How to you calculate how much a smooth sphere is affected by air resistance at different spinning rates? None of the answers to How can you calculate air resistances at different speeds? answer my question. Does spinning affect air resistance? If not, then I know that for sufficiently high Reynold number, it is equal to the density of air times the square of the diameter times the square of the speed times a dimensionless constant. How do you determine the dimensionless constant?
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A smooth ball with a sufficiently high Raynold number will have a turbulent flow of air around it. 12:00 into the YouTube video Turbulent Flow is MORE Awesome Than Laminar Flow, we can see how air flows around a smooth sphere. It appears to have a jet of air dragging behind the sphere nearly exactly the thickness of the sphere moving really slowly. I believe with a high Raynold number, the eddies formed at the surface from viscous resistance are really super tiny and almost all resistance comes from the formation of larger eddies behind the jet of air so it makes almost no difference whether the ball is spinning.
When moving through air in the absence of gravity, its position with time will always be logarithmic. That's because its rate of movement doesn't affect the nature of the flow of air around it. The distance it travels before its speed reduces by a factor of e will always be a fixed constant times its density times its diameter divided by the density of air given that the air has a way lower density and hardly buffers the ball around.
Now as an example, I want to calculate how far a aluminum oxide sphere the diameter of a golf ball will travel in air without gravity before its speed reduces by a factor of e.
According to https://en.wikipedia.org/wiki/Golf_ball#Regulations, a golf ball has a diameter of 42.67 mm. Googling aluminum oxide density gives you a box at the top of the Google page saying that aluminum oxide has a density of 3.95 $g/cm^3$. According to https://en.wikipedia.org/wiki/Density_of_air, air has a density of 1.225 $kg/m^3$ at 15°C. Now we finally have all the tools we need to calculate how far a smooth aluminum oxide sphere the diameter of a golf ball will travel in air in the absense of gravity before its speed reduces by a factor of e.
When the ball is travelling through the air, the rate the ball is losing kinetic energy is the rate that it is adding kinetic energy to the air. to calculate the kinetic energy it is adding to the air, we can treat it like it is building a jet of air behind it moving at its speed getting built at its speed of movement. We know that the sphere is exactly $\frac{2}{3}$ the volume of a cylinder whose thickness and length are its diameter. When the sphere travels its diameter, the amount of kinetic energy it loses is its speed squared times the density of air. The fraction of kinetic energy the sphere loses is 1.5 times the density of air divided by the density of the sphere. Now the fraction of the speed it loses is only half of that. Now the distance a sphere will travel in air without gravity before its speed reduces by a factor of e is 4/3 times its density times its diameter divided by the density of air, probably not absolutely exactly 4/3 but very close, and that distance to the nearest meter is 183 m.
Now what about if you have an aluminum oxide ball with a diameter of 10 m going through hydrogen gas. How far will it go before its speed reduces by a factor of e. Could it be about the distance from Toronto to Montreal, which takes a VIA train travelling at about 34 m/s about 4.5 hours? Yes. Since nitrogen is 14 times denser than hydrogen, all you have to do is take 183 m and multiply it by 10 then multiply it by 14 then divide it by 0.04267 and you get 601,903 m. According to https://www.distancecalculator.net/from-toronto-to-montreal, the distance from Toronto to Montreal is 503,000 m.
Timothy
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