Shopping & Style1 min ago
How is stokes law related to parachutes?
And where can I find some simple information on this?
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For more on marking an answer as the "Best Answer", please visit our FAQ.Sir George's law states: F(drag) = 6*pi*a*coefficient of viscosity*velocity (it would greatly simplify the expanation if I could reproduce the formula for you.) The formula provides a solution to how fast a sphere will drop in a viscous fluid. A parachute would be considered a sphere and the air through which it's passing a viscous fluid. It's found that the speed is highly dependant on the radius of the sphere, hence the pi x a (a being a representation of radius) as well as the viscosity of the fluid, which is found by yet another formula, which I can't reproduce in this forum. The importance in your problem is determining the speed of parachute as it drops through the air which is directly related to its size... which is pretty intuitive, but Stoke's provides the mathmatical solution for you...
the parachute material is like a semi permiable membrain, the harder somthing tries to pass through it the more resistance it occurse. depending on the weight of drag, how much weight is hanging from it, and the diiameter /total area of mass occuring that drag constitutes the rate of desent/amount of drag. the material is also considered to be a great factor because the pass rate of flow through the material is an important factor in the coeficiant/fall rate=resistance