Ascent rate and size of balloon I am part of a school project, Project Stratos to send a balloon to the edge of space (the closer side :P) and was wondering how you would work out the accent rate of a large balloon (roughly 1m^3 of helium with 100g of mass) and the size of it as it increases its Altitude. I am creating a live map (that will be based on predictions rather than its actual location) and want to know the speed it will float up into the atmosphere. Currently we are assuming the ascent rate will be about 5m/s but I doubt that is very accurate and would this speed increase as it gets higher?
Edit: I would also quite like to know the burst height of the balloon.
 A: Ascent rate of a balloon (assuming spherical symmetry) depends on the following forces:
(1) The upward buoyant force $F_B=\frac{4}{3}\pi r^3\rho_{air}g$
(2) The gravitational pull downwards: $F_G=\frac{4}{3}\pi r^3\rho g$
(3) The drag force acting: $F_D\, =\, \tfrac12\, \rho_{air}\, v^2\, C_D\, A$
On a first glance it might appear that the balloon soon reaches a terminal velocity. But the quantities involved in these equations aren't all independent of each other or remain constant. For example the density of air changes with altitude. And the atmospheric pressure drops as you ascent, causing the balloon to increase in volume, thereby increasing the drag on it. Thus to analyse the motion of the balloon carefully, one has to resort to numerical methods and computers. But if you are looking for an approximation the ascent rate could be taken as the terminal velocity and could be obtained by setting,
$$F_B=F_G+F_D$$
leading to,
$$v = \sqrt{\frac{8 g r}{3 C_D} \left( \frac{\rho_{air} - \rho}{\rho_{air}} \right)}$$
A: Your answer using drag assumes the balloon is rising very fast. At the low speeds, typical of a balloon, the dominant resistive force is the viscose downward displacement of the surrounding gas as the balloon rises upwards. This depends linearly on speed, not quadratically. Think lava lamp, not fighter jet.
The laboratory test for this, that can be done in any elementary school, would be to measure the speed of rise with different weights attached to the same helium balloon. One will find the rate of rise drops linearly as the effective density of the balloon (balloon + weight) increases towards atmospheric density. More importantly, the rate of fall, will increase if the density of the balloon is greater than the atmospheric density. This points to the other flaw in your reasoning, in your equation the direction of motion is always positive, or at least undefined, if the balloon is more dense than the air.
