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This is my first post here, but I've been enjoying the site for some time now. I've run into a snag with an equation I'm trying to develop. Its purpose is to help determine power requirements for radio controlled, model helicopter rotors:

P = ( π^{2}Θ^{3}C_{d}ρ A / 2 ) ( R^{4}- r^{4})

P = power (Watts)

π = 3.14159

Θ = rotor head speed (revolutions per second)

C_{d}= drag coefficient (unitless)

ρ = air density (kilograms per cubic meter)

A = rotor blade area (square meters; span times chord)

F_{d}= drag force (Newtons)

V = velocity (meters per second)

R = rotor blade outside (tip) radius (meters)

r = rotor blade inside radius (meters)

τ = rotor head torque (Newton meters)

I started with the relatively well known equation for fixed wing drag:

F_{d}= C_{d}ρ A V^{2}/ 2

Then, knowing that the velocity varies with radius:

V = 2 π r Θ

The following may be obtained:

F_{d}= C_{d}ρ A ( 2 π r Θ )^{2}/ 2

This may be rearranged as:

F_{d}= ( 2 π^{2}Θ^{2}C_{d}ρ A ) r^{2}

Torque equals force times radius, so:

τ = F_{d}r = ( 2 π^{2}Θ^{2}C_{d}ρ A ) r^{3}

It appears to me that this equation only works for a small area at a specific radius. For instance, it might provide a reasonable estimate for a model helicopter's flybar paddle. In order to make it work for large areas over a range of radii, it needs to be integrated with respect to radius, correct? This yields:

τ = ( π^{2}Θ^{2}C_{d}ρ A / 2 ) ( R^{4}- r^{4})

R and r now represent the limits of integration. Since power equals torque times angular velocity:

P = τ Θ = ( π^{2}Θ^{3}C_{d}ρ A / 2 ) ( R^{4}- r^{4})

Since r^{4}<< R^{4}, r^{4}may be neglected (typically) and this becomes:

P = π^{2}Θ^{3}C_{d}ρ A R^{4}/ 2

Now, here's my problem. Unit analysis tells me I have an extra meter in there. I get Watt-meters instead of Watts. I know that it appeared when integrating. Am I integrating incorrectly? Do I need to integrate at all? It's readily apparent that I've missed something obvious, but I'm sure what or where. It's been about 20 years since I last took a calculus class, so go easy on me.

Yes, I know the equation is simplistic. Determing a drag coefficient for a rotor blade isn't easy. However, one could use such an equation to "measure" the effects of changes to head speed and blade dimensions. I.e., if the blade length is increased, how much must the head speed decrease in order to return to the original power level - with the stipulation that the drag coefficient doesn't change significantly.

On second thought, don't go easy on me. Gimme the third degree! I will appreciate any help that comes my way.

Thanks,

Joseph

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# Integration help for helicopter rotor

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