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Unread 02-02-2014, 09:04
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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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