[Closed] Clever science people, a question.

It's possible that it's a useful rule-of-thumb, but the shape of a balloon will affect the volume. A sphere is the most efficient shape for volume per unit of surface area. A flat starfish will have lower volume for the same surface area.

Dumb mathematician here, not very good at the real world.

"Can we assume that 2 different shaped balloons that weigh the same will have the same surface area"
YES, because they contain the same amount of (presumably identical) foil

"... and hence the same volume?"
NOT REALLY, because things with the same surface area don't necessarily have the same volume.

BUT, so long as they're all basically spherical-ish (i.e. you aren't selling helium ballons designed to look like the surface of a lung or the structure of soil ) you should be fine.

I'm a procrastinating science person. Not sure about clever.
If the balloon material has negligable mass, then it's just the volume that matters.
A sphere and a thin disc can have the same volumne, but very different surface area.
Also, surface area and volume do not scale in the same way - see:
https://www.bbc.co.uk/bitesize/guides/zwyfxfr/revision/2

This is proabably not what you want to hear, but if all the shapes are similar, you can proabably get away with surface area as an estimate of the final volume once you work out you need 1 weight for surface area X-Y and 2 weights for Y-Z, etc.

Edit - I type too slow, but think we're all saying the same thing.

Not precisely. Surface area does not reflect volume, it'll depend on the shapes themselves.

However, as a proxy for 2D packages that become 3D when inflated they probably won't be far away, and as weight of uninflated balloon is also a good measure of area in its uninflated 2D form, then you'll probably be close enough.

.

Assuming identical materials (composition, thickness etc) then non-inflated weight should be directly proportional to surface area.

Volume could be wildly different. If it helps visualise, consider 2D shapes. You could draw a square and a capital L shape with the same perimeter length but very different areas.

(And of course, you're filling them with helium so greater volume will make them lighter.)

... I think.

I was with you until "(And of course, you’re filling them with helium so greater volume will make them lighter.)

… I think."

Lighter compared to each other presumably? Not lighter per se.....

🙂

You are all missing a trick here. The helium volume has a very low mass and is unlikely to be significant compared to the mass of the bag. 1 litre of helium gas weighs 0.18 of a gram. You probably cannot measure that accurately in your balloon workshop, just weigh the empty bags and job jobbed.

submerge them in a bucket filled to the brim.

weight the bucket, the bucket full and the bucket semi empty for each shape.

thats your volume.

Admittedly it might take a while to work out all the different shapes

submerge them in a bucket filled to the brim.

weight the bucket, the bucket full and the bucket semi empty for each shape.

thats your volume.

Can't you just push it under and see how much the water level rises? Less Less bit for your arm

Lighter compared to each other presumably? Not lighter per se…..

We're veering dangerously close to the "mass vs weight" argument here. Danger, Will Robinson.

Apologies ,all weights are UNINFLATED.
Parrot 16 g
2 stars 22g

This a tin pot idea but with a suitable digital balance you could weigh them upwards with the scale pointing down above the floating balloons and the difference between the various balloons would be proportional to the difference in helium volume?????

Scientist answer - no. This has been explained above better than I can do. Volume and surface area not linearly related and the ratio changes depending on how "complicated" the shape is.

Looking at those balloons though, they seems to follow the same formula - two halves of a random shape stuck together in such a way that it makes a ~4 inch thick version of the 2D shape.
So if other balloons follow the same style then volume is going to be vaguely proportional to the suface area of the 2D shape when uninflated - and also with the material being equal - proportional to the weight.

One thing thats not clear from the first explanation is your weighing method for helium.

The parrot weighed 16g inflated,the 2 stars 22g. Both needed 2 weights.
A different balloon weighed 28g and needed 3 weights.

assume you mean uninflated? edit - yes you do

parrot was 16g material and 2x8 = 16g of weight to hold it down, so it held enough helium to give 32g of uplift.

two stars were 22g material, 16g weights, so they held enough helium to give 38g of uplift.

3rd mystery balloon 28g material, 24g weight, held enough helium to give 52g uplift.

(Uplift being directly proportional to gas volume)

Weight:Uplift ratio
16:32 0.5
22:38 0.58
28:52 0.54
So fairly close.

Another thing to consider is how much error there is in your "how many weights are needed to hold it" method. If 1 weight doesnt hold it but 2 does, weve assumed that it needs 16g of weight. But really it just needs any more than 8g. and I doubt your hands are sensitive enough to tell.

So allowing the final weight to be only 1g rather than 8, this does the following to the results:

Weight:Uplift ratio
16:25 0.64
22:31 0.71
28:45 0.62

edit - maths

So the balloon weight to volume of gas ratio varies from 0.5 to 0.71 - a 42% increase from best to worst. Is that accurate enough to be useful for your needs of ordering/charging for gas?

Why do you need to know surface area and volume? Is there a question that you're trying to answer with that information? Can you step back and explain what you're trying to achieve then we can address that rather than chase a specific detail.

(sorry if this is teaching granny to suck eggs but as a data analyst people tend to ask for specific details rather than telling me what actual question they need answered, then a week later find out that what they'd asked for doesn’t answer the bigger question so have wasted a load of everyone's time)

Can’t you just push it under and see how much the water level rises? Less Less bit for your arm

well you could but the graduation buckets to the nearest 2ltres is probably not that helpful!

Also if you struggle to push a balloon under water from the top without submerging more than your finger tips i would suggest there is more pressing concerns for you to be dealing with.

But back to the first question as everyone started discussing volumes and geometry...

A slight more obvious example would be the empty mylar* balloon weights 16g empty and has no volume, if you filled it with air it would still weigh the same but the volume has clearly increased.

*which doesn't stretch perceptively at the pressures we are talking about

I'm guessing @zippykona wants to know how much helium they can inhale (to make silly voices) and still have enough left for the party balloons?

We struggle to know how much gas odd shaped balloons take to fill.
We can work it out post inflation by seeing how many 8 g weights it takes to keep it on the ground.
The parrot weighed 16g inflated,the 2 stars 22g. Both needed 2 weights. A different balloon weighed 28g and needed 3 weights.

This in itself is a bad start as your measurement accuracy is only as good as the difference between each known point, for a gauge with 1g increments the accuracy is +/- 1g and for increments of 8g the accuracy is +/- 8g. A far better method would be a very light spring gauge to measure the "lift" of the baloon.

Can we assume that 2 different shaped balloons that weigh the same will have the same surface area and hence the same volume?

For reasons already explained, no.

Another hole in your method is you haven't taken pressure into account, that affects the quantity of gas in the balloon.

Is this a practical question or a school homework thing?

Another hole in your method is you haven’t taken pressure into account, that affects the quantity of gas in the balloon.

valid point I hadn't considered - theres going to be some variance in the pressure put in. Anything just above atmostpheric will make it look full, but theres probably a fair bit more that could go in until the balloon pops. Unless the inflator mechanism from the helium has some sort of cut off valve.

Is this a practical question or a school homework thing?

I think he owns or works in a shop / the shop pictured

How accurate do I need to be?. I can put a sucker on our scales and tie the ribbon to that to guage lift .
Balloons all at same pressure.
The balloon inflator has an auto cut off once it is inflated to a certain psi.

1) Get an infinitely expansile spherical vessel.
2) Empty all the gas out from one balloon into the sphere.
3) Measure the sphere's diameter and calculate the volume.

In the real world, get a normal rubber balloon. Given that you're only interested in relative volumes, assume balloon is roughly spherical and/or expands uniformly and only negligibly pressurises the air inside. Empty the air (use air, not helium for this part) from the parrot into the ballon. Measure balloon diameter. The balloon diameter^3 is proportional to the volume of gas used.

This is the problem with asking boffins to solve problems. They're all so busy out clevering each other they miss the simplest solution. The foil balloons don't stretch much so just fill with water and pour into a measuring jug. Then get the density of the helium for the cut off pressure at room temp and multiply.

Probably more pertinent than why how you can can't, the big question really is why you want or need to, there's probably a quick dirty solution to that without much maths and faff.

How accurate do I need to be?. I can put a sucker on our scales and tie the ribbon to that to guage lift .
Balloons all at same pressure.
The balloon inflator has an auto cut off once it is inflated to a certain psi.

sucker on the scales (and maybe have another dummy weight on there so the scales dont go to a negative value).

Difference between the uninflated empty balloon weight, and its inflated negative weight, is going to be proportional to the volume of helium to inflate it.

No surface area relevance or calculations needed.

How accurate? looking at the figures involved, nearest gram?

The issue so far is that "two weights holds it but 1 doesnt" is you have only ascertained it needs between 8-and-a-bit-grams and 16 grams. Which is quite inaccurate.

Do the manufacturer of the balloons not supply this information? Or can you not just use something like this?

I'm not a clever science person hence going for the easiest option.

Yes I’m going to be that guy…..

It’s about time these horrible bloomin tacky things were banned. Helium is a finite resource and we may run out of it in the next few decades. The balloons take centuries to degrade, they’re often released and can cause great harm to wildlife. I’ve lost count of the number I’ve found

I think he owns or works in a shop / the shop pictured

Ah, I see no pictures, cheers!

How accurate do I need to be?. I can put a sucker on our scales and tie the ribbon to that to guage lift .

Sounds good to me. Don't even need to tie it, just put your weights on top of the string. How sensitive is your scale?

Scales are accurate to a gram.

In fact there’s a word for those who buy such balloons

Parents?

as they are not stretchy just fill it with water and measure the volume that fills it up.

I was going to be that person, but was beaten to it, so I'll just state an interesting fact...it takes 84,000 helium balloons to fill an MRI scanner. Assuming thats boring balloon shaped balloons, can the OP use the advice in the thread to convert this into the equivalent volume of parrot balloons?

As an artist rather than a scientist I wouldn't bother with trying to work out the relative volumes of the different balloons or with trying to accurately weigh one item which itself is very light - If your scales are only accurate to the nearest gram then a gram is really quite a  lot of helium and I doubt two different balloons would vary in volume by many multiples of 5.6 litres. Instead  I'd weigh the gas bottle before and after you've filled a large quantity of the same shaped balloons.

I was going to be that person, but was beaten to it, so I’ll just state an interesting fact…it takes 84,000 helium balloons to fill an MRI scanner. Assuming thats boring balloon shaped balloons, can the OP use the advice in the thread to convert this into the equivalent volume of parrot balloons?

I think we can use science to figure out how much selling irreplaceable helium for frivolous purposes weighs on his conscience 🙂

Houns +1

I'm assuming you actually want to know how much it costs you to fill a balloon and subsequently you should charge?

In which case, split the balloons into 3 or 4 sizes, measure the largest in each category, either by filling with water or displacement in a bucket. Calculate the volume of helium at the filling pressure and charge accordingly. You don't need to measure every balloon accurately.

If you want to know when you will run out fit a pressure gauge to the bottle.

Another idea. Fill each balloon. One by one deflate them under water into an upturned, water filled glass measuring jug and see how much water is displaced by helium in the jug.