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When trying to calculate the lift force generated by a simple rectangular blade, I've found the following equation: $$F = \omega^2 L^2 l\rho\sin^2\phi$$ in which $\omega$ is the angular velocity, $L$ is the length of the helix, $l$ is the width of the helix (both in meters), $\rho$ is the air density at normal conditions, and $\phi$ is the angular deviation of the helix related to the rotating axis. So a 4-helix propeller would lead to $F=4\omega^2 L^2 l\rho\sin^2\phi$ and so on.

However, when I substitute the variables with real values, it seems to me very unreal. For example, if the 4-blades propeller works at $13000 RPM$, $\rho=1.293 kg/m^3$, $\phi=1^\circ$, $l=10^{-2}m$, $L=5\times 10^{-2}m$, leads to $$F=4\times(120\pi\times13000)^2\times(5\times10^{-2})^2\times(10^{-2})\times1.293\times(\sin1^\circ)^2$$$$\therefore F=9.46\times10^5N$$

So in this case, a simple $1cm\times 5cm$ propeller would carry $946 tons$ is surely wrong. So, where is the issue in the formulae?


There was a mistake of my calculation, mr. @Mark Eichenlaub made me see this. The right calculation is $$F=4\times(2\pi\times13000/60)^2\times(5\times10^{-2})^2\times(10^{-2})\times1.293\times(\sin1^\circ)^2$$$$\therefore F=0.072995N$$ which is quite reasonable. So, for lifting a $0.5 kg$ load ($\implies P=9.81\times 0.5=4.9N$) we should get longer blades or change the $\phi$ inclination. Thanks.

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your units do not add up to Newtons. its only Kg/s^2. what is your reference of this formula? – user27665 Jul 30 '13 at 7:41
up vote 1 down vote accepted

One revolution per minute is one sixtieth of a revolution per second. You have it the opposite way.

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You are right! I'm surprised that I could not see this gross error of mine. Thank you. – Rego Apr 1 '12 at 6:04

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