- #1

Jeffsg605

- 6

- 0

[tex]

L = \frac{1}{2} \rho V^2 * C_L * S \\

where \\

\rho\mbox{ = density} \\

V\mbox{ = velocity of a point on the rotor} \\

C_L\mbox{ = lift coefficient} \\

S\mbox{ = surface area swept out by the rotor} \\

[/tex]

So I'm integrating across the length of the rotor. Since [tex]S=\pi r^2 \mbox{ and } V=\omega r[/tex] I get:

[tex]

L= \int_0^r \frac{1}{2}\rho (\omega r)^2C_L\pi r^2dr \\

L= (\frac{1}{2})\rho \omega^2 C_L\pi \int_0^rr^4dr \\

L= \frac{1}{2}\rho \omega^2 C_L\pi \frac{r^5}{5} \\

[/tex]

I'm thinking I must have missed something by now because the units do not make sense. If I solve just for units, I get:

[tex]

\rho = lb/ft^3 \\

\omega = V/R = \frac{1}{min} \\

r = ft \\

[/tex]

The rest have no units.

However this leaves me with L = lb*ft^2/s^2. There is an extra ft in the numerator. Any idea where I went wrong or if this is even the correct approach to calculating rotor lift?

Thanks in advance.