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Re: Catapult math
Here's one of the fun oddities of this year's game.
Let's assume the ball is a sphere (it's close enough for this purpose) Drag on a sphere is based on the Reynolds Number as seen here: http://www.grc.nasa.gov/WWW/k-12/air...ragsphere.html To calculate the Reynolds Number, the formula is: Re = ρvc/μ Where: Re = Reynolds Number ρ = Fluid Density v = Velocity c = Characteristic Length μ = Dynamic Viscosity For air (We're assuming 20deg C): ρ = 1.2041 kg/m^3 μ = 1.983*10^-5 kg/(m-s) For the ball: c = 25 inches = 0.635 m So our formula for Re is now Re = (1.2041 kg/m^3)*v*(0.635 m)/(1.983*10^-5 kg/(m-s)) Re = 3.8558 * 10^4 * v Where v = velocity in m/s I would say our ball is a fairly rough ball given the wrinkles and the material, so look more toward the rough line on the chart. From v = 1 m/s to 5 m/s, we have Reynolds Number values between 3.8558 * 10^4 and 1.9279 * 10^5. We pass straight through the transition between laminar and turbulent flow. So what does this mean? For some of the "hard" shooters, they're going to experience a transition from turbulent to laminar flow as the ball slows down causing potentially unpredictable results. That's one of the reasons I like this game. |
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