If not enough helium is put into the balloon then it cannot lift the weight of balloon and will not rise.
If more helium is added it adds weight, making the balloon full of helium heavier. Assuming that the balloon can take this pressure and not burst it will become too heavy and will not rise.
If the surrounding air is at sea-level atmospheric pressure to what pressure should I fill my helium balloon in order to generate the most lift? Does this vary according the the weight/volume of the balloon itself and if so is there a formula to calculate this? Air and helium are at the same temperature.
I assume the same principals apply to air-ships? Just add weight of engine etc to weight of balloon in the equation?
Cheers,
Andy H
Atmospheric pressure will give it The greatest volume andlowest density I think.
Actually it'll depend on the elasticity of the balloon I guess.
Adding more helium increases the mass and the [i]volume[/i] of the balloon, displacing more air and making it lighter.
This is the difference between mass and weight.
That's what I thought at first Al, but then I imagined a balloon in a helium atmosphere, so helium inside at atmospheric pressure. Tie a knot then put it in air. Not it will have enough lift to pick the weight of balloon up?
Good point about elasticity though, I can't quite get my head around the maths for that!
Graham, this is why I was wondering about pressure, trying to get mass and volume into one number! This is where Al's point about elasticity makes it complicated.
MTQ has sent you in the right direction, I wonder if really you need to consider your understanding of bouyancy. Imagine a flat steel plate 1m squared. Will it float? No. Then use that plate to make a boat, with another 20 plates all welded to it to make a chined hull. It is 20 times heavier (not including the mass of the air) but it floats..
Helium ballons usually aren't very elastic, I guess for that reason.
I agree with Al BTW, just a smidge above atmospheric pressure for max volume, min mass.
If you want the maths, its:
Buoyancy force = Mass of fluid (air) displaced - Mass of object
= p'Vg - m g
Where p'=density of fluid (air), V = Volume of balloon, m=mass of balloon, g=gravitational acceleration
If you neglect the mass of the balloon itself, and just worry about the gas in it, the buoyancy force is
p'Vg - pVg
=Vg(p'-p)
Where p = density of balloon
So the force is directly proportional to the difference between the density of the outside gas (air) and the inside gas (helium). Smaller the internal density, the better. So you want just enough pressure to stop atmospheric pressure from collapsing the balloon.
Helium will have a lower density than air at the same pressure, so the balloon goes up.
Your helium balloon in a helium atmosphere is the same as an air balloon in an air atmosphere - and you can see for yourself that those don't go up (unless you change the density of the internal air by heating it).
Good point, stevomcd.
When andrewh said balloon, I thought of this type.
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So I was thinking of two lines on a graph, one for the increasing pressure and one for the increasing volume.
Wherever the two lines are furthest apart would be the maximum buoyancy, within the balloon's expansion limit.
Like you say, helium balloons are usually the foil type, like this.
[img] 
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Once the balloon has taken shape, any increase in pressure simply increases the mass without making any significant difference to the volume.