - you just have to do an experiment and measure it. When everyone says that the gas leaks out real fast, how fast is fast? Like a normal balloon? Or in a few minutes? I think that the gas leaks out at the same rate (per unit of surface area) as it would in a round balloon of the same initial wall thickness. When looking at how fast Helium leaks out of a 260, as compared to round balloon, the overall rate is not equal. Consider that a round balloon has greater initial lift due to a higher volume : surface ratio, right? (the lift being due to the volume, the major weight being that of the "surface"). Where does the gas escape? Through the surface (the latex). Well, a 260 has greater surface : volume ratio, so it should have a faster relative deflation. Generally, a helium filled 260 sculpture has "landed" and is starting to droop in just a few hours, whereas a round balloon may have an "float" time of 8 - 10 hrs., and have landed, but still round the next day. The difference is that in a Helium-filled round balloon there is so much more initial lift than in a 260 that even after the round balloon loses some Helium, there is enough lift left to keep it afloat. (enough lift left... enough lift left... say that 10 times as quickly as you can :-) Also, consider the *apparent* deflation. A round balloon can loose some volume without being visibly much smaller, whereas a 260 sculpture can begin to "droop" with minimal deflation. In a helium-filled 260, you only have a few tenths of a gram of lift, and that is quickly lost as soon as just a little Helium diffuses out of the balloon and the diameter starts to decrease. I'm guessing that the bigger the balloon, the better it will float? Yes - the more weight it will lift off the ground. Note that at high altitudes in Utah and Colorado, some foil balloons won't float; because of the low air pressure above 5500 feet, the buoyant effect of the helium is decreased. Water vapor in the air (humidity) lowers the molecular weight of the air and decreases the buoyant effect of the helium. Mike Barr writes: Here in the New Orleans area we're 3 ft below sea level and 85-90% humidity is normal. I think most balloon decorators do this from choice, but here you must only use helium and not balloon gas (a mixture of air and helium). We don't sell (round) helium filled latex balloons without Hi-float, because without it, the float time isn't long enough to satisfy anyone. However, we often can't use as much, because of the humidity. In the New Orleans area, latex balloons with the same amount of Hi-float used in Denver won't float (because of the added weight of Hi-float). Watch the outside temp carefully. A fully inflated balloon will pop 4 seconds out the door on a warm day. All of my suggestions are based on experience, so do what satisfies your customers best and makes you the best living. I wonder what the effect of high float in a 260 or a 350 would do if it were filled with helium. My guess is that if would make it too heavy to float, but I know it would seal the pores. So, I don't know. Sounds like a good experiment to test and see what happens. In *general*, I find that the more twists a 260 has, the faster it seems to deflate. What sculptures can I make that will float? You can fill a 260 with helium but there's no guarantee that it will float, because the 260 will only hold a limited amount of helium and usually the weight of the balloon will prevent it from floating. A 350 will float without a problem, although it still won't float for very long. If you want a helium 260 to float in mid-air you've got to fill it up all the way to overcome the weight and then there's nothing to twist/no place for the helium to be pushed to. I know 260's will float at sea level but higher elevations may be different. So you can fly sculptures, but only ones you can make out of full balloons, such as the heart, candycane, spiral, octopus, etc. Try this: inflate a 260 and tie it end to end. (After you've tied the ends together, you can do an apple twist to hide the knot.) Then attach a string or ribbon and you've got a hollow balloon... similar to the look of a Geo only with a much bigger hole in the middle. When I first started twisting I was in a store with a helium tank and I taught the old gentleman there how to make a basic dog. We then got several to float to the ceiling. They were just about completely filled. I love floating sculptures. I do a variety of "floating" sculptures at various functions and have always gotten tremendous response from them. In addition to the traditional "heart on a string", I do planes, hummingbirds, butterflies, helicopters, swans, hats that won't stay put :), a 5' tall floating Goonie Bird, flying fish, and a bunch of others. Let your imagination go! I tie 260's to round helium balloons, so just about anything will fly. Two 260's attached to one 11" balloon will hover (this is going to depend on how stingy they are with the gas at the store, however). I can adjust the balloon's ability to hover by tying on additional ribbon, then snipping it off as the balloon starts to sink. This way, the kid gets to have the fun of a floating sculpture without it disappearing into the stratosphere. I specifically used this to float Larry's bats last Halloween, and used them for party games. One of my favorite things to attach to a hat is a hot air balloonist attached to a round. I make 1/2 of a guy out of a 130 and then I make his basket out of another 130 and attach them to the string of the round balloon and then tie the string to a hat.. It never fails to get a hoot. A "huggy bear" on a helium balloon string has a kind of Whinnie-the-Pooh look. What can I do with floating sculptures? Our family does balloon launches on the 4th of July afternoon, instead of fireworks. The 260's don't have much volume so a small tank fills a lot of them. Just one sculpture at a time, and watch them float out of sight, listening to the cheers from the neighbor's yards. The kids love it, and nobody gets burned. (We spend the 4th at a relatives house on a hillside overlooking where the city does the public fireworks display. It passes the afternoon hours, waiting for dark and the big show.) After dark, if you have a strong spotlight you can follow a balloon quite a ways, and let the neighbors claim a U.F.O. sighting. I've also used a clear Geo Blossom as a prop on somewhat larger airplanes to give them increased lift. I attach a peice of string as to a round balloon. It's fun to watch the kids flying in formation. -------------------------------------------------------------------------------- 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. A 5" round has almost neutral buoyancy in air when inflated with helium. They will be slightly lighter than air (they will rise) if overinflated. They will be slightly heavier than air (they will fall) if underinflated. Experiment to find a size (or combination of sizes) that provides the illusion you are looking for in your application. Note that you can trim off some of the nozzle to lighten balloons that fall. If you need to prepare them ahead of time, and then use them later, you may want to take into account the length of time they will sit before the effect takes place, since they leak helium very fast. An overinflated - slightly lighter than air -helium filled- 5" balloon will only float for an hour or so. make sure that whatever size you've inflated your 5" to with helium that they will reach that "slightly lighter than air" stage before you need them. Always do an experiment first to investigate all the dimensions, including time. (In other words, "Kids, TRY this at home") I have the kids guess what I am making out of balloons, it might take 2 or 3 year olds a while to guess but that's ok, all the guesses can give you ideas for new sculptures. Whoever guesses it wins it. If they win one then they can help other kids guess. For a helium filled balloon to float in midair it would have to have neutral buoyancy. It is _very_ hard to achieve this condition - typically the balloon will either sink or float, but not stop half way. A simple figure made from a 350 MIGHT be able to float, or you could incorporate a round (at least 9-inch) balloon into the sculpture. Tie a LONG ribbon to each figure, and then add enough weight to the balloon so that it barely floats. As the balloon rises, it lifts more of the ribbon from the floor, until the weight of the ribbon equals the lift of the balloon, and a state of equilibrium is reached. Try to get equilibrium with about 6 feet of ribbon. Cut the ribbon off about a quarter inch above the floor at the beginning of your event. The balloons will then rise to the ceiling until they lose enough helium to equal that quarter inch of ribbon. Then they will begin very slowly descending until the ribbon touches the floor. After that they will sink very slowly to maitain the equilibrium between the balloon and the weight of the ribbon. VALVES ON BALLOONS some balloons sold at fairs come equipped with plastic self-sealing valves, so the average person can easily inflate the balloons without having to tie knots. Often, these valves have a string or ribbon attached to them for further time savings. These valves allow re-inflation. the valve will *reduce* floating time, as they don't seal as well as a well-tied knot, and the extra weight of the valve will tend to pull the balloon down quicker! HELIUM We here in Aust. import ALL helium from North America. Would you believe that one major supplier ran out of helium today? So we wait for the next ship to arrive? Hope it isn't too far away! Could have been worse I guess.... at least we had enough gas for the Xmas / New Year sales. Costs us anything from US$0.20c per 11" balloon to $0.45c per balloon. Depending on discount level and supplier. I am just curious. I do not want to start any fueds on the mailing list... so email me DIRECT and details will stay CONFIDENTIAL! I will post information of general interest without being specific to states or company names. The mailing list is not the place to quote specific prices or deals or supplier's names. Example below. Is anyone interested? I hear that helium in N.Z is almost double the Aussie cost. In some parts of SE Asia they are still puting Hydrogen in balloons. (Put out that cigarette)...... because it's cheaper than helium? What about Britain and Europe? Does their helium come from the North Sea oil feilds, or from Poland? Either way, it should be relatively inexpensive. I've also heard that small amounts of helium are now coming through from Saudi. Can anyone confirm? For what it's worth... I beleive that the single biggest factor in determining a countries "balloon market potential" is the price the retailers must pay for helium. Seriously, helium balloons are only afforded by the upper income levels in some countries. In Colorado, much like the price levels I was paying in Missouri, it was around 5-6 cents a balloon tops. Perhaps a little cheaper in Missouri. The biggest source that I'm aware of (outside of the natural helium deposits in and around Texas), is the waste helium that is stripped from Natural Gas during the refining process. I recall reading that we jettison about a TON of it a day. The only thing we seem to be paying for here is the storage, transfer and tank rental. Seems that any country with Oil and Natural Gas should have plenty of Helium, and that would mean Saudi Arabia should be floating away in the stuff. I've heard that Helium is expensive in Europe as well as most countries south/near the equator. HEALTH RELATED Inhaling Helium Can Cause More Than a Funny Voice http://www.kidsource.com/kidsource/content2/news2/helium9_3_96.html INDUSTRIAL BOC Gasses Products Helium http://www.boc.com/gases/products/helium/helium.htm ("Helium is extracted from natural gas deposits. Only a few sources in the world contain a significant proportion of helium and justify its separation. These are in the US, Poland, Algeria and Russia. Because of its high value, helium is the only major industrial gas to be traded internationally") ELEMENTARY Periodic Table of Elements http://www.sci.ouc.bc.ca/chem/web-elements/He.html Helium http://www.crl.com/~peters/periodic/2.htm helium is not flammable, but it is sold as a compressed gas which can become a hazard if handled improperly. Helium is an element. After it is released into our atmosphere, it is so light that it floats off into outer space. Once current supplies trapped underground are exhausted, we'll all be out of luck. The only way I know to "manufacture" it is by the radioactive decay ("alpha particles" are Helium) of elements like Uranium, Thorium, Radium, Radon, Polonium, etc. Kids, don't try this at home or the NRC will come knocking at your door. Hmmmm, it's also possible to produce Helium by fusion of Hydrogen isotopes - the way the sun does it - but it will be many, many years (if ever) before anyone can do that continuously and reliably here on earth. Don't forget that when you are comparing helium prices, purity matters. The purer the helium, the longer the float time. Some helium distributors are selling "balloon grade helium" -- helium mixed with air. Unless you like using lots of "Hi-Float," stick with the good stuff. Perhaps It is only my supplier, and the fact that I heard this as well and chose to ask the tank fillers and sales personel. But the only difference between balloon grade and medical/ industrial grade is that they clean the med/ind tanks of any possible particulates or gas impurities during each fill both grades of helium are filled without mixing from the same tanker truck, which is treated at it's source as though it were to be used only for med/ind use. From my 13 years in the balloon industry, I have found that the change in pressure in a helium tank is almost proportional to the change in volume. So you can assume that when the pressure drops to half you have used up half your gas. (Any variation from this is to small to worry about). Thanks to everone who contributed to this topic on "Helium - Cost and Origin". At last count I had received over 30 direct replies from around the world. I promised not to embarras anyone by disclosing specific deals or mentioning suppliers by name or whereabouts. I hope everyone can accept that I have no intentions of fueling price wars or commercial bickering. If this information educates some of us a little.... and helps us understand the "commerce" of our product, then it could lead to future improvement in our industry. Knowledge breeds innovation! Make sure your gas rep gets to read this summary. His company may decide to get a little more aggressive in the marketplace. ____________________________________________ Answers to the TRIVIA questions:- "Air Liquide" (France) claim to be the world's largest industrial gases company. "BOC Gases" (UK) lay claim to supplying more helium worldwide than any other industrial gases company. Source of Information:- Australian HQ for each company. PRICES Got to be real careful here.... I will assume currency mentioned is LOCAL currency, unless otherwise stated. You have to do the conversions. I am also ASSUMING that everyone that mentioned a COST PRICE has INCLUDED tank rental, delivery charges, sales tax (if applicable) etc. In Australia the COST per 11" latex is from US$0.18cents to 45cents. There are 3 gas companies that import helium to Aust. All of it comes from North America by sea. In Canada the cost to fill an 11" latex balloon with helium is marginally dearer than in the USA.(to be expected) Around 1.5 to 2 cents. My information is that one gas company supplies all other gas companies in certain states or provinces. (is cartel the correct term?) Helium source is Texas. This is also common practice in Australia between the gas companies in certain states. In Latvia (Scandanavian country) all helium is drawn from natural gas deposits in nearby Russia or Sweden.A cylinder that fills 600 x 9" latex costs approx US$107 from Russian suppliers.... but US$148 from Swedish suppliers. In New Zealand and the Pacific Islands..... approx 1.5 x the Australian prices. Source of helium..... the USA. In Britain they want to delay comment till after they change over to the new "Euro Currency". ?????? The truth.....No email received. In the USA.... the cost prices quoted to me ranged from 5cents per 11" latex to near 20 cents per 11" latex. I hope the 5cents was an exagerated brag... and the 20 cents was a mathematical error. The typical prices quoted were 6.5 to 9 cents. No, I won't tell anyone who or where! I do not want to expand upon the above in detail - Please? I don't think I need to go into the "properties of helium" issue. That became a public topic on the list, and repeating it or challenging some of it is not necessary. The balloondeco archives now has a wealth of information on this topic. Knock yourself out! BOC Gases has a web site. I quote from it; "Helium is extracted from natural gas deposits. Only a few sources in the world contain significant proportion of helium and justify it's separation. These are in the US, Poland, Algeria and Russia.Because of it's high value, helium is the ONLY major industrial gas to be traded internationally" Thanks to Sheena Beaverson for the research. The BEST literature I have seen recently on helium production and uses is from "Praxair". The colour folder is called "Helium: Supply is Just The Beginning".Absolutely Amazing! Pick it up and teach yourself (and your business) a thing or two about helium. Their web site is; http://www.praxair.com They have offices all over the world.... and No, we don't buy helium from them! GENERAL Balloon Sculptures don't work well submerged in water. I find the buoyant forces on balloons tend to break apart the sculptures if you try to submerge them. 3/17/99 A 6" heart fully inflated with helium will obtain equilibrium. Meaning: wherever you put the balloon, it stays. I made a 6 balloon flower with 6" hearts and the hearts (really inflated) would hold up the weight of a stem. It waved in the wind very nicely!! A fully inflated 350 will rise vigorously. I haven't tried it yet but I think I could make a cool zeppelin that will stay put. A fully inflated 5" smiley will rise. I have used helium in 9" and 11" rounds and made things like spiders and ghosts with the 260's for legs (having air inflated the 260's) and the round holds the sculpture up in the air for quite a while and it floated around the restaurant I was working. It kind of went with the air currents and seemed to have a life of it's own. You can also attach rounds filled with helium with a string to 260 sculptures and have them hold the sculpture up. We have done this also. Used a round helium filled balloon with fishing line to float spiders around a halloween party. It took a lot of experimentation to get it right. How much helium so they wouldn't sit on the ceiling or drag on the floor. How big a spider sculpture etc. The end effect was great, but it was work to get it right. Of course, you just could do what I do: Slap a piece of tape on the balloon when no one is looking, and say "Hey kids, betcha can't do this. Here, everyone take a needle and try stabbing it through something." The only down-side to the tape "stick" is that this is a technique taught in boy-scout and cub-scout magic books. I've had more than one kid inform me that they learned how to do it in scouts so I let them inspect the balloon and when there is no tape you make
How Balloons are Made
How Latex Balloons Are Made: General Overview
THE NATIONAL LATEX PRODUCTS COMPANY
OVERVIEW OF HOW BALLOONS ARE MADE
Balloons are manufactured from a liquid rubber called latex. The balloon gets its color from the pigment that is added to the latex. Pigments are both organic and inorganic compounds that absorb certain wavelengths of visible light and reflect others. For example, a red balloon is red because the balloon absorbs all the visible light except red frequency light which is reflected back to the eye.
The strength of balloons can be affected by the pigment if the pigment particle is large in size and interferes with the film continuity and if the pigment reacts with any of the other ingredients in the balloon. As far as which color has the most effect on the balloons strength, we have not done any in depth study. Since we use pigments that are water dispersions of very, very, small particle size, and they do not react with any other ingredients in the latex, we do not detect any difference.
The natural rubber latex that we use comes from the sap of the rubber tree , Heveabrasiliensis, that grows in Malaysia. This sap looks like milk and is shipped to America in large ocean tanker ships. Once removed from the tree, the sap is called latex. To make this suitable for balloon production, curing agents, accelerators, oil, color, and water must be added. After these are added, the completed latex is put in an open top tank, and the balloon form, which is in the shape of a balloon, is dipped. Before the form is dipped into latex, it is dipped into a coagulent that causes the rubber particles of the latex to collect on the form. This coagulent is calcium nitrate, water, and/or alcohol. After the coagulent coated form is dried, it is then dipped into the compounded latex. Then the latex coated form passed through a set of revolving brushes that rolls the balloon neck into the bead that is used to aid in the inflation of the balloon. The latex coated form is then washed in hot water to remove any unused nitrate. Following the leaching, the form is put in a 200-220 degrees Fahrenheit oven to cure for 20-25 minutes. When cured, the rubber balloon is removed from the form (stripped).
The following information is taken from the Latex Engineering web site: A good balloon has the following conditions: regular peripheral wall thickness, good end gelation, pinhole free, good bead rolling, and a 'favourable' taste.
The above are controlled by: surface tension control of coagulant and good antisettling of 'chalk', even speed immersion and withdrawal with still liquid surface, correct compound viscosity and correct chemical stability, clean formers and efficient filters good step back of film thickness, well-leached film, and dryness state chemically friendly formula.
An even latex film depends upon a consistent coagulant deposit. This, in turn, depends upon a fast drying time and an even speed of withdrawal from the coagulant (which implies a hot coagulant and former). With small time cycles leading up to the coagulant dip, it is important not to lose heat necessarily after the stripping.
For making a balloon, the formers pass the following stages:
- acid bath needed once every eight hours (or after every round )
- clean water bath of constantly refreshed water
- brushing, especially the former bottom
- warming the bath up to 70 to 80 degrees Celcius
- first coagulant bath, for beading the edge of the balloon
- second coagulant bath for the balloon (the temperature of both coagulant baths is 70 degrees Celcius)
- oven for drying coagulant
- latex dip (former is leaving the bath upright in a good sliding angle - system of batch dipping - to prevent drop forming on the end of the balloon
- oven for setting latex film
- beading
- leaching
- detack tank
- two ovens with a temperature 80 to 90 degrees Celcius, other (higher) temperatures possible
- cooling down by two fans
- stripping by air and rollers
The following questions apply to the manufacturing of latex balloons.
Q: Is the balloon manufacturing process all automated?
A: It is largely automated these days.
Q: How much does a typical balloon cost to produce?
A: Each balloon size and type will have a different cost. It's a combination of the amount of latex used, and how easy it is to automate the process. 260's are harder to make than 11 inch rounds.
Q: After the molds are dipped into the latex, how does the balloon come off its mold? (manual removal or automated process?)
A: It depends on the size and shape of balloon. Some are completely automated. Some are stripped manually (with the help of forced air/water on the form). I believe all current latex manufacturers can strip all of their round balloons automatically.
Q: This question may sound weird, but can you bond/fuse two balloons together, or say, two dried strips of latex together?
A: This is getting into an area I know less about. I can tell you from experience that balloons can stick together if heat is applied, but they dont' hold together very well and can be pulled apart. During manufacture, if you stick them together prior to curing them, they will fuse. This is what happens with the 6-inch hearts and 260's that are stuck together at the tips. the forms are too close together on the racks that get dipped, so after dipping, they sometimes touch each other and join.
I believe inner tubes and tires are butyl rubber (because it is oil resistant), you can probably glue latex balloons together using tire patch cement. It comes in little tubes all the way up to gallon cans. Heat is used in some types of tire patching (hot patches).
How Latex Balloons Are Made: 260 balloons
Making a 260 involves dipping a mold (the same shape as the inside of a 260) into liquid latex. Once they're dipped in liquid latex, they are not allowed to cool. The dipped forms go through a vulcanizing oven, the nozzles are rolled, the balloons are washed, and then they're allowed to return to room temperature and pulled off the mold (stripped). 260's are not easy to produce; the mechanical action of making the 260 affects the final product.
How the latex runs on the mold as you pull it out of the liquid affects the eveness of the wall of the balloon. As you pull the mold out, the viscous latex is going to run (a little or a lot, but it is going to run).
If the mold hangs straight down, the wall of the 260 is thinner at the top, thicker at the bottom. The nozzle will be weaker and fatter, the end of the balloon will be stronger and thinner. If the mold hangs straight down as it dries the balloon will blow up straight. If the mold is turned over as the latex runs, the wall is more even from end to end but one side is a little thicker than the other. This 260 balloon will blow up with a curve. The drip that collected on the end of the mold as it came out of the latex runs down one side.
When you blow up a 260 you can tell how it was made. I assume the older balloon making equipment let the mold hang straight down. Mechanically, it is less expensive. To make balloons that are more even from top to bottom, a manufacturer has to invest in fancy and expensive equipment. To make a really good 260, the mold would need to spin as it turned over. This would give the best chance at an even walled 260.
The quality of the raw latex, how well it has been cleaned, the amount of vulcanization, the type of color and finish, the kind of powder, and variations in temperature and humidity during manufacturing all combine to make every batch of balloons different. How well the manufacturer balances the elements with the tools he has determines his consistency.
Balloon Forms and Molds
Balloon molds are arranged into rows and dipped into liquid latex in assembly line fashion. (see photo above) A balloon mold for a round balloon is shaped like an inverted light bulb. A 260 mold is long and thin. The, which explains the process for creating Geos.
Pioneer Balloon Company display of balloon molds for Qualatex airships, twisting balloons, and novelty balloons.
Pioneer Balloon Company display of balloon molds for Qualatex hearts, donuts, and blossom geos.
How the Rolled Lip on Balloons is Formed
Each balloon mold is the shape and size of the uninflated balloon. For example, a balloon mold for a round balloon is shaped like an inverted light bulb. The molds are arranged into rows and dipped into liquid latex in assembly line fashion. The latex at the top (thin) end of the mold becomes the "lip" when it is rolled down (toward the wide end) by a device which looks like a small motorized brush. As the rows of molds progress down the line, they pass between rotating, cone shaped brushes that are positioned horizontally, one on each side of each row of molds, pointing at the approaching molds. The brushes turn in opposite directions and are positioned so they touch the molds on each side. The point of the brushes start rolling the lip, and the lips continues to form as the row of molds moves along the line from the point to the larger end of the brushes. This occurs while the latex is still uncured, just before it is vulcanized.
Color Issues and Packaging
Balloons are made one color at a time by adding pigment to liquid latex. Pigments are both organic and inorganic compounds that absorb certain wavelengths of visible light and reflect others. For example, a red balloon is red because the balloon absorbs all the visible light except red frequency light which is reflected back to the eye.
The strength of balloons can be affected by the pigment if the pigment particle is large in size and interferes with the film continuity and if the pigment reacts with any of the other ingredients in the balloon. As far as which color has the most effect on the balloons strength, we have not done any in depth study. Since we use pigments that are water dispersions of very, very, small particle size, and they do not react with any other ingredients in the latex, we do not detect any difference.
Pearl Tones.
Pearl tone latex is created by adding crushed mica to the latex. This process can make the latex more brittle, and less twistable. If you want to see proof of this, you have to look no further than at Tilly Pearl 130's.
Gold/Silver/Metallic 260's
Metallic latex is made in the same way as pearl latex.
Agate 260's.
Agate balloons are made by dipping the mold into latex twice.
321 (Bee Body) Balloons A 321 is made by dipping just the tip of the balloon into the latex twice.
After stripping off the molds, they are counted by weight with special precision scales (different colors have slightly different weights) then packaged.
For a bag of assorted colors, a batch of equal quantities of the colors to be assorted are tumbled together, then counted (by weight) and packaged. Because of the tumbling process, there will not be an exact division of colors in each polybag. If you need a specific color, it's best to buy a solid color bag.
The difference in price of different color ballons is due, in part, to the price of the coloring agent. Some colors are more expensive than others. The Standard Colors of 260Q (White, Pink, and Light Blue) less expensive than the Jewel Tone Colors. Solid bags of White, Pink or Light Blue are the same price as a bag of assorted.
Tim Vlamis at Qualatex writes:
Latex Packaging - We are now printing all Qualatex bags with a bar code, description, and packing date code. The date code is known as a "Julian Date Code" and reads as follows: first two digits are the year, the last three digits are the day (1-365). 96031 = January 31, 1996. In proper storage conditions (dark, cool) Qualatex balloons should last for years.
Printing on latex balloons
Printed latex balloons are inflated while the printing takes place, screen printed, then deflated, drummed in rotating industrial dryers to shrink them back to "like new," and packaged. This is why printed latex balloons are so much more expensive than unprinted balloons. Any camera-ready artwork can be imprinted on latex balloons (as long as it does not infringe on copyright).
Balloons are printed using several different methods or types of printing depending upon the quantity, balloon size, delivery date, and other factors invloved in the order. All balloons are printed in an inflated state with the two methods described below.
The first method of printing is a form of offset printing. Ink is appled to the plate which reads right, the plate then transfers the ink to a printing drum, and the image reads wrong. The balloon is then rolled across the printing drum transferring the image to the balloon. The image once again reads right.
The second method of printing is silk screening. This method has the balloon in a flat, uninflated condition for printing. A silk screen, into which an image has been etched, is then laid over the balloon, and ink is forced through the mesh in the screen in the image area and onto the balloon. The non-image area does not permit ink to penetrate, thereby putting ink only in the image area. Silk screening is also done on inflated balloons by using a holding device and slightly compressing the surface of the printing area with the screen prior to applying the ink. This type of screening is done with an automatic machine on small to medium size balloons.
When balloons are inflated for printing, they are only inflated to approximately 75 to 80 percent of the total capacity. This gives the proper tension to the surface of the balloon for ink transfer.
Immediately upon placing the image on the balloon surface, the balloon is released and deflation begins. By the time the balloon is delfated, the ink must be dry to prevent offsetting onto other balloons.
Specialized ink must be used that will bite into the surface of the balloon and yet not go far enough below the surface to penetrate completely throughout to the interior, causing holes in the balloon.
Contrary to popular belief, balloons are not hand stamped in a deflated condition.
Printing alternative: a rubber stamp with quick-drying ink can be used to imprint on an inflated balloon.
Balloon Printing Machines and Services
There are companies that offer balloon printing services. Having balloons custom-printed is much like having T-Shirts silk-screened. The customer supplies the artwork and picks a quantity, and the company fills the order. In recent years, a few companies have introduced services for printing on 260 balloons.
There are also companies which manufacture and sell balloon printing machines to businesses.
How Foil Balloons Are Made
The concept and technology for the "metalization" of plastic sheeting that has given us foil balloons comes directly out of the NASA Space Mission. By the way, all of us should stop referring to foil balloons as Mylar (a trademarked name for a certain type of polyester film mad by DuPont) balloons. The balloon industry refers to them as "foil" balloons, because they are made of nylon sheet, coated on one side with polyethylene and metallized on the other. It's evidently so much harder to make balloons out of aluminized Mylar (and probably so much more expensive) that nobody does it.
Polyesters are "thermosetting" polymers... ie, once formed, subsequent heating won't melt them. Heat them enough and they just burn. For balloon workers, this means that Mylar films cannot be heat sealed. Polyethylene and nylon are "thermoplastic" polymers... ie, subsequent heating will melt them and subsequent cooling will resolidify them. For balloon workers, this means that sheets of "foil balloon material" can be heat sealed together.
The CBA videos said that since it's so much harder to make balloons out of aluminized Mylar, nobody does it. But, if thermoplastic films can be laminated, what would stop somebody from laminating a thermoplastic film onto Mylar to make a heat sealable version? So I looked up Mylar on the web and got some current info straight from the manufacturer. Indeed, DuPont does make a variety of coated Mylars that are heat sealable (because the coating on the Mylar can melt).
The Incredible Balloon webpage says:
In the late 1970's, silver metalized balloons were developed for the New York City Ballet. These balloons are commonly called mylar, but they are actually made from a metalized nylon and are more expensive than latex balloons. Someone wrote me that it is 48 gague (.48 mil) aluminized biaxial nylon w/ a special coating (capron emblem) for heat sealing.
Gary Felix's company was making custom-shaped foil balloons by hand for many years. Gary designed Olympic balloons in the 1984 summer Olympics and also Jesse Ventura's inauguration.
Making Latex Balloons at Home
The man who invented and patented the Geo, Ron Prater from Indiana, made all his prototype balloons at home, and vulcanized them in his kitchen oven (of course, his dad was a chemist at Pioneer Balloon Company...). I have a newspaper article (that was reprinted in a clown magazine) which discusses this.
Regarding making balloons at home, I've watched the hand dipping process and it's a snap. One good person with a few hundred dollars invested could make a gross in about 12 to 16 hours. At that rate, the cost would be prohibitive. - Marvin
Procedure to Manufacture a Latex Dipped Item
- Stir latex coagulent (the talc generally powder settles out). Transfer to vessle suitable for heating. Warm coagulent to 130-140 degrees Fahrenheit (warming is optional and is used mainly for fast production).
- Warm former or form for 3 minutes at 170-175 degrees Fahrenheit.
- Dip warm former into coagulent (coagulent should be under constant agitation).
- Dry coagulent coated former in 170-175 degree Fahrenheit oven for 3 minutes.
- Dip dried former in 70-75 degree Fahrenheit latex.
- Leach latex coated former in 175-180 degree Fahrenheit water for 15 minutes.
- Cure latex on former for 20 minutes at 200-215 degrees Fahrenheit.
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 +--------------------------------> 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 From: TimosV@aol.comSubject: Re: Balloon quality query In response to the question of whether or not we