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kenewbie
Oct27-10, 01:08 AM
Say I want to send a helium balloon into the sky. For all intents and purposes this balloon is circular, with volume V. I will fill it with helium to the same pressure as the air outside it (I assume that this is important). The weight of the balloon and the payload is W.

So, I assume I have to calculate the weight of the helium + balloon, and, somehow, the weight of the air that the balloon replaces. So if the helium and the balloon is g grams lighter than the air would be, then I have g grams of lift?

And how would I go about converting g grams of lift to meters per second?

If anyone can help me out with some ways to calculate this, I would be very grateful. If there are important variables that I have left out, I certainly would appreciate that being pointed out aswell.

k

spancho
Oct27-10, 11:43 AM
To calculate the weight of the gas displaced, you have to use the percent mixture of nitrogen, oxygen, CO2 ect. Im pretty sure that at STP the ratio is 24 liters per one mole of gas particles.

For calculating the rate of assent, since you already know the weight of the craft and the lift, just use Force=Mass*Acceleration

Ergosum
Oct27-10, 11:48 AM
Hi. Your statement is essentially right. You've got a balloon with weight W (including the Helium) that displaces a volume V of air. This air has a weight w, so the effective lifting force is equal to w minus W.

F = w - W

Irrelevant a this basic level: the shape of the balloon, the helium pressure. Of course, at greater pressure the He weights more, but this has already been included in the total weight W.

Now, we've got an object of mass M = W / g subject to a force F, so it acquires an acceleration

a = F / M

And therefore, supposing it starts with zero initial velocity , the speed at any time t is equal to:

v = a * t

Of course, in a more realistic setup the acceleration of the balloon is countered by the friction, when the balloon attains its terminal speed, the friction is equal to the lift, and it ceases to accelerate upwards.

jloram
Oct27-10, 09:55 PM
Is this an elastic balloon or a bag? If elastic, you'll have to put the helium in under pressure, hence extra helium and less lift.