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標題:
help coriolis effect
發問:
At latitude 43(of the central Italy) where 2wsinφ is approximately equal to 10^-4 s-1, a motion of 10m/s would move in an inertia circle of 100 km radius completing an orbit in almost 14 hours. HOW TO CALCULATE THAT? why can't I just use period t=pi/(wsinφ)? PLEASE HELP HELP
Radius of inertia circle R = v/f where v is the speed of the motion, and f is the Coriolis parameter, which equals to 2.w.sin(φ) Hence, the period T to complete an inertia circle with radius R is, T = (2.pi.R)/v = (2.pi).(v/(2wsin(φ)).(1/v) =pi/(wsin(φ)) Substituting values of w = 2.pi/(24 x 3600) and φ = 43 degrees, T = 17.6 hours Your "14 hours" seems to be underestimated. You may refer to the following web-page. Read the section on "Inertia Circles". http://en.wikipedia.org/wiki/Coriolis_effect
其他解答:
help coriolis effect
發問:
At latitude 43(of the central Italy) where 2wsinφ is approximately equal to 10^-4 s-1, a motion of 10m/s would move in an inertia circle of 100 km radius completing an orbit in almost 14 hours. HOW TO CALCULATE THAT? why can't I just use period t=pi/(wsinφ)? PLEASE HELP HELP
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最佳解答:Radius of inertia circle R = v/f where v is the speed of the motion, and f is the Coriolis parameter, which equals to 2.w.sin(φ) Hence, the period T to complete an inertia circle with radius R is, T = (2.pi.R)/v = (2.pi).(v/(2wsin(φ)).(1/v) =pi/(wsin(φ)) Substituting values of w = 2.pi/(24 x 3600) and φ = 43 degrees, T = 17.6 hours Your "14 hours" seems to be underestimated. You may refer to the following web-page. Read the section on "Inertia Circles". http://en.wikipedia.org/wiki/Coriolis_effect
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