# 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.

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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,

$$F_B=F_G+F_D$$

$$v = \sqrt{\frac{8 g r}{3 C_D} \left( \frac{\rho_{air} - \rho}{\rho_{air}} \right)}$$