- #1
Jeffsg605
- 6
- 0
I am trying to calculate the lift generate by a helicopter rotor using the lift equation, which is
[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.
[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.