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How would you measure the mass of a helium balloon?

Clearly, the usual method of putting it on weighing scales won't work on earth because the balloon would float. I also don't think you could put it in a vacuum as, while the balloon wouldn't float, the vacuum would cause the balloon to expand and burst. (I think)

If there are a number of methods, which would be most accurate?

A few conditions

  • You can't weigh a deflated balloon and assume the mass of the helium is negligible.
  • No theoretical calculations of mass from data book values like density and atomic mass.   

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Jon-Richfield says:

It is not easy to get a precise answer, but the principle of one approach is:

Weigh the balloon. You should get a negative result, otherwise don't bother!

Weigh a vessel large enough  to contain the balloon.

Weigh the balloon in the vessel. Do a little arithmetic t make sure that all is in order.

Add a liguid or known density (water?) till the balloon is submerged. Note the weight. Mark the water level.

Remove the baloon, fill vessel up to the mark, then weigh.

A little arithmetic again...

Get the picture?

 

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posted on 2010-12-31 13:44:09 | Report abuse

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Georg says:

Hello Jon,

I see a problem due to  compression of the helim balloon

when immersed in water.

You needed the pressure in the balloon outside and

under submersion, but such "theoretical" calculations

are not allowed (because Rob Stass can't calculate? :=)

Georg

PS

Pressure is unknown,  the question lacks some parameter:

Is it a a small balloon  made from rubber with some excess

pressure or a "real" balloon (Une Charliere), with pressure 

identical to the outside pressure?

PSPS After some thinking, I must say, that even in a

"loose" balloon (Charliere) the pressure of the helium is not

all the same. At the lower boundary its pressure is identical to the

airs pressure, but not at the top end of the helium volume.

Not easy to calculate.

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posted on 2011-01-04 13:24:40 | Report abuse

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Jon-Richfield says:

As I am sure you realise Georg, your reply in turn undersetimates the technical difficulties too. The caution of my legalistic wording "...the principle of one approach is..." was deliberate. Many apparently trivial physical measurements are in practice nearly impossible and one generally uses indirect methods to infer them rather than trying to achieve a direct determination.

Recently a professor of physics pointed out to me that it is barely possible to measure the effects of Coulomb's law on a conductive solid sphere of say 1 metre in diameter. I was quite embarrassed to realise how naive my own assumptions concerning the difficulties had been. After all, the electromagnetic forces involved were far stronger than the gravitational forces measured long ago by Cavendish, so what could the problems be?

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posted on 2011-01-04 15:51:42 | Report abuse

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Paul_Pedant says:

I had a slight rant in another thread (about E = m * c * c) on the fff (furlong/firkin/fortnight) standard for measurements. Now this buoyancy thing has got under my skin nicely.

Suppose you have a nice exact 1kg iron standard weight.

Being as air weighs around 29.1 grams per 23.4 litres (gram-molecular weight plus Avagadro), and the density of iron is 5.5, then your standard weight is buoyed up by 0.23g. Is that part of the standard weight, or not?

It weighs less near the equator (centrifugal force), up a mountain (further from centre of earth), and in good weather (increased air pressure). OK, it's all somewhere down in the 4th or 5th decimal place, but it makes a mockery of "exact" standards.

 

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posted on 2011-01-05 22:18:18 | Report abuse


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miguel2 says:

Weight an air-tight box with the vacuum set inside.

Then, put the helium balloon inside, and make the vacuum again. The balloon with fill the entire volume of the box (which should be somewhat larger than the balloon, but not too much or else the balloon would explode; but you can also include a mesh, a net, or anything that would prevent the balloon growing too big). Weight again.

The difference will be the mass of the balloon including its content, helium or else.

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posted on 2011-01-03 22:02:45 | Report abuse


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Yarrumecurb says:

Neither of the above will work. The fact that helium is lighter than air and is bouyant in air does emphatically NOT mean it has negative mass. So the way which comes to my mind would be to get a completely empty balloon with a leakproof filling mechanism and measure its mass. Then measure the mass of the compressed helium cylinder from which you are going to fill the balloon. Carry out the filling operation without leaking or contaminating the helium, then when the balloon is full re-weigh the cylinder. Add the mass of the helium dispensed to the original mass of the balloon and you have your answer.

Another less accurate method would be to accurately calculate the volume of the filled balloon. Attach it to a sensitive spring balance and measure the upward pull of the connecting string. As air weighs about 1.1 Kg per cubic meter at sea level give or take a few grams for the barometer reading, if the volume of the balloon is (say) exactly a cubic meter and the upward force is 0.1 Kg, the mass of the filled balloon is , yep, you guessed it, a kilogram.

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posted on 2011-01-04 00:12:00 | Report abuse

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Georg says:

Miguels method will work!

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posted on 2011-01-04 13:27:33 | Report abuse


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ukbandit says:

Turn the scales upside down and let the balloon push against it :)

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posted on 2011-01-04 11:25:40 | Report abuse

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Paul_Pedant says:

That just measures the excess lift (the weight of the displaced air, less the combined weight of the helium, balloon and string). That's not what the question is about.

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posted on 2011-01-04 14:02:27 | Report abuse

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Paul_Pedant says:

Apologies for the failure to spot the cheeky tongue. I was distracted by the mention of "negative mass" a couple of posts previously.

Pete is of course well aware that it is the Springbok who thinks the Chris whom I am not is the American whom he is not either.

Irony aside, I did actually wonder whether standard scales would calibrate to zero for upside down (or should that be umop ap!sdn), and whether all the fulcrums (fulcra?) and levers would become displaced.

 

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posted on 2011-01-05 22:34:17 | Report abuse


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Paul_Pedant says:

Mass is not weight.

Place the balloon in a hangar so it is (just) floating. (This just protects it from wind effects - it does not affect the procedure.)

Fix a pulley to the hangar in line with the nose of the balloon. Suspend a large weight (say one tonne for a 100-tonne-ish balloon) so it can drop, say, 5 metres. Run a taut rope from the weight to the nose of the balloon via the pulley.

Drop the weight, and measure the time for the weight to reach the ground, accelerating the balloon with it the same distance (because the rope was taut).

Figure the acceleration a from s = 0.5 * a * t * t, because you measured s and t.

Figure the force F of the weight, which is g (9.81) * 1 tonne.

Figure the mass M of the accelerated bodies, which is (1 + X) tonnes.

F = a * M, hence M, hence X.

QED.

Unless you add slack ropes at the back of the balloon to restrain it, it will promply run into the pulley and deflate, so you will be able to weigh it empty anyway later.

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posted on 2011-01-04 12:01:54 | Report abuse

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Georg says:

Hello Paul,

Your method would work in a vacuum, in air You have to

be aware that a comparable  volume of air is shifted in the opposite

direction if a mass is moved.

Georg

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posted on 2011-01-04 13:20:39 | Report abuse

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Paul_Pedant says:

Not an identical volume of displaced air to the balloon, Georg.

If the balloon is 250 metres long (as Hindenburg was), but my weight only drops 5 metres and therefore moves the airship 5 metres, then the displaced air is only 2% of the volume (and mass) of the airship, which is a small error.

In fact, I would expect (or contrive, with a good choice of weight) an acceleration around 0.1 * g (100mm/sec/sec) and I should therefore be able to plot the displacement every second (with a bit of chalk and a metronome), and plot a graph where I could read off a very good approximation to the initial acceleration, from the slope at time zero.

Actually, my main concern would be that the balloon would appear to accelerate too fast initially, because of a lack of structural rigidity. I would probably want to measure displacement at several positions along its length during the experiment, and analyse them statistically. I might even learn something useful about its rigidity under stress, too.

 

 

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posted on 2011-01-04 13:59:34 | Report abuse

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Paul_Pedant says:

Georg,

I would certainly say if I move an object through twice its length, then double its volume of air has been moved. At each point through which its projected forward area has passed, the air that had previously been there has gone away somewhere and some air has then come back.

The interesting question is "where did it temporarily go?". In water, for example in the case of a displacement boat hull, the fluid is incompressible and has a distinct surface. It is fairly clear the fluid is accelerated backwards, but also that it heaps over quite a wide area of surface, produces standing waves as a wake, and so on.

In compressible air, free to expand in all directions, it is by no means obvious that the air displaced from in front of an object is the same air that fills in behind it. There is essentially an infinite reservoir surrounding the object, and the effects of streamlining and the lack of constraint in flow direction suggest that there is no boundary within which a "comparable air mass" would be constrained.

I'm not clear that an oscillating pendulum bob, of a most unaerodynamic shape, moving through several times its own length each second, in a confined volume of air, has much relevance here.

 

 

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posted on 2011-01-05 23:04:05 | Report abuse


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