[Request] today is a windy day(self)

and I was wondering... how fast should the wind flow to yeet me off the ground? I'm a 1.90 mts male, and my weight is 71 kilograms

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F = m * a = 71 * 9.81 Newtons

F = Area * density_air * v^2

BSA = sqrt(190 cm * 71 kg / 3600) = sqrt(190 * 71) / 60 = 1.935774321097948189801570462700... square meters

71 * 9.81 = (1/60) * sqrt(190 * 71) * 1.229 * v^2

The units will work out.

71 * 9.81 * 60 / (1.229 * sqrt(190 * 71)) = v^2

9.81 * 60 * sqrt(71) / (1.229 * sqrt(190)) = v^2

9810 * 60 * sqrt(71) / (1229 * sqrt(190)) = v^2

(981 * 6 * 100 / 1229) * sqrt(71/190) = v^2

17.110408350829761122935957528669... = v

So a constant wind speed of 17.11 m/s should do the trick. That's about 38 mph. That doesn't seem like a lot, but it is.

it is an average speed of 61 km/h, I thought it would be more tbh, but thanks!