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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 293
OK, 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" inflated
Solving 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?
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:
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.
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.
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.
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.
Can static electricity pop a balloon?
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:
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 , 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.
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.
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.
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.
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!)